Hey, I'm having trouble solving the following problem:

For any constant c in (0,5), suppose {a_n} is a sequence defined recursively as follows.

a_1 = c,

 a_n+1 = sqrt. (5*a_n) for n greater or equal to 1.

a.) Prove sequence is increasing
b.) Show that a_n<5 for all positive integers n.
c.) Prove that lim of a_n exists as n approaches infinity.
d.) Find lim of a_n as n approaches infinity.

Thanks!!!

O.K. For a) You can use the graph of $y=\sqrt{5x}.$

You can find $a_n$ directly. Can you understand it?

Let $x=\lim_{n\to\infty}a_n.$ By part (c), this exists.

Take the limit in the expression $a_{n+1}=\sqrt{5a_n}$ to get

$x=\sqrt{5x}.$

Can you see why? And can you see how to finish it from there?


there are 4 vectors that belong to the span of all complex 2x2 matrices.
please give me an example of a group of 4 (simplest) of these.

thanks.

This question doesn't make sense as written. Do you mean to find four matrices that span the set of all $2\times 2$ matrices? Becuase "belonging to the span of" all $2\times 2$ matrices simply means being any such matrix.

And "group" is probably not the word you meant to use - "set" is more likely.

sorry... you're right
let me express myself better

give an example of four (simplest) "vectors" (each is a *2x2* complex matrix), that span the set of all 2x2 complex matrices

[quote="Kent Merryfield"]$\left\{\begin{bmatrix}1&0\\0&0\end{bmatrix}, \begin{bmatrix}0&1\\0&0\end{bmatrix}, \begin{bmatrix}0&0\\1&0\end{bmatrix}, \begin{bmatrix}0&0\\0&1\end{bmatrix}\right\}$

You didn't mean $4\times 4$ matrices of course - just that the set of $2\times 2$ matrices is a four-dimensional vector space.[/quote]

Ok so please express the following matrix using the four matrices you mentioned above: a11=(1+2i), a12=a21=a22=0.

Calling the four matrices of my set $M_1,M_2,M_3,M_4,$ this is simply $(1+2i)M_1+0\cdot(M_2+M_3+M_4).$

The premise of the problem is such that we are considering the matrices as a vector space over $\mathbb{C}.$ That means we can use complex scalars.


Bottle A contains more soda than Bottle B. Follow these steps:

1) Pour from Bottle A into Bottle B as much soda as B already contains.

2) Pour from B into A as much soda as A now contains.

3) Pour from A into B as much soda as B now contains.

Each bottle now contains 64 ounces. How many more ounces of soda were in Bottle A than Bottle B at the beginning?

Extra credit to whoever can tell me why this can't happen in real life  [/hide]

[hide]Soda A       Soda B

1) a-b        1)   2b

2) 2a-2b    2)   2b-a+b

3) 3a-5b    3)   6b-2a

They are both 64 ounces after the last pour so:

3a-5b = 6b-2a

         a= 11/5b

Substitute:

3(11/5b)-5b=64

                 b=40

Therefore:a=88.

Soda A had 88oz while Soda B had 40oz.[/hide]

You have a little-bitty typo in your step 2:  it should be (2a - 2b), not just 2a.


Prove that : Exist a natural number n satisfy : n! begins with 1995.
(Sorry if my English is not good)

There is a generalization of the problem:
Let ${A_{n}}$ a sequence of positive integers such that:
-$a_{n+1}>a_{n}$
-$a_{n+1}-a_{n}\leq{C}$ for a given constant $C$.
Prove that there exist a k such that $a_{1}.a_{2}...a_{k}$ begins with 1995.

I'm not sure right now, but I think a generalization of the proof of the fact that $a^n$ can begin with any combination of digits when $a$ isn't a power of $10$ works here.

I remember my friend adyo posted a problem that sounds like this: if $ u_{n+1}-u_n$ goes to infinity, but $ u_{n+2}-2*u_{n+1}+u_n$ goes to 0, then the set $ {u_n}$ is dense in [0,1]. It is not an easy problem, tough it was spoiled because of some similarities with a more classical problem: if $ u_{n+1}-u_n$ goes to 0 and u_n goes to infinity, then the same set is dense in [0,1]. Probably  one should concentrate more on that problem...

I still don't know how to solve these problem :(
Could someone please tell me


Given $ a \plus{} b \plus{} c \equal{} 0$,  and none of a,b, and c are equal to 0, find the value of

$ (\frac {b \minus{} c}{a} \plus{} \frac {c \minus{} a}{b} \plus{} \frac {a \minus{} b}{c})(\frac {a}{b \minus{} c} \plus{} \frac {b}{c \minus{} a} \plus{} \frac {c}{a \minus{} b})$

[quote="Dr Sonnhard Graubner"]hello, the result is 9.
Sonnhard.[/quote]

solution please?

hello, my result does also for $ a\equal{}1,b\equal{}2,c\equal{}\minus{}3$ hold.
Sonnhard.

any ideas?

i've tried pretty much everything the last 2 hours.

[quote="unimpossible"][quote="mihail911"]i guess sonnhard's solution does hold but there has to be some other insight. i mean then you can just plug in any random set of three numbers as long as they add up to 0. there's an infinite amount?[/quote]

?? the questions asks for the value of the expression.   :?:[/quote]

well, yes, but what im saying is that if you can choose any 3 values for a, b, and c won't the value of the expression always be different?

hello, your term is equivalent to
$ \frac{a^3\plus{}b^3\plus{}c^3\minus{}a^2b\minus{}a^2c\minus{}ac^2\minus{}ab^2\minus{}b^2c\minus{}bc^2\plus{}3abc}{abc}$. Here you must insert $ c\equal{}\minus{}a\minus{}b$.
Sonnhard.

[hide="the power of multiplication"]
The first term is equivalent to $ \frac{\minus{}(a\minus{}b)(b\minus{}c)(c\minus{}a)}{abc}$
The second term is equivalent to $ \frac{(a\plus{}b\minus{}c)(a\minus{}b\plus{}c)(\minus{}a\plus{}b\plus{}c)\plus{}2(a\plus{}b)(b\plus{}c)(a\plus{}c)}{(a\minus{}b)(b\minus{}c)(c\minus{}a)}$
Multiply, use Sonnhard's idea of substitution to get $ \frac{\minus{}10a^2b\minus{}10ab^2}{\minus{}a^2b\minus{}ab^2}\equal{}10$

[/hide]


function
MATHCOUNTS
Euler
number theory
least common multiple
totient function
relatively prime

Can anyone explain it in basic words? :)  Thanks...

First of all, I don't think this belongs in the mathcounts forum...

Carmichael(n) is the smallest positive number m such that for all a with gcd(a,n) = 1, $ a^m$ is congruent to 1 modulo n. Euler-Fermat's Theorem states that if gcd(a,n) = 1, then $ a^{phi(n)}$ is congruent to 1 modulo n, but there might be a smaller m that satisfies that property.

To further pythag011's post, one can prove that $ m | \phi(n)$ where $ \phi$ is the Totient function.

For a particular $ a$ with $ (a,n) \equal{} 1$, define $ m'$ to be the smallest natural number such that $ a^{m'}\equiv 1 \mod n$. Then $ m'$ is called the order of $ a$ modulo $ n$.

[url=http://en.wikipedia.org/wiki/Carmichael_function]More Info on Wikipedia[/url]

In general, mathcounts won't have problems that need euler's theorem.

Using the Carmichael function is the same as CRT with the Totient function, if you don't feel like learning the Carmichael function.

It's merely a definition, and not even a complicated one at that. The chinese remainder theorem is much more complicated than the Carmichael function to explain, and besides, CRT is for numbers that are ALL relatively coprime, whereas in the Carmichael function, we only require for $ n$ to relatively prime to $ a$.

This topic needs to be moved anyways xD

[quote="math154"]Using the Carmichael function is the same as CRT with the Totient function[/quote]
Almost.  The definition of the Carmichael function contains $ \lambda(2^n) \equal{} 2^{n \minus{} 2}$ for $ n \ge 3$, which you don't get by the totient function.  This is because of a lack of primitive roots and gives you an extra factor of $ 2$ in some cases compared to CRT with the totient function.  For comparison:

$ \varphi(10000) \equal{} 4000$
$ \text{lcm}(\varphi(625), \varphi(16)) \equal{} 1000$
$ \lambda(10000) \equal{} 500$.

[quote="resurrection"]CRT is for numbers that are ALL relatively coprime, whereas in the Carmichael function, we only require for $ n$ to relatively prime to $ a$.[/quote]
What do you mean?  You use CRT on the distinct prime power factors of $ n$, which are obviously relatively prime.

I didn't interpret the meaning of that statement correctly  :blush: . Sorry!


Let $ \pi: \mathbb{R} \times \mathbb{R} \rightarrow \mathbb{R}$, $ (x, y) \mapsto \pi(x, y) \equal{} x$.
Let $ A \equal{} (\mathbb{R}^{\geq 0} \times \mathbb{R}) \cup (\mathbb{R} \times \{0\})$.
Prove that $ f \equal{} \pi|_A$ is a quotient map and that $ f$ is neither an open map or a closed map.

[$ \mathbb{R}^{\geq 0} \equal{} \{ x\in \mathbb{R}: x\geq 0\}$]
[This is Munkres §22.3]

Could you clarify what you mean when you say "$ \mathbb{R}^{\ge 0}$"? Do you mean $ \{ x\in \mathbb{R}: x\ge 0\}$?

After a bit of thought, I think you might have written the problem down wrong. As stated, $ \pi_1$ is not a quotient map (the open set $ U \equal{} \{(x,y)\in\mathbb{R}^2: y > 0\}$ when restricted to $ A$ and mapped by $ \pi_1$ yields $ \{x\in\mathbb{R}: x\ge 0\}$, a closed set that should be open if $ \pi_1$ were quotient map). It's not an open map (as you saw, the open set $ U$ mapped into a non-open set), and it is a closed map. This is all assuming the standard topology on both spaces and the subspace topology on $ A$. I'm pretty sure you didn't mean the quotient topology generated by $ \pi_1$, because then it would be an open and closed mapping. So my best guess is that you wrote down the set $ A$ wrong.

The set $ A$ is correct.  Here is how Munkres [Section 22.2] describes the problem:

Let $ \pi_1: \mathbb{R} \times \mathbb{R} \rightarrow \mathbb{R}$ be the projection on the first coordinate.  Let $ A$ be the subspace of all points $ x \times y$ for which either $ x \geq 0$ or $ y\equal{}0$ (or both); let $ q: A \rightarrow \mathbb{R}$ be obtained by restricting $ \pi_1$.  Show that $ q$ is a quotient map that is neither open nor closed.

I popped open my (virtual  :P ) copy of Munkres and confirmed that you are correct, which confuses me. I believe I can prove that $ q$ is a closed map, but I'm too tired to do that now. So I will do the not open and quotient parts:

Before we begin, notice that $ q^{ \minus{} 1}(U) \equal{} \displaystyle A\cap \bigcup_{x\in U} \{x\}\times \mathbb{R}\rightarrow q^{ \minus{} 1}(U)\mbox{ is open iff }\bigcup_{x\in U} \{x\}\times \mathbb{R}\mbox{ is open}$.

[b]$ q$ is a quotient map:[/b] Let $ U\subset \mathbb{R}$. We wish to show that $ U$ is open iff $ q^{ \minus{} 1}(U)$ is open.

-Supposed $ U$ is open. Then $ U$ is the union of intervals $ (a,b)$ (confusing, I know, but you should be able to tell the difference between the intervals and the coordinates fairly easily). Thus it suffices to prove that $ q^{ \minus{} 1}((a,b))$ is open, which is clear since $ q^{ \minus{} 1}((a,b)) \equal{} \displaystyle \bigcup_{x\in (a,b)} \{x\}\times \mathbb{R} \equal{} \{(x,y)\in\mathbb{R}^2: a < x < b\}$.
-Suppose $ q^{ \minus{} 1}(U)$ is open. Then $ q^{ \minus{} 1}(U) \equal{} \displaystyle A\cap \bigcup_{x\in U} \{x\}\times \mathbb{R}$, with $ U$ open. But then $ U$ is open, so $ q$ is a quotient map.

[b]$ q$ is not open:[/b] Let $ B \equal{} \{(x,y)\in\mathbb{R}^2: y > 0\}$. Then $ B\cap A \equal{} \{(x,y)\in\mathbb{R}^2: y > 0,x\ge 0\}$ is open in $ A$. Then $ q(A) \equal{} \{x\in \mathbb{R}: x\ge 0\}$ is not open in $ \mathbb{R}$, so $ q$ is not an open mapping.

Perhaps, it will be useful for someone who happens upon this post, if I make Roosta's proof more explicit (hopefully makes Roosta feel a bit like G. Perelman as well :) ) and fix some inaccuracies. I am quite new to topology, so please point out, if my rewriting contains errors (or perhaps is totally wrong).

I will denote topology on $ A$ by $ \mathcal T_A$.

[b]$ q$ is a quotient map:[/b] Let $ U\subset \mathbb{R}$. We wish to show that $ U$ is open iff $ q^{ \minus{} 1}(U)$ is open.

-Suppose $ U$ is open. Then $ U$ is the union of intervals $ (a,b)$: $ U \equal{} \displaystyle \bigcup_{i\in I}U_i$ where $ I$ is some (possibly infinite) set.

For one interval $ U_i \equal{} (a,b)$, where $ a,b\in\mathbb{R}$: $ q^{ \minus{} 1}(U_i) \equal{} A\cap\displaystyle \bigcup_{x\in (a,b)} \{x\}\times \mathbb{R} \in \mathcal T_A$, since $ \displaystyle \bigcup_{x\in (a,b)} \{x\}\times \mathbb{R} \equal{} \{(x,y)\in\mathbb{R}^2: a < x < b\}\in \mathcal T_{\mathbb R\times\mathbb R}$.

Now for $ U$, we have $ q^{ \minus{} 1}(U) \equal{} \bigcup_{i\in I}\left[A\cap \bigcup_{x\in U_i} \{x\}\times \mathbb R \right]$, which is open by the (T3) axiom.

-Suppose $ q^{ \minus{} 1}(U)$ is open. Then $ q^{ \minus{} 1}(U) \equal{} \displaystyle A\cap \bigcup_{x\in U} \{x\}\times \mathbb{R}$. But $ U$ must be open, as otherwise $ \displaystyle \bigcup_{x \in U} \{x\}\times\mathbb R \notin \mathcal T_{\mathbb R\times\mathbb R}$. $ \blacksquare$

[b]$ q$ is not open:[/b] Let $ B \equal{} \{(x,y)\in\mathbb{R}^2: y > 0\}$. Then $ B\cap A \equal{} \{(x,y)\in\mathbb{R}^2: y > 0,x\ge 0\}$ is open in $ A$. Then $ q(B\cap A) \equal{} \{x\in \mathbb{R}: x\ge 0\}$ is not open in $ \mathbb{R}$, so $ q$ is not an open mapping. $ \blacksquare$

[b]$ q$ is not closed:[/b] Notice $ L \equal{} \{(x,0)\in \mathbb{R}^2: x < 0\}\in \mathcal T_A$, but also $ C \equal{} \{(x,y)\in\mathbb{R}^2: y > 0,x\ge 0\}\in \mathcal T_A$. Furthermore, $ L$ is closed by definition, since $ C\equal{}A\minus{}L$ is open. However, $ q(L) \equal{} \{x\in \mathbb{R}: x < 0\}$ is not closed in $ \mathbb R$. $ \blacksquare$

Notice that your $ A$ is saturated open and the restriction of quotient maps to saturated open sets are still quotient maps.


Determine all strictly increasing functions $ f: R\rightarrow R$ satisfying relationship $ f(x\plus{}f(y))\equal{}f(x\plus{}y)\plus{}2005$
for any real values of x and y.

[hide="partial"]well if you let $ x\equal{}0$, then we have

$ f(f(y))\equal{}f(y)\plus{}2005$, so it follows that one of the functions is $ \boxed{f(x)\equal{}x\plus{}2005}$. Checking this, both sides are $ x\plus{}y\plus{}4010$, so it is indeed one. [/hide]

[quote="bbgun34"][hide="partial"]well if you let $ x \equal{} 0$, then we have

$ f(f(y)) \equal{} f(y) \plus{} 2005$, so it follows that one of the functions is $ \boxed{f(x) \equal{} x \plus{} 2005}$. Checking this, both sides are $ x \plus{} y \plus{} 4010$, so it is indeed one. [/hide][/quote]

I found only this function too.

Since the RHS is symmetric in $ x$ and $ y$, the left hand side must be too. So for all $ x,y \in \mathbb{R}$:
$ f(x\plus{}f(y)) \equal{} f(y\plus{}f(x))$.
Now, since $ f$ is strictly increasing, $ f$ must be injective. So
$ x \plus{} f(y) \equal{} y \plus{} f(x)$
$ f(x) \minus{} x \equal{} f(y) \minus{} y$
So if there is a function that satisfies the equations, it must be of the form $ f(x) \equal{} x \plus{} c$ for some $ c \in \mathbb{R}$. We'll check if all $ c$ satisfy:
$ x\plus{}y\plus{}2c \stackrel{?}{\equal{}} x\plus{}y\plus{}c\plus{}2005$,
so $ c\equal{}2005$. So the only funtion that satisfies the equation is $ f(x) \equal{} x \plus{} 2005$.


advanced fields
advanced fields unsolved

Does anybody has any idea how to solve an eqution like this? I've tried a lot but without result. \[(1-x^{3})y''-x^{2}y'+9xy = 0\]

It looks to me like you could try the method of Frobenius.

http://mathworld.wolfram.com/FrobeniusMethod.html
http://en.wikipedia.org/wiki/Frobenius_method

Yeah, at the first sight it looks the same, but there is differences which I'm not able ot overcome. Maybe somebody have others proposal? I've got solition with series, but still i hope here it's not necessary.


linear algebra
matrix
linear algebra unsolved

$\frac{1}{n!}$

and the proof is a little bit longer...


analytic geometry
geometry proposed
geometry

Let $ ABC$ be a triangle and point $ P$

a) Through midpoint of $ BC,CA,AB$ draw lines parallel to $ PA,PB,PC$ prove that those lines are concurrent at point $ O$.

b) Let $ A'B'C'$ be cevian triangle of $ P$. Through midpoint of $ PA$ draw line parallell to $ BC$, it cuts $ B'C'$ at $ Q$. $ OB\cap A'C'\equal{}\{B_1\},OC\cap A'B'\equal{}\{C_1\}$ .Prove that $ PQ\parallel B_1C_1$

Note that, it is generalization of the problem on the post [url=http://www.mathlinks.ro/viewtopic.php?t=298925&sid=f1e4264187cecc3d714878864e703c7a]Excircles[/url]

For convenience rename $O \equiv U.$ $P_1$ denotes the midpoint of $PA$ and $\ell_a$  the parallel from $P_1$ to $BC.$ Parallels from the midpoints of $BC,CA,AB$ to $PA,PB,PC$ concur at the complement $U$ of $P$ WRT $\triangle ABC.$ Let $(u:v:w)$ be the barycentric coordinates of $P$ WRT $\triangle ABC.$ Then we have

$U:(v+w:w+u:u+v) \ , \ P_1:(2u+v+w:v:w)$

$\Longrightarrow \ell_a: (v+w)x-(2u+v+w)y-(2u+v+w)z=0$

Thus, $\ell_a$ cuts line $B'C' : -vwx+uwy+uvz=0$ at $Q$ with coordinates

$Q: (u(w-v)(2u+v+w):v(wv-uv+uw+w^2):w(-uv+uw-vw-v^2))$

$PQ:  vw(v+w)^2x+uw(uv-uw-vw-w^2)y+uv(-uv+uw-vw-v^2)z=0$

Coordinates of the infinite point $R_{\infty}$ of  line $PQ$ are then

$R_{\infty}:(u(v-w)(v+w):-v(uv-uw+vw+w^2):w(uw-uv+vw+v^2))$

On the other hand, we have

$BU: (u+v)x-(v+w)z=0 \ , \ CU: (u+w)x-(v+w)y=0$

$A'C': vwx-uwy+uvz=0 \ , \ A'B': vwx+uwy-uvz=0$

Thus, coordinates of $B_1 \equiv BU \cap A'C'$ and $C_1 \equiv CU \cap A'B'$ are then

$B_1:(uw(v+w):v(u^2+uv+vw+w^2):uw(u+v))$

$C_1:(uv(v+w):uv(u+w):w(u^2+uw+vw+v^2))$

Hence, coordinates of the  infinite point $S_{\infty}$ of $B_1C_1$ are given by

$S_{\infty}:(u(v-w)(v+w):-v(uv-uw+vw+w^2):w(uw-uv+vw+v^2))$

Which are identical to the coordinates of the infinity  point $R_{\infty}$ of $PQ$ $\Longrightarrow$ $B_1C_1 \parallel PQ.$

Dear Mathlinkers,
for a) we can think to two parallelogic triangles.
Sincerely
Jean-Louis

Dear Mathlinkers,
for part b) I have found some parallels, but I have not manly idea how to resolve this problem synthetically.
Any ideas?
Sincerely
Jean-Louis

[quote="Luis González"]For convenience rename $O \equiv U \ldots$ [/quote]
In fact, we can take $U$ here as any point on the line joining the midpoint of $BC$ and the complement of $P$.

Dear Mathlinkers,
any ideas for part b)?

Sincerely
Jean-Louis

Dear Mathlinkers,

[url=http://jl.ayme.pagesperso-orange.fr/Docs/Deux%20paralleles.pdf] here [/url]

Sincerely
Jean-Louis


Let $\alpha\in(0;1)$, $n \in N$, $n\geq 2$. Suppose $x_i>0$, $x_1x_2...x_n=1$ Prove the inequality
$\frac{1}{(1+x_1)^\alpha}+...+\frac{1}{(1+x_n)^\alpha}\leq \frac{n}{2^\alpha}$

Nice. I have a proof using Jensen's inequality. Is it your proof too or something else?

I am not sure if Jensen works, so could you post the solution with Jensen. I proved this inequality using Karamata and Shturm. I will post my solution in the evening. By the way the problem can be Generalized for case when $\alpha>1$ and $\alpha<0$ (by reversing the sign).

:oops:  Really sorry. I've just noticed that I did not prove it, I had mistakes. I'll try to prove it in another way. Maybe Jensen doesn't work here, as you said. It would be great to see your proof.


calculus
derivative
function
real analysis
real analysis unsolved

Let n be an integer $\ge 1$ and R be a radius > 0, and write B for the open ball $B_{R}(0)$ in $R^{n}$ (reals).

(1) Let $f: B \to R$ (reals) be a $C^{k}$ function $(1 \le k \le \infty )$ such that f(0) = 0.

Prove that there exist $C^{k-1}$ functions $g_{1},g_{2},...,g_{n}: B \to R$ (reals) such that, for all $x \in B$, $f(x_{1},x_{2},...,x_{n}) = \sum_{j = 1}^{n}{x_{j}g_{j}(x_{1},x_{2},...,x_{n})}$.

(2) Let $h_{1},h_{2},...,h_{n}: B \to R$ (reals) be $C^{k}$ functions $(1 \le k \le \infty )$ such that, for all i, j with $1 \le i,j \le n$ we have $\frac{{\partial h_{i}}}{{\partial x_{j}}}(x) = \frac{{\partial h_{j}}}{{\partial x_{i}}}(x)$, for all $x \in B$.

Prove that there exists a unique $C^{k+1}$ function $f: B \to R$ (reals), with f(0) = 0, such that $\frac{{\partial f}}{{\partial x_{j}}}(x) = h_{j}(x)$ for all $x \in B$ and all j with $(1 \le j \le n)$.


Help me solve this problem. I thing it should be solved using dynamic programming but I can' t figurate how?
 You are given a matrix A size mxn(1<=m<=n<=100) such that a(ij) is from the interval [-50, 50]. You need to chose exactly one number from every row such that if the chosen numbers are a(1, j(1)), a(2, j(2)), .. a(m, j(m)) then it is true that j(1)< j(2)< j(3)< ... < j(m). Problem is to minimize the sum of the chosen numbers.

Well, if you let ans[x][y] be the minimum sum you can get starting from row x, and only using column y or higher, then you have ans[x][y] = min(a[x][i] + ans[x+1][i+1]) for each i >= y, right? I'll leave base cases to you..


Hello everyone,

Strangely and unfortunately, my algebra book does not explain what it means when there is a coefficient in front of a logarithm.

For example, when solving for x:

2 log[size=75]5[/size]X = 3

x = 5^(3/2)

x = 5√5

Evidently, 2 is the index of the radical but why would it be so? Could anyone explain?

Thank you very much.

The coefficient simply denotes multiplication.  That is,

$ 2 \log_{5}x = 3 \Leftrightarrow \log_{5}x = \frac{3}{2}$

$ 2$ is the index of the radical because $ a^{ \frac{1}{n}}= \sqrt[n]{a}$.

[quote="catalan"]Thank you very much for your replies.

However, how would you do algebra with logarithms?

For example, solving for x: log[size=75]2[/size]X + 3 log[size=75]2[/size]4 = 5

Thanks.[/quote]

There are a couple of approaches that all give the same answer.

One way is to isolate the logarithm, exponentiate, then simplify using log/exponentiation properties.  It looks like this:

$ \log_{2}x = 5-3 \log_{2}4 = 5-3 \cdot 2 =-1 \implies$
$ x = 2^{-1}= \frac{1}{2}$

The other is to "combine" logarithms using properties first, then isolate.

$ \log_{2}x+\log_{2}4^{3}= \log_{2}(64x) = 5 \implies$
$ 64x = 2^{5}= 32 \implies$
$ x = \frac{1}{2}$

[quote="catalan"]Thank you very much for your replies.

However, how would you do algebra with logarithms?

For example, solving for x: log[size=75]2[/size]X + 3 log[size=75]2[/size]4 = 5

Thanks.[/quote]

You should learn $ \text{\LaTeX}$. It's easier than changing the font size. :wink:

[hide="Step by step solution"]$ \log_{2}x+3\log_{2}4 = 5$

Now, notice that $ \log_{2}4 = 2$, since $ 2^{2}= 4$

$ \log_{2}x+3\cdot 2 = 5$

$ \log_{2}x+6 = 5$

$ \log_{2}x =-1$

Now, we can convert this to exponential form.

$ x = 2^{-1}= \frac{1}{2}$[/hide]

Thank you very much to all those who responded. I understand it now!

I will also look into Latex. Is there another program other than Mathtype which can convert text to Latex?

if you enclose your latex code in dollar signs, then this forum will automatically convert to latex

[code] $ \log_{5} x $[/code]

gives you

$ \log_{5}x$


...and I don't know about you, but I'm pumped.

How are all o'y'all nationals participants feeling now that it's coming down to the final stretch? Especially you first-timers like me! Every spare grain of thought that I have is spent fantasizing about making it to the countdown round and being on ESPN...Every time I do that I have to snap myself back to reality and tell myself "yeah right," but some part of me always says, "Hey, Brian, you thought that same way about making it to the countdown round at states and making it to nationals, but look at you now, [i]Mr. Second-in-the-State![/i]" You just never know, I guess.

Anyway, feel free to post your excitements, concerns, last-minute worries!

I'm hoping to make it to the Countdown Round. 35th place or above is must :P .

there is no doubt I will
[hide][/hide][hide][/hide][hide][/hide]lose[hide][/hide][hide][/hide][hide][/hide]
nationals but still
[hide]top 25 [/hide]
would be good

Don't worry too much about it. Have fun - store the experience away for later reminiscing about the "good old days", when all the (competitions) were open for you to do.

Having said that, I'm really hoping I can't go to sleep the night before. And the night before. I only win when I am really anxious. Why, I don't know, but I like being excited!

Billy

me two
I came second at state because I slept

I get nervous REALLY EASILY but I dont get affected usually except when I cant solve a problem and then I can feel blood pumping to my face and then I cant solve it. And after I finish taking the test I can solve it in like 10 seconds. Its weird. I know Im gonna
Make top 228 :D At least I hope so

LOL Im practing like crazy I did the entire handbook of 1999 last night. PLEAZZ CAN PPL SCAN OLD NATIONALS BEFORE 1999? IF YOU DONT WANT TO GET THIS SITE IN TROUBLE CAN YOU SEND IT VIA PRIVATE EMAIL???????????????????????????????????????????????????????????????????? My email is biffanddoc@gmail.com (PLEASE PLEASE PLEASE PLEASE PLEASE PLEASE PLEASE PLEASE PLEASE PLEASE PLEASE PLEASE PLEASE PLEASE PLEASE PLEASE PLEASE PLEASE PLEASE PLEASE PLEASE PLEASE PLEASE PLEASE PLEASE PLEASE PLEASE PLEASE PLEASE PLEASE PLEASE PLEASE PLEASE PLEASE PLEASE PLEASE PLEASE PLEASE PLEASE PLEASE PLEASE PLEASE PLEASE PLEASE PLEASE PLEASE PLEASE PLEASE PLEASE PLEASE PLEASE PLEASE PLEASE PLEASE PLEASE PLEASE PLEASE PLEASE PLEASE PLEASE PLEASE PLEASE PLEASE PLEASE PLEASE PLEASE PLEASE )

Everyone at nats gets in the top 228
and i'll try to send you old tests

Where'd you find the 1999 handbook?

I'm pretty psyched too, and looking for some old Sprints/Targets/Countdowns to practice speed and accuracy and get used to the time pressure. If anyone watched the ESPN broadcast of the countdown last year, I'm sure you're fantazing about the part where the commentator says, "And now our number one seed from [insert own town, state]"

you can print the screen and then crop it

i want to make countdown so bad. then i can watch tv in class!! june 3rd is a friday

Roffles. I got a 27/7 (could've been a 29/8 if I knew how to read) on 94-95, but it was probably a fluke. I hope not, because I'm gonna need at least that much to get into Countdown.

I'm practicing my eyes out. I did three Nats rounds today. The first didn't really count, because I was half awake :P .

[quote="biffanddoc"]How do I get a pdf into a image???[/quote]
GhostScript will do this (and a whole lot more).  GhostScript is the tool of choice when you want to read a PostScript file, but don't have a PostScript printer.

Be warned that it takes some effort to install and configure GhostScript.

If you have Windows and if you plan to create a lot of TIFF files, you can set up a TIFF printer.  This is even more effort than simply installing GhostScript.  See [url]http://server3.nethost.co.il/set_tif.html[/url].

Good luck.

[quote="Treething"]I'm leaving in like 45 minutes. Good luck, Treething.[/quote]
Good luck!
BTW, what is the time in USA?

(All AM) 10 ET 9 CT 8 MT 7 PT 6 AT 5 HT

Good luck to all, Im leaving in 5 min. (look for Tommy Gill's name on the top 10 *me* :P  :P )

Last post till the competition.

GO DELAWARE!!!!!!! :P  :lol:  :lol:  :lol:  :)  :)  :lol:  ;)

will the problems be online after the contest?


I got this wrong on a test, but think I am right.

A mechanic works 3 days of twelve hour shifts and then has 3 days off. If today is the mechanics third day off, will he be working 51 days from now?

I chose no, because of no remainder and if there was a remainder I would add to today(my off day).

Yes, he will be working. Suppose today is day number 0. Now, he will work on 1, 2, 3. And he won't on 4,5,6. This sequence goes on. NHow, this means that he works on an ODD MULTIPLE OF 3. Moreover, 51=3*17 which is an odd multiple of 3. So, he will work.

The sequence cycles every 6 days, 51/6 has a remainder of three. Because the mechanic will be working 3 days from now, he will also work 51 days from now.

He returns to his current 'place' every 6 days. 51 is 3 mod 6, and 3 days later he should be working, since it is his third day off today.

So he should be working.

day 1 = last day of not working
51 days from day 1 is 1+51 = day 52

since the cycle repeats every 6 days,
1 = last day of not working = 7 = 13 ... = 1mod6
1+6(8) = 49th day is the last day of not working
at this point you can easily count up to 52

49 = no work
50 = work
51 = work
[b]52 = work[/b]
53 = no work
54 = no work
55 = no work
...

Oh wow. I can't believe I didn't see that.


If I'm working on my Dell laptop and accidentally use the left-right scroll feature on my touchpad...TeXnicCenter crashes...

Why is that? It simply got rid of what I had written... any way to recover it?

You won't be able to get your text back unless you saved it. You can tell TeXnicCenter to save every few minutes (Tools->Options->File).
Report the bug to TeXnicCenter's [url=http://sourceforge.net/tracker2/?func=browse&group_id=36323&atid=416430]bug tracker[/url].


calculus
integration
function
calculus computations

hi
if any of youll could post a solution to this integral id be deeply indebted.(also im sorry, but i dont know latex. i thank youll for your patience :oops: )

evaluate integral [sqrt(cos2x)]/sinx

also i was wondering, is it necessary to involve hyberbolic functions or is there a way without them as well????

once again, thanks a lot :lol:

well rishabh indeed u need to use hyperbolic functions :D

so we need
$\int{\frac{\sqrt{cos 2x }}{sin x }}$ $dx$ $=$ $I$ (say)

now $I$ $=$ $\int \sqrt{cosec^{2}x-2}$ $dx$ $=$ $\int \sqrt{cot^{2}x-1}$ $dx$

let $cot^{2}$ $x$ $-$ $1$ $=$ $u^{2}$
this gives ,

$u\frac{du}{dx}$ $=$ $-\sqrt{u^{2}+1 }(u^{2}+2)$

this gives $I$ $=$ $\int \frac{-u^{2}}{\sqrt{(u^{2}+1)^{3}}+\sqrt{u^{2}+1}}$ $du$
now substitute ,
$u^{2}+1$ $=$ $y^{2}$
this gives ,
$udu$ $=$ $ydy$
make the substitution and u wil get
$I$ $=$ $\int-\frac{1}{\sqrt{y^{2}+1}}$ $dy$  $=$ $-sinh^{-1}y$ $+$ $C$

so finally we get,
 $I$ $=$ $-sinh^{-1}(cot x)$ $+$ $C$ where $C$ is the integration constant

[quote="pankajsinha"][quote="pardesi"]well rishabh indeed u need to use hyperbolic functions :D

so we need
$\int{\frac{\sqrt{cos 2x }}{sin x }}$ $dx$ $=$ $I$ (say)

now $I$ $=$ $\int \sqrt{cosec^{2}x-2}$ $dx$ $=$ $\int \sqrt{cot^{2}x-1}$ $dx$

However there is no need to use hyperbolic functions.

Multiply and divide by sqrt((cotx)^2-1)) and then write the numerator as (cosecx)^2-2 and then break as two integrals                                       I_1=((cosecx)^2)/ sqrt((cotx)^2-1)) and I_2=(-2)/sqrt((cotx)^2-1))
For I_1 put t=cotx and for I_2 put u=cosx
Now the two integrals get reduced to standard formulas[/quote][/quote]

well pankajsinha irrespective of our process our answers must be the same so u should have atlest indirectly used the definition of hyperbolic functions

This debate is a bit silly; hyperbolic functions [i]are[/i] trigonometric functions. We could write the answer in many ways, and arrive at it through many paths- the form in the second post is not a particularly useful one.

[quote="pankajsinha"]well guys i do not have knowledge of hyperbolic functions.what are hyperbolic functions[/quote]

well hyperbolic functions are named so becuse when u write the equation of a rectangular hyperbola(right branch) in polar form u need hyperbolic functions ......it has many "uses" in physics and mathematics.please go to the physics forum and see the post shape of a hanging rope.
well u may also look for it in google.well anyways let define them ,..

$cosh$ $z$ $=$ $\frac{e^{z}+e^{-z}}{2}$
$sinh$ $z$ $=$ $\frac{e^{z}-e^{-z}}{2}$

thanx for the replies everyone,but i found that substituting tanx=t,is the least tedious of all the methods. anyway thanks again for the quick replies :D


inequalities
search
trigonometry
inequalities proposed

Let $x,y,z,t$ be the reals such that: $(1+x^2)(1+y^2)(1+z^2)(1+t^2)=16$, then show that:
$-3 \le xy+xz+xt+yz+yt+zt-xyzt \le 5$

It is the problem 66 in  "old and new inequality" book  :lol:  :lol:

Can you post that book please ...  :)

Davron

you can search on google two names 'Gabriel dospinescu' or 'Vasile cirtoaje' ,you can find the pdf file of that book :)  :)

Yeah, this problem has a trig solution, but the proof above is really short and interesting :lol:

[quote="hungkhtn"]Yes, but the solution in this book is so awful. Here's mine

$S=xy+yz+zt+tx+xz+yt-xyzt$
\[ (S-1)^2=((1-xy)(zt-1)+(x+y)(z+t))^2 \le [(1-xy)^2+(x+y)^2][(1-zt)^2+(z+t)^2]=16  \]

So, we've done.[/quote]
Yes,It's also mine.This solution is nice but not new.It's used in a VMO (2000,I think) problem


Our flight was planned to leave Paris CDG at 4:40 PM... We arrive at the airport at 2 PM. We go to the checkin and get a surprise: No more places in the plane : overbooking  :mad: ... Result: we will leave Paris tomorrow at 1 PM, arrive in Mexico at 5 PM and in Merida at 11 PM.
However, Air France will offer us 800 euros because of the overbooking  :lol:.

The Belgian team travels via CDG too on Monday. Maybe we'll get 800 euros too :-).

Mayber, but the first day of the IMO is on Wednesday  ;)

Wait did they give you each 800 euros? Or did they just give back to whoever is paying your tickets back 800 euros? (in which case I don´t understand why you are so happy  :D )

[quote="Igor"]They said that we will get (individually) 800 euros on a plane ticket or 600 euros cash.[/quote]So I guess the French team is going to buy drinks  in Mexico for everyone :first: :D :D


x,y,z belong to  {Q}

find all x,y,z so

both  x+y+z  & 1/x+1/y+1/z  are nutural.

*(I need to know) :(


superior algebra
superior algebra unsolved

Let $ R$ be a commutative ring and $ I,J$ ideals of $ R$. Count $ Tor_1 ^R(R/I,J)$.

I'm having troubles finding any 'good' projective resolution of $ J$  :|

the long exact sequence gives $ Tor^R_1(R/I,J) \equal{} ker(I \otimes_R J \to IJ)$; I don't think that we can simplify this. sometimes it's $ 0$, sometimes it's $ I \otimes_R J$.


function
search
topology
calculus
calculus computations

Why does Holder continuity imply a function $ f: (a,b) \rightarrow\mathbb{R}$ is constant, when the exponent is greater than 1?

By definition, an function with the Holder condition is defined for an open set. Can a continuous function also have the Holder condition on a closed set?


There are N prisoners.The jailer in one of his crazy moods has decided to line them up in a line so that the ith person can look at only the (i-1) prisoners and their heads in front of him.He then places caps on the head(either black or white in colour). He annonces that the prisoners will be allowed to guess out their cap color aloud and if they guess correctly,they will be freed.Prisoners being the selfless geniuses, they want to maximise the number of people escaping. The prisoners start guessing from the back.The prisoners have met beforehand and worked out a stratergy.
1.Work out a stratergy to save atleast N/2 people.
2Work out a stratergy to save atleast (N-1) people.
3.Do the same problem for k colors,instead of two(white and black).
(RMOTC 07-08 Harikrishna Ramani)

2 has a very good soln.

Make the last person say the color of the hat in front of him, and so on.  Only $ 1$ person dies that way.

@007math
That would save only half of the prisoners.
[hide]The Nth person would say (N-1)th person's color.The (N-1)th person would say whatever Nth person said and escape.
(N-2)th person would say (N-3)th person's color.Then (N-3)th person will escape.Likewise (N-5),(N-7)... will all be saved.
this answers Question 1.Not Question 2.[/hide]

well this ques has been posted at joi earlier

i guessed the answer and got it rite :P

but now site is dead and so is my mind  :rotfl:

Strategy to save N-1 People:

Arbitly assign black 0, white as 1.
Last Person sums the colours of all the caps infront of him modulo 2 and shouts out 0 or 1. He dies in the worst case.
Last but one sums the colours infront of him modulo 2 and uses the answer given by the last person to find his own colour.
and so on..

You have got it right. :)
What about k colors?It's just an extension of this .


induction
combinatorics solved
combinatorics

Suppose n players play some ball sport, passing the ball one to each other. Suppose that if that any player A passed already the ball to some player B, it is not allowed to do this again!
The question is how many consequently "passes" can be made taking into account this rule? An easy reasoning gave as an upper bound: n(n-1) - but are we sure that we can make an arrangement to reach this value?
Can you prove or disprove this?

Clearly, this can be achieved as follows by induction on $n$ :

For $n = 2$ let's consider $A-B-A$.
Now assume that it holds for $n$. Now let $A_1, \cdots, A_{n+1}$ be the persons.
Consider the path $P$ with length $n(n-1)$ for just the persons $A_1, \cdots, A_n$. Clearly, this path uses each of the persons.
Now, form the new path by adding the path $A_i -A_{n+1}-A_i$ when $A_i$ is involved in $P$ for the first time.
This leads to a path $P'$ with length $n(n-1)+2n = n(n+1)$ as desired, and we are done.

Pierre.


Five different CD-ROM games are offered, and one game is randomly included with each box, determine the probability of getting all 5 games if 12 boxes are purchased...

my teacher gives the answer about 24%, and the answer of the book says its 18%, but i get something around 67% by programming... i dont no which one is right...XD

From computation, I would have estimated 68%, personally.  The probability that you never get game $ i$ is $ p \equal{} \left(\frac {4}{5}\right)^{12} \approx 0.0687194\ldots$ for each $ i$.  Thus, the probability that there is some game you never get is surely bounded above by $ 5p \approx 0.34359738\ldots$, so the probability that you get all games is bounded below by $ 1 \minus{} 5p \approx 0.65640261\ldots$.  To get the actual answer doesn't require very much modification of this at all: just do inclusion-exclusion.

This question is usually phrased in terms of coupons (the coupon collector problem); one rendition of it was just posted the other day: http://www.artofproblemsolving.com/Forum/viewtopic.php?t=307230  (but the solution there at present gives only a failed attempt at the inclusion-exclusion argument).

BTW, totally not an open question  :wink:

ok, you find the number of ways where you get all five games divided by the total number of ways you can be sent cds.

For the numerator, you always have the 5 different cds, plus 7 more which can be any combination, so we're interested in the number of combinations of the last 7 videos.  Each of those 7 can be any of the 5 categories of movies, so this is a balls and urns problem.  (7+5-1 choose 4)=330 ways

For the denominator, it's the same thing.  12 cds can go into any of the 5 categories.  (12+5-1 choose 4)=1820 ways

330/1820= about 18.13% or 18%

Only if you believe that there's a $ \frac{1}{3}$ probability that a randomly selected two-child family has one son and one daughter, or that it's equally as likely to get 100 tails in your next 100 coin flips as 50 heads and 50 tails.  (Your solution assumes that every composition is equally likely, e.g., that 12 copies of CD #1 is just as likely as 3 copies of CD #1, 3 of CD #2, and 6 of CD #4.  But this is a very different distribution than the one described.)

I think we can use principle of inclusion-exclusion to find the probability that at least one game is missed:
    $ p(x') \equal{} \binom{5}{1}\left(\frac {4}{5}\right)^{12} \minus{} \binom{5}{2}\left(\frac {3}{5}\right)^{12} \plus{} \binom{5}{3}\left(\frac {2}{5}\right)^{12} \minus{} \binom{5}{4}\left(\frac {1}{5}\right)^{12} \approx 0.32$
Thus, $ p(x) \equal{} 1 \minus{} p(x') \approx 0.68$ And the answer your computer program found is correct


quadratics trigonometry algebra polynomial function

What are the roots of x^3 - 3x^2 + 1 = 0?

$ P(x) \equal{} x^3 \minus{} 3x^2 \plus{} 1\Longrightarrow P(1) \equal{} 0$, so you can factor $ P(x)$ into $ P(x) \equal{} (x \minus{} 1)Q(x)$, where $ Q(x)$ is a quadratic expression.

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

you know that if $ f(x_1).f(x_2) < 0$, then a root of $ f(x)$ lies between $ x_1$ and $ x_2$.

$ f(0) \equal{} 1 \\ f(1) \equal{} \minus{}1 \\ f(\minus{}1) \equal{} \minus{}3 \\ f(2) \equal{} \minus{}3 \\ f(3) \equal{} 1$

so, $ f(x)$ has 3 real roots ... each in the intervals $ (\minus{}1,0), (0,1)$ and $ (2,3)$

but im unable to determine the exact value of the roots.

You are right, 3 real roots . . . . After factoring   it, you can find the 3 roots. I know how to use Cardano's Method but quite adamant because of its length. But i have seen several times in this Portal some mathlinkers can factor it with fewer substitution and that's what i am looking for . . . so far no Altheman, Virgil Nicula, Torajrou, Loup Blanc, darij grinberg to look for. They are very active 2-5 years ago in this FORUM. . . It seems they are on the college playground now FREQUENTLY.

Now I got it.

$ x^3 \minus{} 3x^2 \plus{} 1 \equal{} 0\Longleftrightarrow (x \minus{} 1)^3 \minus{} 3x \plus{} 2 \equal{} 0$, let $ x \minus{} 1 \equal{} t$ the cubic equation can be written as $ t^3 \minus{} 3t \minus{} 1 \equal{} 0$ which is realted to Archimedes.

You can set $ t \equal{} 2\cos \theta$, you will get the desired roots.

Good luck! :)

How can we substitute $ t\equal{}2\cos \theta$ until it is not given that $ t$ lies in the interval $ [\minus{}2,2]$...

Plz tell me if i am wrong..

[b]Kunny[/b] are you refering to this identity due to ARCHIMEDES? cos 3t = 4cos^3 t - 3cost.

Wow, i did not know that this identity is the RESULT of Archimedes. Thanks Kunny.

We go back to the above equation, to find the roots of x^3 - 3x^2 + 1 = 0,

factoring is very hard, we resort to substitution as suggested by kunny, (y^3 - 3y + 1 = 0)

and suggested the substitution (t = 2 cos theta).

Let x = 1/y, then (1/y)^3 - 3(1/y)^2 + 1 = 0,

1/y^3 -3/y^2 + 1 = 0, Simplifying

1 - 3y + y^3 = 0 or y^3 - 3y + 1 = 0.

I will use a different notation. Let y = a cos [i]t[/i]

(a cos t)^3 -3(a cos t) + 1 = 0

a^3 cos^3 t - 3a cos t + 1 = 0; multiply both sides by (4/a^3),

4 cos^3 t - (12/a^2) cos t = - 4/a^3,

compare coefficients with the above identity, (12/a^2) = 3;

then a^2 = 4, or a = 2, -2. Use a = 2 first;

then, cos 3t = -4/(2)^3 = -4/8 = -1/2 = -1/2.

cos 3t = -1/2.

3t = arc cos (-1/2).

t = cos 40, cos 80, cos 120 = -cos 20, cos 40, and cos 80

but y = a cos t, then we have

y = -2cos 20, 2cos 40, 2cos 80, but y = 1/x.

Then x = -1/(2 cos 20), 1/(2cos 40) and 1/(2cos 80)

Thus, the roots are -(1/2)sec 20, (1/2)sec 40, and (1/2) sec 40.

Thanks Kunny, oh, Archimedes Identity bashed it [b]RAGHAV GROVER.[/b]

Using that identity to solve cubic equations is essentially appealing to a special case of Cardano's formula where the radicals take a particular form that can be analyzed with de Moivre's.

Actually, that's interesting enough to pose as a separate question: give necessary and sufficient criteria for the roots of a cubic polynomial with integer coefficients to be expressible in terms of trigonometric functions of nice angles.

thanks for the insight on the problem,[b] t0rajir0u[/b]!

[quote="licu-an"]Thanks KUNNY, this is the substitution that i long for . . .[/quote]

You are welcome! :lol:


advanced fields
advanced fields unsolved
set theory
real analysis

Show that there exists a tournament $ (T,\rightarrow)$ of cardinality $ \aleph_1$ containing no transitive subtournament of size $ \aleph_1$. ( A structure $ (T,\rightarrow)$ is a $ \textit{tournament}$ if $ \rightarrow$ is a binary, irreflexive, asymmetric and trichotomic relation. The tournament $ (T,\rightarrow)$ is transitive if $ \rightarrow$ is transitive, that is, if it orders $ T$.)

[i]A. Hajnal[/i]

The solution below essentially assumes CH (the continuum hypothesis).
[b]Claim 1[/b]: Let $G$ be a tournament with countable number of vertices and $x\not\in V(G)$. $V_j \subset V(G)\,,\, j=1,2,\dots$, such that each $V_j$ be infinite. Denote by $G_j$ the correspondent induced tournament on $V_j$ and suppose $G_j\,,\,j=1,2,\dots$ are transitive.
Then one can extend the tournament $G$ on $V\cup x$, such that each of the induced tournaments on $V_j \cup x\,,\, j=1,2,\dots$ is not transitive.
[b]Proof[/b]: Take two arbitrary vertices $x_1, x_2$ of $V_1$ and choose the direction on $x x_1$ and $x x_2$, so that $\{x,x_1,x_2 \}$ is not a transitive tournament. Remove $x_1,x_2$ and proceed in the same way with $V_2$ instead of $V_1$, and so on. As a result $V_j \cup x\,,\, j=1,2,\dots$ cannot be transitive. $\blacksquare$

Let the vertices of $T$ are ordinals $\{ \alpha \mid \alpha<\omega_1 \}$. We proceed defining directions on $\alpha \beta \,,\, \alpha,\beta < \omega_1$ by transfinite recursion. First, we choose arbitrary directions on $\{\alpha\mid \alpha<\omega_0\}$. Suppose $\omega_0 \leq \alpha <\omega_1$ and we've already difined a tournament on $\{\beta : \beta <\alpha\}$. The set of all transitive subtournaments on $\{\nu : \nu \leq \beta\}$ has cardinality $2^{\aleph_0}$. Assuming $2^{\aleph_0}=\aleph_1$ (here we assume CH), one can enumerate these tournaments for fixed $\beta$ as $V_{\beta, \nu}\,,\, \nu<\omega_1$. Consider $\{V_{\beta, \nu} \mid \beta,\nu < \alpha\}$ which is a set of countable many subtournaments on $\{\nu: \nu <\alpha\}$. We apply claim 1 to it, thus assigning a direction on each $\nu \alpha\,,\, \nu<\alpha$. So, it completes the definition of $(T, \rightarrow)$.
It remains to see why $T, \rightarrow$ satisfies the requirement. Note that if $T_1$ is a countable transitive subtournament of $T$, then it's enumerated somewhere as $V_{\beta,\nu}$ for some $\beta,\nu < \omega_1$. Therefore for each $\alpha\,,\, \alpha> \nu,\beta$ , it's impossible $T_1 \cup \{\alpha\}$ to be transitive tournament. It implies $(T, \rightarrow)$ doesn't contain any transitive subtournament of size $\aleph_1$.


algebra
polynomial
Vieta
quadratics
quadratic formula

If $ x$  and  $ y$ are positive integers such that $ 2xy$  divides $ x^2 \plus{}y^2 \minus{}x$,  prove that $ x$ is a perfect square

I think that Vieta-jumping method works :wink:

[hide]

From $ x^2 \plus{} y^2 \minus{} x \equal{} k2xy$ using quadratic formula follows:

$ x \equal{} \frac{2ky\plus{}1 \pm \sqrt{(2ky\plus{}1)^2 \minus{} 4y^2}}{2}$

So we require $ (2ky\plus{}1)^2 \minus{} 4y^2 \equal{} z^2$ for some integer $ z \ge 0$.

But this is just a pythagorean triple: $ (2ky\plus{}1)^2 \equal{} z^2 \plus{} (2y)^2$.

It's is well known, that all the pythagorean triples can be generated in the following way:

$ z \equal{} (m^2 \minus{} n^2)p, 2y \equal{} 2mnp, 2ky\plus{}1 \equal{} (m^2 \plus{} n^2)p$.

But now as $ gcd(2y, 2ky\plus{}1) \equal{} 1$ we require $ p\equal{}1$.

Plugging these in we get $ x \equal{} \frac{m^2 \plus{} n^2 \pm (m^2 \minus{} n^2)}{2}$ from which we conclude $ x\equal{}m^2$ or $ x\equal{}n^2$.

[/hide]

Set $ d\equal{}gcd(x,y)$, $ x\equal{}da$ and $ y\equal{}db$.

The condition becomes:
$ 2abd^2 | a^2d^2\plus{}b^2d^2\minus{}ad$

From this it follows that:    $ d^2 | ad$,            so $ d | a$

Also:
$ ad | b^2d^2$, $ a | b^2d$ but $ gcd(a,b)\equal{}1$, so $ a | d$

Wich implies $ a\equal{}d$ and $ x\equal{}d^2$.

We are done!

 :)


[quote="osman211"]prove this it cant be x^p+y^p=c^p   x,y,z integer  (p =prime number)[/quote]
Actually,this equation has infinitely many solutions :rotfl: .Take for example $ x \equal{} 3,y \equal{} 4,c \equal{} 5$ and $ p \equal{} 2$  :rotfl:
To be serious,you forgot to mention that $ p\geq 3$. :wink:

If you take p=2, then x,y,z are Pythagorian triples...
For p>2, then the equation has no solutions by Last Fermat's Theorem...
 :rotfl:  :rotfl:  :rotfl:

It's a trivial result of the theory of modular representations of elliptic curves.  :P


[quote="asdf2345"]Have any of you ever turned to occult practices just to find something interesting to do?

Does anyone here believe in the existence of demons and witches, and by that I don't mean the witch of agnesi (no witty jokes ppl)

I like to practice dark rituals at night every day just because I find the world an extremely boring place.

Let me hear your thoughts on the matter. What does everyone else here think of the esoteric and arcane?[/b][/quote]

[color=red][size=200]SLAYER[/size][/color]

seriously,...

You're going to hell.

Confess your sins and be saved now.

[quote="Xevarion"]Any God I could imagine would send you two to down below for continually being disrespectful and annoying towards other people before some random person who does weird harmless stuff in his or her room at night.

From an unattached and objective perspective, what difference is there between occult rituals and video games as a pastime for when you're in your room alone at night? Or even reading literature or math? I don't see any moral distinction; the difference we intuitively feel when we think about it casually arises from the associations we make with each activity.[/quote]

[b]This[/b] is why AoPS gets few new members.

That's just like saying "I see no difference in slaughtering cows for hamburger and murdering humans, at least if they're murdered humanely it shouldn't make a difference." It's moral distinctions and associations that make up the many interconnected webs that support society.

I suppose most people here are pretty down to earth and take a rational viewpoint on the world, maybe because of the influence of mathematics and science? thats just a conjecture...

but this forum needs some ingenuity and fresh concepts, don't you agree? I more or less practice occultism simply out of curiosity as opposed to fervently believing in the existence of supernatural beings and such

its more or less just a way to pass time, no offense to anyone here who actually practices such rituals with seriousness; if anyone does, please post and tell me your thoughts on the matter, as i would love to know more about it

[quote="Treething"][b]This[/b] is why AoPS gets few new members.[/quote]
Maybe this is why AoPS gets few new RT/GFF spammers. New members post in the actual math forums often. But I honestly don't think my personal views drive anyone away from AoPS, and I don't know why you would say that.

[quote="Treething"]That's just like saying "I see no difference in slaughtering cows for hamburger and murdering humans, at least if they're murdered humanely it shouldn't make a difference." It's moral distinctions and associations that make up the many interconnected webs that support society.[/quote]
How does occultism harm anyone? I don't see the relevance of your analogy. I am a strong adherent to Mill's principle of personal freedom to the extent of refraining from harming others, and this principle clearly distinguishes between your analogy and my question.

[quote="Treething"]So how many well-adjusted occult people do you know?[/quote]
I don't know any people who engage in occult practices well. Therefore I don't want to pass unjustified judgment on them. As far as I can tell by the information available to me, there's nothing wrong with this guy [b]asdf2345[/b], for example.

[quote="Treething"]So how many well-adjusted occult people do you know?[/quote]

When will people learn that this sort of statement is totally utterly moot??

Correlation does not imply Causality. :maybe: Durka Der


First of all, congratulations on finally redoing the site.

But I have some minor little aesthetic issues. AoPS has never looked wonderful on my laptop screen, but now the fonts look really nasty. There are somewhere around 5 different font sizes visible at one time: the sidebar size, the forum path, the forum title, the Post a new topic text, the normal text, and the Message Body/Poll/Add Attachment thing. It really isn't very pretty.

Next, the popout menus: The little blue lines between the popout menu links aren't exactly that spectacularly beautiful. If it would be possible to take out those lines, or add them to the sidebar links, it would look smoother, IMO.

And the Math Jam dropdown menu on top makes the whole top bar change size, which is disconcerting. I'm suprised bookaholic hasn't complained yet; maybe she hasn't seen it yet. In any case, it's not smooth. I think the problem is the padding on the sides of the dropdown menu links; if you reduced the padding, it would be fine.

Finally, and this is not quite as important, the View More Emoticons link still makes a new window. It would be nice if all the emoticons were on one page, and the View More Emoticons exposed a hidden <div> or <iframe> with the rest of the emoticons, so we didn't have to close the new windows. :)

Billy

The hidden div solution implies downloading all the emoticons at once, when it's not really necessary in 95% of the time. There are many users with dial-up or other such slow connection and I see no point in delaying them further using something like a hidden div (plus it would be screwed up on IE as pretty much all hidden CSS stuff is :) ).

Dave will probably take care of the rest of the things.

I'm not sure if this is a bug, but this wierd window poped up while I was at http://www.artofproblemsolving.com/Forum/viewtopic.php?p=494541
I got the screen shot but the attachments function says I have reached my limit of 0 MB(:huh:).
Anyways, this is what the window said:

[b]GENERAL ERROR[/b]

SQL requests not
achieved.

[u]DEBUG MODE[/u]

SQL Error: 2006
MySQL server has
gone away

SQL Request: select
user_id from
ad_permissions
where user_id=
18217 and
is_site_admin=1

Line:23
File:
functions_aops.php

It was all exactly in that format in a rectangular window. And there was a blue rectangle under the words GENERAL ERROR. That's the best I can describe it :).

Just means the server can't cope with the hard labour. Don't worry, nothing at hand. :)


prove that
$m^n=\sum_{k=0}^{n}k!S(n,k)\binom{n}{m}$.

where S(n,k) is the sterling numbers typeII.

Hi.

I'm sure you mean [tex]m\choose k[/tex] and not [tex]n \choose m[/tex], don't you?

The LHS is obvously the number of maps from [tex]\{1,2,...,n\}[/tex] to [tex]\{1,2,...,m\}[/tex]. Now, let us count the number of maps [tex]f:\{1,2,...,n\}\to\{1,2,...,m\}[/tex] for which we have [tex]\vert Im(f)\vert=k[/tex]. For constructing such a map, we need to choose the [ŧex]k[/tex] different elements of [tex]\{1,2,...,m\}[/tex] which are in the image of [tex]f[/tex]. This can be done in [tex]n\choose k[/tex] ways. Now, we have to partition the set [tex]\{1,2,...,n\}[/tex] into [tex]k[/tex] subsets so set whenever two elements [tex]i,j\in\{1,2,...,n\}[/tex] are in the same subset of the partition, their image under [tex]f[/tex] is the same. This step can by done in [tex]S(n,k)[/tex] ways. Finally, every subset of the partition is to be assigned to one of the [tex]k[/tex] numbers we have chosen from [tex]\{1,2,...,m\}[/tex]; this can be done in [tex]k![/tex] ways. Multiplying the number of ways in we can do each of the steps and afterwards summing up the values for [tex]k=1,...,n[/tex], we get [tex]m^n=\sum_{k=1}^{n} S(n,k) k! {m\choose k}[/tex].

Hanno


geometry
geometric transformation
conics

[hide="English version"]I would like all the people in this [b]sub[/b]forum to understand that the official language here is Romanian, not English. There are users which do not understand English, or are poor at it, so in order to encourage them to post, we need to keep a stricly Romanian syntax (or at least provide an Romanian translation), except very very special cases.

If you find a problem nice that it is posted in Romanian here, please repost it on the Olympiad section or the College Playground in english and give a link there to the Romanian topic and vice-versa.

Thank you.
[/hide]
[hide="versiunea Romana"]As vrea ca toata lumea sa inteleaga faptul ca acesta este un [b]sub[/b]forum in care limba oficiala este Romana, nu Engleza. Exista utilizatori ai site-ului care nu inteleg limba Engleza, sau o inteleg foarte greu, si pentru a ii incuraja si pe acestia sa raspunda la probleme, este necesar sa incercam sa ne limitam la sintaxa in limba Romana (sau cel putin sa existe si o versiune tradusa in limba romana a postului), exceptand evident cazurile foarte foarte speciale.

Va multumesc.
[/hide]

[quote="cezar lupu"]Sunt total de acord cu tine, dar ce sa fac cand raspund unii utilizatorii in limba engleza? :?
 Mi se pare un pic anormal ca sa le raspund in limba romana :?[/quote]Raspunzi in Romana. Nu e nimic anormal in asta (in [b]acest[/b] subforum).

[quote="mihai miculita"]N-am prea inteles cu ce difera acest subforum de cal pe care am intrat pana acuma?
Tocmai am raspuns (probabil pentru cei care inteleg lb. romana) la o problema nerezolvata
de pe forumul in lb engleza. O inteleg tot vorbitorii de lb. romana, si tot n-am raspuns solicitantului!
Sau poate serveste drept intermediar (in cazul in care considera ca merita) moderatorul?
M. Miculita[/quote]

Probabil va referiti la topicul: [url]http://www.mathlinks.ro/viewtopic.php?t=161923[/url]

In orice caz, limba "oficiala" in toate sub-forumurile de pe site, mai putin [b]National & Local Communities[/b], este limba engleza.  De obicei, atunci cand userii posteaza un mesaj intr-o alta limba pe aceste sub-forumuri, ei sunt rugati sa il editeze in limba engleza sau acestea vor fi sterse. Intrucat probabil nu mai puteti sa vi le editati, am sa vi le editez personal.

PS: Nu am inchis topic-ul din cauza folosirii limbii romane, ci asa cum am precizat problema mai fusese discutata anterior la [url]http://www.mathlinks.ro/Forum/viewtopic.php?p=21311#21311[/url].

M-am referit la topicul: Solutia unei probleme nerezolvate de pe FORUMUL in lb. engleza;
in care dau solutia problemei P3 de la olimpiada Austria_polonia 2002.
M.M.

Legat de conjectura:
ABC and A1B1C1 are simetric respect to a point (daca am inteles bine, acesta este locul unde trebuie sa intervin):
------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------
Let ABC a triangle and P a point in the same plane, A1B1C1 is the simetric of the triangle ABC respect to the point P, the point A belongs to the line La, and similarly Lb, Lc, suppose that La,Lb and Lc concur, define A2 as the intersection of the lines La and B1C1, and similarly B2, C2, prove that A2, B2 and C2 are collinear.
 ------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------
Conjectura din aceasta problema este FALSA!.
v. in acest sens desenul facut cu un program de desenare(euklides.exe) din ATTACHMENT(aici gasiti si enuntul in lb. romana!).

Am adaugat observatia dumneavoastra. Vedeti [url]http://www.mathlinks.ro/viewtopic.php?p=917391#917391[/url].

Dupa ce i-am semnalat, autorului, eroarea din enuntul problemei si autorul a corectat enuntul
in sensul ca: AA2||BB2||CC2, acuma se cade sa-i rezolv aceasta noua problema!

v. solutia mea din ATTACHMENT.

Scuze!,
topicul "Re Bertrand asumption" de pe forumul in limba romana,
am vrut sa-l postez aici! Fiind un raspuns la o solicitare de pe forumul
in lb. engleza.

Pe subforumul GEOMETRY/Geometry Proposed & Own Problems/
la topicul:       angle bisectors and 6 point on a conic,

este enuntata problema urmatoare:
      "in a triangle 3 concurrent cevians form 6 angles and the angle bisectors of these meets the side lines on 6 point. Prove that these 6 points are on a conic."

In ATTACHMENT gasiti enuntul reformulat(in sensul ca am introdus niste notatii) in lb. romana, o figura si solutia pe care eu am dat-o acestei probleme.
Mihai Miculita, ORADEA.

Aceeasi problema mai simplu!

v. ATTACHMENTUL nou!

Nu ştiu cât de on sau off topic este, dar nu am găsit un alt loc mai potrivit pentru următoarea rugăminte: poate să-mi spună cineva care este corespondentul în română al noţiunii "square integrable differentials"?


[quote=chien than]For $a;b;c$ are real numbers .Prove that:
$\frac{1}{(2a-b)^{2}}+\frac{1}{(2b-c)^{2}}+\frac{1}{(2c-a)^{2}}\geq \frac{11}{7(a^{2}+b^{2}+c^{2})}$[/quote]
After substitution $2a-b=x$,... use $uvw$.


trigonometry
analytic geometry
geometry
IMO Shortlist
complex numbers

Let $ABCDEF$ be a convex hexagon such that $\angle B+\angle D+\angle F=360^{\circ }$ and \[ \frac{AB}{BC} \cdot \frac{CD}{DE} \cdot \frac{EF}{FA} = 1.  \] Prove that \[ \frac{BC}{CA} \cdot \frac{AE}{EF} \cdot \frac{FD}{DB} = 1.  \]

just use complex numbers. it's fairly easy.

Is there another proof for this problem please post it if you have ... Darij Grinberg do you have any fantastic proof for this problem or any ides RESPECT  :roll:   :?:

Davron

Let $ABCDEF$ be the labelled in clockwise order. Let $A$ be represented by the complex number $a$ from an arbitary origin, $B$ be represented by $b$ and so forth. Also let $\angle A = \alpha$, $\angle B = \beta$ and so forth. Therefore:
\begin{eqnarray*} (b - a) \times e^{i\beta} \times \frac{|CB|}{|AB|} &=& (b - c) \\ (d - c) \times e^{i\delta} \times \frac{|ED|}{|CD|} &=& (d - e) \\ (f - e) \times e^{i\zeta} \times \frac{|AF|}{|EF|} &=& (f - a) \\ \therefore (b - a)(d - c)(f - e) \times e^{i(\beta+\delta+\zeta)} \times \frac{BC.DE.FA}{AB.CD.EF} &=& (b - c)(d - e)(f - a) \\ \therefore (b - a)(d - c)(f - e) &=& (b - c)(d - e)(f - a) \\ \therefore -(ace + adf - acf - ade - bce - bdf + bcf + bde) &=& abe + acd - abd - ace - bef - cdf + bdf + cef \\ \therefore -(adf - acf - ade - bce + bcf + bde - abf + cde) &=& abe + acd - abd - bef - cdf + cef + abf - cde \\ \therefore -(f - e)(c - a)(b - d) &=& (c - b)(f - d)(e - a) \\ \therefore |f - e||c - a||b - d| &=& |c - b||f - d||e - a| \\ \therefore EF.AC.DB &=& BC.DF.AE \\ \end{eqnarray*}

i dont kidding :D

many years ago , i have solve it by inversion...

u know i didnt want to solve it by that.

but the problem was in my inversive problems notes.

these years i have exams is school and cant come to ML too.

i will promise to send its proof later.

i only posted these sentenses to say that there exists an inversive proof for it. ;)

Hi my dears. I'm here, because I want to post my solutions for this nice, and may be old($1998$) problem.

First I should write the invrsive proof as I promised recently.

Ok…………………………………………………………..

First consider this fact that the circum circles of triangles $ABC, CDE, EFA$ have a common point $P$.

Because, if $P$ is the intersection of circum circles $CDE, EFA$ , then:

$\angle EPC=180-\angle D$

$\angle APE=180-\angle F$

then: $\angle APC=360-(\angle D+\angle F)=\angle B$

and it means that $P$ is on circum circle of $ABC$, too.

Now…………………………………………………….

I will invert this shape with the center $P$ and an arbitrary radius. circum circles of $ABC, CDE, EFA$,

Changes to three line, which each two of them intersects each other at inversion form of $A,E,C$ means

$A',E',C'$. and if $K$ is the radius of inversion,:

$A'B'=K.\frac{BC}{PB.PC}$.

Thus:  $\frac{A'B'}{B'C'}=\frac{AB}{BC}.\frac{PC}{PA}$.and with a similar way we can conclude that:

$\frac{C'D'}{D'E'}=\frac{CD}{DE}.\frac{PE}{PC}$

$\frac{E'F'}{F'A'}=\frac{EF}{FA}.\frac{PA}{PE}$

and if we multiple these equalities we should have:

$\frac{A'B'}{B'C'}.\frac{C'D'}{D'E'}.\frac{E'F'}{F'A'}=(\frac{AB}{BC}.\frac{CD}{DE}.\frac{EF}{FA}).(\frac{PC}{PA}.\frac{PE}{PC}.\frac{PA}{PE})= 1$

	Then according to Menelaus theorem, for triangle $A'C'E'$, the points  $F',D',B',$ should be collinear.

And according to this theorem and for the triangle $A'B'F'$ and the line $E'D'C'$ we know:

$\frac{A'C'}{B'C'}.\frac{B'D'}{D'F'}.\frac{E'F'}{A'E'}=1$

there for:

$\frac{AC}{AE}.\frac{BD}{DF}.\frac{EF}{BC}=(\frac{AC}{BC}.\frac{PB}{PA}).(\frac{BD}{DF}.\frac{PF} {PB}).(\frac{EF}{AE}.\frac{PA}{PF})=\frac{A'C'}{B'C'}.\frac{B'D'}{D'F'}.\frac{E'F'}{A'E'}=1$

and we conclude that: $AC.BD.EF=AE.DF.BC$.

ooooooooooooooooooooooooooofffff, finally it is proved by $INVERSION$.

Really an enjoyable problem.

Very nice, very nice, congratulations to the designer of this problem.

Might have the same basic ideas as the inversion solution, but here's a purely synthetic way:

Let P be the point such that EDP ~ EFA. Note that since $\angle B+\angle D+\angle F = 360, \angle CDP = \angle CBA$. Also, $\frac{PD}{CD}= \frac{AF}{CD}\cdot \frac{DE}{EF}= \frac{AB}{BC}$. Hence CDP ~ CBA. Now using spiral similarity,

$BD \cdot \frac{AC}{BC}= AP = FD \cdot \frac{AE}{FE}$

$\frac{BD \cdot AC \cdot FE}{BC \cdot FD \cdot AE}= 1$

[quote="felixx"][quote="probability1.01"]Might have the same basic ideas as the inversion solution, but here's a purely synthetic way:

Let P be the point such that EDP ~ EFA. Note that since $ \angle B \plus{} \angle D \plus{} \angle F \equal{} 360, \angle CDP \equal{} \angle CBA$. Also, $ \frac {PD}{CD} \equal{} \frac {AF}{CD}\cdot \frac {DE}{EF} \equal{} \frac {AB}{BC}$. Hence CDP ~ CBA. Now using spiral similarity,

$ BD \cdot \frac {AC}{BC} \equal{} AP \equal{} FD \cdot \frac {AE}{FE}$

$ \frac {BD \cdot AC \cdot FE}{BC \cdot FD \cdot AE} \equal{} 1$[/quote] can anyone explain how do you choose point P with more details? I'm not how does it imply thet angles CDP and CBA are equal...[/quote]

It works if the point $P$ is chosen to be outside of the convex polygon near vertex $D$. Since $\angle EDP+\angle PDC+\angle CDE = 360^\circ$, it follows that $\angle PDC= 360^\circ - \angle EDP - \angle CDE = 360^\circ - \angle EFA - \angle CDE = \angle ABC$.

[quote="al.M.V."]err...i'm not darij, and my proof isn't fantastic, but i see a solution with sine law and...that's it.[/quote]

Well...its easy to write like that.Please post the proof al.M.V.To be fair,I don't think there's a proof using the same....and did you do all the calculations.

Ashegh's proof gives us the glimpse of how powerful inversion can be.Well done incredible Ashegh!!

Let $a,b,c,d,e,f$ be the coordinates of $A,B,C,D,E,F$ in the complex plane.By the problem we have
\[ \frac{a-b}{b-c} \cdot \frac{c-d}{d-e} \cdot \frac{e-f}{f-a}=-1. \]
We also have the identity $(a-b)(c-d)(e-f)+(b-c)(d-e)(f-a)=(b-c)(a-e)(f-d)+(c-a)(e-f)(d-b)$ which can be verified by expanding the terms of LHS ans RHS,so we get \[ \frac{b-c}{c-a} \cdot \frac{a-e}{e-f} \cdot \frac{f-d}{d-b}=-1. \] Taking modulus on both sides does the job...

I think this problem is made for using affixes.In my opinion the problem creators could not think of an inversive proof.

[b]sayantancha...[/b], why the first relation happens?
you know that the product of the modules is $1$.
How you deduce that the product of the explicitates is $-1$ ?
And where you involve the angle condition?
Please tell me.

i think i see a proof with jacobi and trig ceva

[quote=sayantanchakraborty]Let $a,b,c,d,e,f$ be the coordinates of $A,B,C,D,E,F$ in the complex plane.By the problem we have
\[ \frac{a-b}{b-c} \cdot \frac{c-d}{d-e} \cdot \frac{e-f}{f-a}=-1. \]
[/quote]

Why do we have this thing and how is the angle condition used here?

[quote=butter67]A beautiful synthetic solution can be found if one sees that the hexagon "almost" tesselates the plane. I can post it if someone requests it. :)[/quote]

Would you please post it? :)

[quote=Taha1381][quote=Valentin Vornicu]this problem was long forgotten, until bugzpoder attained my attention to it. let me give a more detailed hint :)

the condition about the sum of the angles show that the number

N = (a-b)(c-d)(e-f) / (c-b)(e-d)(a-f) is real, where a,b,c,d,e,f are the affixes of their respective vertices denoted with capital letters.

but this means nothing else than N=AB.CD.EF / BC.DE.FA ... I will leave you to make the computations (and judgements) from here on ..[/quote]

Could you explain your answer a bit more?How did you get the condition equelent to that?[/quote]

It's obvious, since for every real number, the argument is congruent to 0 modulo pi

This task appeared on the final of $L$ Polish Math Olympiad.

WLOG $ABCDEF$ clockwise. The problem's conditions translates to $\frac{a-b}{b-c}\cdot \frac{c-d}{d-e}\cdot \frac{e-f}{a-f}=1$. Let $z=\frac{b-c}{c-a}\cdot \frac{a-e}{b-d}\cdot \frac{d}{e}$; we desire to prove $|z|=1$. Now, WLOG $f=0$, so the problem's condition is equivalent to $e=\frac{ad(b-c)}{ab+bd-ad-bc}$, and $z=\frac{b-c}{c-a}\cdot \frac{a-e}{b-d}\cdot \frac{d}{e}$. But substituting our expression for $e$ into this, we get $z=-1$, so $|z|=1$, as desired.

The problem dies immediately if you know this (note: this is a Chinese Handout)
[url]https://wenku.baidu.com/view/e0239e01eff9aef8941e061e[/url]

i stumbled upon the solution with literally no intention to solve it - the problem was there, and now its gone.  maybe i shouldnt do oly problems with no intention of actually solving and just playing around with them anymore now that i think about it

Throw $ABCDEF$ on the complex plane.  The given two conditions can be succinctly rewritten as $$\frac{c-b}{a-b}\cdot\frac{e-d}{c-d}\cdot\frac{a-f}{e-f}=1.\qquad(\spadesuit)$$It is sufficient to prove that $$\color{blue}\frac{b-c}{a-c}\cdot\frac{a-e}{f-e}\cdot\frac{f-d}{b-d}=1.$$Repeatedly using $(\spadesuit)$, \begin{align*}\frac{b-c}{a-c}\cdot\frac{a-e}{f-e}\cdot\frac{f-d}{b-d} &= \frac{(a-b)(c-d)(a-e)(f-d)}{(a-c)(e-d)(a-f)(b-d)} \\ &= \frac{(d-e)(b-c)(c-d)(a-e)(f-d)}{(a-c)(e-d)(f-c)(d-c)(b-d)} \\ &= \frac{(d-e)(c-d)}{(e-d)(d-c)} = 1,\end{align*}done. $\square$

This is very straightforward with complex numbers. The given conditions translate to $\frac{(a-b)(c-d)(e-f)}{(b-c)(d-e)(f-a)} = 1$ precisely. But this implies $$(b-c)(a-e)(d-f) + (c-d)(e-f)(b-d) = (a-b)(c-d)(e-f)-(b-c)(d-e)(f-a) = 0,$$ so $$\frac{(b-c)(a-e)(d-f)}{(c-d)(e-f)(b-d)} = -1,$$ which finishes.

We use complex numbers. Note that
\begin{align*}
0^\circ &= \angle ABC+\angle CDE+\angle EFA \\
&=\text{arg}\left(\frac{a-b}{c-b}\right)+\text{arg}\left(\frac{c-d}{e-d}\right)+\text{arg}\left(\frac{e-f}{a-f}\right) \\
&=\text{arg}\left(\frac{(a-b)(c-d)(e-f)}{(c-b)(e-d)(a-f)}\right)
\end{align*}
So $\tfrac{(a-b)(c-d)(e-f)}{(c-b)(e-d)(a-f)}$ is a positive real number. However, we have
\begin{align*}
1 &= \left(\frac{AB}{BC} \cdot \frac{CD}{DE} \cdot \frac{EF}{FA}\right)^2 \\
&= \frac{|a-b|^2|c-d|^2|e-f|^2}{|c-b|^2|e-d|^2|a-f|^2} \\
&= \frac{(a-b)(c-d)(e-f)}{(c-b)(e-d)(a-f)} \cdot \overline{\frac{(a-b)(c-d)(e-f)}{(c-b)(e-d)(a-f)}}
\end{align*}
Since $\tfrac{(a-b)(c-d)(e-f)}{(c-b)(e-d)(a-f)}$ is a real number, its conjugate is equal to itself, so it is equal to $1$. We can expand this to
\[ace-acf-ade+adf-bce+bcf+bde-bdf=cea-cef-cda+cdf-bea+bef+bda-bdf\]
From both sides, subtract $ace-bdf$ and add $cde-abf$. Move $bde$ and $acf$ from the left to the right, and move $cdf$ and $bea$ to the left. We get
\[bce-cde-bcf+cdf-abe+ade+abf-adf=-abf+abd+bef-bde+acf-acd-cef+cde\]
which implies $(c-a)(e-f)(b-d)=(b-c)(a-e)(d-f)$ and we are done.


geometry
geometric transformation
dilation
ratio
geometry proposed

Let $ ABC$ be a triangle and let $ X_1X_2$ be the tangent to the incircle of $ ABC$ and parallel to $ BC$, and in the same way $ Y_1Y_2$ the parallel to $ AC$ (tangent to the incircle) and $ Z_1Z_2$ the parallel to $ AB$ (tangent to the incircle). If $ r_a, r_b,r_c$ are the radius of the incircles of $ AX_1X_2$, $ BY_1Y_2$, $ CZ_1Z_2$ respectively and $ r$ the radius of the incircle of $ ABC$ , prove that $ r_a \plus{} r_b \plus{} r_c \equal{} r$

This problem is not so difficult.

The dilations transform $ \Delta AX_{1}X_{2} , \Delta BY_{1}Y_{2}, \Delta CZ_{1}Z_{2}$ to $ \Delta ABC$ with the ratio $ k_a, k_b, k_c$
Then $ \frac{r_a \plus{} r_b \plus{} r_c}{r} \equal{} k_a \plus{} k_b \plus{} k_c$
We have $ k_a \equal{} \frac{p\minus{}a}{p} , k_b\equal{}\frac{p\minus{}b}{p} , k_c \equal{} \frac{p\minus{}c}{p}$
And the result follows.

Nice solution mathangel...I know it was not so difficult :blush:


geometry
trigonometry
circumcircle
geometry proposed

[quote="jayme"]Dear Mathlinkers,
let X be the meet point of HD and AC
1.  A, D, I and X are cocyclic; let 0 be this circle.
2.  BD is orthogonal to DX. (http://perso.orange.fr/jl.ayme  vol.1 La droite d'Euler est perpendiculaire à l'axe orthique p. 1).
3.  Consider triangle ADI  and the point B ; triangle EHK and the point A.
4.  Accordling to the Maxwell's theorem (ADI and EHK are parallelogic), HK // BI….
I think that my terminologie is correct?
Sincerly
Jean-Louis[/quote]
Dear Jean-Louis,
Can you give us more details ? I don't see any proofs in your post.

Dear thaithuan_GC and Mathlinkers, you can see at [url]http://www.mathlinks.ro/viewtopic.php?t=216987[/url] in order to find out a solution.

[quote="mr.danh"]Let a triangle ABC, H is its orthocenter, AH meets BC at I. Let D is the symmetric point of I wrt BH. The perp line to CH at H meets AD at E. The perp line from E to HD meets AC at K. Prove that HK is parallel to BC.[/quote]
[quote="thaithuan_GC"]A very nice problem and an excellent solution !
Dear mr.danh and other Mathlinkers, can you post full proof and the relation between these problems. They are really results of creative power.[/quote]
full solution(used lemma of mr.danh at [url=http://www.artofproblemsolving.com/viewtopic.php?t=216987]here[/url] which was solved by Petry )

$S_{XYZ} $ is area of $\triangle XYZ$
$CH$ cuts $AB$ at $Q$,$BH$ cuts $AC$ at $P$.
$X$ is on $AC$ such that $P$ is midpoint of $CX$,$BX$ cuts $PQ$ at $D'$,$QX$ cuts $AD'$ at $Z$
$E'$ is on $AD'$ such that $CE'$ is parallel to $QX$
$C,P,Q,B$ are concyclic $\Longrightarrow \angle AQD'=\angle ACB =\angle D'XP \Longrightarrow A,X,Q,D' $ are concyclic
$ \Longrightarrow \frac{QZ}{XZ}=\frac{S_{AZQ}}{S_{AZX}}=\frac{QA.sin\angle ZAQ}{XA.\sin\angle XAZ}=\frac{QA.D'Q}{XA.D'X} $ $(1)$
apply Menelaus theorem in $\triangle AXB$ (with $P,D',Q$ is colinear)
$ %Error. "Longright" is a bad command.
\frac{QA}{QB}*\frac{D'B}{D'X}*\frac{PX}{PA}=1 $ $(2)$
$\triangle AXB $ is similar with $ \triangle D'QB$
$ \Longrightarrow \frac{AB}{D'B}=\frac{AX}{D'Q} $ $(3)$
from $(1),(2),(3)$ $\Longrightarrow \frac{QZ}{XZ}=\frac{PA*QB}{PX*AB}=\frac{PA*QB}{PC*AB}$ $(4)$
$\frac{HQ}{HC}=\frac{S_{BHQ}}{S_{BHC}}=\frac{BQsin\angle HBQ}{BCsin\angle CBH}=\frac{BQ\frac{AP}{AB}}{CP}$ $(5)$
from$(4),(5)$ $HZ$ is parallel to $AC$
$HZ$ cuts $CE'$ at $T$,$E'H$ cuts $AC$ at $S$
$\frac{SA}{SC}=\frac{ZH}{HT}=\frac{HQ}{HC} \Longrightarrow SH $ is parallel to $AB \Longrightarrow \angle CHE'=90^{\circ} \Longrightarrow E'\equiv E $
$M$ is on $BH$ such that $P$ is midpoint of $HM$ $\Longrightarrow M $ is on $ \odot ABC \Longrightarrow \angle MAC=\angle MBC=\angle MCX \Longrightarrow BX $ is perpendicular to $AM$ at $Y$
$V$ is on $AH$ such that $XV$ is perpendicular to $AH$
$AX.AP=AY.AM=AV.AH $ and $ AM=AH \Longrightarrow AY=AV $ $ (6) $
$BQ.BA=BD'.BX=BI.BC$ and $BX=BC\Longrightarrow BD'=BI \Longrightarrow \frac{BD'}{BX}=\frac{BI}{BC} %Error. "Longright" is a bad command.
D'I $ is parallel to $AC$ and $D'$ is on $BX\Longrightarrow D' \equiv D $
we have
$\frac{AY}{AM}=\frac{AV}{AH} $ (used $ (6) $ ) $ =\frac{AV}{AI}\frac{AI}{AH} =\frac{AX}{AC}\frac{AD}{AZ}=\frac{AD}{AE}$(because $CE$ is parallel to $XZ$)
$ \Longrightarrow XD \parallel ME \Longrightarrow \angle EMA=90^{\circ}$
but $ \angle HDB=\angle HIB=90^{\circ} \Longrightarrow HD\parallel AM $ $\Longrightarrow EM $ is perpendicular to $HD \Longrightarrow \angle AMK =90^{\circ} $
$ K'$ is on $AC$ such that $HK' \parallel BC \Longrightarrow \angle K'HM=\angle MBC=\angle MAK' \Longrightarrow A,M,K',H $ is concyclic
so $ \angle AMK' = 90^{\circ} \Longrightarrow K'\equiv K $
so $ HK \parallel BC $

It is a great problem which has been passed into oblivion .

Let the circumcircle of triangle $ EHK $ intersects $ AC $ twice at $ T $ .
Consider triangle $ TEA $ and the pedal triangle of $ D $ w.r.t. $\Delta ABH $ , $ XYZ $ .
(where $ X $ corresponds $ A $ , etc . )

We see that from
$ \Delta YDZ $ being similar to $ \Delta EHA $  and
$ \Delta ZDX $ to $ \Delta AHT $ ,

$ \Delta EHT $ is similar to $ \Delta YDX $
so $ \angle AKH = \angle HET = \angle DYX = \angle C $ .

Therefore , $ HK $ is parallel to $ BC $ .

Let HD meet BC at L. Let HB meet EK at G.
Let IH meet BL at J and AB meet DH at J'.
let BD meet EH at F
L is the orthocenter of HGK so GL is perp to HK
but L and A are symetric wrt BH and and so are IH and DH so LJ=AJ' and BJ=BJ'.
now since GK and BD are both perp to DH than GH/HB=EH/HF since EF || AB than EH/HF=AJ'/J'B and AJ'/J'B=LJ/BJ so
since so GH/HB=LJ/BJ so GL || HJ and since HJ is perp to BC than so is GL but GL is perp to HK so HK||BC.

Dear Mathlinkers,

http://jl.ayme.pagesperso-orange.fr/Docs/Parallelogie.pdf  p. 12-14.

Sincerely
Jean-Louis


trigonometry
conics
hyperbola
analytic geometry
geometry
geometric transformation
reflection

Hey, should I be able to visualize the shape of polar equations that weren't derived from a rectangular one?

Well, if you're confused, you can plot a few points to see what's going on. Alternatively, you can examine what happens to the length of the radius as you go from $0$ to $2\pi$ radians, which should help you see the graph's shape.

Try this: $r=4\sin 4\theta$ (It should look nice :D)

How about $r=\frac{12}{1+4\sin \theta}$

[hide="They should be..."]A flower and a hyperbola, respectively.  :) [/hide]

Does polar coordinates graph only the positive values of the expression as length of r? Because for the flower one, it seems the 270/4 angles for r are equivalent in length to the 90/4 angles.

All of the angles in 4sin4theta seem to equal an r of 4 even at the angles are equivalent to 270 degrees. I thought polar coordinates only graphed positive values of r?

No, negative values of $r$ are fine.  For example, in polar coordinates, the point $\left(-2,\frac{5\pi}{4}\right)$ is equivalent to $\left(2,\frac{\pi}{4}\right)$.  If you plot the point, it will make sense.

Then how come if I plot cos theta, I only get the r values on the positive side of the x axis?

It's supposed to happen. r=cos(theta) is a circle tangent to the y-axis. For theta between pi/2 and 3pi/2, r is negative. So, for angles that lie in the second and third quadrants, the negative r values reflect the points to the first and fourth quadrants. For example, (-1/2,4pi/3) is point on the graph, which is plotted at (1/2,pi/3).

So, a negative r distance means the r in the opposite direction, or basically theta +/- 180 degrees correct?

That would make sense since the if a vector's size is negative it would be in the opposite direction as that of its positive counterpart.


Let a triangle $ABC$ , the bisector $AD$  , the median $AM$ and a point $P$ on $AD$. Let $E,F$ the projections of $P$ on sides $AB , AC$. If $EF$ meets $AM$ at $K$ , prove that $KP \perp BC$.

[u] Babis[/u]

 [i]( Untill this moment I have not a solution to the problem.Probably it is harder than what I think , but probably easier)[/i]

we can shift BC parallel without changing much...i believe this is a classic problem, i remember part of it showed up on this year's hmmt contest...i did not carefully look at the solution, but there was some trig in there...

Let $\{A,L\}\in AD\cap\odot(ABC)$, $J, I$ the intersections of the parallel through $K$ to $BC$ with $AB, AC$ respectively, and $N, Q$ the projections of $L$ onto $AB, AC$ respectively.
Easy to note, the configurations $E-J-K-P-F-I$ and $N-B-M-L-Q-C$ are homothetic, hence the conslusion (I forgot to mention before, $N, M, Q$ are collinear, because $\Delta BLN\equiv\Delta CLQ$, a.s.a., so $BN=CQ$, also $AN=AQ$ and [b]Menelaos[/b] for $\Delta ABC$ will show that).

Best regards,
sunken rock

Dear Mathlinkers,

you can see
http://www.artofproblemsolving.com/Forum/viewtopic.php?f=46&t=547220

Sincerely
Jean-Louis


algebra
polynomial
calculus
calculus computations

My friend gave me this problem to help him solve in class one day.

$\sum_{n=1}^{\infty}\frac{4^{n}}{(2n)!}$

any ideas on how to begin?
i am thinking taylor polynomial for $e^{x}$ or some sort of variant  but havent gotten anywhere.

[hide="A hint which is less of a joke than you might think"]Differentiate with respect to 2  :wink:
[hide="More seriously"]Consider the series for the exponential, trigonometric or hyperbolic functions.  They can be used to find many, many expressions similar to this one.[/hide][/hide]

$\sum_{k=1}^{\infty}\frac{2^{2k}}{(2n)!}$. Then adjust (but considering in x varialbe) $\sum_{k=0}^{\infty}\frac{x^{2n}}{2n!}-\frac{x^{0}}{0!}$.
(just to refresh your memory : http://en.wikipedia.org/wiki/Taylor_series  :wink: )

thanks for the pointers.
if i did this correctly the answer should be
$\cosh(2)-1$

$=2\sinh^{2}(1).$

Yes, your answer is correct.


[quote="giannis18"]Παιδιά χρειάζομαι βοήθεια με το παρακάτω:
Σε μία άσκηση δίνεται η σχέση $ *$ στο $ R^*$ με $ a*b \equal{} |a|b$ και ζητείται μεταξύ άλλων να δειχθεί ότι υπάρχει ένα αριστερό ταυτοτικό στοιχείο για την $ *$ και ένα δεξιό αντίστροφο για κάθε στοιχείο του $ R^*$.Η ερώτησή μου τώρα είναι:Αφού η $ *$ δεν έχει δεξιό ταυτοτικό στοιχείο (εύκολα επαληθεύεται) πώς γίνεται να έχει δεξιό αντίστροφο για κάιε στοιχείο;Δηλαδή δεν γίνεται να υπάρχει και δεξιό και αριστερό ταυτοτικό στοιχείο και να είναι διαφορετικά;Τότε το δεξί αντίστροφο ενός στοιχείου δεν θα είναι αυτό που πολ/μενο από δεξιά με το στοιχείο δίνει το δεξί ταυτοτικό στοιχείο..;
Ξέρω ότι δεν είναι στα πλαίσια του forum :blush:  αλλά δεν βρίσκω απάντηση πουθενά και θα εκτιμούσα πραγματικά την όποια βοήθεια :help:  :help:[/quote]

Μια χαρά στα πλαίσια του forum είσαι :)

Άσκηση (όχι δύσκολη): να δειχθεί ότι αν υπάρχει και αριστερό και δεξιό ταυτοτικό στοιχείο, τότε αυτά είναι ίσα :)

Hope this helps,

Durandal 1707

Χμ έχεις δίκιο έστω $ e,e'$ δεξιό και αριστερό ταυτοτικό στοιχείο αντίστοιχα τότε $ e*e'\equal{}e'$ επειδή το $ e$ είναι αριστερό ταυτοτικό στοιχείο και $ e*e'\equal{}e$ επειδή το $ e'$ είναι δεξί ταυτοτικό στοιχείο άρα $ e\equal{}e'$.Δεν το είχα σκεφτεί αυτό πριν.
Εν πάσει περιπτώσει όμως μπορεί να οριστεί δεξί αντίστροφο ενώ υπάρχει μόνο αριστερό ταυτοτικό; :maybe:

[quote="giannis18"]Χμ έχεις δίκιο έστω $ e,e'$ δεξιό και αριστερό ταυτοτικό στοιχείο αντίστοιχα τότε $ e*e' \equal{} e'$ επειδή το $ e$ είναι αριστερό ταυτοτικό στοιχείο και $ e*e' \equal{} e$ επειδή το $ e'$ είναι δεξί ταυτοτικό στοιχείο άρα $ e \equal{} e'$.Δεν το είχα σκεφτεί αυτό πριν.
Εν πάσει περιπτώσει όμως μπορεί να οριστεί δεξί αντίστροφο ενώ υπάρχει μόνο αριστερό ταυτοτικό; :maybe:[/quote]

Γιατί όχι? Το $ g'\in G$ είναι δεξί αντίστροφο του $ g\in G$ για την αριστερή ταυτότητα $ e_l$ αν $ gg' \equal{} e_l$.

Το πόσο χρήσιμος είναι αυτός ως ορισμός είναι άλλη ιστορία :)

Cheerio,

Durandal 1707

Απλώς μου φαινόταν παράλογος ένας τέτοιος ορισμός.... :D Όπως και να 'χει ευχαριστώ πολύ για τη βοήθεια :)


If series $\sum_{i=1}^{\infty}u_{i}$ is converges then $\lim_{i\to\infty}u_{i}=0$.

[quote="muckymotter"]This problem was confusing me so any help would be greatly appreciated!

Determine whether or not the following series converges, justifying your answer.

(4/1)-(5/2)+(6/3)-(7/4)+(8/5)-(9/6)+(10/7)-(11/8)+-...[/quote]
$\frac{4}{1}-\frac{5}{2}+\frac{6}{3}-\frac{7}{4}+\frac{8}{5}-...= \sum_{n=1}^{\infty}(-1)^{n+1}\cdot \frac{n+3}{n}$

[hide]Let $a_{n}=(-1)^{n+1}\cdot \frac{n+3}{n}$

$\lim_{n\to\infty}a_{n}$
$=\lim_{n\to\infty}(-1)^{n+1}\cdot \frac{n+3}{n}$
$=\pm 1$

Because $\lim_{n\to\infty}a_{n}= \pm 1 \not = 0$, by the "$n^{th}$ Term Test," $\sum_{n=1}^{\infty}(-1)^{n+1}\cdot \frac{n+3}{n}$ is a divergent series.[/hide]


trigonometry
calculus
calculus computations

show  :|

$ \sum_{\textbf j=1}^\infty\;\;\left(\frac{\sin \;\textbf j}{\textbf j}\right)^{2}\;\;=\;\;\boxed{\frac{\pi-1}{2}}$

Let $ S$ be your sum. Then:

$ S=\sum_{n=1}^{\infty}\frac{1}{2n^{2}}-\sum_{n=1}^{\infty}\frac{\cos 2n}{2n^{2}}$

$ =\frac{\pi^{2}}{12}-\frac{1}{2}\sum_{n=1}^{\infty}\frac{\cos 2n}{n^{2}}$.

Now use the following Fourier expansion:

$ \sum_{n=1}^{\infty}\frac{\cos 2nx}{n^{2}}=\frac{3x^{2}-6\pi x+2\pi^{2}}{12}$, $ 0\leq x\leq \pi$.

Let $ x=1$ in the preceding sum and the result would follow.


[quote="onlylove_math"]This problem is a problem of [/quote]
It is also a very descent type of problem. However, what is "[color=blue]Hungkhtn 's secret in inequality[/color]"

[quote="tranvanluan"]Let $ a;b;c\ge 0$and$ a \plus{} b \plus{} c \equal{} 1$
Prove that
$ \sqrt {a \plus{} b^2} \plus{} \sqrt {b \plus{} c^2} \plus{} \sqrt {c \plus{} a^2}\ge 2$[/quote]
See here:
http://www.artofproblemsolving.com/Forum/viewtopic.php?t=106782

[quote="onlylove_math"][quote="tranvanluan"]Let $ a;b;c\ge 0$and$ a \plus{} b \plus{} c \equal{} 1$
Prove that
$ \sqrt {a \plus{} b^2} \plus{} \sqrt {b \plus{} c^2} \plus{} \sqrt {c \plus{} a^2}\ge 2$[/quote]
This problem is a problem of Hungkhtn 's secret in inequality[/quote]
I have seen VasC's solution for it, it is nice but I think it seems hard to think of. Here is my solution (I like it although it is ugly).
The inequlaity is equivalent to
\[ \sum (\sqrt {a \plus{} b^2} \minus{} b) \ge1
\]

\[ \sum \frac {a}{b \plus{} \sqrt {a \plus{} b^2} }\ge 1
\]
By AM-GM Inequality, we have
\[ \frac {a}{b \plus{} \sqrt {a \plus{} b^2}} \equal{} \frac {a(a \plus{} b)}{b(a \plus{} b) \plus{} (a \plus{} b)\sqrt {a \plus{} b^2}} \ge \frac {2a(a \plus{} b)}{2b(a \plus{} b) \plus{} (a \plus{} b)^2 \plus{} a \plus{} b^2} \equal{} \frac {a(a \plus{} b)}{2a^2 \plus{} 5ab \plus{} 4b^2 \plus{} ca}
\]
Hence, it suffices to prove that
\[ \sum \frac {a(a \plus{} b)}{2a^2 \plus{} 5ab \plus{} 4b^2 \plus{} ca} \ge 1
\]

\[ \Leftrightarrow 4\sum a^4b^2 \plus{} 3\sum a^3b^2c \minus{} 19\sum a^2b^3c \plus{} 16\sum a^4bc \minus{} 12a^2b^2c^2 \ge 0
\]

\[ \Leftrightarrow 4(\sum a^4b^2 \minus{} \sum a^2b^3c) \plus{} 3(\sum a^3b^2c \minus{} 3a^2b^2c^2) \plus{} 15(\sum a^4bc \minus{} \sum a^2b^3c) \plus{} (\sum a^4bc \minus{} 3a^2b^2c^2) \ge 0
\]
which is true by AM-GM Inequality.
We have done.
:)

@Luan: I would like to see your solution if it is different from VasC's solution (the solution in Hungkhtn's book).

[quote="tranvanluan"]Let $ a,b,c\ge 0$ and $ a \plus{} b \plus{} c \equal{} 1.$ Prove that

$ \sqrt {a \plus{} b^2} \plus{} \sqrt {b \plus{} c^2} \plus{} \sqrt {c \plus{} a^2}\ge 2.$[/quote][quote="can_hang2007"]The inequlaity is equivalent to

$ 1\leq\sum_{cyc} (\sqrt {a \plus{} b^2} \minus{} b) \equal{} \sum_{cyc} \frac {a}{b \plus{} \sqrt {a \plus{} b^2} }.$

By AM-GM Inequality, we have

$ \frac {a}{b \plus{} \sqrt {a \plus{} b^2}} \equal{} \frac {a(a \plus{} b)}{b(a \plus{} b) \plus{} (a \plus{} b)\sqrt {a \plus{} b^2}}\ge \frac {2a(a \plus{} b)}{2b(a \plus{} b) \plus{} (a \plus{} b)^2 \plus{} a \plus{} b^2}$[/quote]$ \equal{} \frac {2a(a \plus{} b)}{2a^2 \plus{} 5ab \plus{} 4b^2 \plus{} ca}.$

But $ \left[\sum_{cyc} \frac {2a(a \plus{} b)}{2a^2 \plus{} 5ab \plus{} 4b^2 \plus{} ca} \minus{} 1\right]\prod_{cyc}(2a^2 \plus{} 5ab \plus{} 4b^2 \plus{} ca)$

$ \equal{} \sum_{cyc}ca(b \minus{} c)^2(3b^2 \plus{} 13bc \plus{} 4ca \plus{} 4ab)\geq0.$

can_hang2007's solution is symmetric and nice, I like it.

See also here : http://www.mathlinks.ro/viewtopic.php?t=255880


LaTeX
Galois Theory
superior algebra
superior algebra unsolved

Hi All! Could you please follow the link :

[url=http://www.math.fau.edu/Klingler/alg-num-theory/hmwk5.html]www.math.fau.edu/Klingler/alg-num-theory/hmwk5.html[/url]

Can anyone help me with No. 7? I know it's not that hard, but I can't seem to understand how it works.

Thanks

Zaks

K and L are number fields, L is a normal extension of K, R is the number ring of K, S is the number ring of L. Then :

For each prime Q lying over P,

D is the decomposition group : D(Q|P) = the set of morphisms in the Galois group which leave Q fixed,
E is the inertia group : E(Q|P) = the set of morphisms in the Galois group which leave each element in S fixed modulo Q.

(I'm sorry I don't know how to use BBCode, it would have been much easier)
I hope this clears it

a)
Let $\sigma \in D(V|Q)$, so $\sigma(V)=V$. Then $U = V \cap M = \sigma(V) \cap \sigma(M) = \sigma(V \cap M) = \sigma(U)$, where  $M = \sigma(M)$ since $M$ is normal.
This shows that the mapping is at least well defined.
The injectivity must not be proved, it follows from the injectivity of $Gal(LM|L) \to Gal(M|K)$.

b)
same manner than a)

c)
Let $Q \subset L$ lie over $P$ and let two primes $V_1 , V_2 \subset LM$ over $Q$ be given. Then $V_1 \cap M$ and $V_2 \cap M$ are primes over $P$, thus $V_1 \cap M=V_2 \cap M$, meaning they coincede on $M$. Since they also coincede on $L$ by definition, they must coincede on $LM$.

d)
We want to show that $D(V|Q)$ is trivial. This follows directly since we have an embedding of $D(V|Q)$ into the trivial group $D(U|P)$.

e)
Sorry for not asking before, but what is $\phi(A|B)$ ¿

PS: you can learn $\LaTeX$ here: http://www.mathlinks.ro/Forum/viewtopic.php?t=5261

$\phi$ is the Frobenius automorphism : it has the property $\phi(\alpha) \equiv \alpha^{||P||}$ (mod $Q$)

Thanks!!

I'm sorry for being slow, but I'm not sure I understand how well-definition of the embedding  follows from the conclusion of (a), and how it applies to (b).
thanx


A regular 12-gon is inscribed in a circle of radius 12. The sum of the lengths of all sides and diagonals of the 12-gon can be written in the form $ a\plus{}b\sqrt{2}\plus{}c\sqrt{3}\plus{}d\sqrt{6}$ where $ a$, $ b$, $ c$, and $ d$ are positive integers. Find $ a \plus{} b \plus{} c \plus{} d$.

Is there a way of figuring this out without finding the length of every single diagonal?

[hide="Well, this is a bit faster...hint 1"]
Place the 12-gon in the coordinate plane.

[hide="Hint 2"]
The coordinates of the other vertices are $ \left (12 \cos kx, 12 \sin kx \right )$ where $ x = 30^\circ$ and $ k=1,2,\dots,11$.

[hide="Hint 3"]
Distance formula!  But don't evaluate each of them; sum cleverly.

$ 12 \sqrt{\left (\cos kx - 1 \right )^2 + \sin \left (kx \right )^2} = 12 \sqrt{2-2\cos kx} = 24 \sin \frac{kx}{2}$.

$ \sin 15^\circ + \sin 75^\circ = \sin \left (45^\circ - 30^\circ \right ) + \sin \left (45^\circ + 30^\circ \right )$

[/hide]
[/hide]
[/hide]

Wow, gee, really great stuff.

Thanks, AIME15.


geometry
trigonometry
inequalities
inequalities proposed

nice.here is my solution :)
the inequ is the same  as $2 \sqrt{3} \sin A  \sin B \sin C \leq \sin A \sin B +\sin B \sin C +\sin C \sin A$,just use AM-GM ,and $\sin A \sin B  \sin C \leq \frac{3 \sqrt 3}{8}$ :)

I think this inequality is quite old...
Vasc posted some stronger versions I think.

The inequality is trivial. I can prove a better one:

     ab+bc+ca >= 18Rr. It is known that 2Rr=abc/a+b+c and the inequality reduces to
     (a+b+c)(ab+bc+ca) >= 9abc which is trivial by AM-GM.

This one is stronger

 $2(ab+bc+ca)-(a^2+b^2+c^2) \ge \frac{\sqrt{3}}{4}A$

  ;)

[quote="Mamat"]Prove that:
$ 4\sqrt{3}A \leq ab+ac+bc $ ( a,b,c are sides of triangle, A is area)[/quote]
This is from the First or Second or Third IMO.

[quote="Mamat"]How did you prove that $18Rr \geq 4 \sqrt3 A$[/quote]

  :D  :D  :D

 First,we have:

 $\displaystyle 9R^2 \ge a^2+b^2+c^2 $

 Then

 $9R^2 \ge \frac{(a+b+c)^2}{3}$

 Or

 $R \ge \frac{a+b+c}{3\sqrt{3}} $

 Now because

 $(a+b+c)r=2A$ we have the solution

 More difficult

 Find $\alpha$ Min :

 $R^{\alpha}{r^{2-\alpha}} \ge \frac{2^\alpha}{3\sqrt{3}}A$   :blush:

The proof of the inequality  18Rr>=4 :sqrt: 3 A is very simple.
 By computations the inequality above is equivalent to p<=3 :sqrt:3 R which is due to Mitrinovic.

[quote="shobber"]Mamat wrote:
Prove that:
 ( a,b,c are sides of triangle, A is area)
$4 \sqrt3 A \leq ab+ac+bc$

This is from the First or Second or Third IMO.[/quote]
 IMO problem : $ 4 \sqrt3 A \leq a^2+b^2+c^2$

What Viet Math wrote is wrong.It's not sqrt(3)A/4.In fact it is 4A sqrt(3) and this is was given by Finsler in 1937.
  I have a better one which is due to Blundon. I'll post it if someone is interested.

it can be proved that a sqrt(bc)+b sqrt(ca)+c sqrt(ca)>=18Rr, very easy by using cauchy-schwarz.

in connection to the inequality i have written,the following holds:

     $2(ab+bc+ca)-(a^2+b^2+c^2)\leq18Rr.$

Hi everyone !
We will consider the inequalities about the sides of a triangle ($a,b,c$) and its area ($S$).
1) $ab+bc+ca \ge 4\sqrt{3}S$

$4S=\sqrt{(a+b+c)(a+b-c)(b+c-a)(c+a-b)} \le \sqrt{(a+b+c)abc} \le \sqrt{\frac{(ab+bc+ca)^2}{3}}$

We have a stronger inequality $S \le \frac{\sqrt{abc(a+b+c)}}{4}$.

These inequalities for you :
2) $S \le \frac{3\sqrt[3]{a^2b^2c^2}}{4sqrt{3}} $
3) $S \le \frac{3abc}{4\sqrt{a^2+b^2+c^2}} $

[quote="cezar lupu"]Inequality 3) is trivial. From abc=4RS we get the well known a^2+b^2+c^2<=9R^2. About 2) I'm thinking but I'll post my solution very soon.[/quote]
(2) is weaker than 3 (amgm) so we are done.

i think if u put $x=(b+c-a)$ $y=(c+a-b)$ and $z=(a+b-c)$
the inequality becomes
$\sqrt{3xyz(x+y+z)} \le \frac{(xy+yz+zx)+(x+y+z)^2}{4}$
so since
$\frac{(xy+yz+zx)+(x+y+z)^2}{4} \ge (xy+yz+zx) \ge sqrt{3xyz(x+y+z)}$
by application of the ineq $(a+b+c)² \ge 3(ab+bc+ca)$
!!!!

This inequality is nice:
$ma^2+nb^2+pc^2 \ge 4\sqrt{mn+np+pm}S$
m, n, p > 0.

+With $m=\frac{1}{a^2},n=\frac{1}{b^2},p=\frac{1}{c^2}$ we have (3).
And if let m,n,p be another numbers, we have very very many nice inequalities.


limit
trigonometry
calculus
calculus computations

\[ \mathop {\lim }\limits_{x \to 0} \frac{{\sqrt {1 \plus{} \tan x}  \minus{} \sqrt {1 \plus{} \sin x} }}
{{x^3 }}\]
Halp pls

And exactly the same sorts of tricks will work.  (Well, if you multiply by the conjugate, you'll have to cleverly rewrite $ \tan x \minus{} \sin x$ as $ \tan x \minus{} x \plus{} (x \minus{} \sin x)$ before using the trig limits you already know.)  Why don't you try and then others can correct you as needed?

(The other limit was http://www.artofproblemsolving.com/Forum/viewtopic.php?t=293509 )

[hide]\[ \mathop{\lim }\limits_{x\to 0}\frac{{\sqrt{1\plus{}\tan x}\minus{}\sqrt{1\plus{}\sin x}}}{{x^{3}}}\]

Multiplying and diving by its conjugate we get

$ \lim _{x \to 0 } \frac { \tan x\minus{} \sin x }{ x^3 (( {\sqrt{1\plus{}\tan x}\plus{} \sqrt{1\plus{}\sin x   } )  } }$

$ \equal{}  \lim _{x \to 0 } \frac{\frac{ \sin x}{x} \frac{ ( 1\minus{} \cos x ) } { x^2 } }{ \cos x  ( {\sqrt{1\plus{}\tan x}\plus{} \sqrt{1\plus{}\sin x   } )  } }$

$ \lim _ {x \to 0 } \frac{ ( 1\minus{} \cos x ) } { x^2 } \equal{} \frac{1}{2}$

therefore answer is $ \equal{} \frac{1}{4}$[/hide]


probability
calculus
geometry
3D geometry
pyramid

Three real numbers are chosen randomly between 0 and 1. What is the probability that the first number is greater than the sum of the other two?

[Comment: I got the correct answer using calculus, but I'm sure there's a non-calculus solution.]

think of the 3 numbers in the xyz plane

now consider the cube bordered by planes x=0, y=0, z=0, x=1, y=1, z=1
the first number is in the z axis.
soon, we find that the desired volume is a pyramid with a triangular 1*1 base and a height of 1
next, apply the volume of pyramid formula to find the volume is 1/6 (out of a total of 1) so the probability is 1/6

of course proving the pyramid theorem im sure requires calculus


topology
real analysis
real analysis theorems

Say $ G \subseteq R^m$ and $ f: G \to R$ is differentiable. Assume $ a,b \in G$ are interior points and $ \gamma : [0,1] \to G$ is differentiable on (0,1), continuous on [0,1], and $ \gamma (0) \equal{} a;\gamma (1) \equal{} b$.

$ f \circ \gamma : [0,1] \to R$ is continuous on [0,1] and differentiable on (0,1). By the Lagrange finite increment theorem, $ \exists \varepsilon  \in (0,1)$ such that $ f(b) \minus{} f(a) \equal{} f'(\gamma (\varepsilon )) \circ y'(\varepsilon )$

--

Therefore if G is open and connected and the differential of f is 0 at every point of G then the above shows that $ \forall a,b \in G$ f(b) - f(a) = 0. Therefore f is constant.

Oh nice. My book proved something similar and then used a convoluted way to prove that f must be constant. I think it was just to show different types of proofs though.

You assumed a stronger hypothesis- path-connectedness rather than connectedness. In $ \mathbb{R}^n$, being open and connected implies being path-connected, but this is a nontrivial topological fact.

jmerry, in my book connectedness is defined as path connected.

I understand why my book used a different proof. Pathwise connected means there are continuous paths but not necessarily differentiable, as I have assumed.

It is not apparent to me why pathwise connectedness cannot imply that all paths are differentiable instead of merely continuous, so long as we are working in R(n). Are there any other examples where a differentiable path cannot exist?

If an open set in $ \mathbb R^n$ is connected, then it is connected by smooth curves. But it's probably easier to prove that it's connected by polygonal curves, which is enough for your purpose.

Here's a different approach: first, prove the special case when $ G$ is an open ball (any two points are connected by a straight line segment). Now consider the general case. Pick $ x_0\in G$. Since $ f$ is constant on any open ball contained in $ G$, it follows that the set $ V: \equal{} \{x\in G\colon f(x) \equal{} f(x_0)\}$ is open. But the set $ W: \equal{} \{x\in G\colon f(x)\ne f(x_0)\}$ is also open, as the preimage of an open set. Using the general definition of connectedness ($ G$ is not the union of two disjoint nonempty open subsets) we conclude that $ W \equal{} \varnothing$ which means that $ f\equiv f(x_0)$.


Why is $1000\cdot 1001 \cdot 667 \equiv 2$ mod(3)?

Because we can mod out the individual numbers that are being multiplied instead of modding out the product, we have this:

$1000 \cdot 1001\cdot667 \equiv 2 \pmod{3} \implies 1000 \pmod{3}\cdot1001 \pmod{3}\cdot667 \pmod{3} \equiv 2$

Because $1000 \equiv 1 \pmod{3}$, and $1001 \equiv 2 \pmod{3}$, and $667 \equiv 1 \pmod{3}$, we can multiply them together and get $1\cdot2\cdot1=2 \pmod{3}$.

A mod is an abbreviation for a moderator, someone who makes sure posts are in the right places and stuff.

Here's a good thread discussing mods:

http://www.artofproblemsolving.com/Forum/viewtopic.php?highlight=modular+arithmetic&t=28395

I would try to explain it to you myself, but this thread does a much better job.  Mods can be kind of hard to understand at first (they were for me at least), but you'll soon realize just how important they are and how useful they can be in even the most difficult problems.

BTW : if you want to take mod  3  (or mod 9)  of any number (written in decimal notation) just add all the digits.