trigonometry
logarithms
function
limit
calculus
algebra unsolved
algebra

1) What about expanding [tex]\mbox{tan} x - \mbox{sin} x[/tex] into a serie :)

yeah,i have try it out but there is no result

Hello monster, I can solve the ineq 5.
 Consider the function $ f(x)= \frac{x-\frac{x^3}{6}}{sin(x)}$
 $ f'(x)= (1/6)\frac{ 6sin(x)-3sin(x)x^2-6xcos(x)+x^3cos(x)}{1-cos^2(x)}$
 Consider the function $ g(x)= 6sin(x)-3sin(x)x^2-6xcos(x)+x^3cos(x)$
 Actually if $ g(x) < 0$ then $ f'(x)<0$ give us the desired result.
     Now prove $ g(x) <0$
 $ g'(x) = -x^3sin(x) < 0$ => $ g(x)<g(0)=0$ (q.e.d)

Thấy hay ko?  :D  :D  :D

And this is the solution of problem 6. It's a nice problem.
Since a and b are symmetric we can assume that $ b \geq a$.
By putting $ c=\frac{b}{a}$ we have that
$ \frac{1}{x}ln(a^x+b^x)= ln(a) + \frac{1}{x}ln(1+c^x)$ with $ c \geq 1$
To get the resulf, all we must do now is to show that $ f(x)= \frac{1}{x}ln(1+c^x)$ is strictly decreasing.
$ f'(x)= \frac{xc^xln(c)-(1+c^x)ln(1+c^x)}{(1+c^x)x^2}$
For the decrease of f(x), we must show that $ xc^xln(c)-(1+c^x)ln(1+c^x)<0$
Equivalent to $ (c^x)^{c^x}<(1+c^x)^{1+c^x}$.
This one is true since $ c^x>1$ ( c>=1,x>0)

For the problem 1 ( i think it's the hardest problem), see here :(the 4th post)
http://www.mathlinks.ro/Forum/viewtopic.php?p=273401#273401

This is my sol for the problem 4, it's a boring one.
$ f(x)=\frac{1}{x^2}-\frac{1}{tan^2(x)}-\frac{4}{\pi^2}$
$ f'(x)<0$ $\Leftrightarrow$  $ g(x)=\frac{tan(x)}{\sqrt[3]{1+tan^2(x)}}-x>0$
Of course if $ g'(x)>0$ then $ g(x)>g(0)=0$. I'd prove $ g'(x)>0$
$ g'(x)>0 \Leftrightarrow 3+tan^2(x)>3(1+tan^2(x))^{\frac{1}{3}}$
Equivalent to $ (3+tan^2(x))^3>27+27tan^2(x)$ ( this is true).
Thus, $ f'(x)<0$, so $ f(x)> \lim_{x \to \frac{\pi}{2}}f(x)=0$ (q.e.d)

try them using calculus
I think this is an appropriate way!
They look good!!!! :)  :D

And finally the last is solved.
By putting $ x=a/(1-a)$ the ineq is equivalent to $ e>(1+\frac{1}{x})^x(1+\frac{1}{2x+1})$
or $ 1> xln(1+\frac{1}{x})+ln(1+\frac{1}{2x+1})=f(x)$
This one can be easily proved by diferentiating f(x) to the $ 2^{th}$ derivative.
Cheer! All of them are defeated.

yeah, i have solution for problem 4
by taking logarithm both handsides of the ineq, we have
$ \ln{2}-1<\frac{a}{1-a}\ln{a}+\ln{a+1} $
equivalent to
$ a(\ln{2}+\ln{a})+(1-a)\ln{a+1}-a>\ln{2}-1 $
put $ f(a)=a(\ln{2}+\ln{a})+(1-a)\ln{a+1}-a $
we have ineq  $ f(x)>f(1) $
need to prove that $ f'(a)=g(a)=\ln{2}+\ln{a}-\ln{a+1}-\frac{1-a}{1+a}<0   (0<a<1) $
 we have $ g'(a)=\frac{1}{a}-\frac{1}{1+a}+\frac{1}{{(1+a)}^2}$
                $     =\frac{a+3}{a*(a+1)^2}>0$ (0<a<1)
         so $ f'(a)=g(a)<g(1) =0 $
           $\Longleftrightarrow f(x)>f(1) \forall a\in]0,1[$
problem solved. i feel very happy  :rotfl:  :lol:

Yes, now i try to solve the $ 2^{nd}$ problem :
   For $ x>y>z>0$ we have $ (x-y)(y-z)(x-z)(xy+yz+zx) >0$
       $ %Error. "Loghrightarrow" is a bad command.
\sum_{cyc}\frac{x^2y}{z}>\sum_{cyc}\frac{x^2z}{y}$(1)
And by Cauchy-Schwarts: $ (x^2+y^2+z^2) = (\sum_{cyc}x\sqrt{\frac{y}{z}}x\sqrt{\frac{z}{y}}) \leq \sqrt{(\sum_{cyc}\frac{x^2y}{z})(\sum_{cyc}\frac{x^2z}{y})}$(2)
(1) and (2) lead us directly to the answer.  :lol:

Problem 1 is an easy Langrange exercise. Apply Langrange theorem for the function $ ln(\frac{x}{1-x})$ and then use AM-GM to get the resulf.

I have solved the  problem 1 but not follow "larange", here is my solution
Assume that $ x>y$ , so we have $ x-y>0 $
Multiply both handside of the ineq with $ x-y$, we have
$\ln{\frac{x}{1-x}}-\ln{\frac{y}{1-y}}>4(x-y)$
equivalent to $\ln{\frac{x}{1-x}}-4x>\ln{\frac{y}{1-y}}-4y $
put $ f(t)=\ln{\frac{t}{1-t}}-4t $
we need to prove that $ f(x)>f(y) (1>x>y>0) $
so to get the result,we must show that $ f(t)=\ln{\frac{t}{1-t}}-4t $ is strictly increasing in the open interval $ ]0,1[$
and it is nothing difficult to show that $ f'(t)>0 \forall t\in ]0,1[ $
         --------------$ qua trum $-------------------

Ok, I think your "quatrum" means you need a one-line solution. It's here:
$ \frac{1}{x-y}(\ln{\frac{x}{1-x}}-\ln{\frac{y}{1-y}}) = \frac{1}{c(1-c)}>4$ (0<c<1)

the last problem has been defeated
we have that $ x+2y<\frac{5\pi}{4} $ so $ y<\frac{5\pi}{8}$
=>$\sin{y}>0 $
the ineq equivalent to $x\sin{y}\cos({x+y})-y\sin{x}<0 $
                       <=> $ x(\sin({x+2y})-\sin{x})-2y\sin{x}<0 $
                       <=> $ x\sin({x+2y})-\sin{x}*(x+2y)<0 $
                       <=> $ \frac{\sin({x+2y})}{(x+2y)}<\frac{\sin{x}}{x} $
put $ f(t)=\frac{\sin{t}}{t}---- and ------k=x+2y $
we need to prove that
               $ f(k)<f(x) -----(x,k>0;0<x<k<\frac{5\pi}{4}) $
and it is not difficult to prove that. hehehehe :D  :D  :D
-------------------quatrum-----------------------

Well done! This's a nice proof. I appreciate it!

Good thing Valentin didn't see this, otherwise he'd have said:

"monster_of_god, post them in separate topics! This is now locked."

 :lol:  :lol:  :lol:

Our work is almost, but not completely, done, since we missed a problem ( the problem about $ sin(2x)$. We should try to solve it.

Finally I complete it. It's equivalent to $ f(x)=\frac{1}{sin(2x)}-3x+x^2>0$.
$ f'(x)=-2\frac{cos(2x)}{sin^2(2x)}-3+2x$ (1)
$ f\"(x)=8\frac{cos^2(2x)}{sin^3(2x)}+\frac{4}{sin(2x)}+2>0$ (2)
(2) shows that f is a strictly convex function. So its graph is above its tangents.
We take two tangents $ D_1$,$ D_2$ at $ x=\frac{\pi}{4}$ and $ x=\frac{\pi}{6}$.
They meet each other at an intersection point. By computation we can show this point is above the x-axis. So the gragh of f is above the x-axis, and the resulf follows.

"Who's the god's monster killer? He's the master of darkness." :D  :D  :D[/b]


[quote="silouan"]Could you give your solution . It is unsolved for four days. thanks[/quote]
Let $a+b+c=3.$ then:
$\sum\sqrt{1+\frac{2ka}{b+c}}\geq3\sqrt{1+k}\Leftrightarrow$
$\sum(\sqrt{1+\frac{2ka}{3-a}}-\frac{3k}{4\sqrt{1+k}}(a-1)-\sqrt{1+k})\geq0\Leftrightarrow$
$\sum(4\sqrt{(1+k)(1+\frac{2ka}{3-a})}-(3ka+k+4))\geq0\Leftrightarrow$
$\sum\frac{k(a-1)^2(3ka+8-k)}{(3-a)(4\sqrt{(1+k)(1+\frac{2ka}{3-a})}+3ka+k+4)}\geq0.$ :)


this is an easy problem form the first round of 1994 IRMO.
how many natural solutions does the following equation has?
$2^x+1=3^y$

Maybe this solution is a bit more complicated...

First of all, $2^x+1 \equiv (-1)^x+1 \equiv 0 (\mod 3)$, so $x$ must be odd. Then we have $2^x+1=3^y$ and because $x$ is odd, we can write $(2+1) \cdot (2^{x-1}-2^{x-2}+...-2+1)=3^y$, if $x \neq 1$. (If $x=1$, we get one solution, that is $(1,1)$). This is equivalent to $2^{x-1} - 2^{x-2} + ... - 2 + 1 = 3^{y-1}$. That means $3^{y-1} \equiv (-1)^{x-1} - (-1)^{x-2} + ... - (-1) + 1 \equiv x \equiv 0 (\mod 3)$, so $3|x$. That means we can write $x=6k+3$. Let's put this in the first equation: $2^{3 \cdot (2k+1)} + 1 = 3^y$. Again, we can write $2^{2k} - 2^{2k-1} + ... - 2 + 1 \equiv 2k+1 \equiv 0 (\mod 3)$, because there is no solution for $y=1$. So $3|2k+1$. Thus we can write $2k+1=3l$, so $x=18l+9$. If $y=2$, we get a solution $(3,2)$. If $y \neq 2$, we can repeat the process we were doing before, and get $2l+1=3t$, so $x=54t+27$. But that means $t<l$ and by the method of infinite descent, there is no solution.
Thus the only two solution are $(1,1)$ and $(3,2).$

[quote="Chang Woo-JIn"]This can be solved by following method.

If $x \geq 4$ from $ord_{16}3=4$,  We get $4|y$.

so, $3^y = 1 (mod 5)$. This means that $2^x=0(mod 5)$. contradiction!

Therefore $x< 4$

So, possible solution is $(x,y)=(1,1),(3,2)$

$Q.E.D.$[/quote]

What is ord?

Maybe $ord$ is short for $order$, in which case $ord_{a}p$, where $p$ is prime, would be the the lowest $n$ for which ${a^{n} \equiv 1 \mod{p}}$.  But I think Chang means something different, because by this definition, $ord_{16}3=2$ not 4.

hmm.. Sorry for My Symbol. But This symbol is universally used.

$ord_{m} a$ means the value $t$ s.t lowest power of $a$ satisfies $a^t =1 (mod m)$

 :D

Oh, sorry, I just had the arguments the other way around.  Thanks for clarifying.


limit
calculus
derivative
function
real analysis
real analysis unsolved

Find
\[ \lim_{x\to 0}\lim_{n\to \infty} (e^{\sqrt{x}}+e^{-\sqrt{x}})^{(n)}  \]
Where $f^{(n)}$ is the $n$-th derivative of $f$.

The answer was that the limit is 0. But I even cannot understand the problem.. :?

[quote="xirti"]I do not understand what is the limit of thr n-th derivative$\lim f^{(n)}$... :?  I think that, this limit is a function, name it $g$, and we are suposed to calculate $\lim_{x\to 0} g(x)$..
I also thought to your aproach but I wonder why can you interchange the limits...  :?[/quote]By Ramanujan's argument, for $x > 0$,
$\frac{1}{2}f^{(n)}(x) = \sum_{k=n}^\infty \frac{k!}{(k-n)!(2k)!}x^{k-n} = \sum_{j=0}^\infty \frac{(j+n)!}{j!(2n+2j)!}x^j$.
You are right in that you must first find the limit of these functions as $n \to \infty$. For fixed $x$ therefore
$0 \le \frac{1}{2}f^{(n)}(x) \le \frac{1}{2n}\sum_{j=0}^\infty \frac{1}{j!}x^j$
which should point you in the right direction :)

The function $f$ is the [url=http://mathworld.wolfram.com/Mittag-LefflerFunction.html]Mittag-Leffler function[/url] $E_2$ up to a factor 2.

thank you for clarifying this problem :lol:


[quote="236factorial"]If an edge of a cube is doubled, its surface area is increased by what percent?[/quote]

[hide]Say the original cube was a unit cube. Then the surface area would be $6$

If each edge is doubled, the surface area would turn to $2*2*6=24$.

So it increased by $300\%$[/hide]

Are you sure it was [i]increased[/i] by 400%?  :P

[quote="Drunken_Math"]Since it increased by 18, wouldn't it increase by $300\%$?  :?[/quote]

Yeah, but you put 400....


Let be $ a,b,c\in \left[\frac{1}{3} , 3\right]$ . Show that :

$ \frac {a}{a \plus{} b} \plus{} \frac {b}{b \plus{} c} \plus{} \frac {c}{c \plus{} a}\ge \frac {7}{5}$

[quote="alex2008"]Let be $ a,b,c\in \left[\frac {1}{3},3\right]$ . Show that :

$ \frac {a}{a \plus{} b} \plus{} \frac {b}{b \plus{} c} \plus{} \frac {c}{c \plus{} 1}\ge \frac {7}{5}$[/quote]
Are you sure that the last term is not $ \frac{c}{c\plus{}a}$? :huh:  :oops:

[quote="alex2008"]Let be $ a,b,c\in \left[\frac {1}{3} , 3\right]$ . Show that :

$ \frac {a}{a \plus{} b} \plus{} \frac {b}{b \plus{} c} \plus{} \frac {c}{c \plus{} a}\ge \frac {7}{5}$[/quote]
See here:
http://www.artofproblemsolving.com/Forum/viewtopic.php?t=175423


Show that the only polynomial of odd degree satisfying $p(x^2-1) = p(x)^2 - 1$ for all $x$ is $p(x) = x$

The generalization is here:

http://www.mathlinks.ro/Forum/viewtopic.php?p=247124#p247124

[incomplete solution]
if $P(x)\in \mathbb C[x]$
              $P(x^2-1)=P(x)^2-1$
set $-x$:$P(x^2-1)=P(-x)^2-1$
hence$P(x)^2=P(-x)^2$
then $P(x)=-P(-x)$ or $P(x)=P(-x)$  .
The second result isn't true (because $P$ has odd degree)
and so $P(-x)=-P(x)$
hence there is $ Q(x)\in \mathbb C[x]$ that $P(x)=xQ(x^2)$
.
.
.
.
.
.

First, $P(x)^2=P(-x)^2$ for all $x$. So, either $P(x)=P(-x)$ infinitely often (io) or $P(x)=-P(-x)$ io. The former case is impossible: as the degree of $P$ is odd, then the sign of $P(x)$ and $P(-x)$ are different for sufficiently large $x$. So, the latter case holds; in particular $P(x)+P(-x)=0$. Thus $P(0)=0$, $P(1)=1$ and $P(-1)=-1$. Next, define $a_n$ such that $a_0=0$, $a_{n+1}=\sqrt{a_n+1}$, $n\ge 0$. Clearly $a_n>1$ for all $n\ge 2$. We show $P(a_n)=a_n$ for all $n$. The case $n=0,1$ is clear, let $P(a_n)=a_n$. Taking $x=a_{n+1}$, we find $P(a_{n+1})^2 = a_n+1$, so $P(a_{n+1})=\pm a_{n+1}$. Let $P(a_{n+1})=-a_{n+1}$. Then, taking $x=a_{n+2}$, we get $P(a_{n+2})^2 = 1-a_{n+1}<0$, a contradiction. Hence, $P(a_n)=a_n$ for all $n$. From here, we get $P(x)-x$ has infinitely many roots (any $a_i$), so $P(x)=x$ for all $x$.


look, i understand that the thing im trying to prove is false but for some reason i can't find the mistake in my proof, and im really confused by mods.

i was just sitting one day and i consider if all pythagorean triples were thought of in mod 5. i then came to this weird conclusion that why couldn't 4,5,and 6 be a pythagorean triple? i understand that numerically this is preposterous, but this is my reasoning:

$ 4^{2}\plus{}5^{2}\equal{}6^{2}$
take the mod 5 of each part of the equation, you get
$ 4^{2}\equiv 1\pmod{5}$
$ 5^{2}\equiv 0\pmod{5}$
$ 6^{2}\equiv 1\pmod{5}$

we then get
$ 1\plus{}0\equal{}1\pmod{5}$...which is true?

i understand im wrong, but can anybody point me out where my proof goes askew?

thanks!

You have proved that $ 4^{2}\plus{}5^{2}\equiv{}6^{2}\pmod{5}$, but not $ 4^{2}\plus{}5^{2}\equal{}6^{2}$

Think of mods more as useful restriction that allow us to pinpoint things such that all Pythagorean Triples are multiples of 5, but cant all ways assume the opposite because just because something follows [i]a[/i] restriction doesn't mean it follows [i]all[/i].

Basically:

$ A\equal{}B\Rightarrow A\equiv B (modn)$ is true

but

$ A\equiv B ( modn)\Rightarrow A\equal{}B$ is not.

Like $ 1$ and $ 4$ they both have residue class $ 1$ modulo $ 3$ but they are not equal.

thanks guys!
i see my mistake. this was just a failed attempt of mine at solving a pythagorean triple proof using mods. :blush:  :lol:


Let $f(x)$ be a monic polynomial of degree $n$ with real coefficients. Prove that
\[ \max_{0\le i \le n, i\in {\mathbb Z}} |f(i)| \geq \frac{n!}{2^n}.  \]

I think this math can be solved by using lagrange

I think you need to show how you would do this using Lagrange :)

[hide="hint"]
This might help you http://en.wikipedia.org/wiki/Forward_difference_operator

In particular, what's the n-th forward difference of a polynomial of degree n?
[/hide]


Prove that there exists no non-constant polynomial $ f(x)$ with integer coefficients with $ f(x)$ prime for all $ x$.

This question has been asked multiple times in various forums. As JBL would say, "Search the forums."

[hide="Though to answer"]
[quote="[i]Excursions in Number Theory[/i] by Ogilvy and Anderson"]
Suppose such a polynomial exists:
$ Y(x) \equal{} a_{0} \plus{} a_{1}x \plus{} a_{2}x^{2} \plus{} ... \plus{} a_{n}x^{n}$.
$ \dots$
Since it is assumed to produce primes for all $ x$, select some $ x$, say $ x \equal{} b$, that yeilds prime $ p$:
$ Y(b) \equal{} p$
[/quote]
Then to paraphrase:
Following from the assumption $ Y(b \plus{} mp) \ , \ m \in \mathbb{Z}$ must also be prime.
Finding its value
$ Y(b \plus{} mp) \equal{} a_{0} \plus{} a_{1}(b \plus{} mp) \plus{} a_{2}(b \plus{} mp)^{2} \plus{} ... \plus{} a_{n}(b \plus{} mp)^{n}$.
Multiplying that shows $ p | Y(b \plus{} mp)$. Contradiction. $ \square$.
[/hide]

If you know greek http://www.mathlinks.ro/viewtopic.php?t=126749
If you dont http://mathworld.wolfram.com/Prime-GeneratingPolynomial.html

[url=http://www.mathlinks.ro/Forum/viewtopic.php?search_id=88190084&t=180433]A topic where this was recently discussed[/url]


Find the sum of $1\cdot 1!+2\cdot 2!+3\cdot 3!+\cdots+(n-1)(n-1)!+n\cdot n!$, where $n!=n(n-1)(n-2)\cdots2\cdot1$.

[hide]
note that $(n+1)!=(n!)(n+1)=(n)(n!)+n!$
so it telescopes to...

$%Error. "dsiplastyle" is a bad command.
\sum_{k=1}^{n}k*k!=(n+1)!-n!$
[/hide]

[hide="Kindda obvious hint"]$1\cdot 1!+2\cdot 2!+3\cdot 3!+\cdots+(n-1)(n-1)!+n\cdot n!=$
$1^2 \cdot 0!+2^2 \cdot 1! + 3^2 \cdot 2! + \cdots +(n-1)^2 \cdot (n-2)!+n^2 \cdot (n-1)!$
Which telescopes to...
$\sum_{k=1} ^ {n} = k^2 \cdot (k-1)!$[/hide]

I'm not sure if it helps a lot but I think its an easier way to look at it..........

[quote="sthebedagain"]maokid7
[hide]Don't you mean $(n+1)!-1$?[/hide][/quote]

maokid told us that it telescopes like this; $(n+1)! - n!$ so the final answer would be $(n+1)!-1$ like you said. :lol:

Checking your work is a problem solving tool that should not be neglected. :D

[hide="Solution"]Note that for any positive integer $n,$
\begin{eqnarray*}n\cdot n!+(n-1)\cdot(n-1)!&=&(n^{2}+n-1)(n-1)!\\ &=& (n+1)!-(n-1)!.\end{eqnarray*}

We must now consider when $n$ is odd or even.

If $n$ is odd, \[(n+1)!-(n-1)!+(n-1)!-(n-3)!\cdots-2!+2!-0!.\] If $n$ is even, \[(n+1)!-(n-1)!+(n-1)!-(n-3)!\cdots-3!+3!-1!.\] In both cases, the expression telescopes into $(n+1)!-1.$[/hide]

$(n+1)!-1$ is what I got for the final answer. So I agree with everyone else.

you could also use induction?

[hide]

looking at $n=1,2,3,4$ we get $1,5,23,119$ respectively so we see a pattern

we conjecture $(n+1)!-1$ is the sum, and now we use induction

base case is true and we assume $n$ case is true, now we look at $n+1$ case, so we have:

$1\cdot 1!+2\cdot 2!+\cdots+n\cdot n!+(n+1)\cdot (n+1)! = (n+1)!-1+(n+1)\cdot (n+1)! = (n+2)!-1$

so our induction is complete

[/hide]

Yeah, solving directly and trought induction will both bring the same solution. I think induction is harder to see though.

Induction is a lot easier to see, in my opinion, especially if you approached this problem by doing numerical examples.

If I am not wrong, your download does not give us a solution. All it says is that by induction .... is true.

Good catch Varun; his download is not a proof, merely two lines stating that the sum is ${ (n + 1)! - 1}$.
[hide=" A real proof"]
We notice that $ n \cdot n! = (n + 1)! - n!$
$ \implies \displaystyle\sum_{k = 1}^{n}{k \cdot k!} = \displaystyle\sum_{k = 1}^{n}{(k + 1)! - k!} = (n + 1)! - 1! = \boxed{(n + 1)! - 1}$[/hide]

Note: Why does the (n+1)!-1 in the box look so weird?
Edit: Thanks TachyonPulse.

(n+1)!-1 by telescoping, or just by adding an extra 1x1! in there.


it's 0.07  times 0.2 , squared.
which is .014 squared

0.000196. right?

I also got .000196 (it's just common squares and cube roots (Ex: 343 = $7^{3}$) in decimal form...


when the compound is treated with
a.$ \text{NH}_{2}\text{NH}_{2}(\text{KOH})$
b.$ \text{Zn\minus{}Hg(Conc. HCl)}$

As you might already know, the first reaction, (a), is the Wolff-Kishner reduction which is used to reduce C=O bonds to -CH2-, under alcaline conditons; the second, (b), is the Clemmensen reduction which does the same but under acidic conditions. I was unable to open your picture, so I don't know what is the functional group in lowe left corner of your molecule, but it is expected that the right hand carbonyl will be reduced in both cases.

the lower left corner is $ \text{OCH}_{3}$
now answer it.also whta picture formats does ur computer open
also what about the mail i PMed u

The methoxy group, as all ethers, is stable under alcaline conditions. Under acidic conditions, it could, at most, be converted into an hydroxi group.

That's what I said: the ether group was converted into an hydroxyl group under Clemmensen conditions, in this case methanol and the hydroxy substituted steroid. However, under the strongly acidic conditions, the sterol was converted into the chloride via a SN reaction. The protonated ether could also be directly converted into the chloride.


[hide][quote="Altheman"][quote="felipesa"]Show that the number
\[ \underbrace{11...1}_{n-1}\underbrace{222...25}_{n}  \]

is a perfect square[/quote]

assuming that you mean $\underbrace{11...1}_{n-1}\underbrace{222...2}_{n}5$

this is just $(\underbrace{3...3}_{n-1}5)^2$[/quote]

Yes you must consider \[ \underbrace{11...1}_{n-1}\underbrace{222...2}_{n}5  \]

but the  answer should be a fraction[/hide]


I'm guessing by 0,01 you mean 0.01, but:
we have that, from the first absolute value, that:
2.01>x>1.99
From the second we get that:
-n<(x-2)(x+2)<n
The least value of x we can possibly guess has to be greater than 1.99, so we
know that if we plug in 1.99 for x we'll be able to find the least value of n.
So, plugging this in, we get:
-n<(-.01)(3.99)<n or
n>.0399>-n
Hence, the least value of n is[u] .0399[/u] or [u]C[/u]

[quote="Z-coli"]I'm guessing by 0,01 you mean 0.01[/quote]

In Brazil we use 0,01 and not 0.01.

Forgot to use USA notation.

[hide="Another Solution"]

[list]$ |x-2|<0.01\Longrightarrow 1.99<x<2.01$[/list]Hence

[list]$ 1.99^2<x^2<2.01^2\\
1.99^2-4<x^2-4< 2.01^2-4\\
-0.0399<x^2-4<0.0401$[/list]

Anything wrong?[/hide]


I'm making a list, similar the AoPS resources page, of physics book that would make the average AoPSer stand a good chance of succeeding internationally on physics contests.

I have no experience with physics or physics contests. So I was wondering if one of the more experianced members could help me with my compilation.

Summary of the Request:
Reccomend a list of books, that read sequentially and attentively, would give a good student a complete theorectical backbone for IPho physics. Problem books are easy to come by...

Please help :(

Probably [url=http://www.amazon.com/gp/product/0471332364/102-7962144-6561747?v=glance&n=283155]Fundamentals of Physics[/url] contains a good deal of the core concepts.

By the way, when are the tests?

[quote="Rushil"]The Physics Resources topic in this forum already has a very similar discussion? You might also consider searching for it on the rest of the Physics forum....[/quote]
The physics resources doesn't really contain any detail or pedalogical breakdown.
It doesn't even consider the relative difficulty.

The syllabus is available [url=http://www.ipho2005.com/]here.[/url]

If you create a resources page for physics,i think you can search on google for Indian,australian,etc physics olympiads at all levels and integrate them all at one place.I'll search for some...

Some books-1)Resnick and Halliday (Walker and Krane edition)
2)200 puzzling physics problems
3)Richard Feynman Lectures

[quote="shadysaysurspammed"]
200 puzzling physics problems
[/quote]

This is one of the best books for preparing for the IPHO...my favourite physics question all come from there!


group theory
abstract algebra
superior algebra
superior algebra unsolved

If the group $ G$ has order $ 4n\plus{}2$, then there exists a non-trivial normal subgroup $ G^{\prime}$, where $ G^{\prime}\neq G$.

If we let $ G$ act on itself by right multiplication, we notice that the kernel is trivial, hence $ G$ is isomorphic to a subgroup $ H$ of $ S_{\left|G\right|}$.

Furthermore, $ H \not \subseteq A_{\left|G\right|}$, since an element $ g$ of order $ 2$ induces an odd permutation on $ G$. Indeed, the elements of $ G$ can be divided in pairs, each of them being interchanged by letting $ g$ work on $ G$. Since there are an odd number of such pairs, the cycle decomposition of the permutation induced by $ g$ has an odd number of cycles.

Now it follows that $ [H : A_{\left|G\right|} \cap H] \equal{} 2$, hence $ G$ has a normal subgroup of index $ 2$.

Yes, that is the same solution as mine.

$ \underline{G \hookrightarrow S_{|G|}: }$ $ \varphi: G\rightarrow S_{|G|}, \varphi: g\mapsto L_g,$ where $ L_g: S_{|G|}\rightarrow S_{|G|}, L_g: g\mapsto ga.$

$ \underline{L_g\in S_{|G|}: }$
$ L_g(a)\equal{}L_g(b)\Leftrightarrow ga\equal{}gb \Leftrightarrow a\equal{}b$ (Cancelation law). So $ L_g$ is injective.
Because $ |L_g(S_{|G|})|\equal{}|S_{|G|}|$, it is a bijection.

$ \underline{\varphi\; \mathrm{homomorphism: }}$
$ \varphi(ab)\equal{}\varphi(a)\varphi(b)\Leftrightarrow L_{ab}\equal{}L_aL_b \Leftrightarrow L_{ab} (g) \equal{}L_aL_b (g), \forall g\in S_{|G|}\Leftrightarrow \newline
\Leftrightarrow L_{ab}(g)\equal{}abg\equal{} L_a(bg)\equal{}L_a L_b (g)$.

$ \underline{\varphi\; \mathrm{injective}: }$ $ \varphi(a)\equal{}\varphi(b)\Leftrightarrow L_a\equal{}L_b \Leftrightarrow L_a(g)\equal{}L_b(g), \forall g\in S_{|G|} \Leftrightarrow ag\equal{}bg \Leftrightarrow a\equal{}b.$

In our problem $ \varphi: G \rightarrow S_{4n\plus{}2}, \varphi: g \mapsto L_g$.

Since $ G$ is even, there exists an element $ g$ of order $ 2$ in $ G$. Then also $ \mathrm{order}\varphi(g)\equal{}2$, since $ \mathrm{order}(\varphi(g))\leq \mathrm{order}(g)$, but $ \varphi$ is an injection, therefore the order of $ \varphi(g)$ is not $ 1$.

Therefore $ \varphi(g)$ is the product of disjunct transpositions. Except that $ \varphi(g)$ does not have a fixed point, otherwise $ L_g(a)\equal{}ga\equal{}a$ for some element $ a\in G$. Then $ g$ is unity. Contradiction.

That means $ \varphi(g)$ is  a product of $ 2n\plus{}1$ transpositions, therefore an odd permutaon. That means $ \varphi(G)\cap A_{4n\plus{}2}\neq \varphi(G)$ is normal and nontrivial in  $ \varphi(G)$. Then $ \varphi^{\minus{}1}({\varphi(G)\cap A_{4n\plus{}2}})\neq G$ is normal and nontrivial in $ G$. (We have to narrow $ S_{4n\plus{}2}$ to $ \varphi(G)$ first.)


Enuntul acestei probleme este urmatorul:

"Diagonalele patrulaterului inscriptibil ABCD se intersecteaza in punctul K.
Cercul circumscris triunghiului ABK intersecteaza a doua oara dreptele BC si AD
in punctele M si N. Demonstrati ca: |KM|=|KN|."

SOLUTIE(MM): [Notez cu m(XYZ) masura unghiului XYZ]
Pe o figura in care B este pe [MC] si N pe [AD], avem:
BKNM-inscriptibil--->m(MNK)=m(KBC);     (1)
ABCD-inscriptibil--->m(KBC)=m(KAN);     (2)
  m(KAN)=m(NMK)(subantin arcul NK).      (3)
Din (1), (2) si (3)--->m(MNK)=m(NMK)--->|KM|=|KN|.

Interesante problemele, dar ar fi bine sa invatati sa folositi LaTeX, totul ar arata mult mai bine si ar fi mai usor de citit. Si nici nu e greu.


inequalities
Cauchy Inequality
inequalities unsolved

Let $ a,b,c\in R^ \plus{}$. Prove that

$ \sum_{cyc} \frac {ab(a^3 \plus{} b^3)}{a^2 \plus{} b^2}\geq \sqrt {3abc(a^3 \plus{} b^3 \plus{} c^3)}$

Can you post a proof for the lemma,

$ \frac{x(b\plus{}c)}{y\plus{}z} \plus{} \frac{y(c\plus{}a)}{z\plus{}x} \plus{} \frac{z(a\plus{}b)}{x\plus{}y} \geq \sqrt{3(ab\plus{}bc\plus{}ca)}$ ?

I've had no success in proving the lemma..

[quote="qwerty414"]Can you post a proof for the lemma,

$ \frac {x(b \plus{} c)}{y \plus{} z} \plus{} \frac {y(c \plus{} a)}{z \plus{} x} \plus{} \frac {z(a \plus{} b)}{x \plus{} y} \geq \sqrt {3(ab \plus{} bc \plus{} ca)}$ ?

I've had no success in proving the lemma..[/quote]
 suppose that a+b+c=1 and after that use cauchy inequality


I've got two questions, but I'll post the second after I get the first one cleared up.  I got the answer to this, but I just listed the first 7 terms of the sequence and saw the pattern.  I can't figure out a logical way to think about why this is, though.  Help?

An elf can climb a staircase by jumping either 1 or 2 stairs at a time.  If [tex]e_n[/tex] represents the number of different ways the elf can reach the [tex]n[/tex]th stair of the staircase, define the sequence {[tex]e_n[/tex]} recursively.

[tex]e_n = e_{n-1} + e_{n-2}[/tex]

because if the last jump is 2 steps, then there are [tex]e_{n-2}[/tex] ways, and if the last jump is 1 step, then there are [tex]e_{n-1}[/tex] ways.

Ohhh, I get it :)  Thanks.  Now this one... I don't even know where to start :(

Suppose that the binary "alphabet" contains only two letters, A and B.  Then there would be two "words" of length 1: A, B; four "words" of length 2: AA, AB, BA, BB; and eight "words" of length 3: AAA, AAB, ABA, ABB, BAA, BAB, BBA, BBB.  Define {[tex]w_n[/tex]} recursively if [tex]w_n[/tex] represents the number of "words" of length n that do [u]not[/u] contain the BB pattern.

[tex]w_n = w_{n-1} + w_{n-2}[/tex]

because for every sequence of length n-1, you just add an A.  You can also add a B for every sequence of length n-1 ending in A.  The number of sequences of length n-1 ending in A is equal to however many sequences of length n-2 ending in anything.


geometry
circumcircle
geometric transformation
reflection
similar triangles
geometry proposed

Let $ \omega$ be the circumcircle of triangle $ ABC$ and $ P,Q$ be the points on $ \omega$.Let $ P_{1},P_{2},P_{3}$ be a symmetric points to $ P$ w.r.t $ BC,AC,AB$ respectively.Denote $ A'\equal{}QP_{1}\cap BC$,$ B'\equal{}QP_{2}\cap AC$,$ C'\equal{}QP_{3}\cap AB$.Prove that $ A',B',C'$ are collinear.

The points $ A', B', C'$ are the points such that if a ray of light is fired in the direction from $ Q$ to any of them, it will "bounce off" the corresponding side to form a line containing $ P$.  So then it seems that, based on the similar triangles formed, we can draw the Simson line wrt Q and use Menelaus a lot.  Just an idea.

Here is an image to this unsolved problem:
[img]http://img411.imageshack.us/img411/830/asan3kw6.png[/img]

Could you K81o7 continue your proof,or maybe somebody else will post his own solution?
I don't know why,but i think that this is not hard problem,and there must be easy solution :roll:
Thank you.

This problem is unsolved for about 3 months.Please,let me know if someone trying to solve it,or tried to solve it,and prove or find out some useful properties to solve this problem. :(
I hope we can find synthetic solution,especially without using complex numbers.
Best regards,Erken.

Almost one year left since the time when I've posted the problem,but it is still unsolved.
Sometime ago I've finally managed to solve this problem,but I feel that my solution is still too messy.I would like to post the moments which were used by me to start that solution and to end it.
[hide="How to Start:"]Try to reflect the point $ Q$ w.r.t the sides of triangle $ \triangle ABC$
[/hide]
[hide="How to End:"]Use the well-known fact which tell us that $ \overline{P_1P_2P_3}$ and $ \overline{Q_1Q_2Q_3}$ intersect at $ H$,orthocenter of $ \triangle ABC$.
[/hide]

Dear Mathlinkers,
the line A'B'C' is the Seimiya's line od P and Q wrt ABC.
You can see my synthetic proof and more on my website
http://perso.orange.fr/jl.ayme  (vol.3 Un orthopôle sur la droite de Seimiya  and La droite se Seimiya-Ayme-Turner)
Sincerely
Jean-Louis


[quote="pizza1512"]I can't get my head around trying to solve this particular surd simultaneous equation so could someone please help me by showing some methods to solving this. Thanks.

$ 5x\minus{}3y \equal{} 41$ [1]
$ (7\surd2)x \equal{} (4\surd2)y \equal{} 82$ [2][/quote]
[hide="Solution"]Take the second equation and divide it by $ \sqrt{2}$. Now you have $ 7x\plus{}4y \equal{} 41\sqrt{2}$.

Now change them to $ 20x\minus{}12y \equal{} 164$ and $ 21x\plus{}12y \equal{} 123\sqrt{2}$.

Add them to get $ 41x \equal{} 164\plus{}123\sqrt{2}$, and solve to get $ x \equal{} 4\plus{}3\sqrt{2}$. Plug $ x$ into one of the equations to get $ y \equal{}\minus{}7\plus{}5\sqrt{2}$.

The solution is $ \boxed{\left(4\plus{}3\sqrt{2},\minus{}7\plus{}5\sqrt{2}\right)}$.[/hide]

[quote="Quevvy"][quote="pizza1512"]I can't get my head around trying to solve this particular surd simultaneous equation so could someone please help me by showing some methods to solving this. Thanks.

$ 5x\minus{}3y \equal{} 41$ [1]
$ (7\surd2)x \equal{} (4\surd2)y \equal{} 82$ [2][/quote]
Do you mean $ (7\surd2)x\plus{}(4\surd2)y \equal{} 82$ ?[/quote]

Oops... I typed it incorrectly... yep it should be $ (7\surd2)x\plus{}(4\surd2)y \equal{} 82$

You can use "\sqrt{}" instead of "\surd" for better-looking square roots. Other commands can also be learned [url=http://www.artofproblemsolving.com/LaTeX/AoPS_L_About.php]here[/url].

lol , I was thinking, what? It shouldn't be =. Now the + makes complete sense. :p

$ (7\sqrt{2})x\plus{}(4\sqrt{2})y \equal{} 82$ :)


Find all pairs $(k,m)$ of positive integers such that $k^{2}+4m$ and $m^{2}+5k$ are both perfect squares.

[hide]Let's put $k^{2}+4m=(k+p)^{2},\; m^{2}+5k=(m+q)^{2}$. Then $p,q\in\mathbb{N}^{+}$. Transforming these equations we get:
$4m=2kp+p^{2}$
$5k=2mq+q^{2}$.
Plugging the first into the second and then the second into the first gives
$5k=2q\cdot \frac{2kp+p^{2}}{4}+q^{2}$
$4m=2p\cdot\frac{2mq+q^{2}}{5}+p^{2}$
which is equivalent to
$(5-pq)k= \frac{p^{2}q}{2}+q^{2}$.
$\left(4-\frac{4}{5}pq\right)m= \frac{2pq^{2}}{5}+p^{2}$
Since $k$ and $\frac{p^{2}q}{2}+q^{2}$ are positive numbers, also $5-pq$ is positive, i.e. $pq\leqslant 4$. Thus $pq\in\{1,2,3,4\}$. From the second equation: $m=\frac{2pq^{2}+5p^{2}}{20-4pq}$, so $2|5p^{2}$. Hence $2|p$. The only possible pairs $(p,q)$ are $\{2,1\}$, $\{2,2\}$, $\{4,1\}$. After evaluating $(k,m)$ for each pair we see that the pairs $(k,m)$ we got satisfy the given conditions. Therefore the final result is $(k,m)\in \{(1,2),(8,9),(9,22)\}$. [/hide]


symmetry
function
geometry
geometric transformation
reflection
probability
expected value

Prove that $ \sum_{t\equal{}1}^{n} 2t {n \choose t}^2 \equal{} n {2n \choose n}$ .

[hide="Solution"] Consider the set of lattice paths from $ (0, 0)$ to $ (2n, 0)$ in steps of the form $ (1, 1)$ and $ (1, \minus{} 1)$.  There are $ {a \choose b}$ paths from $ (0, 0)$ to $ (a, a \minus{} 2b)$, so partition the paths according to which point of the form $ (n, n \minus{} 2t)$ they pass through; this is the familiar combinatorial proof that

$ \sum_{t \equal{} 0}^{n} {n \choose t}^2 \equal{} {2n \choose n}$.

Since the paths have a reflectional symmetry across the $ x$-axis, it then follows that the expected value of $ n \minus{} 2t$ is $ 0$, hence the expected value of $ t$ is $ \frac {n}{2}$, which is equivalent to the desired result. [/hide]

A generalization was recently posted on the xkcd forums [url=http://echochamber.me/viewtopic.php?f=3&t=39539]here[/url].

t0r, you have a 1 in a lower index of summation that really wants to be a 0.

This is almost not worth mentioning, but the desired result is easily seen to be an algebraic consequence to Vandermonde's identity:
\begin{align*} \sum_{t \geq 1} 2t \binom{n}{t}^2 & = \sum_{t \geq 1} 2n \binom{n - 1}{t - 1}\binom{n}{t} \\
& = 2n \binom{2n - 1}{n - 1} \\
& = n \binom{2n}{n}. \end{align*}
(And of course we could do it with generating functions, as well.)

I see.. thanks. Is there a way, though, to prove it using binomial expressions? For example, we can prove that $ \sum_{k\equal{}0}^n {n \choose k}^2 \equal{} {2n \choose n}$ by considering $ (1\plus{}x)^n (1\plus{}x)^n \equiv (1 \plus{} x)^{2n}$ and looking for the coefficient of $ x^n$.

Right, with generating functions ;)  I think the thread t0r links to has one such solution, but here it is anyhow: $ \sum_{t \equal{} 0}^n \left(2t \binom{n}{t}\right) \cdot \binom{n}{t}$ is the coefficient of $ x^{n \minus{} 1}$ in $ \left(2\cdot \frac{d}{dx}(1 \plus{} x)^n\right) \cdot (1 \plus{} x)^n$.  But that's just $ 2n (1 \plus{} x)^{2n \minus{} 1}$, so it's also equal to $ 2n \binom{2n \minus{} 1}{n} \equal{} n \binom{2n}{n}$.  (The manipulations of this solution are pretty much identical to the manipulations in my solution.)

Nice! I couldn't find the right generating function.

Hmm... I tried to find a clean committee-forming-style proof, but ended up down JBL's Vandermonde route. However, I did find a combinatorial proof that $ k\binom{n}{k}\equal{}n\binom{n\minus{}1}{k\minus{}1}$  :dry:

[hide]$ k\binom{n}{k}$ counts the number of ways to choose $ k$ of $ n$ objects, disregarding order, then picking one of those $ k$ objects (the objects here are distinct). We could also fix our chosen object in our resulting set, and fill in the gaps around it with other objects in $ \binom{n\minus{}1}{k\minus{}1}$ ways, where there are $ n$ ways to choose that object. This means that $ k\binom{n}{k}\equal{}n\binom{n\minus{}1}{k\minus{}1}$.

Then our sum is $ \sum_{k\equal{}0}^{n}2k\binom{n}{k}^2\equal{}2n\sum_{k\equal{}0}^{n}\binom{n\minus{}1}{k\minus{}1}\binom{n}{k}$, blah blah blah Vandermonde.[/hide]

[quote="grn_trtle"]Hmm... I tried to find a clean committee-forming-style proof[/quote]
I gave a proof of the generalization on the xkcd thread I linked to using linearity of expectation.  I think a committee-forming proof would be essentially the same.

Edit:  [hide="Here it is, basically"] Pick two committees $ A, B$ of the same size out of a group of $ n$ people and pick a president in $ A$.  The total number of ways to do this is as follows:  line up two copies of the group of $ n$ people and pick a subset of size $ n$; there are $ {2n \choose n}$ ways to do this.  Declare the people picked in the first copy committee $ A$ and declare the complement of the people picked in the second copy committee $ B$.  There are, on average, $ \frac {n}{2}$ possible presidents in $ A$; the rigorous way to prove this is essentially the reflection argument I gave. [/hide]

We can use pairing to reduce the problem to a known one...

Let $ S\equal{}\sum_{k\equal{}0}^n2t\binom{n}{t}^2\equal{}\sum_{k\equal{}0}^n2(n\minus{}t)\binom{n}{n\minus{}t}^2$

Then $ 2S\equal{}\sum_{k\equal{}0}^n2n\binom{n}{t}^2$

$ S\equal{}n\cdot\sum_{k\equal{}0}^n\binom{n}{t}^2$

As a comment, can we evaluate the sum where $ t$ is replaced by $ t^a$ where $ a$ is a natural number (of course the answer is that it is not that hard, but i'm curious what methods work best here, recursion, generating functions etc).

Small typo: Your sums should be $ \sum_{t\equal{}0}^{n}$. I do the same thing all the time ;)

[quote="Altheman"]can we evaluate the sum where $ t$ is replaced by $ t^a$ where $ a$ is a natural number (of course the answer is that it is not that hard, but i'm curious what methods work best here, recursion, generating functions etc).[/quote]
As is usually the case in these situations, it's much easier to replace $ t$ with $ {t \choose a}$ and interpolate from there.  This is true of the generating functions approach and the recursive approach and the probabilistic approach (although I don't understand the probabilistic approach well for higher moments).

Here is a direct [hide="combinatorial proof"]

Suppose you have a class of $ n$ boys and $ n$ girls. Let's say you want to choose a group of $ n$ students with one boy among them as their leader.
To count the no. of ways in which you can do this:
Choose $ i$ boys and $ n\minus{}i$ girls with one of the $ i$ boys as their leader. This can be done in $ i \binom n i\binom n {n\minus{}i}$ ways.
But $ \binom n {n\minus{}i} \equal{} \binom n i$, so the no. ways to make a selection is $ \sum_{i\equal{}1}^n i {\binom n i}^2$

Counting in another way, you can choose a leader from one of the $ n$ boys and then choose $ n\minus{}1$ kids from the remaining $ 2n\minus{}1$. So, $ n \binom {2n\minus{}1} {n\minus{}1}$ ways.

So, $ 2 \sum_{i\equal{}1}^n i {\binom n i}^2 \equal{} 2n \binom {2n\minus{}1} {n\minus{}1} \equal{} n{2n\choose n}$

I didn't read the other proofs so forgive me if I have repeated someone's solution.[/hide]


Let $ \mathbb U$ and $ \mathbb W$ be the linear subspaces of $ \mathbb R^4$ such that $ \mathbb U=<\{(1,0,1,2),(0,-1,0,1)\}>$ and $ \mathbb W=\{(a,b,c,d)\in\mathbb R^4|3d-a-b=0\,\wedge\,4c+d-b=0\}.$ Find $ (\mathbb U+\mathbb W)^\perp.$

[hide="My work"]
\begin{eqnarray*}
  \mathbb W&=&\left\{ \left( a,\frac{a+12c}{2},c,\frac{a+4c}{2} \right)\in \mathbb{R}^{4},\forall \,a,c\in \mathbb{R} \right\} \\
 & =&\left\{ a\left( 1,\frac{1}{2},0,\frac{1}{2} \right)+c\left( 0,6,1,2 \right),\forall \,a,c\in \mathbb{R} \right\} \\
 & =&<\left\{ \left( 1,\frac{1}{2},0,\frac{1}{2} \right),\left( 0,6,1,2 \right) \right\}> \\
 & =&<\left\{ (2,1,0,1,),(0,6,1,2) \right\}>.
\end{eqnarray*}

Thus we get $ \mathbb U+\mathbb W=\{(1,0,1,2),(0,-1,0,1),(2,1,0,1,),(0,6,1,2)\},$ provided that these are linearly independent.

Hence $ (\mathbb U+\mathbb W)^\perp=\left\{ (x,y,z,u)\in \mathbb{R}^{4}\left| \begin{aligned}
   \left\langle (x,y,z,u),(1,0,1,2) \right\rangle &=0 \\
  \left\langle (x,y,z,u),(0,-1,0,1) \right\rangle &=0 \\
  \left\langle (x,y,z,u),(2,1,0,1) \right\rangle &=0 \\
  \left\langle (x,y,z,u),(0,6,1,2) \right\rangle &=0
\end{aligned} \right. \right\}$

Is this right?
[/hide]

You are on the right track but your answer can be simplified a lot.

[hide="Quick Answer"]
Indeed, the 4 vectors in $ \mathbb{U}\plus{}\mathbb{W}$ are linearly independent since their determinant is nonzero (7, I believe), but doesn't this mean that $ \mathbb{U}\plus{}\mathbb{W} \equal{} \mathbb{R}^4$? Therefore, $ (\mathbb{U}\plus{}\mathbb{W})^\perp \equal{} \{(0,0,0,0)\}$.[/hide]


trigonometry
induction
geometry unsolved
geometry

We have cosx + cosy + cosz = sinx + siny + sinz = 0 ... and we must prove:
cos(n*x) + cos(n*y) + cos(n*z) = sin(n*x) + sin(n*y) + sin(n*z) = 0

I tried hard to solve it but i didn't  :(  .. help me please! .. i think we must start from cosx<0 , cosy>0, cosz>0.  :help:

[quote="St3faN"][quote="kunny"]Let $A(\cos nx,\ \sin nx), \ B(\cos ny,\ \sin ny),\ C(\cos nz,\ \sin nz)$, consider the barycentre of $\triangle{ABC}$.[/quote]

Can you be more explicit?[/quote]

Sorry that this didn't make sence.

[quote="Kouichi Nakagawa"]
Is a condition of a problem right?[/quote]

oh :wallbash: .. I forgot about it : n>3. and n isn't a multiple of 3.

Good!  Now consider
\[\frac{1}{z_{1}}+\frac{1}{z_{2}}+\frac{1}{z_{3}}.\]

[quote="nayel"]complex numbers(de moivre's formula) maybe.

Let $z_{1}=\cos x+i\sin x$
$z_{2}=\cos y+i\sin y$
$z_{3}=\cos z+i\sin z$
we have $z_{1}+z_{2}+z_{3}=0$

now someone can complete? :wink:[/quote]

well, $\frac{1}{z_{1}}+\frac{1}{z_{2}}+\frac{1}{z_{3}}=\cos x+\cos y+\cos z-i(\sin x+\sin y+\sin z)=0$

now can it be shown by induction that $z_{1}^{n}+z_{2}^{n}+z_{3}^{n}=0$? :?:
if so then $\frac{1}{z_{1}^{n}}+\frac{1}{z_{2}^{n}}+\frac{1}{z_{3}^{n}}=0$

so $\cos nx+\cos ny+\cos nz=0=\sin nx+\sin ny+\sin nz$ and we will be done.

thanks for your help, nsato.

EDIT: sorry i found some mistakes in my induction.

but if
We have sinx + siny + sinz =  sin3x + sin3y + sin3z = 0  prove
cos2xcos2ycos2z>=0


For each positive integer $n$, find positive integer solutions $x_{1}, x_{2}, ..., x_{n}$ to the equation\[\frac{1}{x_{1}}+\frac{1}{x_{2}}+...+\frac{1}{x_{n}}+\frac{1}{x_{1}x_{2}...x_{n}}= 1.\]

How are you supposed to make eclipses?

[code] \ldots  \text{ or } \cdots [/code] gives $\ldots \text{ or }\cdots$

[hide="For the problem:"]
I remember seeing this before. Start with smaller cases, and once you find $x_{1}$ and $x_{2}$, I think you can define a recursion.  :) [/hide]

There might be more to it than just "induction"

Use induction to prove that $x_{n+1}=\left(\prod_{i=1}^{n}x_{i}\right)+1$.


calculus
integration
superior algebra
superior algebra unsolved

Find integral closure of $ \mathbb{C}[x]$ in $ \mathbb{C}(x)(\sqrt[d]{1\minus{}x^d})$. I suspect that it must be $ \mathbb{C}[x](\sqrt[d]{1\minus{}x^d})$. How do you prove this?

I didn't check the answer, but you can use Serre's criterion for normal.


geometry
circumcircle
geometric transformation
trigonometry
modular arithmetic
Gauss
cyclic quadrilateral

Given triangle $ ABC$. Let $ M$ be an arbitray point on A-altitude. $ CM\cap (BC) \equal{} \{K\}, BM\cap (BC) \equal{} \{L\}$, $ CM\cap (AMB) \equal{} \{X\}, BM\cap (ACM) \equal{} \{Y\}$.
 Prove that the midpoints of $ BC, XY$ and $ KL$ are collinear.

Could you please watch the problem? I think there is a typo, since $ CM\cap (BC) \equal{} \{C\},BM\cap (BC) \equal{} \{B\}$, or did you use the notation $ (XY)$ to say the circle with diameter $ XY$ :oops:

Yes, $ (BC)$ is a circle with diameter $ BC$.

$ O,F,P$ is collinear
$ \Leftrightarrow FP\perp KL$
$ \Leftrightarrow \vec {FP}.\vec {KL}\equal{}0$
$ \Leftrightarrow (\vec {LX}\plus{}\vec {KY})(\vec {LH}\plus{}\vec {HK})\equal{}0$
$ \Leftrightarrow \vec {LX}.\vec {HK}\plus{}\vec {KY}.\vec {LH}\equal{}0$ $ (1)$
from $ (\vec {LX},\vec {HK})\equiv \minus{}(\vec {KY},\vec {LH})(mod 2\pi)$
$ (1)\Leftrightarrow LX.HK\equal{}KY.LH$
$ \Leftrightarrow LX.HB\equal{}KY.HC$
$ \Leftrightarrow HB.CX\equal{}HC.BY$
$ \Leftrightarrow \frac {HB}{HC}\equal{}\frac {AX}{AB}$
True.
Done.

$ \bullet$ Let be the point $ E\equiv BK\cap CL$, which lies on the line of the $ A$-altitude $ AD$ of the given triangle $ \bigtriangleup ABC$, as well.

So, based on the below [b][size=100]Lemma[/size][/b], we have that $ \angle BEX = \angle CEY$ $ ,(1)$

From $ (1)$ $ \Longrightarrow$ $ \angle BEY = \angle CEX$ $ ,(2)$

From $ (2)$ and because of $ \angle EBY = \angle ECX$, we have that the triangles $ \bigtriangleup EBY$ and $ \bigtriangleup ECX$ are similar.

From this similarity and because of $ (1)$ we have that $ \frac {KX}{KC} = \frac {LY}{LB}$ $ ,(3)$

$ \bullet$ We consider now the non-convex quadrilateral $ BCXY$ and based on the [b][size=100]ERIQ theorem[/size][/b] ( = [b]E[/b]qual [b]R[/b]atios [b]I[/b]n [b]Q[/b]uadrilateral ), because of $ (3)$, we conclude that the midpoints of the segments $ BC,\ KL,\ XY$ are collinear and the proof is completed.

[b][size=100][color=DarkBlue]LEMMA. - A triangle $ \bigtriangleup ABC$ is given and let $ P$ be, an arbitrary point on its $ A$-altitude $ AD.$ We denote as $ M,\ N,$ the points of intersections of the circumcircles $ (O_{1}),\ (O_{2})$, of the triangles $ \bigtriangleup BHP,\ \bigtriangleup CHP$ respectively, where $ H$ is the orthocenter of $ \bigtriangleup ABC.$ Prove that $ \angle BAM = \angle CAN$.[/color][/size][/b]

Kostas Vittas.

PS. I will post here later, the proof of the above [b]Lemma[/b] I have in mind.

[quote="vittasko"][b][size=100][color=DarkBlue]LEMMA. - A triangle $ \bigtriangleup ABC$ is given and let $ P$ be, an arbitrary point on its $ A$-altitude $ AD.$ We denote as $ M,\ N,$ the points of intersections of the circumcircles $ (O_{1}),\ (O_{2})$, of the triangles $ \bigtriangleup BHP,\ \bigtriangleup CHP$ respectively, where $ H$ is the orthocenter of $ \bigtriangleup ABC.$ Prove that $ \angle BAM = \angle CAN$.[/color][/size][/b][/quote]
[b][size=100]PROOF OF THE LEMMA.[/size][/b] - We draw the circumcircles $ (K),\ (L),$ of the triangles $ \bigtriangleup AHB$ and $ \bigtriangleup AHC$ respectively and let be the points $ X\equiv (K)\cap CM$ and $ Z\equiv (L)\cap BN.$

It is easy to show that $ \angle XAB = \angle XHB = \angle ZHC = \angle ZAC$ $ \Longrightarrow$ $ \angle XAB = \angle ZAC$ $ ,(1)$

$ \bullet$ Let be the points $ Q\equiv AC\cap (O_{2})$ and $ R\equiv AB\cap (O_{1})$ and from $ (AQ)\cdot (AC) = (AP)\cdot (AH) = (AR)\cdot (AB)$, we conclude that the quadrilateral $ BCQR$ is cyclic and so, we have that $ \angle AQR = \angle B$ $ ,(2)$

Also, from the cyclic quadrilateral $ BCEF,$ we have that $ \angle AEF = \angle B$ $ ,(3)$

From $ (2),\ (3)$ $ \Longrightarrow$ $ \angle AQR = \angle AEF$ $ \Longrightarrow$ $ QR\parallel EF$ $ ,(4)$

It is easy now to show that $ QN\parallel AZ$ $ ,(5)$ from $ \angle QNH = \angle QCH = \angle AZH$ and similarly, we have that $ RM\parallel AX$ $ ,(6)$

From $ (4),\ (5),\ (6)$ $ \Longrightarrow$ $ \frac {FM}{MK} = \frac {FR}{RA} = \frac {EQ}{QA} = \frac {EN}{NZ}$ $ \Longrightarrow$ $ \frac {KM}{MX} = \frac {EN}{NZ}$ $ ,(7)$

Hence, because of $ (7)$, from the similar right triangles $ \bigtriangleup FAX,\ \bigtriangleup EAZ,$ we conclude that $ \angle MAB = \angle NAC$ and the proof of the [b][size=100]Lemma[/size][/b] is completed.

Kostas Vittas.

Thanks you dear [b]Elementarry[/b] and [b]Kostas Vittas[/b] for your nice solutions.
Here is my solution:
$ \vec{EF} \equal{} \frac {1}{2}(\vec{KX} \plus{} \vec{LY}), \vec{OE} \equal{} \frac {1}{2}(\vec{BK} \plus{} \vec{CL})$. So we need to prove that $ \vec{KX} \plus{} \vec{LY} \equal{} k(\vec{BK} \plus{} \vec{CL})$.
A translation $ T_{\vec{LK}}: Y\rightarrow P, C\rightarrow Q$. Denote $ R, S$ the midpoints of $ XP, BQ$
So we only need to show that $ R,K,S$ are collinear.
Note that $ \angle BKX \equal{} \angle QKP \equal{} 90^o$. Let $ SK\cap PX \equal{} \{R'\}, BK\cap CL \equal{} \{J\}$
$ \frac {\sin \angle SKQ}{\sin \angle SKB} \equal{} \frac {BK}{KQ} \equal{} \frac {BK}{LC} \equal{} \frac {\sin \angle BJA}{\sin \angle CJA}$
But $ \angle SKQ \plus{} \angle SKB \equal{} \angle BJC \equal{} \angle BJA \plus{} \angle CJA$ so $ \angle SKQ \equal{} \angle BJA, \angle SKB \equal{} \angle CJA (1)$
$ R'\equiv R \Leftrightarrow \frac {KX}{KP} \equal{} \frac {\sin \angle R'KP}{\sin \angle R'KX} \equal{} \frac {\cos \angle SKQ}{\cos \angle RKS}$
$ \Leftrightarrow \frac {\cos \angle SKQ}{\cos \angle RKS} \equal{} \frac {KX}{LY} \equal{} \frac {BK.\cot \angle BXK}{CL.\cot \angle CYM} \equal{} \frac {BK}{CL}.\frac {\cot \angle BAM}{\cot CAM}$
$ \equal{} \frac {\sin \angle BJM}{\sin \angle CJM}.\frac {\cot \angle BJM}{\cot \angle CJM} \equal{} \frac {\cos \angle BJM}{\cos \angle CJM}$
It's right from $ (1)$.
We are done.

Dear Livetolove and Mathlinkers,
in order to have perhaps another approach of your nice problem consider the circle going through B, M, C and noted U his center.
Prouve: UM is parallel to OFP.
Sincerely
Jean-Louis

Based on the [b][size=100]Lemma[/size][/b] I mentioned before, we can have another more elementary approach as follows :

We consider ( see the [b]schema t=306006[/b] ) the similar right triangles $ \bigtriangleup KEX,\ \bigtriangleup LEY,$ erected on the side-segments $ EK,\ EL$ respectively, of the triangle $ \bigtriangleup EKL,$ outwardly to it.

So, it is well known that the line segment $ OQ$ as the midperpendicular of $ KL,$ bisects the segment $ XY$ and the proof is completed.

Kostas Vittas.

This is my solution. It is based on the same idea with Mr.[b]vittasko[/b]

Denoted by $ E\equiv BK\cap CL, P\equiv EB\cap (AMB), Q\equiv EC\cap (AMC)$. It is clearly that $ \overline{EP}\cdot\overline{EB}\equal{}\overline{EA}\cdot\overline{EM}\equal{}\overline{EQ}\cdot\overline{EC}$ so $ B,C,P,Q$ are concyclic. Hence $ (BL,BQ)\equiv (CP,CK) \pmod{\pi}$. Moreover, $ (BP,BL)\equiv (CP,CL) \pmod{\pi}$ so $ (BL,BQ)\equiv (CP,CK) \pmod{\pi}$. On the other side $ (XP,XC) \equiv (BK,BL) \equiv (CK,CL) \equiv (YB,YQ) \pmod{\pi}$. Therefore the triangles $ \triangle PXC$ and $ \triangle QYB$ are similar. This yields $ \frac{KX}{KC} \equal{} \frac{LY}{LB}$ so the midpoints of $ BC,KL,XY$ are collinear. Q.E.D.

Dear Mathlinkers,
another idea : the line OFP goes also through the midpoints of the segment joining the centers of (AMB) and (ACM).
Sincerely
Jean-Louis

Use [b]Lemma[/b]:
$ \frac {AE}{AB} \equal{} \frac {CK}{CD}$ and $ \frac {LA}{LC} \equal{} \frac {NE}{NK} \equal{} \frac {MB}{MD}$
then $ L,N,M$ are collinear.
Apply we $ O,F,P$ are collinear.

Dear Mathlinkers,
sorry my last result is wrong!!!
Sincerely
Jean-Louis

Another problem:
Given triangle $ ABC$ and altitude $ AA_1$. Let $ P$ be an arbitrary point on $ AA_1. BP\cap AC\equal{}\{B_1\}, CP\cap AB\equal{}\{C_1\}$
$ A_1B_1\cap (AB)\equal{}X, A_1C_1\cap (AC)\equal{}Y$. Prove that $ A$, the midpoint of $ B_1C_1$ and the midpoint of $ XY$ are collinear.

[quote="livetolove212"]Another problem:
Given triangle $ ABC$ and altitude $ AA_1$. Let $ P$ be an arbitrary point on $ AA_1. BP\cap AC \equal{} \{B_1\}, CP\cap AB \equal{} \{C_1\}$
$ A_1B_1\cap (AB) \equal{} X, A_1C_1\cap (AC) \equal{} Y$. Prove that $ A$, the midpoint of $ B_1C_1$ and the midpoint of $ XY$ are collinear.[/quote]
[quote="Elementaryyy"]It is not true.[/quote]
I think that [b]livetolove212[/b] thought about a different result.

Surely, it is well known that the line connecting the midpoints of the segments $ XY,\ B_{1}C_{1}$ he mentioned, passes through the midpoint of the segment $ AA_{1}.$

Best regards, Kostas Vittas.

[quote="Elementaryyy"]It is not true.[/quote]

Why is it not true?
Actually, this problem is [url=http://www.mathlinks.ro/viewtopic.php?p=1474889#1474889]Vietnam TST 2009-problem 5[/url] and I see that it is also an application of [b]ERIQ theorem[/b] [b]Mr. Kostas Vittas[/b] has mentioned.

[quote="vittasko"]I think that livetolove212 thought about a different result.

Surely, it is well known that the line connecting the midpoints of the segments $ XY, B_1C_1$ he mentioned, passes through the midpoint of the segment  $ AA_1$

Best regards, Kostas Vittas.[/quote]

Dear [b]Mr. Kostas Vittas[/b], I think it is not true.

[quote="livetolove212"]
Why is it not true?
Actually, this problem is [url=http://www.mathlinks.ro/viewtopic.php?p=1474889#1474889]Vietnam TST 2009-problem 5[/url] and I see that it is also an application of [b]ERIQ theorem[/b] [b]Mr. Kostas Vittas[/b] has mentioned.

[quote="vittasko"]I think that livetolove212 thought about a different result.

Surely, it is well known that the line connecting the midpoints of the segments $ XY, B_1C_1$ he mentioned, passes through the midpoint of the segment  $ AA_1$

Best regards, Kostas Vittas.[/quote]

Dear [b]Mr. Kostas Vittas[/b], I think it is not true.[/quote]
See fig:$ A,J,M$ is not collinear.
Dear  [b]Mr. Kostas Vittas[/b] $ J,K,M$ is on $ Gauss$ line of $ AC_1HB_1$.

[quote="livetolove212"]$ (AB)$ is the circle with diameter $ AB$.[/quote]
 :ewpu: Sorry for my blindness dear [b]livetolove212[/b]. I was sure that you thought something different ( than the [b]Gauss line[/b] ) but, I didn't see your notation of $ (AB)$ as the circle with diameter $ AB$.

Best regards, Kostas Vittas.

Dear Mr. [b]Kostas Vittas, Elementarry[/b] and Mathlinkers,
Sorry for my careless style.
But, any idea for this problem? With another expression of [b]Vietnam TST 2009-problem 5[/b], I think we would have a shorter solution for it.

Use [b]Lemma[/b] I posted:
A line through $ A$ parallel $ BC$ cut $ A_1C_1,A_1B_1$ at $ R,S$.
From $ A_1(R,A,S,C)\equal{}\minus{}1$ we have $ AR\equal{}AS$.
And not hard to prove that $ \frac {YC_1}{YR}\equal{}\frac {XB_1}{XS}$ then from [b]Lemma[/b]
we have $ A,I,J$ are collinear.


I'm a total beginner at Linux, and I want to install it one a brand new slave hard drive.  I have a beginners book, and they recommend [url=http://www.mandriva.com/en/download/mandrivaone]Mandriva[/url] but the problem is, I don't know which version to download.  I'm a total noob, so I want a graphical interface.  The book talks about KDE, so I think I should download KDE, except it talks about "non-free drivers."  What exactly does this mean?

Since I wasn't sure, I just downloaded the Purely Free version.  Will I still be able to use the KDE window manager?

Again, I'm totally new, so any beginner's tips would be useful.  Also, I may switch to Ubuntu, but I want to get good at Linux first, and everyone tells me Mandriva is the way to go for beginners.

If I remember correctly KDE comes with Mandriva, so you won't need to download anything. Non-free drivers are just the official Nvidia or ATI drivers, which should automatically be installed. Go on and jump in; the best way to learn isn't from the book but from trying.

I actually started with Ubuntu and it's rather easy. Once you have the disk (LOOONG download), the installation is fairly easy. It should be able to automatically resize your windows partition (if you have one) to make space for its partitions.

I don't know anything about Mandriva, and the link you provided didn't really help, but Ubuntu has a nice GUI and applications that can do a lot of the administrative stuff that you'd often do in a terminal, but to get good with the terminal you have to use the terminal. Just don't play around with superuser too much ;). Personally, I've always been told that Ubuntu is for beginners and Gentoo is for the more experienced.

[quote="mysmartmouth"]Mandriva is a pain.  I can use the Live CD, but it causes problems when I try to install to the external hard drive.  I'm gonna go ahead and format the hard drive to kill Mandriva, and then try again with Ubuntu.  If I do that, will it automatically kill the bootloader Mandriva installed on my Windows drive?

Also, I think I am lucky enough to have a modem that works with Linux![/quote]

It'll update the bootloader, but you'll still have a (non-working) Mandriva entry in the bootloader. (Incidentally, Ubuntu will automatically format for you :))

If you have to have dialup, congrats on getting a good modem!

Gentoo is for people who have lots of time and/or have specialized hardware or want to learn more than they ever wanted to know about their computer. Ubuntu is kinda like macs in that it just works, but Gentoo has more stuff you can easily install. Further, Gentoo is faster once compiled.

Hm, Ubuntu wouldn't install.  I totally uninstalled everything Linux after much trial and tribulation.  I am going to buy a new computer when Vista comes out, and then convert this to exclusively Linux.

But I came out of this with some Linux experience, as well as extensive troubleshooting experience (see [url=http://seansoni.blogspot.com]my blog[/url]).  Also, I learned how to use the command line more effectively.

Getting started with Ubuntu was very easy for me...their forums(http://www.ubuntuforums.org) were a huge help in setting it up and stuff.  All you have to do is enter a search for your specific problem and they usually have a few threads on how to help you out with it.

Now, I use strictly Ubuntu and I don't think I'll be going back to Windows again.  I'd recommend Ubuntu as a beginner distro because of its friendly GUI interface that helps you adjust to linux.

I like both Ubuntu and SuSE, but I personally think Mandriva sucks and it's way too hard to install anything. (K)ubuntu is the best distro I've used so far (I've used ubuntu, suse, mandriva, gentoo, and maybe others that I don't remember right now).

I agree with Chess64. I started with SuSE several years ago, and have since used Gentoo, CentOS, Slackware and Kubuntu. Kubuntu is easily the nicest distro I've used - really painless installation (especially given that up until that point I'd been using Gentoo everywhere, which has an install process that can last literally days), stable system and nice package management.

Each distro has its strong points, so I suppose it's really a personal choice. SuSE has a nice config tool (YaST), Gentoo (despite what a lot of people say) does indeed help you learn more about the workings of the OS, as long as you're willing to do more than mindlessly follow the install process (and if you're very patient :P) and CentOS is nice if you want to play around with server-side stuff. Kubuntu (and Ubuntu I suppose) is generally easy to use, and lets you get on with whatever you want to do, without much hassle.

Good luck whichever you choose :)


Este problema esta horrible. O por lo menos asi esta mi solucion.
El máximo valor que toma la suma es $100^{2}$.

[b]Demostracion:[/b]
Sea $s = \sum_{i=1}^{n}a_{i}^{2}$
Sea $t=\sum_{i=3}^{n}a_{i}$
Sea $u=\sum_{i=3}^{n}a_{i}^{2}$

Lo primero es que $a_{2}\le 50$ pues es menor o igual que $a_{1}$.
Fijemos primero $a_{2}$. Notemos que la suma $a_{1}^{2}+a_{2}^{2}$ se maximiza si hacemos $a_{1}$ maximo es decir $a_{1}= 100-a_{2}$.

Ahora si $98a_{2}\le 100$ es decir $a_{2}\le \frac{50}{49}$ tenemos que:
$s\le (100-a_{2})^{2}+99a_{2}^{2}= 100^{2}-200a_{2}+100a_{2}^{2}\le 100^{2}$ con igualdad si $a_{2}=2$ que no se puede pues $2> \frac{50}{49}$ o si $a_{2}=0$

Entonces podemos asumir que $a_{2}> \frac{50}{49}$.
Sea $k=\lfloor \frac{100}{a_{2}}\rfloor$
Sabemos que $t=100$, de lo contrario podemos empezar a sumarle valores a los $a_{i}$ ($i\ge3$) y obtenemos sumas mayores de los cuadrados(hasta que la suma de los $a_{i}$ sea $100$).
Ahora probaremos que:
$u \le ka_{2}^{2}+(100-ka_{2})^{2}$

Sabemos que:
$a_{i}\le a_{2}$ para $i\ge 3$ (de ahora en adelante $i,j \ge 3$)
Ahora haremos smoothing para probar lo que queremos.
Si $a_{i}+a_{j}\le a_{2}$ tomamos $a'_{i}= a_{i}+a_{j}$ y $a'_{j}=0$ y $a'_{i}^{2}+a'_{j}^{2}=(a_{i}+a_{j})^{2}\ge a_{1}^{2}+a_{2}^{2}$(notemos que el orden de los $a_{i}$ para $3\le i\le 100$ no importa, lo unico que importa es que $a_{2}\ge a_{i}\ge 0$. Al final se pueden ordenar).
Si $a_{i}+a_{j}= a_{2}+\alpha$ con $a_{2}> \alpha \ge 0$ ponemos:
$a'_{i}=a_{2}$ y $a'_{j}=\alpha$ y tenemos que:
$a_{2}^{2}+\alpha^{2}= (a_{2}+\alpha)^{2}-2a_{2}\cdot \alpha \ge (a_{i}+a_{j})^{2}-2a_{i}\cdot a_{j}= a_{i}^{2}+a_{j}^{2}$ pues $2a_{i}\cdot a_{j}\ge 2a_{2}\cdot\alpha$ pues suman lo mismo pero $a_{i}, a_{j}$ se acercan mas al promedio.
Con esto queda probado lo que queriamos, que es tener la maxima cantidad de $a_{i}= a_{2}$ (que son $k$) y el resto en un solo $a_{j}$.
Llamemos $a_{2}= a$ para que sea mas corto de escribir.
Lo que queremos probar ahora es:
$(100-a)^{2}+a^{2}+ka^{2}+(100-ak)^{2}\le 100^{2}$ $\iff$
$50+\frac{(k^{2}+k+2)a^{2}}{200}\le a(k+1)$
Pero sabemos que $k=\frac{100}a-r$ con $0\le r < 1$
Reemplazando obtenemos que lo anterior se cumple si y solo si:
$a(r^{2}-r+2) \le 100$ despues de expandir y cancelar terminos.
Pero $r^{2}-r \le 0$ pues $0\le r <1$
Entonces tenemos que:
$a(r^{2}-r+2) \le 2a \le 100$ que es cierto pues $a\le 50$

Por lo que tenemos que el minimo $s$ es en efecto $100^{2}$ y la igualdad se da solamente en los siguientes dos casos:
$a_{1}=100$ y $a_{2}=a_{3}=...=a_{100}=0$
$a_{1}=a_{2}=a_{3}=a_{4}=50$ y $a_{5}=a_{6}=...=a_{100}=0$
QED

Nadie tiene una forma mas bonita de solucionar este problema? He estado pensandolo y no se me a ocurrido nada. :mad:


geometry
perimeter
rectangle
calculus
calculus computations

I need help with another EVP.

A window has perimeter 10 m and is the shape of a rectangle with the top edge replaced by a semicircle. Find the dimensions of the rectangle if the window admits the greatest amount of light.

Thanks!

$ P\equal{}(\pi \plus{}2)r\plus{}2h$

$ A\equal{} \frac{\pi}{2} r^2 \plus{}2rh$

$ 2h\equal{}10\minus{}(\pi\plus{}2)r$

$ A(r)\equal{}\frac{\pi}{2}r^2\plus{}10r\minus{}(\pi \plus{}2)r^2\equal{}\minus{}(\frac{\pi}{2}\plus{}2)r^2\plus{}10r$

$ A'(r)\equal{}\minus{}(\pi\plus{}4)r\plus{}10$, $ A'(r)\equal{}0 \to r\equal{}\frac{10}{\pi\plus{}4}\equal{}h$

so, we get height above with width twice that. (for the rectangular dimensions)


In how many ways can you locate 17 different books in 3 different boxes?

I dont understand how to resolve this exercise. I have the solution 129140163

[url=http://www.mathlinks.ro/Forum/viewtopic.php?t=51376&search_id=837198339]Take a look here[/url]

thanks kunny, so for the first part is variations with repetition.

What about the second part:

How many ways if all the boxes must be full?

The solution: 128746950

Apparently is 3^17 - 3(2^17 - 1)

I dont understand that number 3 multiplying (2^17 - 1) ?? why??


This problem has been posted before on the forum, but I would like to see a different proof, preferably without substitutions.

Given $a,b,c,x,y,z\geq 0$ and $x+y+z=1$, prove that:

$ax+by+cz+2\sqrt {(ab+bc+ca)(xy+yz+zx)}\leq a+b+c$

Thanks!

Any ideas?  I am finding this problem quite frustrating.

$a+b+c=\sqrt{\left(a^2+b^2+c^2+2(ab+bc+ca)\right)\left(x^2+y^2+z^2+2(xy+yz+zx)\right)} \ge ax + by + cz + 2\sqrt{(ab+bc+ca)(xy+yz+zx)}$
Cauchy!

I don't know whether to laugh at myself or thank you...so I'll just do both.  Thanks.


logarithms
algebra
functional equation
calculus
calculus computations

Show that the functional equation

$ f\left(\frac{2x}{1+x^{2}}\right) = (1+x^{2})f(x)$

is satisfied by

$f(x) = 1+\frac{1}{3}x^{2}+\frac{1}{5}x^{4}+\frac{1}{7}x^{6}+\cdots$ with $|x| < 1$.

Simply check that $\frac{\arctan(iz)}{iz}=\frac{-1}{2z}\ln \frac{1-z}{1+z}$, verifies the functional equation.


graph theory
combinatorics unsolved
combinatorics

Let a set $X$ be given which consists of $2 \cdot n$ distinct real numbers ($n \geq 3$). Consider a set $K$ consisting of some pairs $(x, y)$ of distinct numbers $x, y \in X$, satisfying the two conditions:

[b]I.[/b] If $(x, y) \in K$ then $(y, x) \not \in K$.
[b]II.[/b] Every number $x \in X$ belongs to at most 19 pairs of $K$.

Show that we can divide the set $X$ into 5 non-empty disjoint sets $X_1, X_2, X_3, X_4, X_5$ in such a way that for each $i = 1, 2, 3, 4, 5$ the number of pairs $(x, y) \in K$ where $x, y$ both belong to $X_i$ is not greater than $3 \cdot n$.

This is a graph theory problem, the numbers are vertices and the pairs are edges. For each $x\in X$ let $d(x)$ be the degree (number of edges) of $x$. If $U,V$ are subsets of $X$ let $d(U,V)$ be the number of edges between $U$ and $V$. For each decomposition of $X$ into components, an edge $xy$ is called intercomponent if $x$ and $y$ belongs to different components. Let $X = (A,B,C,D,E)$ be the decomposition such that the number of intercomponent edges are maximal. For each $a \in A$ we claim that $d(x,A) \leq d(x,B)$. If not, moving $a$ to $B$ would increase the number of intercomponents. Similar for $B,C,D,E$. Thus $5d(x,A) \leq d(x,A) +d(x,B)+d(x,C)+d(x,D)+d(x,E) \leq 19$ hence $d(x,A) \leq 3$ for each $a \in A$.
Therefore, $d(A,A) \leq 3n$. Same argument for other components.

This is basically the same argument that shows up more commonly when we want only 2 disjoint sets (the max-cut problem):
http://www.artofproblemsolving.com/Forum/viewtopic.php?f=41&t=392464


Prove: If there are 6 people at a party, then either 3 of them knew each other before the party or 3 of them were complete strangers before the party.

In fact you can do better: there are at least 2 such triples.

[quote="hoshattack"]The sum of a+b=6.[/quote]

Consider a party with three pairs of friends.  If by "a" you mean "the largest group of people that know each other" and by "b" you mean "the largest group of people that don't know each other," then $ a \equal{} 2, b \equal{} 3, a \plus{} b \equal{} 5$.

The general problem is far from common sense, and is in fact an outstanding problem in graph theory.  See [url=http://mathworld.wolfram.com/RamseyNumber.html]Ramsey numbers[/url].

[quote="diophantient"]This is equivalent to saying that $ K_{6}$ has a monochromatic triangle, right?[/quote]

Exactly.  Assuming that you mean each edge in $ K_6$ is colored one of two colors and a monochromatic triangle is three vertices where each edge is the same color.  Or with pleurestique's addition, that there are two such triangles.

[hide="solution"]put it on a graph, red=friends, blue=non-friends. consider any vertex A, by pigeonhole some 3 edges, say AB,AC,AD going out of it have to be the same color, say red. if any of BC,CD,DB are red then we get a red triangle so they're all blue, but then we get a blue triangle.[/hide]

[quote="hoshattack"]Wouldn't that just be common sense?  Since there are a total of six people, if 3 people didn't know each other, then 4 people would have know each other.  The sum of a+b=6.  Either a or b must be 3 or greater.

Or is it something way different?[/quote]

Pigeon hole principle:

If we have n+1 pigeons and want to keep them in n boxes(jaja), no matter how we do this there will always at least 2 pigeons in one box.

Traduced: common sense.

 :rotfl:

This is just a ver specif way to define it, and I'm very far to be saying this is useless.

 :)


Do there exist 10 distinct integers, the sum of any 9 of which is a perfect square?

does integer mean >= 0 ?

If not :

-3050936351, 48427561, 392801329, 431065081, 435316609, 435789001, 435841489, 435847321, 435847969, 435848041

is a solution !

[hide]

Using linear algebra :

A*X = Y where

A =

(1,1,...,1,0)

(1,1,...,0,1)

..

(0,1,...,1,1)

A is inversible (is it the right word ? = not singular, det != 0), and B=A^1 is "routine" to evaluate. Also det(A)=9.

Then take V = (3^2,3^4, ..., 3^20)

X = B*V is then a solution.

If it's an olympiad level problem, no doubt there is an easy (and less ugly   ) solution. At least we now know a solution exists.[/hide]


How to do this ?

A plane has been partitioned into $n$ regions by 3 sets of parallel lines such that no three meet in a point. What is the least number of lines required such that $n>2001$  ?

Hint: Consider what happens when you add one new line to an arbitrary partition.


Given $ a, b, c \geq\ 0.$ Prove that:
$ (a\plus{}b\plus{}c)^3 \geq\ 27abc \plus{} 6\sqrt[]{3}(a\minus{}b)(b\minus{}c)(c\minus{}a)$

Try $ a \equal{} \min\{a,b,c\},$ $ b \equal{} a \plus{} x$ and $ c \equal{} a \plus{} y.$
 If your inequality is true then this way helps to you. :wink:

I know. But can U check for me it is trues or not. If it's trues, it's very nice althought easy  :)

Equality occurs when $ a \equal{} b \equal{} c > 0$ or $ a \equal{} k(3 \minus{} \sqrt []{3}); b \equal{} k(3 \plus{} \sqrt []{3}); c \equal{} 0$ and any cyclic.


calculus
integration
calculus computations

I know this must be an easy problem to be in a BC calculus exam, but I'm just drawing a blank:

A particle moves along the x-axis so thats its acceleration at any time t is a(t) = 2t - 7. If the initial velocity of the particle is 6, at what time t during the interval 0<= t <= 4 is the particle farthest to the right?

$v(t)=\int a(t) dt=t^2-7t+Const.$, from $v(0)=6$, we have $v(t)=t^2-7t+6=(t-1)(t-6)$.
Let $x(t)$ be the dislocation at the time $t$, $x'(t)=v(t)$, for $0\leqq t\leqq 4$,yielding the answer $t=1$.


modular arithmetic
number theory unsolved
number theory

Prove that there are infinite prime numbers of the form $ 4n\minus{}1$, where $ n \in N$.

$ P_{1}\equal{}7,...,P_{n}\equal{}4k\plus{}3$,$ i>j$ $ \implies$  $ P_{i}>P_{j}$, $ (P_{i}\equiv \minus{}1\pmod{4})$
$ \implies$  $ 4.(P_{1}.P_{2}...P_{n})\plus{}3\equal{}q$
if  $ q$  is prime  $ \implies$  $ q>P_{n}$  contradiction!
if  $ q$  is coprime $ \implies$  ,$ q$ has divisor same form.But $ P{i}\nmid q$  contradiction!

[quote="Obel1x"]Prove that there are infinite prime numbers of the form $ 4n \minus{} 1$, where $ n \in N$.[/quote]

Let $ p_1 ,p_2 , ...,p_n$ be all prime numbers.

Suppose that number $ 4 p_1 p_2 ...pn \minus{} 1$  is composite

Then it has at least one prime divisor $ 1 < p_i \leq n$ , but number $ 4p_1 p_2 ...p_n \minus{} 1$ is not dividable by any of $ p_i$ ,contradiction


[quote="Beginner"]I've posted this problem in computational calculus... section, but nobody solved it. I consider this is hard.
find $y(x)$ so that
\[ (2+2x \sqrt{y})x dx + (2+x^2 \sqrt{y})y dy =0  \][/quote]

Maybe try to find an integrating factor to make it exact.

I don't think that help.
let $k(x)$ be the integrating factor
then we need to find $k(x)$ so that
\[ \frac{\partial (2xk(x)+2x^2 k(x) \sqrt{y})}{\partial y} = \frac{\partial (2yk(x)+x^2 y k(x) \sqrt{y})}{\partial x}  \]
this yield
\[ x^2 k(x) \sqrt{y} = k'(x) y^2 (2+x^2 \sqrt{y}) + 2x k(x) y^2 \sqrt{y}  \]  :ninja:
 :(  :?
i also try using subtitutioun $u=x^2$ and $v=y^2$ but i don't know how to continue
anybody solve it?

NOTE: My friend from other faculty ask me to compute this ODE, maybe this appears in his modeling of problem, and maybe there is initial value for this ODE so that it can solve using numerical method.


[quote="José"]There are $210$ peoples in a gym. Half of the women and the third part of the boys are, each, at a bycicle. If there are $85$ bycicles occupied, how many boys and girls are there at the gym?[/quote]

[hide]Call the number of girls $x$ and the number of boys $y$
.Thus, $x+y=210$. Half the girls and a third of the boys on bicycles can be represented as $\frac{x}{2}+\frac{y}{3}=85$.
Solving both as yields x=90, y=120.[/hide]

[quote="daermon"][quote="mtms5467"][quote="José"]There are $210$ peoples in a gym. Half of the women and the third part of the boys are, each, at a bycicle. If there are $85$ bycicles occupied, how many boys and girls are there at the gym?[/quote]
What is the ratio of boys to girls?[/quote]

you don't need to know the ratio, otherwise it would be trivial  :wink:[/quote]
Oh. Whoops. :oops:

[hide="answer"]Call the number of boys $a$ and the number of girls $b$. We can set up the equations $\frac{1}{2}b+\frac{1}{3}a=85$ and $a+b=210$. Solving we find $a=120$ and $b=90$.[/hide]

[quote="José"]There are $210$ peoples in a gym. Half of the women and the third part of the boys are, each, at a bycicle. If there are $85$ bycicles occupied, how many boys and girls are there at the gym?[/quote]

ok so w=women and b=boys

1/2 w + 1/3 b = 85
w+b=210
w= 210-b

(210-b)/2 + 1/3 b =85
105- 1/2b + 1/3 b= 85
1/6b=20
b=120
g= 90

( Is this right :?: )

210(you said bluntly boys AND girls)

or if you meant the num. of boys and the num. of girls....[hide]120 boys, 90 girls.[/hide]

[quote="José"]There are $210$ peoples in a gym. Half of the women and the third part of the boys are, each, at a bycicle. If there are $85$ bycicles occupied, how many boys and girls are there at the gym?[/quote]

[hide="..."]
There are 90 girls, and 120 boys, but you didn't exactly ask a question...

mtms5467, the ratio is 4:3...[/hide]

I think there are 90 girls and 120 boys. Is that right? :?:

Yes, it's right. I think everybody knows the answer, so it's useless to post another solution

[quote="José"]There are $210$ people in a gym. Half of the women and the third part of the boys are each at a bicycle. If there are $85$ bicycles occupied, how many boys and girls are there at the gym?[/quote]

[hide]
$b+g=210$
$b/3+g/2=85$
$b+3g/2=255$
$b+g=210$
$g/2=45$
$g=90$
$b=210-90$
$b=120$[/hide]


number theory
prime numbers
relatively prime

If the quotient of a/b has a period of "c" figures and (b-1)/c results to be a non-integer number, then "b" is NOT a prime number.  "a" and "b" are two integers, primes between them.

For instance, from 2/57 results the quotient 0,035087719298245614 which has a period of 18 figures and because the result of (57-1)/18 is 3,1 which is a non-integer, then 57 is not prime.

I have to mention that this conjecture doesn't determine if “b” is prime but if “b” is not prime, since it does not succeed in finding all numbers that are not prime.
The non-prime numbers that the conjecture does not find on an interval between 3 and 5000 (from which all multiples of 2 and 5 have been excluded) are: 9 - 33 - 91 - 99 - 259 - 451 - 481 - 561 - 657 - 703 - 909 - 1233 - 1729 -  2409 - 2821 - 2981 - 3333 - 3367 - 4141 - 4187 - 4521 -  …

The Carmichael numbers are a subset of the whole non-prime numbers which the conjecture can not identify.
I would appreciate any suggestions or demonstrations regarding this conjecture. Thank you all in advance.

This conjecture is true- it's a special case of Fermat's "little" theorem. As a computational method, it's entirely contained in the idea of pseudoprimes (base 10 in this case).

Assume that $ b$ is relatively prime to $ 10$ and $ a$ is relatively prime to $ b$; the alternative isn't interesting. The period length $ c$ is the smallest positive $ k$ for which $ 10^k\equiv 1\mod b$. If $ b$ is prime, $ 10^{b\minus{}1}\equiv 1\mod b$. Since the only $ k$ with $ 10^k\equiv 1$ are the multiples of $ c$, $ c|b\minus{}1$.

[quote="jmerry"]This conjecture is true- it's a special case of Fermat's "little" theorem. As a computational method, it's entirely contained in the idea of pseudoprimes (base 10 in this case).

Assume that $ b$ is relatively prime to $ 10$ and $ a$ is relatively prime to $ b$; the alternative isn't interesting. The period length $ c$ is the smallest positive $ k$ for which $ 10^k\equiv 1\mod b$. If $ b$ is prime, $ 10^{b \minus{} 1}\equiv 1\mod b$. Since the only $ k$ with $ 10^k\equiv 1$ are the multiples of $ c$, $ c|b \minus{} 1$.[/quote]

Could you show me with a little example how can you find the unknown digit "x" of 2/7=0,285X14, developing all the operations? Thank you.

[quote="Explorer"]Could you show me with a little example how can you find the unknown digit "x" of 2/7=0,285X14, developing all the operations? Thank you.[/quote]

I already answered this question in your [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=201696]other thread[/url].  Here is a computational demonstration:

We want to examine $ k \equal{} \frac{10^6 \minus{} 1}{7}$.  We can easily compute that $ k \equal{} 142857, ak \equal{} 285714$ which gives your answer directly, or we can do the following:

Since $ 10^6 \equiv 1 \bmod 11$, we know that $ 11 | k$.  Then $ 11 | ak \equal{} 285X14$, which leaves remainder

$ (4 \plus{} X \plus{} 8) \minus{} (1 \plus{} 5 \plus{} 2) \equiv 4 \plus{} X \bmod 11$

which gives $ X \equal{} 7$.

[quote="t0rajir0u"][quote="Explorer"]Could you show me with a little example how can you find the unknown digit "x" of 2/7=0,285X14, developing all the operations? Thank you.[/quote]

I already answered this question in your [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=201696]other thread[/url].  Here is a computational demonstration:

We want to examine $ k \equal{} \frac {10^6 \minus{} 1}{7}$.  We can easily compute that $ k \equal{} 142857, ak \equal{} 285714$ which gives your answer directly, or we can do the following:

Since $ 10^6 \equiv 1 \bmod 11$, we know that $ 11 | k$.  Then $ 11 | ak \equal{} 285X14$, which leaves remainder

$ (4 \plus{} X \plus{} 8) \minus{} (1 \plus{} 5 \plus{} 2) \equiv 4 \plus{} X \bmod 11$

which gives $ X \equal{} 7$.[/quote]

Sorry, I have posted in the wrong thread. Thanks a lot for making me notice that and also for your answer.


It is common to develop the real numbers axiomatically. That is, we take $\mathbb{R}$ to be a set which satisfies the axioms of an ordered field, and which also satifies one additional axiom. What should that one additional axiom be? I offer seven possible choices:

1. Every nonempty subset that is bounded above has a least upper bound.

2. Suppose $A\cup B=\mathbb{R}, A\cap B=\emptyset, A\ne\emptyset, B\ne\emptyset$, and for all $a\in A,b\in B, a<b$, then either $A$ has a greatest element or $B$ has a least element.

3. Monotone bounded sequences converge.

4. Cauchy sequences converge.

5. If $I_1\supset I_2\supset I_3\supset\cdots$ are nested closed bounded intervals, then $\bigcap_{n=1}^{\infty}I_n\ne\emptyset$.

6. Bounded sequences have convergent subsequences.

7. Every decimal expansion converges to a real number, and every real number has at least one decimal expansion.

(Note: the second phrase in #7 doesn't look like completeness, but without it, we wouldn't have the Archimedean property.)

The most common textbook path through this would start by assuming 1 as an axiom. I would consider the following implications "textbook obvious":
$1\implies 3\implies 5\implies 6\implies 4, 1\implies 2, 3\implies 7.$

I have two challenges:

First, to show that all seven of these statements are equivalent; that we could start with any one as the axiom and prove the other six as theorems.

Second, to prove as many of the 42 direct implications as possible without going through one of the other statements. Clearly, you can't do all of these, but which ones can you do?

Any construction of the real numbers should lead fairly quickly to one of these properties.

What about the Dedekind method  ( Dedekind slice )of constructing real numbers??

 ( A real number is a set of rational numbers with some properties )

The dedekind methode is very similar to/exactly the same as the second axiom.
(Dedekind axioms: a dedekindian slice is a (ordered) pair $(A,B)$ of two sets of rational numbers that fulfil $A\cup B=\mathbb{Q}, A\cap B=\emptyset, A\ne\emptyset, B\ne\emptyset, \forall a \in A, b \in B : a<b$ and $B$ has no least element).
But I think there is an other problem: not all these axioms guarant the Archimedean property (for example the 4.)


If in the formula $ C \equal{} \frac {en}{R\plus{}nr}$, $n$ is increased while $ e$, $R$ and $r$ are kept constant, then $C$:

$\textbf{(A)}\ \text{Increases} \qquad
\textbf{(B)}\ \text{Decreases} \qquad
\textbf{(C)}\ \text{Remains constant} \qquad
\textbf{(D)}\ \text{Increases and then decreases} \qquad\\
\textbf{(E)}\ \text{Decreases and then increases}$

[hide="solution"]e, R, and r are constants, so we arbitrarily assign values to them.  To keep it simple, we make all equal to 1.

$ C = \frac{1*n}{1+n*1}$

$ C = \frac{n}{1+n}$
testing n=1, 2, 3, we get 1/2, 2/3, 3/4, respectively, or 6/12, 8/12, 9/12.
With the numerator just 1 less than the denominator, C is progressing towards 1.

Thus, C increases, and the answer is A.  [/hide]

[quote="Tecnogram888"][hide="solution"]e, R, and r are constants, so we arbitrarily assign values to them.  To keep it simple, we make all equal to 1.

$ C = \frac{1*n}{1 + n*1}$

$ C = \frac{n}{1 + n}$
testing n=1, 2, 3, we get 1/2, 2/3, 3/4, respectively, or 6/12, 8/12, 9/12.
With the numerator just 1 less than the denominator, C is progressing towards 1.

Thus, C increases, and the answer is A.  [/hide][/quote]
What if $ R=0$? Or what if $ R<0$?

[hide="sol."]en/R+nr-en+e/R+nr+r=enR+ern^2+enr-enR-ern^2-er-enr/(R+nr)(R+nr+r)
=-er/(R+nr)(R+nr+r).
So, it increases if er is pos., decreases if er is neg.[/hide]

Sorry for revival, but I'm not quite sure I understand the answer. Can anyone clarify?

$C=\frac{e}{r+\frac{R}{n}}$ which, clearly, is increasing with $n$.

Not if R is negative, e is positive, etc.

How would you solve this without testing the cases of

Negative.  Negative

Positive.  Positive

Negative positive

Positive negative

And so what exactly is the correct answer?

The official answer key says B, but i think the problem isn't clearly stated as it doesnt say where n starts

$C(R+nr) = en$
$CR = en-Cnr$
$CR = n(e-Cr)$
$\frac{CR}{e-Cr} = n$
Assume R is positive for now.
$\frac{e-Cr}{C} \cong \frac1n$
I use congruent to mean in/decreases in the same direction.
Note as C increases, the numerator decreases, and the denominator increases, so $\frac1n$ decreases, so n increases.
However, n decreases if R is negative.

One could argue that C is correct if e=0 or R=0


Does anyone know of a good trigonometry/complex number book?

I used a regular precal book when I learned precal a few years ago (it was Larson's). Most books do not properly cover problem solving so I reccomend ACoPS and the USSR problem book for good introductory chapters into easy trig/complex numbers problems. Or you could wait until the AoPS book. :)

103 trigonometry problems by titu andreescu

Depends what you want- if its a problem book you're after, 103 Trigonometry Problems is a gem. For a challenging introductory text however, good trigonometry-centric books are far and few these days. You could go with the classics which are as good as ever- like the Michigan University reprint of S L Loney's Trigonometry (volumes 1 and 2, where 2 introduces complex numbers); or Dover's edition of E W Hobson's book.

book S L Loney's Trigonometry  link: http://rapidshare.com/files/141478225/Loney-Plane_trigonometry.rar.html

there is also a blog called IIT dreams where you can find like a whole bunch of stuff.


probability
combinatorics proposed
combinatorics
TST
China TST

Let $ S$ be a set that contains $ n$ elements. Let $ A_{1},A_{2},\cdots,A_{k}$ be $ k$ distinct subsets of $ S$, where $ k\geq 2, |A_{i}| \equal{} a_{i}\geq 1 ( 1\leq i\leq k)$. Prove that the number of subsets of $ S$ that don't contain any $ A_{i} (1\leq i\leq k)$ is greater than or equal to $ 2^n\prod_{i \equal{} 1}^k(1 \minus{} \frac {1}{2^{a_{i}}}).$

[quote="darij grinberg"]The proof is flawed. $ H_{A\cup B} \equal{} H_A\cup H_B$, not $ H_{A\cup B} \equal{} H_A\cap H_B$. Mind the difference between "do not contain $ A$" and "do not intersect $ A$".

Fang-jh: Which year's TST is this?

I don't see any reason to require the sets $ A_i$ to be different, by the way. The condition $ a_i\geq 1$ is equally useless.

  darij[/quote]
it is problem 3 of the fourth quiz of Chinese TST 2008  (March 22,2008)

[quote="darij grinberg"]Sorry for a potentially stupid question, but what is a "quiz"? Is it just a round of the TST, or is it something easier than the usual TST's?

  darij[/quote]
There are 6 quizzes in total, it is easier than the usual TST.

It directly follows from Kleitman lemma (which is itself quite subtle):

If $ U$ and $ V$ are two collections of subsets of a set $ S$, each of them closed under taking subsets (i.e. $ A\in U$, $ B\subset A$ implies $ B\in A$) then $ |U\cap V|\ge |U|\cdot |V|/2^{|S|}$.

That is, events ``to belong to $ U$'' and ``to belong to $ V$' for randomly chosen subset of $ S$ correlate positively (or how to say it?)

Incorrect, sorry:

So let us consider a random subset $X$ of $S$ that is constructed by placing each element of $S$ in $X$ with probability $\frac{1}{2}$. Now, consider the probability that some specific $A_i$ is completely contained in $X$. That means for each $x \in A_i$, we must have placed $x$ in $X$, and this happens with probability $\frac{1}{2^{a_i}}$. Thus, the probability that $A_i$ is not completely contained in $X$ is $1 - \frac{1}{2^{a_i}}$. Thus, the probability that no $A_i$ is contained in $X$ is simply $ \prod_{i = 1}^k(1-\frac{1}{2^{a_{i}}}). $ But there are $2^n$ total subsets of $S$, and then the result follows. $\blacksquare$

[quote=JBL][quote="TheStrangeCharm"]Thus[/quote] These are not independent events![/quote]

Sure,you can't multiply them like this!

Sorry for reviving
But is there any solution with expected value?

Take a random subset $U$ of $S$. Let $E_i\,,\,i=1,2,\dots,k$ be the event that $U$ doesn't contain $A_i$. We will prove
$$P(E_1\cap E_2\cap\dots\cap E_k)\geq \prod_{i = 1}^k(1 - \frac {1}{2^{a_{i}}}).$$
The proof goes by induction. We start with the equality:
\[P(E_2\cap\dots\cap E_k)=P(E_1)P_{E_1}(E_2\cap\dots\cap E_k)+P(\overline{E_1})P_{\overline{E_1}}(E_2\cap\dots\cap E_k).\]
It follows:
\begin{align*}(*)\,\,\,\,\,\,\,\,\, P(E_1\cap\dots\cap E_k) &= P(E_1)P_{E_1}(E_2\cap\dots\cap E_k) = \\
&= P(E_2\cap\dots\cap E_k) - P(\overline{E_1})P_{\overline{E_1}}(E_2\cap\dots\cap E_k)\\
&\geq P(E_2\cap\dots\cap E_k)- P_{\overline{E_1}}(E_2\cap\dots\cap E_k). \end{align*}

Let's consider the probability $P_{\overline{E_1}}(E_2\cap\dots\cap E_k)$. Let us denote $A'_i := A_i\setminus A_1\,,\,i=2,3,\dots,k$. We want to estimate the probability the random set $U$ to contain $A_1$ and $U_1 :=U\setminus A_1$ not to contain any $A'_i\,,\,i=2,\dots,k$. To each such $U_1$ corresponds $2^{a_1}$ sets $U :=U_1\cup U'$, where $U'\subset A_1$ not containing any $A'_i\,,\,i=2,\dots,k$.
It means:
$$P_{\overline{E_1}}(E_2\cap\dots\cap E_k) =2^{-a_1}P(E'_2\cap\dots\cap E'_k) \leq  2^{-a_1}P(E_2\cap\dots\cap E_k) $$
Here the event $E'_i,i=2,\dots,k$ means the random set $U$ not to contain $A'_i$.

Now, we use the induction hypothesis, so $P(A_2\cap\dots\cap A_k)\geq \prod_{i = 2}^k(1 - \frac {1}{2^{a_{i}}})$. Then $(*)$ yields:
\[P(A_1\cap\dots\cap A_k)\geq \prod_{i = 2}^k(1 - \frac {1}{2^{a_{i}}}) (1-\frac{1}{2^{a_1}})=\prod_{i = 1}^k(1 - \frac {1}{2^{a_{i}}}).\]
which completes the induction step.

$\textbf{Solution :}$

Induction on $\sum_{i\not= j} |A_i\cap A_j|$. Consider larger set $S':= S\cup\{n+1\} = [1, n+1]$ and same subsets $A_1,\ldots , A_k$ of it. So it's enough to prove that number of subsets of $S'$ that don't contain any $A_i$ is greater than or equal to $2^{n+1}\prod_{i=1}^k(1-\frac{1}{2^{a_i}})$. We can assume that $\exists a\in A_1\cap A_2$. Consider new subsets of $S'$ : $\{n+1\}\cup A_1\setminus\{a\}, A_2, A_3, \ldots , A_k$. Consider next numbers :

1) Let $X$ be number of subsets of $S'$ that don't contain any $A_i$.

2) Let $X_{n+1}^a$ be number of subsets of $S'$ that don't contain any $A_i$ and also don't contain element $n+1$ but contain element $a$.

3) Let $X_{a}^{n+1}$ be number of subsets of $S'$ that don't contain any $A_i$ and also don't contain element $a$ but contain element $n+1$.

4) Let $X_{n+1, a}$ be number of subsets of $S'$ that don't contain any $A_i$ and also don't contain elements $n+1, a$.

5) Let $X^{n+1, a}$ be number of subsets of $S'$ that don't contain any $A_i$, but contain elements $n+1, a$.

6) Let $Y$ be number of subsets of $S'$ that don't contain any $A_i$ for $i\not= 1$ and don't contain set $\{n+1\}\cup A_1\setminus\{a\}$.

7) Similarly define numbers $Y_{n+1}^a, Y_{a}^{n+1}, Y_{n+1, a}, Y^{n+1, a}$.

By induction hypothesis we know that $Y \geq 2^{n+1}\prod_{i=1}^k(1-\frac{1}{2^{a_i}})$, so it's enough to prove that $X\geq Y$.

$\textbf{Lemma 1 :}$

We get that $X_{a}^{n+1} + X_{n+1}^a\geq Y_{a}^{n+1} + Y_{n+1}^a$.

$\textbf{Lemma 2 :}$

We get that $X_{n+1, a} = Y_{n+1, a}$.

$\textbf{Lemma 3 :}$

We get that $X^{n+1, a} = Y^{n+1, a}$.

Proofs of lemmas 1-3 is simple exercise. So, finally, we get that $X = (X_{a}^{n+1} + X_{n+1}^a) + X_{n+1, a} + X^{n+1, a}\geq (Y_{a}^{n+1} + Y_{n+1}^a) + Y_{n+1, a} + Y^{n+1, a}\geq 2^{n+1}\prod_{i=1}^k(1-\frac{1}{2^{a_i}})$. $\Box$

Here's a solution using a nuke.

[b] Theorem: [/b] ([url="https://en.wikipedia.org/wiki/FKG_inequality"]FKG Inequality[/url]) Let $X_1, \dots, X_n$ be decreasing functions on $\{0, 1\}^n$ (or any lattice). Then $\mathbb{E}[\prod X_i] \ge \prod \mathbb{E}[X_i].$

Now to finish use this inequality for the variables $X_i(S) = 0$ if $A_i \subseteq S$ and $1$ otherwise.

Keeping stuff low-powered, I guess.
-------
We begin with the key lemma.

[color=#c00][b]Lemma.[/b][/color] If we replace any element $x$ in one of the $X_i$s with an element $y$ that does not appear in any of the other $X_i$s, then the number of subsets of $\{1,2,\cdots,n\}$ not containing any of the other $X_i$s does not increase.

[color=#c00][i]Proof.[/i][/color] WLOG the $X_i$ having $x$ replaced with $y$ has index $1$. Consider the $X_i$: $X_1\setminus \{x\}, X_2, X_3, \ldots$. There are two classes of subsets of $\{1,2,\ldots,n\}$ that work for $(X_1\setminus \{x\})\sqcup \{y\}, X_2, \cdots, X_m$: Ones that also work $X_1\setminus \{x\}, X_2, X_3, \ldots$, and ones that do not.

Notice that all of the subsets that work for $X_1\setminus \{x\}, X_2, X_3, \ldots$ also work for both $(X_1\setminus \{x\})\sqcup \{y\}, X_2, \cdots, X_m$ and $X_1,X_2,\cdots X_m$.

Now, we look at the subsets that do not work for \[(X_1\setminus \{x\}), X_2, X_3\cdots, X_m\] but do work for $X_1, X_2, \cdots, X_m$. These subsets are exactly those which contain $X_1\setminus \{x\}$ that do not contain $x$ and $X_2, X_3, \cdots, X_m$.

Similarly, we look at the subsets that do not work for \[(X_1\setminus \{x\}), X_2, X_3\cdots, X_m\] but do work for $(X_1\setminus \{x\})\sqcup \{y\}, X_2, X_3,\cdots, X_m$. These subsets are exactly those which contain $X_1\setminus \{x\}$ that do not contain $y$ and $X_2, X_3, \cdots, X_m$.

Thus, it suffices to show that there are at least as many subsets that contain $X_1\setminus \{x\}$ that do not contain $x$ and $X_2, X_3, \cdots, X_m$ than those that contain $X_1\setminus \{x\}$ that do not contain $y$ and $X_2, X_3, \cdots, X_m$. But notice that if a subset contains both $x$ and $y$ or neither $x$ nor $y$ and works for the second case, it also works for the first case. If a subset $S$ contains only $x$ and works for the second case, then $(S\setminus \{x\})\sqcup \{y\}$ also works for the first case. Thus, the first is at least as large, and the proof of the Lemma is complete.
---------
Now, we show the problem for $\sum_i |X_i|\leq n$ and $|X_i|$ fixed for each $X_i$.
[color=#c00][i]Proof.[/i][/color] If we have two equal numbers in two of the $X_i$, then we know that we can replace one of them with a number that does not appear in any of the $X_i$. By the size assumption, we won't run out of numbers, so the minimum number of subsets not containing any of the $X_i$ is attained when all of the $X_i$ are disjoint. Then, by a simple counting argument (e.g. using proportions, similar to the proof of the formula of Euler's Totient Function), we have the desired bound of \[2^n\cdot\prod_{i=1}^{m}\left(1-\frac{1}{2^{|X_i|}}\right).\]
------
Finally, we show the problem for arbitrary fixed $|X_i|$. Let $N$ be a number larger than $\sum_{i}|X_i|$. Then, we have, for any choice of $X_i$,
\begin{align*}
\text{[\# subsets of $[n]$ not containing any $X_i$]} &= \frac{1}{2^{N-n}}\text{[\# subsets of $[N]$ not containing any $X_i$]}\\
&\geq \frac{1}{2^{N-n}}\cdot 2^N\cdot\prod_{i=1}^{m}\left(1-\frac{1}{2^{|X_i|}}\right)\\
& =2^n\cdot\prod_{i=1}^{m}\left(1-\frac{1}{2^{|X_i|}}\right)
\end{align*}
as requested. $\blacksquare$

Ha! This will be a good example to illustrate a basic spirit of probablistic method - making counting simpler!


number theory
least common multiple
algebra
polynomial
IMO
IMO 1981

[b]a.)[/b] For which $n>2$ is there a set of $n$ consecutive positive integers such that the largest number in the set is a divisor of the least common multiple of the remaining $n-1$ numbers?

[b]b.)[/b] For which $n>2$ is there exactly one set having this property?

[hide]Let $k = \prod p_{i}^{e_{i}}$ be the largest element of the set.  Then $k$ divides the least common multiple of the other elements of the set iff the set has cardinality of at least $\max \{ p_{i}^{e_{i}}\}+1$, since for any of the $p_{i}^{e_{i}}$, we must go down at least to $k-p_{i}^{e_{i}}$ to obtain another multiple of $p_{i}^{e_{i}}$.  In particular, there is no set of cardinality $3$ satisfying our conditions, because each number larger than or equal to $3$ must be divisible by a number that is larger than two and is a power of a prime.

For $n > 3$, we may let $k = \operatorname{lcm}[n-1, n-2] = (n-1)(n-2)$, since all the $p_{i}^{e_{i}}$ must clearly be less than $n$ and this product must also be larger than $ n$ if $n$ is at least $4$.  For $n > 4$, we may also let $k = \operatorname{lcm}[n-2, n-3] = (n-2)(n-3)$, for the same reasons.  However, for $ n = 4$, this does not work, and indeed no set works other than $\{ 3,4,5,6 \}$.  To prove this, we simply note that for any integer not equal to $6$ and larger than $4$ must have some power-of-a-prime factor larger than $3$.

Q.E.D.[/hide]

Let $A: (a, a+1, a+2, \cdots, a+n-1)$ such that $a, n \in \mathbb{Z}^{+}.$ In other words, this is our desired set.

[b]Lemma: $\exists A$ such that $A:{a, a+1, a+2, \cdots, a+n-1}$ where $a, n \in \mathbb{Z}^{+}$ and $a+n-1|lcm(a, a+1, \cdots, a+n-2)$ if and only if $(n-1)! | a+n-1.$[/b]

If we take the polynomial $P(a) = a(a+1)(a+2)\cdots(a+n-2)$ and find the remainder when divided by $a+n-1,$ we simply find $\frac{a+n-1}{P(-n+1)}.$ If this remainder is equal to zero, then we can solve for $a.$ If $a$ is an acceptable value, we can then construct an appropriate set $A.$ Note that

$P(-n+1) = (-n+1)(-n+2)(-n+3)\cdots(2)(1) = (-1)^{n-1}(n-1)!$
$\Leftrightarrow \frac{a+n-1}{P(-n+1)}= \frac{a+n-1}{(-1)^{n-1}(n-1)!}$

Thus, if $(n-1)!|a+n-1,$ we have that there exists a valid set $A$ with cardinality $n.$ $\blacksquare$

We will now commence proving parts A and B; this lemma will be used in both parts.

[hide="Part A"]
[b]Claim: For $\all n \ge 4,$ there exists such a set $A.$[/b]
[b]Proof:[/b]

If $a = (n-1)!-(n-1)$ and $a > 0,$ or

$(n-1)!-(n-1) > 0 \Leftrightarrow (n-1)! > (n-1) \Leftrightarrow n \ge 4,$

there exists such a sequence, according to our lemma. Such a sequence would be: $A: ((n-1)!-n+1, (n-1)!-n+2, (n-1)!-n+3, \cdots, (n-1)!+1).$

Thus, we can say that a valid sequence exists for $\all n \ge 4.$ $\blacksquare$

[/hide]
[hide="Part B"]
[b]Claim: Only for $n = 4$ does a unique $A$ exist.[/b]
[b]Proof:[/b]

According to our lemma, we have that $(n-1)! | a+n-1.$ The largest such $a$ that satisfies this is $(n-1)!-(n-1).$ The next largest that satisfies this is $(n-2)!-(n-1).$ However, since we only want one solution, we let $(n-2)!-(n-1) \le 0 \Leftrightarrow n \le 4.$

Since $2 < n \le 4$ and $n = 3$ doesn't yield any valid $A,$ the only such $n$ is $n = 4.$ The unique sequence $A$ for $n=4$ is $A: (3, 4, 5, 6).$ $\blacksquare$

[/hide]

Could someone grade me out of seven points and give me style points as well? I would really appreciate it  :)

Basically for a set to satisfy the property, the largest number must not have any prime powers that are not divisible by any of the other $n-1$ numbers.

For a given $n$, let $k_n$ be the largest element.

The answer for part $a)$ is $\boxed{n\ge 4}$.

Part A1: Show that there is no such set for $n=3$.
Suppose there was. Then the largest element must not have a prime power that is not in the first two elements. So all of it's prime factors are either divisible by the first or second element, which implies they must all be $2$. But $\nu_2 (k_n)>\nu_2 (k_n-2)$ in this case, a contradiction. $\blacksquare$

Part A2: Show that there always exists a set for $n\ge 4$.
We let $k_n=\prod_{i=1}^{x} p_i^{a_i}$ for nonnegative integers $a_i$ and $p_i$ is the $i$th prime. We let $p_i^{a_i}<n$ for all $i$. It's clear that this works because any $p_i^{a_i}$ divides some other element as $p_i^{a_i}<n$. Such a product of prime powers  exists for $n\ge 4$ because the product of all distinct primes less than $n$ is always greater than $n$ (not that this is not the case for $n=3$). $\blacksquare$

The answer for part $b)$ is $\boxed{n=4}$.

Part B1: Show that $\{3,4,5,6\}$ is the unique set for $n=4$.
Again let $k_4=\prod_{i=1}^{x} p_i^{a_i}$. Clearly all the $p_i$ are either $2$ or $3$. The highest possible power of $2$ is $1$ and the highest possible power of $3$ is also $1$. But we also need $k_4\ge 4$, for the set to contain only positive integers. So we must have $a_1=a_2=1$. Thus, $6$ is the only possible value of $k_4$. $\blacksquare$

Part B2: Show that there exist more than $1$ set for $n>4$.
Let $k_n=\prod_{i=1}^{x} p_i^{a_i}$. Let $p_x$ be the largest prime less than $n$. Then set all $a_i$ to $1$. This works as $k_n=\prod_{i=1}^x p_i>n$. Now since $n>4$, we can also set $a_1=2$, which also works. $\blacksquare$


It's true by Muirhead ($(6,3,0) \succeq (5,2,2)$) or more simply by rearrangement ?

I'm wondering somewhat about that. Can Muirhead be used with non-symmetric sum?

Indeed you're right, Muirhead can't be used here...


determine if $ n!\minus{}1$ is composite for infinitely many $ n$'s, where $ n$ is a positive integer

$ (p \minus{} 2)! \equiv 1 \mod{ p }$.

Is it an open question if $ n! \minus{} 1$ or $ n! \plus{} 1$ is prime for infinitely many $ n$'s?