If a = 1 and b = 2 and so on  then what word would = 1000000 if we multiply them all together :?

eg egg would be 5 * 7 * 7 = 245 i.e the 5th letter of the alphabet and the 7th letter  of the alphabet wice all multiplied together

huh?  please explain the question a little more.

[hide]1000000's factors under 26 are 1, 2, 4, 5, 10, 20, 25,contributing to the letters a, b, d, e, j, t, and y. Now it's time to unscramble.  [/hide]

bated,bayed,beady are what i can find
bated:2*1*20*4*5=800
deadbeat,abetted,debated,jetbead,abated,batted,beaded,etc.

even counting the number of such letter sequences that ARENT real words is probably above this forum?  it's pretty hard!  some thought tells you though that the sequence has to be at least 5 letters: y,y,y,t,p which obviously don't form any [english] words.

Can you at least tell us what the word relates to. Math? Science? Movies?

[quote="nat mc"]I got some:
beethey
payyeed
yepayed
beheadyy
yebheady
behead bye[/quote]
None of them seems to be word except 'behead bye'. And my Google Toolbar agrees me. :D


The digits of a positive integer $n$ are four consecutive integers in decreasing order when read from left to right. What is the sum of the possible remainders when $n$ is divided by $37$?

[hide]The possibilities of $N$ are $9876, 8765, 7654, 6543, 5432, 4321, 3210$ so summing these remainders together, we get $\boxed{217}$ You'll notice a pattern with the remainders, it's $34,33,32,....28$[/hide]

[hide]
First note that there are only 7 possibilities, which are easy to find so I won't write them.  We find the remainder upon dividing the smallest, 3210, by 37 is 28.  Then to get the next one we need to add 1111, which has remainder 1 when divided by 37, so the remainder of 4321 will be 29.  So the answer is $28 \cdot 7 + \frac{6 \cdot 7}{2} = 217$
[/hide]

Is this really an aime question? It is so easily brute forceable, it is almost like a mathcounts one.....

[hide]n can be $9876, 8765, 7654, 6543, 5432, 4321,$ or $3210$
The remainder of $3210/37$ is $28$, and then to get the next number, we have $3210+1111$. 1111mod37=1, so 4321mod37 is 29. To get the next number, we add 1111 again, so the remainder is 30, and so on, until we find that the remainders of all 7 are $29, 30, 31, 32, 33, 34,$ and $35$. Adding them, we get 217. [/hide]

AIME #1's don't tend to be too difficult.


The two polynomials $ P(x)\equal{}x^4\plus{}ax^3\plus{}bx^2\plus{}cx\plus{}d$ and $ Q(x)\equal{}x^2\plus{}px\plus{}q$ take negative values on an interval $ I$ of length greater than 2, and nonnegative values outside $ I$ . Prove that there exists $ x_0 \in R$ such that $ P(x_0)< Q(x_0)$

[quote="mayhem"]The two polynomials $ P(x) \equal{} x^4 \plus{} ax^3 \plus{} bx^2 \plus{} cx \plus{} d$ and $ Q(x) \equal{} x^2 \plus{} px \plus{} q$ take negative values on an interval $ I$ of length greater than 2, and nonnegative values outside $ I$ . Prove that there exists $ x_0 \in R$ such that $ P(x_0) < Q(x_0)$[/quote]

Let $ I\equal{}(k_1,k_2)$, then clearly $ k_1,k_2$ are roots of both $ P(x)$ and $ Q(x)$ $ \Rightarrow Q(x)|P(x)$
Also roots of $ P(x)$ are not in $ I , (\minus{} \infty , k_1) , (k_2 , \plus{} \infty)$ (or $ P(k_3)\equal{}0<Q(k_3)$ )

[b]Case 1:[/b] $ P(x)\equal{}(x\minus{}k_1)^2(x\minus{}k_2)^2$, then $ P(x) \ge 0 \forall x \in I$, contradiction.
[b]Case 2:[/b] $ P(x)\equal{}Q(x)[(x\plus{}m)^2\plus{}n^2]$, where $ m,n \in R$
Assume that $ (x\plus{}m)^2 \le 1 \forall x \in I$, then $ \minus{}1 \le x\plus{}m \le 1 \forall x \in I \Rightarrow$ length of $ I \le 2$ , contradiction. So $ (x_0\plus{}m)^2\plus{}n^2 \ge (x_0\plus{}m)^2>1$ for some $ x_0 \in I$
As $ Q(x_0)<0$ we have $ P(x_0)\equal{}Q(x_0)[(x_0\plus{}m)^2\plus{}n^2 \ge (x_0\plus{}m)^2]<Q(x_0) (1)\equal{}Q(x_0)$ , done.


Breite ola ta zeygh (x,y) pragmatikwn arithmwn pou einai lyseis tou systimatos:

                       $ x^4\plus{}2x^3\minus{}y\equal{}\minus{}\frac{1}{4}\plus{}\sqrt{3}$
                       $ y^4\plus{}2y^3\minus{}x\equal{}\minus{}\frac{1}{4}\minus{}\sqrt{3}$

Ksexasa na pw oti to onoma mou einai Kostas kai eimai kainourgios sto forum.
Exete kanei oloi sas poly kali douleia edw pera!!
Synexiste etsi!

hello, from the first equation we get $ y\equal{}x^4\plus{}2x^3\minus{}\frac{1}{4}\plus{}\sqrt{3}$, putting this in the second equation we have $ (x^4\plus{}2x^3\plus{}\frac{1}{4}\minus{}\sqrt{3})^4\plus{}2(x^4\plus{}2x^3\plus{}\frac{1}{4}\minus{}\sqrt{3})^2\minus{}x\equal{}\minus{}\frac{1}{4}\minus{}\sqrt{3}$. After a ugly algebraic computation i have got
$ x\equal{}\frac{1}{2}\sqrt{3}\minus{}\frac{1}{2}$ and $ y\equal{}\minus{}\frac{1}{2}\sqrt{3}\minus{}\frac{1}{2}$.
Sonnhard.

You got the correct solution.Another solution is the following:

             By adding the two equations we get:

              $ (x^4\plus{}2x^3\minus{}x\plus{}\frac{1}{4})\plus{}(y^4\plus{}2y^3\minus{}y\plus{}\frac{1}{4})\equal{}0$

Now the difficult part in this solution is to recognize that this is a sum of two squares:

              $ (x^2\plus{}x\minus{}\frac{1}{2})^2\plus{}(y^2\plus{}y\minus{}\frac{1}{2})^2\equal{}0$

And now,you can get the solution.


quadratics
algebra
quadratic formula
number theory proposed
number theory

[hide="my solution"]
By quadratic formula we get:
$ x_1 \equal{} 1 \plus{} \sqrt {1 \plus{} 2007y^2}$
$ x_2 \equal{} 1 \minus{} \sqrt {1 \plus{} 2007y^2}$

Since $ 1 \plus{} 2007y^2 \equal{} k^2$ does have solutions in $ Z$,for $ y \equal{} \plus{} 5, \minus{} 5$ and $ k \equal{} 224$ see it here:
http://www.mathlinks.ro/viewtopic.php?p=1626175#1626175 . Substituting,
we get $ x_1 \equal{} 225,x_2 \equal{} \minus{} 223$
some of the solutions are $ (x,y) \in Z \equal{} (225,5);(225, \minus{} 5);( \minus{} 223,5);( \minus{} 223, \minus{} 5)$ [/hide]


algebra
polynomial
inequalities
triangle inequality
superior algebra
superior algebra solved

Hi, I found a problem (but not one of its solution), and I will greatly appriciate if somebody can either give me a needed insight or a solution, here goes:

Let $T(z)$ be a polynomial with integer coefficients, and let $P(z) = z^n - 1$. Suppose $P(z)$ and $T(z)$ share a common root and that for any root ,$u_i$, of $P$, $|T(u_i)|$ $\leq$ $1$, prove that $P(z)$|$T(z)$.

Don't be so impatient. It's been less than a day- lots of problems wait.

In any case, it isn't very hard. Let $Q(z)=a_{n-1}z^{n-1}+a_{n-2}z^{n-2}+\cdots+a_1z+a_0$ be the remainder when $T$ is divided by $P$. $Q$ is a polynomial of degree no more than $n-1$ with integer coefficients, and $Q(u_i)=T(u_i)$. We want to show that $Q$ is the zero polynomial.

By the standard interpolation formula, $Q(z)=\sum_iQ(u_i)\prod_{j\neq i}\frac{z-u_j}{u_i-u_j}=\sum_iT(u_i)\frac{z^{n-1}+u_iz^{n-2}+\cdots+u_i^{n-2}z +u_i^{n-1}}{nu_i^{n-1}}$.
Equating coefficients, $a_j=\sum_iT(u_i)\frac{u_i^{n-j}}{n}$. By the Triangle inequality, $|a_j|\le\sum_i\frac{|T(u_i)|}{n}\le\frac{n-1}{n}<1.$ Since $a_j$ is an integer, $a_j$ is zero and $Q$ is identically zero.

My apologies to whom it may concern,
       I did not mean to be impatient, forgive me, (Struggled for about a week with his problem). At any rate, thank you so much for your solution, Its an awesome solution.

Once again, my apologies.


limit integration induction function search logarithms inequalities

Find the necessary and sufficient condition for the sequence $a_{n}=c^{a_{n-1}}$ ,$a_{0}=1$ to be convergent. ($c$ is a real positive number)

It is obviuous that the case $c=1$ satisfies our demands so it remained to consider the case $c\neq1$.

That case can be decomposed into the following two:

1) The case $c>1$. In this case $(a_n)$ is increasing. Since $\exists\,\lim a_n\Leftrightarrow\sum\limits(a_{n+1}-a_n)=\sum (c^{a_n}-a_n)<\infty\Leftrightarrow\int\limits^{+\infty}(c^x-x)dx<\infty$ and $\int\limits^{+\infty}(c^x-x)dx=+\infty$, we conclude that $\lim a_n$ doesn't exist.

2) The case $c<1$. It is easy to see that $a_{2n+1}<a_{2n}$ for $n\ge0$. So if we show that $(a_{2n})$ is decreasing and $(a_{2n+1})$ is increasing, we will be able to consider the case $c<1$ as a solution. Note that it is enought to prove one of these statements since $\frac{a_{2n+2}}{a_{2n}}=\frac{c^{a_{2n+1}}}{c^{a_{2n-1}}}=c^{a_{2n+1}-a_{2n-1}}$.

So let's concentrate our attention on considering $b_n: =a_{2n}$. We can prove the statement $\forall\,n\ge0\;\;b_n>b_{n+1}$ using mathematical induction: first of all note that $b_0=1>b_1=c^c$, now if we know that $b_n>b_{n+1}$ for some $n\ge0$ then $b_{n+1}=c^{\,c^{b_n}}>c^{\,c^{b_{n+1}}}=b_{n+2}$.

Thus the answer is $0<c\le1$.

I think you've made a mistake because $\sqrt{2}$ is bigger than one but when $a_n={\sqrt{2}}^{a_{n-1}}, \lim a_n =2$.

You are right.  :) Indeed, the proof in 1) is invalid since in our case $\sum(c^{a_n}-a_n)<\infty$ isn't the same as $\int^{+\infty}(c^x-x)<\infty$ because $x\mapsto c^x-x$ isn't decreasing function (so I hastened, as always). Anyway, it seems the case $c>1$ should be analysed separately because of the peculiarity: $(a_n)$ increases.

Suppose $x: =\lim a_n$ exists then it is necessary that the equation $c^x=x$ be solvable. So let's find all $c>1$ for which it is true.

Introduce $f(x): =c^x-x$. Since $c^x-x>0$ for great $x$, one have to search $c$ so that $\min f(\mathbb R)\le0$. But as $c>1$, $\min f(\mathbb R)=f\left(-\frac{\log\log c}{\log c}\right)=\frac{1+\log\log c}{\log c}\le0\Leftrightarrow c\le e^{1/e}$.

Inversely, our sequence with $c=e^{1/e}$ is majorized by $e$ (and hence converges) since $e^{1/e}<e$ (e.g. for $n=2$, $a_n=\left(e^{1/e}\right)^{e^{1/e}}<\left(e^{1/e}\right)^e=e$).

Thus the answer is $0<c\le e^{1/e}$.

Have I missed something?  :)

[quote="Vladimir Lubyshev"]So if we show that $(a_{2n})$ is decreasing and $(a_{2n+1})$ is increasing, we will be able to consider the case $c<1$ as a solution. [/quote]
How can you prove that $\lim a_{2n}=\lim a_{2n+1}$?

We can prove $\lim a_{2n}=\lim a_{2n+1}$ for example as follows. Denote $x: =\lim a_{2n}$ and $y: =\lim a_{2n+1}$ (they exist due to boundedness and monotony).

Now $a_{2n}=c^{c^{a_{2(n-1)}}}$ and $a_{2n+1}=c^{c^{a_{2n-1}}}$ so $x=c^{c^x}$ and $y=c^{c^y}$. But the equation $t=\left(c^c\right)^t$ has a unique solution since $t\mapsto t$ is strictly increasing and $t\mapsto\left(c^c\right)^t$ is strictly decreasing because $c^c<1$ for $0<c<1$. Thus $x=y$.

[quote="mathematica"][quote="Vladimir Lubyshev"]Have I missed something?  :)[/quote]
Yes, you didn't prove that when $0<c\le e^{1/e}$, the sequence $a_n=c^{a_{n-1}}$ is convergent, you only proved that the equation $c^x=x$ has an answer, which is not suffiecient for convergence. (You just proved that the sequence $a_n=e^{\frac{a_{n-1}}{e}}$ is convergent.)[/quote]

The point is that maximal range for $c$ is the interval $(0,e^{1/e}]$. Since we have proved the convergence for $c\in(0,1]$ we are working with $I: =(1,e^{1/e}]$. You can justify the convergence in this case due to the following items:

1) For $c\in I$ our sequence increases so it converges iff it is upper bounded.
2) Let $(a_n)$ correspond to $c$ and $(a'_n)$ correspond to $c'>c$ then $a_n<a'_n$. So if we prove that our sequence converges for $c=\max I=e^{1/e}$ (what we have done), we will prove that it converges for $(1,e^{1/e}]$ and hence all the interval $(0,e^{1/e}]$ works.

can you prove that when $c=e^{\frac{1}{e}},\lim a_n=e$?

Since $m: =\min f(\mathbb R)$ is attained at the unique point $x_*(c)=-\frac{\log\log c}{\log c}$ and in the case $c=e^{1/e}$ $m=0$, the point $x_*(e^{1/e})=e$ is the unique solution of the equation $f(x)=0\Leftrightarrow c^x=x$.

You may want to compare your answer with that in [url]http://www.artofproblemsolving.com/Forum/viewtopic.php?&t=48441[/url] (though no proof is given there)

I have viewed $(a_n)_{0\le n\le n_0}$ on my computer for different $n_0\in\mathbb N$ and various $0<c<1$ (and certainly $c<e^{-e}$) and the test corroborates that $(a_{2n})$ decreases, $(a_{2n+1})$ increases and that $a_{2n+1}<a_{2n}$. This means that $x: =\lim a_{2n}$ and $y: =\lim a_{2n+1}$ always exist. So if there is a mistake, it must be embodied in the strict inequality $y<x$.

When viewing $a_n$ for odd and even $n$ I noted that they tend to different "limits". But I think computer lies here because of rising inaccuracy.

$x=\left(c^c\right)^x$ and $y=\left(c^c\right)^y$ and $c^c<1$ for $0<c<1$ so $t=\left(c^c\right)^t$ must have a unique solution and hence $x=y$. Anyhow, if there is a mistake, it should be here...

Yes, it is here. The correct equation is $t=c^{(c^t)}$ ;).

Can you find those $c\in(0,1)$ for which the equation $c^{c^t}=t$ have a unique solution?

Yes, it is not that hard. Let $f(t)=c^{c^t}-t$. Then $f'(t)=c^{c^t}\cdot c^t\cdot \log^2 c-1$. Note that the maximum of the expression $c^u u$ is attained at $u=-\frac 1{\log c}$ and equals $-\frac 1{e\log c}$. Thus, $f'(t)\le - e^{-1}\log c-1$. If $c\ge e^{-e}$, the right hand side $\le 0$ and $f$ is decreasing, so it may have only one root (in the case $c=e^{-e}$ you should argue that the maximum of the derivative is achieved at just one point, so $f'(t)<0$ everywhere except one point). It is possible to show that for $c<e^{-e}$ there are several roots but it is neither necessary nor sufficient for solving the original problem. Remember, our task is to show that $a_n$ does not converge to the unique root of the equation $t=c^t$ for $c<e^{-e}$. Do you know any simple conditions on an injective mapping $F: \mathbb R\to \mathbb R$ with a fixed point $x_*$ such that the Piquard iterations can never converge to $x_*$ unless you start them at $x_*$? ;)

I don't know a general theorem concerning the question but the case when $F\colon\mathbb R\to\mathbb R$ is (strictly) increasing can be analyzed in detail.

Denote $g(x): =F(x)-x$ and for a neighbourhood $V$ of $x_*$, $V_-: =\{x\in V\mid\,x<x_*\}$, $V_+: =\{x\in V\mid\,x_*<x\}$. Let $(x_n)$ be a Piquard sequence then lets consider the only two cases

1. $x_*$ is unique.
  (a) $g\mid_{V_-}\cdot g\mid_{V+}<0\Rightarrow\lim x_n=x_*$;
  (b) $g\mid_{V_-}\cdot g\mid_{V+}>0\Rightarrow\lim x_n=x_*$ only if $x_0\le x_*$ in the case $g\mid_{V_-}>0$ and if $x_*\le x_0$ in the case $g\mid_{V_-}<0$.
2. $x_*$ isn't unique $\Rightarrow$ it is not mandatory that $\lim x_n=x_*$.

The case when $F$ is (strictly) decreasing is harder. In this case Piquard sequence $(x_n)$ can even be non-convergent. Take for example $F(x)=1-x$ then $x_{2n}=x_0$, $x_{2n+1}=1-x_0$ so $\lim x_n$ exists in the only case of $x_0=\frac12$ = fixed point of $F$.

[quote="Vladimir Lubyshev"]I don't know a general theorem concerning the question [/quote]
OK, here is the one that can be used here: if $|F'(x_*)|>1$, then the Picard iterations starting from any point other than $x_*$ can never converge to $x_*$. Such a point is called a repelling point in the theory of dynamical systems and the idea is that, for small $|x-x_*|>0$, we have $|F(x)-x_*|>|x-x_*|$.  Can you finish the proof now?

$x_*=e^{-1}$ for $c=e^{-e}\Rightarrow x_*<e^{-1}$ for $c<e^{-e}$.

$F(x): =c^x$ then $c<e^{-e}\Rightarrow |F'(x_*)|=|c^{x_*}\log c|=|\log x_*|>1$. Thus $x_*$ is repelling for all $c\in(0,e^{-e})$.

I will think tomorrow about why one can derive from it that $c<e^{-e}$ don't work...

So there is a neighborhood $V$ of $x_*$ such that $|F(x)-x_*|>|x-x_*|$ for $x\in V$.

Since $c<x_*$, we can take any $x\in V$ with $c<x<x_*$ and obtain that $a_{2n}-x_*>F^{2n-1}(x)-x_*=|F^{2n-1}(x)-x_*|>|x-x_*|>0$ so $\lim a_{2n}\neq x_*$ and hence $\lim a_n$ doesn't exist.

[b]P.S.[/b] There was a misprint before (instead of $2n$ and $2n-1$ there was $2n+1$).

[quote="Vladimir Lubyshev"]we can take any $x\in V$ with $c<x<x_*$ and obtain that $a_{2n+1}-x_*>F^{2n+1}(x)-x_*=|F^{2n+1}(x)-x_*|>|x-x_*|>0$ [/quote]
I do not quite understand how you obtained the last inequality but, perhaps, I'm missing something obvious :?.

I am sorry, I have been perplexed by indices  :wacko:  :) . Here is the rigth one:
\[ a_{2n}-x_*>F^{2n-1}(x)-x_*=|F^{2n-1}(x)-x_*|>|x-x_*|.  \]

Proof:

$c<x\Rightarrow a_2=c^c>c^x=F^1(x)\Rightarrow a_3<F^2(x)\Rightarrow a_4>F^3(x)\Rightarrow\ldots\Rightarrow a_{2n}>F^{2n-1}(x)\Leftrightarrow a_{2n}-x_*>F^{2n-1}(x)-x_*$.

$x<x_*\Rightarrow F^1(x)=c^x>c^{x_*}=x_*\Rightarrow F^2(x)<c^{x_*}=x_*\Rightarrow F^{3}(x)>c^{x_*}=x_*\Rightarrow\ldots\Rightarrow F^{2n-1}(x)>x_*\Rightarrow F^{2n-1}(x)-x_*=|F^{2n-1}(x)-x_*|>|x-x_*|$.

So $\lim a_{2n}\neq x_*$ and hence $\lim a_n$ doesn't exist.

[quote="Vladimir Lubyshev"]
$F^{2n-1}(x)>x_*\Rightarrow F^{2n-1}(x)-x_*=|F^{2n-1}(x)-x_*|>|x-x_*|$.
[/quote]
I still don't quite see why this implication is valid as written. Anyway, the theorem I cited has a simple proof: suppose that $a_n=F^n(a_0)\to x_*$. Then there exists $N$ such that all $a_n$ with $n\ge N$ belong to $V$ (in your notation). But then $|a_N-x_*|\le |a_{N+1}-x_*|\le\dots\le |a_n-x_*|$ for all $n>N$. Since $a_N\ne x_*$ (otherwise all previous $a$'s equal $x_*$ too because of the injectivity of $F$), we get that $a_n$ do not tend to $x_*$ - contradiction.

Where can I find some good material about Picard Iterations?

[quote="fedja"]Yes, it is not that hard. Let $f(t)=c^{c^{t}}-t$. Then $f'(t)=c^{c^{t}}\cdot c^{t}\cdot \log^{2}c-1$. Note that the maximum of the expression $c^{u}u$ is attained at $u=-\frac 1{\log c}$ and equals $-\frac 1{e\log c}$. Thus, $f'(t)\le-e^{-1}\log c-1$. If $c\ge e^{-e}$, the right hand side $\le 0$ and $f$ is decreasing, so it may have only one root (in the case $c=e^{-e}$ you should argue that the maximum of the derivative is achieved at just one point, so $f'(t)<0$ everywhere except one point). It is possible to show that for $c<e^{-e}$ there are several roots but it is neither necessary nor sufficient for solving the original problem. Remember, our task is to show that $a_{n}$ does not converge to the unique root of the equation $t=c^{t}$ for $c<e^{-e}$. [/quote]
I checked with Mathematica and I found out that the equation $c^{(c^{t})}=t$ has a unique solution for all $c>0.065989$(approximately).

[quote="mathematica"]
I checked with Mathematica and I found out that the equation $c^{(c^{t})}=t$ has a unique solution for all $c>0.065989$(approximately).
[/quote]
Which is $e^{-e}$ (exactly). :)
[quote="mathematica"]
But the sequence still converges for positive numbers smaller than this amount[/quote]
What sequence? :? The original one does not converge but tends to a cycle of length $2$.


I understand how to find the answer but how would you prove the following problem only using what is given (my professor did this problem but I can't remember what she did).  Also I wasn't sure where this should go:

Assume that if $ x \in \mathbb{R}$, then $ \exists ~n \in \mathbb{N} ~n>\equal{}3$  so that $ n > x$ and $ x > 0$ then $ 0 < \frac {1}{n} < x$.

Given $ A_{n} \equal{} \left[\frac {1}{n}, 2 \plus{} \frac {1}{n}\right]$, find $ \bigcup_{n \equal{} 1}^{\infty} A_{n}$ and prove your answer.

[hide="Solution"]
The answer is $ (0,3]$.

Suppose $ 1 \leq x \leq 3$. Then $ x \in A_1$, so $ x$ is in the union of the infinite family.

Suppose $ 0 < x < 1$. Using your lemma, there exists an $ n$ such that $ 1/n < x$. But since $ x < 1$, $ 1/n \plus{} 2 > x$. Therefore $ x \in A_n$.

Suppose $ x > 3$ and that $ x$ is in the infinite family. Then for some $ n \in \mathbb{N}$, $ 2 \plus{} 1/n > 3$, or $ n < 1$. Contradiction.

Suppose $ x \leq 0$ and that $ x$ is in the infinite family. Then for some $ n \in \mathbb{N}$, $ 1/n \leq 0$. Since $ n$ is positive, it is valid to multiply by it on both sides, giving $ 1 \leq 0$. Contradiction.
[/hide]

Thanks for that.  I see what you did but the answer isn't $ (0,3]$, it is $ (0,7/3]$.  It did help to point me in the right direction though.

You're clearly wrong about what the answer is: $ 3 \in A_1 \subset \bigcup_{n \equal{} 1}^\infty A_n$ and $ 3 \not\in (0, 7/3]$.

Er, did you mean to say $ n \equal{} 3$ instead of $ n \equal{} 1$ then?

Yeah, $ n\equal{}3$.  Copied it down wrong sorry

On the last part you say suppose $ x < \equal{} 0$ and then use the lemma.  Is there any way to do this without using the lemma since it states that $ x > 0$ or is this justified to use?  If so, please explain.

There are lots of ways to reject nonpositive $ x$, e.g. write "Every $ A_i$ is a subset of $ {\bf R}_{> 0} \equal{} (0, \infty)$, so their union must be so as well."


geometry perimeter algebra polynomial inradius incenter AMC

A right triangle has perimeter 32 and area 20.  What is the length of its hypotenuse?

I think this is pretty standard for the AMCs.
Anyone have a different solution from the following?

[hide="Solution"]Let the legs be of length $ a,b$ and the hypotenuse have length $ h$.

We are given that $ a\plus{}b\plus{}h\equal{}32$ and $ ab\equal{}40$.
We wish to find the value of $ h$.

Note that $ (a\plus{}b)^2\equal{}a^2\plus{}b^2\plus{}2ab\equal{}80\plus{}h^2$ (by the Pythagorean theorem).
We then have that $ a\plus{}b\equal{}\sqrt{80\plus{}h^2}$
Thus, $ h\plus{}\sqrt{80\plus{}h^2}\equal{}32$
$ h^2\plus{}80\equal{}h^2\plus{}1024\minus{}64h$
$ 64h\equal{}944$
$ h\equal{}\frac{59}{4}$[/hide]

I had something like that; just different substitutions of $ ab$.

indeed it is. you're solution is probably the most straightforward and obvious. btw, good job on the amc 10!

I was mad because this looked like a really classic problem, yet I wasn't able to think of anything except Heron's formula.

Thanks mihail.  I guess heron's would work, but it would take too long.  Usually there's another way to solve these problems.

Urgh.
I didn't know how to do this question, so I was trying to use guess and check to find the legs... but as it turns out the legs are irrationals  :| so this is the only question I left blank.
Yeah, after the test I cheated using my calculator's equation solver and got 14.75 aka 59/4.

[quote="xpmath"][hide="hehe"] After using the solution Lyndon put up, I decided to make sure it was right and brute forced it. Basically, let the legs be x and y, so you get that the hypotenuse squared+2xy, which is the hypotenuse plus, um, 40 I think, is a perfect square. Yay for brute force.[/hide][/quote]

Who's Lyndon?

Aggg...
I wasted like 5 minutes on this problem.
Finally, I got to the point where I was able to find the only h from the 5 choices that gave a rational perimeter so I plugged it in and it worked...
I can't believe I didn't just square the whole thing.

[quote="Altheman"]$ A \equal{} rs$ implies that $ r \equal{} \frac {5}{4}$. But in a right triangle, $ r \equal{} s \minus{} a$, so $ a \equal{} 16 \minus{} \frac {5}{4} \equal{} \frac {59}{4}$.[/quote]

I suspected there'd be a solution along these lines.  Nice find :)

The most general solution I found was to substitute the elementary symmetric polynomials $ u \equal{} ab, v \plus{} a \plus{} b$.

[quote="123456789"][quote="Altheman"]$ A \equal{} rs$ implies that $ r \equal{} \frac {5}{4}$. But in a right triangle, $ r \equal{} s \minus{} a$, so $ a \equal{} 16 \minus{} \frac {5}{4} \equal{} \frac {59}{4}$.[/quote]

Where did you learn that the inradius of a right triangle is the semi-perimeter minus the hypotenuse? And how is this proven?[/quote]
Yeah I'm curious to know how this is proven too. I thought there might be some simple way that existed but I didn't know about (just as usual :() but that simple...wow.

This is really an elegant solution...

A proof:
A personal notation: in a general triangle inscribed by a cricle, a vertex's [i]tangent length[/i] is the distance from the vertex to a point of tangency on its adjacent side.
If we call these lengths $ a',b',c'$, respectively, set a system:
$ a' \plus{} b' \equal{} c$
$ b' \plus{} c' \equal{} a$
$ c' \plus{} a' \equal{} b$
Substitute the half-perimeter with $ s$, we can easily obtain the solution:
$ (a',b',c') \equal{} (s \minus{} a,s \minus{} b,s \minus{} c)$.
For a right triangle, note that the incenter, right angle vertex, and the two points of tangency on legs form a square. Thus the inradius equals to the right angle vertex's length, which is $ s \minus{} a$ for hypotenuse $ a$.

I somehow (and luckily) used guess and check to solve this, and increased all of the terms by 4 accordingly.

[quote="timwu"]For a right triangle, note that the incenter, right angle vertex, and the two points of tangency on legs form a square.[/quote]

This is the important point.  We immediately derive the relation $ r \equal{} \frac{a \plus{} b \minus{} c}{2}$.  Additionally, substituting the Pythagorean triple-generating formulae $ a \equal{} 2mn, b \equal{} m^2 \minus{} n^2, c \equal{} m^2 \plus{} n^2$ we can derive the little-known formula

$ r \equal{} n(m\minus{}n)$.

This formula can be used to find all integer right triangles of a given inradius.  It's not exactly widely applicable, but it's nice to know :)

[quote="timwu"]This is really an elegant solution...

A proof:
A personal notation: in a general triangle [b]inscribed by a cricle[/b], a vertex's [i]tangent length[/i] is the distance from the vertex to a point of tangency on its adjacent side.
[/quote]

I think you mean that the circle is inscribed in the triangle  :wink: Then yes, that is a nice solution.  You can also then find the length of $ a',b',c'$... Which is sometimes useful.


haha i must be stupid... i cant factor the thing... well... i guess if x will stand for the sum of all the values of x... you can say that the sum of all the roots of that equation is 1 (i think, i used (-b/a)). then that would mean 1-(1/1)=0... but i guess that is a long shot.

edit- i dunno if this will help, but i think that the product of all the roots is -1. I think that (-c/a) tells you the product of the roots of a third degree equation.

heh would you mind showing your work please?

[hide]We can rewrite $x^{3}-x^{2}+x=\frac{1}{3}$ as $3x^{3}-3x^{2}+3x-1=0$.  We want to find $1-\frac{1}{x}$.  Let this value equal $y$.  Cubing both sides of this equation, we get:
$1-\frac{3}{x}+\frac{3}{x^{2}}-\frac{1}{x^{3}}=y^{3}$.  Multiply both sides of this equation by $x^{3}$ and we have $x^{3}-3x^{2}+3x-1=y^{3}x^{3}$.

Now we have the two equations:
$3x^{3}-3x^{2}+3x-1=0$
$x^{3}-3x^{2}+3x-1=y^{3}x^{3}$.

Subtracting these equations gives us $2x^{3}=-y^{3}x^{3}$, so $-2=y^{3}$, and $y=\sqrt[3]{-2}$.  Since $y$ is the desired expression, our answer is $\sqrt[3]{-2}$.[/hide]

all right here is the proof: (this is gonna look bad, because i don't latex, maybe some one can rewrite this)

here is the original equation:
(x^3)-(x 2 )+x-(1/3).
lets call the coiffitients a,b,c. so i won't have to deal with numbers, till the end. so our equation is this:
x3 + ax2 + bx + c = 0
now i define t, such that: x = t - a/3.
The substitution x = t - a/3 eliminates the quadratic term; in fact, we get the equation:
t3+Pt+Q=0, where P=b-(a 2 /3), and Q=c+[(2a3-9b)/27]
This is called the depressed cubic.

Suppose that we can find numbers u and v such that

u3-v3=Q, and u.v=p/3
A solution to our equation is then given by

t = v − u, (as can be checked by directly substituting this value for t in the original equation above).
now, if you solve for "v" you get:
"v"=P/(3u). know plug this into the u3-v3=Q equation, you get:
u=[(Q/2)-( :sqrt: [(Q 2 /4)+(P3/27)]]^1/3
Since t = v − u and t = x + a/3, we find
X=(P/3u)-u-(a/3)
which will be equal to
X=[(-432^(1/3))/18]+[2/(54^(1/3))]+(1/3) = 1/(1+ 2^1/3)=.442493334
and 1-(1/x) will equal to -2^(1/3) :D

[quote="dangvh"]edit- i dunno if this will help, but i think that the product of all the roots is -1. I think that (-c/a) tells you the product of the roots of a third degree equation.[/quote]

Yes, the product of the roots is $\frac {-c}{a}$.  However, you need to get a polynomial equal to zero before you can use this formula.  So you have to subtract the $\frac 1 3$ over to get it equal to zero, then use $\frac {-c}{a}$ to find the product of the roots is $\frac 1 3$.  Although you are right, this doesn't help us at all  :)

or you can get mathematica to solve to for you!

rearranging the terms of the equation , we get

(1 - 1/x ) 3  = -2 = 2e  i:pi:   where i  = :sqrt: -1

Substituting 1-1/x by  z,

we have

z3  = -2 = 2e  i:pi:
the solutions of which are

 z = 2 1/3e i(2k+1):pi:/3   where k = 0,1,2

So the solutions are
1-1/x =  2 1/3(1/2 + i :sqrt: 3/2) ,

            2 1/3(1/2 + i :sqrt: 3/2)

           - 2 1/3


limit
real analysis
real analysis unsolved

To solve this problem , maybe you need to prove that : $ n^x-(n-1)^x \to 0 $ if $ x <1 $ , x is not 0 .


$ The AB\parallel{}DC$ and $ AB>DC$  in trapezium $ ABCD$.If $ P$ and $ Q$ be the mid-points of the diagonals $ AC$ and $ BD$,prove that $ 1.PQ\parallel{}AB$
2. $ PQ$=$ \frac{1}{2}(AB\minus{}DC)$.

Thanks again.

1) P and Q are both halfway distant from AB and CD since they are the midpoint of the diagonals, thus they are both at the same distance from AB and CD, and thus PQ is parallel to each of them
2) Draw the line through PQ, let it intersect AD and BC at M and N. As we said before, line MPQN is parallel to both AB and CD and equidistant from each, so $ PM\equal{}(1/2)CD$, $ QN\equal{}(1/2)CD$, and MN, the midsegment, is $ \frac{AB\plus{}CD}{2}$, so $ PQ\equal{}MN\minus{}PM\minus{}QN\equal{}(1/2)(AB\minus{}CD)$

Producing $ CQ$ and $ DP$ to meet $ AB$ at $ Y$ and $ X$ resp.
In $ \triangle DCQ$ and $ \triangle QYB$
$ DQ = QB$; $ \angle DQC = \angle YQB$; $ \angle DCQ = \angle BYQ$

$ \triangle DCQ \cong \triangle QYB$
$ CQ = QY$
since $ P$ and $ Q$ are mp. of $ AC$ and $ CY$
thus, $ PQ \parallel AB$

also, $ DCBY$ is a pa^gm
since $ Q$ bisects $ CY$ and $ BD$
$ DC = BY$
As, $ PQ = \frac {AY}{2}$
      $ = \frac {AB - BY}{2}$
      $ = \frac {AB - DC}{2}$

Thanks a lot.I have already troubled you enough.

@dgreenb801, my proof for PQ||AB was precisely same as yours.

Both solutions were excellent,But easy for you people.

Thanks again.

Joining $ B \minus{} P$ and $ A \minus{} Q$
$ [APB] \equal{} \frac {[ABC]}{2}$
$ [AQB] \equal{} \frac {[ADB]}{2}$
$ [ADB] \equal{} [ABC]$
Thus, $ [APB] \equal{} [AQB]$
since area of triangles standing on $ AB$: $ \triangle APB$ and $ \triangle AQB$ are equal
$ PQ \parallel AB$


solve in R the following system:

xy+[x][y]=[z]
x[y]+[x]y=z

Solution:

[x]y+x[y]-xy-[x][y]=z-[z] which is equivalent to:

(x-[x])([y]-y)=z-[z] or (x-[x])(y-[y])+(z-[z])=0

because alpha>=[alpha] , alpha from R we get that
z=[z] and y=[y] or z=[z] and x=[x].

if x=0 then z=0 and y is arbitrary.
if x=[x] then x is an integer and z=x(y+[y]).
and if x<>0 then z/x=[y]+y. we have that {y}+2[y]={z/x}+[z/x] from which we get:
[z/x]==0(mod 2) and {y}={z/x} and [y]=1/2 [z/x]
we get that y=1/2[z/x]+{z/x}.
so we get the solutions:
x=o,y=beta,z=0, beta is a real number;
x=p, y={q/p}+1/2[q/p], z=q where q is from Z,p from Z\{0} and 0=[q/p].

hope it is correct!
tschuess! :D :D


1. Establish that if an $ m \times n$ matrix has a right inverse, then $ m \le n$.

2. Is there a matrix $ A$ that has one and only one right inverse but has no left inverse?  Prove your claim.

1. is answered here [url]http://www.mathlinks.ro/viewtopic.php?t=209895[/url]
2. Suppose $ A$ does not have a left inverse, then $ m < n$, and since $ AX \equal{} I_m$ is consistent it follows that it has infinitely many solutions, contradicting our hypothesis.


it's a simple application of cauchy's inequality :D

by cauchy ,it suffices to prove that (c²+d²)²>=ac+bd
but:    (a²+b²)(c²+d²)>=(ac+bc)²
hence:(c²+d²)²>=ac+bd.

Seif.


Be careful with your symmetric sum notation -- $ [4,1,1]\equal{}\sum_{sym} a^4bc$.

This inequality can be shown just from Schur's and AM-GM (or Muirhead):

$ 3\sum_{sym} \frac{1}{2}(a^5b\plus{}b^5a)\ge 3\sum_{sym} a^3b^3$ by AM-GM.
$ \sum_{sym} \frac{3}{4}a^5b\plus{}\frac{1}{4}ab^5\ge \sum_{sym} a^4b^2$ by AM-GM.
$ abc\sum_{sym} a(a\minus{}b)(a\minus{}c)\ge 0\Longrightarrow \sum_{sym} a^4bc\plus{}\sum_{sym} a^2b^2c^2\ge 2\sum_{sym} a^3b^2c$ by Schur's.

Adding these together, we get the desired result.

[quote="samath"]Be careful with your symmetric sum notation -- $ [4,1,1] \equal{} \sum_{sym} a^4bc$.

[/quote]

no, by convention $ \sum_{sym} a^4bc \equal{} a^4bc \plus{} b^4ca \plus{} c^4ab$ (it doesn't count $ a^4bc$ and $ a^4cb$ as different), whereas $ [4,1,1]$ is the double of that  :!:

e.g. when you write $ \sum_{sym}a^{2}b^{2}c^{2}$ it is really not clear how many terms this is supposed to have. If you want to write the sums in an 'homogenous' way, better write your last line as $ [4,1,1]\plus{}[2,2,2]\ge 2[3,2,1]$. :)


Find the number of the second digit from the right most for $11^{12^{13}}.$

[hide]The second right-most digit in powers of 11 repeats every 11 powers.

$12^{13}\equiv x\bmod{11}$

$1^{13}\equiv x\bmod{11}$

$x=\boxed{1}$[/hide]

[hide]
We'll find it mod 100 then take the first digit.

$\phi(100) = 40$
From a bit of calculation, $12^{13}= 32 \pmod{40}$

$11^{32}= \boxed{2}1 \pmod{40}$
[/hide]

Well first of all, where did you pull that statement from? (That second from left repeats in 11's)

I doubt it is true, maybe the first time a certain number you tested repeats is 11... but it is actually 1 cycle every 40, right? (or a divisor of 40, it is the order of 11 mod 100)

[quote="me@home"]Well first of all, where did you pull that statement from? (That second from left repeats in 11's)[/quote]
$11^{1}=11$
$11^{2}=\ldots21$
$11^{3}=\ldots31$
$11^{4}=\ldots41$
$11^{5}=\ldots51$
$11^{6}=\ldots61$
$\vdots\qquad\qquad\vdots$

After finding my error...

[hide]The second right-most digit in powers of 11 repeats every 10 powers.

$12^{13}\equiv x\pmod{10}$

$2^{13}\equiv x\pmod{10}$

$2\equiv x\pmod{10}$

$x=\boxed{2}$[/hide]

[quote="me@home"][quote="JavaMan"][hide]
We'll find it mod 100 then take the first digit.

$\phi(100) = 40$
From a bit of calculation, $12^{13}=12((4^{3})^{2})=(12)(16)=14^{2}-2^{2}= 32 \pmod{40}$

$11^{32}= (121)^{1}6=\boxed{0}1 \pmod{40}$
[/hide][/quote]

Just showed some steps and corrected your mistake[/quote]

Your last step should be $\pmod{100}$ :)

Oh, right :blush:

Well having established the property mod 40, we look for a property mod 25:
$11^{2}=-4$

Thus $11^{32}=4^{16}=2^{32}=16(128^{4})=16(81)=96=-4 \mod{25}$
which we can then solve for... but I think I messed up since $\phi(25) = 20 \not = 15$ which is sort of implied...


In the triangle $ABC$, the lines $AE$ and $CD$ are the bissectors of $\angle BAC$ and $\angle ACB$ respectively; $E \in BC$ and $D \in AB$. For which amplitude of the angle $\angle ABC$ we have :

a) $|AD| + |EC| = |AC|$
a) $|AD| + |EC| > |AC|$
a) $|AD| + |EC| < |AC|$

[hide="Solution"]Let $a=BC, b=AC, c=AB, \beta=\angle ABC$, and let $\rho$ denote either of the relations $=,<,>$ we want to explore.

We'll keep in mind that $a^2+c^2=b^2+2ac\cos\beta$. From the known property of angle bisector, we have ${AD={b\over{a+b}}c}$ and ${EC={b\over{b+c}}a}$, so the problem becomes

$\begin{eqnarray*}{{bc}\over{a+b}}+{{ab}\over{b+c}} &\rho& b\\ {c\over{a+b}}+{a\over{b+c}} &\rho& 1\\ c(b+c)+a(a+b) &\rho& (a+b)(b+c)\\a^2+c^2 &\rho& ac+b^2\\b^2+2ac\cos\beta &\rho& ac+b^2\\ 2\cos\beta &\rho& 1\\ \cos\beta &\rho& {1\over 2}$

Hence, case (a) is valid for $\beta=60^\circ$, case (b) for $\beta<60^\circ$ and case (c) for $\beta>60^\circ$.[/hide]


I just found this FireFox addon, and I wanted to show it to you guys - https://addons.mozilla.org/en-US/firefox/addon/4082

I find it pretty useful.  There are buttons that you can click and preview your LaTeX formulas to make it easier to type.

Hard to explain, so check it out for yourself!  :)


Prove that if an analytic function fails to assume a constant value over all the interior points of a region, then it is not constant throughout any neighborhood of any interior point of that region; in addition, $ |f(z)|$ will not be constant in any such neighborhood.

I think this is the same question as 2 here: http://www.artofproblemsolving.com/Forum/viewtopic.php?p=1233591#1233591


What would you prefer a future based on your parents social position (for example herediting and living on that only) or on your own merits skills and abilities and why.

I think I'd split it 1 to 3 or, you know, a bit of both, because your parents' lives and what you learn from them is part of who you are. But your own actions are much more important bacuase they define what kind of person you are, not your parents. I guess I like the second choice better if I had to pick one.

Most likely the latter.  I sincerely appreciate my parents and their effort, but wherever I go in life, I want it to be by way of my own effort.

[quote]I agree with RobinHood3000, mainly becuase of that warm feeling you get when I see that you did something all by yourself.[/quote]
Hahahaha me too, it makes me feel proud! :D


inequalities
real analysis
real analysis unsolved

ƒ(x) is defined and continuous on (1;+∞). Assume  0<a<1 and c>0 and │ƒ(x)│<c(x^a) , for all x>1 .Prove:
For any T>0 there is X(n)→+∞ s.t   lim(ƒ(X(n)+T)- ƒ(x))=0. (as n→+∞)

Choose T, and build $x_n$ terms one by one. Suppose $x_1,...,x_n$ built with $|f(x_i+T)-f(x_i)|<1/i , x_i>i$ then consider $g(x)=f(x+T)-f(x) on [n,+\infty]$, if |g(x)|>1/(n+1) for any x you have either g(x)>1/(n+1) or g(x)<-1/(n+1). From here it is easy to conclude that f cannot verify the inequality you mentioned. $x_{n+1}$ is built and the sequence too.


conics
geometry
circumcircle
parabola
ratio
geometric transformation
homothety

1) Let $ P_1P_2P_3....P_n$ be a bicentric polygon with incircle $(I)$ and circumcircle $(O).$ $(I)$ touches the sides of the polygon through $A_1,A_2,A_3, ... ,A_n.$ Prove that  lines $A_1A_2,$ $ A_2A_3,$ $A_3A_4,..., $ $A_nA_1$ bound a polygon circumscribed in a conic.

2) Assume that $ ABCD$ is a bicentric quadrilateral with incircle $(I)$ and circumcircle $(O).$  $O_1, O_2, O_3$ and  $ O_4$ denote the circumcenters of $ \triangle OAB,$ $\triangle  OBC,$ $\triangle OCD$ and $ \triangle ODA.$ Show that the parabola that passes trough $ O_1, O_2, O_3$ and  $ O_4$  has focus $O,$  if and only if the ratio between the areas of $(I)$ and $(O)$ is $ 2\varphi \minus{} 1,$ where $\varphi$ is the golden number.

1) Lines $ A_1A_2, A_2A_3, A_3A_4,$ etc. are polars of $ P_1, P_2, P_3 $ etc WRT $(I)$ $\Longrightarrow$ Sidelines of the polygon $A_1A_2A_3A_4...A_n$ are tangent to the polar conic $\mathcal{K}$ of $(O)$ with respect to $ (I).$ The foci of $\mathcal{K}$ are the centers of the inverse $(O')$ of $(O)$ under the inversion WRT $(I)$ and the center of the homothetic circle $(O'')$ of $(O')$ under the homothety with center $I$ and coefficient $2$ $\Longrightarrow$ $IO$ is the focal axis of $\mathcal{K}.$

2) $ \frac {r^2}{R^2} \equal{} 2\varphi \minus{} 1$ $\Longleftrightarrow$ $(I)$ passes through $O,$ due to [url=http://www.cut-the-knot.org/Curriculum/Geometry/Fuss.shtml]Fuss theorem[/url].

Inversion WRT $(O)$ takes lines $ AB, BC, CD, AD$ into circles $\odot OAB,$ $\odot OBC,$ $\odot OCD,$ $ \odot ODA$ and $(I)$ tangent to $AB, BC, CD, AD$ into a tangent line $ \tau$ to the circles $\odot OAB,$ $ \odot OBC,$ $\odot OCD,$ $\odot ODA$ $\Longrightarrow$ Parabola $\mathcal{K}$ passing through $ O_1, O_2, O_3$ and $O_4$  has focus $ O$ and directrix $\tau.$


integration
function
calculus
inequalities
real analysis
real analysis solved

f:R->R continuous s.t. for any a<>b we have 1/(b-a)* \int {a, b} f(x)dx >= f(a). Prove that f is constant.

I realize this isn't too challenging, but I'll post some more.

Lets show that f is constant on any interval  [a;b]

[b]Lemma:[/b] If a continuos function doesn't change its sign on a compact interval, and its integral on that interval is 0, then the function is 0 on that interval.

[b]Proof of lemma[/b]: You do that :D :D

So lets consider an interval [a,b] with a<b.
From Weierstrass there exist x0 in [a,b] so that f(x0)=max(f(x)) with a<=x<=b.

So we have three cases:

[b]1)[/b] x0=a. then we have that (using the mean theorem ):

 \int (from a to b) f(x)dx= f (  \varepsilon )(b-a)<=f(a)(b-a).

we know that:  1/(b-a)* \int {a, b} f(x)dx >= f(a) which is equivalent to:

 \int {a,b} f(x)dx>=f(a)(b-a).

From the two inequalities we get that:

 \int {a,b} f(x)dx=f(a)(b-a) which is equivalent to the following fact:
 \int {a,b} [f(x)-f(a)]dx=0

from the fact that: f(x)<=f(a) for any x in [a,b] from the lemma we have that: f(x)-f(a)=0. d.h. f is constant in [a,b].

[b]2)[/b]: x0=b. Is the same with the first case!

[b]3)[/b]: x0 is in (a,b).

From the mean theorem we have that:
 \int {a,x0} f(x)dx = f (  \varepsilon 1) (xo-a)<=f(x0)(x0-a).

We know that:  1/(x0-a)* \int {a, x0} f(x)dx >= f(x0) which is equivalent to:  \int {a,x0} f(x)dx>=f(x0)(x0-a).
so we get that:  \int {a,x0} f(x)dx=f(x0)(x0-a) or :
 \int {a,x0} [f(x)-f(x0)]dx=0 and we have that f(x)<=f(x0) for x in [a,x0] and from the lemma we get that f(x)=f(x0) for any x in [a,x0].
you do the same for x in [x0,b] so from the two subcases of 3) we get that f(x)=f(x0) for any x in [a,b].

problem solved!!! :D :D
cheers!!! :D :D

Thx. I thought you said you didn't know integrals  :D


geometry
circumcircle
function
modular arithmetic
induction
LaTeX

1) Find the value of

(1^4+2007^4+2008^4)/(1^2+2007^2+2008^2)

2) Find all solutions in positive integers [i]x,y,z[/i] to the simultaneous equations.

[i]x+y-z=12
x^2+y^2-z^2=12[/i]

3) Let ABC be a triangle with an obtuse angle at A. Let Q be a point (other than A,B, or C) on the circumcircle of the triangle, on the same side chord BC as A, and let P be the other end of the diameter through Q. Let V and W be the feet of the perpendiculars from Q onto CA and AB respectively. Prove that the triangle PBC and AWV are similar. (Note: the circumcircle of the triangle ABC is the circle which passes through the vertices A,B and C.)

4) Let S be a subset of the set of numbers {1,2,3,...,2008} which consists of 756 distinct numbers. Show that there are two distinct elements [i]a,b[/i] of S such that [i]a+b[/i] is divisible by 8.

5)Let P be an internal point of the triangle ABC. The line through P parallel to AB meets BC at L, the line through P parallel to BC meets CA at M, and the line through P parallel to CA meets AB at N. Prove that
[i]BL/LC X CM/MA X AN/NB (is equal or smaller than) 1/8[/i]
and locate the position of P in the triangle ABC when equality holds.

6) The function [i]f[/i] is defined on the set of positive integers by [i]f(1)=1, f(2n)=2f(n) and nf(2n+1)=(2n+1)(f(n)+n)[/i] for all values where n is equal or bigger than 1.
(i) Prove that [i]f(n)[/i] is always an integer.
(ii) For how many positive integers less than 2007 is [i]f(n)=2n[/i]

PS: I'm sorry I do not have any programs to write the equations above in a more mathematical format that would be more pleasing for the eye.

4. for $ 0\leq i \leq 7$, let $ S_i$ be the set of elements in $ \{1,...,2008\}$ that are $ i\pmod{8}$. now suppose that $ X\subseteq \{1,...,2008\}$ is a set such that no two distinct elements have a sum divisible by $ 8$. then $ X$ has at most $ 1$ element in each of $ S_0$ and $ S_4$, and at most $ 251$ elements in each of $ S_1\cup S_7$, $ S_2\cup S_6$, $ S_3\cup S_5$. this is because $ |S_i| \equal{} 251$ for $ i\neq 0,4$. thus $ X$ has at most $ 2\cdot 1 \plus{} 3\cdot 251 \equal{} 755$ elements. done.

6. rewrite (ii) as $ f(2n \plus{} 1) \equal{} 2f(n) \plus{} \frac {f(n)}{n} \plus{} 2n \plus{} 1$. we prove that $ f(k) \equal{} kn_k$, where $ n_k$ is the number of 1's in the base 2 representation of $ k$. the base cases are easy to check. assume that the statement is true for all $ n < m$. write $ m \equal{} 2^{a_1} \plus{} 2^{a_2} \plus{} \cdots \plus{} 2^{a_r}$, where $ a_1 < a_2 < \cdots < a_r$. if $ a_1\geq 1$, we have by the induction hypothesis that $ f(m) \equal{} 2f(2^{a_1 \minus{} 1} \plus{} \cdots \plus{} 2^{a_r \minus{} 1}) \equal{} 2\cdot \frac {m}{2}\cdot r \equal{} mr$, so the formula holds for $ m$. if $ a_1 \equal{} 0$, we have by the induction hypothesis, and the first sentence above, that $ f(1 \plus{} 2(2^{a_2 \minus{} 1} \plus{} \cdots \plus{} 2^{a_r \minus{} 1})) \equal{}$ $ 2f(2^{a_2 \minus{} 1} \plus{} \cdots \plus{} 2^{a_r \minus{} 1}) \plus{} (r \minus{} 1) \plus{} m \equal{} 2\cdot(r \minus{} 1)\cdot\frac {m \minus{} 1}{2} \plus{} (r \minus{} 1) \plus{} m \equal{} mr$, so again the formula holds.

so for the second part we need to find the number of $ n\leq 2006$ that have exactly 2 $ 1$'s in their base-2 representations. since $ 2006$ starts $ 11...$ and has $ 11$ digits in its representation, the answer is $ \binom{11}{2} \equal{} 55$

2. substituting $ z \equal{} x \plus{} y \minus{} 12$ in the second equation and simplifying, we have

$ x^2 \plus{} y^2 \minus{} (x \plus{} y \minus{} 12)^2 \equal{} 12 \implies \minus{} 144 \minus{} 2xy \plus{} 24(x \plus{} y) \equal{} 12 \implies$

$ \implies xy \minus{} 12x \minus{} 12y \equal{} \minus{} 78 \implies (x \minus{} 12)(y \minus{} 12) \equal{} 66$. (*)

factoring $ 66$ and keeping in mind that we must have $ x \plus{} y > 12$ (since $ z > 0$), (*) easily gives all solutions.

3. another easy one. since quadrilateral $ ABPC$ is cyclic, $ \angle WAV \equal{} \angle BPC$. since $ WAVQ$ is cyclic, $ \angle WVA \equal{} \angle WQA$. but $ \angle QAP \equal{} 90$, so $ \angle WQA \equal{} 90 \minus{} \angle WAQ \equal{} \angle BAC \equal{} \angle BCP$. done.

i think for question 2, instead of 60, its 66?

you're right. i've fixed it. it's also better with 66 cuz we have fewer factors :)

kokming, click on LaTeX at the left hand side of the screen and you can learn to present the problems in LaTeX. Seems like all the problems (except #5) have been solved!

for 5) denote by $ [XYZ]$ the area of $ \triangle XYZ$ . and let $ [BPC] \equal{} x,[APC] \equal{} y$ and $ [APB] \equal{} z$ . we have $ \frac {AM}{MC} \equal{} \frac {[ABC]}{[BPC]} \minus{} 1 \equal{} \frac {y \plus{} z}{x}$ and $ \frac {CL}{BL} \equal{} \frac {[ABC]}{[APB]} \minus{} 1 \equal{} \frac {x \plus{} y}{z}$ and $ \frac {BN}{AN} \equal{} \frac {[ABC]}{[BPC]} \minus{} 1 \equal{} \frac {x \plus{} z}{y}$ so $ \frac {CL}{BL}.\frac {AM}{MC}.\frac {BN}{AN}. \equal{} \frac {y \plus{} z}{x}.\frac {x \plus{} z}{y}.\frac {x \plus{} y}{z} \ge 8$.

for 6) $ \star$ we can prove easily by induction that $ f(n)$ is an integer number .$ f(1) \equal{} 1$ is true , suppose that  $ f(1),f(2),....f(n)$  are integers ( so $ k$ must divide $ f(k)$ for $ k \in${$ 1,...,n$}$ )$.if $ n \plus{} 1$ is even then $ f(n \plus{} 1) \equal{} 2f(\frac {n \plus{} 1}{2})$ is an integer ,if it is odd  let $ n \plus{} 1 \equal{} 2p \plus{} 1$ then $ p \le n$ and$ f(n \plus{} 1) \equal{} f(2p \plus{} 1) \equal{} (2p \plus{} 1)(1 \plus{} \frac {f(p)}{p})$ is an integer .
$ \star$ now ,if $ b$ is odd $ > 1$ then $ f(b) \ge 2b$ since for  $ b \equal{} 2l \plus{} 1$ we have  $ f(b) \equal{} (2l \plus{} 1)(1 \plus{} \frac {f(l)}{l}) \ge 2(2l \plus{} 1) \equal{} 2b.$
anf if $ n$ is not power of $ 2$,let $ n \equal{} 2^sb$ $ (b$ is odd $ )$ then $ f(n) \equal{} f(2^sb) \equal{} 2^sf(b) \ge 2^{s \plus{} 1}b \equal{} 2n.$
now if $ f(n) \equal{} 2n$ then$ f(b) \equal{} 2b$ so $ f(l) \equal{} l(b \equal{} 2l \plus{} 1)$ ,so $ l$ must be power of $ 2$, let $ l \equal{} 2^q$ then  $ n \equal{} 2^sb \equal{} 2^s(2l \plus{} 1) \equal{} 2^s(2^{q \plus{} 1} \plus{} 1)$. so the number of such $ n$ is the number of positive integers $ s$ and $ q$ such that $ 2^s(2^{q \plus{} 1} \plus{} 1) \le 2007 \le 2^{11}$ which it is $ \binom{11}{2}.$


trigonometry
function
ratio
continued fraction

I was looking through an old problem book and found a continued fraction that I don't know how they derived.  Can someone help me understand the following:

[tex]
\coth{\frac{1}{k}}=k+\frac{1}{3k+\frac{1}{5k+\frac{1}{7k+\frac{1}{\ddot}}}}
[/tex]

[quote="The_Cow_King"]by coth they mean hyberbolic cotangent, like cosh and sinh?[/quote]

To answer your question simply: yes.

coth (1/k) = 1/tanh(1/k) = (e^(1/k)+e^(1/k) )/(e^(1/k)-e^(1/k) )

I know that stuff like sine and cosine can be represented by series (called Fourier series I think, calc stuff), so I would not doubt that hyperbolic trig functions can be represented as some infinite sum or continued fraction, since they have many of the same properties/identies (hyperbolic trig functions and regular trig functions).  I would start by thinking of ways to represent e as a continued fraction or infinite series or something like that.  Mind you I don't really know what I am talking about.

I am somewhat familiar with the notation (coth, continued fraction, fourier series, etc.).  I am mainly looking for a derivation of the continued fraction itself along with an explanation/generalization for similar functions and how they can be expressed as continued fractions.  I hope someone can help.

I am interested to see an answer to this question.  ComplexZeta probably knows since he seems to know the answer to every obscure math question...

I seem to have developed a reputation I don't deserve. Anyway, I don't see any way to prove this result. I agree that it's a neat one. I could possibly find by someone else if I were to look on the web and in books, but that's not within the spirit of the question, in my opinion. (I guess the secret is out; I don't know everything. ;))

Progress!

[tex]\displaystyle \cosh(x)= 1 + \frac{x^2}{2!} + \frac{x^4}{4!} + \ldots[/tex]
[tex]\displaystyle \sinh(x) = x + \frac{x^3}{3!} + \frac{x^5}{5!} + \ldots[/tex]

Everywhere other than 0, which is where Coth(1/x) is defined, we have
[tex]\displaystyle \cosh(\frac{1}{x})= 1 + \frac{x^{-2}}{2!} + \frac{x^{-4}}{4!} + \ldots[/tex]
[tex]\displaystyle \sinh(\frac{1}{x}) = x^{-1} + \frac{x^{-3}}{3!} + \frac{x^{-5}}{5!} + \ldots[/tex]

Thus,

[tex]\displaystyle \coth(1/x) = \frac{1 + \frac{x^{-2}}{2!} + \frac{x^{-4}}{4!} + \ldots}{x^{-1} + \frac{x^{-3}}{3!} + \frac{x^{-5}}{5!} + \ldots}[/tex]

Now, do the first term of the long division -- it's 1/(1/x) = x.  Reciprocate the result and do the first term of the next long division -- it's (1/x)/((1/2 - 1/6)/x^2) = 3x.  And so on.

A proof?  Well, no.  Also, you might have to be finicky about convergences and such.  (It does converge pointwise everywhere except at 0, where the original function is undefined anyhow. And I think it converges uniformly outside of any fixed open ball around 0, but I'm not sure.)  But I think manipulating these series is definitely possible to make it work beyond just calculating a few terms like I've done.

A question for the great TexMasters -- how do I insert line breaks within the [ tex ] tags?  Or is it not possible?

[quote="JBL"]A question for the great TexMasters -- how do I insert line breaks within the [ tex ] tags?  Or is it not possible?[/quote]

Use \quad for medium space breaks and \qquad for larger space breaks. (There are also commands for subtle smaller space breaks, but I assume you're asking how to make these bigger ones).

Thanks JBL.  That demonstration was straightforward and easy to follow.  I'm sure it wouldn't be a leap to use a similar process to derive other functions as continued fractions either.

Well, as long as they could be expressed as a ratio of two functions with some kind of power series, yes.  I'm not sure if what I wrote can be proven, but it looks like it might work inductively or something like that.  It does look quite nice, though, even unproven :)  [/"taking pride in own handiwork"]

Yes applying the same process, you can derive

[tex]\cot{\frac{1}{x}}=x-\frac{1}{3x-\frac{1}{5x-...}}[/tex]


I have 5 marbles numbered 1 through 5 in a bag.  Suppose I take out two different balls at random.  What is the expected value of the sum of the numbers on the balls?

The least sum is $ 1\plus{}2\equal{}3$. The highest sum is $ 5\plus{}4\equal{}9$. The total value of all sums is $ 3\plus{}4\plus{}5\plus{}...\plus{}8\plus{}9\equal{}45\minus{}3\equal{}42$. The number of ways to pick $ 2$ is $ \dbinom{5}{2}\equal{}10$. Thus, our answer is $ \frac{42}{10}\equal{}4.2$

[quote="akhil0422"]The least sum is $ 1 \plus{} 2 \equal{} 3$. The highest sum is $ 5 \plus{} 4 \equal{} 9$. The total value of all sums is $ 3 \plus{} 4 \plus{} 5 \plus{} ... \plus{} 8 \plus{} 9 \equal{} 45 \minus{} 3 \equal{} 42$. The number of ways to pick $ 2$ is $ \dbinom{5}{2} \equal{} 10$. Thus, our answer is $ \frac {42}{10} \equal{} 4.2$[/quote]

That actually is wrong!

There are $ \binom{5}{2} \equal{} 10$ different pairs of marbles can be drawn, and the expected value of the sum is the average of the sums of each pair. This is
\[ \frac {1}{10}((1 \plus{} 2) \plus{} (1 \plus{} 3) \plus{} (1 \plus{} 4) \plus{} (1 \plus{} 5) \plus{} (2 \plus{} 3) \plus{} (2 \plus{} 4) \plus{} (2 \plus{} 5) \plus{} (3 \plus{} 4) \plus{} (3 \plus{} 5) \plus{} (4 \plus{} 5)) \equal{} \frac {60}{10} \equal{} \boxed{6}.\]

sorry reviving an old topic

There is a much easier way to solve this problem. You find the average of possible values, in this case $ 3$, and you multiply is by $ 2$, and you get the solution, $ 6$.

I got the expected value of one ball drawn and simply doubled it.
EDIT:  It is kind of like what Latex said, but different in that he said average and I said expected value (which happen to be the same).

Proof Exercise:  Why does doubling the expected value of one ball work?

The answer is actually[hide][hide][hide] 6[/hide][/hide][/hide]

[quote=Argonauts16][hide=Answer]8.5[/hide]
That is alcumus solution.[/quote]

Nope, this is right.

[quote=rubixsolver][quote=MathArt4][hide=Correct Solution] The expected value for the first marble is $3$. The same expected value is for the second draw.
$$3+3=\boxed{6}$$[/quote]

No, its 8.5[/quote]

Nope, it isn't. Alcumus says it is [hide=answer.]$6$

[quote=rubixsolver]No it's 8.5, I had this problem before[/quote]

It must have been a different variation.

[quote=rubixsolver]No it's 8.5, I had this problem before[/quote]

You did the problem in which it was the expected value of the products.

It's six people and if you need help then just use  a probability table

[quote=Argonauts16]The answer is 8, you must have done a different problem.[/quote]

The answer to this problem is not $8$. It is [hide=Answer.]$6$

[quote=Argonauts16]I did this on alcumus and they said the answer was 8.[/quote]

If you take a look at the past 20 posts, you will see that it is not $8$ or $8.5$.

[quote=Argonauts16]@MathArt4 I literally posted that right after I finished the problem. It is 8.[/quote]

It must be a different version

@Argonauts16 you might have have done a different problem

Yeah rubixsolver got a different version and got a different answer

[quote=AWaltz]Sorry that this is relating to a post a while previous,
[quote=MathArt4]The expected value for the first marble is $3$. The same expected value is for the second draw.
$$3+3=\boxed{6}$$[/quote]
Did you take the median of the possible numbers?[/quote]

The median is $3$.

[hide][hide][hide]It is 6 and the median is 3[/hide[/hide][/hide]

[quote=AWaltz]I don't understand why the median is more likely.[/quote]

It isn't more likely.

[quote=AWaltz]But then why is the expected value 3?[/quote]

It is because it is the middle number on the die.
oops, not good at explaining

[quote=AWaltz]But then why is the expected value 3?
[quote=MathArt4]It isn't more likely[/quote][/quote]

They said the median was 3 if I understood correctly

[hide][hide][hide=Alcumus Solution]There are $\binom{5}{2} = 10$ different pairs of marbles can be drawn, and the expected value of the sum is the average of the sums of each pair. This is \begin{align*}
\frac{1}{10}((1+2)+(1+3)+(1+4)+(1+5)+(2+3)&\\
+(2+4)+(2+5)+(3+4)+(3+5)+(4+5))&=\frac{60}{10} = \boxed{6}. \end{align*}[/hide][/hide][/hide]

The problem is that some have the version where that you multiply the two numbers, and some have the version where you add the two numbers. Sorry for bumping the thread.

can't we find the average of the five (3) and double that to get the answer?
is it correct?

All of you guys are over-complicating this:
Least Outcome is [b]3[/b]
Highest Outcome is [b]9[/b]
The expected outcome is average of 3 and 9 = [b][u][i][i]6[/i][/i][/u][/b]

Note: This strategy works in this case only because the balls (1,2,3,4,5) are in a linear progression.

I really don't get this. See attached. I put in the right answer... TWICE, and it did not work.

[quote=Wukong]huh?
you probably copy pasted something wrong like a weird tab or whatever
idk
it might be bug
no idea.[/quote]

Prob a bug, like the last time.

[quote=Samchooo]please don't put Alcumus solutions as well as bump 4 year old threads! Also, please hide your answer! :)[/quote]

First point: bump as much as you want, but use common sense.
Second point: Yes, please do.

[quote=Wukong]"last time"???[/quote]

It has happened b4, and that time too, you mentioned it and you were the second one to comment. It pertained to square roots. Maybe I can fish it out.


calculus
integration
geometry
rectangle
geometric transformation
parallelogram
inequalities

I have post CWMO 2001 and 2004 problems, I'll post CWMO 2003 problem here, which is also in Chinese.
Have a good day!

I am sure you are able to translate text into english. Why did you not do it?

There are a total of 8 questions, with 4 questions to be done over 4 hours, over 2 days.

Day 1

1) Place the numbers $1, 2, 3, 4, 5, 6, 7, 8$ at the vertices of a cuboid such that the sum of any $3$ numbers on a side is not less than $10$.  Find the smallest possible sum of the 4 numbers on a side.

2) Let $a_1, a_2, \ldots, a_{2n}$ be $2n$ real numbers satisfying the condition $\sum_{i=1}^{2n-1} (a_{i+1} - a_i)^2 = 1$.  Find the greatest possible value of $(a_{n+1} + a_{n+2} + \ldots + a_{2n}) - (a_1 + a_2 + \ldots + a_n)$.

3) Let $n$ be a given positive integer.  Find the smallest positive integer $u_n$ such that for any positive integer $d$, in any $u_n$ consecutive odd positive integers, the number of them that can be divided by $d$ is not smaller than the number of odd integers among $1, 3, 5, \ldots, 2n - 1$ that can be divided by $d$.

4) Given that the sum of the distances from point $P$ in the interior of a convex quadrilateral $ABCD$ to the sides $AB, BC, CD, DA$ is a constant, prove that $ABCD$ is a parallelogram.

Day 2

5) The sequence $\{a_n\}$ satisfies $a_0 = 0, a_{n+1} = ka_n + \sqrt{(k^2 - 1)a_n^2 + 1}, n = 0, 1, 2, \ldots$, where $k$ is a fixed positive integer.  Prove that all the terms of the sequence are integral and that $2k$ divides $a_{2n}, n = 0, 1, 2, \ldots$.

6) A circle can be inscribed in the convex quadrilateral $ABCD$.  The incircle touches the sides $AB, BC, CD, DA$ at $A_1, B_1, C_1, D_1$ respectively.  The points $E, F, G, H$ are the midpoints of $A_1B_1, B_1C_1, C_1D_1, D_1A_1$ respectively.  Prove that the quadrilateral $EFGH$ is a rectangle if and only if $A, B, C, D$ are concyclic.

7) The non-negative numbers $x_1, x_2, \ldots, x_5$ satisfy $\sum_{i=1}^5 \frac{1}{1+x_i} = 1$.  Prove that $\sum_{i=1}^5 \frac{x_i}{4 + x_i^2} \leq 1$.

8) $1650$ students are arranged in $22$ rows and $75$ columns.  It is known that in any two columns, the number of pairs of students in the same row and of the same sex is not greater than $11$.  Prove that the number of boys is not greater than $928$.

I hope my translation is correct, because I haven't been reading mathematics in Chinese for quite some time already :blush: .

Valiowk!
You are great! Thank you very much!

Thanks, Valiowk, for posting all these problems!

For problem 6):

In fact, it's pretty easy to see that the points E, F, G, H are the inverses of the points A, B, C, D in the incircle of the quadrilateral ABCD, and thus, the points A, B, C, D are concyclic if and only if the points E, F, G, H are concyclic. But the quadrilateral EFGH is a parallelogram (the Varignon parallelogram of the quadrilateral $A_1B_1C_1D_1$), and hence, the points E, F, G, H are concyclic if and only if this parallelogram is a rectangle. Hence, the points A, B, C, D are concyclic if and only if the quadrilateral EFGH is a rectangle.

  Darij

[quote="zhaoli"]I have post CWMO 2001 and 2004 problems...[/quote]
Do you mean you posted CWMO 2002 and 2004? I didn't see CWMO 2001...

CWMO 2001 problems are posted at
http://www.mathlinks.ro/Forum/viewtopic.php?t=15596

in the 4th, can we take P on the boundary?

Yes, we can, because sum of distances is a continuous function of P.

Valiowk!
Please, translate problems from
http://www.mathlinks.ro/Forum/viewtopic.php?t=15596

Thank you!

[quote="guest"]for the 7. take xi = cot(ai/2) then u get sum(cos ai) = 3 and
the ineq comes to proving sum{[1-(cos ai)^2]/[5 + 5(cos ai)^2 - 6cos ai]} is not bigger than 1. this can be proved using jensen inequality. Can someone maybe verify if im correct or not ?[/quote]

oh, sorry we take xi = cot(ai/2)^2

the official solution of 7, as far as i remember, let 1/(1+xi) = yi and use some simple quadratic equation techinques to arrive at the ans

Most beutiful solution for pr.7 I saw recently on forum, but I don't remember topic's title.
Main trick is to pove simple inequality:
\[\frac{x}{x^2+4}\leq\frac{1}{20}+\frac{3}{4(1+x)},\]
which is equivalent to $(x-4)^2(x+4)\geq 0$.

For problem 7.

You can prove following lemma.

[b]Lemma.[/b] Suppose lines $m$ and $n$ are not parallel, then locus of points $P$ s.t. $P$ lies in interior of angle $A$ (see Figure 1) and $%Error. "dist" is a bad command.
(P,m)+\dots(P,n)=const$ is a segment $BC$, $BC\perp l$, where $l$ is a bisector of $A$.

Suppose $ABCD$ is not a parallelogram. We may assume $AB\not\|CD$. Consider  inside of $ABCD$ segment $P_1P_2$ s.t. $P_1P_2\perp l$, where $l$ is bisector of angle between $AD$ and $BC$ (if these lines are not parallel), or $P_1P_2\|AD$, if $AD\|BC$. In both cases:
\[%Error. "dist" is a bad command.
(P_1,AD)+%Error. "dist" is a bad command.
(P_1,BC)=%Error. "dist" is a bad command.
(P_2,AD)+%Error. "dist" is a bad command.
(P_2,BC)=%Error. "dist" is a bad command.
(P,AD)+%Error. "dist" is a bad command.
(P,BC)\qquad \forall P\in[P_1P_2].\]
It follows that $%Error. "dist" is a bad command.
(P,AB)+%Error. "dist" is a bad command.
(P,CD)=const$ $\forall P\in[P_1P_2]$, but since $AB\not\|CD$ we obtain $P_1P_2$ is perpendicular to bisector of angle between $AB$ and $CD$. Clearly, it is impossible.

My solution of prob. 4:
First, prove a lemma that is already given in Myth's post.
Second, choose a line l perpendicular to the bisector of angle A.
for any P on l, we have d(P,AB)+d(P,AD) is a constant.
Thus d(P,BC)+d(P,CD) is a constant too.
Again by the lemma we have l is perpendicular to the bisector of Angle C.
since the bisectors of A and C are parallel, we can then deduce that Angle B=Angle D.(they both equals to $\pi-\frac{A}2-\frac{C}2$)
similarly, angle A=angle C.
thus ABCD is a parallelogram.

P.S. I ranked 4th in CWMO 2002 and 3rd in CWMO 2003........

hello i think  i have a solution for problem 3 here goes:
well first of all we only have to consider all d which odd and not greater than 2n-1. that said put d=2x-1 so lets observe the first n odd numbers mod d. note that the sequence of numbers is periodic with length 2x-1. for example the odd numers mod 5 is:1.3,0,2,4... so now what we have to work is the worst possible case meaning when we have a sequence of odd numbers where the first element =2 mod d. note that the period mod d has 0 in position x in the 2x-1  numbers and that 0 ocurrs only once in the period. so we have to only look at the worst possible case which i described before. in this case we need 2n+x-2 numbers which is our solution. am i correct??? please answer :blush:

ohh sorry i forgot to conclude:
our minimum is achieved when x is maximum to cover all bases so our answer is when x=n. and then 3n-2 is the answer.
am i right??

could give a correct solution??? :blush:

I'm too lazy to translate the sol (from chin to eng..)
so maybe whoever chinese member may translate it for me...

http://gifted.hkedcity.net/Gifted/ActReview/CWMO2003/solutions.pdf

(1) First, we prove that u_n  >= 2n – 1.
Since u_1  >= 1, we may assume n  >= 2.
In {1, 3, …, 2n – 1}, there is only 1 number divided by 2n – 1.
While in {2(n + 1) – 1, 2(n +2) – 1, …, 2(n + 2n – 2) – 1}, which have 2n – 2 numbers, there is no number divided by 2n – 1.
So u_n  >= 2n – 1.

(2) Second, we prove that u_n  <= 2n – 1.
The case of d is even is trivial.
We only need to consider the case that d is odd and 1  <= d  <= 2n – 1.
Consider 2n – 1 odds: {2(a + 1) – 1, 2(a +2) – 1, …, 2(a + 2n – 1) – 1}.
Let s, t be integers such that
(2s – 1)d  <= 2n – 1 < (2s + 1)d,
(2t – 1)d < 2(a + 1) – 1  <= (2t + 1)d,
then in {1, 3, …, 2n – 1}, the number of integers divided by d is s.
So if we can prove that (2(t + s) – 1)d  <= 2(a + 2n – 1) – 1, then we are done.
In fact, (2(t + s) – 1)d = (2t – 1)d + (2s – 1)d + d  <= 2(a + 1) –1 + 2n - 1 + 2n – 1 = 2(a + 2n – 1) – 1.
So u_n  <= 2n – 1.

Thus, the answer is u_n = 2n – 1.

Translated from the book "走?IMO----数学奥林匹克试题集锦(2004)"

Prob. n°5
We write the relation between a_n and a_{n+1} and between a_n and a_{n-1}
[tex]a_{n+1}-ka_n=\sqrt{(k^2-1)a_n^2+1}[/tex]
[tex]a_n-ka_{n-1}=\sqrt{(k^2-1)a_{n-1}^2+1}[/tex]
squaring and subtracting the second one we get:
[tex]a_{n+1}^2-a_{n-1}^2=2ka_n(a_{n+1}-a_{n-1})[/tex]
[tex]a_{n+1}=2ka_n-a_{n-1}[/tex]
and simply we can conclude that if a_0=0 and a_1=1 are integers then a_n is integer for all n natural numbers. We can prove that 2k|a_{2n} by induction. In fact we have 2k|a_0 and if 2k|a_{2n} then [tex]2k|a_{2(n+1)}=2ka_{2n+1}-a_{2n}[/tex].


Can anyone please tell me what a 'quant' is? What kind of work they do? What kind of education is required? What are some of the best universities to attend in order to become a quant? How much do they get paid? Please suggest links to any websites that contain info about qunat.
Any help will be appreciated.

There are some threads in this forum about quants.  I suggest reading those.  Here are some things pertaining to quants and related careers:

http://en.wikipedia.org/wiki/Quantitative_analyst
http://www.ams.org/careers/ewepsic.html
[url=http://www.mualphatheta.org/Mathematical_Log/Issues/0402/MAO_Mathematical_Log_Careers_Spring_02.htm]Arbitrage Trading[/url]


$ \lim_{x\to\infty}$x.(sin.(pi/x))=?
i need to see solution way.Thanks for all help..

$ \lim_{ x \to \infty } \pi \frac {sin \frac {\pi}{x}}{\frac {\pi}{x}} \equal{} \pi$
 note  $ \lim_{ x \to 0} \frac {sinx}{x} \equal{} 1$


playing an instrument will certainly help your chances.  top schools want to see you as a well-rounded, multi-talented individual, unless you are super super super great in one area.

but you don't want to be "too" well-rounded. i.e. you want to have a passion you stick with and have gotten better at instead of just doing a bunch of things you think colleges are looking for. that's not the fashion right now. also, i can't believe i actually have to say this, but geezus you guys, VOLUNTEERING. hello, ever heard of it? it's not that hard, it's a lot of fun, and being a regular volunteer opens you up to a lot of things:
1. volunteering in general looks good on a resume
2. more, and better, recommendations
3. some scholarships require a commitment to service
4. hello, its FUN
5. and rewarding
6. but mostly fun! (okay, im working on a habitat house tomorrow and i'm super excited about it, you'll have to excuse me)

If ALL other things are equal about two college applicants, the one who has pursued musical instrument practice and progressed to developing some playing ability will look better than the one who has not. (However, I can't convince my own son of this, so obviously people don't all have the same opinion in this matter.) But that's mostly just because playing a musical instrument shows you use some of your "free" time to do something challenging that takes steady dedication and practice. You might be better off, if you are the kind of person who hangs out here on AoPS, to simply reach a higher level in math competitions. Or you could, as Bleumoose has pointed out, do any number of different kinds of volunteer work. In other words, don't do something to look good to colleges--do something that is important to YOU. (And that is why my son wins the argument about practicing piano. I think it's a good idea, but he keeps busy doing other things.)

Specifically about piano playing, I can say that the Ivy League colleges get LOTS of applications from kids who have taken piano lessons for years. I like for young people to take piano lessons--my wife is a piano teacher by occupation. But there isn't any shortage of piano players on any Ivy League campus. Surely more piano players are rejected than accepted each year at each of the most selective colleges in the country. If you devoted a lot of time and effort to studying piano already, definitely mention that on your application, and that will look good. But if you are (for example) a high school junior, and just now are starting to take piano lessons only to have a better-looking college application, you can probably find something more personally satisfying and more appealing to colleges to do.

Best wishes on your college applications.

I had heard an admissions officer said something to this effect “I supposed it’s good that we attract so many applicants who are expert pianists. For once I would like to see an expert chef, or something else neat."

luckily, i'm musically inept, so i HAD to make some other use with my time.  :blush:

[quote="1/(ln x)"]does anyone know if playing an instrument (i.e. piano) will affect your chances of going to an ivy league?

what other factors do ivy leagues look at?[/quote]

If you are applying from China (mainland) it is more unusual than if you are a Chinese applying from the US, where it is routine for Asian mothers to push their children into music lessons.  The admissions benefit of merely knowing how to play is probably small, but there is a much larger "life benefit" both at university and especially later in life if you have musical training.   That is the better reason for studying it, if you have time and interest.   Many people kick themselves as adults for not learning music as children.

The same is true for many sports, cooking, languages, etc.     There is much less time to reach a high level of proficiency as one gets older and has job or family absorbing one's energies.

I wasn't so great with piano. My mother wanted me to play, but it just wasn't right for me.

Now, I find entertainment and challenges through math. Even my community service is math related.  :)


Trời ơi tìm mãi mới thấy nơi nói Tiếng Việt . Chào các anh , em là lính mới , tiếng anh thì ngu quá trời , có anh nào có file nào về từ điển Toán học Anh Việt không ? Cho em với, cảm ơn nhiều!

Theo mình nghĩ bạn nên mua một cuốn từ điển Toán học Anh - Việt ở ngoài thị trường để tra cho tiện, Cuốn của nhà xuất bản Khoa học kỹ thuật xuất bản năm 1993 ấy. Nó đầy đầy đủ hơn các file trên mạng.

Còn nếu bạn cần để tra trên máy thì xài thử cái này cũng được:


abstract algebra
search
articles
modular arithmetic
number theory

questions: i wentto my first jam and saw mod which later i was explained meaning module... i understand that it's kind of like base numbers but could someone please explain in plain english?

Search the math forum; there will be a ton of results. :)

Here's the wikipedia article on modular arithmetic,

[url=http://en.wikipedia.org/wiki/Modular_arithmetic]en.wikipedia.org/wiki/Modular_arithmetic[/url]

If $a\equiv b \pmod n$, then $n|a-b$ (n divides a-b).

Masoud Zargar


geometry
parallelogram
geometry unsolved

Hello,

[quote]Let ABC be a triangle with AB<BC<AC.

And let A', B' and C' be three points such that $ (AA') \bot (BC)$, $ (BB') \bot (AC)$, $ (CC') \bot (AB)$, AA'=BC, BB'=AC, and CC'=AB.

We suppose that $ \angle AC'B\equal{}\frac{\pi}{2}$

Prove that $ C' \in (A'B')$[/quote]

I have allmost solved this problem :

I proved that the triangles A'BC' and C'AB' isosceles. So I only need to prove that A'BC'=C'AB'=pi/2. And to do this, I think i should prove that (C'B) // (B'A) and (C'A) // (A'B). But i can't :(

[url=http://img187.imageshack.us/my.php?image=geoym8.png][img]http://img187.imageshack.us/img187/375/geoym8.th.png[/img][/url]

Mhdi

Construct parallelogram B’ADB and D’AA’B
Then we have △ABC≌△AA’D   △ABC≌△B’DB
So AB’=AC’ BC’=BA’ and BD⊥AC’ D’A⊥BC’
We have AB’∥BD   BA’∥AD’   =>  B’A⊥AC’   BC’⊥BA’
∠AC’B’+∠AC’B+∠BC’A’=45+45+90=180
=> A’,  B’, C’ are collinear

Sorry but could you explain why BD⊥AC’ ?  :roll:

notice  △ABC≌△B’DB  with AA'⊥BC  AD⊥AC AB⊥A'D
and BD and AC’are  corresponding sides
so BD⊥AC’


hi, im having trouble finding the sum of this series

sigma ((4^n)+(2^n))/(5^n)  n=1 to infinity

i tried the usual method trying to find r and a but i can never get a constant r, do i have to transform the series to something else? if so how? Thanks.

Split the series into two: $\sum_{n=1}^\infty \frac{4^n}{5^n}$ and  $\sum_{n=1}^\infty \frac{2^n}{5^n}$.


i have a signature.  however, some of my posts dont carry my signature.  whats going on?

There's a checkbox under the post editing box that says "Attach signature"... check it.

Or you can check Yes under "Always attach my signature" in your profile to automatically include a signature on every post, including PMs.

i did check it.  Some of my posts have a signature attached and some dont.

Maybe some of the posts that you posted before you checked "Yes" is missing the signature...you can always go and edit it to attach the signature, and quite frankly I think that's the only way to fix an already existing post without a signature.

Hello! reviving an old topic but i've seen someone use asymptote in their signature. How do you do this?

They made their signature before that rule was made (you can also see a couple of users with quote tags in their signature, etc...).

EDIT: Ignore this, I'm wrong.

[quote="NeoMathematicalKid"]Actually, PythonNut redesigned his signature to include Asymptote only 2 or 3 months ago.[/quote]

Exactly what I was wondering about.

You can delete your signature (or uncheck "Attach a signature" in each post) and create your own every time you post. For example, look at the signature here.

_________________
So you can change your "signature" like this.

Images, however, aren't allowed in signatures. Plus your BBCode is wrong.

PythonNut's real signature does have that thing.

it has asy2 tags but they show up as asy tags.

[quote="Ramanujam297"]Is Geogegra allowed?[/quote]
Yes, it's allowed. It won't work though, so there's no use in putting it in your signature at all. ;D

[quote="ewolf1230"]wait how do I set up a signature[/quote]
[quote="ewolf1230"]wait how did I get that......[/quote]
[quote="ewolf1230"]ah. go to profile. okay. I figured it out.[/quote]
Sorry for backseat moderating, but please edit your posts in the future.

Ignore my previous post, it's wrong.

@above: Saying "Sorry for backseat moderating" is not a valid excuse for backseat moderating. It's even worse because i) you know that you are backseat moderating, ii) you know that you shouldn't.


[quote="arqady"][quote="filomen"]If

$ a,b,c > 0$

then

$ (a \plus{} b \plus{} c)(\frac {1}{a} \plus{} \frac {1}{b} \plus{} \frac {1}{c})\ge \frac {1}{3}[14 \plus{} 4\frac {a^{2} \plus{} b^{2} \plus{} c^{2}}{ab \plus{} bc \plus{} ca} \plus{} (4 \minus{} \frac {ab \plus{} bc \plus{} ca}{a^{2} \plus{} b^{2} \plus{} c^{2}})^{2}]$[/quote]
The following stronger inequality is also true.
$ a,b,c > 0$

then

$ (a \plus{} b \plus{} c)(\frac {1}{a} \plus{} \frac {1}{b} \plus{} \frac {1}{c})\ge \frac {1}{3}[9 \plus{} 9\frac {a^{2} \plus{} b^{2} \plus{} c^{2}}{ab \plus{} bc \plus{} ca} \plus{} (4 \minus{} \frac {ab \plus{} bc \plus{} ca}{a^{2} \plus{} b^{2} \plus{} c^{2}})^{2}]$[/quote]

Dear Arquady, thank you very much. very nice....


I've seen the inequality $2r\le R$ several places but I've had trouble locating a proof. Could somebody please post one (or give me a hint to finding it myself)?

OI^2=R(R-2r), equality when OI=0, we know <OAI=1/2|<B-<C|, but that is equal to 0, so <A = <B = <C, equality when ABC is equilateral

[quote="Altheman"]OI^2=R(R-2r)[/quote]

Hm? Where are you getting that?

To prove $OI^{2}=R(R-2r)$, see http://mathforum.org/library/drmath/view/55106.html

To prove $R\geq 2r$ without using $OI^{2}=R(R-2r)$, see http://www.lclark.edu/~mathsci/euler.pdf


Draw a diagram in which the intersection of angle AEF and angle DPC is ray ED

In order to have the angles share one ray, points F=D and E=P.


modular arithmetic
number theory unsolved
number theory

Some diagonals of a n-sides polygon are satisfying 3 following conditions:
-	There is no diagonal which cuts another one.
-	The polygon is divided into triangles
-	The number of the diagonals start from each vertex is an even
Prove that 3|n

Do what Myth explained [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=27314]here[/url], in step $1$, post number $2$, to color the triangles of the triangulation chessboard-style. It's easy to see that we can choose this coloring so that the triangles having one side in common with the polygon are all white.

After that, orient the edges of the white triangles in a counterclockwise direction. It's now easy to see that all the triangles in the triangulation are oriented, and all the edges of the polygon are oriented in the same direction (i.e. you can go around the polygon on its edges, traversing each edge correctly, in the direction in which it's oriented).

After this, do what I tried to explain in the same thread, in post number $5$: when we go around a cycle in a counterclockwise direction, we get a number of the cycle, by adding $1$ for each edge we traverse correctly, and $-1$ for each edge we traverse in the direction opposite to its own orientation. Then look at this number $\pmod 3$. We call this residue $\pmod 3$ the boundary of the cycle. Since each cycle is the sum of triangles, and each one of those has boundary $0$, it means that each cycle has boundary $0$. However, the cycle consisting of all the edges of the polygon has boundary $n\pmod 3$, and the conclusion follows.

You may find a solution of a slighty more precise version of this problem in the course I wrote on graph for our olympiad preparation (last exercice of the paper) :

http://www.animath.fr/cours/graphes.pdf

Pierre.


1) Outline a synthesis of p-allylanisole.

2) When p-allylanisole is treated with chlorine, two products with molecular formula $ \ce{C10H12OCl2}$ are obtained. What are their structures?

3) Would the same result in (b) also be obtained with p-allylchlorobenzene?

umm....base catalysed hydrolysis of phenol with acetic acid followed by heating with AlCl3 (fries rearrangement) and then treatment with formaldehyde in presence of base (aldol condensation) and then reduction of carbonyl gp to get 3-(para hydroxy phenyl) propanol. And then SOCl2 followed by treatment with alc. KOH to get the allyl group. Reaction with sodium metal followed by CH3Br to get required product.

[quote="Hell_ever"]followed by heating with AlCl3 (fries rearrangement)[/quote]

At what temperature?

[quote="Hell_ever"]and then treatment with formaldehyde in presence of base (aldol condensation)[/quote]

That doesn't look like a good idea, since we have a phenol.

at low temp or room temp approximately.

Well we can protect the OH and then carry out aldol condensation.

PS: In my first statement i meant acid catalysed esterification and not hydrolysis.

OR, instead of protecting OH gp, one can react with sodium and then CH3Br to get the methoxy gp in second step itself instead of last step.

[quote="Hell_ever"]one can react with sodium and then CH3Br[/quote]

No, I don't think so.

Phenols are slightly acidic, so it is enough to treat them with sodium hydroxide and then add the alkyl halide.


let's spark some political controversy...

Yes!  Both my parents are Taiwanese but I was born in the US.  Look at my flag...it's Taiwan's!

[url=http://www.taipeitimes.com/News/edit/archives/2004/11/20/2003211838]Great Firewall of China[/url] (news story)

Well, it doesn't seem like Taiwan's going to have enough clout to take back China by diplomacy or force, so it looks like independence is all we can hope for...both my parents are from Taiwan as well...

I'm a major advocate of people's liberation. The major trouble of the past century has been that people have not had their own nation. Palestinians would not have invented modern terrorism if they had their nation. The Basques would not use terrorism if they had their homeland. The war in the Balkans would not have occured if 6 very different peoples were not in one country. The P-IRA would not be necessary. Consider the hardships imposed on colonies when the European Empires expanded. There are tons of other reasons why people deserve to have their own nation!

Ya, I should have my own country.

Edit: Okay, I guess that was too random. The world would be chaotic if all the people that wanted their own country got their wish.

i think north korea should be part of south korea

:evil: taiwan is part of china,we will fight for it!

<adds Chinese (pardon me, [i]Taiwanese[/i]) accent, just for effect>

You Communists stole our country!!  You should be lucky we don't make more trouble for you!!

[quote="Danbert"]Another question:  Should Hong Kong be independent of China (and Britain, for that matter)  ?[/quote]
It sounds ridiculous for Hong Kong to become independent.......especially since China has given quite a great amount of economic benefits to Hong Kong recently to please the unpleasant among the public because of the delay of the reform of electoral method......well, at least most Hong Kong people wouldn't like independence now.

李登辉knows that if he wont succeed in getting independence.  About half of the Taiwanese (maybe slightly less) are against independence.  A lot of them just doesnt care.  Recently he trys to change every company with the world "China" to Taiwan... But he got warned by APEC organization
The only problem is that if Taiwan is united with China, then it will lose its democracy form of government

[quote="siuhochung"][quote="Danbert"]Another question:  Should Hong Kong be independent of China (and Britain, for that matter)  ?[/quote]

No i don't think so.
I don't think taiwan should be separated from China either.
It's occupied by guo ming party around 50 years ago, and it's still a part of china. It sounds nonsense to me if it's independent.[/quote]

i agree with u.

[quote="Myself"]Why would the US put troops in Taiwan?  Because it doesn't want China to attack!  Why would China put troops by the US?  To prevent the US from attacking Mexico or Canada?[/quote]
OK do you really think US' intentions are to prevent China from attacking Taiwan?  I highly doubt the US is [i]that[/i] self-sacrificing.  And besides if China seriously decides to attack Taiwan, the US will definitely back out like how it did in the Vietnam War.  I never said that China would place troops by US, that was just a hypothetical situation.

[quote="tokenadult"]Frank Hsieh pointed out that the American states all get to vote for the American president. Who gets to vote for the national leaders of China?[/quote]

With a population of 1.5 billion, there is simply no way to have everyone vote in China. Just look at the confusion in Florida in 2004.

[quote="tokenadult"] Taiwan's people don't want to give up voting power to get the kind of deal that Hong Kong got after retrocession.[/quote]

China has a "one country, two systems" policy, which means that Hong Kong and Macao got to keep most of their former government system, although because it is now a part of China, the highest authority ultimately resides with Beijing. HK and Macao are highly autonomous, and even kept their own currency. If you talk to some people from Hong Kong, which I did, most of them will tell you there wasn't much change after Britain returned HK to China. The current leader of HK used to be a government official when the British ruled HK.

[quote="bubala"]I"m not taking a stance on the issue because I don't know very much about it, but why are you Chinese so anti-Taiwanese?

By the way, it won't look very good to the world if a big country like China smashed a little country like Taiwan to kingdom come.  Trust me.  The U.S. and Japan would especially not like it.

Thus, if China conquered Taiwan without really good reasons, it might spark a world war.  That would not be good.[/quote]

Bubala, there is no one in Mainland China who are anti-Taiwan. Taiwanese people are also Chinese people, so there's no reason for the Chinese to hate their own. What Chinese people are against is the independence of Taiwan.

Like others have mentioned, the Taiwan problem is very complicated historically. Taiwan is not just some small country or small island off the coast of China. Taiwan is not a country, and there has always been only one China, so it's fully justified if China must take down Taiwan in the case that Taiwan declares independence. By take down I mean the groups that are pushing for independence. Keep in mind these are Chinese people on the island, and the military does not want to harm any Chinese civilians. They want to keep the cities intact and take out only military facilities in the case of a conflict. There's no point in getting back a bombed out island where everyone is dead. The Chinese people want to see a unified, harmonious, and prosperous China, with Taiwan as a province.

[quote="joejia"][quote="A+MATH"]
Blah! Go back to their mother land and suffer under communists. That sure sounds a whole lot better than being an independent democratic country.[/quote]
A+MATH, have you ever visited China or lived there? About 1.5 billion people live there, and they are not all "suffereing under the communists." Learn more about China before you make such claims.[/quote]
Good point.
I live in China and I enjoy my life. I have been to Australia once and lived with my cousin there, I did not see any difference between the lives in the two countries.

Yea I think people are making snap judgements of life in China.  It may have been a little difficult in beginning of the 20th century, but now it isn't much different from life in America, and considering how fast China has been developing in the past few years, life will definitely get better.

[quote="LynnelleYe"][quote]Why would the US put troops in Taiwan? Because it doesn't want China to attack![/quote]
That sounds extremely naive to me somehow. How can you be so sure the motives of the U.S. are always beyond reproach?[/quote]

about US placing troops in other country

there r troops in S. Korea but when the relationship between N. and S. Korea got bad (when N. Korea first declare they had nukes)....the US backed their troops from the 38th parallel.

there r troops placed in German....i really don't see y

there r troops placed in alot of the airport in the middle east that was "borrowed" from those countries during the Iraq warfare but up to today those airports have not being returned therefore the motive of american troops in the middle east is uncertain....

[quote="bubala"]
Just because China is a big, powerful nation now doesn't mean that it gets to be imperalistic and conquer all the surrounding placees.

Of course, I'm not saying the U.S. isn't imperalistic--it is.  We tried to build our own empire in the mid-19th century, and the world didn't like it very much.  In fact, California and the Southwest all belonged to Mexico before we said "Oh, this is ours" and basically stole it.  Today, we still have a very unilaterist attitude, and again, the world doesn't like it.

The United States used to belong to Great Britain. Should Great Britain go to war with us?
China used to belong to Japan.  Should Japan go to war with China?
Taiwan used to belong to China.  Should China go to war with Taiwan?[/quote]

Great Britian had already had war with American but America is a different case cause it's culture is a blend of many different nations instead of the US being all British ancestry and plus after the war their relationship got a lot better....

China never belonged to Japan....Japan invaded China and they failed....GET UR HISTORY STRAIGHT

Taiwan doesn't used to belong to China it still belong to China....agreed by most gov. except 39 of them who r getting sweet deals from Taiwan....

It is funny that all of u claim that China is a big powerful country picking on Taiwan...

throughout history the winners r always right and the losers r always wrong
*wouldn't General Mcarthur be on a Japanese military court trial if the Japs have wrong
*wouldn't ppl like Hilter be hailed as god for centuries to come
*the North have always been higher than the south during the post civil war period
but wouldn't the south be able to change the northern view if they had won....

*If the national party stayed in China and Mao ran to Taiwan would the US help Mao....it is not a matter of big nation picking on smaller ones it is about countries that r pro-US or countries that r going to be a future threat to US

like i don't see the US getting on Israel and Indian's case on developing Nukes and even helps them but they have to pick on Iran and N. Korea cause they r not their "friends"....if US's real purpose is to maintain world peace it is like N. Korea and S. Korea isn't the only two country who can be bombed with nukes.....there r Indian and Pakistan,
Israel and palenstan, Iran and probably US....

notice how "People's Republic of China" has 2 inherent lies in it's name.

question: should taiwan separate from china?

answer: doesn't matter as long as it doesn't start a nuclear war.  seriously, both parties need to get over this thing.  nationalism is dead.  i'm sorry.

gosh I don't want a possibility of World War III.  ;)

I still don't understand why everybody is making a big fuss out of this issue. Being called Chinese or Taiwanese will not affect you personally at all. I have a friend who is  Taiwanese, but uses Chinese half the time even in her SAT-12 essay. The modern Taiwanese government was founded as the government in exile of China. But you know people are greedy. Power & politics come in the last decade. Politics have instigated this debate. & now we have a slippery border strife.

My friend is offended when he's called Chinese because he's taiwanese. He's such an idiot. Is taiwanese language different from chinese?

accidentally posted twice, don't know how to delete this one

I don't think that this is much of a question.  I'm personally against Taiwan's independence.  However, there's not much of a chance that Taiwan can declare independence, even with the May 28th, 2008 deadline Chen Shui-Bian has set.  Taiwan's secession from the PRC would set a precedent for Tibetan secessionists, Muslim secessionists in Xin Jiang and others.  So the questions is not simply whether the Taiwan should seceed, but the other parts of China also.  I assume that China would never let this happen without a war, for even a peaceful settlement would spark controversies in other parts of the China.

There's also been some posts about China being communist and therefore Taiwan should be independent because of that.  Honestly, after i read posts like that, I expected their next post to be something like, "Clinton is a witch!  Burn him!"  The entire communist ideal is villified by American propaganda.  I can't forsee anyone saying, "Germany is socialist! They have no right to rule Bavaria." if Bavaria ever decides to seceed.  Ideally, it's a better system than America's democracy, just like socialism.  Though the ruling party in China is communist, they are leaning in a ever more socialist way.  The only thing communist about China now is the CPC's name.  Also, I want to note that a democracy is not better for the common person.  If you were a farmer, as long as you were left to live your everyday life, would you care that it was a dictator or an elected president ruling?  It's only when the government gets overly oppressive, and any government can, that a certain regime becomes evil.

As far as the other countries that "belonged" to China in the past and are now independent like Vietnam and Korea, this is not true either.  There always has been a tributary system among China and its bordering countries.  They always maintained sovereignty, thought may have been heavily influenced by the Chinese, throughout history and gave China tributes in exchange for military protection.  The only exception is Mongolia.  This area was taken over by the Soviet Union, and seceeded from them, not China.  There is no major successful secessionist precedent set in the history of China.

Just like JOEJIA said, Taiwan was never an independent country.  No one is conquering another country in the China-Taiwan issue.  This isn't much of an imperialistic issue.  Yes, the KMT did go to Taiwan after the Communist Party took over control of the mainland.  Most of the world recognized Taiwan as the ruling government over Mainland China.  Because the world shifted their views to Beijing being the legitimate capital doesn't mean Taiwan acheieved independence.  An example would be that even after the Mongol Dynasty (Yuan) was overthrown in China, Mongolia still remained under the succeeding dynasty's rule.

If the US ever puts troops in Taiwan, you can bet its not out of self sacrifice.  Protecting democracy is important, but I would probably think the answer is to protect the US' self-interest.  I'm sure a lot of people have seen Hotel Rwanda.  Why does the US go to "liberate" Iraq and protect the kruds from Saddam's genocide but leave the Tutsis in Rwanda to die?  The genocide lasted 100 days and was ultimately ended with no US intervention.  Label me a cynic, but I'd say its because Rwanda has no oil reserve in this gradually more oil-starved world.

236factorial, the Taiwanese language is Hakka, its as different as Cantonese is to Mandarin.  I think your friend is more offended that you labeled him as one of those "war-mongering, barbaric, immoral communists" he was brought up to believe.  Propaganda does miracles, not only for dictatorships and oligarchies, but for democratic countries too.


topology
linear algebra
matrix
function
linear algebra unsolved

Prove that the set of diagonalisable matrices is dense in the set of all matrices.

Over what field? If you don't specify something, an arbitrary field is assumed, and that doesn't really work here; we need a topology. Every field can be given the discrete topology- and this is false for the discrete topology.

Let's go ahead and assume the field is $ \mathbb{C}$ and the topology of of $ M_n(\mathbb{C})$ is the usual metric topology of $ \mathbb{C}^{n^2}.$

Let $ A$ be a non-diagonalizable matrix. By Schur's lemma, there is a unitary matrix $ U$ such that $ U^*AU \equal{} T,$ with $ T$ upper triangular. A necessary (not sufficient) condition for $ A,$ and hence $ T,$ to not be diagonalizable is for there to be some duplication in the set of diagonal elements (eigenvalues) of $ T.$ Given any $ \delta > 0,$ we can find $ E \equal{} \text{diag}(\delta_1,\delta_2\,\dots,\delta_n)$ such that the $ \delta_i$ are all different and $ |\delta_i| < \delta.$ Then (assuming we chose $ \delta$ small enough), $ T \plus{} E$ has distinct eigenvalues and is thus diagonalizable, so that $ U(T \plus{} E)U^*$ is diagonalizable and approximates $ A$ arbitrarily well. (Here we're implicitly using that multiplication, either left or right, by a fixed matrix is a continuous function in our topology.)

Of course, over $ \mathbb R$ this is false: there are no diagonalizable matrices in a small neighborhood of $ \begin{pmatrix}0 & 1 \\
\minus{} 1 & 0\end{pmatrix}$


Tina randomly selects two distinct numbers from the set $ \{1,2,3,4,5\}$ and Sergio randomly selects a number from the set $ \{1,2,...,10\}$. The probability that Sergio's number is larger than the sum of the two numbers chosen by Tina is

$ \textbf{(A)}\ 2/5 \qquad \textbf{(B)}\ 9/20 \qquad \textbf{(C)}\ 1/2\qquad \textbf{(D)}\ 11/20 \qquad \textbf{(E)}\ 24/25$

i am pretty sure this can be done with balls and urns but i dont know what it is. dont ask

umm
for 10
1/10*1
1/10*9/10
1/10*8/10
1/10*6/10
.....
its not that hard to count

add the probablilites and get 2/5

[quote="redcomet46"]Tina randomly selects two distinct numbers from the set $ \{1,2,3,4,5\}$ and Sergio randomly selects a number from the set $ \{1,2,...,10\}$. The probability that Sergio's number is larger than the sum of the two numbers chosen by Tina is

$ \text{(A)}\ 2/5 \qquad \text{(B)}\ 9/20 \qquad \text{(C)}\ 1/2\qquad \text{(D)}\ 11/20 \qquad \text{(E)}\ 24/25$[/quote]

Sergio has to choose 4 at least.
so you can find all sets.
(Tina, Sergio) = (1&2, 4)    (1&3, 5)  (1&4, 6) (1&5, 7).............
                               , 5)
                               , 6)
                         ..... , 10)

           like it..

im a little confused on this problem. im not sure i understand the solutiosn that are being posted.

this is how i did it:
There is 125/250 chance that sergio's number is larger or equal to the sum of Tina's numbers. There are 1+2+3+4+5+4+3+2+1+0=25 possibilities in which the numbers are equal. 125/250-25/250=100/250=2/5 (A)
I made the mistake of overlooking DISTINCT and still got it right :lol:

That's because probabilities cancel out.
The probability if they are distinct or not distinct is the same,  because the numerators and denominators balance each other out...

Wait how is the probability 125/250....like where do you get that?

You can also use generating functions for this problem.

[hide=transformation]
Subtract $6$ from each person.

Then Tina is basically picking from $\{-2,-1,0,1,2\}$ and Sergio is picking from $\{-5,-4,\cdots,4\}.$

Then Tina's possible sums are $-3,-2,-1,0,1,2,3$ and Sergio's are $-5,-4,-3,\cdots,3,4.$ We can split this into two cases.

Case 1: Picks $-5,-4,4$

Happens with probability $\frac{3}{10},$ succeeds with probability $\frac{1}{3}.$ So in total $\frac{1}{10}.$

Case 2: Does not do that

Happens with probability $\frac{7}{10}.$

Let $T$ be the probability they tie. Then the probability they succeed is $\frac{1-T}{2}.$ Note Sergio's probabilities are equally spread, so the chance they tie is $\frac{1}{7}.$

[hide=If you are not convinced of this, more details]
Let $P(n)$ be the probability Tina picks $n.$ Then the probability they pick the same number is $\frac{1}{7}(P(-3)+P(-2)+\cdots+P(2)+P(3))=\frac{1}{7}(1)=\frac{1}{7}.$
[/hide]
So succeeds with probability $\frac{3}{7}.$ So in total $\frac{3}{10}.$

Answer is $\frac{1}{10}+\frac{3}{10}=\frac{2}{5}.$

[/hide]

[quote=Mudkipswims42][hide=Expected value]The expected value of Tina's sum is $3+3=6$.  The probability that Sergio picks larger than a 6 is $\dfrac{4}{10}=\boxed{\dfrac{2}{5}}$[/quote]
[url=https://artofproblemsolving.com/community/c7h2754990p24074865]Misleading, see here.[/url]


The bisectors of the angles of $\triangle ABC$ cut $BC,CA,AB$ in $D,E,F$. Prove that
\[ \frac1{AB \cdot CE} + \frac1{BC \cdot AF} + \frac1{CA \cdot BD} = \frac1{r \cdot R} . \]

[i]Traian Tămîian[/i]

Let $AB=c,BC=a,AC=b$, then using the angle bisector theorem about the ratio of the segments in a triangle whenan angle bisector intersects the side:
$CE=\frac{ab}{a+c}$
$AF=\frac{bc}{a+b}$
$BD=\frac{ac}{b+c}$
Also:
$Area=pr=\frac{abc}{4R}$
So:
$Rr=\frac{abc}{2(a+b+c)}$
Substituting  the above identities in the expression, get:
LS$=\frac{a+c}{abc}+\frac{a+b}{abc}+\frac{b+c}{abc}= \frac{2(a+b+c)}{abc}=$RS.
QED