[quote="mathVNpro"]Let $ f,g$ be continuous: $ [0,1]\longrightarrow [0,1]$ such that:
$ f(g(x)) = g(f(x))$, for all $ x\in [0,1]$. Suppose $ f$ is strictly increase.
Prove that: There exists $ a\in [0,1]$ such that $ f(a) = g(a) = a$.[/quote]

Nice problem !

Since $ g(0)\geq 0$ and $ g(1)\leq 1$, and $ g(x)$ continuous $ \exists u\in[0,1]$ such that $ g(u)=u$

$ g(f(u))=f(g(u))=f(u)$ and so $ f(u)$ is also a fixed point of $ g(x)$

Let then the sequence $ a_0=u$ and ${ a_{n+1}=f(a_n})$ $ \forall n\geq 0$. We have $ g(a_n)=a_n$ $ \forall n\geq 0$
If $ a_1\geq a_0$, $ a_n$ is a non decreasing sequence (since $ f(x)$ is strictly increasing).
If $ a_1\leq a_0$, $ a_n$ is a non increasing sequence (since $ f(x)$ is strictly increasing).

In either case :
Either $ \exists p$ such that $ a_{p+1}=a_p$, so $ f(a_p)=a_p=g(a_p)$ Q.E.D
Either such $ p$ does not exist and so $ a_n$ has a limit $ L$ in $ [0,1]$. Then, since $ g(a_n)=a_n$, $ f(a_n)=a_{n+1}$ and $ f(x)$ and $ g(x)$ are continuous, we have $ f(L)=L=g(L)$ Q.E.D.

[quote="pco"][quote="mathVNpro"]Let $ f,g$ be continuous: $ [0,1]\longrightarrow [0,1]$ such that:
$ f(g(x)) = g(f(x))$, for all $ x\in [0,1]$. Suppose $ f$ is strictly increase.
Prove that: There exists $ a\in [0,1]$ such that $ f(a) = g(a) = a$.[/quote]

Nice problem !

Since $ g(0)\geq 0$ and $ g(1)\leq 1$, and $ g(x)$ continuous $ \exists u\in[0,1]$ such that $ g(u) = u$

$ g(f(u)) = f(g(u)) = f(u)$ and so $ f(u)$ is also a fixed point of $ g(x)$

Let then the sequence $ a_0 = u$ and ${ a_{n + 1} = f(a_n})$ $ \forall n\geq 0$. We have $ g(a_n) = a_n$ $ \forall n\geq 0$
If $ a_1\geq a_0$, $ a_n$ is a non decreasing sequence (since $ f(x)$ is strictly increasing).
If $ a_1\leq a_0$, $ a_n$ is a non increasing sequence (since $ f(x)$ is strictly increasing).

In either case :
Either $ \exists p$ such that $ a_{p + 1} = a_p$, so $ f(a_p) = a_p = g(a_p)$ Q.E.D
Either such $ p$ does not exist and so $ a_n$ has a limit $ L$ in $ [0,1]$. Then, since $ g(a_n) = a_n$, $ f(a_n) = a_{n + 1}$ and $ f(x)$ and $ g(x)$ are continuous, we have $ f(L) = L = g(L)$ Q.E.D.[/quote]

Nice solution [b]pco[/b], much shorter than mine. [b]Here is mine approach[/b]:
Denote by $ h(x) = g(x) - x$. Therefore, $ h(x)$ will be continuos on $ [0,1]$ and $ h(0) = g(0)\ge 0$;
$ h(1) = g(1) - 1\le 0 \Longrightarrow$ [b]Exist[/b] $ x_0\in [0,1]$ such that $ h(x_0) = 0\Longrightarrow g(x_0) = x_0$.
[u][b]Case $ 1$[/b][/u]: $ f(x_0) = x_0$. Put $ a = x_0 \Longrightarrow f(a) = g(a) = a$ (Done!)
[u][b]Case $ 2$[/b][/u]: $ f(x_0)$ is different from $ x_0$.
Consider a sequence $ \{x_n\}$: $ x_1 = f(x_0)$; $ x_{n + 1} = f(x_n)$, $ n\ge1$.
If $ f(x_0) < x_0$ then $ \{x_n\}$ strictly increases, otherwise, $ \{x_n\}$ strictly decrease.
It is so obvious that $ x_n\in [0,1]$, hence in both probabilities, $ \lim\limits_{n\rightarrow \infty}x_n = a\in [0,1]\Longrightarrow$
${ f(a) = f(\lim\limits_{n\rightarrow \infty}x_n}) = \lim\limits_{n\rightarrow \infty}f(x_n) = \lim x_{n + 1} = \lim x_n = a$, $ (1)$.
In the other hand, $ g(x_n) = g(f(x_{n - 1})) = g(f(x_{n - 1}))$
$ \Longrightarrow g(x_1) = f(g(x_0)) = f(x_0) = x_1\Longrightarrow....\Longrightarrow g(x_n) = f(g(x_{n - 1}) = f(x_{n - 1}) = x_n$.
$ \Longrightarrow g(a) = g(\lim x_n) = \lim g(x_n) = \lim x_n = a$, $ (2)$.
Combine $ (1),(2)$, we get $ f(a) = g(a) = a$.
Our proof is completed. :lol:

[quote="mathVNpro"] Nice solution [b]pco[/b], much shorter than mine. [/quote]

Thanks. But our two solutions are exactly the same, even though i wrote mine in a shorter way. :)

[quote="mathVNpro"]Let $ f,g$ be continuous: $ [0,1]\longrightarrow [0,1]$ such that:
$ f(g(x))\equal{}g(f(x))$, for all $ x\in [0,1]$. Suppose $ f$ is strictly increase.
Prove that: There exists $ a\in [0,1]$ such that $ f(a)\equal{}g(a)\equal{}a$.[/quote]
I found a problem without condition "$ f$ is strictly increase". Please tell me how I can solve it.


vector
algebra
polynomial
linear algebra
linear algebra unsolved

Let $ T$ be a linear operator on the finite-dimensional vector space $ V$. Suppose $ T$ has a cyclic vector. Prove that if $ U$ is any linear operator which commutes with $ T$, then $ U$ is a polynomial in $ T$.

Since $ T$ has a cyclic vector, we know the minimal and characterstic polynomials for $ T$ are the same, but I don't see how that helps.

Any hints? Thanks in advance.

If $ T$ has a cyclic vector $ v$, show that the behavior of $ U$ is determined by the vectors $ v, Uv, U^2 v, U^3 v, ...$.  Then think about how you would write these vectors in terms of $ v, Tv, T^2 v, T^3 v, ...$.

I think I got it. I'm not sure about your statement "the behavior of $ U$ is determined by the vectors $ v,Uv,U^2v,U^3v,\ldots$." I think the behavior of $ U$ is determined by just knowing $ Uv$. How's this for a proof of the problem:

Let $ n\equal{}\text{dim } V$ and $ v$ be a cyclic vector for $ T$. Then $ v,Tv,\ldots,T^{n\minus{}1}v$ span $ V$. First consider the space of all polynomials in $ T$. These operators all commute with $ T$ and the space is $ n$-dimensional (because the minimal polynomial for $ T$ has degree $ n$.) Now consider the space of all operators that commute with $ T$. This space contains the previous one as a subspace. We prove that these two spaces are the same by proving that this space is also $ n$-dimensional.

Let $ U$ be an operator that commutes with $ T$. Since $ v,Tv,\ldots,T^{n\minus{}1}v$ span $ V$, the behavior of $ U$ is determined by the vectors $ Uv,UTv,\ldots,UT^{n\minus{}1}v$. Since $ U$ commutes with $ T$ these vectors are the same as $ Uv,TUv,\ldots,T^{n\minus{}1}Uv$. Thus knowing the vector $ Uv$ determines $ U$. Since knowing one vector determines $ U$ and our vector space is $ n$-dimensional, the space in question is also $ n$-dimensional and we're done.

[quote="BugRug"]I think the behavior of $ U$ is determined by just knowing $ Uv$.[/quote]
Well, yes, since this determines the vectors $ U^2 v, ...$.  Anyway, yes, that's essentially what I was thinking of.

See also [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=241005]this thread.[/url]


Put dollar signs $ \$$ instead of [tex] at the beginning and at the end of the equation.
Then the equation will become $ x^4\plus{}1\equal{}x\plus{}3$. And between, why didn't you write $ x^4\equal{}x\plus{}2$?^^

Hi vitoriogauss.

$ x^{4}\plus{}1\equal{}x\plus{}3$

$ x^{4} \minus{} x \minus{} 2\equal{}0$

$ f(x) \equal{} x^{4}\minus{}x\minus{}2 \Rightarrow\, f(\minus{}1) \equal{} 0$

Dividing $ f(x)$ by $ (x\plus{}1)$,

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

Now $ g(x) \equal{} x^{3}\minus{}x^{2}\plus{}x\minus{}2\Rightarrow\, g'(x)\equal{} 3x^{2}\minus{}2x\plus{}1 \equal{} 3(x\minus{}\frac{1}{3})^{2} \plus{} \frac{2}{3}>0$

so there so only one other real root. (We could also draw the graph.)

$ g(1) \equal{} \minus{}1 <0$ and $ g(2) \equal{} 4 >0$

So this root is between $ x\equal{}1$ and $ x\equal{}2$.

$ g(1.3) \equal{} \minus{}0.193..., g(1.4) \equal{} 0.184...., g(1.35) \equal{} \minus{}0.012125$

We could keep refining this or use Cardan's formula (which I can't remember but can be found on the internet) to get the exact root or visit the website Wolfram alpha (which doesn't work on my computer (it says done but with errors on page and displays a blank page) but I had it working before).

I hope this helps you.


Ten pirates have got their hands on a hoard of 100 gold pieces, and wish to divide the loot between them. They are democratic pirates, in their own way, and divide as follows. The fiercest pirate makes a proposal about the division, and everybody votes - one vote each including the proposer. If 50% or more are in favour, the proposal passes and is implemented forthwith. Otherwise the proposer is thrown overboard and the procedure is repeated with the next fiercest pirate. Which proposal will maximize the fiercest pirate's gain?

The fiercest pirate claims 96 gold pieces, offering one each to the 3rd, 5th, 7th, and 9th fiercest. Numbers 2, 4, 6, 8, and 10 will vote against this proposal, but numbers 3, 5, 7, and 9, who would receive nothing from #2 if they threw #1 overboard, will support it, giving the plan the 50% that it neads.

The argument works by induction up from the bottom. If it ever came down to just #9 and #10, #9 would claim all 100. Thus, if it were #8, 9, and 10, #8 could offer one piece to #10 to get him to vote for a plan to give 99 pieces to #8. And so on.

Very good!!! It's correct. Have you seen this type of puzzle before?


geometry
rectangle
circumcircle
incenter
geometry unsolved

For any triangle $\triangle XYZ$, let P, Q, R be the midpoints of the circumcircle arcs YZ, ZX, XY opposite to the vertices X, Y, Z. The incenter I is the intersection of the circles (P), (Q), (R) with radii $r_P = PY = PZ, r_Q = QZ = QX, r_R = RX = RY$.

The quadrilateral $A_1A_2A_3A_4$ is cyclic and $O_1, O_2, O_3, O_4$ are the midpoints of the arcs $A_4O_1A_1, A_1O_2A_2, A_2O_3A_3, A_3O_4A_4$. Hence, $B_1 \equiv S_1 \cap S_2, B_2 \equiv S_2 \cap S_3, B_3 \equiv S_3 \cap S_4, B_4 \equiv S_4 \cap S_1$ different from $A_1, A_2, A_3, A_4$ are the incenters of the overlapping triangles $\triangle A_1A_2A_4, \triangle A_2A_3A_1, \triangle A_3A_4A_2, \triangle A_4A_1A_3$. It is well known that these 4 incenters form a rectangle in a cyclic quadrilateral.


inequalities
symmetry
inequalities proposed

Let $x_1,x_2,...,x_n$ $(n \geq 3)$ be non-negative numbers. If $m\in\{3,4,...,n\}$, then
$\frac{m-1}{2}\sum x_1^3+\frac 3{m-2}\sum x_1x_2x_3 \geq \sum x_1x_2(x_1+x_2)$,
with equality if $m$ or $m-1$ of $x_1,x_2,...,x_n$ are equal and the others are zero.

[b]Remark1.[/b] The inequality is equivalent to
$(\sum x_1)^3+\frac{3m}{m-2}\sum x_1x_2x_3 \geq \frac{3m-1}{m-1}(\sum x_1)(\sum x_1x_2)$.
[b]Remark2.[/b] These equivalent inequalities are valid for $m\in\{3,4,...,n\}$, but are not valid if
$m\in (3,n)$ is not integer. ;)

if my calculation does not have any mistake , the question is equivalent to prove

$\sum x_1(mx_1-1)((m-1)x_1-1)\geq 0$  for $\sum x_1=1$  :)

[and from this form , we can see why the equality holds when $m$ or $m-1$ of the term is equal but other is $0$ .]

To complete the prove , let $f(x)=x(mx-1)((m-1)x-1)$ and $f''(x)=6m(m-1)x-2(2m-1)$ .By using half concave , half convex theorem [url]http://www.artofproblemsolving.com/Forum/viewtopic.php?t=64933[/url]

the expression is minimal when $k$ term is $\frac{1}{k}$ while $n-k$ term is $0$ . (also $k\leq \frac{3m(m-1)}{2m-1}$ must hold )

So we have the minimal is $k\cdot\frac{1}{k}\frac{(m-k)((m-1)-k}{k^2}=\frac{(m-k)((m-1)-k)}{k^2}$

Since $m,k$ is integer , the minimal is when $k=m-1$ or $k=m$ .

 :)

[quote="Lovasz"]Excuse me, the sum symbol $\sum$ is cyclic sum or symmetric sum?[/quote]

I believe it is symmetry sum  :)

[quote="Vasc"]Excellent approch, Shyong. :)

[quote="shyong"] To complete the prove , let $f(x)=x(mx-1)((m-1)x-1)$ and $f''(x)=6m(m-1)x-2(2m-1)$ .By using half concave , half convex theorem [url]http://www.artofproblemsolving.com/Forum/viewtopic.php?t=64933[/url]
the expression is minimal when $k$ term is $\frac{1}{k}$ while $n-k$ term is $0$ . [/quote] Why?
We can have also $n-k-1$ terms equal to 0, one term $x\in (0,\frac{2m-1}{3m^2-3m})$ and k terms equal to $\frac{1-x}{k}$, where $1 \leq k \leq n-1$. We still must prove that $f$ is not minimal for this case.[/quote]

ok , let us "really" complete it :P

in this cases , we have the minimum be

$\frac{(1-x)(m(1-x)-k)((m-1)(1-x)-k)}{k^2}+x(1-mx)(1-(m-1)x$

we see that $(m-1)(1-x)\in (m-1+1/3m-2/3,m-1)$ and $m(1-x)\in (m-1+1/3+1/(3m-3),m)$

if we pick $k<m-1$ or $k>m-1$ the front part of the expression is always positive .So to minimize , we pick $k=m-1$ hence the expression become

$\frac{-(1-x)(1-mx)(m-1)x}{(m-1)^2}+x(1-mx)(1-(m-1)x)=\frac{x(m-2)(1-mx)^2}{m-1}>0$
so this case still doesnt give us the minimum $0$ .  :)

Anything left out?


Hey guys!

Cobb's this Saturday!
On Halloween! :spider:

*mad props for finally using that smiley semi-legitimately*

Who's going? :)

I believe that the competition starts around 9:00.

So how'd everyone do?  I couldn't go yesterday cuz I had a quiz bowl competition.

JV:
1. Chamblee 890
2. Walton 847
3. Northview 734
4. Atlanta International 699
5. Wheeler 628

Varsity:
1. Walton 1035
2. Northview 829
3. Pace 721
4. Chamblee 689
5. Westminster 658
5. Milton - It's possible that a scoring error was made against Milton so they are 5th also. No way to confirm the error unfortunately.

Varsity Individual:
1. Claus! :-)
2. Abubakar
3. Edward
4. Allen
5. Oliver
6. Robert
7. Yubo
8. Nathan
9.  Michael Hopkins - Pace
10. Arvind

What about JV?  I heard Denny got 1st.

Congrats to everyone who won, btw.  Congrats especially to Claus.  You are following in the exalted footsteps of Heng!  lol, jk, but seriously, congrats!

EDIT: Anyone have the test questions?


Twelve teams participated in a soccer tournament. According to the rules, one team gets $ 2$ points for a victory, $ 1$ point for a draw and $ 0$ points for a defeat. When the tournament was over, all teams had distinct numbers of points, and the team ranked second had as many points as the teams ranked on the last five places in total. Who won the match between the fourth and the eighth place teams?

I'm not sure this is a complete solution as I assume that (i) each pair of teams plays exactly once, and (ii) there was a winner in the match between the 4th and 8th placed teams (i.e. no draw). Anyway, here goes:

First, some facts:
- There are a total of $\binom{12}{2} = 66$. Since each match contributes exactly 2 points, the total scores of all the teams is $66 \times 2 = 132$.
- The maximum possible score for any team is $11 \times 2 = 22$ (i.e. all wins). Hence, the maximum possible score for the 2nd placed team is $21$.

Divide the teams into 2 groups: Group A consisting of the 1st to 7th place teams and Group B consisting of the 8th-12th place teams (i.e. the 5th lowest ranked teams).
There are $\binom{5}{2} = 10$ games where both teams are from Group B. Hence, the sum of the scores for the teams in Group B is at least $10 \times 2 = 20$.
This means that the 2nd placed team must have scored either 20 or 21. This in turn implies that for matches where one team comes from Group A and one team comes from Group B, all of these matches ended up with the team from Group A winning except possibly for one game which ended in a draw.

Hence, assuming that there was a clear winner in the match between 4th & 8th place, the winner was the 4th placed team.

The assumption that there's a clear winner between 4th and 8th place teams is not necessary. If the second place scores 21 points, then the first place scores 22 points. Thus the first place defeats everyone, including the second place. But 21 points is only possible as 10 wins and 1 draw, so the second place cannot have lost, contradiction. So the second place scores 20 points; thus any match between Group A and Group B gives a win for the team in Group A.


algebra
polynomial
superior algebra
superior algebra unsolved

Show that $n \mid [K(a^\frac{1}{n}): K]$.

If $\mu_{2n}$ denotes the $2n$-th roots of unity, show that if $a \in K$ is a $2n$-th power in $K(\mu_{2n})$, it is a $n$-th power in $K$.

If $K(\mu_{2n})/K$ is cyclic, show that $a$ is a $2n$-th power in $K$.

[quote="steven128"]Show that $n \mid [K(a^\frac{1}{n}): K]$.[/quote]

this does not hold in general (e.g. $K = \mathbb{Q}, a= 4, n = 2$, then $K(a^\frac{1}{n})=K$). or what do you mean by $K(a^\frac{1}{n})$?

the case $a=0$ is uninteresting, let's assume $a \neq 0$. thoughout, $a^{1/n}$ means a fixed root of $X^{n}-a$ in $\overline{K}$.

first, let $p$ be a prime. it is well-known that $X^{p}-a$ is irreducible over $K$ or splits as $(X-a^{1/p})^{p}$ over $K$. hence $[K(a^{1/p}) : K]$ is $1$ or $p$.

if $n=p^{m}$ is a prime power, we have $[K(a^{1/p}): K] = [K(a^{1/p}): K(a^{1/p^{m}})] \cdot [K(a^{1/p^{m}}): K]$. the left hand side divides $p$, hence also the factors on the right hand side. especially we get $[K(a^{1/p^{m}}): K] \mid p \mid p^{m}$.

not let $m,n$ be coprime integers with the property $[K(a^{1/m}): K] \mid m, [K(a^{1/n}): K] \mid n$. then the same holds for their product $m \cdot n$. to see this, choose $u,v \in \mathbb{Z}$ s.t. $um+vn = 1$. then $a^{1/(m \cdot n)}= (a^{1/n})^{u}(a^{1/m})^{v}$. besides, we have $(a^{1/(m \cdot n)})^{m}= a^{1/n}$ and similary for $a^{1/m}$. thus, $K(a^{1/(m \cdot n)}) = K(a^{1/m}, a^{1/n})$. now $K(a^{1/m}$ and $K(a^{1/n})$ are extensions of [i]coprime[/i] of degree over $K$; an easy application of the degree formula yields $[K(a^{1/m}, a^{1/n}) : K] = [K(a^{1/m}): K] \cdot [K(a^{1/n}): K] \mid m \cdot n$.

combining these result, we get $[K(a^{1/n}): K] \mid n$ for every integer $n$.

i think some hypothesis is missing...
take $K=\mathbb{Q}$, $a=1$, $n=3$, $\omega = e^{2\pi i/3}= a^{1/n}$.
$K(a^{1/n})/K$ has degree $2$...
maybe something like "if $a$ has an n-th root in $K$, then $a^{1/n}\in K$"?

anyway, -oo-, i don't understand your proof..
first of all, as my example two lines ago shows, it's not true that either $x^{p}-a$ is irreducible, or it splits as $(x-a^{1/p})^{p}$...
then, i think that $K \subset K(a^{1/p}) \subset K(a^{1/p^{m}})$, so your use of the multiplicativity of degrees is wrong...

i'd try it another way: call $d$ the biggest divisor of $n$ such that $a$ is a $d$-th power in $K$. so actually $K(a^{1/n}) = K(b^{d/n}) = K(b^{1/n'})$, where $n'|n$. and then i'd try to show that its minimal polynomial is $x^{n'}-b$... but here i got a bit stuck..
i'll think about it...

somehow  I assumed $char(K)=p$ ;). but $p$ varies, so the proof is wrong :-)


ok, that would be huge  :maybe:.
Something like:
$2^{6}*3^{4}*5^{2}*${the other primes lower than 100 with the highest exponent possible}

Basically it is the result of the multiplication of the primes lower than 100 with the greatest possible exponent possible.

note that the (highest exponent possible) would be 1 for primes greater than 10, since 100=$10^{2}$...

So, it would be $2^{6}\times3^{4}\times5^{2}\times7^{2}\times\text{(all other primes less than 100)}$

There are only 25 primes less than 100  :roll:

so it would be

$2^{6}\cdot3^{4}\cdot5^{2}\cdot7^{2}\cdot11\cdot13\cdot17\cdot19\cdot23\cdot29\cdot31\cdot37\cdot41\cdot43\cdot47\cdot53\cdot59\cdot61\cdot67\cdot71\cdot73\cdot79\cdot83\cdot89\cdot97$

 :D

What?  Aren't both of those the same thing (1-100)?

[quote="k8reindeer"][quote="physics"]Find the lcm of first 100 natural number.[/quote]

they are the same thing

I think it's the first 100 natural numbers, not all the natural numbers less than or equal to 100.[/quote]

[quote="jli"][quote="k8reindeer"][quote="physics"]Find the lcm of first 100 natural number.[/quote]

I think it's the first 100 natural numbers, not all the natural numbers less than or equal to 100.[/quote]they are the same thing[/quote]
o. sorry, got confused  :blush:


function
induction
limit
algebra unsolved
algebra

Let $f(x)$ is a continous function defined on $[0,1]$ satisfying the conditions
i) $f(0)=0; f(1)=1$
ii) $6f(/frac{2x+y}{3})=5f(x)+f(y)$ for all $x\geq y; x,y/in [0,1]$
Calculate $f(/frac{8}{23})$

[quote="nmt_2006"]Let $f(x)$ is a continous function defined on $[0,1]$ satisfying the conditions
i) $f(0)=0; f(1)=1$
ii) $6f(\frac{2x+y}{3})=5f(x)+f(y)$ for all $x\geq y; x,y\in [0,1]$
Calculate $f(\frac{8}{23})$[/quote]
$x=0,y=1$ give $f(\frac{1}{3})=\frac{1}{6}$, $x=1,y=0$ give $f(\frac{2}{3})=\frac{5}{6}$. By induction we can calculate $f(\frac{a}{3^{k}})$ for any integer $0\le a\le 3^{k}$.
But I think, that don't exist continiosly function.
From i,ii) we have $f(1-x)=1-f(x)$ and $f((1-\theta )x+\theta y)+f(\theta x+(1-\theta )y)=f(x)+f(y)$ Therefore, if f is continiosly $f(\frac{x+y}{2})=\frac{f(x)+f(y)}{2}\Longrightarrow f(x)=x$.
Therefore can'not be define $f(\frac{8}{23})$.

But rule ii) can only be applied when $x \geq y$.

Without verifying that such an $f$ necessarily exists, we can calculate what $f(\frac{8}{23})$ would have to be.  Letting

$x_{0}= 1$
$x_{1}= 0$
$x_{3k+2}= \frac{x_{3k+1}+2 x_{3k}}{3}$
$x_{3k+i}= \frac{2x_{3k+i-1}+x_{3k-2}}{3}$ for $i = 0, 1$,

we claim that $\lim_{n \to \infty}x_{n}= \frac{8}{23}$.  Let $x = \lim_{n \to \infty}x_{n}$.  Note that applying a linear transformation $T$ that takes $x_{0}, x_{1}$ to $x_{3}, x_{4}$, respectively, will preserve the above recurrence relations, so $T$ will take $x_{i}$ to $x_{i+3}$ for all $i$, so $T$ will take $x$ to $x$.  Note that $x_{3}$ = $\frac{4}{9}$ and $x_{4}$ = $\frac{8}{27}$, so if $T(x) = mx+b$,

$\frac{8}{27}= b$
$\frac{4}{9}= m+b$
$T(x) = \frac{4}{27}x+\frac{8}{27}= x$
$x = \frac{8}{23}$.

Letting $y_{n}= f(x_{n})$ and $y = f(x)$, by continuity we see that $y = \lim_{n \to \infty}y_{n}$.  By rule ii) we have

$y_{0}= 1$
$y_{1}= 0$
$y_{3k+2}= \frac{y_{3k+1}+5 y_{3k}}{6}$
$y_{3k+i}= \frac{5y_{3k+i-1}+y_{3k-2}}{6}$ for $i = 0, 1$,

So we find $y$ similarly to how we found $x$;  let S be a linear transformation that takes $y_{0}$ and $y_{1}$ to $y_{3}$ and $y_{4}$ respectively.  Since $y_{3}$ and $y_{4}$ are $\frac{25}{36}$ and $\frac{125}{216}$ respectively, we have

$S(y) = (\frac{25}{36}-\frac{125}{216}) y+\frac{125}{216}$
$\frac{191}{216}y = \frac{125}{216}$
$y = \frac{125}{191}$.

Let $a,b\in (0,1), c,d\in R$. There are infinetely many functions  $f(x), \ f(0)=c,f(1)=d$ (continium) satisfyed condition
(1) $f(ax+(1-a)y)=bf(x)+(1-b)f(y) .$
If $a\not =b$ there are not continiosly nonconstant functions satisfyed (1).
If $a=b$ exist only one continiosly function with define f(0),f(1): $f(x)=c+x(d-c)$.
It is easy to prove, because from (1) $f(a^{k}(1-a)^{m})=c+b^{k}(1-b)^{m}(d-c) \ \forall k,m\in Z$.
Therefore for all $y$ we have $x_{n}\to \frac{8}{23}$, suth that $f(x_{n})\to y.$  y can be $+\infty$ and $-\infty$.

Let $f(x)$ is a continous function defined on $[0,1]$ satisfying the conditions
i) $f(0)=0; f(1)=1$
ii) $6f(/frac{2x+y}{3})=5f(x)+f(y)$ for all $x\geq y; x,y/in [0,1]$
Calculate $f(\frac{8}{23})$


In a 30-year period, what is the maximum possible number of
leap years?

We have that $ \frac{30}{4}\equal{}7.5$, which says that there are least $ 7$ leap years. However, if we let the first leap year in our count be the first or second year, then we get an extra leap year and so the maximum possible is $ \boxed{8}$.


Based on a problem from a Singapore Math book - about M level.

Ying-Ying has $\$10$ more than Ming-Ming. Ying-Ying bought 12 plates and spent all her money. Ming-Ming bought 4 plates and 8 cups at $\$0.65$ each and spent all her money. How much money did Ming-Ming have to start with?

Ying-Ying's money = $y$
Ming-Ming's money = $m$
[hide="Then,"]$y=m+10$
$m-8\cdot{0.65}=m-5.2$ <- money left after buying the cups
$\frac{y}{12}=\frac{m-5.2}{4}=$ <- money per plate
Cross multiply - $4y=12m-62.4$ then $y=3m-15.6$
Using substitution, $m+10=y=3m-15.6$ then $25.6=2m$ $m=12.8$
So Ming-Ming has $\$12.80$ and Ying-Ying has $\$22.80$[/hide]

[hide=" :lol: "]We only want how much money Ming-Ming had. So let that value be $\$x$. Ying-Ying has $\$x+10$. So:

$x+10=12p$
$x=4p+8c$

($p$ are plates and $c$ are cups)

We know that $c=\$0.65$, so:
$x=4p+8c=4p+5.2$

$x-5.2=4p$
$3x-15.6=12p$

Substituting 12p into the first equation,
$x+10=3x-15.6$
$2x=25.6$
$x=12.80$

So Ming-Ming originally had $\$12.80$ to begin with.[/hide]

This should go in Mathcounts section, it is a bit hard if you take it in one slice, but just substitute Ming-Ming's money for the other girl's and you easily get your answer.


conics
hyperbola
analytic geometry
algebra
difference of squares
special factorizations

A point whose coordinates are both integers is called a lattice point.  How many lattice points lie on the hyperbola $x^2-y^2=2000^2.$

[hide]This factors into $(x+y)(x-y)=2000^2=2^85^6$. The two factors $x+y$ and $x-y$ differ by $2y$, so have the same parity, so both are even.

Divide by 4, $\left( \frac{x+y}{2} \right) \left( \frac{x-y}{2} \right)=2^65^6$. The number of solutions where $\frac{x+y}{2}$ and $\frac{x-y}{2}$ are both positive are given by the number of positive divisors of $2^65^6$ which is $7^2=49$.

Since $\frac{x+y}{2}$ and $\frac{x-y}{2}$ could also both be negative, there are a total of $\boxed{98}$ points.[/hide]

[hide]By the difference of squares, $(x+y)(x-y)=2000^2=2^85^6$. Because $x+y$ and $x-y$ multiply to an even number and have the same parity, both are even.

Dividing by 4, $\left( \frac{x+y}{2} \right) \left( \frac{x-y}{2} \right)=2^65^6$. The number of positive solutions $\displaystyle \frac{x+y}{2}$ and $\frac{x-y}{2}$ are given by the number of positive divisors of $2^65^6$, which is $7^2=49$.

However, since both fractions could also be negative, we have twice as many lattice points for a total of $\boxed{098}$ points.[/hide]

[quote="vvr1590"]why couldn't you apply the same logic to

(x+y)(x-y) = 2000^2

and say that the number of solutions to that for x+y and x-y is the number of factors of 2000^2, which is 63?[/quote]
This is because $x+y=x-y+2y$, so both $x+y$ and $x-y$ must have the same parity (even or odd), which means that many of the factors of $2000^{2}$ don't actually work.

[quote="vvr1590"]aha, thanks. But how do we know that there's no such restriction on

1/2(x+y) and 1/2(x-y)

?[/quote]
Their difference is just $\frac{x+y}{2}-\frac{x-y}{2}=y$, which can be any integer, so there are no restrictions.

[quote="4everwise"][hide]By the difference of squares, $ (x \plus{} y)(x \minus{} y) \equal{} 2000^2 \equal{} 2^85^6$. Because $ x \plus{} y$ and $ x \minus{} y$ multiply to an even number and have the same parity, both are even.

Dividing by 4, $ \left( \frac {x \plus{} y}{2} \right) \left( \frac {x \minus{} y}{2} \right) \equal{} 2^65^6$. The number of positive solutions $ \displaystyle \frac {x \plus{} y}{2}$ and $ \frac {x \minus{} y}{2}$ are given by the number of positive divisors of $ 2^65^6$, which is $ 7^2 \equal{} 49$.

However, since both fractions could also be negative, we have twice as many lattice points for a total of $ \boxed{098}$ points.[/hide][/quote]

Hey, sorry for reviving an extremely old post, but why did you take out $ 2^2$ again?

[quote="mr.k_74"]Hey, sorry for reviving an extremely old post, but why did you take out $ 2^2$ again?[/quote]

$ x\plus{}y$ and $ x\minus{}y$ are both even, which places a restriction on them which we do not want. Dividing both by $ 2$ gets us the product of two integers with no restrictions, so we can claim the answer is just the number of divisors.

[quote=Kamran011][quote=NikoIsLife][quote=Kor_4origner]Umm.. I can't understand why you guys take out 2^2 in the solution...

Did you consider x+y/2 and x-y/2 as an positive integers?[/quote]

At first, we are asked to solve for the number of integer solutions $x$ and $y$ which satisfy $(x+y)(x-y)=2000^2$.

Since we already know that $x+y$ and $x-y$ are even integers, we can simply take out the $2^2$ to get:
$$2^2\left(\frac{x+y}2\right)\left(\frac{x-y}2\right)=2000^2\implies\left(\frac{x+y}2\right)\left(\frac{x-y}2\right)=1000^2$$
Now, since we don't have any restrictions on $\frac{x+y}2$ and $\frac{x-y}2$ anymore, this means that $\frac{x+y}2$ and $\frac{x-y}2$ can now be any positive integers we want! Because of this, we let $a=\frac{x+y}2$ and $b=\frac{x-y}2$.

The problem is now equivalent to counting the number of positive integer solutions $(a,b)$ in:
$$ab=1000^2$$
which makes things a lot easier.[/quote]

You are right but why is it necessary to divide the expression by 4?[/quote]
It's not exactly "necessary," but it only makes things slightly easier.

Because if we had to solve for the number of solutions $x$ and $y$ in
$$(x-y)(x+y)=2002^2$$
It would be slightly more complicated, because you would have to take into account that $x-y$ and $x+y$ are both even.

At first, you might think that "the answer is the number of even positive divisors $2002^2$ has," then that's wrong. Because its possible that $x-y$ is even, but $x+y$.

The reason why we divided by $4$ is so that we wouldn't have any restrictions on how to count the number of solutions anymore, which prevents things from being complicated.

This was kind of a scuffed problem for me…

$x^-y^2= (x+y)(x-y) = 2^8 \cdot 5^6.$ From inspecting the parities of $x+y,$ we see that both $x+y$ and $x-y$ are even. Therefore, $\frac{x+y}{2} \cdot \frac{x-y}{2} = 2^6 \cdot 5^6.$ Since $2^6 \cdot 5^6$ have $7 \cdot 7$ positive divisors and both $x$ and $y$ can be negative, the answer is $\boxed{098}.$


inequalities
rearrangement inequality
inequalities proposed

$ a\sqrt {ab} \plus{} b\sqrt {bc} \plus{} c\sqrt {ca}\le a\cdot \frac {a \plus{} b}{2} \plus{} b\cdot \frac {b \plus{} c}{2} \plus{} c\cdot \frac {c \plus{} a}{2}$

$ \equal{}\frac {1}{2}(a^2 \plus{} b^2 \plus{} c^2 \plus{} ab \plus{} bc \plus{} ca)\leq a^2 \plus{} b^2 \plus{} c^2\leq a^4 \plus{} b^4 \plus{} c^4 \plus{} \frac {3}{4}$

$ \Longleftrightarrow \left(a^2 \minus{} \frac {1}{2}\right)^2 \plus{} \left(b^2 \minus{} \frac {1}{2}\right)^2 \plus{} \left(c^2 \minus{} \frac {1}{2}\right)^2\geq 0$.

The equality holds when $ a \equal{} b \equal{} c \equal{} \frac {1}{\sqrt {2}}$.

[quote="alex2008"]If $ a,b,c > 0$ , show that :

$ a\sqrt {ab} \plus{} b\sqrt {bc} \plus{} c\sqrt {ca}\le a^4 \plus{} b^4 \plus{} c^4 \plus{} \frac {3}{4}$[/quote]
Obivously $ a^4 \plus{} b^4 \plus{} c^4 \ge a^3b \plus{} b^3c \plus{} c^3a$. So we just need:
$ a^3b \plus{} b^3c \plus{} c^3a \plus{} \frac {3}{4} \ge a\sqrt {ab} \plus{} b\sqrt {bc} \plus{} c\sqrt {ca}$
Let $ x \equal{} a\sqrt {ab}, y \equal{} b\sqrt {bc}, z \equal{} c\sqrt {ca}$.
Then we just have to prove:
$ x^2 \plus{} y^2 \plus{} z^2 \plus{} \frac {3}{4} \ge x \plus{} y \plus{} z$.
But $ x^2 \plus{} \frac {1}{4} \ge x$ is true by AM-GM, and doing the same for $ y,z$ we are done :) $ x^2 \plus{} \frac{1}{4} \equal{} \frac {2x^2 \plus{} \frac {1}{2}}{2} \ge \sqrt {(2x^2) \cdot \frac {1}{2}} \equal{} |x| \ge x$

kunny, I think that it's not so obvious, but follows from AM-GM applied from the following numbers:

$ a^4 , a^4, a^4, b^4$
$ b^4 , b^4, b^4, c^4$
$ c^4 , c^4, c^4, a^4$

So, we have summing all:

$ 4(a^4\plus{}b^4\plus{}c^4) \geq 4(a^3b\plus{}b^3c\plus{}c^3a)$

[quote="Mr.Bikrone"][quote="kunny"][quote="Mathias_DK"]
Obivously $ a^4 \plus{} b^4 \plus{} c^4 \ge a^3b \plus{} b^3c \plus{} c^3a$[/quote]

Why does the inequality hold Obviously?[/quote]
I think it was obviously proved by the Rearrangement Inequality[/quote]
Actually the AM GM that msecco gave is only an explanation of the rearrangement ineq, so it does not matter. :rotfl:

[quote="kunny"][quote="Mathias_DK"]
Obivously $ a^4 \plus{} b^4 \plus{} c^4 \ge a^3b \plus{} b^3c \plus{} c^3a$[/quote]

Why does the inequality hold Obviously?[/quote]
The sequences $ (a^3,b^3,c^3)$ and $ (a,b,c)$ are similarily sorted, and hence the result follows. (It can also easily be proved by AM-GM or even easier by Hölder)

Holder's solution is more nice:
$ (a^4\plus{}b^4\plus{}c^4)^3(b^4\plus{}c^4\plus{}c^4)\geq(a^3b\plus{}b^3c\plus{}c^3a)^4$
then ...


logarithms
integration
calculus
function
complex analysis
advanced fields
advanced fields unsolved

We [b]DEFINE[/b] logarithm over complex plane as follows:

\[\ln z = \int_1^z \frac{1}{z} dz\]

where the integral path does not cross the negative part of real axis, i.e., it does not cross ${x\in %Error. "mathh" is a bad command.
{R} | x\leq 0}$, thus $\ln z$ is defined uniquely.

Now consider

\[\frac{d}{dz} (\ln (e^z) - z) = \frac{1}{e^z} \cdot e^z - 1 = 0\]

thus

$\ln(e^z) - z$ is a constant, say $\ln(e^z) - z = c$

But put $z=0$ and $z=2\pi i$, we get $c=0$ and $c=-2\pi i$ respectively.

What's wrong?

The function ln(ez) is not holomorphic for z = k*pi*i ,  k<>0 ..
 :cool:

can you explain it in details? why isn't it holomorphic?

Let k be an integer  , k <> 0 .   {Ln(ex+y*i) - Ln(ex+k*pi*i)}/((y-k*pi)*i) = y/(y-k*pi)

Manifestly  this one doesnt  converge when y --> k*pi

 :cool:


inequalities
induction
quadratics
function
inequalities proposed

For a quadratic function $f(x)=x^2+ax+b,$ answer the following questions.

Let $x_1,\ x_2,\ \cdots\cdots,\ x_n$ be arbiterary real numbers. Prove the following inequality in terms of induction for $n\geq 2 .$

\[ f\left(\frac{1}{n}\sum_{k=1}^n x_k \right)\leq \frac{1}{n}\sum_{k=1}^n f(x_k). \]

this is the same as to prove jensen's inequality, since $f(x)$ is convex... so cauchy induction must work fine...
for $n=2$ it can easily be verified $(1)$ suppose it's true for $n$

from the inductive hypothesis and the inequality for $n=2$ we have that
$\frac{1}{n}\sum_{k=1}^n f(x_k)+\frac{1}{n}\sum_{k=n+1}^{2n} f(x_k)\geq f\left(\frac{1}{n}\sum_{k=1}^n x_k \right)+f\left(\frac{1}{n}\sum_{k=n+1}^{2n} x_k \right)\geq 2f\left(\frac{1}{2n}\sum_{k=1}^{2n} x_k \right)$
so if it's true for $n$ then it's true for $2n \ \ (2)$

now, take $x_n=\sum_{k=1}^{n-1} x_k$, and it can easily be shown that this implies that if it's true for $n$ then it's true for $n-1 \ \ \ (3)$.

the conclusion follows from $(1),(2)$, and $(3)$


Is there a way to generate cubic triples? (Like for Pythagorean triples:
$m^2-n^2,2mn,m^2+n^2$) I know solutions exist, like (1,6,8), but I just can't find a way to generate them. I'm stupid, I know...

Note: A cubic triple is a^3+b^3+c^3=d^3. :oops:

[quote="arqady"]$(3m^2+5mn-5n^2)^3+(4m^2-4mn+6n^2)^3+(5m^2-5mn-3n^2)^3=$
$=(6m^2-4mn+4n^2)^3.$ :)[/quote]

Are  $m$ and $n$ arbitrary integer values? Could you please post the whole solution?

In terms of only two variables?   :huh:  Can you show that this produces all primitive cubic quadruples?

[quote="jasamper88"] Could you please post the whole solution?[/quote]
[quote="t0rajir0u"]  Can you show that this produces all primitive cubic quadruples?[/quote]
I do not think, that it exists :D


In the set of $ n$ positive interger numbers $ S\equal{}\{1,2,...,n\}$. $ (a,b,c,d)$ is $ 4$ numbers which is taken from $ S$. $ (a,b,c,d)$ is call [i]good[/i] if we have $ a\plus{}b\equal{}c\plus{}d$. How many [i]good[/i] $ (a,b,c,d)$?

Let f(m) = the number of solutions of a+b=m with a,b in S.
Then you're after
$ \sum_{m\equal{}2}^{2n} f(m)^2.$

f(m) takes the values 1,2,3,...,n,...,3,2,1 on this sequence of m - might have this wrong :)

so you want

$ 2\sum_{j\equal{}1}^{n\minus{}1} j^2 \plus{} n^2 \equal{} (n\minus{}1)n(2n\minus{}1)/3 \plus{} n^2 \equal{} 2n^3/3 \plus{} n/3$
-evo

[quote="evoluciona"]Let f(m) = the number of solutions of a+b=m with a,b in S.
Then you're after
$ \sum_{m \equal{} 2}^{2n} f(m)^2.$

f(m) takes the values 1,2,3,...,n,...,3,2,1 on this sequence of m - might have this wrong :)

so you want

$ 2\sum_{j \equal{} 1}^{n \minus{} 1} j^2 \plus{} n^2 \equal{} (n \minus{} 1)n(2n \minus{} 1)/3 \plus{} n^2 \equal{} 2n^3/3 \plus{} n/3$
-evo[/quote]

Can you give a clearer explantion to your proof? It seems interesting!  :idea:

I have seen many solutions, in my book and in the forum , and those give different answers....

such as  n(n-2)(n-5)/24  ,etc.


A point is taken inside a $\triangle{ABC}$ and lines are drawn through the point from each vertex, thus dividing the triangle into $6$ parts. Four of the parts have the areas shown. Find $%Error. "area" is a bad command.
{ABC}$.

Let $x$ be the area of the triangle on the left side of the triangle with the area 84, and let $y$ be the area of the triangle on the right side of the triangle with the area 84.

Using area ratio, we can set up two equations:
$\frac{x+84}{y}=\frac{70}{35}\Longleftrightarrow x-2y=-84$
$\frac{x+84}{40}=\frac{y+35}{30}\Longleftrightarrow 3x-4y=-112$

Solving for $x$ and $y$, we obtain $x=56$ and $y=70$, so the area of $\triangle ABC$ is $56+84+70+40+30+35=\boxed{315}$

frt and shobber you are right.
 there is no general rule for this, well not that I know of, I do it without any :huh:

[quote="gotztahbeazn"]wow thats cool

but can anyone explain to me like the general rule of thumb for "area ratios" and stuff??

like what rule tells us we can just add up the areas like that and put them into ratios[/quote]

What do you know about the ratio of the areas of two triangles which share an altitude?

Read my post in the Geometry Formula and Theorems. (the link under my answer). Also, I used area ratios to prove Pappus theorem in that forum also. Area ratio is an extremely useful method in solving geometry problems.


Suppose that  $ D$ is the set of all even numbers from Natural.

If we define the set $ F(r) \equal{} \{y | y\in D \; , n \plus{} 1 \leq y \leq n \plus{} r \}$ , for some $ r\in \mathbb{N}$ and fixed $ n \in \mathbb{N}$.

Find the Supremum and limit of Supremum of $ F(r)$ as $ r$ goes to $ \infty$

 :D

:what?: Seems like a poorly concocted homework exercise aimed at checking understanding of the notions of the supremum and the limit to me. The sets in question are finite, so the supremum is the same as the maximum and the second part is about as trivial as the first if you know the definitions. Am I missing something? :?

Lol :rotfl:

next time I'll write everything using notations.

But I meant by the limit here is what is the value of limit of the sequence of $ F(r)$.

 :P  :D

[quote="Hidden Scofield"]
next time I'll write everything using notations.
[/quote]
It won't make the problem more interesting ;)
[quote]
But I meant by the limit here is what is the value of limit of the sequence of $ F(r)$.
[/quote]
Be careful: $ F(r)$ are sets, $ \sup F(r)$ are numbers. The problem as you posed it originally asks about the limit of the latter.


What is the number of $2n$-tuples $(a_1,...,a_{2n})$ satisfying the following conditions:
1) $\forall _{m \in [1,2n]} a_m \in \{-1,1\}$
2) $\forall _{m \in [1,2n]} \sum_{i=1}^m a_i \geq 0$
3) it is given $k<n$ such that there is subsequence of $a_1,...,a_{2n}$: $a_{i_1},...,a_{i_k}$ where all terms are equal to $-1$. (in other words, some $a_i$s are already chosen to be $-1$ - of course choose was made to not contradict the rest of the conditions).

How to interpret the 3rd condition?

3') $i_1, \dots, i_k$ are given, i.e., the sequence has -1's at $k$ specified positions

3'') $i_1, \dots, i_k$ are not given, i.e., it is known that the sequence has at least $k$ -1's but no specific positions are given.

I assume 3''.

The number of sequences $a_1, \dots, a_n$ with exactly $m$ -1's is $f(m+1,n-m)$ where $f$ is defined as in [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=359917#p359917]this thread[/url].

We know that $f(m+1,n-m) = F(m,n-m) = \binom{n}{m}\frac{n-2m+1}{n-m+1}$.
And to get the answer we need is to sum up $f(m+1,2n-m)$ over $m=k..n$.
\[ \sum_{m=k}^{n} f(m+1,2n-m) = \sum_{m=k}^{n}  \binom{2n}{m}\frac{2n-2m+1}{2n-m+1} = 2 \binom{2n-1}{n} - \frac{k}{2 n - k + 1}\binom{2n}{k}. \]


Happy new year . I wish positive energy to all people  :D

$2006+1$ wishes for a new, fun-filled year.

$3^{2}\cdot 223$  wishes for a happy new year  :lol:   .

Also I want to mention that we are the third biggest subforum here in mathlinks at posts and we are goind to be 2 nd .
I hope next year will be first.  :lol:
Go <<ελληνικά>>

[quote="silouan"]$3^{2}\cdot 223$  wishes for a happy new year  :lol:   .
[/quote]

Silouan,  :lol:

PS: Αν είχαμε άλλους 2 Σιλουανούς, θα ήμασταν πρώτοι με διαφορά!

HAPPY NEW YEAR!!!

DON't SPEND IT ON AOPS!!!

TO INFINITY AND BEYOND!! THE FUTURE!!!! :lol:

Best Wishes for a best 2007 with health and successes

AND A NEW CURE FOR AIDS!!!!

Think of all those poor people that have died from that EVIL DISEASE!  :(

HAPPY NEW YEAR!!!! AND A GREAT 2007!

They should use more money on the research for this

Happy Card{ :) , :idea: } Card{n>2, x^n+y^n=z^n, x,y,z,n Z} CardP({n>2, x^n+y^n=z^n, x,y,z,n Z}) Card{A,K,Q,J,10,9,8}


Είχα καιρό να προτείνω άσκηση και βάζω τώρα αυτή την οποία θεωρώ ωραία και έξυπνη .

Έστω μία ακολουθία $ a_n$ , $ n\geq 1$ θετικών όρων με $ a_n\rightarrow 0$ . Έστω τώρα ακολουθία $ \{ a_n'\}$ ώστε ο $ a_n'$ να ισούται είτε με $ a_n$ είτε με $ 0$ .
Αποδείξτε ότι το σύνολο των αθροισμάτων $ \sum_{n = 1}^{\infty} a_n'$ είναι υπεραριθμήσιμο .

[quote="silouan"]Είχα καιρό να προτείνω άσκηση και βάζω τώρα αυτή την οποία θεωρώ ωραία και έξυπνη.[/quote]
Όντως έξυπνη άσκηση.

[hide]
Δείξτε ότι υπάρχει υπακολουθία $ a_{i_1},a_{i_2},\ldots$ ώστε $ a_{i_n} \geq 3a_{i_{n \plus{} 1}}$ για κάθε $ n$.
[/hide]

Μου δημιουργήθηκε μια απορία. Μήπως θα ήταν πιο πονηρό να μας ζητήσεις να δείξουμε ότι υπάρχει ακολουθία $ \{a_n'\}$ ώστε $ \sum{a_n'}$  να είναι άρρητος; Ή μήπως βλέπει κανείς καμιά λύση αυτού του ερωτήματος που δεν χρησιμοποιεί αριθμησιμότητα;

Dem, είσαι σίγουρος για τη φορά της ανισότητας που προτείνεις? :maybe:

Cheerio,

Durandal 1707

[b]Edit[/b]: Όσο για το ερώτημα με τον άρρητο, δεν ξέρω αν θα σκεφτόμουν ποτέ να δείξω πρώτα υπεραριθμησιμότητα, nice thought ;) Τώρα, αν βγαίνει ανεξάρτητα... η λύση είναι προφανής αν υπάρχει ένας άρρητος, οπότε μπορούμε να θεωρήσουμε ότι η $ a_n$ αποτελείται μόνο από ρητούς. Σε αυτήν την περίπτωση, νομίζω μπορείς να φτιάξεις τον άρρητο ως εξής:
1. πάρε τον πρώτο όρο $ a'_1\equal{}a_{k_1} \equal{} \frac {m_1}{n_1} < \frac {1}{2}$.
2. πάρε τον δεύτερο όρο $ a'_2\equal{}a_{k_2}\equal{}\frac {m_2}{n_2} < \frac {1}{n_1}$ και $ n_2 > n_1$
3. $ a'_3\equal{}a_{k_3}\equal{}\frac {m_3}{n_3} < \frac {1}{n_2}$ και $ n_3 > n_2$
...
Έτσι, το άθροισμα φτιάχνει κάτι που να προσομοιάζει αρρήτους τύπου $ 0.101001000100001...$ σε δυαδική βάση. Βέβαια, πρέπει να προσέξεις ώστε να μην υπάρχουν επαναλαμβανόμενα ψηφία, αλλά νομίζω ότι αρκούν οι παραπάνω παρονομαστές... δεν είμαι και σίγουρος όμως γιατί δεν το έχω κοιτάξει πολύ αναλυτικά και μπορεί να χάνω κάτι (πιθανότατα :P).

[quote="Durandal"]Dem, είσαι σίγουρος για τη φορά της ανισότητας που προτείνεις? :maybe: [/quote]

Η φορά της ανισότητας είναι σωστή. Το 3 πήγε σε λάθος τόπο.  :oops: Θα το διορθώσω.

[quote="Durandal"]
Όσο για το ερώτημα με τον άρρητο, δεν ξέρω αν θα σκεφτόμουν ποτέ να δείξω πρώτα υπεραριθμησιμότητα, nice thought ;) Τώρα, αν βγαίνει ανεξάρτητα... η λύση είναι προφανής αν υπάρχει ένας άρρητος, οπότε μπορούμε να θεωρήσουμε ότι η $ a_n$ αποτελείται μόνο από ρητούς. Σε αυτήν την περίπτωση, νομίζω μπορείς να φτιάξεις τον άρρητο ως εξής:
1. πάρε τον πρώτο όρο $ a'_1 \equal{} a_{k_1} \equal{} \frac {m_1}{n_1} < \frac {1}{2}$.
2. πάρε τον δεύτερο όρο $ a'_2 \equal{} a_{k_2} \equal{} \frac {m_2}{n_2} < \frac {1}{n_1}$ και $ n_2 > n_1$
3. $ a'_3 \equal{} a_{k_3} \equal{} \frac {m_3}{n_3} < \frac {1}{n_2}$ και $ n_3 > n_2$
...
Έτσι, το άθροισμα φτιάχνει κάτι που να προσομοιάζει αρρήτους τύπου $ 0.101001000100001...$ σε δυαδική βάση. Βέβαια, πρέπει να προσέξεις ώστε να μην υπάρχουν επαναλαμβανόμενα ψηφία, αλλά νομίζω ότι αρκούν οι παραπάνω παρονομαστές... δεν είμαι και σίγουρος όμως γιατί δεν το έχω κοιτάξει πολύ αναλυτικά και μπορεί να χάνω κάτι (πιθανότατα :P).[/quote]

Πρέπει να προσέξεις όμως να συγκλίνει το άθροισμα. (Όπως το περιγράφεις ίσως σου βγει 1/2 + 1/3 + 1/4 + ...) Κάτι παρόμοιο όμως πρέπει να δουλεύει. Το ωραίο είναι πως με αριθμησιμότητα (φαίνεται να) είναι αρκετά πιο απλό.

[quote="Demetres"]Πρέπει να προσέξεις όμως να συγκλίνει το άθροισμα. (Όπως το περιγράφεις ίσως σου βγει 1/2 + 1/3 + 1/4 + ...) Κάτι παρόμοιο όμως πρέπει να δουλεύει. Το ωραίο είναι πως με αριθμησιμότητα (φαίνεται να) είναι αρκετά πιο απλό.[/quote]

Ναι, μάλλον θέλει δυνάμεις του $ 2$ σε κάθε βήμα και κάτι που να απαγορεύει τις επαναλήψεις ψηφίων... both seem easy enough, αλλά βαριέμαι να κάνω τις λεπτομέρειες ;)

Sil, ωραίο πρόβλημα :)

Cheerio,

Durandal 1707

Να βάλω και την λύση που είχα προτείνει συμπληρωμένη:

Παίρνουμε $ a_{i_1} \equal{} a_1$. Επειδή $ a_n \to 0$, μπορώ να βρώ $ i_2 > i_1$ ώστε $ a_{i_1} \geq 3a_{i_2}$ (η ανισότητα είναι σωστή αυτή τη φορά  :) ). Ομοίως βρίσκουμε $ i_1 < i_2 < i_3 < \ldots$  ώστε $ a_{i_n} \geq 3a_{i_{n\plus{}1}}$ για κάθε $ n$.

Τώρα, όλα τα αθροίσματα της μορφής $ \sum\limits_{n\equal{}1}^{\infty}{a'_{i_n}}$ είναι διαφορετικά άρα μπορούμε να τα θέσουμε σε 1-1 αντιστοιχία με το $ \{0,1\}^{\mathbb{N}}$ το οποίο είναι υπεραριθμήσιμο.


Let $ a,b,c>0$ and $ ab\plus{}bc\plus{}ca\equal{}abc$. Prove that: $ \frac{{\sqrt {b^2  \plus{} 2a^2 } }}
{{ab}} \plus{} \frac{{\sqrt {c^2  \plus{} 2b^2 } }}
{{bc}} \plus{} \frac{{\sqrt {a^2  \plus{} 2c^2 } }}
{{ca}} \geqslant \sqrt 3$

[quote="thanhnam2902"]Let $ a,b,c > 0$ and $ ab \plus{} bc \plus{} ca \equal{} abc$. Prove that: $ \frac {{\sqrt {b^2 \plus{} 2a^2 } }} {{ab}} \plus{} \frac {{\sqrt {c^2 \plus{} 2b^2 } }} {{bc}} \plus{} \frac {{\sqrt {a^2 \plus{} 2c^2 } }} {{ca}} \geqslant \sqrt 3$[/quote]

$ \frac{1}{a}\equal{} x$ , $ \frac{1}{b}\equal{} y$  , $ \frac{1}{c}\equal{} z$

$ \Leftrightarrow x\plus{}y\plus{}z\equal{}1$

$ \sqrt {x^2 \plus{} 2y^2 } \plus{} \sqrt {y^2 \plus{} 2z^2 } \plus{} \sqrt {z^2 \plus{} 2x^2 } \ge \frac{1}{\sqrt{3}}((x\plus{}2y)\plus{}(y\plus{}2z)\plus{}(z\plus{}2x))\equal{} \sqrt{3}$


function
integration
calculus
limit
complex analysis
algebra
functional equation

Why is
$\int_0^{\infty}t^xe^{-t}dt$ defined for $x\leq -1$? Shouldn't it diverge?

For $-1<x<0$ (or $-1<\text{Re }x<0$), the integral is improper at zero. That doesn't make it divergent- try a comparison to $t^{x}$.

For $\text{Re }x<-1$, we're in the realm of complex analysis, and the definition of the gamma function depends on the notion of analytic continuation.

[Edit- fixed a reading mistake]

It does diverge for $x \leq -1$ However $\Gamma (x)$ can be defined for all $x \neq 0, -1, -2, \ldots$ by analytic continuation.

[quote="0714446459923"]So instead of being defined by that integral, it is defined by $(x+1)\Gamma(x)=\Gamma(x+1)$ for $x\leq -1$?[/quote]

Close,  but it should be noted that the integral you gave originally is not $\Gamma (x)$, it is for $\Gamma (x+1)$.
The integral equation for $\Gamma (x)$ is $\int_0^{\infty} t^{x-1}e^t \ dt$, which is defined for $x>0$.

Your functional equation is also wrong. It should be $x\Gamma(x) = \Gamma (x+1)$.

The values for $x \leq 0$, $x \neq 0, -1, -2, \ldots$, can be definied by the equation $\Gamma(x-1)=\frac{\Gamma(x)}{x-1}$.

[quote="Beginner"]There is another form of the expresion of Gamma function which is ussually taking as definition in some calculus books that is
\[ \Gamma(x) = \lim_{n \rightarrow \infty} \frac{n^{x-1} n!}{(x)(x+1) \cdots (x+n-1)}  \]

The integral form can be obtained from this expresion.

Now you can see why it is not defined when $x=0, -1, -2, \cdots$.[/quote]

Your equation here is slightly incorrect. It should be $\Gamma(x) = \lim_{n \rightarrow \infty} \frac{n^{x} n!}{(x)(x+1) \cdots (x+n)}$. But your comment about why it is undefined for x = 0, -1, -2, etc. is still correct.

The confusion about notation is because of slightly non-intuitive definition: $n!$ is actually $\Gamma(n+1)$. Make sure you keep your indices straight!  ;)

I always wondered why $\Gamma(x)=(x-1)!$ Is there any particular reason for the -1?

[quote="Eric Weisstein"]It is related to the factorial by
$\Gamma(n)=(n-1)!$
a slightly unfortunate notation due to Legendre which is now universally used instead of Gauss's simpler $\Pi(n)=n!$ (Gauss 1812; Edwards 2001, p. 8). [/quote]

From [url=http://mathworld.wolfram.com/GammaFunction.html]Mathworld[/url].


quadratics
modular arithmetic
number theory unsolved
number theory

Prove that for all $ p\in\mathbb{P}$
exist $ x,y,z\in\mathbb{N}$,$ x^{2}\plus{}y^{2}\plus{}z^{2}\minus{}w.p\equal{}0$ for  $ 0<w<p$

[quote="cnyd"]Prove that for all $ p\in\mathbb{P}$
exist $ x,y,z\in\mathbb{N}$,$ x^{2} \plus{} y^{2} \plus{} z^{2} \minus{} w.p \equal{} 0$ for  $ 0 < w < p$[/quote]

If you use the common notation $ 0\notin \mathbb N$, this is wrong : let $ p\equal{}2$ and so $ w\equal{}1$ and it exists no positive integer $ x,y,z$ such that $ x^2\plus{}y^2\plus{}z^2\equal{}2$

[quote="cnyd"]Prove that for all $ p\in\mathbb{P}$
exist $ x,y,z\in\mathbb{N}_0$,$ x^{2} \plus{} y^{2} \plus{} z^{2} \minus{} w.p \equal{} 0$ for  $ 0 < w < p$[/quote]
If $ \minus{} 1$ is quadratic residue mod $ p$ then :
$ \exists$ $ a\in[1,\frac {p \minus{} 1}2]$ such that $ a^2 \equal{} \minus{} 1\pmod p$ and so $ 0^2 \plus{} 1^2 \plus{} a^2 \equal{} 0\pmod p$ and obviously $ 0 < a^2 \plus{} 1 < p^2$ and so $ 0 < w < p$

If $ \minus{} 1$ is not quadratic residue mod $ p$, then :
Let $ a \equal{} \min\{x\in\mathbb N$ such that $ x$ is not a quadratic residue mod $ p\}$.
$ a \minus{} 1$ and $ \minus{} a$ are then quadratic residue and we have $ a \minus{} 1 \equal{} b^2\pmod p$ and $ \minus{} a \equal{} c^2\pmod p$
Then $ 1 \plus{} b^2 \plus{} c^2 \equal{} 0\pmod p$ and, obviously, $ b,c\neq 0\pmod p$ and we can choose $ b,c\leq\frac {p \minus{} 1}2$ so that $ 0 < 1 \plus{} b^2 \plus{} c^2\leq 1 \plus{} 2\left(\frac {p \minus{} 1}2\right)^2 < p^2$ and so $ 0 < w < p$

it's almost like this:
for all $ n \in N$ there exist $ b,c \in N_0$ such that $ n | c^2\plus{}b^2\plus{}1$ if and only if $ 4 \nmid n$.
it's in PEN,A39.

[quote="cnyd"]why $ a \minus{} 1$ is quadratic residue for $ \minus{} 1$  is not  quadratic residue mod$ p$[/quote]

Because of the definition of $ a$ : Let $ a \equal{} \min\{x\in\mathbb N$ such that $ x$ is not a quadratic residue mod $ p\}$.

Since $ a$ is the minimum (and is obviously $ >1$), then $ a\minus{}1$ is a quadratic residue.

another version:$ p$ is odd prime
Let us consider sets: $ I_1\equal{}\{0,1,4,..., (\frac{p\minus{}1}{2})^2\}$ and $ I_2\equal{}\{\minus{}1\minus{}0,\minus{}1\minus{}1,\minus{}1\minus{}4,..., \minus{}1\minus{}(\frac{p\minus{}1}{2})^2\}$. Note that all remiders in $ I_1$ by modulo $ p$ are different and the same thing in $ I_2$.
Because $ |I_1|\plus{}|I_2|\equal{}\frac{p\plus{}1}{2}\plus{}\frac{p\plus{}1}{2}\equal{}p\plus{}1>p$ there must be elements with the same remainder, and we have that exists such $ 0\le a,b\le\frac{p}{2}$, that:$ a^2\equiv \minus{}1\minus{}b^2(\mod{p})$ and finally:$ p|a^2\plus{}b^2\plus{}1\le \frac{p^2}{2}$

[quote="greenvert"]another version:$ p$ is odd prime
Let us consider sets: $ I_1 \equal{} \{0,1,4,..., (\frac {p \minus{} 1}{2})^2\}$ and $ I_2 \equal{} \{ \minus{} 1 \minus{} 0, \minus{} 1 \minus{} 1, \minus{} 1 \minus{} 4,..., \minus{} 1 \minus{} (\frac {p \minus{} 1}{2})^2\}$. Note that all remiders in $ I_1$ by modulo $ p$ are different and the same thing in $ I_2$.
Because $ |I_1| \plus{} |I_2| \equal{} \frac {p \plus{} 1}{2} \plus{} \frac {p \plus{} 1}{2} \equal{} p \plus{} 1 > p$ there must be elements with the same remainder, and we have that exists such $ 0\le a,b\le\frac {p}{2}$, that:$ a^2\equiv \minus{} 1 \minus{} b^2(\mod{p})$ and finally:$ p|a^2 \plus{} b^2 \plus{} 1\le \frac {p^2}{2}$[/quote]

Quite nice and OK. You just have to add "$ 1\plus{}1\plus{}0\equal{}2$" in order to solve for $ p\equal{}2$, and it's finished.


induction
real analysis
real analysis unsolved

f:[0,1]->R, and f is $ C^m [0,1]$. Moreover, $ f^{(k)}(x) \neq 0$ for any $ k \le m$
prove that $ m! (max|f|) \ge min|f^{(m)}|$

You need to adjust that constant. Consider $ f(x)\equal{}(x\minus{}1/2)^m$. Or, if $ m$ is even, $ f(x)\equal{}(x\minus{}1/2)^m\minus{}1/2^{m\plus{}1}$.

sorry for the mistake.
I added something and I think it is right now.

[quote="Skyward_Sea"]No one could solve this?[/quote]
Er, in fact it's too naive...

By induction, $ |f'(0) \minus{} f'(1)| > M/(m \minus{} 1)!$, let $ g(x) \equal{} f'(kx)/k^{m \minus{} 1}$ we can get  $ M/(m \minus{} 1)! < |g(1) \minus{} g(0)| \equal{} |f'(k) \minus{} f'(0)|/k^{m \minus{} 1}$, i.e., $ |f'(k) \minus{} f'(0)| > M/(m \minus{} 1)! * k^{m \minus{} 1}$. Since $ f'$ is monotic, All the $ f'(k) \minus{} f'(0)$ has the same sign, gets the result by integrate them.


linear algebra
matrix
linear algebra unsolved

If $ A \in M_{n}(\mathbb{R})$ and $ A^2\equal{}A$

we have: $ P^{\minus{}1}AP\equal{}\begin{bmatrix}E_{r} & 0\\ 0 & 0\\ \end{bmatrix}$[color=red] (Is this right :D )[/color]

where $ P \in GL_{n}(\mathbb{R})$

and $ r\equal{}rank(A)$

OK, if $ B \in M_{n}(\mathbb{R})$ and $ rank(B)\equal{}rank(A)$

then can we figure out $ P^{\minus{}1}BP$ ?

That's not nearly enough information about $ B$ to say anything useful or interesting.

[quote]Is this right[/quote]

Yes. An idempotent matrix of rank r is similar to a diagonal matrix with r 1's in the diagonal.

For B, in my opinion we can say only that there are invertible matrices Q,R such that

$ B \equal{} Q \begin{bmatrix}E_{r}& 0\\ 0 & 0\\ \end{bmatrix} R$.

In other words, A and B are equivalent to the same matrix, and as the equivalence of matrices is a transitive relation, A and B are equivalent: we have A = S B T for some invertible S,T.


algebra
polynomial
complex numbers
algebra unsolved

Let $P(x)$ be a polynomial with complex coefficients such that $P(0)\neq 0$. Prove that there exists a multiple of $P(x)$ with real positive coefficients if and only if $P(x)$ has no real positive root.

Proof for the forward direction:

Let $p(x)$ denote a complex-coefficient polynomial satisfying the given conditions. If the multiple of $p(x)$ has no real positive roots, then $p(x)$ has no such roots either, so we need only show that the multiple has no positive real roots. This result follows directly from Descartes's Rule of Signs and the fact that all coefficients are positive. Thus, $p(x)$ has no positive real roots either.

Does the polynomial $p(x) = (x-i)(x-i)$ contradict the converse?

Pay attention, mathisfun!

$Q(x)$ is a multiple of $P(x)$ if exists a polinomial $R(x)$ s.t. $Q(x)=P(x)R(x)$

Sorry! I did not know that the multiplier could be nonconstant. There goes my solution...

I have, I think, part of the proof for the reverse direction; someone out there could definitely improve or finish it:

Assume WLOG that all roots are complex.

Write $p(x)$ in factored form: $p(x) = (x-\alpha_1)(x-\alpha_2)...(x-\alpha_n)$

Let $\alpha_i'$ denote the complex conjugate of $\alpha_i$.

$(x-\alpha_1')...(x-\alpha_n')p(x) = (x-\alpha_1)(x-\alpha_1')(x-\alpha_2)(x-\alpha_2')...(x-\alpha_n)(x-\alpha_n') \\ \\ = (x^2-(\alpha_1 + \alpha_1')x^ + |\alpha_1|^2)...(x^2-(\alpha_n + \alpha_n')x+|\alpha_n|^2)$

Clearly this polynomial has real coefficients. However if the real parts of some of the roots are positive, then the polynomial will have negative coefficients. I am still attempting to find a  suitable factor to negate those negatives.

Just one of the old shortlist problems that solves this problem instantly :
If $f(x)$ is a real polynomial such that $f(x)>0$ for all $x\geq0$ we have for some $n$ the coefficients of $(1+x)^nf(x)$ are positive.


It  are known  $ \tan\frac{x}{2}\equal{} \frac{\tan x\plus{}a\minus{}1}{\tan x\plus{}a\plus{}1}$ there are equation  solution. Do proof  $ \mid{a}\mid\ge1$

[quote="Pirkuliyev Rovsen"]It  are known  $ \tan\frac {x}{2} \equal{} \frac {\tan x \plus{} a \minus{} 1}{\tan x \plus{} a \plus{} 1}$ there are equation  solution. Do proof  $ \mid{a}\mid\ge1$[/quote]

Let $ t\equal{}\tan(\frac x2)\neq 1$. The equation becomes $ t\equal{}\frac{\frac {2t}{1\minus{}t^2}\plus{}a\minus{}1}{\frac {2t}{1\minus{}t^2}\plus{}a\plus{}1}$ $ \equal{}1\minus{}\frac2{\frac {2t}{1\minus{}t^2}\plus{}a\plus{}1}$ $ \iff$ $ a\equal{}\frac{1\plus{}t^2}{1\minus{}t^2}$ $ \equal{}\frac 1{\cos(x)}$

Hence the result.


I want to write in the note area (int the left of the body). How?

[quote="stevem"]By default margin notes go on the right [code]\marginpar{righthand side} [/code]
Both sides would be [code]\marginpar{left-side}{right-side}[/code]
In your case to get left-side notes use [code]\marginpar{}{This is my note}[/code]

If all your notes are on the left side you can make this slightly easier by first saying
[code]\reversemarginpar[/code] and from then on use[code]\marginpar{This is my note}[/code][/quote]

Exactly what I want.

Thanks


Four pairs of socks lie in a row on a clothesline. How many arrangements in a row are there if each sock is not adjacent to its matching one?

We have to use set theory on this one, i think the answer is [hide]864[/hide]? I've done a similar problem in the Australian Mathematics Competition.

When you say Set Theory, u mean Inclusion-Exclusion right?

Yeap, Inclusion-Exclusion.

You can do it this way: First, there are $ \frac{8!}{2^{4}}\equal{} 2520$ ways to arrange the socks including the permutations where the  (we divide by $ 2^{4}$ because the socks in each pair can be swapped with no change to the permutation). Similarly, there are $ 4\cdot\frac{7!}{2^{3}}\equal{} 4\cdot 640 \equal{} 2520$ ways where one pair is together. Then there are $ 6\cdot\frac{6!}{2^{2}}\equal{} 1080$ ways where two pairs is together (note this includes the number of ways where one pair is together in some instances). And there are $ 4\cdot\frac{5!}{2^{1}}\equal{} 96$ ways there three pairs are together (including the number of ways where two pairs are together), and $ \frac{4!}{2^{0}}\equal{} 24$ ways where all four pairs are together.

So the answer is $ 2520\minus{}2520\plus{}1080\minus{}240\plus{}24 \equal{} 864$ ways, as stated before  :P

[quote="brightzhu"][quote="aidan"]

 $ 4\cdot\frac {5!}{2^{1}} \equal{} 96$ [/quote]
Why is this a 96?[/quote]

Typo mistake, it's 120  :blush: But the final answer is correct

It's actually 240, which is what you (correctly) subtracted while doing inclusion-exclusion.


integration
calculus
calculus computations

Could someone help me with this problem?
$ u_{tt} \equal{} c^2 u_{xx}$

Idid some work, and this is the answer:
$ u(x,t) \equal{} \frac{1}{2}\left[g(x\minus{}ct) \plus{} g(x\plus{}ct)\right] \plus{} \frac{1}{2c} \int_{x\minus{}ct}^{x\plus{}ct} h(\xi) d\xi$


Let ABC be a triangle, and let A1,B1,C1 be on BC,CA,AB, respectively, with none of A1,B1,C1 coinciding with a vertex of ABC. Show that if

AB+BA1=AC+CA1,  AB+AB1=BC+CB1,    and    AC+AC1=BC+BC1

then
       Area(ABC) >= 4.Area(A1B1C1)

If I well understand, A1B1C1 is the extouch triangle of ABC (A1 = common point of BC with the A-excircle...)
Thus area(ABC)/area(A1B1C1) = 2R/r
See, for instance

http://mathworld.wolfram.com/ExtouchTriangle.html

Kind regard. JPE


hi guys. i've got another dumb question for you. cus i don't know how to use these code thingys on the forum. how do make the hiding codes say things besides [hide][/hide]? yeah. dumb, but i just don't know. so...please?

[code][hide=click this text]this text will be shown[/hide][/code]

Again, [url=http://www.artofproblemsolving.com/Forum/index.php?f=145]here[/url] might be a better place to post that question.

Just hover above "Hide" at the top of the place where you type. Underneath it, it says the code.


A necklace is formed by alternating 3 black beads with 5 white beads. If the finished necklace has 48 beads, how many of the beads are white?

$ \frac{5}{8}$ of the beads are white. So out of $ 48$ beads there are $ 48\cdot\frac{5}{8}\equal{}\boxed{30}$ white beads.


When rating a solution, what exactly are you supposed to be rating? I've been rating whether or not the solution makes sense to me and how well I feel it is written, but is that right?

That's how I've been doing it. I mean, how else are you supposed to rate a solution?

This is what I typically rate problems on (though I will admit I haven't rated any recently):

1. Is it clear
2. Does it fully explain everything
3. It is the fastest way to solve the problem

[quote="bryanxqchen"]I just do what it tells me to rate.
@above poster. Does it really matter if it is fast? Well, to some people yes. But as long as you learn its good.[/quote]

Well, I would like to see the optimal solution, but I don't rate the solution solely on that.

I rate it on its efficiency also, but also on how easy it is to understand.


calculus
integration
trigonometry
inequalities
real analysis
real analysis unsolved

Hello, I encountered such an integral:
$ \int_0^{\plus{}\infty} \frac{x^\alpha}{1 \plus{} x^\beta |\sin x|^\gamma} \mathbf{d}x$
and I dunno for what $ \alpha, \beta, \gamma$ does the integral coverage :huh:
Could you help ? :)

Hint:
Let's interval $ (0,\plus{}\infty)$ divide by parts $ [\pi k, \pi (k\plus{}1)]$ and estimate integral on interval $ [\pi k, \pi (k\plus{}1)]$ by using inequality :$ \frac{2}{\pi}x<\sin x <x$ whe $ x \in (0,\frac{\pi}{2})$  :)


function
integration
real analysis
real analysis solved

Let ${g_n}$ be a sequence of Riemann integrable functons from $[0,1]$ to $\mathbb{R}$ such that $|g_n(x)\leq1|$ for all $n,x$. Define $G_n=\int_{0}^{x}g_n(t)dt$. Prove that a subsequence of ${G_n}$ conveges uniformly

It follows from the Arzela-Ascoli Theorem. The sequence $(G_n)_n$ is bounded (uniformly) in the $\sup$ metric on $C([0,1])$ and also equicontinuous. The theorem states that in these conditions, the family has compact closure, or, equivalently, every sequence has a convergent subsequence. Since a convergent subsequence in the $\sup$ metric is a uniformly convergent subsequence of functions, we have the conclusion.

Yeah, thanks , grobber, it is what i thought too. As far as i understand equicontinuity of the functions we can show using the fact that all of them satisfyies Lipschiz condition on $[0,1]$


Which AoPS class covers conics - general forms, solving, etc.? I lack almost all conics skills, as I have learned recently, and can't really find any good book/article on them to learn from.

Chapter 5 in AoPS Volume 2 covers Conics...


Given a natural number $n\ge 3$, prove that there exists a set $A_n$ with the following two properties:
1) $A_n $ consists of $n$ distinct natural numbers
2) For any $a\in A$, the remainder of the product of all elements of $A_n\backslash \{a\}$ divided by $a$ is $1$.

Help me !!! I have tried hard to solve this problem but...  :(
I think it 's hard problem

[quote="NoName"]Simply, let $a_1 = 2, a_2 = 3, a_{m+1} = 1+\prod_{i=1}^m a_i$ for natural $m \ge 2$.
Then $\{a_1, a_2, ..., a_{n-1}, a_n-2\}$ is a set having the desired properties.[/quote]
i don't think i'ts true. Try a look at case:$m=4$.
we have $a_1=2, a_2=3, a_3=5, a_4=29$. Then $a_1.a_2.a_4=2.3.29 \neq 1 (mod 5)$.

[quote=mavropnevma]You don't pay attention ... Only the $n$-th element is the product of the previous [b]minus[/b] $1$; all other are the  product of the previous[b] plus [/b]$1$. So for $n=3$ the numbers are $2,3,5$, and for $n=4$ the numbers are $2,3,7,41$. The proof is eminently correct.[/quote]

Ummm wait why should we change the process for $m=4$? for $n=3$ the set $N={2,3,5}$ is perfect. for $n=4$, just take $a_4=2 \cdot 3 \cdot 5 +1=31$ and notice that since the reminder upon division by any of the original terms is $1$, the original congruences stay unaffected. One can easily check by hand that this is true.
One can simply induct using this idea. Thus done. In my opinion the problem is too easy for the 4th spot  :D unless the problems aren't ordered by difficulty. Any contestant would probably first solve $n=3$ easily in less than 2 minutes since the most obvious goes of ${2,3,5}$ works and after less than a minute would realise that since these numbers are relatively prime, multiplying them and adding $1$ destroys the problem. Did i say anything wrong ( feel free to correct :roll:  :love: )?


Let $ p$ is a prime number and $ p|2^{2^n}\plus{}1$ . Prove that : $ 2^{n \plus{} 2}|(p \minus{} 1)$

No, not very hard. Inf act easy using orders.
But to the point:
a) This title is completely useless, it says nothing about the problem. Seehttp://www.mathlinks.ro/viewtopic.php?t=135914 .
b) It was posted often before. But as it is shorter than searching: consider the multiplicative order of $ 2 \mod p$.


How many whole numbers are between $ \sqrt[3]{10}$ and $ \sqrt[3]{200}$?

cube root of 10 is less than 3, grater than 2
cube root of 200 is less than 6, greater than 5
so 3, 4, 5, 6 are in between, answer is 4

Actually, $ \sqrt[3]{200} < 6$, so the answer is $ \boxed{3}$.

Whoops....that's what I put, but 6 somehow snuck in there....


Hi
what is AOPS phone number?
I tried what it says under contact info (619-659-1612) only to hear the message that the number is no longer in service.
if you know their current phone # please let me know

That's the correct number.  Try calling us again.

Try adding 1 before the number. I've been on a few phones that tell you a number isn't in service unless you're dialling it like an international call.


calculus
derivative
calculus computations

So here it is... seems simple, I just can't isolate the y' when solving for the derivative:

Find the derivative of y with respect to x at y=2

y^2= (x+y)/(x-y)

Thanks in advance....

Great, thanks for your help... just a question regarding the answer though... on the second last line, how did you eliminate the "dy" on the right hand side of the equation... should it not have been subtracted along with the x?


calculus
integration
limit
trigonometry
function
real analysis
real analysis unsolved

Calculate the limit $ \lim_{n \rightarrow \infty} \int_{0}^{\frac{\pi}{3}} \frac{\sin^n{x}}{\sin^n{x} \plus{} \cos^n{x}}$
Help!  :)

$ \int_0^{\frac {\pi}3}\frac {\sin^nx\,dx}{\sin^nx \plus{} \cos^nx} \equal{} \int_0^{\frac {\pi}3}\frac {dx}{1 \plus{} \cot^nx} \equal{} \int_0^{\frac {\pi}4}\frac {dx}{1 \plus{} \cot^nx} \plus{} \int_{\frac {\pi}4}^{\frac {\pi}3}\frac {dx}{1 \plus{} \cot^nx}$

We have $ \frac {1}{1 \plus{} \cot^nx}\le 1,$ which is an integrable function on this finite interval, so the Dominated Convergence Theorem applies and we may interchange limit and integral.

On $ \left[0,\frac {\pi}4\right),$ we have $ \frac {1}{1 \plus{} \cot^nx}\to 0$ and on $ \left(\frac {\pi}4,\frac {\pi}3\right],$ we have $ \frac {1}{1 \plus{} \cot^nx}\to 1.$ So the limit is

$ \int_0^{\frac {\pi}4}0\,dx \plus{} \int_{\frac {\pi}4}^{\frac {\pi}3}1\,dx \equal{} \frac {\pi}{12}.$

(I used the DCT as a convenience. More elementary justifications are certainly possible.)

[quote="Kent Merryfield"]
On $ \left[0,\frac {\pi}4\right),$ we have $ \frac {1}{1 \plus{} \cot^nx}\to 0$ and on $ \left(\frac {\pi}4,\frac {\pi}3\right],$ we have $ \frac {1}{1 \plus{} \cot^nx}\to 1.$[/quote]
How do you see so fast these conclusions?

If $ \cot x > 1$, $ \cot ^n x \to \infty$ as $ n \to \infty$, while if $ 0 \leq \cot x < 1$, $ \cot^n x \to 0$.

Yes, I can get that, but I don't fully understand why takin' $ \left[0,\frac{\pi}{4}\right)$ since $ \cot x$ is not defined at $ x\equal{}0$ but defined at $ x\equal{}\frac\pi4.$

About $ 0:$

I realized that I was being perhaps a little sloppy with my language there, but there's not really a problem. The original function, expressed in sines and cosines, is defined at $ 0$ and is equal to $ 1,$ no matter what $ n$ is. So the statement about the limit works at that point.

About $ \frac{\pi}4:$

That's the one point where the sequence of functions does not converge to either $ 1$ or $ 0.$ Instead, the function is equal to $ \frac12$ there no matter what $ n$ is.

I was speaking Lebesgue language, and I used the words "almost everywhere." I really don't care what happens at this single point, which is a set of measure zero.

----

Since this sequence of continuous functions converges to a discontinuous limit, the convergence cannot possibly be uniform - at least not uniform on $ \left[0,\frac{\pi}3\right].$ That limits somewhat our choice of arguments for why we can interchange the limit and the integral.


geometry
geometric transformation
dilation

:juggle: A is a white crystalline, ionic compund that is widely used as a fertiliser and as an explosive in mining operations.It decomposes explosively above 300C to give two colourless odourless diatomic gases B and C, and H2O. At lower temperatures like 200C A decomposes to give a triatomic gas D and water.D is used as a mild anaesthetic and a propellant for whipped cream.Like A, D is thermodynamically unstable and decomposes on heating to B and C. Compound A is prepared commercially by reacting gaseous E with F. E has distinctive, irritating, pungent odour and can be produced (along with some by products) by the action of dilute aqueous NaOH on A, upon heating. Commercially, E is prepared by reacting B and hydrogen in the presence of an iron oxide catalyst at high temperature (400C) and high pressure ( say 250 atm).F is a strong acid and an oxidising agent.F is prepared as follows;
$ \Rightarrow$ Reaction of E with C at 850c in the presence of catalyst, produces a colourless diatomic gas G with H2O. :bruce:
$ \Rightarrow$ Further reaction of G with C produces a brown triatomic gas H which exists in equlibrium with colourless I. I is isoelectronic with Cr2O4$ ^2\minus{}$ ( the oxalate ion).
$ \Rightarrow$ H disproportionates in H2O to produce F and G with the latter being recycled. :maybe:

A) Identify compunds A to I.
B) Write balanced equations for each of the reactions.

[hide]PLS GIVE ME AN ANSWER TO THE QUESTIONS with expln.It is unsolved in my book.[/hide] :diablo:

i hate chem. questions like these make chem boring.

:oops: In above question,I forgot to add that D on reaction with atomic oxygen also produces G. :spider:

a is ammonium nitrate. b is nitrogen. c is oxygen.d is nitrous oxide. e is ammonia. f is nitric acid. g is nitric oxide. h is nitrogen dioxide. i is dinitrogen tetroxide

cheap question  :P

[b][u]Chemistry is seriously interesting. Its the way you look at it.Just think it as a detective puzzle to find out the compounds.[/u][/b]
@Sagar: How can D be nitrous oxide when it is given that it is a triatomic gas? Pls post all of the compunds including the rxns pls.

I know know, I managed decently with chem keeping that thought, but seriously....

nitrous oxide (n2o) aka  :rotfl:  gas

nh4no3->n2+o2+h2o
nh4no3->n2o+h2o
n2o->n2+o2
nh3+hno3->nh4no3
naoh+nh4no3->nh3
n2+h2->nh3
nh3+o2->no+h2o
no+o2->no2
no2<->n2o4
h2o+no2->hno3+no

sry not balanced

Good work Sagar :coolspeak:
[hide]I seriously hate this kind of questions[/hide]

[quote="sumanth0811"]no more chem in my life
please do not start such topics...i feel suicidal[/quote]

Same here.. I can't mug at all after 10th, lost all marks of inorg. chem in JEE :(

plz don't say that.....if there was no chem i wud surely not qualify in jee :noo:

such questions r very easy 2 score, like that methylene blue question in jee
the above question too cud b cracked by just two phrases, explosive fertilizer and propellant for whipped cream

[quote="Sagar_Vaze"]plz don't say that.....if there was no chem i wud surely not qualify in jee :noo:[/quote]

That's wid you buddy.. If there was no chem I wud've been in top 500.. I got less in Chem in all the exams, be it board, JEE, AIEEE, BITS.. I love maths.. and Chem is not at all gonna help me in my carrier.. I dunno why they don't ask C++/Java for a guy wanna do CS, instead of asking sucking Chem..

[hide]Srry, bas "Dil k armaan aasuon mein beh gaye"..
Hope u have heard the song.. :P [/hide]

i suck at chem but at jee i did it better than physics
i screwed up physics big big time....if not for physics maybe 500rank wud be possible :(


combinatorics unsolved
combinatorics
Permutation patterns

Numbers $ 1,2,3,...,2004$ are arranged in a row such that each number is either greater than all of the numbers to its left or less than any of the numbers to its right.How many such arrangements are there?

[hide="Quick solution, assuming I am not misunderstanding the statement of the problem"]
The last element of the permutation must be 2004.  Suppose the last element were not 2004.  Then, 2004 must have occurred somewhere to the left of the last element, which is clearly not greater than 2004.  There aren't any elements to the right of the last element either, so it fulfills neither of the two criteria.

Now, each of the first 2003 numbers will be less than at least one of the numbers on its right (because of the 2004) no matter what, so there are $ \boxed{2003!}$ permutations.
[/hide]

Probably the "any" should also be interpreted as "all," i.e., each entry is either larger than everything to its left or smaller than everything to its right.  This is equivalent to saying that the permutation contains no three term decreasing subsequence, or that it avoids the permutation pattern 321.


geometry
geometric transformation
reflection
trigonometry

If the platform moveswith unform velocity, the particle should be projected with $ \theta \equal{} \frac {\pi}{2}$. else I don't see any reason why it should hit $ B$.

hello
well find the relative velocity of the particle wrt to sq.platform.
if we wants to it to strike at $ B$ the particle should not have any horizontal velocity.
resolve $ u$
so $ ucos\theta\equal{}v$
or $ \theta\equal{}cos^{\minus{}1}(\frac{v}{u})$
thank u

[quote="kabi"]hello
well find the relative velocity of the particle wrt to sq.platform.
if we wants to it to strike at $ B$ the particle should not have any horizontal velocity.
resolve $ u$
so $ ucos\theta \equal{} v$
or $ \theta \equal{} cos^{ \minus{} 1}(\frac {v}{u})$
thank u[/quote]
The particle is moving with the platform with $ v$ when projected. So it will go away straight without any deflection in the direction it is fired. How do you expect it to "turn" in order to hit $ B$???

answer of kabi is right.but i agrees with rajiv.as plateform is moving with uniform speed it should be 90.

[quote="iit2009"]a smooth square plateform ABCD is moving towards right with a uniform speed v.at what angle theta particle be projected from A with speed u so that it strikes B.[/quote]
The particle can hit $ B$ only if the platform moves with the said $ v$ after its projection, and this should be mentioned in the question. Hence the time taken by the particle to travel $ a$ is: $ \frac{a}{u \sin \theta}$. This time should be same as the time required for $ B$ to reach the particle along line $ BC$. Solve the equations for $ t$, you'll get the answer. I think kabi has done the same thing...

Rajiv
In the frame of square ,if we have a velocity at the $ \theta\equal{}cos^{\minus{}1} {\frac{v}{u}}$ the particle just has a vertical velocity thus it will strike $ B$.
Thank u

[quote="kabi"]Rajiv
In the frame of square ,if we have a velocity at the $ \theta \equal{} cos^{ \minus{} 1} {\frac {v}{u}}$ the particle just has a vertical velocity thus it will strike $ B$.
Thank u[/quote]

Yes you are right. But the condition I mentioned in the previous post is a necessary one, else, the particle will have a velocity $ v\minus{}u$ wrt reference frame of the platform, with $ \theta$. (I'm talking about you (or anyone else, maybe me too :rotfl: ) to be sitting on the platform and observing.)

iit2009,if it has to reach b then it should be projected diagonally, or else depending on the relative velocity of the square with respect to the particle the particle can either reach a or c

[quote="aadil"]iit2009,if it has to reach b then it should be projected diagonally[/quote] Diagonally??? :huh:

[quote="Immanuel Bonfils"]I think Rajiv considered the particle moving with the platform before it is projected, as a matter of fact, a natural assumption.[/quote]

For the particle, $ t_1 \equal{} \frac {a}{u \sin \theta }$

For table, $ t_2 \equal{} \frac {a \cot \theta}{v}$

For the particle to hit $ B$, $ t_1 \equal{} t_2$.

Equating, we get $ \theta \equal{} \cos^{ \minus{} 1} \frac {v}{u}$

Which is exactly what [b]kabi[/b] did. And [b]aadil[/b], for this to be a diagonal projection, $ u \equal{} \sqrt {2} v$.  :D

[quote="aadil"]sorry,my mistake[/quote]

Never mind... :D  :wink:


function
limit
real analysis
real analysis unsolved

how do you prove that the limit of a function is unique?

Assume two limits $L_1$ and $L_2$. Try $\epsilon=\frac{|L_1-L_2|}2$.

I'll do it for sequences rather than for functions because that makes the proof a little easier to see, but the essentials behind this proof would be the same in all cases.

Suppose $\lim_{n\to\infty}x_n=a$ and $\lim_{n\to\infty}x_n=b$

Given $\epsilon>0,\,\exists\,N_1$ such that if $n>N_1,\,|x_n-a|<\frac{\epsilon}2.$

Similarly, $\exists\,N_2$ such that if $n>N_2,\,|x_n-b|<\frac{\epsilon}2.$

Choose an $n>\max(N_1,N_2).$ Then $|a-b|\le|a-x_n|+|x_n-b|< \frac{\epsilon}2+ \frac{\epsilon}2=\epsilon.$

Since we now have that $|a-b|<\epsilon\,\forall\,\epsilon>0.$ Thus $a=b.$

If you're looking for broad generalizations, the topological property of $\mathbb{R}$ that we explicitly used was that it is Hausdorff.


In a triangle ABC, D is midpoint of BC . If $\angle ADB = 45 ^{\circ}$ and $\angle ACD = 30^{\circ}$, determine  $\angle BAD.$

[hide][hide] From $BD=CD$ we have  $\frac{BD}{AD}=\frac{CD}{AD}$. Let $\angle BAD=x$ by sine theoreme

$\frac{sin x}{sin(x+45)}=\frac{sin 15}{sin 30}$ or $2sin x\cos 15=sin (x+45)$

$sin(x+15)+sin(x-15)=sin(x+45)$

$sin(x+15)=sin(x+45)-sin(x-15)=2sin30cos(x+15)=cos(x+15)$

$x=30$[/hide][/hide]

[hide="Here is an another trigonometrical solution."]$x=m(\widehat {BAD})\Longrightarrow m(\widehat {DAC})=15^{\circ},\ B=135^{\circ}-x\Longrightarrow$

$1=$ $\frac{DB}{DC}=\frac cb\cdot \frac{\sin x}{\sin 15^{\circ}}\Longrightarrow$

$\sin C\sin x=\sin B\sin 15^{\circ}\Longrightarrow$

$\sin 30^{\circ}\sin x=\sin (45^{\circ}+x)\sin 15^{\circ}\Longrightarrow$

$\sin x=2\sin (45^{\circ}+x)\sin 15^{\circ}\Longrightarrow$

$\cos (90^{\circ}-x)=\cos (30^{\circ}+x)-\cos (60^{\circ}+x)\Longrightarrow$

$\sin (30^{\circ}-x)=\cos (30^{\circ}+x)-\cos (90^{\circ}-x)\Longrightarrow$

$\sin (30^{\circ}-x)=2\sin 60^{\circ}\sin (30^{\circ}-x)\Longrightarrow$

$\sin \left(30^{\circ}-x\right)\left(1-\sqrt 3\right)=0\Longrightarrow \boxed {\ x=30^{\circ}\ }\ .$[/hide]

anybody for a synthetic solution? i think that one exists!

Ok, here's my solution.
[hide]
Let $E$ be a point on $AC$ such that $ED=BD=CD$.
Since $\triangle CDE$ is an isosceles triangle, $\angle CED=\angle ECD=30^\circ$.
Thus $\angle ADE=\angle ADC-\angle EDC=135^\circ -120^\circ =15^\circ$.
Then $\angle EAD=\angle CED-\angle ADE=30^\circ -15^\circ =15^\circ$.
Since $\angle EAD=\angle ADE$, we have $AE=ED=BD=CD$.

Also, since $BD=ED$ and $\angle BDE=45^\circ +15^\circ=60^\circ$, $\triangle BDE$ is an equilateral triangle, and hence $BE=BD=DE$.
Therefore, $AE=BE$.

$\angle AEB=180^\circ-\angle BED-\angle DEC=180^\circ -\angle 60^\circ-\angle 30^\circ =90^\circ$, so $\triangle AEB$ is an isosceles right triangle.
Hence $\angle BAD=\angle BAE-\angle DAE=45^\circ -15^\circ =\boxed{30^\circ}$.

[/hide]

[hide="here is what I had"]Draw a perpendicular from $B$ to $H$. Then $\angle HBC=90-30=60$. By the 30-60-90 triangle, $HB=\frac{1}{2}*BC=BD$. Therefore, $\triangle BDH$ is equilateral. By the exterior angle theorem, $\angle CAB=15$. Also $\angle ADH=60-45=15$. Then using the equal base angles, $AH=HD=BD=DC$. Then $\triangle AHB$ is isosceles right so $\angle BAH=45 \implies\angle BAD=45-15=30$[/hide]

AD/sin 30=CD/sin 15
BD=(sin 15/sin 30)AD
Let angle BAD=x
BD/sin x=AD/sin(135-x)
sin15/sin30=sin x/sin135-x
simplifying we get-
($\sqrt{3}$-1)/2=sinx/(sinx+cosx)
and finally we get
tanx=1/$\sqrt{3}$
x=30
angleBAD=30

[quote=SatisfiedMagma][quote=div5252]AD/sin 30=CD/sin 15
$BD$=($\sin{15}$/$\sin{30}$)$AD$
Let angle $\angle BAD=x$
$$BD/\sin{x}=AD/\sin{(135-x)}$$
$$\sin15/\sin30=\sin x/\sin135-x$$
Simplifying we get-
$$(\sqrt{3}-1)/2=\sin{x}/(\sin{x}+\cos{x})$$
and finally we get
$$\tan{x}=1/\sqrt{3} \implies x=30$$
$$\angle BAD=30$$[/quote][/quote]
Well the art of the problem lies on solving it geometrically,just draw $BE perpendicular on AC$ and that will magically give many things,circumcentre, a$ 30-60-90 triangle$,and a lot of things I won't tell which ones,checking this completes our solution.resulting$angleA=30°$.

[hide]Construct an equilateral triangle on $\triangle CDE$, Notice that AC would be perpendicular to DE Because of the 30-30 bisection, Let F = intersection of DE and AC Then it's not hard to see that $\triangle CFD$ is congruent to $\triangle CFE$, By our construction we know that $BD=DC=DE$ which implies that $\angle BEC = 90$ so we get $\angle BEC = 30$ , We also notice that by cpct in our earlier congruence we get $FD=FE$ , which by sas shows $\triangle AFD$ is congruent to $\triangle AFE$ which means angles AED = ADE=75 also $AEB= AED - BED = 75-30=45$, we are given $ADB = 45$ so this implies AEDB is a cyclic quadrilateral!, Now $BAD = BED = 30$[/hide] NOt that clean diagram  :P But here it is

[hide=Solution using Trigonometry (Sine Rule)]Let $\angle BAD=x$. $\angle ABD=135^{\circ}-x$. $\angle CAD=15^{\circ}$.
By sine rule,
$\frac{BD}{DA}=\frac{sinx}{sin(135^{\circ}-x)}=\frac{DC}{DA}=\frac{sin(15^{\circ})}{sin(30^{\circ})}=\frac{\sqrt{3}+1}{\sqrt{2}}=\sqrt{2}\frac{sinx}{(sinx+cosx)}$.
$(\sqrt{3}-1)(sinx+cosx)=2sinx$.
$\sqrt{3}+1=1+cotx$.
$cotx=\sqrt{3}$.
$x=60^{\circ}$. [/hide]
Please refer the diagram attached.
Hope this helps :)

$\angle ADB= 45^{\circ}$, $\angle ACB= 30^{\circ}$
Constructing $BX \perp AC$ such that $X$ lies on $AC$.
As $\Delta BXC$ is a right-angled triangle and $D$ is the mid-point of $BC$, $D$ is the circumcentre of $\Delta BXC$.
$\Rightarrow \angle XDC= 120^{\circ} \Rightarrow \angle ADX= 15^{\circ}$. Also, $\angle XBC= 60^{\circ}$.

$\angle XDB= \angle ADB+ \angle ADX= 45^{\circ}+15^{\circ}=60^{\circ}$. As $D$ is the circumcentre of $\Delta XBC$, $XD=CD=BD$.
$\angle XBD= \angle XDB = 60^{\circ}$, and $XD=BD \Rightarrow \Delta XBD$ is equilateral, $DX=BX=BD$

By angle-sum in $\Delta ADC$, we get $\angle DAC= 15^{\circ} =\angle XDA \Rightarrow \Delta XDA$ is isosceles $\Rightarrow AX=DX$.
But $DX=BX \Rightarrow AX=BX$. By angle-sum in $\Delta AXB$, we get $\angle XAB= 45^{\circ} \Rightarrow \angle BAD= \angle XAB- \angle XAD= 45^{\circ}-15^{\circ} = \boxed{30^{\circ}}$


This was question 1 on last years Balkan MO: http://www.mathlinks.ro/Forum/viewtopic.php?p=498977#p498977.
(It was also rather well known before then).

Too bad I couldn't solve it in the competition.   :(

[quote="TRAN THAI HUNG"]My teacher give this problem for me.Can you do it for me? :oops:  :(
Prove that:
$\frac{1}{a^{2}+bc}+\frac{1}{b^{2}+ac}+\frac{1}{c^{2}+ab}\leq\frac{a+b+c}{2abc}$ $\ (a,b,c>0)$[/quote]
$\frac{1}{a^{2}+bc}+\frac{1}{b^{2}+ac}+\frac{1}{c^{2}+ab}\leq\frac{a+b+c}{2abc}\Leftrightarrow$
$\Leftrightarrow\frac{1}{2}\cdot\left(\frac{a+b+c}{abc}-\frac{1}{ab}-\frac{1}{ac}-\frac{1}{bc}\right)+\sum_{cyc}\left(\frac{1}{2bc}-\frac{1}{a^{2}+bc}\right)\geq0\Leftrightarrow$
$\Leftrightarrow\sum_{cyc}\frac{a(a^{2}-bc)}{a^{2}+bc}\geq0\Leftrightarrow\sum_{cyc}\left(\frac{a(a-b)(a+c)}{a^{2}+bc}-\frac{a(c-a)(a+b)}{a^{2}+bc}\right)\geq0\Leftrightarrow$
$\Leftrightarrow\sum_{cyc}(a-b)\left(\frac{a(a+c)}{a^{2}+bc}-\frac{b(b+c)}{b^{2}+ac}\right)\geq0\Leftrightarrow$
$\Leftrightarrow\sum_{cyc}\frac{(a-b)^{2}(a^{2}+b^{2}+ac+bc)}{(a^{2}+bc)(b^{2}+ac)}\geq0.$ :wink:

By the way $\frac{1}{a^{2}+bc}+\frac{1}{b^{2}+ac}+\frac{1}{c^{2}+ab}\leq\frac{a+b+c}{2abc}\Leftrightarrow$
$\Leftrightarrow\sum_{sym}\left(a^{4}b^{3}-a^{4}b^{2}c\right)+\sum_{cyc}\left(a^{5}bc-a^{3}b^{3}c\right)\geq0,$ which is trivial.


Let $ABCDEF$ be a convex haxagon. $AB \cap EF=P$, $EF \cap CD=Q$, $CD \cap AB=R$. And $DE \cap BC=S$, $DE \cap FA=T$, $FA \cap BC=U$.
Prove: If $AB: RP=CD: QR=EF: PQ$, then $BC: US=DE: ST=FA: TU$.

See http://www.mathlinks.ro/Forum/viewtopic.php?t=4822 .

  darij


is it possible to show all $n \in \mathbb{Z}$ like this:
$n=\pm{1^2}\pm{2^2}\pm{3^2}\pm...\pm{m^2}$ :?:

See NoName's and my idea for a generalisation (with proof): http://www.mathlinks.ro/Forum/viewtopic.php?t=30998


algebra
polynomial
search
algebra unsolved

Determine all polynomials $p(x)$ with real coefficients such that $p((x+1)^3)=(p(x)+1)^3$ and $p(0)=0$.

we take $ (a_n)_{n\in\mathbb{N}}: \ a_0\equal{}0,a_{n\plus{}1}\equal{}(a_n\plus{}1)^3$
let the sequence $ (b_n)_{n\in\mathbb{N}}: \ b_n\equal{}P(a_n)$
we have $ b_0\equal{}a_0$ and $ b_{n\plus{}1}\equal{}(b_n\plus{}1)^3$ so $ \forall n: \ a_n\equal{}b_n$
so $ \forall n\in\mathbb{N}: \ P(a_n)\minus{}a_n\equal{}0$ and the set $ \{a_n: \ n\in\mathbb{N}\}$ is infinit
so $ P(X)\minus{}X\equal{}0$

[quote="aviateurpilot"]we take $ (a_n)_{n\in\mathbb{N}}: \ a_0 \equal{} 0,a_{n \plus{} 1} \equal{} (a_n \plus{} 1)^3$
let the sequence $ (b_n)_{n\in\mathbb{N}}: \ b_n \equal{} P(a_n)$
we have $ b_0 \equal{} a_0$ and $ b_{n \plus{} 1} \equal{} (b_n \plus{} 1)^3$ so $ \forall n: \ a_n \equal{} b_n$
so $ \forall n\in\mathbb{N}: \ P(a_n) \minus{} a_n \equal{} 0$ and the set $ \{a_n: \ n\in\mathbb{N}\}$ is infinit
so $ P(X) \minus{} X \equal{} 0$[/quote]

can someone prove why if the set is infinite then P(x)=x?

Because a polynomial of degree n has at most n roots (except we have a constant polynomial) If we have that a polynomial has infinite roots it must be the constant polynomial equal to zero. Since this polynomial is P(x)-x  we obtain P(x)=x as desired

The theorem he used is the fundamental theorem of algebra and is very well known, if you search you will find tons of proofs

Daniel

To be sure, the fundamental theorem of algebra is much deeper and harder to prove than the fact used here.
Here, we used that a polynomial cannot have [b]more [/b]roots than its degree.
The fundamental theorem of algebra says that a polynomial cannot have [b]less [/b]roots than its degree (at least over the complex numbers).
We really don't need this here.

Here is a slightly more interesting problem: Can you find all polynomials $p$ with $p((x+1)^3)=(p(x)+1)^3$ (i.e. we remove the condition $p(0)=0$)?


i propose it last week and i decide to give it to my two friends at Le Hong Phong High school for the gifted, $nttu$ and $darkmaster$ :P

for a,b,c>0 and a+b+c=1 prove that

$2 \leq 9abc+\sum\frac{ab}{c}+2(ab+bc+ca)$ :D

good luck to everyone ;)

Thanks for this problem .
  I don't have enough time now , so I don't have a nice solution .
   Hope this one true :
 This problem is equivalent to :
 $ (ab+bc+ca)^2+2abc(ab+ac+bc)+9a^2b^2c^2 \geq 4abc $ (1)
 Indeed ,
  $ (1) \iff a(b-c)^2(3a-2bc+6abc)+b(c-a)^2(3b-2ac+6abc)+c(a-b)^2(3c-2ab+6abc)\geq 0 $
 Put $ A= a(3a-2bc+6abc) $
  We need to prove : $ A(b-c)^2+B(c-a)^2+C(a-b)^2 \geq 0 $
  We can prove : $ A+B+C \geq 0 , AB+AC+BC \geq 0 $
   This is enough to follow the conclusion   ;)

  This is not a full solution , just my first thought . I will return with another solution  :)


limit
integration
trigonometry
function
calculus
calculus computations

Find the limit of sequence
$ L = {\lim }\limits_{n \to  + \infty } \int\limits_0^1 {\frac{{\sin ^n x}}{x}} dx$

Write $ u_n\equal{}\int_0^1\left(\frac{\sin x}{x}\right)\sin^{n\minus{}1}x\,,dx.$

Note that $ [0,1]$ is a subset of the first quadrant, so all of these functions are nonnegative. Note also that $ \frac{\sin x}{x}$ can be extended to be a continuous function on $ [0,1]$ and that $ \frac{\sin x}{x}\le 1.$ Then, $ \sin^{n\minus{}1}x\le \sin^{n\minus{}1}1,$ where $ \sin 1\equal{}r$ is a particular positive number smaller than $ 1.$

We get that $ 0\le u_n\le r^{n\minus{}1},$ so $ \lim_{n\to\infty}u_n\equal{}0.$

You could also say that the integrands converge uniformly to zero and quote the fact that uniform convergence allows one to interchange limit and integral; that's in effect what I did above - show the details of the uniform convergence.


A large aquarium of height $5.00m$ is filled with fresh water to a depth of $2.00m$. One wall of the aquarium consists of thick plastic of $8.00m$ wide. By how much does the total force on that wall increase if the aquarium is next filled to a depth of $4.00m$.

Consider a small strip of thickness $dx$ and at a depth $x$ from the surface of the waterThe pressure at this point is $pgx$ and the area of this element is $adx$. Force due to this element is
$dF = (pgx)(adx)$ wherre a is the width of the base.

$F_{total}= \int^{h}_{0}pagx \,dx$
                  = $pag$ $\frac{h^{2}}{2}$
substituting the values given
$\delta$F = $pag(y^{2}-x^{2})/2$where x and y are the initial and final height of the water column.
Answer - $188160 N$
[ I have assumed standard values of p = 1g/cc and g = 10 m/s^2]


arithmetic sequence
algebra proposed
algebra

Prove that the sequence $ (a_n)$ defined by $ a_1\equal{}1; a_n\equal{}a_{n\minus{}1}\plus{}a_{[\frac{n}{2}]}$ for $ n\equal{}2,3,4,....$ contains infinitely many integers divisible by $ 7$.
Note: $ [\frac{n}{2}]$ denotes the biggest integer not bigger than $ \frac{n}{2}$.

Suppose $ a_{k}$ is the largest number in the sequence divisible by 7. All the next equations are modulo 7. From $ a_{k}\equal{}0$ we get $ a_{2k}\equal{}a_{2k\minus{}1}\equal{}a_{2k\plus{}1} \equal{} d$ if this number is zero than we get a contradiction. But now the numbers from $ a_{2(2k\minus{}1)\minus{}1}$ to $ a_{2(2k\plus{}1)\plus{}1}$ form arithmetic progression with difference d and length 7. Thus at least one of these numbers is divisible by 7. Contradiction.
It remains to find at least one number in the sequence divisible by 7 and this should not be that hard.


Find all polynomials $P(x)$ such that

$xP(x-1) = (x-15)P(x)$

I think this works:

[color=blue]Problem[/color].Find all polynomials like $P(x)$ with complex coeffiencets such that \[\forall x \in C : xP(x-1)=(x-15)P(x)\] Solution. \[P(x)=a*\prod_{i=0}^{14}(x-i)\] for some constant $a \in C$
At first note that $p(0)=p(1)=p(14)=0$ hence \[P(x)=x(x-1)(x-14)P_{1}(x)\] for some $p_{1}(x) \in C[x]$.

Replace $P(x)$ in our Hypothesis and we will have \[(x-14)p_{1}(x)=(x-2)p_{1}(x-1)*\] Note that $p_{1}(2)=p_{1}(13)=0$ so \[p_{1}(x)=(x-2)(x-13)p_{2}(x)\] for some $p_{2}(x) \in C[x]$
Replace $p_{1}(x)$ in $*$ and we will reach \[\forall x \in C : (x-13)p_{2}(x)=(x-3)p_{2}(x-1)\] in the same way \[(x-12)p_{3}(x)=(x-4)p_{3}(x-1)\] Continue this process till you reach the equation \[(x-8)p_{7}(x)=(x-8)p_{7}(x)\] which means $p_{7}(x)= a$ for some $a \in C$ now note that all of the numbers $0,1,2,...,14$ are roots of $P(x)$.


inequalities
function
complex analysis
complex analysis unsolved

Suppose that the radius of convergence of $\sum_{n\geq 0}{a_n z^n}$ is infinity and that it's sum, noted $f(z)$ verifies $|f(x)|\leq 1$ for all $x$ real and $|f(z)|\leq (1+|z|^n)e^{ a|Im z|}$ for some nonnegative constants $a,n$. Then prove that $|f(z)|\leq e^{a|Imz|}$

I tried Schwarz's lemma unsuccessfully (it is not accurate enough). Any hint Harazi ?  :D

Hint (given in the exam): let $z=x+iy$ and $s>0$ fixed. Consider the function $g_s$ defined for $Z$ complex different from $-\frac{i}{s}$, $g_s(Z)=(1-isZ)^{-n-1}e^{i\cdot a\cdot |y|\cdot Z} f(x+Zy)$. Apply the maximum modulus principle on a convenable semi-disque and deduce that $|g_s(i)|<1$.


Express the value of $ \frac{3}{14} \div \frac{9}{28}$ as a common fraction.


Six scientists are working of a secret project. They wish to lock up the documents in a cabinet so that the cabinet can be opened when and only when three or more of the scientists are present. What is the amllest number of keys each scientist must carry?

I think its equal to $ 5 \choose 2$ $ =10$...
[hide="another question:whats the minimum number of locks?"]
$ 6 \choose 2$ $ =15$[/hide]