integration
function
calculus
derivative
calculus computations

Can anyone help me out with the following question?

Thanks a lot!

With a Fourier transform question, there's always the question of convention: where do you put the $2\pi$ and which signs do you use? Not knowing your conventions, I'll use mine:

$\widehat{f}(\xi)=\int_{-\infty}^{\infty}f(x)e^{-2\pi i\xi x}\,dx$

Note that your function $f$ is real and even which means that $\widehat{f}$ will be real and even; that $f$ is compactly supported, which will mean that $\widehat{f}$ will be $C^{\infty}$ (in fact, entire); and $f$ is continuous but $f'$ has jump discontinuites, from which we infer that $\widehat{f}(\xi)=O(\xi^{-2})$ as $|\xi|\to\infty.$

Now let me find some derivatives:

$f'(x)=-2x$ on $(-1,1)$ (and zero otherwise)

$f''(x)=-2\chi_{[-1,1]}+2\delta(x+1)+2\delta(x-1)$

$f'''(x)=-2\delta(x+1)+2\delta(x-1)+2\delta'(x+1)+2\delta'(x-1)$

$\widehat{f'''}(\xi)=-2e^{2\pi i\xi}+2e^{-2\pi i\xi}+4\pi i\xi(e^{2\pi i\xi}+e^{2\pi i\xi})$

${=-4i\sin(2\pi\xi}+8\pi i\xi\cos(2\pi\xi)$

Now, divide by $(2\pi i\xi)^{3}$ to get $\widehat{f}:$

$\widehat{f}(\xi)=\frac{2\sin(2\pi\xi)-4\pi\xi\cos(2\pi\xi)}{4\pi^{3}\xi^{3}}$

That's real and even. Examination of the power series shows that it is analytic at zero and that $\widehat{f}(0)=\frac43,$ which is what it should be.

Your second question should be accounted for by Plancherel's Theorem

$\int_{\infty}^{\infty}|\widehat{f}(\xi)|^{2}\,d\xi= \int_{-\infty}^{\infty}|f(x)|^{2}\,dx.$

[And yes, I suspect my $2\pi$ conventions are different than yours. That can be coped with with suitable changes of variable.]


hello all!

I would like to make to make latex documents. the only problem is, I don't have an implementation on my computer, because at the moment i don't have one.  :rotfl: is there an online thing that i can use? Also, I will recieve douments. So i will need to be able to edit these somehow..

Thanks in advance for your suggests!

This site has the TeXer, which you can use to turn math mode expressions into image files. You can then embed these pictures in your documents. For the full typesetting features, you need an implementation.

yes, the thing is though, that if i install a program, the functionality would not be complete, because i believe it would have to install certain files into the system (fonts or something). Actually, I don't need it to work completely, i just need it to read recieved files and be able to edit those files, then send them to someone.

i don't know with this limited functionality (i.e. be able to type files but not compile them) that i can send stuff to others?

You only need a simple editor to edit tex files, nothing else as they are just text documents. The problem is that until you compile them you won't know if you'd made any mistakes. There are several online compilers you can use and they are listed in the second set of bullet points at  [url=http://sixthform.info/steve/wordpress/?p=59]Online Latex[/url].

For windows, use notepad or something else.  Goto Start->All Programs->Accessories->It should be in a category or standing out.  Much easier to do it then to explain it
Mac) TextEdit works great!  Applications->TextEdit unless they changed stuff in Leopard.
:D

hm. but can notepad open up the format that latex is in? (btw which format is latex docs?)

As has been said, .tex files are pure text documents. You can read them and edit them with any text editor, but you can only compile to pdf (or other display format) with an implementation.
Much like programming languages, a $ \text{\LaTeX}$ implementation comes with a text editor designed to work nicely with the raw files; mine colors things like dollar signs, \, and comments (%). None of these features are actually necessary; they're just conveniences.

thank you all for your suggestions, actually, i got an implementation running on a public maschine. now i know that if it breaks i should simply install it onto a memory stick!  :wink:


abstract algebra
superior algebra
superior algebra unsolved

Let $ G$  be a group and $ Z(G)\equal{}\{x \in G \vert xy\equal{}yx, \forall y \in G\}$.Let p be a prime number.Prove that:
     a) If $ \vert G \vert \equal{}p^{3}$ then $ \forall y \in G$ we have $ y^{p} \in Z(G)$.
     b) If $ \vert G \vert \equal{} p^{4}$ then the set $ A\equal{} \{ x y^{p} x^{\minus{}1} y^{\minus{}p}\}$ has maximum n elements.

a) $ Z(G)$ is nontrivial (since $ G$ is a nontrivial $ p$-group), and if $ Z(G) \equal{} G$, the claim is clear. otherwise $ G$ is not abelian, hence $ G/Z(G)$ not cyclic, so $ |Z(G)| \equal{} p$ and $ G/Z(G) \cong  \mathbb{Z}/p \times \mathbb{Z}/p$. that's it.

Sorry...my mistake...It is P elements: :D


In proofs we usually label equations something like this:

x^2 = 75 + x              (*)

so that we could refer to them later.  However, as you can see, it deletes all my spaces when I want to do that, which makes the resulting equation a bit confusing.  Is there a way to get around this problem?

Press ALT + 0160 for a "forced" space. Example:

x^2 = 75 + x                                          (*)

(Hint: You can also copy the character and using paste with ALT+v for fast pasting)

The only way to fix that would be to use dots like ・ or .
You could also use spoiler/black letters to try and line up lines, but it takes a lot of time and sometimes it doesn't work that well.
Ex.
2x+3y=z・・・・・・・・①
or if you don't have any special software this
3x+t=2y............1
would do.
Also,
[spoiler]QQQQ[/spoiler]you can write over here
[spoiler]QQQQ[/spoiler]but as you can see, it's kind of hard to line it up.

Hope I helped!


induction
algebra
polynomial
algebra unsolved

Find all polynomials $P \in \mathbb{R}[X]$ such that
\[(x-4)P(x+1)-xP(x)+20=0,\]
for all $x \in \mathbb R$.

:blush:  Ok! But i think your method not works with following:
Let $a,b\in\mathbb{R}$, find all $P\in\mathbb{R}[x]$ such that \[xP(x-a)\equiv (x-b)P(x).\]
Happy hunting!  :wink:

[quote="mathisfun1"]It's straightforward to show via induction that $P(x) = 5$ for all positive integers. Hence $P(x)=5$.[/quote]
Sorry, But you can't have $P(5)=5$ :P

Since $P(4) = 5$, $4P(5)-20-4P(5)+20=0$, so we can certainly make $P(5)=5$. But I think you mean that $P(5)$ does not have to equal 5. The problem arises in my induction: If $P(n) = 5$, then $nP(n+1)-5n-4P(n+1)+20 = 0 \Rightarrow P(n+1) = \frac{5(n-4)}{n-4}= 5$. This breaks down when $n=4$. If $P(5) \ne 5$, then let $P(x) = a_{n}x^{n}+a_{n-1}x^{n-1}+...+a_{1}x+a_{0}$. Comparing the leading terms of $x(P(x+1)-P(x))$ and $4P(x+1)$, we find that $n=4$. So $P(x) = a_{4}x^{4}+a_{3}x^{3}+a_{2}x^{2}+a_{1}x+a_{0}$. $P(1)=P(2)=P(3)=P(4)=5$, etc.

[hide="following the hint"]
From $(x-4)Q(x+1)=xQ(x)$, we immediately see that 1 and 4 are roots by plugging in $x=0,4$. Then plug in $x=1,3$ to find that 2 and 3 are also roots. Therefore, there exists a polynomial $C(x)$ such that
\[Q(x)=(x-1)(x-2)(x-3)(x-4)C(x)\]
and the equation becomes $C(x+1)=C(x)$. This means $C$ is constant (since $C(x)-C(0)$ has roots at all nonnegative integers $x$, so has infinitely many roots and is therefore the zero polynomial). So we obtain the solution
\[P(x)=5+k(x-1)(x-2)(x-3)(x-4)\]
for any constant $k$.
[/hide]

[quote="N.T.TUAN"]Final, what is your answer?[/quote]
Using Scorpius119's ending: $P(1) =P(2)=P(3)=P(4) = 5 \Rightarrow P(x)-5 = a_{4}x^{4}+a_{3}x^{3}+a_{2}x^{2}+a_{1}x_{1}+a_{0}-5 = k(x-1)(x-2)(x-3)(x-4)$.
 :blush: To be honest, I didn't see the more efficient ending -- I was thinking about substituting back into the original equation and getting a system of equations.

Ok, Now we can continue for :
[quote="N.T.TUAN"]:blush:  Ok! But i think your method not works with following:
Let $a,b\in\mathbb{R}$, find all $P\in\mathbb{R}[x]$ such that
\[xP(x-a)\equiv (x-b)P(x). \]
Happy hunting!  :wink:[/quote]

[quote="N.T.TUAN"]Ok, Now we can continue for :
Let $a,b\in\mathbb{R}$, find all $P\in\mathbb{R}[x]$ such that
\[xP(x-a)\equiv (x-b)P(x). \]
[/quote]

1) if $a=0$ and $b=0$, any $P(x)$ is solution
2) if $a=0$ and $b\neq 0$, $P(x)=0$ is the only solution
3) if $a\neq 0$ and $b=0$, $P(x-a)=P(x)$ $\forall x\neq 0$ $\Rightarrow $ $P(x)=c$
4) if $ab\neq 0$ :
$x=0$ in $xP(x-a)=(x-b)P(x)$ $\Rightarrow $ $P(0)=0$
If $P(r)=0$ with $r\neq 0$, then $P(r-a)=0$ $\Rightarrow $ all roots of P are $0,a, 2a, \cdot, ua$ for some u of $\mathbb{Z}^{+}$
$x=b$ in $xP(x-a)=(x-b)P(x)$ $\Rightarrow $ $P(b-a)=0$
If $P(r)=0$ with $r\neq b-a$, then $P(r+a)=0$ $\Rightarrow $ all roots of P are $b-a,b-2a, b-3a, ..., b-ka$ for some k of $\mathbb{N}$

Hence we must have $b=ka$ and all roots are $0, a, 2a, ..., (k-1)a$ $\Rightarrow $ $P(x)=c\prod_{i=0}^{k-1}(x-ia)^{n_{i}}$
Identification in initial equation gives $n_{i}=1$

As a result, we have :
a) if $a=0$ and $b=0$, then $P(x)$ is any
b) if $a=0$ and $b\neq 0$, then $P(x)\equiv 0$
c) if $a\neq 0$ and $b=0$,  then $P(x)\equiv c$
d) if $ab\neq 0$ then it must exist k in $\mathbb{N}$ such that $b=ka$ and solution is $P(x)=c\prod_{i=0}^{k-1}(x-ia)$

--
Patrick

$E=\{k\in \mathbb{C}| \ p(k)=0\}$

$a=b=0$ => $S=R[X]$
$a\neq 0,b=0$ => $P\equiv c$.
if $ab\neq 0$
if $|E-a\mathbb{N}|\neq 0$ such that $P(k)=0$
then $\forall (s,k)\in \mathbb{N}\times (E-a\mathbb{N}),$ $p(k-sa)=0$ so, $p\equiv 0$
if $p\not\equiv 0$
$E\subseteq a\mathbb{N}$
we can suppose that $am=MAX(E)$ so, it's evident that $E=\{am,a(m-1),a(m-2),..,0\}$.
then $p(x)=c\prod_{i=0}^{m}(x-ia)^{n_{i}},\ c\in\mathbb{R}^{*}$
and $x\prod_{i=1}^{m+1}(x-ia)^{n_{i-1}}=(x-b)\prod_{i=0}^{m}(x-ia)^{n_{i}}$
so $x-b=x-(m+1)a\ (b\ must\ be\ in\ a\mathbb{N}^{*})$( and $n_{i}=n={i-1}$ for $i=1..m$
finally.
$p(x)=cx\prod_{i=1}^{m}(x-ia)^{n}\ ,with\ (c,m,n)\in \mathbb{R}\times \mathbb{N}^{*}\times \mathbb{N}$


induction
combinatorics solved
combinatorics

Let t(n) be the minimum size of the sequence $a_1, a_2,...,a_{t(n)}$ such that:
for every permutation $\pi$ of (1, 2, ...,n) there exist $i_1 <i_2 <...<i_n$ satisfying
$\pi =(a_{i_1}, a_{i_2}, ...,a_{i_n})$
1. Find t(4)
2. Prove that t(n)$\geq \frac {(n-1)(n+4)}{2}$
3. Is there any number better than $\frac {(n-1)(n+4)}{2}$

I think this was given at an old and good All-Soviet Union Olympiad

Hmmmm... I was sure that this problem will appear on Mathlinks some day....
Ok, let's prove some result about it, which are better than desired in the statement.
I will post more than one message because of the probable length of the entire thing....

Let $n$ be a positive integer. A sequence $U_1,...,U_k$ will be called '$n$-universal' if it contains each permutation of ${1,2,...,n}$ as a subsequence.
Let $t(n)$ be the minimal length of an universal sequence.

First, as far I Know, the exact value of $t(n)$ is unknown.
It is conjectured that for $n \geq 3$, we have $t(n) = n^2 - 2n + 4$.

Kleitman and Kwiatkowski [1] have proved that $t(n) \geq n^2 - cn^{ \frac {7} {4} + \epsilon}$ for all $\epsilon > 0$, where $c > 0$ is a constant.
Using the result proved below, it gives that $t(n) \sim n^2$.

Claim 1 :
$t(1) = 1$ and $t(n) \geq \frac {n(n+1)} {2} + n - 2$ for $n \geq 2$.

Proof :
For $n=1$ it is trivial, so let's give $n > 1$.
Let $S$ be an $n$-universal sequence, and let $l(S)$ be its length.
- If there exists $k \in \{ 1,...,n \}$ which appears only once in $S$. Then, since $S$ is $n$-universal, it has to contains at least one $(n-1)$-universal sequence before the appearance of $k$ and at least one after.
In that case : $l(S) \geq 1 + 2t(n-1)$. (1)

- If each integer $k \in \{ 1,...,n \}$ appears at least twice in $S$ : Let $k$ be the one for which the rank of the first appearance is the greatest. This rank is then at least $n$. After this appearance, we have to find at least an $(n-1)$-universal sequence and the second appearance of $k$.
In that case $l(S) \geq n+1 + t(n-1)$. (2)

It follows from (1) and (2) that $t(n) \geq min(1+2t(n-1), n+1 + t(n-1)$.
If $t(n-1) \geq n$ this becomes $t(n) \geq n+1 + t(n-1)$. (3)

Since clearly, $t(1)=1$ and $t(2)=2$, it follows that $t(3) \geq 7$.
In another hand, $1,2,1,3,1,2,1$ is a $3$-universal sequence, thus $t(3) \leq 7$.
Therefore $t(3)=7$.

Now, suppose that $n \geq 3$ is given and that $t(n) \geq \frac {n(n+1)} {2} + n - 2$.
Then $t(n) \geq n+1$ so that, from (3) :
$t(n+1) \geq n+2+t(n) \geq n+2+ \frac {n(n+1)} {2} + n-2 = \frac {(n+1)(n+2)} {2} + (n+1) - 2$, which proves the result by induction.

Claim 2 : $t(4) = 12$.

Proof :
From claim 1, we have $t(4) \geq 12$.
In another hand, direct checking shows that
$2,1,3,4,1,2,3,1,4,2,1,3$ is $4$-universal, hence result.

Pierre.

Here comes the big part :

Claim 3 :
For all $n$, we have $t(n) \leq n^2 - 2n + 4$.

Proof.
The result is true for $n \leq 4$ so we assume that $n > 4$.
Let $U$ be the sequence $5,6,...,n$.
We define the sequence $S$ by alternating  the terms of the sequence $2,3,4,2,3,4,2,...$ and $1,U,1,U,1,...$ which gives :
$S=2,1,3,|U,4,1,2,|U,3,1,4,|U,2,1,3,|U,4,1,2,|...$ (*)

where the sequence ends after $n-1$ segments of 4 symbols beggining by $U$. Since $U$ has length $n-4$ it follows that $S$ has length $n^2 -2n + 4$.
It remains to prove that $^S$ is $n$-universal.

Let's denote the $k^{th}$ segment by $Ub_k1d_k$ (the first three terms are not a segment).
Then, for $1 \leq k \leq n-1$, we have $\{ b_{k-1},b_k,d_{k-1} \} = \{b_k,d_{k-1},d_k \} = \{ 2,3,4 \}.$ (**)
Let $A_k,B_k,C_k,D_k$ be the sequences containing the first terms of $S$ and ending in the $k^{th}$ segment bu the last term of $U$, by $b_k$, by $1$, by $d_k$, respectively.
A sequence of $k$ distinct terms chosen in $\{1,...,n \}$ is called a $k$-sequence.
We will prove by induction the four following sentences :
$A$ : '$A_k$ contains each $k$-sequence which does not end by $b_k$'.
$B$ : '$B_k$ contains each $k$-sequence.'
$C$ : '$C_k$ contains each $(k+1)$-sequences which does not contains $1$ and does not end by $d_k$.'
$D$ : '$D_k$ contains each $(k+1)$-sequences which contains $1$.'

First note that from (*), it is clear that $A_1,B_1,C_1,D_1$ satisfy the conditions $A,B,C,D$.
Now suppose that for a given $k \in \{2,...,n-1 \}$, the sequences $A_{k-1},B_{k-1},C_{k-1},D_{k-1}$ satisfy the conditions $A,B,C,D$ respectively.
Then :
1) Let's give a $k$-sequence which does not end by $b_k$. We have :
$A_k = A_{k-1}b_{k-1}1d_{k-1}U = B_{k-1}1d_{k-1}U$.
If the term before the last of the $k$-sequence is $b_{k-1}$, then all the terms of the $k$-sequence except the last are contained in that order in $A_{k-1}b_{k-1} = B_{k-1}$ from the induction hypothesis. But each term other than $b_k$ and $b_{k-1}$ is contained in $1d_{k-1}U$. Thus $A_k$ contains the $k$-sequence.
- If the term before the last is not $b_{k-1}$ :
Since $A_{k-1}$ satisfies $A$, the first $k-1$ terms of the $k$-sequence are conteined in $A_{k-1}$. And the last is necessarly in $b_{k-1}1d_{k-1}U$ from (**).
Thus, in both cases $A_k$ satisfies $A$.

2) Since $A_k$ contains $B_{k-1}$, it now follows easily that $B_k$ satisfies $B$.

3) Let's give a $(k+1)$-sequence which contains 1 but does not end by $d_k$.
- If the last term is 1 : Each of preceding terms are in $B_k$ from 2).
Since $C_k = B_k1$, the $(k+1)$-sequence is contained in $C_k$.
- If the last term is not 1 and the term before the last is not $d_{k-1}$ :
We have $C_k = C_{k-1}d_{k-1}Ub_k1$
Since $C_{k-1}$ satisfies $C$ the first $k$ terms are in $C_{k-1}$. From (**), the last is in $d_{k-1}Ub_k1$ thus the $(k+1)$-sequence is contained in $C_k$.
- If the last term is not 1 and the term before the last is $d_{k-1}$ :
Now, we use that $^C_k = D_{k-1}Ub_k1$. Since $D_{k-1}$ satisfies $D$, the first $k$ terms are in $D_{k-1}$ and the last is in $Ub_k$, from (**).

Then, in each case, $C_k$ satisfies $C$.

4) Now, since $D_k = C_kd_k$ it clearly satisfies $D$, and we are done!

Now it suffices to use $D$ for $k=n-1$, which directly gives that $S$ is $n$-universal, which proves the claim.

Reference :
[1] D.J. Kleitman, D.J. Kwiatkowski, 'A lower bound on the length of a sequence containing all permutations as subsequences', Journal of Comb. Theory, serie A, 21 (1976), p. 129-136.

Pierre.


1. $\sin(25)\sin(35)\sin(85)=\cos(x)/a$

what is x+a?

2. If $\tan(11x)=\tan (34 \deg)$ and $\tan(19x)=\tan(21 \deg)$ find $\tan (5x)$-exact value

3. If $2\sin{x}+2=\sqrt{2}\cos{x}$ find $\sec{x}+\tan{x}$

[quote="b-flat"][quote="asdff"]skimnc, did you take into consideration the + or - part?[/quote]

What do you mean?  The sign of $\sin x$, etc should not matter in this problem (no pun intended).[/quote]

what I am saying is that the answer can be ${\frac{\sqrt{2}}{2}}$OR $-{\frac{\sqrt{2}}{2}}$
because cosine can be positive or negative (or secant in this case)

no, his solution is fine: the only trouble may occur if $\cos{x}$ is zero, and $\sin{x}=-1$, in which case the expression would be indeterminate (although its limiting value might? be zero)

Actually I solved it the dumb way and got

(3sinx+1)(sinx+1)=0, meaning
sinx=-1, but then secx is undefined
or sinx=-1/3, meaning cos x=$\sqrt{8}/3$ or $-\sqrt{8}/3$

but $-\sqrt{8}/3$ is extraneous (the original equal would have a positivce number on one side and negative on the other)

so only $\sqrt{8}\3$ works, giving us one answer only

[quote="pianoforte"][hide="problem 3"]$\implies \sin x+1=\frac{\sqrt{2}}{2}\cos x$
$\frac{\sin x+1}{\cos x}=\frac{\sqrt{2}}{2}=\sec x+\tan x$.[/hide][/quote]

dude we covered it already.

[quote="t0rajir0u"][quote="jasonwingofcarmelin"]1. $\sin(25)\sin(35)\sin(85)=\cos(x)/a$

what is x+a?[/quote]

Obviously this problem does not actually have a unique solution, since for any $|a| \ge \sin 25 \sin 35 \sin 85$ there exists a corresponding $x \in \mathbb{R}$.[/quote]

sorry x and a are positive integers what is the minimum of x+a.

-infinite, if that's what it wants. Because you can have like Cos(-...), which is the same as cos(-15), and divided by 4.

[i]Positive[/i] integers.  Now the question is valid.


I finished 12th std CBSE in May 2007. I wish to pursue a career in mathematics.

Is BMath at ISI Bangalore a good option for me? Is there any age limit for taking the entrance exam?

I guess there will be some age limit but it would be sth like 22-23 years old for B. Math ( no official data with me)
There are people here who have joined after 1 year break and 2-year break after CBSE 12th exam too...

So if you want to, you can apply this February when their prospectus and admission form comes out...

wow... a post after such a long break...... Hope you continue your participation here......

Soumya welcome back to AoPS.   Are you having holidays?

Hi can anyone tell me what are the expected cutoffs for selection into the BMath program?

which is a better option ISI kolkata or bangalore?


[quote="José"]Well, I know that IOI is a training programm for high-school students, right? I'd like to have more info about it. Any detail will be appreciated. Thanks.[/quote]

First learn the secrets from this [url=http://www.google.co.uk/support/?hl=en]source.[/url] Then browse this very sub-forum and finally you may have a look [url=http://olympiads.win.tue.nl/ioi/index.html]there.[/url]

in iran team,Nima Ahmadi pour collected a silver medal in 2006 ioi and a gold medal in 2006 imo.
he had also gotten a silver medal in 2005 imo
he is here in mathlinks.you can see this iraian genius :winner_first:  :winner_second:  :winner_second:

I met nima at mexico, he is a nice guy.
But Iran had a gold medalist at IOI2006 too!
You guys are really doing well everywhere!


Baker Bob made a gross of donuts.  He sells some to his normal customers, but he has promised to send the remaining donuts to the school's bake sale.

When he gets ready to pack the donuts, he finds there are 4 donuts remaining when he puts them in cartons holding a dozen, and he finds there are 9 donuts remaining when he puts them in cartons holding a baker's dozen.

He decides it would look most professional if every carton were full, so he puts some donuts in the smaller cartons and some donuts in the larger cartons, making sure every carton is full.

How many cartons of donuts did Baker Bob deliver to the school bake sale?

Note:  A gross is 144 items.  A baker's dozen is 13 items.

Alright, I did it the stupid way, perhaps someone can post a better solution with better usage of gcf or w/e.

[hide]
Answer 8 containers

My solution :blush:
possible choices for the total number or doughnuts

t=total
x=boxes of dozens
y=boxes of baker's dozens

12x=4=t
13y+9=t
t<144

so t can be
using the first given, subsituting for x 16,28,40,52,64,76,88,100,112,124,136
using the second given, subsituting for y 22,35,48,61,74,87,100,113,126,139

Considering they both have 100 in common, that must be the number of dougnuts given to the school.

100=13+13+13+13+12+12+12+12=4*13+4*12

4+4=8

Answer: 8 containers [/hide]

EDIT:corrected answer

 :blush:  :blush:  :blush:  :blush:  :blush:  :blush:

[hide="A Shortcut"]Because the number of donuts gives a remainder of 4 when divided by 12, the number of donuts must be a multiple of four. 13 is one more than a multiple of four, as is 9, so for 13x+9=12x+x+8+1=4(3x+2)+x+1 to be a multiple of four, x must be one less than a multiple of four, so we would only have to compute the values of 13x+9 for which x is one less than a multiple of four.[/hide]

[quote="hiitsme"][hide] ...
13y+9=t
...
using the second given, subsituting for y
21,34,47,60,73,86,99,112,125,138[/hide][/quote]
Better double-check that second list of values, hiitsme.


Which of you here have gone to the Indian IMO Training Camp this year and in the previous years? Are there any Indian IMO organisers/coordinators here!? Please , speak up!! We can all gain from you!  :D

yap, im an imotc guy, but trust me, u can gain nothing from me.
im pretty dumb!

Hello,
    Are you the Venkata(from Jharkhand, I beleive) who got through INMO last year when you were in Class IX? Only very few make it to IMOTC from Class IX,not more than one or two every year.Anyway, you have three more years to have a shot at clearing IMOTC and making it to INMO.Congratulations.Did you see the INMO paper this year?How many could you solve?My brother from Class IX here in Chennai manged two to two and a half.
Sathej

Hello,
    Sorry.It must have read 'I believe'.
Sathej

oops duz somebody know my name out here?
anyways jokes apart, i ve seen the inmo paper.
seemed easier than last yr.
i sat for some 1.5 hrs. and could solve 3.5 sums.maybe cud have solved 1/2 more if i had spent some more time.
anyways it doesnt matter whether one is a 9ther or a 12ther at the camp.
making it to the team matters(which i wont this yr tooo :( )
anyways ur brother will most probably b in.
what abt u?it wud be cool if 2 brothers make it to the camp at the same time.
hope to c ya guys there.


Please , can someone see, if there is any typo at problem  X.232 from RMT (4/2007).

 I can not solve it .

 babis

No, it's OK. Write everithing in terms of $ R$ and angles. The condition reduces to $ \sin (B\minus{}C)\equal{}0.$

[quote="enescu"]No, it's OK. Write everithing in terms of $ R$ and angles. The condition reduces to $ \sin (B \minus{} C) \equal{} 0.$[/quote]

 Thank you Bogdan !

 I friend told me he found an counterexample , so I stopped the effort.

 (for example : $ Z_A\equal{}1, Z_B\equal{}i, Z_C \equal{} \frac{1}{5}(4\plus{}3i)$

 These numbers satisfy the given conditions , but do not form a isosceles triangle)

 Babis

Yes, you are right! The triangle should be acute angled...


calculus
integration
trigonometry
calculus computations

Evaluate the integral:

[img]http://www.artofproblemsolving.com/Forum/latexrender/pictures/2c5024a7f13b17d317e894fd78b5b566.gif[/img]

ok so im going to use integration by parts: thus

u = cos5x
du = -sin 5x * 5
--------------------------
dv = x dx
v = x

but for somereason im getting screwed up somewhere with the rest of this: i must be assigning things wrong. any ideas?

i think you messe4d up the concept of integration by parts, in this case, is very simple, just take:
$\int xcos5x dx=\frac{xsen5x}{5}-\int \frac{cos5x}{5}dx$
what is very simple to evaluate.

I do not quite comprehend how you arrived at what you did.

[img]http://www.artofproblemsolving.com/Forum/latexrender/pictures/745c1015d7bf61aca56f8287fe5c47cb.gif[/img]

isn't that the form of integration by parts? so that
u = cos5x
and
dv = xdx
[b]thus:[/b]
du = -sin5x * 5 dx
and
v = x

Brambila set u = x and dv = cos 5x.  It should be

\[\int x\cos 5x\,dx=\frac{x\sin 5x}{5}-\int\frac{\sin 5x}{5}\,dx,\]
which is easy to evaluate.

[quote="Kelth"]dv = xdx
[b]thus:[/b]
v = x[/quote]

You might want to rethink this step.


function
integration
algebra proposed
algebra

Find all real-valued continuous functions defined on the interval $(-1, 1)$ such that $(1-x^{2}) f(\frac{2x}{1+x^{2}}) = (1+x^{2})^{2}f(x)$ for all $x$.

Let $F(x)=\int_{0}^{x}f(t)dt$, then equation equivalent to $F(\frac{2x}{1+x^{2}})=2F(x)$. Let $q(x)=\frac{2x}{1+x^{2}}$, then
$F(x)=\frac{1}{2}F(q(x))=\frac{1}{4}F(q(q(x))=...=\frac{1}{2^{n}}F(q^{(n)}(x))$.
Because $q(x)>x$ increase, exist unique $r=r(x)\in R$, suth that $x=q^{(r)}(\frac{sign x }{2})$. It give solution $F(x)=\frac{1}{2^{r(x)}}$.

$\varphi: (-1,1)\to \mathbb{R},x\mapsto{2x\over 1+x^{2}}$
Then $\varphi'(x)={2(1-x^{2})\over (1+x^{2})^{2}}>0$ for all $x\in (-1,1)$ and $\varphi(x)\mathop{\longrightarrow}_{x\to-1}-1$ and $\varphi(x)\mathop{\longrightarrow}_{x\to 1}1$ and $\varphi(x)-x={x(1-x^{2})\over 1+x^{2}}$.
So $\varphi$ is a bijection $(-1,1)\to (-1,1)$ (continuous and strictly increasing)
and $\varphi(x)>x$ for $x\in (0,1)$ and $\varphi(x)<x$ for $x\in (-1,0)$.
$1-\varphi(x)^{2}={(1-x^{2})^{2}\over (1+x^{2})^{2}}$ for all $x\in (-1,1)$,
and this gives $(1-\varphi(x)^{2})f(\varphi(x))=(1-x^{2})f(x)$ for all $x\in (-1,1)$.
Now $\varphi^{-n}(x)=\underbrace{\varphi^{-1}(\cdots \varphi^{-1}(x)\cdots)}_{n \text{ times}}\mathop{\longrightarrow}_{n\to\infty}0$ for all $x\in (-1,1)$,
so $(1-x^{2})f(x)=(1-(\varphi^{-n}(x))^{2})f(\varphi^{-n}(x))\mathop{\longrightarrow}_{n\to\infty}f(0)$.

So solutions are $f(x)={c\over 1-x^{2}}$ for all $c\in\mathbb{R}$.


Is anyone following the USAICO (this year's version of USACO camp)? List of participants and some daily updates here: http://oldweb.uwp.edu/academic/mathematics/usaco/2005/camp05/camp05.htm

Heh, I wish I could make it.  It is only about 15 minutes from my house.  I am just not good enough at algo programming; I never did very well during those 3-4 hour contests, always just wanted to get up and move around after the first couple hours.

if u have more years then try hard from now :)


AMC USA(J)MO USAMO geometry MATHCOUNTS videos Gauss

Congrats to:

Ravi Fernando
Michael Zhao (AOPS USER)
wooYoung Park
George Yu
Sam LIte (SEMI USER)
Max Schindler (HECK YEAH)
Victor Wang (AOPS USER)
Xiao Shi

I am also finding it funny Max and Victor are the same grade but go to HS and MS respectively.

BTW, CA had more than 100 qualifiers this year.

Darn. I am not on this list.

(2 in KS, Stephen Rong and Kiran Kota (inactive aopser))

Hm, any relation to [url=http://starwars.wikia.com/wiki/Rahm_Kota]General Rahm Kota[/url]??

Darn our school almost beat your whole state in USAMO qualifiers (8, several people off by <1 AMC question)

[quote="hjoon0125"]that is A LOT[/quote]
What? No, we got beaten by Saratoga, which had 9. Darn if only those 2 people who got 127.5/7 got 132/7 :( Actually darn, if the people who made it last year made it this year (more than half the people who made it last year didn't make it this year) then we would be at 12, which would be pretty cool. Well I guess it just shows that one silly little school in the Bay Area is better than all of Missouri :P

Consider it another way.  We Missourians have had to suffer in schools that frankly don't care about AMC and have no legacy at all.  Therefore, our accomplishments are that much better than yours.

Also our state has more middle-schooler qualifiers than your state combined.  :roll:

Also your state has 1000000000000 times more Asians. :wink:  :P

R? C?

Oh real, complex?

And on the contrary piano, I already have gotten one  :P

contrary to the title, this thread has no legitimacy whatsoever

we may therefore safely ignore this thread (and relegate it to the pits of oblivion)

And you have fallen into the pit of oblivion.

Goodbye, rating!

[quote="Mewto55555"]It is all interconnected...math make people think of asians, and asians make Lyndon think of....let's not go there![/quote]

wait. what. who's Lyndon? seriously guyz. oh wait that's that math dude in my state. right

Oh dude really?
YOu know LYNDON?

I hear he's perverted and n00bish. WHat is your opinion of him?

yeah how is lyndon ji cough*lyndon bi* like?

Yeah man dude that kid sucks at math so hard. he can't even do like usamo #3s which are insanely trivial.

I wonder if resurrection notices our sarcasm....


trigonometry
function
complex analysis
complex analysis unsolved

Can $\sin{z}\in \mathbb{R} \backslash [-1,1]$ if $z \in \mathbb{C}$?

By Picard's Little Theorem, if $f$ is a nonconstant analytic function, then $f(\mathbb{C})=\mathbb{C}$ or  $\mathbb{C}\setminus\{z_0\}$ for some $z_0$.  So the answer is 'yes'.

Let $z=\frac{\pi}2+it.$ Then $\sin z=\cosh t,$ which takes on every real value greater than $1.$

For large negative reals, try $z=-\frac{\pi}2+it.$

In general, $\sin(x+iy)=\sin x\cosh y+i\cos x\sinh y.$


Eli throws five darts at a circular target and each one lands within one of the four regions. The point value of a dart landing in each region is indicated. What is the least score greater than five points that is not possible when the point values of the five darts are added together. (the values are 1, 2, 4, and 6)

Is there an easy way to do this type of problem? Sorry if its been asked before...


Do you like sudoku? It has nothing to do with math but with logic. You can play every day [b][color=blue][url=http://www.miniclip.com/sudoku/sudoku.htm]Here[/url][/color][/b]
[url=http://www.miniclip.com/sudoku/sudoku.htm]hide[/url]

Wrong forum...look here: http://www.artofproblemsolving.com/Forum/index.php?f=360

i wanted to say yes....but...there's no yes  :?


rotation
geometry
geometric transformation

I understand that it has two double bonds, but there is a single bond between C2 and C3 that can rotate and make the molecule not planar. So why does my book say it is planar and therefore not a chiral molecule?

Yes and hence i think there is restrictre rotation  :) .
And btw, can single bonds rotate?  :maybe:

Obviously single bonds rotate. :D
But in this case both the carbons which are involved in that single bond are already $ sp^2$ so I think there is restricted rotation due to angle strain and thus the carbons would like to maintain an angle of $ 120^0$ so that's why it is planar and without free rotation :maybe:

I knw that chemrock but do they rotate for an conjugated compound like this was my question. Sry for not beong clear  :)

I have explained it for u da before ur message :huh:

bythecliff, in a conjugated system, even the single bonds have a partial double bond character due to delocalisation of pi electrons. hence the bond order C-C bond over here is greater than one. even if it were not a planar molecule, the molecule would not be chiral.
and chemrock, i dont think there are two forms which can be isolated, there is only one form but the rotation is slightly restricted.
and at room temp, rotation abt the single bond should be possible

1. The molecule is planar because all carbon atoms are sp2: with this hibrydization scheme, all three bonds in each carbon lie in a plane with approximate angles of 120º between each bond.

2. There are two forms for an open chain diene like this: the s-cis and the s-trans, which are more or less intercoverted with (or without) some heating.

carcul, can the two forms be isolated at room temp.
and then, is that why you require heating(to convert transoid form to cisoid form ) for diels alder reaction in many cases.

The two forms are easily interconverted, so isolation is not possible, as far as I know. Yes, that's why in some Diels-Alder reactions some heating is required.


How do you write 90 degrees with the little circle thing?

insert into the of  the file begining

\DeclareRobustCommand{\celsius}{%
  \ifmmode^\circ\mathrm{C}\else$ ^\circ$C\fi}

or for just degree
\DeclareRobustCommand{\Dgree}{%
  \ifmmode^\circ\else$ ^\circ$\fi}

Genick Bar-Meir
http://www.potto.org
where free textbooks roam!


[quote="Werwulff"][quote="nthd"]Prove that$\forall x,y,z \ge 0$
                            $\sum(\sqrt{x^2+yz}-\sqrt{y^2+zx})^2\le\sum(x-y)^2$
 :)  :)[/quote]
It is trivial and thus boring to show that this inequality is equivalent to
\[ \sum 2xy\leq\sum\sqrt{(x^2+yz)(y^2+zx)}  \]
which can easily be proved using cauchy-schwartz and AM-GM:
\[ \sum\sqrt{(x^2+yz)(y^2+zx)}=\sum\sqrt{x^2+\sqrt{yz}^2}\sqrt{\sqrt{zx}^2+y^2}\geq \sum(x\sqrt{zx}+\sqrt{yz}y)= \\=\sum(x\sqrt{zx}+\sqrt{zx}z)\geq\sum 2\sqrt{x\sqrt{zx}\sqrt{zx}z}=\sum 2xz  \]
Nice ineq  :)  :)  :)[/quote]Thank you :)
I think we can find all $k$ such that
$\sum(\sqrt{x^2+kyz}-\sqrt{y^2+kzx})^2\le \sum(x-y)^2$


90% cred ca are loc intr-un triunghi ascutiunghic, nu stiu sigur insa:

$1+4\cos A \cos B \cos C\geq 2\sum{\cos A \cos B}$

Reamintesc o problema de Marian Tetiva:
Daca $x,y,z\geq 0$ si $x^{2}+y^{2}+z^{2}+2xyz=1$, atunci \[4xyz+1\geq 2\sum xy.\]
Punem $x=\cos A$, $y=\cos B$, $z=\cos C$, intr-un triunghi ascutitunghic $ABC$ si obtinem problema :)

lol, e acelasi lucru. unde e inegalitatea asta? in old & new ?

Cred ca glumesti :huh:
Este in doua articole(pe care le ai si tu)

aaaa, gata am gasit-o in cel cu inegalitatea lui gerretsen. ce tare. eu ti-am zis cum am ajuns la ea, pornind de la alta inegalitate, chiar nu o mai vazusem pe asta.

[quote="pohoatza"][color=darkred]Daca un triunghi $ABC$ este neobtuzunghic, atunci $1+4\cos A \cos B \cos C\geq 2\sum{\cos A \cos B}$[/color][/quote]
[color=darkblue]Vom demonstra o inegalitate algebrica (irationala), la prima vedere simpla, din care se poate
deduce inegalitatea propusa, dar si alte inegalitati interesante, folosind substitutii inspirate.

$1.\blacktriangleright$ $\boxed{\{x,y,z\}\subset [0,\infty)\Longrightarrow \prod (x+y)+4xyz\ge 2\cdot\sum xy\sqrt{(x+z)(y+z)}}\ \ (*)\ .$

Intr-adevar, $(x+y)(y+z)(z+x)+4xyz=6xyz+\sum xy(x+y)=$

$\sum [xy(x+y)+2xyz]=$ $\sum xy[(x+z)+(y+z)]\ge 2\cdot \sum xy\sqrt{(x+z)(y+z)}\ .$

$2.\blacktriangleright$ In inegalitatea $(*)$ vom particulariza pentru $\{\begin{array}{c}x: =\frac{\cos A}{a}\ge 0\\\\ y: =\frac{\cos B}{b}\ge 0\\\\ z: =\frac{\cos C}{c}\ge 0\end{array}\|$ $\Longrightarrow$

$\{\begin{array}{c}\prod (\frac{\cos A}{a}+\frac{\cos B}{b})+4\cdot\prod\frac{\cos A}{a}\ge 2\cdot\sum \frac{\cos A\cos B}{ab}\sqrt{(\frac{\cos A}{a}+\frac{\cos C}{c})(\frac{\cos B}{b}+\frac{\cos C}{c})}\\\\ a=b\cos C+c\cos B\ ,\ b=a\cos C+c\cos A\ ,\ c=a\cos B+b\cos A\end{array}$

$\Longrightarrow$ $\prod\frac{c}{ab}+\frac{4\cdot \cos A\cos B\cos C}{abc}\ge 2\cdot\sum\frac{\cos A\cos B}{ab}\sqrt{\frac{b}{ac}\cdot\frac{a}{bc}}$

$\Longrightarrow$ $1+4\cdot \prod\cos A\ge 2\cdot\sum\cos A\cos B\ .$

$3.\blacktriangleright$ In inegalitate $(*)$ vom particulariza pentru $\{\begin{array}{c}x: =p-a>0\\\\ y: =p-b>0\\\\ z: =p-c>0\end{array}\|$ $\Longrightarrow$ $\boxed{\ \sum \frac{\sqrt{ab}}{p-c}\le 2\cdot(1+\frac{R}{r})\ }\ ,$ adevarata in orice triunghi $ABC\ .$

[b][u]Observatie.[/u][/b] Tinand seama de relatia $HA=2R\cos A$ intr-un
triunghi neobtuzunghic, relatia propusa este echivalenta cu inegalitatea
$\boxed{\ 2R^{3}+HA\cdot HB\cdot HC\ge R\cdot (HA\cdot HB+HB\cdot HC+HC\cdot HA)\ }\ .$
In demonstratia acesteia poti folosi identitatile (in triunghi neobtuzunghic)
$\boxed{\ \sum a\cdot HA=4S\ ,\ a\cdot HB+b\cdot HA=2Rc\ }$ si inegalitatea
$\boxed{\ HA\cdot HB\le R^{2}\cdot\frac{c^{2}}{ab}\ }$ care se poate obtine usor astfel :
$(2Rc)^{2}=(aHB+bHA)^{2}\ge 4\cdot aHB\cdot bHA$. Aceasta ultima inegalitate
este echivalenta cu inegalitatea cunoscuta $\boxed{\ \sin 2A\sin 2B\le \sin^{2}C\ }$.[/color]


AMC AIME MATHCOUNTS USA(J)MO USAMO calculus MIT

turns out, that i actually got a 5. i don't care cuz even if i had gotten 6 or 7, i wouldn't have made it?? btw, who was the other 8th grader (besides youkow) who got a 6. was it justin? seriously, mathcrazed, tell youkow that he needs to get an Aops account. DUUUUUUUUDDE, sachin and david got 14's DANG.

if you mean the username "justinahmann", its probly an imposter.

lol so are those stats telling me that i got only 1 less AIME probelm than Nathan Benjamin?  Dang, I thought he was "god-like" in 8th grade when he owned the mathcounts countdown.

hmm lol i think the 10th graders for this state were pretty impressive, 5 people with 8+

[quote="mathcrazed"]wtf. no. it's his actual username. i see him everyday, like seriously. everyday. I would know =D[/quote]

but he's like never on.

hey dude cognos,off topic...but nevertheless, i forgot what you said, did you say that your school allowed you to take physics next year or not?

I official got a 1.

>.<

even though i put 7 right answers.

well dosen't matter how good at math you are if you're stupid.

[quote="cognos599"]i guess, i know yixin took it at our school this year. so i am assuming that at least he will be doing it next year, so i am pretty sure my school will let me. but i will cross that bridge when i get there.[/quote]

dang my school wont let me take physics C next year...

even though ive already learned calc and physics C is an intro course (not to mention people in 12th and 11th now have taken physics C, they are just implmeneting the rule that we cant take physics C in 11th grade this year)...^so yea unimpossible, what you told me last year about physics at our school was right...but oh well, one of my friends who always wants to take C as an 11th grader is pretty adamant and good at getting his point across, so hopefully, he can convince the principal...

dude unimpossible, dang, what happened?
oh yea and if you didnt move, our USAMO qualifiers in our school history would double with the number of qualifiers we will have next year alone at our school

[quote="the future"]dang my school wont let me take physics C next year...[/quote]
Go tell them to gtfo. I took phys c in 11th, ez as s.h.i.t. lol i heard that freed is getting pissed cuz lyndon wants to take multi as a frosh. hahah, ya you show them lyndon, chs sucks a.s.s.

for us, i don't think that there are any pre-requisites for physics, but calculus is a co-requisite.

^ wait you mean national 9th grader?

because i dont think there was one in indiana that got one...

lol and my posts are probably going to go to like 800 by next week...

cause im spending like 2 hours doing math every night now...for the next week

yeah national 9th grader.. mehhh another beast who'll beat me to the chase when applying for MIT.

doing math on AoPS?

omg my mp3 broke (partly my fault since I was the one who played around with it formatting it from FAT32 to Mac OS Journaled)

[quote="undefined117"]dude I envy that 9th grader who got a 14..

@the future, I think I should stop aiming for MIT too :lol:[/quote]

A 9th grader got a 14???  :o
That's crazy!

hmm mathcrazed are you taking calc at carmel or something?

[quote="mathcrazed"]lol i haven't even met the math teacher but she already hates me? great......... I think just to piss her off, i'll fail all my hw and ace tests. =P[/quote]

lol hw isnt collected
havent done a single hw this year...
just review the notes and book the day before the test, its probably more effective than doing all the hw because the memory time will be shorter...eh calc is a bunch of mem anyways, only the free response questions make you think about what you are doing...

nah she doesnt hate you, shes actually pretty funny and easy going, but she probably just hates the system of how 9th graders can take calc (hmm havent asked youkow, but did they allow him to skip?), lol she makes fun of everyone, if someone in your class next year has like a funny name or something, she probably wont pick on you as much.

dude thats like saying i hate you because you are like 100x better than me at math.

and how does everyone know your dad?  like everyone talks about him in my 10th grade classes...odd, that includes some people i know for certain that have never even met you or your dad...

oh man, every math teacher in chs knows dr g-unit.

stupidkid, you going to arml right?
even though youll be done with high school.

Got moderately pwned.  Still laughing at my stupidity

i took 2007 and 2006 geometry this year and i got the scores 29 and 31 respectively. on the 2007, all 3 were dumb mistakes, like one the last one i thought it was a cylinder instead of a cone and mistakes like that. on the 2006 i also made a dumb mistake, idk what it was. i had better get a perfect tomorrow or else.

btw MathAndKnowledge is doing comprehensive and he thinks he will get 28+, and mustit thinks that mathandknowledge will get 15-, i think he will get in between 15 and 28 so ya.

GOOD LUCK TO ALL 3 PPL(exaggeration)  STATEWIDE WHO ARE TAKING IT.

hehe 28+ are you serious? i have no idea what happened but i got 7 wrong I think.. I'd forgotten everythingggg I'd learned in precalc it was devastating.

what class is Mathandknowledge taking?

oh I see.

what the CRAPP cognos!!!@#% GIBBIT


geometry
perimeter
trigonometry
trig identities
Law of Cosines

Triangle $ ABC$ has side lengths $ AB \equal{} 5$, $ BC \equal{} 6$, and $ AC \equal{} 7$. Two bugs start simultaneously from $ A$ and crawl along the sides of the triangle in opposite directions at the same speed. They meet at point $ D$. What is $ BD$?

$ \textbf{(A)}\ 1 \qquad \textbf{(B)}\ 2 \qquad \textbf{(C)}\ 3 \qquad \textbf{(D)}\ 4 \qquad \textbf{(E)}\ 5$

[hide="Solution"]The perimeter of the triangle is 5 + 6 + 7 = 18, so each bug crawled 9 units.
Starting from Side A, the first bug crawls 5 units to point B, and then 9 - 5 = $\boxed{4}$ units to point D.
Choice D
[/hide]

[hide="But not impossible"] Let $\angle CDB = \theta$.  By Law of Cosines,

$5^{2}= 4^{2}+AD^{2}-8AD \cos \theta$
$7^{2}= 2^{2}+AD^{2}+4AD \cos \theta$

So

$123 = 24+3AD^{2}\implies AD = \boxed{ \sqrt{33}}$ [/hide]

it cant be a root. the five choices are up there

t0rajir0u was clearly responding to [quote=worthawholebean]The first time I tried this problem I misread it and thought it said AD -- which would make the problem significantly more difficult.[/quote]

If you want to find $\overline{AD}$, then you should just apply Stewart's Theorem. It's a direct application.


ratio
geometry
symmetry
geometry unsolved

Points X, Y, Z lie respectively on the sides BC, AC, AB  of the triangle ABC. We know that AX bisects BY in K, BY bisects CZ in L, CZ bisects AX in M, find Area(KLM)/Area(ABC).

Using parallel projection, project the triangle $\triangle ABC$ into an equilateral triangle $\triangle A'BC$ (BC stays in place), leaving line segment ratios and area ratios invariant. $X \in BC,\ Y \in CA,\ Z \in AB$ are carried into $X \in BC,\ Y' \in CA',\ Z' \in A'B$ and the triangle $\triangle KLM$ into an equilateral triangle $\triangle K'L'M'.$ Then skip primes from the notation.

Denote $k = KL = LM = MK$ and $x = AM = MX = BK = KY = CL = LZ.$ The quadrilaterals $AMLY \cong CLKX$ are congruent by symmetry and the triangles $\triangle AMC,\ \triangle MXC$ have equal areas, having equal bases AM = MX and common vertex C. Hence the triangle $\triangle KLM,\ \triangle CLY$ have equal areas and

$k^{2}= KL \cdot LM = CL \cdot LY = xy = x(x-k)$

$\left(\frac{x}{k}\right)^{2}-\frac{x}{k}-1 = 0,\ \ \ \frac{x}{k}= G$ or $-\frac{1}{G}$

where $G = \frac{1+\sqrt{5}}{2}$ is the golden ratio. We have to take the positive root $\frac{x}{k}= G.$ Denote the triangle areas $\sigma = |\triangle ABK| = |\triangle BCL| = |\triangle CAM|$ and $\tau = |\triangle KLM| = |\triangle AMZ| = |\triangle BKX| = |\triangle CLY|.$ Then

$\frac{|\triangle KLM|}{|\triangle ABC|}= \frac{\tau}{3 \sigma+\tau}$

Areas of the triangles $\triangle s BCY,\ \triangle BCL$ are in ratio of their bases BY, BL:

$\frac{\sigma+\tau}{\sigma}= \frac{BY}{BL}= \frac{2x}{x+k}$

$1+\frac{\tau}{\sigma}= \frac{2G}{G+1}= \frac{2}{G}$

$\frac{\tau}{\sigma}= \frac{2}{G}-1 = \frac{1-(G-1)}{G}= \frac{G-1}{G^{2}}= \frac{1}{G^{3}}$

$\frac{|\triangle KLM|}{|\triangle ABC|}= \frac{\tau}{3 \sigma+\tau}= \frac{1}{3G^{3}+1}= \frac{1}{3G^{2}+3G+1}= \frac{1}{6G+4}$

A symmetrical expression is obtained by fooling with the golden ratio:

$\frac{|\triangle KLM|}{|\triangle ABC|}= \frac{1}{2}\cdot \frac{1}{G [3+2(G-1)]}= \frac{1}{2}\cdot \frac{G-1}{2G+1}= \frac{1}{2}\cdot \frac{1-\frac{1}{G}}{1+G}= \frac{1}{2}\cdot \frac{3-\sqrt{5}}{3+\sqrt{5}}$

[color=darkred][b]Lemma.[/b] Let $ABC$ be a triangle and let $X\in (BC)$ , $Y\in (CA)$ , $Z\in (AB)$ be the points for which know $XB=x\cdot XC$ , $YB=y\cdot YA$ , $ZA=z\cdot ZB\ .$ Denote the intersections $L\in BY\cap CZ$ , $M\in CZ\cap AX$ , $K\in AX\cap BY\ .$ Then $[KLM]=\frac{(1-xyz)^{2}}{(1+x+xy)(1+y+yz)(1+z+zx)}\cdot [ABC]\ .$ Particularly, $x=y=z$ $\Longrightarrow$ $[KLM]=\frac{(x-1)^{2}}{x^{2}+x+1}\cdot [ABC]\ .$[/color]

[color=darkblue][b]Proof of the proposed problem.[/b] Using the notations from the above lemma, $MA=MX$ , $KB=KY$ , $LC=LZ\ .$ Apply the Menelaus' theorem to the transversals $\overline{AKX}$ , $\overline{BLY}$ , $\overline{CMZ}$ for the triangles $BYC$ , $CZA$ , $AXB$ respectively :

${\|\begin{array}{ccccccc}\overline{AKX}\ ,\ \triangle BYC & \Longrightarrow & \frac{AY}{AC}\cdot \frac{XC}{XB}\cdot \frac{KB}{KY}=1 & \Longrightarrow & \frac{XC}{XB}=\frac{AC}{AY}& \Longrightarrow & \frac{1}{x}=1+y\\\\{ \overline{BLY}\ ,\ \triangle CZA}& \Longrightarrow & \frac{BZ}{BA}\cdot\frac{YA}{YC}\cdot \frac{LC}{LZ}=1 & \Longrightarrow & \frac{YA}{YC}=\frac{BA}{BZ}& \Longrightarrow & \frac{1}{y}=1+z\\\\ \overline{CMZ}\ ,\ \triangle AXB & \Longrightarrow & \frac{CX}{CB}\cdot\frac{ZB}{ZA}\cdot\frac{MA}{MX}=1 & \Longrightarrow & \frac{ZB}{ZA}=\frac{CB}{CX}& \Longrightarrow & \frac{1}{z}=1+x\end{array}\Longrightarrow$
$x=y=z=\alpha\ ,$ where $\alpha^{2}+\alpha-1=0$ and $\alpha =\frac{-1+\sqrt 5}{2}\ .$ Apply the particular case of the above lemma : $\frac{[KLM]}{[ABC]}=$ $\frac{\alpha^{2}-2\alpha+1}{\alpha^{2}+\alpha+1}=$ $\frac{(1-\alpha )-2\alpha+1}{(1-\alpha )+\alpha+1}=$ $\frac{2-3\alpha }{2}$ $\Longrightarrow$ $\boxed{\ [KLM]=\frac{7-3\sqrt 5}{4}\cdot [ABC]\ }\ .$[/color]


$X$ is a set that $X\subset \mathbb{R}$ and $X$ is not $\oslash ,\mathbb{R}$ and
$\forall a\in \mathbb{R}$,denote $X_{a}=\{x+a|x\in X\}$.
prove that $\{X_a|a\in \mathbb{R}\}$ has infinite member. :)

I don't think I understand this question.

I understand it as: a nontrivial (see above for the exact ones) subset of the reals has $\infty$ many different translates by real numbers.

$X$ is a subset that $X \subset \mathbb{R}$, for example $X=(0,1)\cup[100,500)$,and $X$ is not $\oslash,\mathbb{R}$.for all members of $\mathbb{R}$ like $a$ we make a set .the name of this set is$X_{a}$and $X_{a}=\{x+a|x \in X\}$.we must prove  $Y=\{X_{a}|a\in \mathbb{R}\}$ that contains all sets $X_{a}$ ,has infinit  member,it means that we have infinite different $X_{a}$. :)


How do I balance this chemical equation using the method of redox ?

Could anyone explain me, I'm not very good.

C2H6O + O2 = CO2 + H2O

C2H6O + O2 = CO2 + H2O

First you know that you need 6 H from C2H6O, meaning you multiply the entire H2O molecule by 3. So now we have:

C2H6O + O2 = CO2 + 3 H2O

Now the Carbons (C) aren't balanced, so we multiply the CO2 molecule by 2:

C2H6O + O2 = 2 CO2 + 3 H2O

We have 3 Oxygens (O) on the reactant side and 7 on the product side. So we multiply O2 by 3 to balance:

C2H6O + 3 O2 + 2 CO2 + 3 H2O

And voila.

Or you could look at the formal charges (oxidation numbers) of each atom. Each bond represents a pair of electron (and a double bond represents two pairs of electron). Formally assign each pair of electrons totally to whichever of the atoms in the bond is more electronegative; if the two atoms are the same, assign one electron in each bond to each atom. Oxygen is more electronegative to carbon, which is in turn more electronegative than hydrogen.

So each hydrogen would almost always have a formal charge of $ \plus{}1.$ The only exception would be $ \ce{H_2}$ (hydrogen gas) in which case you give each a charge of $ 0.$

Oxygen would generally have a formal charge of $ \minus{}2.$ The exceptions here are $ \ce{O_2}$ (oxygen gas) in which case we have $ 0,$ and also peroxides (but there are no peroxides in this reaction.)

Carbon then takes up all the slack, carrying formal charges ranging from $ \minus{}4$ in methane $ (\ce{CH_4})$ to $ \plus{}4$ in carbon dioxide $ (\ce{CO_2})$ or carbonate ion $ (\ce{CO_3^{2\minus{}}})$ and everything in between.

You didn't say what $ \ce{C_2H_6O_2}$ is - it could be ethanol or it could be dimethyl ether. If it's ethanol, the two carbon atoms have formal charges of $ \minus{}3$ and $ \minus{}1;$ if it's dimethyl ether, they're $ \minus{}2$ and $ \minus{}2.$ In the end, since we're burning the whole thing up anyway, it doesn't really matter.

The only oxidation numbers (formal charges) that change in this are the carbons and some of the oxygens. The carbons must go from $ \minus{}3,\minus{}1$ or $ \minus{}2,\minus{}2$ to $ \plus{}4$ each in the carbon dioxide molecules. For the time being, assume we have one $ \ce{C_2H_6O}$ molecule and hence two $ \ce{CO_2}$ molecules. So the total oxidation numbers of the two carbon must increase by $ \plus{}12$ (an average of $ \plus{}6$ per carbon atom.) The counterbalance is some of the oxygen. The oxygen in the alcohol/ether starts at $ \minus{}2$ and ends at $ \minus{}2$ so that doesn't change, but each oxygen atom from $ \ce{O_2}$ goes from $ 0$ to $ \minus{}2.$ So we need 6 oxygen atoms - 3 oxygen molecules - to balance the change in the two carbon atoms.

So: one molecule of alcohol/ether, and three oxygen molecules. That gives us 6 hydrogen atoms, which don't change in formal charge, but they have to go somewhere: hence 3 water molecules in the product. If we've done everything right , the oxygen atoms should balance by themselves. Let's check this:
\[ \ce{C2H6O} \plus{} \ce{3O2}\rightarrow \ce{2CO2} \plus{} \ce{3H2O}\]
Count the oxygens: $ 1\plus{}3\cdot 2\equal{}7$ on the left, $ 2\cdot 2\plus{}3\equal{}7$ on the right. Yes, it works.

Professor Merryfield, formal charge and oxidation number are different concepts.

What Luiseduardo really wanted was this:

1) Calculate oxidation numbers: -2 (C, is C2H6O), 0 (O, in O2), +4 (C, in CO2), and -2 (O, in H2O). So (as we knewed from the start) carbon is oxidized and oxygen is reduced.

2) Balance separatly the oxidation and reduction half reactions:

$ C_2H_6O \plus{} 3H_2O \longrightarrow 2CO_2 \plus{} 12e^\minus{} \plus{} 12H^\plus{}$

$ O_2 \plus{} 4e^\minus{} \plus{} 4H^\plus{} \longrightarrow 2H_2O$

3) Now add the above equations after multiplying the second one by 3. One gets

$ C_2H_6O \plus{} 3O_2 \longrightarrow 2CO_2 \plus{} 3H_2O$.


Oh I have posted this problem before but I didn't get a real solution.This time please help me.
Let ABCD is a quadrilateral, M is a point on CD(different C,D).Prove that MA+MB< max{CA+CB;DA+DB}
(Note {CA+CB;DA+DB} is the biggest value in the two) :blush:

Math is my life :!:

This involves calculus:

Note that AM+BM is a function of CM (among other variables).
Let $\theta_A$ be $\angle AMC$ and $\theta_B$ be $\angle BMC$.

It can be seen that $\frac{\partial{AM}}{\partial{CM}}=sec(\theta_A)$ and $\frac{\partial{BM}}{\partial{CM}}=sec(\theta_B)$.

So, $\frac{\partial(AM+BM)}{\partial{CM}}=sec(\theta_A)+sec(\theta_B)$.

Find the second partial derivative:
$\frac{\partial^2(AM+BM)}{\partial{CM^2}}=sec(\theta_A)tan(\theta_A)+sec(\theta_B)tan(\theta_B)$ which is postive for all $0<\theta_A,\theta_B<\pi$.

Hence, all relative extrema on the interior of the interval of CM's possible values are minima, not maxima.

At least one of the interval's endpoints (CM=0 or CM=CD) must possess the only maximum(maxima), in which case either M is at C or M is at D, respectivally.  Thus, the only maximum(maxima) of AM+BM is(are) AC+BC or AD+BD.

However, M cannot be at C or D (remember - it must be between), so AM+BM is doomed to always be  less than either AC+BC or AD+BD.

(I did leave out something: the first derivative is not continuous, but when it isn't, it always goes from negative to positive anyway, ruling out a maximum.)


Each side of an equilateral triangle is 2 units long. Express in simplest radical form the number of square units in the area of the triangle.

The formula for the area of an equilateral triangle with side x is x^2sqrt 3/4
4sqrt3/4=
[b]sqrt3[/b]

Please $\LaTeX$ify it; I don't quite get what you're saying. Also, how would $4\sqrt{\frac{3}{4}} = \sqrt{3}$?

It seemed pretty clear to me.

He meant: $\frac{4\sqrt{3}}{4}$ which is equal to $\sqrt{3}$


Given positive real numbers $a,b,c$, prove that \[\frac{(b+c)^{2}}{a^{2}+bc}+\frac{(c+a)^{2}}{b^{2}+ca}+\frac{(a+b)^{2}}{c^{2}+ab}\geq 6.\] I'm afraid if this was proposed by other one and posted already :(

proposed by Darij [url=http://www.mathlinks.ro/Forum/viewtopic.php?t=60128]here[/url] :)


You are to design a “vehicle” that will safely transport an egg from the third floor of a building to the ground.  Your vehicle should be as light and as fast as possible.  Any ideas?

A parachute of the right size may do it.
Or you have to use the stairs in the building?
A hot air balloon will do it too. As the air cools down it will go down.
And it's apparent weight will be close to zero....


If x^2 + y^2 - 30x - 40y + 242 = 0, then the largest possible value of y/x can be written as m/n ,
where m and n are relatively prime, positive integers. Determine m + n.

[color=red]Warning. This has been edited to suit a typo[/color]

[hide="Hint"]Consider a geometric representation[/hide]

[hide="Solution"]
$ x^2 \plus{} y^2 \minus{} 30x \minus{} 40y \plus{} 24^2 \equal{} 0 \Longleftrightarrow (x \minus{} 15)^2 \plus{} (y \minus{} 20)^2 \equal{} 7^2$

Consider the line through the circle, passing through the origin, $ y \equal{} mx$. We want to maximise $ m$. If the line $ l_1$ passes through the circle, then we can steepen the line until it is tangent.

[img]http://i144.photobucket.com/albums/r184/mozzerhbh/AIME-Mock-2.png[/img]

Therefore we must find a tangent. (And check the obvious that $ 7^2 < 15^2 \plus{} 20^2$)

This requires us to find $ m$ such that the simultaneous equations have one solution.

$ \begin{cases} y \equal{} mx \\
(x \minus{} 15)^2 \plus{} (y \minus{} 20)^2 \equal{} 7^2 \end{cases}$

Substituting

$ (x \minus{} 15)^2 \plus{} (mx \minus{} 20)^2 \minus{} 7^2 \equal{} 0$

If there is one solution $ \Delta \equal{} 0$. Therefore

$ \minus{} 176m^2 \plus{} 600m \minus{} 351 \equal{} 0$ Therefore $ m \equal{} \frac {117}{44}$ (or $ \frac {3}{4}$)

Therefore $ m \plus{} n \equal{} 117 \plus{} 44 \equal{} \boxed{161}$
[/hide]

There was a typo in the question, which would explain the incorrect result.  :)

[quote="Hustla25"]If x^2 + y^2 - 30x - 40y + 24^2 = 0, then the largest possible value of y/x can be written as m/n ,
where m and n are relatively prime, positive integers. Determine m + n.[/quote]

thank you for your help

----and sorry about the typo
but please continue to answer the rest of the problems posted


[quote="Jakey"]Can anyone help me with these problems? And please show workings...

1) How many divisions are necessary to determine if 283 is prime?

2) For which integers k does x^k have the largest coefficient in the expansion of (x+3)^50.

3) What is the probability of having 4 boys and 4 girls in a family of 8 children?

4) Determine which is larger sqrt 10 + sqrt 17 or sqrt 53.

5) In a race of 1760 meters, A beats B by 330 meters, A beats C by 460 meters. If the contestants continue to run at the same uniform rates, by how many meters will B beat C?

thats all..thanks and i hope you could show some workings.[/quote]

[hide="1"]Find the square root of 283[/hide]
[hide="2"]Binomial Expansion[/hide]
[hide="3"]How many different possibilities of children are there/ How many of them are 4:4[/hide]
[hide="4"]Set them equal (adding x to one side) Solve for x[/hide]

Number 5:
I'm taking the differences in distance is compared when A finishes.
The difference in distance between B and C is proportional to the amount of time elapsed (since their speed are uniform), and is also proportional to the amount of distanced already traveled, so we have
$ (330/1430)*130\plus{}130$


linear algebra
matrix
induction
algebra unsolved
algebra

Let $a_{n}$ be the number of $n$-digit integers formed by $1,2$ and $3$ which do not contain any consecutive 1's. Prove that $a_{n}$ is equal to $(\frac{1}{2}+\frac{1}{\sqrt{3}})(\sqrt{3}+1)^{n}$ rounded off to the nearest integer.

I'm lazy, and I won't calculate the constants, but here is my proof. [hide]Firstly, some definitions:
$A_n$: the set of $n$-digit integers formed by $1$, $2$ and $3$ which do not contain any consecutive $1$'s. $a_n=|A_n|$.
$B_n$, $C_n$ and $D_n$: the partitions of $A_n$ with last digit $1$, $2$ and $3$ respectively. $b_n=|B_n|$ and so on.

(1) Obviously, $a_n=b_n+c_n+d_n$.
(2) By construction, it's easy to see that
$b_n=a_{n-1}-b_{n-1}$, $c_n=d_n=a_{n-1}$.
(3) Using (1) and (2), $a_n=3a_{n-1}-b_{n-1}$.
(4) So, we've got the recursive system:
$a_n=3a_{n-1}-b_{n-1}$,
$b_n=a_{n-1}-b_{n-1}$,
with the initial conditions $a_1=3$ and $b_1=1$.
(5) Putting (4) in a matricial form and calculating the $n$-th power of the matrix (by diagonalization or induction), we get a solution
of the form $a_n=k_1{\lambda_1}^n+k_2{\lambda_2}^n$. The constants can be calculated with some initial values.
If $|\lambda_2|<1$, soon (for every $n>n_0$) we have $a_n=round(k_1{\lambda_1}^n)$.[/hide]


How many perfect cubes are there in the sequence $ 1^{1},2^{2},3^{3},...,100^{100}$

[hide]All perfect cubes would be of the form a^3x.  So, all multiples of 3 and perfect cubes from 1 to 100 should be counted.

Perfect cubes: 1, 8, 27, 64

33 + 4 - 1 = [b]36[/b] (subtracting one because 27 is counted twice)[/hide]


I remember that in one of the previous AoPS classes (Either Intermediate Trig or WOOT) a way of solving polynomials when certain values for other polynomials using the same values are given was talked about, something similar to the following:

a is a positive integer (odd?)

Given that x^a + y^a + z^a = some value and x + y + z = some value, we are able to find the value of x^10 + y^10 + z^10 by using a property that is apparently related to x^a + x^-a.

If someone could point out where I can find it, or simply explain how to do it here, I would be quite grateful.

When the round is over, could someone explain this. It sounds interesting (I'm not taking this for the USA thingy (since I'm not american))

Thanks

[quote="timwu"]This sounds very similar to one problem on current round's USAMTS.[/quote]

Do they remove past references to problem types similar to current USAMTS problems? If so, then that explains why I couldn't find it in my class transcripts.


A regular 14-gon with side $a$ is inscribed in a circle of radius one. Prove  \[ \frac{2-a}{2 \cdot a} > \sqrt{3 \cdot \cos \left( \frac{\pi}{7} \right)}. \]

We have
$a^2=2-2\cos{\frac{\pi}{7}}=4\sin^2{\frac{\pi}{14}} \leftrightarrow a=2\sin{\frac{\pi}{14}}$, and
$\sin{\frac{\pi}{14}}<\frac{\pi}{14}<\frac{32}{140}$.
So  $2a\sqrt{3\cos{\frac{\pi}{7}}} \leq 3a^2+\cos{\frac{\pi}{7}}=6-5\cos{\frac{\pi}{7}}$,
and so it is enough to prove that
$6-5\cos{\frac{\pi}{7}}<2-a \leftrightarrow 4+a<5\cos{\frac{\pi}{7}}=5-10\sin^2{\frac{\pi}{14}}
\leftrightarrow \newline \leftrightarrow10\sin^2{\frac{\pi}{14}}+2\sin{\frac{\pi}{14}}-1<0$,
which is obvious by noting that  $\sin{\frac{\pi}{14}}>0, \frac{32}{140}<\frac{\sqrt{11}-1}{10}$.


Ok

how to play GHOST

somebody starts out by saying a letter.

like b

then, the next person says another letter, and tries to continue the word.

but, if you spell a word, then you get a letter.

first time-G
next time-H
and so on, until u spell GHOST.

then you died.

and are out of the game

youmay consult dictionaries to help youu find words.

also, if you think the pereson before you is not thinking of a word, you may challenge them, and ask them to tell you the word, and definition.

NO SWEAR WORDS, stuff like that.

order of ppl are in the order u sign up.

do not post letter until we start the game

signups begin now

8 ppl tops

[quote="eggylv999"]join

wasnt there another version of this?[/quote]
Yes, there was, but abandoned due to little participation.

Whatever, I join.

i will join too

4got to include taht in my topic game

1 spot left!

I just took the last spot
Auroraak joins

[quote="bluecarneal"]I guess I should start then....
[b]a[/b][/quote]

Not in the game, but - isn't "a" a word?

4got to add

words=3 letters or more

so "a" does not qualify

"add" quailifys

but "ad" doesnt

thanks James4|

asb

suggestion:if you do not post in five days, you are out

[quote="Joe10112"]Ok

also, if you think the pereson before you is [b]not[/b] thinking of a word, you may challenge them, and ask them to tell you the word, and definition.

[/quote]

Is the not supposed to be there?

correction:

if you think the player before you just put a random letter there, and is not thinking of a word, then you may challenge tem

after dojo is me

skip me

i wont play

i will just be the overseer for theis game


geometry
circumcircle
Euler
power of a point
radical axis
geometry proposed

Suppose $ABC$ is a triangle. And points $A, D$, are symmetrical to $BC$.$AK$ is the diameter of

the circumcircle of triangle $ABC$. $BK, CK$ intersect lines $AC, AB$ at points $M, N$

, respectively. Prove that:

 circle with diameter $AN$ , and circle with diameter $AM$, and circumcircle of triangle $ADI_{a}$,

are three coaxal circles.

( coaxal circles means : circles with the same radical axis )

[quote="vittasko"]The intersection point P (other than A), of the circles with diameters the segments AM, AN, lies on the segment MN  ( well known and easy to prove ). So, if Ia, is the center of the excircle of ABC, with respect to the angle A, then the circumcircle of the triangle APIa, is intersected from the altitude of ABC through A, at a point so be it D, not always symmetrical of A, with respect to the sideline BC.

     Sorry, if I am understanding wrong about the point Ia.

     Kostas Vittas.[/quote]

good,what do u think about this problem?

is it easy or hard?

 :wink:

[quote="Ashegh"]good,what do u think about this problem?

is it easy or hard?[/quote]

He says that it is wrong ;) .

And I can confirm that, unless you mean by $I_{a}$ something different from the A-excenter of $\triangle ABC$.

  darij

[quote="darij grinberg"]

... He says that it is wrong ;) ... .[/quote]
did he? :?:
[quote="darij grinberg"]... And I can confirm that, unless you mean by $I_{a}$ something different from the A-excenter of $\triangle ABC$.

  darij[/quote]

no he means A-excnter. :wink:

[quote="Amir.S"][quote="darij grinberg"]

... He says that it is wrong ;) ... .[/quote]
did he? :?: [/quote]

Read his post again ;)

Ashegh, please correct the problem, there must be some typo.

  Darij

[quote]... Sorry, if I am understanding wrong about the point Ia. ... [/quote]

He said that he is not sure that $I_{a}$ is A-excenter.

Then what is $I_{a}$? Please clarify your problem...

  darij

oooooooooooh,

i thought $I_{a}$ is a wellknown sign in geometry.

the problem is true surely. :wink:

$I_{a}$ is A_excenter of triangle $ABC$.

[hide]i inverted euler line of triangle $ABC$, round point$A$, and created this

problem. and i really want to know that is it hard, without using inversion

or not?[/hide]

Oh, I really confused…

Why should it be wrong? Im sure about it.plz read this solution it is completely true.

Let me invert the shape round point $A$, and with an arbitrary radius.
......................................................................................................................................................................

Theorem one:

Suppose$ABC$ is a triangle with circumcenter$O$,and $A,D$ are two symmetrical points with respect

To line$BC$.prove that after an inversion round point $A$, $D$ turns into the circum center of triangle

$AB'C'$.(which $B',D',M',C'$,are the inverse of the points$B,D,M,C$,respectively.

Ok, let me start to invert it:

Points $B,M,C$ are collinear, then $AB'M'C'$ is cyclic.

$\angle AMB=\angle AMC=90$, then: $\angle AB'M'=\angle AC'M'=90$

it shows us $AM'$, is the diameter of triangle $AB'C'$.

we know that, $AM.AM'=AD.AD'$, then devide this equality at $2$.

We have: $\frac{AD}{2}.AD'=AM.\frac{AM'}{2}$

But, $\frac{AD}{2}=AM$, then we conclude that: $AD'=\frac{AM'}{2}$

Because of $AM'$is diameter, and $D'$is midpoint of $AM'$, then $D'$ is circumcenter of triangle

$AB'C'$, and it is done.

……………………………………………………………………………………………………………

theorem two:

suppose $ABC$ is a triangle and $I_{a}$, is center of A_excircle(excircle is tangent to lines$AB,AC$, at$P,Q$

respectively)prove that after inversion round point$A$, $I_{a}$, turns into $I$ of triangle $AB'C'$.

let me start another inversion to prove this nice theorem:

as I havd been said, I don’t invert all the shape,I invert just a part of it.

$\angle PBI_{a}=180-\angle ABI_{a}$

$\angle CBI_{a}=\angle ABI_{a}-\angle ABC$

and because, $\angle PBI_{a}=\angle CBI_{a}$

then: $180-\angle ABI_{a}=\angle ABI_{a}-\angle ABC$($1$)

I want to know that how do the angles of($1$), changes, after inversion.

According to($1$), we have:

$180-\angle AI'_{a}B'=\angle AI'_{a}B'-\angle AC'B'$

then: $\angle I'_{a}AB'+\angle I'_{a}B'A=\angle AI'_{a}B'-\angle AC'B'$

then: $\angle I'_{a}AB'+\angle I'_{a}B'A=\angle I'_{a}AC'+\angle I'_{a}B'C'$

and because,$\angle I'_{a}AB'=\angle I'_{a}AC'$,

then:$\angle I'_{a}B'A=\angle I'_{a}BC'$, and it shows us that $B'I'_{a}$, is bisector of $\angle AB'C'$.

and finally after too boring calculation, we proved that $I'_{a}$ is $I$ of triangle$AB'C'$

……………………………………………………………………………………………………………

ok, lets come back to our problem,

if I invert the shape, according to the two lemma,

I claim that $D$, turns into $D'$(circumcenter of triangle $AB'C$)

And $I_{a}$, turns into $I$ of triangle$AB'C'$.

After that,$\angle ACN=\angle ABM=90$

It means: $\angle AN'C'=\angle AM'B'=90$

Circles with diameters$AM,AN$, turn into lines $B'M',C'N'$ respectively.

Then the intersection of them is $H$ of triangle$AB'C'$.

Then if Circles with diameters$AM,AN$, and circumcircle of triangle$ADI_{a}$, are three coaxal circles,

They should have another  common intersection except$A$.

It means that points$A,D,I_{a}$, and the intersection of Circles with diameters $AM,AN$, should be cyclic.

And according to it,we should prove that in the inversive shape,

$D',I'_{a}$ and intersection of lines $B'M',C'N'$ are collinear.

But $D'$ is $O$ of triangle $AB'C'$.

And $I'_{a}$ is $I$ of triangle $AB'C'$.

And intersection of lines $B'M',C'N'$ is $H$ of triangle $AB'C'$.

But the colinearity of these three points is clear.

Because in each triangle, $O,H,I$ are collinear(Euler line)

The problem is solved.

Some times the Phenomenons which we see, aren't trust.

And some times the Phenomenons which we believe not in them,are really trust.

It shows us that sometimes our eyes tell lie to us. :D

[quote="darij grinberg"][quote="Ashegh"]Because in each triangle, $O,H,I$ are collinear(Euler line)[/quote]

...

  Darij[/quote]
 :rotfl:

oh oh oh,,, im so sorry for it.

i meant, what is the relation betwin the angles of triangle$ABC$,

if that three circles are coaxal :oops:

i really know, what euler line is :rotfl:


geometry
3D geometry
sphere
calculus
integration

We have a fixed grounded conductive sphere of radius $R$. We put a tiny charge $q$ in distance $d$ from the center of the sphere.$(d\gneqq R)$. Find the time in which the tiny charge gets to the surface of the sphere.

I got integral like this but I can't solve it now :(

$\int{\frac{dx}{\sqrt{ln(x^{2}-R^{2})}}}$

Is it Ok?

Nice problem. I got \[t = \frac{d^{2}-R^{2}}{4q}\sqrt{\frac{m}{kq}}.\] Is that ok?

Well I didnt get the same answer, its much more complicated. :maybe:
Did you take into account that the magnitude and place of the mirror charge is changing?

Drat... I did the case assuming that the sphere is at all times at a very large distance from the point charge :mad:

Now I see the equations... They appear solvable, but I'm not solving that unless I have to :) Since you've already done it, will you post the answer?


this problem is from Mathcounts warmup 10, year 03-04.

Noel earns Unparseable_latex_formula20 to take his girlfriend to the movies. if he continues this same spending pattern, how many weeks will it take Noel to first exceed $125 in his savings account?

the answer is supposed to be [hide] 13 [/hide] but i keep getting 14.

This was talked about here a couple of weeks back. The answer given was wrong.


pigeonhole principle
calculus
integration
number theory
relatively prime

[quote="aleph0"]Does "having relationship of multiple or factor to each other" mean that they are relatively prime or that neither is a factor of the other?[/quote]

Yes.

Neither integral multiple nor divisor , one to another.

Choose $ n \plus{} 1, n \plus{} 2, ..., 2n$. The smallest multiple of $ n \plus{} 1$ is $ 2n \plus{} 2$, so neither is a factor of the other.

[quote="vinskman"]From $ {1,2,3,4,....,2n}$ ,prove that  there exist n integers such that no any pair of numbers having relationship of multiple or factor to each other.

:sleeping:
:starwars:[/quote]

Well... you would try to pick all the odd numbers... because you can't have more than one even number in the group... But then how would that work? In a set of 2n, you have n odd numbers, but what if some of the odds were not relatively prime (like 7 and 21)

Unless I misinterpreted the question or am downright wrong.  :blush:

[b]Edit:[/b][i] I didn't see ffao's post above.[/i]

Do you mean that, for all $ n$, there exists a subset of $ \{ a_1, a_2, \ldots a_n \}$ of the set $ \{ 1 \ldots 2n \}$ such that $ \forall i,j \in [1,n]$, $ \frac {a_i}{a_j} \not \in \mathbb{N}$?

[hide="Answer"]The subset $ \{ n \plus{} 1, n \plus{} 2, \ldots 2n \}$ always works.[/hide]

I've seen a closely worded problem where the quantifiers are switched:  for any subset of $ n$ elements of $ \{ 1, 2, ... 2n \}$, there exists a pair of elements such that neither divides the other.

I've seen another similar-sounding problem: for any $ n \plus{} 1$-element subset of positive integers $ \le 2n$, there esists a pair of elements such that one divides the other.


Here is the problem:
Find the counterexample for this statement:
If we have
$lim\left_{x \rightarrow a} f(x)=b$
$lim\left_{x \rightarrow b} g(x)= c$
then
$lim\left_{x \rightarrow a} f(g(x))$ exists

[quote="Ravi B"]In the last line, did you mean $g(f(x))$, not $f(g(x))$?[/quote]
Yes, you are right, Ravi B

Let $g(x)=\begin{cases}0&x\ne0\\1&x=0\end{cases}.$

Let $f(x)=0.$

Then $\lim_{x\to0}f(x)=0$ and $\lim_{x\to0}g(x)=0$ but

$\lim_{x\to0}g(f(x))=1\ne0.$

The way to rule out this example would be to strengthen the hypothesis on $g$ from having a limit at $b$ to being continuous at $b.$

If $\lim_{x\to a}f(x)=b$ and $g$ is continuous at $b,$ then we do have that $\lim_{x\to a}g(f(x))=g(b).$

Kent's example shows that the limit of $g(f(x))$ doesn't have to equal $c$.  The original question was to show that the limit doesn't have to exist at all.  Fortunately, we can modify Kent's example to show this stronger statement.

Let's keep $g$ the same as Kent's.  Let's change $f$ as follows: \[f(x) = \begin{cases}0 & \textrm{if }x \textrm{ is rational;}\\ x & \textrm{if }x \textrm{ is irrational.}\end{cases}\] Then $\lim_{x \to 0}f(x)$ is still $0$.  Furthermore, we have \[g(f(x)) = \begin{cases}1 & \textrm{if }x \textrm{ is rational;}\\ 0 & \textrm{if }x \textrm{ is irrational.}\end{cases}\] Thus $\lim_{x \to 0}g(f(x))$ doesn't exist.


I'm sorry if this isn't related, but I'm trying to create a schedule for 30 teams, in which each team plays 17 teams 6 times each, and plays 12 teams 5 times each.

Is that possible?

If so, is there a way to generate them according to some sort of formula?

Thanks.

This sounds very much like a graph theory problem? Unfortunately I really have no expertise in graph theory  :P.

123456789, just so you know, the equation in your signature is not always true. :wink:

I'm sorry, but I'm not that advanced in mathematics (only taking like Precalculus right now), so I don't know anything about the graph theory...

I'm trying to create a program for 30 teams, in which team plays 17 teams 6 times each, and plays 12 teams 5 times each.

If anybody has any clue about how I can proceed on this problem using like a formula or something like that, I would appreciate it.


calculus
derivative
function
real analysis
real analysis unsolved

problem5: f has continuous second derivative in [0,2] and f(0)=f(1)=f(2)=0. prove that for every natural n exists c_n in (0,2) such that n^2f(c_n)+f''(c_n)=2nf'(c_n). is there a functions with the properties above for which for every natural n the number c_n is unique.


[quote="IrinaSharova"]Please find the solution of the equation $ \frac{dy}{dx}+y^{2}+\frac{x}{1-x}y=\frac{1}{1-x}$ then satisfies $ y=2$ at $ x=0$.

Thank you for helping. :P[/quote]

This is a Riccati's equation. To solve it, you can find a partial solution $ y_{1}$. By substituing $ y=y_{1}+\frac{1}{z}$, you have a simple equation and you must find $ z$.

It's easy to see that this equation have a partial solution $ y_{1}=\frac{1}{x}$.

Just wondering what is partial solution?

[quote="marshell"]It's easy to see that this equation have a partial solution $ y_{1}=\frac{1}{x}$.[/quote]

I also wonder how to find the partial solution.

Thank you for helping. :P

hello, i have found this "nice" solution
$ f(x)\equal{}\frac{1\plus{}e^{x}}{x\plus{}e^{x}}$.
Sonnhard.


This is a word game with the letters {A,B,C,D,E}

- The initial word is AB
- You can add a C to a word ended in B
- From a word Ax (x is a word) you can form a new word Axx
- You can change BBB for D
- You can change DDD for E
- You can erase EE
- You can also change EE for C
- You can add D to a word ended in C

Is it possible to form ACDDDD?

It's supposed to show how you can transform AB into ACDDDD, following those rulles! :oops:

Here is the answer:

ABBB

ADBBB

ADDBBB

ADDD

AE

AEBBBBBBBBB

AEDDD

AEE

AC

ACD

ACDBBBBBBBBB

ACDDDD

You can't put ABBB. The rulle says you can transform AB (x=B) into ABB. Then you got ABB (x=BB) and if you continue doing this you'll get ABBBB...

Keep trying :lol:

[quote="stevenmeow"]eheh right
this is quite math intensive

for convenience, the nth line of instruction given will be called the nth rule, so starting with AB would be the 1st rule

assign values to letters except A, b/c there will be only 1 A ever (no rule creates an A except the 1st rule)

B=1, C=18, D=3, E=9 (A=0)

the final word has value 30

you start with value 1 and here is the effect each rule has on the value:
1st rule: used once, starts you off with a value of 1
2nd rule: adds 18
3nd rule: doubles value
4th rule: nothing
5th rule: nothing
6th rule: subtracts 18
7th rule: nothing
8th rule: adds 3

now we can prove ACDDDD is IMPOSSIBLE!!!!!
excluding the first rule (which can only be used once), there is no rule that can change a number from being a number that is not divisible by 3 to a number that is divisible by 3
yes that is confusing but read it a few times and you'll understand it
since you start off with 1, a number not divisible by 3, the value will never become divisible by 3 and therefore can never become 30

phew...
btw i think this doesnt belong in classroom math unless im wrong and there is a solution[/quote]

Nice job!  I was thinking about the same solve: B=1 C=18...  But couldn't to formulate it as a solve....  :roll:

[quote="myruggy89"]i agree with nick 42. it is more of a logic question than a mathamathics[/quote]

So in what subject there can be logic questions? In math!  :wink:


induction
inequalities unsolved
inequalities

The first one:
$(1/2).(3/4)...((2n-1)/(2n))\geq 1/(2\sqrt(n))$ can be proved by induction, because the induction stpe is equivalent to:
$\frac{2n+1}{2n+2}\ge \frac{\sqrt{n}}{\sqrt{n+1}}$ and that is equivalent to $n+1\ge 0$, which is true. QED


prove that:
$ \frac{1}{sin2x} \plus{} \frac{1}{sin4x} \plus{} \frac{1}{sin8x} \plus{}....\plus{}\frac{1}{sin2^nx} \equal{} cotx\minus{}cot^nx$

hello, in the case with $ n\equal{}1$ we get
$ \frac{1}{\sin(2x)}\equal{}\cot(x)\minus{}\cot(x)$?
Sonnhard.

Maybe ...
[quote="aadil"]Prove that $ \sum_{k\equal{}1}^n\frac {1}{sin\left(2^kx\right)} \equal{} \cot x \minus{} \cot \left(2^nx\right)$ .[/quote]

[hide]Easy to prove that $ \frac1{\sin2\alpha}\equal{}\cot\alpha\minus{}\cot2\alpha$ from which the identity follows easily.[/hide]

[quote]Maybe ...
aadil wrote:
[quote]Prove that  .  $ \sum_{k\equal{}1}^{n}\frac{1}{sin\left(2^{k}x\right)}\equal{}\cot x\minus{}\cot\left(2^{n}x\right)$[/quote][/quote]
yes,thats what i meant

Let $ a_{n}\equal{}\frac{1}{\sin (2^{n} x)}$ and $ b_{n}\equal{}\cot x\minus{}\cot (2^{n}x)$ .Note that $ a_{n}\equal{}b_{n}\minus{}b_{n\minus{}1}$ .Thus the sum is telescopic and equals to

$ \sum_{k\equal{}1}^{n} a_{k}\equal{}\sum_{k\equal{}1}^{n} b_{k}\minus{}b_{k\minus{}1}\equal{}b_{n}\minus{}b_{0}$


group theory
abstract algebra
superior algebra
superior algebra unsolved

Suppose that $Y$ is a cyclic group.  How might I go about proving that each $X$ such that $X\subseteq Y$ is a cyclic subgroup?

I know that since $Y$ is cyclic, there exists $y\in Y$ such that $\left<y\right> = Y$.  Does it have to do with the fact that a cyclic subgroup will ensure closure under the binary operation (in this case)?

Obviously, if the subgroup is cyclic then there exists an element $a$ that has the property $a^{n}=e$ for some $n$.  Also, for each $a$ there must exist $a^{k}$ such that $a^{k+1}=a^{n}=e$.  Finally, if there are two elements $a^{i}$ and $a^{j}$, then $a^{i}a^{j}=a^{i+j}$, which is in $X$ since $a$ is a generator and $X$ is cyclic.

Is this a valid / complete justification?

Consider minimum positive integer n s.t $y^n \in X$.
Now $X=<y^n>$ can be shown using the fact that if $y^k$ s.t   k  not a multiple of n$\in X$ then say $k= nq + r$,which implies $y^r \in X$ which is a contradiction to minimality of n.


:maybe: [hide="I may have missunderstood the problem"]

In plane, there's only one point wich equidists from $ A$, $ B$ and $ C$(the three of them at same time): the circuncircle of triangle $ ABC$.

In 3-D, there's only one straight line wich satisfies this, the perpendicular to plane $ ABC$, through the circuncircle of triangle $ ABC$.

The other possible interpretation:

The number of straight lines wich equidist from $ A$ and $ B$, $ B$ and $ C$ or $ C$ and $ A$ is three.

In wich case those are the perpendiculars through the midpoints of $ AB$, $ BC$ and $ CA$. (as $ A$, $ B$ and $ C$ are different points those are different lines)
[/hide]

 :blush: Indeed, I misunderstood it.

[quote="LoboSolitario"]Let $ A$, $ B$ and $ C$ noncollinear points.
Prove that the number of straight lines on the plane $ ABC$ equidistant of $ A$, $ B$ and $ C$ is $ 3$.[/quote]

[asy]size(250); pair A,B,C,D,F,G,P,Q,R; A=origin; B=(100,0); C=(30,70); draw(A--B--C--cycle); D=midpoint(A--B); dot(D); F=midpoint(B--C); dot(F); G=midpoint(C--A); dot(G); draw(D--F--G--cycle); label("A",A,W); label("B",B,E); label("C",C,N); label("D",D,S); label("E",F,NE); label("F",G,NW); P=(0,35); Q=(30,35); R=(100,35); draw((-20,35)--(120,35)); draw(A--P); draw(C--Q); draw(B--R); dot(P); dot(Q); dot(R); label("P",P,SW); label("Q",Q,SE); label("R",R,SE); label("AP=CQ=BR",(120,35),E);[/asy]

The three lines passing through the midpoints of $ AB,BC,CA$ satisfy the given condition. And it's not difficult to see that no other line satisfies.


if $k=1987^z$, then : [url=http://www.mathlinks.ro/Forum/viewtopic.php?t=22161]look at that![/url] :D


probability
expected value
conditional probability
probability and stats

Can any one solve this question? Thanks

2) A monkey types on a 26-letter keyboard that has lowercase letters only.
Each letter is chosen independently and uniformly at random from the
alphabet. If the monkey types 1,000,000 letters, what is the expected
number of times that sequence "proof" appears?

let $ A_i$ be the random variable that is equal to $ 1$ is the word "proof" begins at the position $ i$, and $ 0$ otherwise. You are trying to find the expectation of \[ N\equal{} A_1 \plus{} A_2 \plus{} \ldots \plus{} A_{999996}\] Can you conclude from here ?

I am still confused, It looks like you rephrased the problem,
let Ai be the random variable that is equal to 1 is the word "proof" begins at the position i, and 0 otherwise.
N = A1 + A2 + A3 + ....+ A999996
So:
1. If A1 = 1, then A2,A3,A4,A5 must be 0,similarly,If A1 = 2, then A3,A4,A5,A6 must be 0,.........and so on
2. if A999996 = 1, others must be 0 then N = 1, similarly, if A999996 = A999992 = 1, others must be 0, then N = 2  ... and so on
3. but, if A1 = 1, N could be max 200000, if A2 = 1, then N could be max 199999
4.the P(A1 = 1) is 1/2,
P(A2 = 1) = P(A2=1|A1=0), N could be max 199999
  P(A3 = 1) = P(A3=1|A2=0, A1=0), N could be max 199998
  P(A4 = 1) = P(A4=1|A3=0, A2=0, A1=0), N could be max 199997
  P(A5 = 1) = P(A5=1|A4=0, A3=0, A2=0, A1=0), N could be max 199996
  P(A6 = 1) = P(A6=1|A5=0, A4=0, A3=0, A2=0), N could be max 199995
  P(A7 = 1) = P(A7=1|A6=0, A5=0, A4=0, A3=0), N could be max 199994
                     ……….. and so on
5.  P(A2=1|A1=0) = P(A2=1,A1 = 0) / P(A1=0)
               = P(A1=0|A2=1)P(A2=1) / P(A1=0|A2=1)P(A2=1)+P(A1=0|A2=0)P(A2=0)
               = (1*0.5) / (1*0.5)+(0.5*0.5)
               =2/3

The above is what I can think and I am not sure they are correct, I still do not know how to solve it. could you please tell me more? thanks

Why would you want to make this a conditional probability problem?  (As you've noticed, that makes everything complicated.)  Instead, can you compute the expectation of $ A_1$?  What about the expectation of $ A_2$?  Etc.  Now, what do you know about the expectation of a sum of random variables?

Sorry, I am a beginner. All i know is that expectation is the summation over multiplication of random variables and their probability.
Thanks alot for the helps,

The expectation of A is

So the E(A) is sum(A=1 to 999996)[1 * 2^(999996-A)/2^999996 ]

I am not sure if this is correct,

I find it very implausible that you were asked this question without first having learned a theorem about the expectation of linear combinations of random variables.  Perhaps you should look over your notes?  (And while you're doing that, you can still answer my other questions: what's the expected value of $ A_1$?  What's the expected value of $ A_2$?  What's the expected value of $ A_{5628}$?)

Ah, I see you've been editing.  I don't know what thought process led to your current expression, so I don't know if there are any ideas that can be recovered there :(

Really, I suggest you start by answering the following simpler questions:
What is the expected value of $ A_1$?
What is the expected value of $ A_2$?
What is the expected value of $ A_{5628}$?
Or even more simply, what is the probability that $ A_1 \equal{} 1$?  What is the probability that $ A_2 \equal{} 1$?  What is the probability that $ A_{5628} \equal{} 1$?
(Note I'm not asking for a conditional anything.)

Thanks for the reply.   :lol:

I do not understand why they are independent? Because if A1 = 1, then A2-A4 must not be 1
So, if it is not conditional, I am guessing the probabiliy that A1 =1 is 0.5

is it correct?

thanks

They aren't independent!  But, it is still possible to compute the (non-conditional) expected value of each of them!  For example, let $ X$ be the random variable that tells me how many heads I get when I flip a fair coin twice and $ Y$ be the random variable that tells me how many tails I get.  $ X$ and $ Y$ are very non-independent, but yet it's easy to calculate that the expectation of $ X$ is equal to $ 1$ (just ignore $ Y$ and do the computation) and similarly it's easy to compute that the expectation of $ Y$ is equal to $ 1$ (just ignore $ X$ and do the computation).

To compute the expectation of $ A_1$, you don't need to know anything about the random variables $ A_2, \ldots$, you just need to apply the definition of expectation to the event measured by $ A_1$.  Similarly, for $ A_2$, etc.  All these random variables have nice, simple definitions -- can you compute their expected values from their definitions?


[i]Moderator edit: Please never, ever post problems from summer program admission quizzes on any AoPS forum. We will have to notify adminstrators of this summer program that this question was compromised, and that is very unfair to the applicants who did their own work on this problem. [/i]

wow, tough problem.  Maybe this should be in intermediate instead or something?

[i]Moderator edit: If you see something like this happen again on Getting Started, please private message tokenadult right away.[/i]

It's actually not as tough as it appears at first.  At first site I thought it was Intermediate or above also!  But I think its within the scope of Getting Started.

Yes, this problem looks insanely hard. Upon [b]further inspection[/b] you will find that this problem is very easy. It would not have been that easy if the problem asked for the tens digit or the first digit of that number :D  :D  :D .

BTW, even if I present an answer, the proof technique is too advanced for this forum. It will have to appear in the higher forums (sry, forum policies cannot be broken).

[i]Moderator edit: Thanks for thinking about keeping forums on the proper level. If you see a situation like this again, please private message tokenadult, and I will use moderator powers to make appropriate adjustments in the thread.[/i]

Basically I just meant that these tricks are higher than getting started level :) I was just referring to problems in the getting started forum.

[i]Moderator edit: Thanks for thinking about the appropriate level of this forum. Alas, now it is the season when North American summer programs have application quizzes out to applicants, so please kindly let me know by private message to tokenadult if you see a discussion of a problem that seems too hard for the person asking about the problem: that may be an attempt to cheat on an application quiz.[/i]

Sure - But it will be better if we moved it into intemediate topics  form  (before we get into trouble with the moderators).

[i]Moderator edit: Thank you for thinking about the appropriate level of the forum. Here is another thing to think about when a thread looks much too hard for its intended level: as happened last year, during this season some kids are working on application quizzes for summer programs, and some can't resist the temptation to cheat by asking about intermediate steps (as apparently happened in this thread) to check answers for the quiz questions. Let's all be alert to that possibility. Please private message tokenadult if you see a thread asking about a problem that seems to take unusual amounts of thought and steps for a beginning problem-solver: it might be an application quiz problem.[/i]

[quote="SnowStorm"][problem statement][/quote]

Hmm, Is this legal? Because this is the first question on the Mathpath Qualifying Exam!!!

[i]Moderator edit: It is NOT legal, and I am glad that you pointed this out. As you see, I have had to extensively edit this thread. Please private message tokenadult if you ever see any situation like this on any AoPS forum I moderate, and please private message the board administrators IMMEDIATELY if you see any doubtful  thread during summer program application season. Thanks for pointing this out.

There, does everyone see what happened here? When a thread is getting answers mostly from adults in a lower-level forum, suspect that something might be wrong, and let a moderator know about it right away by private message. Also please private message board adminstrators right away if you see current competition problems or program application problems discussed anywhere on AoPS. Thanks for your help.[/i]

[quote="SnowStorm"]Really?  A friend from school told me this question... Maybe this thread should be deleted.[/quote]

[i]Moderator's edit: Most of the thread has been deleted now, and now I need to private message moderators of another forum that they have some deleting to do. I also need to private message the board administrators and email the MathPath staff about this. Don't just post problems from a "friend" unless you know where the problems came from. Posts on AoPS are most helpful if they indicate a source of a problem.[/i]


[hide="Just a note"]I don't believe this is what belongs in the Linguistics forum. Linguistics is the formal study of the structure of languages and how they relate to each other, not how best to use language as a work of art.  :)  But that's just what I think.[/hide]

Hey AIME15,

I think you should start your essay with words that are stronger and appeal more to the heart. Also, don't make it like a one paragraph essay, make it many paragraphs. That way, the reader will have time to actually digest what you're saying and it will make your introduction stronger.

-rts2007

Some children [i]I know[/i]? Don't weaken by making your evidence anecdotal. Either appeal to the heart or appeal to facts, but don't appeal to you the author's experience unless you're an expert.


calculus
integration
trigonometry
real analysis
real analysis unsolved


geometry
circumcircle
inradius
geometry unsolved

Let $ ABC$ be a triangle with $ BC\equal{}a,CA\equal{}b,AB\equal{}c$,let $ O$ and $ R$ be respectively its circumcenter and circumradius,let $ I_a,I_b,I_c$ be respectively the centers of its escribed circles in angles $ A,B,C$ and let $ r$ be the inradius of $ \triangle ABC$.Prove that
$ \frac{1}{2R}\leq\frac{OI_a}{(a\plus{}b)(a\plus{}c)}\plus{}\frac{OI_b}{(b\plus{}c)(b\plus{}a)}\plus{}\frac{OI_c}{(c\plus{}a)(c\plus{}b)}\leq\frac{1}{4r}$

nobody can solve this problem,darji can you help me?


geometry
induction
analytic geometry
rotation
geometry open

Dear friends, T[0] = a given triangle ABC and, by induction, T[n+1] is the contact triangle of the incircle of T[n]. Obviously, as circumradius(T[n+1])/circumradius(T[n])<=1/2, the vertices of T[n] converge to the same point.
Is it possible to localize this point? Has this point rationnal - or algebraic, or analytic - barycentric coordinates in term of a,b,c (sidelenghts of ABC) or in term of angles A,B,C?
Kind regards from France.
Jean-Pierre Ehrmann

Not only the vertices of the triangle T(n) converge to the same point, the triangles T(2n) and T(2n+1) converge to  2 equilateral triangles mutually rotated by :pi:/3. Consider the cevians of the triangle T(0) through the vertices of the triangle T(n). They converge to cevians concurring at the limiting point, identical to the common center of everything of the 2 limiting equilateral triangles. Therefore, they pairwise form angles :pi:/3 and consequently, the limiting point is the Fermat point of the triangle T(0) and even the common Fermat point of the entire sequence of T(n).

I don't think so : T[0] and T[1] have different Fermat points.
In any case, I don't understand what you mean with your two equilateral triangles. Their six vertices should be the same point : the common limit point we are looking for.
In fact, I suspect that the limit point is not analytic and, thus, it is impossible to localize this point.
Kind regards


1)  If $ f(\frac{x\minus{}1}{x\plus{}1}) \plus{}f(\frac{\minus{}1}{x}) \plus{} f(\frac{1\plus{}x}{1\minus{}x}) \equal{}x$
For every $ x$ different from $ 0$,$ 1$,$ \minus{}1$, where $ f: R\minus{}>R$. Find an explicit formula for $ f(x)$.

2) Find all functions $ f: R\minus{}>R$ such that $ f(x\plus{}y)\minus{}f(x\minus{}y)\equal{} f(x)f(y)$ for every $ y$ and $ x$ real numbers.

Please give an step by step solution and explain it very clear. Thanks! :)

For the 2.

For $ x\equal{}y\equal{}0$ we take $ f(0)\equal{}0$.

Setting $ x\equal{}0$ we take $ f(y)\minus{}f(\minus{}y)\equal{}f(0)f(y) \Longleftrightarrow f(y)\equal{}f(\minus{}y)$.

Now $ y\minus{}\minus{}>\minus{}y$ the equation becomes:
$ f(x\minus{}y)\minus{}f(x\plus{}y)\equal{}f(x)f(\minus{}y) \Longleftrightarrow f(x\plus{}y)\minus{}f(x\minus{}y)\equal{}\minus{}f(x)f(y)$.

So $ f(x)f(y)\equal{}\minus{}f(x)f(y),\forall x,y \in \mathbb{R}$

So setting $ y\minus{}\minus{}>x$ we take $ 2f^2(x)\equal{}0 \Longleftrightarrow \boxed{f(x)\equal{}0}$.

both of them are pretty easy.for the first one the only thing you have to do,is to make the following:$ x\minus{}>\frac{x\minus{}1}{x\plus{}1}$,$ x\minus{}>\frac{\minus{}1}{x}$,$ x\minus{}>\frac{x\plus{}1}{1\minus{}x}$ and solve the 4x4 system which is trivial. :P  :)


Does there exist a positive integer $ n>1$ such that $ n$ is a power of $ 2$ and one of the numbers obtained by permuting its (decimal) digits is a power of $ 3$ ?

[hide="question"]

Um, $ 2 \equiv \minus{}1 \pmod {3}$.  $ 2^a \equiv \pm 1 \pmod {3}$.  So $ 2^a$ does not divide $ 3$.  Regardless of permuting the digits, the number will still not be divisible by 3 because the divisibility rule relies only on the sum of the digits.  So the number will not be divisible by 3, much less be a power of 3.

Am I reading the question wrong?  The problem seems too easy.[/hide]

[quote="azn6021023"][hide="question"]

Am I reading the question wrong?  The problem seems too easy.[/hide][/quote]

The problem is OK and I agree that is very easy.


$x,y,z$ are positive real such that :

$2xyz=3x^2+4y^2+5z^2$

Prove that $3x+2y+z \ge 36$

Try another fun problem like this in Vietnam team selection test 2001
[quote="orl"]Let’s consider the real numbers $a, b, c$ satisfying  the condition

\[ 21 \cdot a \cdot b + 2 \cdot b \cdot c + 8 \cdot c \cdot a \leq 12. \]

Find the minimal value of the expression

\[ P(a, b, c) = \frac{1}{a} + \frac{1}{b} + \frac{1}{c}. \][/quote]

Note the numbers $2,3,4,5$ and $3,2,1$ and the result.! It's really nice and funny. :)

This is only coincident that one can constitute easily by variating constant procedure.
To Hung: remember to bring me the book you have borrowed since 2004 :D :)

Hi. I'll send you, Nmtruong.

The general Problem is:

$x,y,z \ge 0$ satisfy  that:
$ax^2+by^2+cz^2 =xyz$

Then $x+y+z \ge \frac{(1+\sqrt{1+a\alpha})(1+\sqrt{1+b\alpha})(1+\sqrt{1+c\alpha})}{\alpha}$

$\alpha$ is a positive real satisfy:

$\frac{2}{1+\sqrt{1+a\alpha}}+\frac{2}{1+\sqrt{1+b\alpha}}+\frac{2}{1+\sqrt{1+c\alpha}}=1$

It's not difficult!

But I can't understand, in the first problem :

$3x^2+4y^2+5z^2=2xyz$

We have $3x+2y+z \ge 36$

The equal hold with $a=b=c=6$?!!!
Hi, thanks God!


These days I'm going through old Sovjet olympiads,
and I discovered the following nice and simple geometry problem:

Let ABC be an isosceles triangle with |AB| = |AC|.
Let P be the orthogonal projection of the midpoint M of [BC] on AB.
Let Q be the midpoint of [MP].
Prove that AQ and PC are orthogonal.

[i]Sovjet olympiad 1962[/i]

Let H be the orthogonal projection of C onto [AB]. It is easy to see that P is the midpoint of [HB].
Moreover <APM = 90 = <CHB. End <MAP = <MAB = 90 - <CBA = <BCH. it follows that the triangles AMP and CBH are similar.
Let s be the similitude such that s(A) = C, s(M) = B and s(P) = H.
Then, s(Q) = P, since s perserves the midpoint.
It follows that s( (AQ) ) = (CP). (1)
But in another hand, s((AM)) = s((CB)), from which we deduce that the angle of s is \pm 90.
Using (1), the result follows...

Pierre.

A vectorial solution is also straightforward.


Let x,y,z positive real numbers such that x+y+z=1.
Prove that

 xy+yz+zx  \leq 1/4 +3xyz

it's a classical application of Schur :D:D
we have \sum x(x-y)(x-z) \geq 0  that is \sum x3 + 3xyz \geq \sum x2 y + xy2 and further more

( \sum x )3 \geq 4 ( \sum x2 y + xy2  ) + 3xyz that is

( \sum x )3 + 9xyz \geq 4 \sum x \sum xy and because \sum x = 1 we have

1 + 12 xyz > 1 + 9xyz \geq 4 \sum xy that is

1/4 + 3xyz > 1/4 + 9/4 xyz \geq \sum xy and we are done. so manlio your inequality is rather weak. :)


algebra
polynomial
calculus
derivative
search
algebra proposed

For $ f\in\mathbb{R}[X]$ with $ \deg f \equal{} n$ and $ f(x)\geq 0\ \forall x\in\mathbb{R}$, prove that $ \sum_{k \equal{} 0}^{n}f^{(k)}(x)\geq 0\ \forall x\in\mathbb{R}$, where $ f^{0}(x) \equal{} f(x)$.

I placed this problem in this section since it requires very little knowledge of calculus.

Check this out: http://www.mathlinks.ro/viewtopic.php?search_id=1657165616&t=159238
(It's not [i]exactly[/i] the same problem, but the solutions can be adapted.)

Just to give a slightly different argument, define the polynomial $ g$ by
\[ g(x) \equal{}\sum_{k\equal{}0}^{\infty}f^{(k)}(x).\]
As perfect_radio showed in the link, we have $ g' \equal{} g\minus{}f$, and so $ g'(x)\le g(x)$.  Define $ h$ by $ h(x) \equal{} e^{\minus{}x}g(x)$.  By differentiating, we get $ h'(x)\le 0$, so $ h$ is decreasing.  Because $ g$ is a polynomial, $ h(x)$ approaches 0 as $ x$  approaches infinity.  Thus $ h$ is nonnegative.  That means $ g$ is nonnegative.


quadratics
calculus
integration
number theory unsolved
number theory

Find the set of all ordered pairs $(s,t)$ of positive integers such that \[t^{2}+1=s(s+1).\]

I have changed the problem statement from $ t^{2}+1=s(s+t) $ to $ t^{2}+1=s(s+1) $.

http://www.imocompendium.com/othercomp/Tur/TurMO94.pdf says it is $t$.
A Turkish monthly math journal says it is $1$.

s(s+1)=$t^2+1$ . so , $s^2+s-(t^2+1)$=0. now we treat it as a quadratic in s. for having integral solution , we must have that D=$1-4.1.-(t^2+1)$=$(1+4[t^2+1])$=$4t^2+5$ is a perfect square.
so .let $4t^2+5=y^2$ , which is trivial to solve using standard factorization. :lol:

multiply the equation by 4,we get
4t^2+4=4s^2+4s
=>{2s+1}^2-{2t}^2=5
=>{2s+2t+1}{2s-2t+1}=5.1
=>2s+2t+1=5,2s-2t+1=1
=>s=t=1  is the only solution in positive integers

[hide=Sol]If $t,s\geq 2$, then we can see the following:
If $s\geq t$, then $s(s+1)\geq t(t+1) = t^2 + t>t^2+1$.
Next, if $s<t$, then $s(s+1)<(t-1)(t) = t^2-t<t^2+1$.
Therefore $t,s$ cannot be greater or equal to $2$.
If $s$ was $1$, then we have $t^{2} + 1 = 2$, so $t=1$. Similarly, if $t=1$, then we also get that $s = 1$. Therefore, our only solution is $(s,t) = (1,1)$.[/hide]

[hide=S]$t^2+1=s\left(s+1\right)$
$t^2+1=s^2+s$
$t^2+1+\frac 14=s^2+s+\frac 14$
$t^2+\frac 54=\left(s+\frac 12\right)^2$
$\frac 54=\left(s+\frac 12\right)^2-t^2$
$\frac 54=\left(s+t+\frac 12\right)\left(s-t+\frac 12\right)$
$5=\left(2s+2t+1\right)\left(2s-2t+1\right)$

$\Longrightarrow 2s+2t+1=5, 2s-2t+1=1\implies (s, t)=\boxed{(1, 1)}$
$\Longrightarrow 2s+2t+1=-5, 2s-2t+1=-1\implies (s, t)=(-2, -1)\notin\mathbb N$
$\Longrightarrow 2s+2t+1=1, 2s-2t+1=5\implies (s, t)=(1, -1)\notin\mathbb N$
$\Longrightarrow 2s+2t+1=-1, 2s-2t+1=-5\implies (s, t)=(-2, 1)\notin\mathbb N$[/hide]