function
inequalities
integration
trigonometry
calculus
real analysis
real analysis unsolved

Let $ p\geq 2$. Prove that:

$ \sqrt[p]{\frac{\Gamma^{2}\left(\frac{p}{2}\right)}{\Gamma(p-1)}}\geq \frac{1}{2}$,

where $ \Gamma$ is the Gamma function.

It is equivalent to the inequality $ B \left( \frac{p}{2}, \frac{p}{2}\right) \; = \; \frac{\Gamma\left(\frac{p}{2}\right)^{2}}{\Gamma(p)}\; \geq \; \frac{1}{p 2^{p}}.$

Using beta function identity,

$ B \left( \frac{p}{2}, \frac{p}{2}\right) \; = \; 2 \int_{0}^{\frac{\pi}{2}}\sin^{p-1}\theta \cos^{p-1}\theta \; d\theta \; = \; \frac{4}{2^{p}}\int_{0}^{\frac{\pi}{2}}\sin^{p-1}\phi \; d\phi$

where $ \phi = 2\theta$. Apply $ \sin x \geq \frac{2}{\pi}x$ for $ 0 \leq x \leq \frac{\pi}{2}$ to the last integral and we have

$ B \left( \frac{p}{2}, \frac{p}{2}\right) \; \geq \; \frac{2\pi}{p 2^{p}}\; > \; \frac{1}{p 2^{p}}$

as desired.


trigonometry
geometry
angle bisector
trig identities

[quote="Ignite168"]If $A=30^\circ$ and $B=15^\circ$, find the value of $(1+\tanh)(1+\tanh)$.

[hide="Hint on how to do it without Trig"]I used the Angle Bisector Theorem. I'm sure knowing Trig Identities would make this problem go A LOT faster but they're not necessary.[/hide][/quote]

haha if you quote your post, Ignite, then it would have tan..okay i see

$\begin{eqnarray*}(1+\tan A)(1+\tan B)&=& (1+\tan 30)(1+\tan 15)\\ &=& \left(1+\frac{\sqrt{3}}{2}\right)(1+2-\sqrt{3}) \\ &=& \left(\frac{2+\sqrt{3}}{2}\right)(1+2-\sqrt{3}) \\$

ahh im too tired

Just a note: if you want to show something like $\tan A$, DO NOT post it as [b]\tanA[/b].  It will read that as though [b]tanA[/b] is a function.  You must type it as [b]\tan A[/b].  In dollar signs, of course.

[quote="Ignite168"]If $A=30^\circ$ and $B=15^\circ$, find the value of $(1+\tan{A})(1+\tan{B})$.

[hide="Hint on how to do it without Trig"]I used the Angle Bisector Theorem. I'm sure knowing Trig Identities would make this problem go A LOT faster but they're not necessary.[/hide][/quote]

[hide="Computationally intensive way"]$\tan{A}=\frac{\frac{1}{2}}{\frac{\sqrt{3}}{2}}=\frac{\sqrt{3}}{3}$

$\tan{B}=\frac{\frac{\sqrt{6}-\sqrt{2}}{4}}{\frac{\sqrt{6}+\sqrt{2}}{4}}=\frac{\sqrt{6}-\sqrt{2}}{\sqrt{6}+\sqrt{2}}=\frac{8-4\sqrt{3}}{4}=2-\sqrt{3}$

$\frac{3+\sqrt{3}}{3}\times (3-\sqrt{3})=\boxed{2}$[/hide]

And once you're tired of computation...

[hide="A better approach"] $\tan 45 = \frac{\tan 15+\tan 30}{1-\tan 15 \tan 30}= 1 \Leftrightarrow$
$\tan 15 \tan 30 = 1-\tan 15 \tan 30 \Leftrightarrow$
$(1+\tan 15)(1+\tan 30) = \boxed{2}$ [/hide]


What digit must be placed in the blank to make the four-digit integer $ 20\_7$ a multiple of 11?

what is the divisibility rule for 11? if u know that, u can just apply it here. :rotfl:

Let $ a$ be the missing digit.  Then \[ 2000\plus{}10a\plus{}7\equiv 0\pmod{11}\Rightarrow a\equiv \boxed{5}\pmod{11}\]

Since this is in a state countdown, it helps to know the divisibility rule for $ 11$, which is that a number $ a_1a_2a_3a_4...a_n$ is divisible by $ 11$ if and only if \[ a_1\minus{}a_2\plus{}a_3\minus{}a_4\plus{}\cdots \plus{}(\minus{}1)^{n\minus{}1}a_n\] is divisible by $ 11$.  So here, we get $ 2\minus{}0\plus{}a\minus{}7\equiv 0\pmod{11}$, which gets you the same result.

Or you could know that 2002 is a multiple of 11 (it is 1001x2) meaning that to get a 7 on the end, we get 5 11's making 2057.


So I told you to go home and prove this:

Triangle BAC is a right triangle with A being the right angle. Draw the altitude AD to the hypotenuse Prove that BD*DC = AD^2

similar triangles ftw

lol, why am I here, I don't live in oklahoma...

eh.. I kind of get it, but I don't know which Similarity Theorem I used to determine this.... or do you compare it with bigger triangle first??

[hide]I compared 2 small similar triangles, $ \triangle BAD$, and $ \triangle ACD$ and set up a ratio like this:

$ \frac{BD}{AD}\equal{}\frac{AD}{DC}$ and we cross multiply to get $ AD^2\equal{}BD \cdot DC$[/hide]

I think my equations should be right, but can't really prove where I got it from...

Prove:
   $ \sigma(n) \le H_n \plus{} \ln(H_n)e^{H_n}$
where $ \sigma(n)$ is the sum of the divisors and $ Hn$ is the nth harmonic number. :)

[quote="Chinaboy"]Good job, DK!

Worked on those area problems that you missed last week? *cough*[/quote]

Okay, if I can find my paper...it's somewhere in my room..but I can find it.

Can you fully explain how those 2 triangles end up to be similar?

I knew that they have to end up to be similar in the first place..somehow..
I think you went over this with me once... :?:

[quote="Chinaboy"]The two non-right angles of a right triangle add up to 90 degrees. And you have many right triangles...[/quote]

So you don't use Similar Congruency, like AA or something like that?

Well, yes, you have to find the similar triangles first, but finding them isn't all that hard. There are loads of them in the problem. The hard thing is to find the right similar triangles to use.


calculus integration limit trigonometry logarithms inequalities function

Prove the following inequality.

\[ 1\leq \lim_{n\to\infty} \int_{\frac{1}{n}}^{\frac{\pi}{2}} \frac{\pi \sin x}{x(\pi \minus{}x)}\ dx<3\].

I think that we can agree that on the interval $ \left [ 0, \frac{\pi} {2} \right ]$ we have

$ \frac{2}{\pi}x \leq \sin x \leq x$

Now, multiplying everything with

$ \frac{\pi}{x(\pi\minus{}x)}$

we get

$ \frac{2}{\pi\minus{}x} \leq \frac{\pi \sin x}{x(\pi\minus{}x)} \leq \frac{\pi}{\pi\minus{}x}$

Now, integrating from $ \frac{1}{n}$ to $ \frac{\pi}{2}$ we get:

$ \minus{}2\ln(\pi\minus{}x)\Big|^\frac{\pi}{2}_\frac{1}{n}\leq \int^\frac{\pi}{2}_\frac{1}{n}\frac{\pi \sin x}{x(\pi\minus{}x)}\,dx \leq \minus{}\pi\ln(\pi\minus{}x)\Big|^\frac{\pi}{2}_\frac{1}{n}$

$ 2\ln\left(\pi\minus{}\frac{1}{n}\right)\minus{}2\ln\left(\frac{\pi}{2}\right)\leq \int^\frac{\pi}{2}_\frac{1}{n}\frac{\pi \sin x}{x(\pi\minus{}x)}\,dx \leq \pi \ln \left (\pi \minus{} \frac{1}{n} \right) \minus{} \pi \ln \left ( \frac {\pi} {2}\right)$

Letting $ n \rightarrow \infty$ we get

$ 2\ln 2 \leq \lim_{n\rightarrow \infty}\int^\frac{\pi}{2}_\frac{1}{n}\frac{\pi \sin x}{x(\pi\minus{}x)}\,dx \leq \pi \ln 2$

$ 1<1.38629... \leq \lim_{n\rightarrow \infty}\int^\frac{\pi}{2}_\frac{1}{n}\frac{\pi \sin x}{x(\pi\minus{}x)}\,dx \leq 2.17759...<3$

[quote="lambda"]$ \lim_{n\to\infty}\int_{\frac{1}{n}}^{\frac{\pi}{2}}\frac{\pi\sin x}{x(\pi-x)}\ dx =\int_{0}^{\frac{\pi}{2}}\frac{\sin x}{x}\ dx +\int_{0}^{\frac{\pi}{2}}\frac{\sin x}{\pi-x}\ dx=\int_{0}^{\pi}\frac{\sin x}{x}\ dx  =\cdots$(substitution:$[x:=\pi-x]$)[/quote]
And therefore:
\[\int_0^\pi\frac{\sin x}x dx=\int_0^\pi\int_0^\infty e^{-tx}\sin x\ dtdx
=\int_0^\infty\int_0^\pi e^{-tx}\sin x\ dxdt=\\=\int_0^\infty\Im\left(
\frac{e^{x(-t+i)}}{-t+i}\Big|_0^\pi\right) dt=\int_0^\infty\frac{1+e^{-\pi t}}{t^2+1} dt=\frac\pi2+\int_0^\infty\frac{e^{-\pi t}}{t^2+1} dt.
\]
Now since $0\le\frac1{t^2+1}\le1$ one  immediately gets the estimates
\[1.57...=\frac\pi2<\int_0^\pi\frac{\sin x}x\ dx<\frac\pi2+\int_0^\infty e^{-\pi t}\ dt=\frac\pi2+\frac1\pi=1.889...\]

Another solution: As previous posts we can easily show that

\[ \lim_{n\to\infty}\int_{\frac{1}{n}}^{\frac{\pi}{2}}\frac{\pi\sin x}{x(\pi-x)}\ dx = \int_{0}^{\pi}\frac{\sin x}{x}\ dx \]

Now using the well-known inequalities $1-\frac{x^2}{6} \leq \frac{\sin x}{x} \leq 1-\frac{x^2}{6}+\frac{x^4}{120},$ we get

\[  \int_{0}^{\pi} 1-\frac{x^2}{6}\; dx \leq \int_{0}^{\pi}\frac{\sin x}{x}\ dx \leq \int_0^{\pi} 1-\frac{x^2}{6}+\frac{x^4}{120} \; dx \]

\[ 1.419 \approx \pi - \frac{{\pi}^3}{18} \leq \int_{0}^{\pi}\frac{\sin x}{x}\ dx \leq \pi - \frac{{\pi}^3}{18}+\frac{{\pi}^5}{600} \approx 1.929 \]

First of all, since $\frac{\pi sin(x)}{x(\pi-x)}\geq 0$ for all $0\leq x\leq \frac{\pi}{2}$, by the monotone convergence theorem, we have that
${lim_{n}{\int_{\frac{1}{n}}}^{\frac{\pi}{2}}\frac{\pi sin(x)}{x(\pi-x)} dx=\int_{0}}^{\frac{\pi}{2}}\frac{\pi sin(x)}{x(\pi-x)} dx $
Consider the following function:
$g(x)=sin(x)/x$ for $0\leq x\leq \frac{\pi}{2}$.
It is well known that this real valued function is a decreasing map such that $g(0)=1$ and $g( \frac{\pi}{2})= \frac{2}{\pi}$.
By using this inequality, lets compute
$\int_{0}^{\frac{\pi}{2}}{\frac{\pi sin(x)}{x(\pi-x)} dx}\leq \pi \int_{0}^{\frac{\pi}{2}}{\frac{dx}{\pi-x}}=\pi ln(2)<4ln(2)$
which is approximately $2,17$.
Finally it is clear that $\pi - x\leq \frac{\pi}{2}$
and that $2\leq \pi\frac{sin(x)}{x}$
hence $2=\int_{0}^{ \frac{\pi}{2}}{\frac{4}{pi}dx}\leq \int_{0}^{\frac{\pi}{2}}{\frac{\pi sin(x)}{x(\pi-x)} dx}$

[quote="toshiyuki"]hence $2=\int_{0}^{ \frac{\pi}{2}}{\frac{4}{pi}dx}\leq \int_{0}^{\frac{\pi}{2}}{\frac{\pi sin(x)}{x(\pi-x)} dx}$[/quote]
But the value of the integral is $\int_{0}^{\frac{\pi}{2}}{\frac{\pi sin(x)}{x(\pi-x)}dx}=1.85...$


function
logarithms
floor function
algebra
domain
functional equation
algebra unsolved

Find all functions $f: \mathbb R^{+}\to \mathbb R^{+}$ such that for all real $x$, we have
\[f(x)^{2}=f(x^{2})+2.\]

haven't seen it, but i'll try to explain an idea for a solution.
[hide]
its' easy to see that $f(1)=2$ just by pluging $1$into the equation and that $f(x)>\sqrt{2}$.
$f(x)=\sqrt{2+f(x^{2})}\geq \sqrt{2+\sqrt{2}}$ since the limit of $\sqrt{2+\sqrt{2+...}}=2$ we get that $f(x)\geq 2$ for all $x$
now you need an infinite partition of $R$ where each subset is closed under the operation of squaring and square-rooting or equivalently
if $x$ is in $A$ then $x^{2^{n}}$ is in $A$ for all integers $n$
for any such subset $A$ pick one number $a_{0}$ and let $f(a_{0})$ be an arbitrary value greater than $2$. from that you can calculate every value for $a$ from $A$ by solving the recurence relation $a_{n+1}+2=a_{n}^{2}$ which holds for every integer $n$ and all $a_{n}$ are positive...
djuro gave me an idea for this one.
set $f(a_{0})=x+\frac{1}{x}$ for $x>1$ then $f(a_{n})=x^{2^{n}}+\frac{1}{x^{2^{n}}}$ for any integer $n$
since $x=\frac{f(a_{0})+\sqrt{f(a_{0})^{2}-4}}{2}$ we now have a direct formula for $a_{n}$

[/hide]

Hmm..., goc, just a little thing, function is $f: R^{+}\to R^{+}$ and therefore $f(0)$ does not exist, as I can see it.

thanks. doesn't change much, totaly forgot bout it...
for mornik
nadam se da vjezbas(cini mi se bar da vjezbas posto si konstantno na forumu ;) )...ak trebas kaj,javi se

I see the solution $f(x) = x+\frac{1}{ x}$.

ja sam inace smislio taj zadatak al ga ne znam rijesit.

satja

Well, there may be other solutions, but anyhow you should have posted it in Algebra: Proposed&Own if you came up with the problem yourself.

@goc: A nešto malo radim. Nemam kaj doma radit pa sjedim na forumu.

More generally, the function $f(x) = x^{c}+x^{-c}$ works, where $c$ is a real constant.

djuro, did you want to assume that the function $f$ satisfies some continuity property?  Otherwise, you could get crazy examples by choosing different $c$ above for different $x$.

[quote="goc"]he doesn't know anything about continuity yet, he's a student of mine... :)
try to solve it with assumption of continuity...[/quote]

Let's make a good problem then :)
I want not less than 50% of Copyright.

p.s. jel continuous znaci neprekidna?

Da. Pretty much all the nice functions are continuous, even though I just know goc will find one which is discontinuous and try to convince us it's really nicer than, say, $f(x)=x^{2}$.  :D
http://en.wikipedia.org/wiki/Continuous_function

[quote="djuro"]Find all functions $f: R^{+}\to R^{+}$ such that
\[f(x)^{2}=f(x^{2})+2. \]
I'd like to know has anyone seen this before?[/quote]

Using the idea of Goc, it is possible to show that a general solution of this functional equation is :

Let $u(x)$ and $v(x)$ any two funtions defined in $[0,1)$ and whose values are strictly positive. Then $f(x)$ is defined as :

For $x>1$, $f(x)=u(\{\frac{\ln(\ln(x))}{\ln(2)}\})^{2^{\lfloor\frac{\ln(\ln(x))}{\ln(2)}\rfloor}}+u(\{\frac{\ln(\ln(x))}{\ln(2)}\})^{-2^{\lfloor\frac{\ln(\ln(x))}{\ln(2)}\rfloor}}$
For $x=1$, $f(x)=2$
For $0<x<1$, $f(x)=v(\{\frac{\ln(-\ln(x))}{\ln(2)}\})^{2^{\lfloor\frac{\ln(-\ln(x))}{\ln(2)}\rfloor}}+v(\{\frac{\ln(-\ln(x))}{\ln(2)}\})^{-2^{\lfloor\frac{\ln(-\ln(x))}{\ln(2)}\rfloor}}$

Where $\{y\}$ means fractional part of $y$

[quote]p.s. jel continuous znaci neprekidna?[/quote]

Jeste.

Would the problem be better if one adds more conditions?
Is this problem worth something?

I ask because it's mine :) I put Copyright on this prob.

[quote="behemont"]pco, how do you come up with something like that?[/quote]

1) $f(x)\geq 2$ $\forall x>0$
We have $f(x)>0$. Then, if it exists $a>0$such that $0<f(a)<2$, then we have $0<f(a^{2})<f(a)<2$ and the sequence $a_{n}=f(a^{2^{n}})$ is a strictly decreasing sequence whose each element is in $(0,2)$. So this sequence has a limit which verifies $L=L^{2}-2$ and so $L=-1$ or $L=2$, which is impossible if $0<f(a)<2$.
So $f(x)\geq 2$ $\forall x>0$

2) $f(x)=a(x)+\frac{1}{a(x)}$ with $a(x)\geq 1$ and $a(x)^{2}=a(x^{2})$.
For any $y\geq 2$, it exists a unique real $z\geq 1$ such that $y=z+\frac{1}{z}$. So it exists a unique function $a(x)\geq 1$ such that $f(x)=a(x)+\frac{1}{a(x)}$.
Then $f(x)^{2}=f(x^{2})+2$ becomes $a(x)^{2}+\frac{1}{a(x)^{2}}=a(x^{2})+\frac{1}{a(x^{2})}$.
Then, since there is a unique way of splitting a number $\geq 2$ in a sum of a number $\geq 1$ and its inverse, we have $a(x)^{2}=a(x^{2})$

3) Solutions of $a(x)\geq 1$ and $a(x)^{2}=a(x^{2})$ $\forall$ $x>0$
Since $x>0$, we can write $a(x)=b(\ln(x))$ and we have $b(\ln(x))^{2}=b(2\ln(x))$ $\forall$ $x>0$ and so $b(x)^{2}=b(2x)$ $\forall$ $x$
Then, we can distinguish 3 situations :

3.1) $x>0$. Let then $b(x)=c(\frac{\ln(x)}{\ln(2)})$ and we have $c(\frac{\ln(x)}{\ln(2)})^{2}=c(1+\frac{\ln(x)}{\ln(2)}$ and so $c(x+1)=c(x)^{2}$
So $c(x+n)=c(x)^{2^{n}}$ and so we can define $c(x)=u(x)$ for $x\in[0,1)$ ($u(x)$ is any function defined on $[0,1)$ and $\geq 1$) and then :
$c(x)=c(\lfloor x\rfloor+\{x\})=c(\{x\})^{2^{\lfloor x \rfloor }}=u(\{x\})^{2^{\lfloor x \rfloor }}$

3.2) $x=0$ $\implies$ $b(0)^{2}=b(0)$ and, since $a(x)\geq 1$ and $b(x)\geq 1$ : $b(0)=1$.

3.3) $x<0$. Let then $b(x)=c(\frac{\ln(-x)}{\ln(2)})$ and we have $c(\frac{\ln(-x)}{\ln(2)})^{2}=c(1+\frac{\ln(-x)}{\ln(2)}$ and so $c(x+1)=c(x)^{2}$
So $c(x+n)=c(x)^{2^{n}}$ and so we can define $c(x)=v(x)$ for $x\in[0,1)$ ($v(x)$ is any function defined on $[0,1)$ and $\geq 1$) and then :
$c(x)=c(\lfloor x\rfloor+\{x\})=c(\{x\})^{2^{\lfloor x \rfloor }}=v(\{x\})^{2^{\lfloor x \rfloor }}$

4) General solution of the requested functional equation :
Let $u(x)$ and $v(x)$ any two functions $\geq 1$ and defined in $[0,1)$ :
$x>1$, $f(x)=u(\{\frac{\ln(\ln(x))}{\ln(2)}\})^{2^{\lfloor \frac{\ln(\ln(x))}{\ln(2)}\rfloor }}+u(\{\frac{\ln(\ln(x))}{\ln(2)}\})^{-2^{\lfloor \frac{\ln(\ln(x))}{\ln(2)}\rfloor }}$
$x=1$, $f(x)=2$
$x<1$, $f(x)=v(\{\frac{\ln(-\ln(x))}{\ln(2)}\})^{2^{\lfloor \frac{\ln(-\ln(x))}{\ln(2)}\rfloor }}+v(\{\frac{\ln(-\ln(x))}{\ln(2)}\})^{-2^{\lfloor \frac{\ln(-\ln(x))}{\ln(2)}\rfloor }}$

5) We can retrieve the solution $f(x)=x^{a}+\frac{1}{x^{a}}$ : Just take $u(x)=v(x)=e^{a2^{x}}$

[quote="djuro"]Find all functions $f: R^{+}\to R^{+}$ such that
\[f(x)^{2}=f(x^{2})+2. \]
I'd like to know has anyone seen this before?[/quote]

About your second question : this equation is in the very frequent class of equations like : $f\circ u=v\circ f$ where $u(x)$ and $v(x)$ are bijective functions.

You must first check that if $u(x)$ has fixed points, $v(x)$ has too  (since $u(x)=x$ $\implies$ $v(f(x))=f(x))$
For such an equation, you have $f\circ u^{\circ n}=v^{\circ n}\circ f$ (where $h^{\circ n}=h\circ h\circ h\cdots \circ h$ $n$ times).
Then the key point is to find (if possible) a subset $S$ of the domain of $f(x)$ such that any real $y$ may be written in a unique way as $y=u^{\circ n}(x)$ with $x\in S$ and $n\in\mathbb{Z}$

If this is possible, an if we call $n(x)$ (values in $\mathbb{Z}$) and $seed(x)$ (values in $S$) the two numbers such that $x=u^{\circ n(x)}(seed(x))$, and if we have $h(x)$ any function defined in $S$, then :

Obviously $n(u(x))=n(x)+1$ and $seed(u(x))=seed(x)$ and then :

$f(x)=v^{\circ n(x)}(h(seed(x)))$

Trivial examples :

1) $f(x+1)=f(x)+7$
We have $u(x)=x+1$ and $v(x)=x+7$
We can take $S=[0,1)$, $n(x)=\lfloor x\rfloor$ and $seed(x)=\{x\}$ and we have the general solution $f(x)=h(\{x\})+7\lfloor x \rfloor$ with $h(x)$ beeing any function defined in $[0,1)$
Obviously, in this case, we can write $k(x)=h(x)-7x$ and then $f(x)=k(\{x\})+7x$

2) $f(x+1)=(f(x))^{2}$
We have $u(x)=x+1$ and $v(x)=x^{2}$
We can take $S=[0,1)$, $n(x)=\lfloor x\rfloor$ and $seed(x)=\{x\}$ and we have the general solution $f(x)=h(\{x\})^{2^{\lfloor x \rfloor}}$ with $h(x)$ beeing any function defined in $[0,1)$ and with values $\geq 0$
For example, with $h(x)=a^{2^{x}}$, we have the simple solution $f(x)=a^{2^{x}}$

3) $f(x^{3})=f(x)+9$
We have $u(x)=x^{3}$ and $v(x)=x+9$
Impossible since $u(x)$ has fixed points $-1$ and $+1$ but $v(x)$ has no fixed points.

....
In your case, we have $f(x^{2})=f(x)^{2}-2$
We have $u(x)=x^{2}$ and $v(x)=x^{2}-2$ and $u(x)$ is a bijective function  $R^{+}\to R^{+}$
...


Let $a_1,..,a_n,b_1,...,b_n$ be real numbers. Let $x_{ij}$ be the number of indices $k$ such that $b_k \geq max(a_i,a_j)$.  Suppose that $x_{ij}$>0 for any $i$ and $j$. Prove that we can find an even permutation $f$ and an odd one $g$ such that $\sum x_{if(i)}/x_{ig(i)} \geq n$.

In the mean time, I have found out what are odd/even permutations.

But what about the solution of this problem?


Each face of a cube is painted either red or blue, each with probability $\frac{1}{2}$. The color of each face is determined independently. What is the probability that the painted cube can be placed on a horizontal surface so that the four vertical faces are all the same color?

[hide]The top, bottom, and one of the verical faces can be any color. Then the other three vertical faces must be the same as the first. So the probability is $(\frac{1}{2})^{3}=\frac{1}{8}$[/hide]

that's not one of the choices.

the choices are 1/4, 5/16, 3/8, 7/16, and 1/2   :blush:

Well...

[hide]
If all the faces are the same color (2 ways), the condition can be satisfied.

The condition can also be satisfied if 5 faces are one color and the other face is the other color.

There are $2\cdot\binom{6}{5}=12$ ways in this case to color the cube.

Finally, you can have 4 faces of the same color and 2 [i]opposite[/i] faces of the other color.

There are 3 pairs of opposite faces, so there are $2\cdot 3=6$ ways in this case to color the cube.

The probability is

$\frac{2+12+6}{64}=\boxed{\frac{5}{16}}$.[/hide]


projective geometry
geometry
perpendicular bisector
geometry proposed

Let circles $k_1$ and $k_2$ intersect in $P$ and $Q$ and let $A,\ C\in k_1$ and $B,\ D\in k_2$ such that $A-B-C-D$ and $AB$ is a perpendicular bisector of $[PQ]$. Let $X$ be an arbitraty point on $PQ$ and $R,\ S$ be intersection points, other than $C$ and $B$, of $XC$ and $XB$ with $k_1$ and $k_2$, in that order. Let $AX\cap RP=\{T_1\}$, $AS\cap RD=\{T_2\}$, $XS\cap PD=\{T_3\}$. Prove that $T_1$, $T_2$ and $T_3$ are collinear.

[i]Remark.[/i] The notation $A-B-C-D$ means that the points $A$, $B$, $C$, $D$ are collinear in that order.

[quote="Bojan Basic"]Let circles $k_1$ and $k_2$ intersect in $P$ and $Q$ and let $A,\ C\in k_1$ and $B,\ D\in k_2$ such that $A-B-C-D$ and $AB$ bisects $[PQ]$. Let $X$ be an arbitraty point on $PQ$ and $R,\ S$ be intersection points, other than $C$ and $B$, of $XC$ and $XB$ with $k_1$ and $k_2$, in that order. Let $AX\cap RP=\{T_1\}$, $AS\cap RD=\{T_2\}$, $XS\cap PD=\{T_3\}$. Prove that $T_1$, $T_2$ and $T_3$ are collinear.

[i]Remark.[/i] The notation $A-B-C-D$ means that the points $A$, $B$, $C$, $D$ are collinear in that order.[/quote]

Are you sure about this problem? On my sketch, it doesn't hold. The only case when we really can assure that the points $T_1$, $T_2$ and $T_3$ are collinear is when the line through the points A, B, C, D coincides with the line joining the centers of the circles $k_1$ and $k_2$; in other words, it is the case when the segments AC and BD are diameters of the circles $k_1$ and $k_2$. In this case, the solution of the problem is easy: After [url=http://www.kalva.demon.co.uk/imo/isoln/isoln951.html]the IMO 1995 problem 1[/url], the lines AR, DS, PQ are concurrent; in other words, the lines XP, AR and SD are concurrent. Hence, after the Desargues theorem, applied to the triangles XAS and PRD, it follows that the three points $XA\cap PR$, $AS\cap RD$ and $SX\cap DP$ are collinear. But these three points are just your points $T_1$, $T_2$ and $T_3$: in fact, $XA\cap PR=T_1$, $AS\cap RD=T_2$ and $SX\cap DP=T_3$. Hence, the points $T_1$, $T_2$ and $T_3$ are collinear. $\blacksquare$

  Darij

$AB$ is a perpendicular bisector of $[PQ]$, I'd made a typo (corrected now), sorry about that. Nice solution, congratulations.


limit
logarithms
real analysis
real analysis solved

Find $ \lim_{n \to \infty}(2 \sqrt[n]{x}-1)^n $ , where $ x >0 $

But it is seems easy.Just take its logarithm:
$n\ln{(2\sqrt[n]{x}-1)}{\sim}n(2\sqrt[n]{x}-2){\sim}2\ln{x}$.
So the original limit is $e^{2\ln{x}}=x^{2}$.

[quote="LevonNurbekian"][quote="Matnomi"][quote="LevonNurbekian"]
$n\ln{(2\sqrt[n]{x}-1)}{\sim}n(2\sqrt[n]{x}-2){\sim}2\ln{x}$.
[/quote]
Could you give me a short description of each step ?[/quote]
In each step I used famous limit $\lim_{x{\rightarrow}0}{\frac{\ln{(1+x)}}{x}}=1$ i.e.
$\ln{(1+x)}{\sim}x$,when $x$ is goes to zero.Since $\sqrt[n]{x}{\rightarrow}1$ when $n$ goes to $\infty$
$\ln{(2\sqrt[n]{x}-1)}{\sim}(2\sqrt[n]{x}-2)=2(\sqrt[n]{x}-1){\sim}2\ln{(\sqrt[n]{x}-1+1)}=\frac{2\ln{x}}{n}$.[/quote]

you get $\frac{2\ln{x}}{n}$ at the end, what happen to the $n$???  how do you get it to become $2\ln{x}$ ??

Nice solution ,  Levon !
   I have another solution for this problem .
 $ (\sqrt[n]{x}-1)^2 \geq 0 \to x^2 \geq (2\sqrt[n]{x}-1)^n $
 On the other hand :
  $ (2\sqrt[n]{x}-1)^n = x^2(1+(\frac{2}{\sqrt[n]{x}}-\frac{1}{\sqrt[n]{x^2}}-1))^n \geq x^2(1-n.\frac{(\sqrt[n]{x}-1)^2}{\sqrt[n]{x^2}}) $
   We have :
$ x= (\sqrt[n]{x}-1+1)^n \geq n(\sqrt[n]{x}-1) $
  So $ (2\sqrt[n]{x}-1)^n \geq x^2(1-\frac{x^2}{n\sqrt[n]{x^2}}) $
 This infers $ \lim_{n\to \infty}(2\sqrt[n]{x}-1)^n= x^2 $  ;)


Let p(x) be a 1999-degree polynomial with integer coefficients that is equal to +-1 for 1999 different integers x. Show that p(x) can't be written as the product of 2 polynomial with integer coefficients.

There must be another hypothesis otherwise it is untrue for $p(x)=x^2$.

Pierre.

Also $p(x)=(x^2-3x+1)^2$ satisfies $p(0)=p(1)=p(2)=p(3)=1$.
It is known that if $p(x)=(x-a_1)(x-a_2)\dots (x-a_n)-1$ where $a_i$ are distinct integers is irreducible.

Suppose $p(a_{1})=\ldots=p(a_{k})=1$ and $p(b_{1})=\ldots=p(b_{m})=-1,$ where
$a_{i},b_{j}$ are distinct integers and $k+m=1999$. Then $p(x)=(x-a_{1}
)\ldots(x-a_{k})f(x)+1=(x-b_{1})\ldots(x-b_{m})g(x)-1,$ for some integer
polynomials $f$ and $g.$ It follows that

\[
(x-b_{1})\ldots(x-b_{m})g(x)-(x-a_{1})\ldots(x-a_{k})f(x)=2.
\]
Plugging $x=a_{1},\ldots,a_{k}$ we obtain that $2$ has at least $k$ distinct
divisors. Similarly, 2 has at least $m$ distinct divisors. Since $k+m=1999,$
this is a contradiction.

It follows that there exist $a_{1},\ldots,a_{1999}$ such that either
$p(a_{i})=1$ for all $i,$ or $p(a_{i})=-1$ for all $i.$ Suppose $p(a_{i})=1$
for all $i$ and $p(x)=f(x)g(x)$ for some nonconstant integer polynomials $f$
and $g.$ We have $f(a_{i})g(a_{i})=1,$ hence $f(a_{i})-g(a_{i})=0,$ for all
$i.$ But $\deg(f-g)<1999,$ hence $f-g=0.$ It follows that $p=f^{2},$ a
contradiction since $p$ has an odd degree.

Yes, if the degree is supposed to be odd, the conclusion follows easily (I assume that you meant that the two polynomials were supposed to be nonconstant, otherwise choose $Q(x)=1$...)

Assume that $P(x) = Q(x)R(x)$ where $Q,R$ are two nonconstant integer polynomials with respective degrees $n,q,r$. Wlog we may assume that $q \leq r$ and $n$ is odd. Since $p+q = n$ we deduce that $q < \frac n 2$. (1)
Now, let $x_1, \cdots, x_n$ be distinct integers such that $p(x_i)= \pm 1$ for $i=1, \cdots , n$.
Thus $Q(x_i) = \pm 1$ for all $i$. It follows that $Q(x)-1$ or $Q(x)+1$ has at least $\frac n 2$ distinct roots, which is impossible from (1).

Pierre.

P.S. Orlando, I win  :P  (1801)

Yes, but the same argument works if $\deg p=2k,$ with $k\geq3.$ Suppose $p=f^{2}$ and $p\left(  a_{i}\right)  =1,$ for $i=1,\ldots,2k.$ Then $f\left( a_{i}\right)  =1$ for $k$ values of $i$ and $f\left(  a_{i}\right)  =-1$ for
the other $k$ values. Assume that $f\left(  a_{1}\right)  =\ldots=f\left( a_{k}\right)  =1$ and $f\left(  a_{k+1}\right)  =\ldots=f\left( a_{2k}\right)  =-1.$ Then
\[
a\left(  x-a_{1}\right)  \ldots\left(  x-a_{k}\right)  -a\left(  x-a_{k+1}
\right)  \ldots\left(  x-a_{2k}\right)  =2.
\]
for some integer $a$. We obtain that if $k\geq3,$ then $2$ has at least $5$ distinct divisors.

As I posted before, the assertion is not true for $k=2.$


Each of the numbers 1, 2, ... , 10 is colored red or blue. 5 is red and at least one number is blue. If m, n are different colors and m+n ≤ 10, then m+n is blue. If m, n are different colors and mn ≤ 10, then mn is red. Find all the colors

???? the colors are red and blue  :D

Pierre.

I think Peter means to find the color associated with each number.

WWWoops! :blush:

Indeed Fierytycoon, that's what i meant :D

This problem is very nice (even if it is said to be easy).

Let b1 the first blue number.

If 1 is red then 1*b1 is red, impossible so 1 is blue.
If r1, r2, .., r5, .. are the red numbers, then r1+1, ..., r4+1 are blue (r5 may be 10). So we have at least 5 blue numbers, since there can't be more then 5, there are 5.

So 1 is blue, there are 5 red (r1..r5) and five blue numbers (b1..b5) and we can't have two consecutive red numbers.

Now 10 can't be blue otherwise 2 red numbers would be consecutive. So 10 is red, so 9 is blue. Likewise 8 can't be blue, ....

I find that only possible distribution is B R B R B R B R B R which is checked to be correct.

:D belenos, you have solved a slight variant of this problem

My problem says "5 is red" with which I mean: "Number 5 is red" and not "5 numbers are red".

So you got a solution for this version too? :)


[quote="xzlbq"][size=150]Let a,b,c be positive numbers,prove that:
$ {\frac {a \plus{} t}{b \plus{} t}} \plus{} {\frac {b \plus{} t}{c \plus{} t}} \plus{} {\frac {c \plus{} t}{a \plus{} t}}\leq {\frac {a }{b}} \plus{} {\frac {b}{c}} \plus{} {\frac {c}{a}}$

BQ[/size][/quote]
Because it's equivalent to
$ (t^3\plus{}(a\plus{}b\plus{}c)t^2)\sum_{cyc}(a^2c\minus{}abc)\plus{}t\sum_{cyc}(a^3c^2\minus{}a^2b^2c)\geq0.$

[quote="xzlbq"][size=150]Let a,b,c be positive numbers,$ t \geq 0$prove that:
$ {\frac {a \plus{} t}{b \plus{} t}} \plus{} {\frac {b \plus{} t}{c \plus{} t}} \plus{} {\frac {c \plus{} t}{a \plus{} t}}\leq {\frac {a }{b}} \plus{} {\frac {b}{c}} \plus{} {\frac {c}{a}}$

BQ[/size][/quote]

without loss of generality, we can assume that $ c \equal{} \min\{a,b,c\}$, and we have

\[ \frac {a}{b} \plus{} \frac {b}{c} \plus{} \frac {c}{a} \minus{} \left( {\frac {{a \plus{} t}}{{b \plus{} t}} \plus{} \frac {{b \plus{} t}}{{c \plus{} t}} \plus{} \frac {{c \plus{} t}}{{a \plus{} t}}} \right)\]
\[ \equal{} \left( {\frac {1}{{ab}} \minus{} \frac {1}{{\left( {a \plus{} t} \right)\left( {b \plus{} t} \right)}}} \right)\left( {a \minus{} b} \right)^2 \plus{} \left( {\frac {1}{{ac}} \minus{} \frac {1}{{\left( {a \plus{} t} \right)\left( {c \plus{} t} \right)}}} \right)\left( {a \minus{} c} \right)\left( {b \minus{} c} \right),\]
so, ineq clear true.

[quote="xzlbq"][size=150]Let a,b,c be positive numbers,$ t \geq 0$prove that:
$ {\frac {a \plus{} t}{b \plus{} t}} \plus{} {\frac {b \plus{} t}{c \plus{} t}} \plus{} {\frac {c \plus{} t}{a \plus{} t}}\leq {\frac {a }{b}} \plus{} {\frac {b}{c}} \plus{} {\frac {c}{a}}$

BQ[/size][/quote]

another solution:

let

$ f(t) \equal{} \frac {a}{b} \plus{} \frac {b}{c} \plus{} \frac {c}{a} \minus{} \left( {\frac {{a \plus{} t}}{{b \plus{} t}} \plus{} \frac {{b \plus{} t}}{{c \plus{} t}} \plus{} \frac {{c \plus{} t}}{{a \plus{} t}}} \right),(t \ge 0)$

then we have

$ f'(t) \equal{} \frac {a \plus{} t}{(b \plus{} t)^2} \plus{} \frac {b \plus{} t}{(c \plus{} t)^2} \plus{} \frac {c \plus{} t}{(a \plus{} t)^2} \minus{} \frac {1}{a \plus{} t} \minus{} \frac {1}{b \plus{} t} \minus{} \frac {1}{c \plus{} t},$

easy to prove that

$ \frac {a \plus{} t}{(b \plus{} t)^2} \plus{} \frac {b \plus{} t}{(c \plus{} t)^2} \plus{} \frac {c \plus{} t}{(a \plus{} t)^2} \ge \frac {1}{a \plus{} t} \plus{} \frac {1}{b \plus{} t} \plus{} \frac {1}{c \plus{} t},$ (can use only AM-GM)

thus, we get $ f'(t) \ge 0$, so we have

$ f(t) \ge f(0) \equal{} 0,$

and get the result.

Remark: this prove can be work to $ n$ variable.  :maybe:


linear algebra
matrix
algebra
polynomial
combinatorics unsolved
combinatorics

A $2n\times 2n$-table is filled with $+1$'s and $-1$'s, and the sum of all numbers in the table is not negative. Prove that we can select $n$ rows and $n$ columns such that the $n\times n$ table formed by these columns and rows has sum at least $n$ (that is, the sum of all its numbers is at least $n$).
I have solved this for odd $n$.

It is All-Russia Olympiad'95, 8-th problem for 9 grade.

I think I have a valid solution. I present very briefly it's main steps
Call table the interesections of some rows with some columns. (m n) is the coeff of [x^n] in (1+x)^m.
For n odd, it's obvious. Take a 2n*n table with nonnegative sum.
The n rows with biggest sums satisfy the condition (exercise!)
For n even, observe the lemma:
if we have 2n even numbers a/1/..a/2n/ with the sum of positive at least n+k and total sum at least n-k we can select n of them to sum at least to n (exercise!)
Suppose the affirmation of the problem is not true. We can suppose the sum of numbers in the whole matrix is 0.
Applying lemma to the rows or colums, we have the sum of positive rows(
with positive sum) is at most n-1, so we have at least n+1 nule rows (the same for columns).
Now if we select n columns and divide the matrix in two 2n*n tables, by applying the lemma to both of them (letting the row to be our a1..an),
we conclude the sum of positive n-rows is at most 2(n-1).Summing over all possibilities we have the sum of all positive subrows of length n is at most
(2n n)*2(n-1).
But in our (n+1) nule colums this sum is sum of (2i*(n n/2+i)^2)*(n+1)
By using the identity (2n n) = sum of (n i)^2 and little estimation, also some estimation for positive rows(exercise!) we conclude it's bigger that
(2n n)*2(n-1) for n \geq 4 or 6.
We are done

It is quite tangled proof. I can't check it at the moment.

Just a little correction.
Write (2n n)*(n-1) instead of (2n n)*2(n-1) because the number of partitions of a set with 2n elements into two with n elements is (2n n)/2, not (2n n).
About the proof, I can say that at All-Russian Olympiads there some problems with longer and trickier solutions

Let me write you the solution in details:
call table the pairwise intersections of some rows with some columns.
Call sum (of table,row,column) the sum of numbers in it.
Call (n m) the binomial coefficient
For n odd, choose first n columns and color then red. the other color black.
We have a 2n*n red table and a 2n*n black table. In one of them the sum is non-negative.
Choose 'it, then order decreasingly its rows a/1/>=..>=a/2n/.
Obviously they are odd, so non-zero
If a/n+1/>=1 then a/1/+..a/n/>=n so we have our n*n table.
If a/N+1/<=-1 then a/n+1/+..a.2n/<=-n. Sunce the sum is nonnegative, a/1/+..a/n/>=n.

The case n even is much harder
Lemma 1. if we have even numbers a/1/>=..>=a/2n/ with sum(a/i/)>=n then a/1/+..a/n/>=n.
-proof:The sum of positive a/i/ is at least n. If all a/1/..a/n/ are positive or 0,
then the sum is at least n. If not, then a/n+1/<0 so we continue like 5 lines above

Lemma 2. if we have even numbers a/1/>=..>=a/2n/ with sum (a/i/) at least -k,
sum of positive a/i/ at least n+k, 0<=k<=n, then a/1/+..a/n/>=n.
-proof: If all a/1/..a/n/ are positive or 0, it's like above.
If else a/n+1/+..a/2n/<=-2*n. If a/1/+..a/n/<n we have sum(a/i/)<-n contradiction

Lemma 3. sum( (n i)^2)=(2n n)
-proof: just compare the coefficient of x^n in the polynomial (1+x)^2n=(1+x)^n*(1+x)^n

Now pass to our problem. Soppose the assertion of the problem is false.
Suppose the sum of the whole matrix is 0.
If we find n columns with total sum at least n, in this 2n*n table denote r/i/ the sum of row i
The r/i/ satisfy lemma 1 so we find our n*n table.
So we can't find n columns with total sum at least n.
If c/i/ is the sum of columns, by lemma 2 the sum of positive c/i/ is at most n-1,
so we have at most n/2-1 positive r/i/, so at most n/2-1 negative r/i/ so at least n+2 r/i/=0
Now let's partition the 2n columns in two n-element subsets. Color the first subset's columns red,
the other black. We get two 2n*n tables. We have shown that the sum of any of them is at most
n-1 above. So the sum of any of them  is at leat 1-n.
We can apply lemma 2 to the table with non-positive sum, and in the table with positive sum
decreasing the sum to 0. We obtain the sum of positive n-rows of the two tables is at most
n+n+(n-1)=3n-1.
Summing over all (2n n)/2 such partitions, we have
the sum of all sub-rows with n elements of the matrix is at most (3n-1)/2*(2n n).
But in any of zero columns the sum of it positive subrows if sum( (n n/2+i)^2*2i)(0<=i<=n/2)
Now summing for the at least (n+2) subrows and comparing it with lemma 3
"i ciuti-ciuti povertetsea" we  conclude it's bigger than (3n-1)/2*(2n n) for n>=10 so
we reach contradiction. For n<=8 we can prove it manually.
The proof is finished.

Can someone verify it? I tried to check the solution half year ago, but I didn't succeed then.  :(

Neither did I .Anyone with shorter proof???? :)  :)

The official solution use the fact that $n$ is even.
Since the sum of all numbers in the table is nonnegative, there is a row containing at least $n$ units. Rearrange columns of the table in order to obtainunits in the first $n$ cells of this row. Let $A$ be a table formed by first $n$ colums and $B$ be a table formed by remaining $n$ columns.
Let $A_1$ be $n$ rows of $A$ with the maximal sums, $A_2$ be remaining rows of $A$. If sum of numbers in $A_1$ is at least $n$ then we are done.
Suppose the contrary. Let's prove that for all rows in $A_2$ the sums are negative. Indeed, if $A_2$ contains a row with nonnegative sum, then all sums in $A_1$ are nonnegative too, but $A_1$ contains a row consisting of $n$ units. It means that the total sum in $A_1$ is at least $n$. Contradiction.
Therefore, all row sums in $A_2$ is at most $-2$ ($n$ is even!). It follows that the sum of all numbers in $A$ is less than $n+n\cdot (-2)=-n$. It follows that the total sum in $B$ is greater than $n$.
Let $B_1$ be $n$ rows of $B$ with maximal sums and $B_2$ be $n$ remaining rows. Let's prove that the total sum in $B_1$ is at least $n$. Indeed, it is true if all sums in $B_2$ are nonpositive, and if $B_2$ contains at least one positive row than all rows in $B_1$ are positive too. QED

a silly question.   What does QED stand for?

[quote="symmetricmean"]What does QED stand for?[/quote]

The name of a mathematical club in Bavaria, Germany.  :D  :rotfl:

There is also a second meaning for "QED", namely "quod erat demonstrandum". An abbreviation one puts at the end of a proof.

  Darij

Oh Myth great proof, Thank you very much ;)  ;)  ;)


Let p/q = 1 - 1/2 + 1/3 - 1/4 + ... - 1/2k + 1/(2k+1)
If (p,q)=1 is it possible to determine k so that p is divisible by 1997?

That wasn't hard to get Bugi, but tough thing is to find the connection of the sum and prime number 1997. It is because it was proposed in year 1997. It could be any other prime number..


I THINK IT'S BETTER NOT TO COMMENT ON THE TIBET ISSUE. AFTER ALL A LOT HAPPENS IN SOUTH-EAST ASIA WHEN IT COMES TO TOP LEVEL POLITICS. WHETHER THE CHINESE LEAVE TIBET OR NOT, A LOT DEPENDS UPON ITS RELATIONS  WITH INDIA, PAK ETC.LET'S NOT CROSS THE LINE BECAUSE WE ARE NOT ALLOWED TO SPREAD COMMUNAL HATRED. SO PLEASE DO NOT OVERSTEP  WHILE REPLYING. MATTERS MAY JUST FLARE UP.AND YES, PEOPLE IN THE WEST ARE NOT AS WELL-EQUIPPED TO COMMENT SPOT-ON ON THIS TOPIC AS PEOPLE IN INDIA.

this is a horrible thread, there is no poll here, and you are off to a horrible start to your posting career here.

congratulations I guess! (thread locked)


geometry
3D geometry
tetrahedron
real analysis
real analysis solved

find a dense subset of R^3 so that 4 points
are never on a plane.

I think the set of points of the form $(e^{r_1},e^{r_2},e^{r_3})$ works, where $r_i$ are rationals. We could also consider $(2^{r_1\sqrt 2},2^{r_2\sqrt 2},2^{r_3\sqrt 2})$, with $r_i$ rational. We use the determinant formula for the volume of a tetrahedron and the transcewndence of the number $2^{\sqrt 2}$ (from Gelfond's theorem). It might work, but I'm not at all sure.

I dont think it works. I think that the points $ A(e,e,e), B(e,e^2,e^2),C(e,e^3,e^3), D(e,e^4,e^4) $ are on the same plane.

You're right.. It was pretty stupid of me to write that :).

It's interesting problem,
First took a contable dense set $\{d_n\}\subset\mathbb{R}^3$ and consider the oepn balls $B_n=B_{1/n}(d_n)$ where
\[B_r(a)=\{x\in\mathbb{R}^3\,/\, |x-a|<r\}\]
Note that if we take $x_n\in B_n$ then $\{x_n\}$ is dense.
(for $x\in\mathbb{R}^3$ and $\varepsilon>0$ there is a $m\in\mathbb{N}$ such that  for $n\geq m$ ,  $1/n<\varepsilon/2$ and for the density of $\{d_n\}$ there is a $k\geq m$ such  $d_k\in B_{\varepsilon/2}(x)$ then $x_k\in B_k\subset B_{\varepsilon}(x)$).
Now lets construct the set $\{x_n\}$: take $x_1=d_1,x_2=d_2,x_3=d_3$ and for $n>3$

\[x_{n+1}\in B_{n+1}-\bigcup_{1\leq i<j<k\leq n}S(x_i,x_j,x_k)\]
Where $S(a,b,c)$ in the plane generated by $a,b,c$. They are well define since any ball cannot be contained in a finite union of planes (neither contable).
Then $\{x_n\}$ is dense and for construction there is no plane that contains 4 points of it.

This construction works in $\mathbb{R}^n$ for $n\geq 2$ (that is in $\mathbb{R}^n$ there is a dense set with no $n+1$ points of it in any hyperplane) anyway it'll be nice see and explicit example of such set.


Can you post a complete solution ?, I don't see it easily as you!

[quote="skywalker"]Given $x_{i}>0, i=1,...,n$ and $n\ge 1$ any $\lambda \ge n^{n-1}-1$ we have
$\sum_{i=1}^{n}\frac{x_{i}}{\sqrt[n-1]{x_{i}^{n-1}+\lambda\prod_{ k\not=i}x_{k}}}\ge\frac{n}{\sqrt[n-1]{\lambda+1}}$
Now let $n=3$ we get for $x,y,z>0$ and $\lambda \ge 8$ prove that
$\frac{a}{\sqrt{a^{2}+\lambda bc}}+\frac{b}{\sqrt{b^{2}+\lambda ca}}+\frac{c}{\sqrt{c^{2}+\lambda ab}}\ge \frac{3}{\sqrt{1+\lambda}}$
Now let $\lambda=8$ we get 2nd of 42 IMO.[/quote]
How about :

for $a,b,c \geq\ 0$ and prove that
$\frac{a}{\sqrt{a^{2}+\lambda bc}}+\frac{b}{\sqrt{b^{2}+\lambda ca}}+\frac{c}{\sqrt{c^{2}+\lambda ab}}\ge min(\frac{3}{\sqrt{1+\lambda}},2)$

$\lambda \geq\--1$

 :D


integration
trigonometry
calculus
calculus computations

Here is a fun little problem.

$\int\sqrt{1+\sin(2\theta)}d\theta$

put ${sin(2\theta) = 2 sin(\theta) cos (\theta)}$ and
insted of 1 write ${ sin^{2}(\theta)+cos^{2}(\theta)}$
the term becomes ${\int (sin(\theta)+cos(\theta) )d\theta}$ which easily done

Well, $\int|\cos\theta+\sin\theta|\,d\theta.$

You do need to pay attention to the $\pm$signs.


Money invested in the Sunshine Investment Club has tripled every 10 years. If $ \$100$ is invested in the account and increases at the same rate, how many dollars will be in the account after 30 years?


modular arithmetic
number theory
least common multiple

Consider $ A\equal{}2 \sqrt{2\plus{}\sqrt{3}}\minus{}\sqrt{9\minus{}2\sqrt{18}}\minus{}\sqrt{3}$. Prove that $ A^{2008}\minus{}16$ is divisible by $ 255$.

Hi, I cannot see clearly why $ A\equal{}\sqrt 2$.Any explonation would be lovely.
  Bye !

[quote="Anavel_Gato"][hide]$ A \equal{} \sqrt {2}(\sqrt {3} \plus{} 1) \minus{} (\sqrt {6} \minus{} \sqrt {3}) \minus{} \sqrt {3} \equal{} \sqrt {2}$,so $ A^{2008} \minus{} 16 \equal{} 2^{1004} \minus{} 16$.

$ A^{2008} \minus{} 16\equiv ( \minus{} 1)^{1004} \minus{} 1 \equal{} 0\pmod{3}$,
$ A^{2008} \minus{} 16 \equal{} 4^{502} \minus{} 16\equiv ( \minus{} 1)^{502} \minus{} 1 \equal{} 0\pmod{5}$ and
$ A^{2008} \minus{} 16 \equal{} 16^{251} \minus{} 16\equiv ( \minus{} 1)^{251} \plus{} 1 \equal{} 0\pmod{17}$

hence $ A^{2008} \minus{} 16$ is divisible by $ 3\times 5\times 17 \equal{} 255$.[/hide][/quote]
One very small but needed-to-be-caught error in your solution (not to be picky or anything, but I'm just warning those out there who would use a solution like yours for an official math contest): you have to state that $ \text{lcm}(3,5,17) \equal{} 255$ (or alternatively, $ \gcd(3,5,17) \equal{} 1$).  Merely the fact that $ 3\cdot5\cdot17 \equal{} 255$ is not enough.
And as seen by enndb0x's post, you definitely need more explanation as to why $ A \equal{} \sqrt {2}$ (I myself can see why, but other solution readers may not be able to).

I know that you didn't intend your solution to be perfectly rigorous and complete or anything (so don't take my suggestions offensively), but the things you wrote in your solution show some common pitfalls in solution-writing.  Remember that in any algebraic computation, the reader should not have to "think" about how you got from one step to the next; there should be enough steps so that you have done the thinking for the reader and the reader can simply follow along.

[quote="enndb0x"]Hi, I cannot see clearly why $ A \equal{} \sqrt 2$.Any explonation would be lovely.
  Bye ![/quote]

Do you know that
$ \sqrt {a \plus{} b \plus{} 2\sqrt {ab}} \equal{} \sqrt{(\sqrt{a} \plus{} \sqrt{b})^2} \equal{} \sqrt {a} \plus{} \sqrt {b}$?

[quote="MathAndKnowledge"]I know that you didn't intend your solution to be perfectly rigorous and complete or anything (so don't take my suggestions offensively), but the things you wrote in your solution show some common pitfalls in solution-writing.[/quote]
Most solutions on the forums here are not written to the level of completion that an olympiad grader would expect; it's not like we're competing here, and people can ask questions about sketchy parts. Also keep in mind there are lots of people for whom English is not their first language. Not allowing them to omit anything makes it harder for them.

@[b]MathAndKnowledge[/b]
I know the importance of what you say,but I'm sometimes lazy :P

anyway,andersonw's solution is smarter :roll:


geometry
cyclic quadrilateral
geometry unsolved

Let $ ABCD$ be a cyclic convex quadrilateral s.t. $ AB\times{CD}\equal{}AD\times{BC}$. Has $ ABCD$ a special geometric property ?

[quote="loup blanc"]Let $ ABCD$ be a cyclic convex quadrilateral s.t. $ AB\times{CD} \equal{} AD\times{BC}$. Has $ ABCD$ a special geometric property ?[/quote]
A cyclic quadrilateral with such a condition like that is called [b]harmonic quadrilateral[/b].
It also have other properties:
$ \bullet$ For any point $ X$ on the $ (ABCD)$ then $ X(ACDB) \equal{} \minus{} 1$. Let $ X\longrightarrow \{A,B,C,D\}$ then $ XA$ or $ XB$ or $ XC$ or $ XD$ becomes the tangents from $ A,B,C,D$ wrt $ (ABCD)$.
$ \bullet$ The tangents from $ C,A$ of $ (ABCD)$ and $ BD$ are concurrent. Similarly for $ B,D$

[quote="Mathias_DK"][quote="loup blanc"]Let $ ABCD$ be a cyclic convex quadrilateral s.t. $ AB\times{CD} \equal{} AD\times{BC}$. Has $ ABCD$ a special geometric property ?[/quote]
What does $ AB \times{CD}$ mean? Is it the length of $ AB$ times the length of $ CD$, or is it something like the complex product?[/quote]
It is the first meaning.


Given $\frac{x}{y} = \frac{2}{3}$ and $\frac{y}{z} = \frac{3}{2}$, what is the value of $\frac{x}{z}$?

[hide]The answer is one because:
2y=3z and 2y=3x

Therefore, z=x[/hide]


calculus
integration
calculus computations

Let $a$ be positive numbers.For $|x|\leq a,$ find the maximum and minimum value of $\int_{x-a}^{x+a}\sqrt{4a^{2}-t^{2}}\ dt.$


MATHCOUNTS
geometry
perimeter
analytic geometry
number theory
prime numbers

ok ive decided to be evil :twisted: :twisted: :twisted:

im gonna post these questions when nobody is on... so if you're lucky you'll get to solve some problems :twisted:

im goin to post 10 questions at a time, post your answers SHOWING ALL WORK.

NO WORK, NO CREDIT. and please dont post junk or other unnecessary **** on this thread.

suggestions: pm (private message) me

[size=200]6th to 8th graders only!!![/size] PLZ... BE TRUSTWORTHY!

1) What is the value of a if 3^(-a) * 2^a * 5^a = 0.0081?

2) Joey always rides his bicycle to school, and he begins his ride at the same time every day. If he averages 10 miles per hour, he arrives at school 2 minutes late, but if he averages 15 mph, he arrives 1 min. early. How many miles does Joey ride to school? Express your answer to the nearest tenth.

3) What is the greatest product obtainable from 2 integers whose sum in 1998?

4) What is the sum of the least and the greatest three-digit prime numbers that are less than 200?

5) How many positive integers less than 100 have reciprocals with terminating decimal representations?

6) I have some tables and chairs. If I place 2 chairs at each table, I have 1 extra chair. If I place 3 chairs at each table, I have one table with no chairs. What is the sum of the number of tables and the number of chairs?

7) What is the largest possible integer among 4 distinct positive integers whose arithmetic mean is 20?

8) 2 letters are selected without replacement from the 10 letters in the word CALCULATOR. What is the probabability, expressed as a common fraction, that the letters are the same?

9) Find the perimeter of a triangle determined by the points (1,3), (1,7), and (4,3). (no work needed)

10) Find the coordinate of the point halfway between -11/6 and 3/5 on the number line. Express your answer as a fraction.

NO PROBLEMS WILL BE RESTATED. THESE PROBLEMS WERE COPIED EXACTLY OUT OF THE PACKETS AND BOOKS I GOT THEM FROM. NO COMPLAINTS OR YOU WILL BE EJECTED FROM THE CONTEST.

WHEN I POSTED THESE PROBLEMS, I WAS THE ONLY USER ON, SO DON'T COMPLAIN... GOOD LUCK!  :mrgreen:

1) .0081 equals 3^4*10^-4, which equals 3^4*5^-4*2^-4. So a=

-4.

3) Since the greatest product of 2 such integers is when they form a square (there has been a lot posted about this in the MATHCOUNTS Problem Series forum), the 2 integers are 1998/2= 999. 999*999=998001.

4) First, divide 199 by all primes less than its square root and find that it is prime. Next, do the same to 101. It is prime and their sum is 300.

6) We can set up equations C-1=2T and C=3(T-1). Solve as systems and get T=4. Using the equations, C=9. Their sum, 13, is the answer.

7) The sum of the integers divided by 4 is the arithmetic mean. Multiply both sides by 4, and the sum of the integers is 80. If we set the smallest 3 integers to as small values as possible (1,2,3), we have 80=1+2+3+x. Solve. x=74.

8) Number of favorable outcomes:6. (1st C, then second C, second C then 1st C....)Number of possible outcomes: 10*9=90.

6/90=2/30.

9)[hide] Oviously, the distance from (1,3) to (4,3) is 4, and the distance from (1,3) to (1,7) is 7. By Pytahgorean, then, the last leg is sqrt 65. So the perimeter is[/hide] 11+:sqrt: 65.

10) This means the same as the average. The sum is -37/30, and this over 2 is -37/60.

#5.  21  Integers with only factors of 2 &/or 5 will have terminating decimal representations, becuase 1/2 and 1/5 are the only prime numjbers under 100 whose reciprocals terminate.  Thus: 7 powers of 2 and 3 possible powers of five (2^0-2^6 & 5^0-5^2) total of 21.

2) Trial and error. The diffrence in time for 10 miles is 20 mins, for 5 10, for 3, 6, for 2 4 and for 1 2. It is in between 1 and 2. We guess that it is exactly between the 2, so we check 1.5. Lo and behold, the diffrence is 3 minutes, so it is 1.5 miles long.

ok, topper got all of them right except for 9.

but im not sure if your answer to #5 is correct... i have a different answer... anyone else have a different answer?

also, topper, are you in the 6th to 8th grades? just checking... otherwise it would be unfair...

9) [hide]The length from 1,3 to 4,3 is 3 and the length from 1,3 to 1,7 is 4.  We get the most basic Pythagorean triple 3, 4, 5.

Ergo the perimeter is 12 units.[/hide]

P.S. I am in eight grade, and I don't think you need trial and error in number two.

[quote="notfunny"]also, topper, are you in the 6th to 8th grades? just checking... otherwise it would be unfair...[/quote]
I believe so since he was in the UNL Math Day...

#2)  We were learning about this in Honors algebra.

]If he went 10 mph, he would travel for x+1/30 hours.  If he went 15mph, he would travel for x-1/60 hours.  Ergo, this all leaves:
10(x+1/30)=15(x-1/60) :arrow: 10x+1/3=15x-1/4 :arrow: 1/3=5x-1/4 :arrow: 7/12=5x :arrow: 7/60=x.

Then plug x into either equation.  10(7/60)+1/3=7/6+1/3=3/2
then double check 15(7/60)-1/4=7/4-1/4=3/2.

The answer is 1 1/2 miles.  I know you got the same answer, topper, but this is another way to solve it

5)  The prime numbers that have terminating reciprocals are 2 and 5.  The highest number with base 2 is 2^6 or 64, which leaves seven recipricals.  With 5 the largest number less then 100 is 80 or 5*2 :^4: , which leaves with five.  With 5 :^2: the largest number less than 100 is 50 or 5 :^2: *2 leaving two.  So 6+5+2 is 13.

Ergo, the answer is 13 terminating reciprocals.

Hey I'm 12 years old, okay? (7th)

I qualified to attend UNL Math Day by scoring high enough on a entrance exam.

frodo gets 9.

i still have a different answer to #5, but sometimes the answer key can be wrong...

i still have a different answer... mysticterminator could you pm me your answer?

[quote]
5) How many positive integers [b]less than[/b] 100 have reciprocals with terminating decimal representations?[/quote]

I previously said seven but put six instead, so 7+5+2=14

i guess da key is wrong... chinaboy put up 14 first... but there was no work... and then frodo showed work... who do you think i should give the point to?

next set of problems: tommorow, around 9:30 - 10:00 AM, PST.

BE THERE!


Hi, 100 + 7x = 9 (mod 11) .
Is there any way to simplify or to find x = ?(mod 11)??

clearly, $ 7x\equiv\minus{}91\pmod {11}$

$ \implies7x\equiv8\pmod{11}$

$ \implies7x\equiv63\pmod{11}$

$ \implies\boxed{x\equiv9\pmod{11}}$

Once you had 7x=-91 (mod)11 we then have x=-13=9 (mod11)

Thats slightly quicker =P

Ah, true. I was too busy worrying about my $ \text{\LaTeX}$ to actually use my brain. :D

You can add 55, because 55 is divisible by 11.

[quote="dreamer.ande"]How do u get from 7x = 8 (mod 11) to 7x = 63(mod 11)[/quote]

Basically 63 is the only number equal to 8 (mod 11) that is also divisible by 7. You can do this if you assume x is an integer.


Today was my first day of college classes and I am thinking about how to organize my class materials (notes, homework, handouts, etc.) I am trying to decide whether to use notebooks or binders for my math and physics classes. I want to be able to save my notes and homeworks and handouts, basically all my class materials, throughout the years.

Pros for notebooks:
(1) I like them better
(2) The notes are more compact
(3) The notes have less of a chance of being ripped out (e.g. ripped out of a three-ring binder) and getting lost.

Cons for notebooks:
(1) I will have to use a separate system for homework. If I use a notebook for homework, then I will have to rip pages out of notebooks to hand in my homework to the teacher, and once you start ripping pages out notebooks lose their purpose. Such an alternative system might be a notebook for notes, a notebook for to-be-assigned homework, a folder for already-assigned-and-graded homework, and a folder for handouts. Yikes!
(2) I will have to store homework and handouts elsewhere. My physics materials won't be all in a single unit.

Pros for binders:
(1) Everything (notes, homeworks, handouts, etc.) for a class is all in one place. Very convenient!

Cons for binders:
(1) Not as cool as notebooks
(2) Things can get ripped and fall out of the three rings that bind them, consequently getting lost.
(3) Binders can become very fat and obnoxious.

So, what system do you guys use to be organized? Is there a time-tested organizational system for mathematics and the sciences? Or is it just really a matter of preference and how much you want to save throughout the years?

Here's a suggestion: have a separate notebook (or notebooks) for taking notes, and keep homework in a binder with dividers. Before each exam, rewrite all of your notes into a "master notebook" very nicely and carefully. It's a great way to review. Also, then you have a nice copy of your notes, but at the same time you can use the note-taking notebook for scratch work on homework, you can tear out pages, you can cross stuff out and rewrite without feeling bad, etc. And then your homework is conveniently all together, and since you don't use it nearly as much as if you were taking notes, the binder won't break and it won't get ridiculously fat.

You can also get those notebooks that have holes punched in the pages, so you can tear a page out and then put it in a binder. That way, you could do homework in your notebook and then transfer it to the homework binder later.

I always type all of my homework, so I don't really have the same issue as you do, but I do take all of my notes in notebooks.

I used notebooks (with unruled paper) when I was an undergraduate, and that was a good solution. But now I have found a way better solution: get a binding machine and make your own notebooks!


[quote="Ubemaya"]what's the formula for a sphere again?

this problem looks hard...[/quote]
It isn't that bad.
[hide]$108\sqrt{3}\pi-36 \pi$ Unless I made a careless mistake. I don't feel like doing the humdrum math stuffs.[/hide]

wait what do you mean by inside the outer sphere doesn't that just mean its volume or do you mean inside the outer sphere but outside the cube and for outside the inner sphere do you mean outside the inner sphere but inside the cube??  oh yes and Ubemaya $\frac{4r^{3}\pi}{3}$

Okay I'm going to make a random guess.

[hide]

$576\pi\sqrt{3}$
[/hide]

[hide]radius of larger sphere: $3\sqrt{3}$

radius of smaller sphere: 3

so just find the space between them which is
$108\pi\sqrt{3}-36\pi$[/hide]
edit:added the pi thing

You have to express your answer in decimal form.

bpms is correct, I think ragnarok forgot to distribute the $\pi$.

[quote="13375P34K43V312"]A cube has side length 6 cm. The cube has a sphere inside it so that the sphere is tangent to each face of the cube. A second sphere is outside the cube such that each vertex of the cube is tangent to the inside of the second sphere. How many cubic centimeters are in the space outside the inner sphere and inside the outer sphere? Express your answer to the nearest cubic centimeter and use $\pi=3.14$[/quote]
[hide]
the diagonal of the cube is $6\sqrt{3}$ centimeter long
a side of the cube is $6$ centimeter long
the inner sphere has a radius of $3$ cm
the outer sphere has a radius of $3\sqrt{3}$ cm
the inner sphere has a volume of $36\pi cm^{3}$
the outer sphere has a volume of $108\pi\sqrt{3}cm^{3}$
the inner sphere has a volume of $113 cm^{3}$
the outer sphere has a volume of $339 cm^{3}$
edit-sorry, i didnt see that it was asking for the volume in between, not the volume of the two spheres.
the space in between has a volume of $226 cm^{3}$
[/hide]


calculus
trigonometry
calculus computations

I'm a 4th grader by age and I'm trying to study calculus, so I don't know much about it. How do you derive the fact that if
$y=arc \sec x$, then
$\frac{dy}{dx}=\frac{1}{x\sqrt{x^2-1}}$

Draw a triangle. Make $y$ the angle and name the sides so that you get the given equation. Do $\sin$ of each side and then take the derivative. Figure out what the result means using your triangle.


Let $d(n)$ denote the number of positive divisors of $n$ and let
$e(n)=\left[2000\over n\right]$ for positive integer $n$. Prove that
\[d(1)+d(2)+\dots+d(2000)=e(1)+e(2)+\dots+e(2000).\]

[hide="Hint"]$d(n)$ mean　the number of lattice points in $xy=n$.
$e(n)$ mean the number that counted these lattice points among lengthwise direction.[/hide]

In general: $ d(1)\plus{}d(2)\plus{}\dots\plus{}d(n)\equal{}e(1)\plus{}e(2)\plus{}\dots\plus{}e(n)$.
Using this equality we can solve the problem 3 of the Shortlist IMO 2006. :)


[quote="mehmetcantu"]find a,b natural numbers;which satisfyies property $ a^2 \plus{} b^2|a^3 \plus{} b$ and $ a^2 \plus{} b^2|b^3 \plus{} a$.[/quote]
we have:
$ a^2\plus{}b^2|a(a^2\plus{}b^2)\minus{}(a^3\plus{}b)\equal{}b(ab\minus{}1)$
$ a^2\plus{}b^2|b(a^2\plus{}b^2)\minus{}(b^3\plus{}a)\equal{}a(ab\minus{}1)$
Let's $ (a;b)\equal{}k$ we have that:
$ (ab\minus{}1;k^2)\equal{}1$
$ k^2|a^2\plus{}b^2|b(ab\minus{}1)$ $ \rightarrow$ $ k^2|b$
$ k^2|a^2\plus{}b^2|a(ab\minus{}1)$ $ \rightarrow$ $ k^2|a$
therefore $ k\equal{}1$
therefore $ a^2\plus{}b^2|ab\minus{}1$
if $ ab>1$ then
$ a^2\plus{}b^2 \ge 2ab >ab\minus{}1$ contradiction
therefore $ ab\equal{}1$ $ \rightarrow$ $ a\equal{}b\equal{}1$

thanks for the solution :) this problem is from first round of turkey 2008 it is a multiple choice exam and i did it correctly without a proof :D

[quote="mehmetcantu"]find a,b natural numbers;which satisfyies property $ a^2 + b^2|a^3 + b$ and $ a^2 + b^2|b^3 + a$.[/quote]

Let $ d=gcd(a^3+b,a+b^3)$
$ \{ \begin{array}{ll}d | a^3+b \\ d | a+b^3 \\ \end{array}  \Rightarrow \{ \begin{array}{ll}d | a(a^2b^2-1) \\ d | b(a^2b^2-1) \\ \end{array} \Rightarrow d|gcd(a,b)(a^2b^2-1)$
We also have $ gcd(a,b)^2 |a^2+b^2|d|gcd(a,b)(a^2b^2-1) \Rightarrow gcd(a,b) | a^2b^2-1 \Rightarrow a=b=1$

 :roll:  :P


trigonometry
calculus
integration
calculus computations

a. $ \int_{0}^\infty\;\frac{\cos\,x}{x}\;\textbf dx$

b.  $ \int_{0}^\infty\;\frac{{e}^{\minus{}x}}{x}\;\textbf dx$

Quick answer: they both diverge.

$ \int_1^{\infty}\frac {e^{ \minus{} x}}x\,dx$ converges absolutely by comparison to $ \int_1^{\infty}e^{ \minus{} x}\,dx,$ and

$ \int_1^{\infty}\frac {\cos x}{x}\,dx$ doesn't converge absolutely, but it can be shown to converge conditionally (by an integration by parts argument).

But both $ \int_0^1\frac {e^{ \minus{} x}}x\,dx$ and $ \int_0^1\frac {\cos x}{x}\,dx$ are greater than $ \int_0^1\frac {c}x\,dx$ for some constant $ c > 0$ and hence both are infinite.

I would say that both integrals equal $ \plus{} \infty.$

It is possible to compute a finite value for $ \int_1^{\infty}\frac{e^{\minus{}x}\minus{}\cos x}x\,dx$ but that wasn't the question you asked. Should it have been the question you asked?

[size=150]What am I asking  is how to solve those problems using Laplace.... Thanks!!!! [/size]

[quote="Kent Merryfield"]Please stop shouting. You're being willfully blind. If there's nothing to solve, then it really doesn't matter what technique you're not solving them with.[/quote]

will you please help me?

:lol:  Thanks

Look, you've been told everything there is to tell about both questions you asked if one is to understand them literally. I'm too lazy to try to guess what you might really mean, so just go, check the statement of the problem, and post it exactly as it is written in your textbook or whatever place you took it from. Then we may be able to tell you more :).


Given $ (O)$ and $ A$ is the point outside $ (O)$. Let $ AM,AN$ be the tangents from $ A$ to $ (O)$. Let $ d_1,d_2$ be $ 2$ arbitry lines through $ A$ intersects $ (O)$ at $ X,T$, $ Y,Z$ respectively, such that $ X$ is between $ A,T$; $ Y$ is between $ A,Z$. Let $ G$ be the intersection of $ XZ$ and $ YT$. Prove that: $ G,M,N$ are collinear.

问题一种可利用性。

the result is easy using the pole and polars theory.
[b]If a line passes through a point , then the polar of this point goes through the pole of the line[/b]

Clearly the polar of G passes through A. Hence the polar of A must go through G, and the polar of A is nothing but the line that joins the points of contact M and N of the tangents from A to the circumference.

What you say is good. The question is, can you solve it without using POle and POlar?

I guess that i could demostrate it with harmonic divisions  :lol:   but it is equivalent to using polars


Solve
$ x^4\plus{}x^3\plus{}x^2\plus{}x\plus{}1\equal{}0$
[hide]All four roots are imaginary[/hide]

This is the 5th [url=http://mathworld.wolfram.com/CyclotomicPolynomial.html]Cyclotomic Polynomial[/url], so its roots are primitive 5th roots of unity.

For a more elementary solution, multiply by $ x\minus{}1$ and apply De Moivre's theorem.


Tryouts for the Chicago Area All-Stars Math Team, which will compete as the Chicago ARML teams in Iowa City, Iowa on June 4th and June 5th, will be held from 4 to 10pm at Evanston Township High School on January 28th, 2010.

ARML is a contest that is both team and individual based. There is an Individual Round (10 problems presented in pairs, generally in increasing difficulty), a Team Round (10 questions, 15 members per team (Chicago has 4 teams)), a Relay Round (2 sets of questions with 3 questions each; 3 members; members finish their problem with the given information and/or the information from the person in front of him/her and passes his/her answer to the next person), and a Power Round (proof-style team contest)

The tryout test will consist of a pseudo-Individual Round with about 24 questions, with (brief) time for dinner in between. The top 30ish scores will be announced at the end of the day; these will be the A/B teams. Nomination forms for the D team (sophomores or lower, first years) will also be distributed.

The tryouts are free and open to anybody in high school or younger.

@coachfritz, If I'm wrong about any of this, feel free to correct me  :wink:

darn, I couldn't go to ARML this year due to my stubborn parents registering me for the June 5 SAT math II :(  :(  :(

but for those of you trying out, good luck

Are the other two teams (not the A/B) also in the B division or are they unofficial? Also, are there typically a lot of cuts?

@gauss1181

There are exceptions for the SAT date.  The ARML competition is one of them since, as of now, it is unknown whether or not you'd qualify for the team.

http://www.arml.com/sat_2010.php

http://www.arml.com/SAT_Alternate_Test_Date_Flyer_2009_2010.pdf

Either way, you are more than welcome to come to tryouts.

@echen  As far as teams go, we are still trying to fine tune the process.  A/B are about the top 30-.  We will likely take all participants who do better than the 31st best scorer the night of tryouts (total score only, no tiebreaker mechanism).    More will be added from the rest to fill out B after tryouts.  C and D are purely invitational.   C will be first-timers and repeat participants whom we think can help the A/B teams in the future.  It is highly unlikely a senior or a middle school student would be chosen for C.  C will likely have the top freshman and sophomore scorers not on A/B and a few juniors not making A/B.   D  is made up of freshman/sophomore first-timers based on tryouts scores and a recommendation from the math team coach.  They'd also be expected to compete for the A/B teams in the future.

There are no real "cuts" once you are invited unless you do not do the homework and don't come to practice.  We often have a few too many people for B, C, and D.   These extras get to compete as part of an alternate team at the ARML competition.  Don't worry about what division a team competes in since everyone gets the same test.

coach fritz

I'll be at the tryouts. I really like ARML problems and the ARML tryout problems, especially since it's noncalc. The time limit always kills me though. I wish we got all 24 problems at once.

So who's on the team? I assume there have been a few practices already.

Are we even allowed to name names on here?

I checked on Arml.com for any info about any state team. Not there.

That would be because
(1) http://www.arml.com is not the Chicago ARML team site and would not list individual members of each team, and
(2) The teams are not decided yet.


geometry
circumcircle
trigonometry
inequalities
function
triangle inequality
trig identities

ABC is  an acute  triangle. Prove that $  \frac{5\sqrt{3}}{2}\frac{a+b+c}{3}\leq\frac{h_a+h_b+h_c}{3}+6R  $
Thanks

Why do you want ABC to be acute?

[color=blue][b]Problem.[/b] For any triangle ABC with circumradius R and altitudes $h_{a}$, $h_{b}$, $h_{c}$, prove that $\frac{5\sqrt3}{6}\left(a+b+c\right)\leq\frac{h_{a}+h_{b}+h_{c}}{3}+6R$.[/color]

[i]Solution.[/i] By the Extended Law of Sines, $a=2R\sin A$, $b=2R\sin B$ and $c=2R\sin C$. Now, the altitude $h_{a}$ is known to equal $h_{a}=b\sin C$; thus, $h_{a}=2R\sin B\sin C$. Similarly, $h_{b}=2R\sin C\sin A$ and $h_{c}=2R\sin A\sin B$. Thus, the inequality in question,

$\frac{5\sqrt3}{6}\left(a+b+c\right)\leq\frac{h_{a}+h_{b}+h_{c}}{3}+6R$,

becomes

$\frac{5\sqrt3}{6}\left(2R\sin A+2R\sin B+2R\sin C\right)$
$\leq\frac{2R\sin B\sin C+2R\sin C\sin A+2R\sin A\sin B}{3}+6R$.

Upon division by $\frac{R}{3}$, this becomes

$5\sqrt3\left(\sin A+\sin B+\sin C\right)\leq 2\left(\sin B\sin C+\sin C\sin A+\sin A\sin B\right)+18$.

Denote $s=\sin A+\sin B+\sin C$ and $t=\sin^{2}A+\sin^{2}B+\sin^{2}C$. Then,

$s^{2}-t=\left(\sin A+\sin B+\sin C\right)^{2}-\left(\sin^{2}A+\sin^{2}B+\sin^{2}C\right)$
$=2\left(\sin B\sin C+\sin C\sin A+\sin A\sin B\right)$.

Hence, the inequality

$5\sqrt3\left(\sin A+\sin B+\sin C\right)\leq 2\left(\sin B\sin C+\sin C\sin A+\sin A\sin B\right)+18$

rewrites as $5\sqrt3 s\leq\left(s^{2}-t\right)+18$. Equivalently, $t\leq s^{2}-5\sqrt3 s+18$.

Now, in order to prove this, we will show that $t\leq\frac94\leq s^{2}-5\sqrt3 s+18$.

In fact, $t\leq\frac94$ rewrites as $\sin^{2}A+\sin^{2}B+\sin^{2}C\leq\frac94$ and is a well-known triangle inequality (see [url=http://www.mathlinks.ro/Forum/viewtopic.php?p=120219#120219]http://www.mathlinks.ro/Forum/viewtopic.php?t=17684 post #2[/url], for instance). Meanwhile, $\frac94\leq s^{2}-5\sqrt3 s+18$ simplifies to $s^{2}-5\sqrt3 s+\frac{63}{4}\geq 0$. But $s^{2}-5\sqrt3 s+\frac{63}{4}=\left(\frac{7\sqrt3}{2}-s\right)\left(\frac{3\sqrt3}{2}-s\right)$, so it remains to prove that $\left(\frac{7\sqrt3}{2}-s\right)\left(\frac{3\sqrt3}{2}-s\right)\geq 0$. But since the function $f\left(x\right)=\sin x$ is concave on the interval [0°; 180°], while the angles A, B, C of triangle ABC lie in this interval, the Jensen inequality yields

$\sin A+\sin B+\sin C\leq 3\cdot\sin\frac{A+B+C}{3}=3\cdot\sin\frac{180^{\circ}}{3}$      (since A + B + C = 180° by the sum of angles in triangle ABC)
$=3\cdot\sin 60^{\circ}=3\cdot\frac{\sqrt3}{2}=\frac{3\sqrt3}{2}$,

what rewrites as $s\leq\frac{3\sqrt3}{2}$, and thus $\frac{7\sqrt3}{2}-s\geq 0$ and $\frac{3\sqrt3}{2}-s\geq 0$, what yields $\left(\frac{7\sqrt3}{2}-s\right)\left(\frac{3\sqrt3}{2}-s\right)\geq 0$. This solves the problem.

  darij


1.Find x,y belong (0;pi/2) such as sinx+siny=sinxy
2. Prove that: sinA+sinB/2+sinC/3 < 1+\sprt{2}/2
3. Prove that:cosA+cosB/2+cosC/3 <= 3.  \sprt{3}/2

Sorry people because English is very bad

1) $ xy\equal{}0$ implies $ x\equal{}y\equal{}0$, which is a solution in the (closed) interval. From now on assume $ xy>0$. If $ x,y > \frac{\pi}{6}$ holds, then $ \sin xy >2\cdot\sin\frac{\pi}{6}\equal{}1$, a contradiction. Wlog assume $ x\leq\frac{\pi}{6}$. If $ xy\leq\frac{\pi}{2}$, then we get from $ \sin xy \geq \sin y$ the contradiction $ xy\geq y\Longrightarrow x\geq 1$. Therefore, we have $ \frac{\pi}{2}<xy$, which yields the contradiction $ \frac{\pi}{2}\geq y>\frac{\pi}{2x}\geq 3$

As for 2),3) I wonder whether $ A,B,C$ denote angles in a triangle ?!


what don't you get? the definition of a locus? or what?

[quote="bigman"]A point P inside a regular tetrahedron ABCD is such that PA = PB = rt(11) and
PC = PD = rt(17). What is the side length of ABCD?

in the answer pamphlet it says: Let X be the midpoint of side AB and let Y
be the midpoint of side CD. Note that the locus of points
P inside of ABCD such that PA = PB and PC = PD is
the line segment XY (minus its endpoints).

i don't get this. Can someone explain[/quote]Here is a definition of a [url=http://mathworld.wolfram.com/Locus.html]locus[/url]. In this case, your conditions are $ PA\equal{}PB$ and $ PC\equal{}PD$ with $ P\in XY$ minus endpoints.


[quote="Urc"]I'm not very good at trigonometry, and it's been hard learning it, so I'm wondering, is there an easy way to learnt he unit circle and stuff, cause I only know BASIC trig, like sine is o/h[/quote]
Unit Circle?? You mean radial trigonometry??
I can suggest You a Book.
Buy Plane Trigonometry Part-1 By S.L. Loney. Its the perfect guide.

[quote="fishythefish"]Well, you might also want to be more specific.

Do you take a trig class?
Are you looking for a review?
Do you want to start from scratch?
etc...

Because I could tell you right now that the unit circle is the set of points in the xy-plane that are 1 unit from the origin.

That probably doesn't help you at all, though.[/quote]

Not a review, more of like...a place to learn trig? When I say unit circle, I mean what it does and stuff, I don't understand any trig.

You can never go wrong with AoPS books... as long as you have the patience and will to make your way through them...

Get familiar with the identities (MEMORIZE THEM) after you are comfortable with the very basics (unit circle). I hated the unit circle angle measure calculations though (like sin 7pi/6 since it doesn't come as quickly to me as pi/6) but unit circle helps.

[quote="Identity"][quote="elcric"]I was wondering:

Is $ \arctan$ the same as $ \tan^{ \minus{} 1}$?

Thanks![/quote]

yes[/quote]

Just wondering

But why is it called arctan?

And are arcsin and arccos the same as well? Cuz I learned trig, but for some reason, we never covered the "arc-"s...

[quote="isabella2296"]Yes, they are; I believe "arc" is just an alternative way of saying it. Please correct me if I'm wrong.[/quote]

 :thumbup:  :P


Prove for every real numbers a,b and c:

a/(b+c) + b/(c+a) + c/(a+c)  \geq  3/2

And prove for evey real numbers a,b,c and d:

a/(b+c) + b/(c+d) + c/(d+a) + d/(a+b)  \geq  4

The second inequality is false. Just say a=b=c=d=1 :maybe:
(I think you mean \geq 2)

(check your sources better, Xander, man)

I may be wrong but I think the first one got posted before yet.
Well anyway there's so much of these :D I'll try it asap

PS: can anyone tell me if this thing with n variables and thus n terms  \geq n/2    is true? I "think" it'll be, but I'm to too tired to think of it right now :D

besides, I suggest it's c/(a+b) and not c/(a+c) ??? :?

Yes, Peter you're right, the second was a little mistake, I mean  \geq 2 and not 4. Now try again ;-)

Xander, learn to log in automatically ;)

Besides, are you well prepared for wednesday's 2nd round?

[quote="Lagrangia"]These two ineqs are already solved on the forum...[/quote]

yes that was my initial thought as well :D
but all I found back was a/(a+b) sum, not a/(b+c) sum

But I may have overlooked it, I didn't really search.

[quote="Arne"]Try to log in when posting Xander :D

Hm, the problem is not quite true in general :D It's a famous 'unsolved' problem.

The generalization (with n/2) holds for n = 3, 4, 5, 6, 7, 8, 9, 10.
The generalization is false for n = 14, 16, 18, 20, 22, 24, and for all n \geq 26.
The problem remains unsolved for n = 11, 12, 13, 15, 17, 19, 21, 23, 25.

In 1969, a Russian student proved that

a1/(a2 + a3) + a2/(a3 + a4) + ... + an/(a1 + a2) \geq cn/2

where c = 0.989.[/quote]

:) Interesting!!

But unsolved or not, I guess modern computers can calculate wether it's true or not, can't they? :?

:D well I though everything was possible like that

 but you're right, numbers around 15-20 would take like a year on modern computing speed :?

[quote="Arne"]
In 1969, a Russian student proved that

a1/(a2 + a3) + a2/(a3 + a4) + ... + an/(a1 + a2) \geq cn/2

where c = 0.989.[/quote]

Hey is that exactly 0.989 ??? :? That looks incredible. Anyone able to explain where that constant comes from?


A circle is centered at (5, 15) and has a radius of 130 units. Point Q (x, y) is on the circle, has integer coordinates, and the value of the x-coordinate is twice the value of the y-coordinate. What is the maximum possible value for x?

Try referring to the [url=http://mathcounts.org/Document.Doc?id=222]2009 Chapter Solutions[/url].

the equation of the circle is:

(x-15)^2+(y-15)^2=130
Now, they said that the x coordinate is twice the y coordinate,
so the point we're looking for is (2y, y)

By putting this value in the original equation, we end up with two values of y: 4 and 6 (i think)

the greater value is 6, so the x value is 2 x 6 = 12

Final Answer: 12


What is the 100th term of the arithmetic sequence $ \frac{3}{4},5.25,\frac{39}{4},14.25,\ldots$? Express your answer as a decimal to the nearest hundredth.

Terms are of form $ 4.5n\minus{}3.75$ so it's $ 4.5\times{100}\minus{}3.75\equal{}446.25$

Or you could solve it a different way. This is my unique way of solving these kind of problems.

Step 1: Find the common difference.

$ 5.25 \minus{} 0.75 \equal{} 4.5$

Step 2: Multiply the common difference by the number of terms minus one.

$ 99 \cdot 4.5 \equal{} 445.5$

Step 3: Take this number and add it to the first term.

$ 445.5 \plus{} 0.75 \equal{} 446.25$

And we're done! $ \boxed {446.25}$


A regular hexagon of side $ 7$ is surrounded on each side by regular hexagons of the same size adjacent to its sides. What is the perimeter of the resulting figure?

[quote="rubik"]The resulting figure's perimeter is basically 6 hexagons of side 7, each missing a side. This is $ 5 *6 * 7 \equal{} \boxed{210}$.[/quote]
I think you're over counting some sides twice. The new hexagons have some sides that overlap, because the original was a regular polygon. Particularly six sides of length 7 have been over counted twice and must be subtracted because of overlap.

$ 210\minus{}84\equal{} \boxed{126}$


Let $ a,b,c$ be the real number such that $ a\plus{}b\plus{}c\equal{}2$ and $ ab\plus{}bc\plus{}ca\equal{}1$.
Find the maximum value of $ M\equal{}a^3(a\plus{}1)\plus{}b^3(b\plus{}1)\plus{}c^3(c\plus{}1)$

[quote="SUPERMAN2"]Let $ a,b,c$ be the real number such that $ a \plus{} b \plus{} c \equal{} 2$ and $ ab \plus{} bc \plus{} ca \equal{} 1$.
Find the maximum value of $ M \equal{} a^3(a \plus{} 1) \plus{} b^3(b \plus{} 1) \plus{} c^3(c \plus{} 1)$[/quote]

By $ pqr$ we have : $ p \equal{} 2$ , $ q\equal{}1$ , $ r \le \frac{4}{27}$;

$ M \equal{} a^3(a \plus{} 1) \plus{} b^3(b \plus{} 1) \plus{} c^3(c \plus{} 1) \equal{} \frac{\minus{}p^4\plus{}8p^2q^2\plus{}2q^4}{9} \plus{}4pr\plus{}pq^2\plus{}3r\equal{}$ $ 4\plus{}11r \le 5\frac{17}{27}$

Equality holdes for $ a\equal{}b\equal{}\frac{1}{3}$ , $ c \equal{} \frac{4}{3}$.


linear algebra
combinatorics unsolved
combinatorics

In a set with $ n$ elements, one considers $ n\plus{}1$ subsets such as each one of these subsets has an odd number of elements.
 prove that the intersection of two of these subsets has an even number of elements.

As posted, I don't think the statement of the problem is correct.

For example,  take a set of size $ 2k\plus{}1$, and then take the collection of all subsets of the form
\[ \{1\}\cup\bigcup_{j\in T}\{2j, 2j\plus{}1\},\]
where $ T$ is any subset of $ \{1, 2,\dots k\}.$  There are $ 2^{k}$ such sets, and they satisfy the conditions of the problem.

If we instead changed the problem to require us to show that the intersection of two of these subsets have an \textbf{odd} number of elements, then we would obtain a result due to Berlekamp known as the 'Oddtown' theorem which, as far as I know, does not have an elementary proof, though there is a very elegant proof using
[hide]linear algebra over $ \mathbb{F}_{2}^{n}$[/hide]


Hi, I  like to draw using pastel and would like to share with you some of my pastel works here

http://dhonykh.artistportfolio.net/index.php?secret=141&artist_id=14730

Quite nice.  I enjoy drawing and I find pastel is nice to use if you want a desired effect.  I find that if you want to try to re-create a Monet, pastels work really, really well.  Really anything in the impressionist style.

[size=150]wow that's really good. [/size]i drew a landscape in art this year too. but the image isn't on my comp.

What grade are you in and what class is this for?  Or is it a leisure activity?


The easiest way to this this problem is to pick a side length for an edge and figure out the ratio, as all such figures are similar anyway. Since the volume of a cone required dividing by $ 3$, it is to our help to set a side length as a multiple of $ 3$. $ 6$ is probably the best choice here. In that case, the radius of the cone is $ 3$, and the height of it is equal to a side length, so it measures $ 6$. The volume of the cone is $ \frac{1}{3}(3^2)(6)\pi=\frac{1}{3}(9)(6)\pi=\frac{1}{3}(54)\pi=18\pi$. The volume of the cube is simply $ 6^3$, or $ 216$. Now, all we need to do is to put these two values into a ratio and simplify the fraction. $ 18\pi/216=\boxed{\pi/12}$.


geometry
parallelogram
geometry unsolved

Given $ ABCD$ is a quadrangle.$ AC$ cuts $ BD$ at $ O$.$ OH \perp AB;OK \perp CD;OE \perp BC;OF \perp AD(H \in AB;K \in CD;E \in BC$$ ;F  \in AD)$
Suppose that $ OH \equal{} OK;OE \equal{} OF$
Prove that $ ABCD$ is a parallelogram.

If $ AB \parallel CD$ ( or $ AD \parallel CD$ ), it's easy to get the result.
If $ AB$ intersects $ CD$ at $ E$, $ AD$ intersects $ BC$ at $ F$: we have $ EO$ is the bisectors of $ \angle AED$, but note that $ E(ADOF) \equal{} \minus{} 1$, so we have $ EF \perp EO$. Similarly, $ FE \perp FO$, so $ O,E,F$ are collinear, which is impossible.

what means the symbol E(ADOF) ???  :blush:


This topic doesn't belong here.
Anyway are you serious about solving that equation?

[hide="Solution"]$200 = 125e^{-0.4t}+65 \Rightarrow  e^{-0.4t}= \frac{200-65}{125}= 1.08 \Rightarrow -0.4t = \ln(1.08) = 0.077 \Rightarrow  t =-0.1925$
[/hide]

Sorry, Carcul. I guess I had a bad hair day.


geometry
3D geometry
sphere
trigonometry
inequalities
triangle inequality
real analysis

Assume that it is known that for every pair of points on the sphere that aren`t
diametrically opposite, there is a unique shortest piecewise smooth curve which
connects the points. Prove that the curve is an arc of a big circle.

one can add a question: prove that there exists a unique shortest path ...

Is it related to geodesic (riemanian geometry) ?

We can prove it by "elementary" means :) (I mean, without using any differential geometry). Let me restate the problem:

Given two points on a sphere, the shortest continuous curve of bounded variation connecting them is an arc of a great circle of the sphere (the uniqueness part when the two points are not antipodes follows from this).

A notation first: given points $X,Y$ on the unit sphere $\mathbb S^2$, $|X-Y|$ represents the distance in $\mathbb R^3$ between them, while $\rho(X,Y)$ is the length of the shortest arc of a great circle of $\mathbb S^2$ connecting $X,Y$. Let $f: [0,1]\to\mathbb S^2$ be a path of length $0<\ell<\infty$ between $A=f(0)$ and $B=f(1)$. Let $\Delta=(0=x_0<x_1<\ldots<x_{n-1}<x_n=1)$ and $A_i=f(x_i),\forall i\in\overline{0,n}$. Take some $\varepsilon>0$. If the pairs of consecutive $x_i$'s are close enough, then for all $i\in\overline{1,n},\ A_{i-1},A_i$ will be so close to each other that $1\le \frac{\rho(A_{i-1},A_i)}{|A_{i-1}-A_i|}<1+\varepsilon$, since $\rho(X,Y)$ is the angle between the rays $OX,OY$ ($O$ is the origin in $\mathbb R^3$), while $|X-Y|$ is twice the sine of half of that angle, and $\frac{\sin x}x$ goes to $1$ as $x\to 0$. On the other hand, we can find a division $\Delta$ of $[0,1]$ with the pairs of consecutive $x_i$'s close enough to each other s.t. $\frac{\sum_{i=1}^n|A_{i-1},A_i|}\ell\in(1-\varepsilon,1+\varepsilon)$.

In conclusion, we can find divisions $\Delta$ of $[0,1]$ for which the great arcs of minimal length connecting $A_{i-1},A_i$ make up a path which approximates $f$ arbitrarily well in length. Now, proving the triangle inequality for great arcs (i.e. $\rho(X,Z)\le\rho(X,Y)+\rho(Y,Z)$, with equality iff $Y$ lies on the smallest great arc connecting $X,Z$) requires only a simple geometric argument, and if we apply this to all the paths $A_0,A_1,\ldots,A_n$ constructed above (which approximate $f$ arbitrarily well), then we're done.


limit
probability
real analysis
real analysis unsolved

I have a question, is there any criterion to find if the convergent series has rational or irrational limit ?
(Obvious example: $\lim_{n\to \infty}\sum_{i=1}^{n}\frac{1}{n^{2}}$- irrational,  $\lim_{n\to \infty}\sum_{i=1}^{n}\frac{1}{n(n+1)}$ - rational)

No.

There aren't really very many series we know how to sum, and rationality is usually just as hard as finding an exact sum.

And while we're at it, let's take a series with entirely irrational terms and a rational sum. Say,

$\sum_{k=0}^{\infty}\frac{(-1)^{k}\pi^{2k+1}}{2^{2k+1}(2k+1)!}=1.$

No, there's no criterion. Of course, it's a lot easier for a series to have an irrational limit than a rational limit, but that's just a probability question: a randomly chosen convergent series (by some rules for "randomly chosen series" that I don't want to get too specific about) would result in a randomly distributed real number over some range, and for any continuous probability distribution of real numbers, the probability that the number is rational is zero and that it's irrational is one.

I don't know any examples of series whose sum is rational but unknown.  :maybe: However, there are series whose sum can be proved to be irrational even though there is no closed form for the sum: see [url=http://en.wikipedia.org/wiki/Liouville_number]Lioville numbers[/url]. These are series with rational terms that converge very quickly. Their partial sums provide very good rational approximations to the total sum of the series. Surprisingly*, this implies that the total sum is irrational.

(*) This is not really surprising: distinct rational numbers with not-too-large denominators cannot be very close to each other.


integration
function
inequalities
logarithms
factorial
calculus
calculus computations

Does $ \frac {\Gamma'(n)}{\Gamma(n)} \equal{} \minus{} \gamma \plus{} \int_{0}^{1}\frac {1 \minus{} x^n}{1 \minus{} x}dx$ and if so, how do I prove it?

Also, something I've been wondering
it looks like for 0<x<1,
$ 1 \minus{} \gamma > \frac {d}{dx}ln(\Gamma(x)) > \minus{} \gamma$
and since
$ ln(\Gamma(1)) \equal{} 0$
$ \gamma \minus{} 1 < ln(\Gamma(a)) < \gamma$
so
$ e^{\gamma \minus{} 1} < \Gamma(a) < e^{\gamma}$
where a approaches 0.
yet the gamma function actually increases without bound as a approaches 0

what did I do wrong?

[quote="0714446459923"]Does $ \frac {\Gamma'(n)}{\Gamma(n)} = - \gamma + \int_{0}^{1}\frac {1 - x^n}{1 - x}dx$ and if so, how do I prove it?[/quote]
actually, it is  $ \frac {\Gamma'(n)}{\Gamma(n)} \; = \; - \gamma + \int_{0}^{1}\frac {1 - x^{n - 1}}{1 - x}\;\textbf dx$

start with weierstrass. it says, for a complex number $ z$,

$ \dfrac{1}{\Gamma(z)}\; = \;z\;e^{\gamma\,z}\prod_{k = 1}^\infty\;\left(1 + \dfrac{z}{k}\right)e^{ - \dfrac{z}{k}}$

take natural log of both sides, then differentiate w.r.t. $ z$

$ - \dfrac{\Gamma'(z)}{\Gamma(z)}\; = \;\dfrac{1}{z} + \gamma + \sum_{k = 1}^\infty\;\left(\dfrac{1}{k + z} - \dfrac{1}{k}\right)$

if the $ \mathbb Im$ component of $ z$ is zero, then we have

$ - \dfrac{\Gamma'(x)}{\Gamma(x)}\; = \;\dfrac{1}{x} + \gamma + \sum_{k = 1}^\infty\;\left(\dfrac{1}{k + x} - \dfrac{1}{k}\right)$

furthermore if $ x$ is confined to the positive integers only, you can write

$ - \dfrac{\Gamma'(n)}{\Gamma(n)}\; = \;\dfrac{1}{n} + \gamma + \sum_{k = 1}^\infty\;\left(\dfrac{1}{k + n} - \dfrac{1}{k}\right)$

which you can easily rewrite as

$ \dfrac{\Gamma'(n)}{\Gamma(n)}\; = \; - \gamma + \sum_{k = 1}^{n - 1}\;\dfrac{1}{k}\qquad \ldots\quad (1)$

$ \Rightarrow \dfrac{\Gamma'(n)}{\Gamma(n)}\; = \;\boxed{ - \gamma + \textbf H_{n - 1}}$

note: $ \int_{0}^{1}\;\dfrac {1 - x^{n - 1}}{1 - x}\;\textbf dx$ is nothing but $ \textbf H_{n - 1}$ in disguise.

[quote="0714446459923"]
Also, something I've been wondering
it looks like for 0<x<1,
$ 1 - \gamma > \dfrac {d}{dx}\ln(\Gamma(x)) > - \gamma$[b] [color=blue](***)[/color][/b]
and since
$ \ln(\Gamma(1)) = 0$
$ \gamma - 1 < \ln(\Gamma(a)) < \gamma$[b] [color=red](***)[/color][/b]
so
$ e^{\gamma - 1} < \Gamma(a) < e^{\gamma}$
where a approaches 0.
yet the gamma function actually increases without bound as a approaches 0

what did I do wrong?[/quote]

first off, your inequality (shown in blue stars) is incorrect. $ \dfrac {d}{dx}\ln(\Gamma(x))$ cannot be positive for $ x$ between 0 and 1.
second, what you have (shown in red stars) is in itself wrong. you probably meant to say
$ \gamma - 1 < \dfrac{\Gamma'(a)}{\Gamma(a)} < \gamma$.
even that is incorrect, as the relation (shown in blue stars) was incorrect to begin with.

$ \forall x\in(0,1),\quad\Gamma(x)$ is strictly (monotone) decreasing, with $ \Gamma(x)\; > \;0$.
it therefore follows that $ \dfrac {d}{dx}\ln(\Gamma(x)) \; < \;0\qquad \forall x\in(0,1)$

we can do slightly better. looking at equation $ (1)$ in the earlier part, we immediately see

$ \dfrac {d}{dx}\ln(\Gamma(x)) \; < - \gamma \qquad \forall x\in(0,1)$

$ \Rightarrow \Gamma'(x) \; < - \gamma \qquad \forall x\in(0,1)$

I figured I was off by 1 in the initial formula or something, but when I tried it with my calculator, it seemed to work.

also the difference between
$ \int_{0}^{1}\frac{1\minus{}x^n}{1\minus{}x}dx$ and $ H_n$ is that the first one works for noninteger n as well. Right?

[quote="0714446459923"]...also the difference between
$ \int_{0}^{1}\frac {1 \minus{} x^n}{1 \minus{} x}dx$ and $ H_n$ is that the first one works for noninteger n as well. Right?[/quote]

This is another example of analytic continuation. Define $ H_z\equal{}\int_{0}^{1}\frac {1 \minus{} x^z}{1 \minus{} x}dx$ for any complex $ z$ such that $ \mbox{Re}\, z>\minus{}1$. This definition agree with the original definition for integer $ z$ and furnish values for complex valued $ z$ as well;  the gamma function did the same thing for the factorial (but with a shift).

Note: the above definition is equivalent to $ H_z\equal{}\gamma \plus{}\frac{\Gamma^{\prime}(z\plus{}1)}{\Gamma(z\plus{}1)}$ for $ \mbox{Re}\, z>\minus{}1$.


[hide="B"]$w$ must be either $1$ or $2$.

[b]Case 1:[/b] $w=2$. $x+2y+3z=8$. $z$ must be $1$. Now, $x+2y=5$. This means either $x=3$ and $y=1$ or $x=1$ and $y=2$. The latter gives a smaller product. $xyzw=1\cdot2\cdot1\cdot2=4$

[b]Case 2:[/b] $w=1$. $x+2y+3z=12$. $z$ can be $1$, $2$, or $3$.

[b]Sub-Case 1:[/b] $z=1$. $x+2y=9$. Maximizing $y$, we get $x=1$ and $y=4$. $xyzw=1\cdot4\cdot1\cdot2=8$

[b]Sub-Case 2:[/b] $z=2$. $x+2y=6$. Maximizing $y$, we get $x=2$ and $y=2$ ($x$ cannot be $0$). $xyzw=2\cdot2\cdot2\cdot2=16$

[b]Sub-Case 3:[/b] $z=3$. $x+2y=3$. Only one possibility: $x=1$ and $y=1$ ($x$ cannot be $0$). $xyzw=1\cdot1\cdot3\cdot2=6$

Case 1, with $x=1$, $y=2$, $z=1$, and $w=2$, yields the smallest possible value of $xyzw$, which is $\boxed{4}$.[/hide]

[quote="rjt"](a) Find, with proof, the greatest possible value of $xyzw$ subject to the condition that $x,y,z,w$ are positive reals with $x+2y+3z+4w=16$.

(b) Find, with proof, the smallest possible value of $xyzw$ such that $x,y,z,w$ are positive integers with $x+2y+3z+4w=16$.[/quote]

[hide="a"]AM-GM?
\[\frac{x+2y+3z+4w}{4}\ge \sqrt[4]{2\cdot 3\cdot 4\cdot xyzw}\]

\[\frac{32}{3}\ge xyzw \]
$x=4$, $y=2$, $z=\frac43$, $w=1$  :P [/hide]

[quote="ch1n353ch3s54a1l"]$x=8$, $y=2$, $z=\frac23$, $w=1$  :P [/quote]

Looks good, except I think you mean $x = 4$ and $z = \frac{4}{3}$ :wink:.

i_like_pie: But in those sub-cases, $w=1$?

Oops...

[hide]The smallest value is $\boxed{3}$ when $x=1$, $y=1$, $z=3$, $w=1$.[/hide]

[quote="rjt"][quote="bpms"][quote="rjt"](a) Find, with proof, the greatest possible value of $xyzw$ subject to the condition that $x,y,z,w$ are positive reals with $x+2y+3z+4w=16$.

(b) Find, with proof, the smallest possible value of $xyzw$ such that $x,y,z,w$ are positive integers with $x+2y+3z+4w=16$.[/quote]
[hide="a"]
$\frac{x+2y+3z+4w}{4}\geq 2\sqrt[4]{6xyzw}$
$\frac{8}{3}\geq xyzw$[/hide][/quote]

$8 \neq 2^{4}$

We all have bad days, today must be yours :P[/quote]
$\frac{16}{6}=\frac{8}{3}$

[quote="ch1n353ch3s54a1l"]You can't factor the two out of the radical...it's a fourth root not a square root[/quote]
:ewpu:
Yes, I agree with rjt.


What is the least whole number that is divisible by 7, but leaves a remainder of 1 when divided by any integer 2 through 6?

$ x\equiv 1\pmod{60},x\equiv 0\pmod{7}\Rightarrow x\equal{}301$

If the number leaves a remainder of 1 when divided by 2,3,4,5, or 6, then it must be 1 more than a multiple of $ LCM(2,3,4,5,6)=60$, in other words, in the form $ 60k+1$.

This value must be divisible by 7, hence
\begin{align*}
60k+1 \equiv 0 \text{ (mod 7)} &\Rightarrow 4k+1 \equiv 0 \text{ (mod 7)} \\
&\Rightarrow 4k\equiv 6 \text{ (mod 7)} \\
&\Rightarrow 8k \equiv 12 \text{ (mod 7)} \\
&\Rightarrow k\equiv 5 \text{ (mod 7)} \\
\end{align*}

Letting $ k=5$, we get $ 60(5)+1=\boxed{301}$

EDIT: ok, basically the same thing as Bugi's, but with a little more explanation :P

Well you know that to leave a remainder when divided by 2, the number has to be odd and to leave a remainder of 1 when divided by 5 it has to end in 1 or 6.  Hence it must end in 1.  It's now easy to see that 301 is the first number that meets all criterion.

Another way to solve $ x\equiv0\pmod7$ and $ x\equiv1\pmod{60}$ is to realize that $ x\equiv\minus{}119\pmod{60}$ and $ x\equiv\minus{}119\pmod7$ are the same conditions, giving $ \gcd(60,7)\minus{}119\equal{}\boxed{301}$ as the smallest (since it is less than $ 420$).


mean - 8.25
std dev - .35
What should be the should the margin of error be for a 90% confidence interval

Thanks

[hide]
$\mu: 8.25$
$\sigma: .35$

alright to find the margin of error we always need $n$
in this case, you did not even give if it's a $z$ or a $t$ confidence interval.

Assuming it's a $z$ interval, first of all we do not need to know the mean for the margin of error.

First find $z*$ for the $90%
$ which is 1.645.

its $z*$ in which * means critical.
assuming $n$ is $25$ just multiply
$1.645 * \frac{.35}{\sqrt25}$

That is the margin of error for the $z$-interval.[/hide]


I need the first equation simplified, the second equation solved, and the third factored. I've been trying to solve these three forever today.

 1) 6+√3 ÷ 2-2√3 (those are radical signs)
2) √k+9-√k=√3 (the k+9 is all under the radical sign)
3) (2k+3)2-(2x+3)(y-2)-20(y-2)2 (the "2" after (2k+3) ans (y-2) is suppose to be a square sign.

I thought for number 1) I should use the conjugent (sp) but I wasn't sure if I foil or not and I got three different answers that don't look right. Number 2) I have no idea, I tired to make the √k by itself on one side that then square everything, but when I plugged the number I got back in the equation it didn't work so I don't know what I'm doing. Help would be MUCH appreciated.

[hide="#2"]
$\sqrt{k+9}=\sqrt{k}+\sqrt{3}\implies k+9=k+2\sqrt{3k}+3$

$\implies 2\sqrt{3k}=6\implies \boxed{k=3}$.[/hide]

I really didn’t expect anyone to answer one of my questions, sometimes people don’t like to help. I’m just in a chapter at school where I just can’t figure out how to solve the problems and it gets me so frustrated. I just wanted to say thanks again.

yes, but I messed-up and the "k" should be an "x"

[quote="rnwang2"][quote="ckck"][hide="3"]$(2k+3)^{2}-(2x+3)(y-2)-20(y-2)^{2}=4k^{2}+12k+77y-2xy+4x-20y^{2}-65$[/hide]
Hope I didn't mess up factoring[/quote]
You did the opposite of factoring.

This is a little tricky, but if you set $a=2x+3$ and $b=y-2$, you get

$a^{2}-ab-20b^{2}=(a-5b)(a+4b)$.

[hide]
$\implies (2x+3-5(y-2))(2x+3+4(y-2))$

$=\boxed{(2x-5y+13)(2x+4y-5)}$.[/hide][/quote]

Oh he said factoring. I get mixed up. :blush:

[quote="littlefirefly"]then do I get (2x-5y+13)(2x+4y-5) smaller by foliling, or am I thinking of something else other than factoring?[/quote]
No, that is the final factored form.

wow, you're 100 times smarter than I am, lol. Now I feel stupid... thanks again.


1.a,b,c>0
prove that
$ \sum \frac{{(ab)}^2}{(a\plus{}b)^2}\geq \frac{9abc}{4(a\plus{}b\plus{}c)}$
2.a,b,c>0
prove that:
$ \frac{(\sum ab)^2}{\sum (a\plus{}b)^2}\geq \frac{9abc}{4(a\plus{}b\plus{}c)}$

[quote="tuandokim"]1.a,b,c>0
prove that
$ \sum \frac {{(ab)}^2}{(a \plus{} b)^2}\geq \frac {9abc}{4(a \plus{} b \plus{} c)}$
2.a,b,c>0
prove that:
$ \frac {(\sum ab)^2}{\sum (a \plus{} b)^2}\geq \frac {9abc}{4(a \plus{} b \plus{} c)}$[/quote]

The first Inequality is Iran TST 1996 Inequality (after setting $ x\equal{}ab, y\equal{}bc, z\equal{}ca$)  :)

The second, try $ a\equal{}b\equal{}1, c\equal{}5$   :wink:


The bus ticket in a public transport system are numbered from 000000 to 999999 . A ticket called " Happy Ticket " if the sum of its firt 3 number is the same as that of the last 3 number . Work out the possibilities of finding a " Happy Ticket " on the purchase of a variable ticket .

[hide]Let $ S(x)$ be the sum of the digits of the 3-digit number $ x$

We calculate $ S(x), \ \forall x=000,001,...,999$.
It is $ 0\leq S(x)\leq 27$

We write the results in the table $ N$
In such way, that $ N[k]$ indicates how many solutions have the equation $ S(x)=k$ in $ x\in\{0,1,2,...,999\}$

Of course $ \sum_{k=0}^{999}{N[k]}= 1000$

Now, suppose that the first 3 digits are $ a=x_{5}x_{4}x_{3}$ and the last three digits are $ b=x_{2}x_{1}x_{0}$

If $ a=000$ then we have $ N[0]$ out of $ 1000$ possibilities to find $ S(b)=0$ at the last 3 digits
If $ a=001$ then we have $ N[1]$ out of $ 1000$ possibilities to find $ S(b)=1$ at the last 3 digits
.....
and so on

And we should add all these quantities $ P(a=k)\cdot \frac{N[k]}{1000}$, for $ k=0,...,999$

$ \sum_{k=0}^{999}{\left(P(a=k)\cdot\frac{N[k]}{1000}\right)}= \sum_{k=0}^{999}{\frac{1}{1000}\cdot \frac{N[k]}{1000}}= \frac{1}{10^{6}}\cdot\sum_{k=0}^{999}{N[k]}= \frac{1}{1000}$[/hide]


A rectangular field is to be enclosed by a fence and divided into two parts by another fence.  Find the maximum area that can be enclosed and seperated in this way with 800 m of fencing.
Rough Sketch of Fencing:
(a rectangular field seperated into two parts)
[code]
____________________________
|           |              |
|           |              |
|           |              |
|___________|______________|
[/code]

[quote="Aznwithabrain"]through easy guess and check I obtained
[hide]
$160*160=25600 meters$[/hide]

Edit:
its wrong
i think it's just
[hide]
$200*200=40000$[/hide][/quote]

Explanation?
[hide="Explanation"]Let the vertical distances be $x$ and the horizontal distances (4 of them) be $y$. So we know that:

$4y+3x=800$

We want to maximize $2xy$. Applying the AM-GM,

$800 = 4y+3x\geq 2\sqrt{12xy}$

Thus,

$4\cdot 12xy\leq 800^{2}$

$2xy\leq \frac{800^{2}}{24}= \frac{8000}{3}$

So the answer is $\frac{32000}{3}$.[/hide]

EDIT: I fixed it, Quatto. I feel stupid... (I am)

oh
i guess vishalarul is right
I'm totally wrong

That's incorrect.  The AM-GM states that $\frac{a+b}{2}=\sqrt{ab}$, but you forgot the division by 2 on the Left Hand Side.  By the way, you don't need the AM-GM.  Instead, you could do as follows:
[hide] $3x+4y=800\Rightarrow 4y=800-3x\Rightarrow 2y=400-1.5x$.  So, $2xy=2y(x)=(400-1.5x)(x)=400x-1.5x^{2}$.  Thus, we want the maximum value of $400x-1.5x^{2}$.  Applying the vertex formula, we find that the value of x to maximize this equation is $\frac{-400}{-1.5(2)}=\frac{400}{3}$.  Thus, $2y=200$.  Thus, $2xy=\boxed{\frac{80000}{3}}$.  [/hide]


Let $ a;b;c>0$.Prove that:
$ \sum_{cyc} \frac{ab}{a\plus{}3b\plus{}2c} \leq \frac{a\plus{}b\plus{}c}{6}$
How many proof?

[quote="friendlist"]Let $ a;b;c > 0$.Prove that:
$ \sum_{cyc} \frac {ab}{a \plus{} 3b \plus{} 2c} \leq \frac {a \plus{} b \plus{} c}{6}$
How many proof?[/quote]
By Cauchy Schwarz Inequality, we get
\[ \sum \frac{ab}{a\plus{}3b\plus{}2c} \equal{}\sum \frac{ab}{(a\plus{}c)\plus{}(b\plus{}c)\plus{}2b} \le \sum \frac{1}{9}ab\left( \frac{1}{a\plus{}c}\plus{}\frac{1}{b\plus{}c}\plus{}\frac{1}{2b}\right) \equal{}\frac{1}{9}\left( \sum \frac{ab}{c\plus{}a}\plus{}\sum \frac{ab}{b\plus{}c} \plus{}\frac{1}{2}\sum a\right) \equal{}\frac{1}{9} \left( \sum \frac{ca}{b\plus{}c}\plus{}\sum \frac{ab}{b\plus{}c} \plus{}\frac{1}{2}\sum a\right) \equal{}\frac{1}{6}\sum a\]
We have done. :)


Let $ ABCD$ be a quadrilateral and $ (I)$ be the incircle of it.Let $ E$ be the intersection of two diagonals $ AC$ and $ BD$.Let $ M,N,P,Q$ be the incenter of triangle$ EAB,EBC,ECD,ED$ respectively.Prove that $ M,N,P,Q$ are cyclic.