1) When I "build" a file using TeXnix center, I always get a "badbox" message when using two newline commands. I don't have any problem with doing this, so how do I remove these annoying messages?

2) Is there a way to change the size font used between dollar signs? What I mean is, if say, you have $ y= \frac{\frac{x-h}{2a^{2}}}{5\pi}$, how would you enlarge the size so that the fraction isn't so hard to read?

1) You shouldn't ever be using 2 \newline commands hence the bad box errors. To start a new paragraph just leave a blank line between them. If you really must have extra vertical space you can use \bigskip or \vspace or \\*[[i]amount of space[/i]]

2) In a document, use either \dfrac instead of \frac or \displaystyle at the start of the formula.

I renamed one of my .tex files and now when I "build" the document, my asy pictures don't show up in the pdf format. When I run asy, it gives me some error message that basically says it cannot load module"filename". How on earth do I fix this?

You have to be careful that you don't rename your file and end up with the wrong file extension. Have a look in My Computer/Windows Explorer and see what the filename is called. Ensure that you can see file extensions (Tools->Folder Options->View-> Untick the 'Hide extensions for known file types' box) anc check your file isn't called something like filename.tex.tex

I actually found the problem was with the filename I chose to save with. It seems that asymptote only reads files where there is no space between words and/or no uppercase letters.

I was trying to save my file as "The Study of Functions" but when I resaved it simply as "functions", it worked. Odd...

[quote="Prime factorization"]I actually found the problem was with the filename I chose to save with. It seems that asymptote only reads files where there is no space between words and/or no uppercase letters.[/quote]That's true of many LaTeX related programs. Avoid filenames with spaces as much as you can.


You use huron's formula to find the area of a triangle, when you only know the length of the 3 sides. The formula is:
s=(a+b+c)/2
A=sqrt((s)(s-a)(s-b)(s-b))


Determine the prime numbers $a,b,c$  so that the number $A = a^{4}+b^{4}+c^{4}-3$ is also prime.

$3|A$ if $a \neq b \neq c \neq 3$ so WLOG $a=3$.
$4|A$ if $a \neq b \neq c \neq 2$ so WLOG $b=2$.
$A=94+c^4$ we know that if $c>5$ then $c^4 \equiv 1 \mod 5$ and $5|A$, So $c=5$
and $A=719$. And I do'nt knwo wheter it is prime or not ;)


geometry
geometric transformation
reflection
circumcircle
trigonometry
ratio
homothety

A nonisosceles triangle $A_{1}A_{2}A_{3}$ is given with sides $a_{1}$, $a_{2}$, $a_{3}$ ($a_{i}$ is the side opposite to $A_{i}$). For all $i=1,2,3$, $M_{i}$ is the midpoint of side $a_{i}$, $D_{i}$ is a foot of the bisector of $A_{i}$, $T_{i}$ is the point where the incircle touches side $a_{i}$, and the reflection of $T_{i}$ in the interior bisector of $A_{i}$, yields the point $S_{i}$. Prove that the lines $D_{2}D_{3}$, $T_{2}T_{3}$, $S_{2}S_{3}$ and $M_{1}S_{1}$ are concurrent.

Hi,I can prove that  $S2S3,T2T3,D2D3$  are concurrent.

$D3S3$ meet $D2S2 at P$
Then Let $S2S3$ meet $D2D3$ at $I1$
$T2T3$ meet $D2D3$ at $I2$

$\frac{PS3}{D3S3}\cdot\frac{D3I1}{I1D2}\cdot\frac{D2S2}{PS2}=1$

$\frac{A1T3}{T3D3}\cdot\frac{D3I2}{I2D2}\cdot\frac{D2T2}{A1T2}=1$

As $PS3=PS2$     $A1T3=A1T2$

$D3S3=D3T3$    $D2S2=D2T2$

Hence $\frac{D3I1}{I1D2}=\frac{D3I2}{I2D2}$

Hence $I1,I2$ is the same point.

$ D_{1}S_{1},\ D_{2}S_{2},\ D_{3}S_{3}$ are tangents to the triangle incircle $ (I),$ in fact, the common internal tangents of the incircle and the excircles other than the triangle sides. Let $ D_{2}S_{2}$ meet $ A_{3}A_{1}$ at $ B$ and let $ D_{3}S_{3}$ meet $ A_{1}A_{2}$ at C. The (concave) quadrilateral $ BD_{2}CD_{3}$ is tangential and $ D_{2}D_{3}$ is one of its diagonals (BC is the other one). $ T_{2}S_{2}T_{3}S_{3}$ is its contact cyclic quadrilateral with diagonals $ T_{2}T_{3},\ S_{2}S_{3}.$ Since these 2 quadrilaterals have a common diagonal intersection X, the lines $ BC,\ D_{2}D_{3},\ T_{2}T_{3},\ S_{2}S_{3}$ are concurrent at X.

By simple angle addition, the lines $ S_{2}S_{3}\parallel A_{2}A_{3},\ S_{3}S_{1}\parallel A_{3}A_{1},\ S_{1}S_{2}\parallel A_{1}A_{2}$ are parallel, hence the triangles $ \triangle S_{1}S_{2}S_{3}\sim \triangle A_{1}A_{2}A_{3}$ are centrally similar. Since $ \triangle S_{1}S_{2}S_{3}$ is inscribed in the incircle $ (I, r)$ and $ \triangle A_{1}A_{2}A_{3}$ in the circumcircle $ (O, R),$ the value of their similarity coefficient is $ \frac {r}{R}.$ Let $ D_{2}S_{2}$ meet $ D_{3}S_{3}$ at P, forming an isosceles triangle $ \triangle PS_{2}S_{3}$ with $ PS_{2} \equal{} PS_{3} \equal{} r \tan A_{1}.$ The ratio $ \frac {XS_{2}}{XS_{3}},$ in which the concurrency point X divides the segment $ S_{2}S_{3},$ can be calculated by Menelaus theorem for the triangle $ \triangle PS_{2}S_{3}$ cut by the line $ D_{2}D_{3},$

$ \frac {D_{2}P}{D_{2}S_{2}}\cdot \frac {XS_{2}}{XS_{3}}\cdot \frac {D_{3}S_{3}}{D_{3}P} \equal{} 1$

Assume $ a_{2} < a_{1} < a_{3}$ (The cases $ a_{1} > a_{2}, a_{3}$ and $ a_{1} < a_{2}, a_{3}$ are treated similarly, just with appropriate sign changes.)

$ D_{2}S_{2} \equal{} D_{2}T_{2} \equal{} A_{1}T_{2} \minus{} A_{1}D_{2} \equal{} (s \minus{} a_{1}) \minus{} \frac {a_{2}a_{3}}{a_{3} \plus{} a_{1}} \equal{} (s \minus{} a_{2})\ \frac {a_{3} \minus{} a_{1}}{a_{3} \plus{} a_{1}} > 0$

$ D_{3}S_{3} \equal{} D_{3}T_{3} \equal{} A_{1}T_{3} \minus{} A_{1}D_{2} \equal{} (s \minus{} a_{1}) \minus{} \frac {a_{2}a_{3}}{a_{2} \plus{} a_{1}} \equal{} (s \minus{} a_{3})\ \frac {a_{2} \minus{} a_{1}}{a_{2} \plus{} a_{1}} < 0$

$ \frac {XS_{2}}{XS_{3}} \equal{} \frac {D_{2}S_{2}}{D_{2}P}\cdot \frac {D_{3}P}{D_{3}S_{3}} \equal{} \frac {D_{2}S_{2}}{D_{3}S_{3}}\cdot \frac {PS_{3} \plus{} D_{3}S_{3}}{PS_{2} \plus{} D_{2}S_{2}} \equal{}$

$ \equal{} \frac {a_{3} \minus{} a_{1}}{a_{2} \minus{} a_{1}}\cdot \frac {s \minus{} a_{2}}{s \minus{} a_{3}}\cdot \frac {(s \minus{} a_{3})(a_{2} \minus{} a_{1}) \plus{} (a_{2} \plus{} a_{1})\ r \tan A_{1}}{(s \minus{} a_{2})(a_{3} \minus{} a_{1}) \plus{} (a_{3} \plus{} a_{1})\ r \tan A_{1}}$

Substituting $ r \equal{} (s \minus{} a_{3}) \tan \frac {A_{3}}{2} \equal{} (s \minus{} a_{3}) \frac {\sin A_{3}}{1 \plus{} \cos A_{3}},$ $ r \equal{} (s \minus{} a_{2}) \tan \frac {A_{2}}{2} \equal{} (s \minus{} a_{2}) \frac {\sin A_{2}}{1 \plus{} \cos A_{2}}$ and then $ \sin A_{1} \equal{} \frac {a_{1}}{a_{3}}\sin A_{3},\ \sin A_{1} \equal{} \frac {a_{1}}{a_{2}}\sin A_{2}$ in the numerator and denominator, respectively,

$ \frac {XS_{2}}{XS_{3}} \equal{} \frac {a_{3} \minus{} a_{1}}{a_{2} \minus{} a_{1}}\cdot \frac {a_{2}}{a_{3}}\cdot \frac {a_{3}(a_{2} \minus{} a_{1}) \cos A_{1} \plus{} a_{1}(a_{2} \plus{} a_{1}) (1 \minus{} \cos A_{3})}{a_{2}(a_{3} \minus{} a_{1}) \cos A_{1} \plus{} a_{1}(a_{3} \plus{} a_{1}) (1 \minus{} \cos A_{2})}$

Using $ a_{2} \equal{} a_{3}\cos A_{1} \plus{} a_{1}\cos A_{3},$ $ a_{3} \equal{} a_{2}\cos A_{1} \plus{} a_{1}\cos A_{2}$ in the numerator and denominator, respectively (that's how $ h_{2}, h_{3}$ altitudes divide $ a_{2}, a_{3}$),

$ \frac {XS_{2}}{XS_{3}} \equal{} \frac {a_{3} \minus{} a_{1}}{a_{2} \minus{} a_{1}}\cdot \frac {a_{2}}{a_{3}}\cdot \frac {a_{2}a_{3}\cos A_{1} \minus{} a_{1}a_{2}\cos A_{3} \plus{} a_{1}^{2}}{a_{3}a_{2}\cos A_{1} \minus{} a_{1}a_{3}\cos A_{2} \plus{} a_{1}^{2}}$

Substituting the appropriate cosine theorem formulas

$ \frac {XS_{2}}{XS_{3}} \equal{} \frac {a_{3} \minus{} a_{1}}{a_{2} \minus{} a_{1}}\cdot \frac {a_{2}}{a_{3}}\cdot \frac {a_{2}^{2} \plus{} a_{3}^{2} \minus{} a_{1}^{2} \minus{} a_{1}^{2} \minus{} a_{2}^{2} \plus{} a_{3}^{2} \plus{} 2a_{1}^{2}}{a_{2}^{2} \plus{} a_{3}^{2} \minus{} a_{1}^{2} \minus{} a_{3}^{2} \minus{} a_{1}^{2} \plus{} a_{2}^{2} \plus{} 2a_{1}^{2}}$

$ \boxed{\frac {XS_{2}}{XS_{3}} \equal{} \frac {a_{3} \minus{} a_{1}}{a_{2} \minus{} a_{1}}\cdot \frac {a_{3}}{a_{2}}}$

For $ a_{2} < a_{1} < a_{3},$ the result is negative, so that X lies inside the segment $ S_{1}S_{2}.$ Now we show that the line $ M_{1}S_{1}$ cuts the segment $ S_{2}S_{3}$ in the same ratio, which is much simpler. A circle $ (M_{1})$ with radius $ M_{1}T_{1} \equal{} \frac {a_{3} \minus{} a_{2}}{2}$ is perpendicular to both the the incircle $ (I)$ and excircle $ (E_{1})$ and the 9-point circle $ (N)$ passes through its center $ M_{1}.$ Inversion in $ (M_{1})$ carries $ (I), (E_{1})$ into themselves and $ (N)$ tangent to both $ (I), (E_{1})$ into the line $ D_{1}S_{1},$ the common internal tangent of the incircle $ (I)$ and the excircle $ (E_{1})$ other then the triangle side $ A_{2}A_{3}.$ $ D_{1}M_{1}$ is a tangent to $ (I)$ at $ S_{1},$ hence $ S_{1}$ is the image of the Feuerbach point F, the tangency point of $ (I),\ (N).$ Thus

$ (M_{1}F \minus{} S_{1}F) \cdot M_{1}F \equal{} M_{1}S_{1}\cdot M_{1}F \equal{} M_{1}T_{1}^{2} \equal{} \frac {(a_{3} \minus{} a_{2})^{2}}{4}$

Since F is the external homothety center of $ (I),\ (N),$ the central similarity of the triangle $ \triangle S_{1}S_{2}S_{3}$ inscribed in $ (I)$ to the medial triangle $ \triangle M_{1}M_{2}M_{3}$ inscribed in $ (N)$ follows and also the central similarity $ \triangle S_{1}S_{2}S_{3}\sim \triangle A_{1}A_{2}A_{3}$ used above. Consequently, we have $ \frac {2r}{R} \equal{} \frac {S_{1}F}{M_{1}F}.$ Calculating $ S_{1}F$ from these 2 equations yields

$ S_{1}F^{2} \equal{} \frac {(a_{3} \minus{} a_{2})^{2}}{4}\cdot \frac {4r^{2}}{R(R \minus{} 2r)} \equal{} (a_{3} \minus{} a_{2})^{2}\cdot \frac {r^{2}}{d^{2}}$

where d = IO. By cyclic exchange for $ a_{2} < a_{1} < a_{3},$

$ \boxed{S_{1}F \equal{} (a_{3} \minus{} a_{2}) \cdot \frac {r}{d},\ S_{2}F \equal{} (a_{3} \minus{} a_{1}) \cdot \frac {r}{d},\ S_{3}F \equal{} (a_{1} \minus{} a_{2}) \cdot \frac {r}{d}}$

and F is on the arc $ S_{2}S_{3}$ of $ (I)$ opposite to $ S_{1}.$ (The expression for $ S_{1}F$ does not follow cyclic exchange, so that the 1st barycentric coordinate of F in $ \triangle S_{1}S_{2}S_{3}$ is negative, the 2nd and 3rd are positive.) Let $ M_{1}S_{1}\equiv S_{1}F$ meet $ S_{2}S_{3}$ at Y (inside the segment $ S_{2}S_{3}$).

$ \angle YFS_{2}\equiv \angle S_{1}FS_{2} \equal{} \angle S_{1}S_{3}S_{2} \equal{} \angle A_{3}$

$ \angle YFS_{3}\equiv \angle S_{1}FS_{3} \equal{} \angle S_{1}S_{2}S_{3} \equal{} \angle A_{2}$

Areas of the triangles $ \triangle YFS_{2},\ \triangle YFS_{3}$ with the common side $ YF$ and baseline $ S_{2}S_{3}$ are in the ratio of their bases. Taking the direction $ YS_{2}$ positive and $ YS_{3}$ as negative,

$ \boxed{\frac {YS_{2}}{YS_{3}} \equal{} \minus{} \frac {\frac {1}{2}\ YT \cdot S_{2}F\ \sin A_{3}}{\frac {1}{2}\ YT \cdot S_{3}F\ \sin A_{2}} \equal{} \frac {a_{3} \minus{} a_{1}}{a_{2} \minus{} a_{1}}\cdot \frac {a_{3}}{a_{2}}}$

As a result, $ \frac {YS_{2}}{YS_{3}} \equal{} \frac {XS_{2}}{XS_{3}},$ which means that $ Y \equiv X$ are identical. Could somebody come up with a soultion using polars ?


If a new ``Supertrain'' is advertised to cut the travel time between two cities by $ 75\%$, and the time of the old train trip was $ 5$ hours, $ 20$ minutes, what will the Supertrain's time be? Express your answer in hours as a common fraction.

5 hours and 20 minutes is $ \frac{16}{3}$ hours.

If we cut the time of travel by 75%, we are now at 25%, or $ \frac{1}{4}$. the amount of time.

$ \frac{16}{3} \times \frac{1}{4} \equal{} \frac{4}{3}$ hours.


Let $a_1 = 25, a_2 = 48$ and for all $n\geq 1$, let $a_{n+2}$ be the remainder from dividing $a_n+a_{n+1}$ by $100$. Find the remainder from dividing $a_1^2+a_2^2+...+a_{2000}^2$ by $8$.


Find $ x,y\in R$ such that
$ log x \plus{} \frac {log(xy^8)}{(log x)^2 \plus{} (log y)^2} \equal{} 2$
$ log y \plus{} \frac {log(\frac {x^8}{y})}{(log x)^2 \plus{} (log y)^2} \equal{} 0$

[hide]Put $ a: = \log x, b: = \log y$ to get

\begin{eqnarray}a + {a + 8b\over a^2 + b^2} = 2 \\
b + {8a - b\over a^2 + b^2} = 0\end{eqnarray}

Easy checking shows that neither $ a = 0$ nor $ b = 0$ yields a solution, hence $ ab\neq 0$. Therefore we can multiply the first equation by $ b$, the second by $ a$ and add them up to get

$ 2ab + 8 = 2b\iff a = {b - 4\over b}\quad(*)$

Now $ (2)$ yields

\begin{eqnarray*} & & a^2b + b^3 + 8a - b = 0 \\
&\iff & {b^2 - 8b + 16\over b} + b^3 + {8b - 32\over b} - b = 0 \\
& \iff & b - 8 + {16\over b} + b^3 + 8 - {32\over b} - b = 0 \\
& \iff & b^3 - {16\over b} = 0 \\
& \iff & b^4 = 16 \\
&\iff & b = \pm 2\end{eqnarray*}

From $ (*)$ we find pairs $ (a,b)\in\{(3, - 2),( - 1,2)\}$. Checking shows that both satisfy $ (1)\land(2)$, hence the solutions to the initial system are $ (x,y)\in\{(1000,0.01),(0.1,100)\}$[/hide]


I found the following result :
If $a\leq x_1\leq...\leq x_n\leq b$ ,$c\leq y_1\leq...\leq y_n\leq d$, $z_1,...,z_n$ a permutation of $y_1,...,y_n$ and $F: (a,b)\times(c,d) \longrightarrow \mathbb{R}$ such that $\frac{\partial^2}{\partial x\partial y}F\geq 0$ (or $\frac{\partial}{\partial x}F$ is incresing on y)then:
$\sum_{i=1}^n F(x_i,y_i)\geq \sum_{i=1}^n F(x_i,z_i)\geq \sum_{i=1}^n F(x_i,y_{n-i})$
I will post the proof in the case $n=2$.
In fact, we have to prove that $F(x_1,y_1)+F(x_2,y_2)\geq F(x_1,y_2)+F(x_2,y_1)$ which is equivalent to: $F(x_1,y_1)-F(x_1,y_2)\geq F(x_2,y_1)-F(x_2,y_2)$. Now, let $f(x)=F(x,y_1)-F(x,y_2)$, $f'(x)=\frac{\partial}{\partial x}F(x,y_1)-\frac{\partial}{\partial x}F(x,y_2)\leq0$, so $f(x_1)\leq f(x_2)$.
I think this inequality is useful, if we take $F(x,y)=xy$ we have Chebyshev, and if $F(x,y)=-ln(x+y)$ we have :[url=http://www.mathlinks.ro/Forum/viewtopic.php?t=17280]here[/url]

Have you found any other applications of this generalization?

ps: do the numbers need to be bounded?


Let $ m,n$ be given positive natural numbers.
(a)
Find the number of ordered tuples of positive natural numbers $ (a_1,a_2,...,a_n)$ such that $ 1 \leq k \leq n: a_k \mid m$ and for distinct $ i,j$ we have $ \gcd(a_i,a_j)\equal{}1$.
(b)
Generalize...

Let $m=p_1^{d_1} p_2^{d_2} \cdots p_s^{d_s}$ be the prime decomposition of $m$. Then the answer to (a) is:
\[\prod_{j=1}^s (1+n d_j) = \tau(m^n).\]


Stanford college Duke Harvard Princeton Putnam

I got accepted into both (rejected by Harvard and Princeton), and I'm having a hard time choosing between the two... I've heard that Duke has an amazing undergraduate math program (they do really well on Putnam, too), but I also know that Duke's graduate math courses aren't so hot.

On the other hand, I don't really see a strong undergraduate math department at Stanford, but their graduate math seems pretty decent.

Any advice?

Thanks!

Hey jay how are you doing?

I'm sorry that I cant help you that much but I wish you best of luck.
wherever you go, you'll do fine.

In my opinion, strength of math department is surely important, but since you'll most likely live there for four years, the atmosphere of each school is also important.
as far as I know, they are about the same size and located in suburban area.

have you visited both? maybe one might click your mind?

Congratulations. Do you expect to be doing graduate-level courses while you are still completing your undergraduate degree? That may answer your question right there.

My general advice to people who have more than one college to choose from is to go as far from their home region as they can. You are from the east, as I recall, so Stanford would involve more learning about another region for you. (But on the contrary, maybe ethnically Chinese people, as I think you also are, don't need any more familiarity with California.   :P  ) Indeed you can hardly go wrong either way. But my selection factor in such a case would be to go where you will have the most incidental learning experiences from doing the unfamiliar OUTSIDE of math class.

Good luck in making that fine decision you have before you.

Thanks all.  Hi Geehoon.

Yeah, hopefully I'll be able to take a few graduate courses while I'm there.  I just wasn't sure how the undergrad math and grad math actually fit together.

Thanks for all your input, I really appreciate it.


What is the combinatorial interpretation and proof of following identity?

$\prod_{n=0}^{\infty} (1-q^{2n+2} ) (1+q^{2n+1} )^2 = \sum_{n=-{\infty}}^{\infty} q^{n^2}$

Hmm.. I think it can be solved if we think of "number of PARTITIONS into ak+b parts"

I think I got a "combinatorial interpretation" :

Define

$E_{e} (n)$ is Number of combinations such that sum of selected numbers is n, and restriction for selection is "[b]choose even number even times[/b], and same odd number is allowed fot 2 times and not allowed for even numbers.

$U_{o} (n)$ is Number of combinations such that sum of selected numbers is n, and restriction for selection is "[b]choose even number odd times[/b], and same odd number is allowed fot 2 times and not allowed for even numbers.

Then,
$E_{e} (n) = U_{o} (n)$ unless $n=k^2$
$E_{e} (n) = U_{o} (n) + 2$ if $n=k^2$ , $k$ is not zero.


Please everyone have a look at a proposed proof of Riemann Hypothesis I found in the web.Please provide some feedback and forward it to anyone interested..

http://www.aronpalmer.com/riemann_proof.htm

Some remarks:

Not sure you need to know much about the area, the last bit about the zeta function being expressable in an even positive decreasing
function is pretty definitely correct, and the first bit about not being able to have zeros for s> 0 is pretty definately correct, the bit I am worried
about is going from the one sided transform to the two sided transform and being able to show that Odd(s) + Even(s) is a convergent polynomial.

Thanks,
Nick

greeting ! this is the link that help you to find what you want exactly , pleas see B.gunes topic , he used even and odd eta

Dirichlet series , " three approach to show Riemann hypothesis"

http://empslocal.ex.ac.uk/people/staff/mrwatkin//zeta/RHproofs.htm

g.luck


algebra polynomial function algorithm domain

Hey, Im a new guy at the forum, pleased to meet everyone here. I was dealing with a hard olympiad problem for which i have not been able to find a solution for, id appriciate some help, here goes:

If p(x) and q(x) are monic polynomials that are irredeucible  over the rationals, and have roots a, and b, respectively, such that a+b is rational, prove that the polynomial  p(x)^2 - q(x)^2 also has a rational root.

Thanks

Let r=a+b, then r/2 is a rational root of p^2(x)-q^2(x).

Spoiler:

[hide]Let f(x)=p(r-x), then f(b)=p(r-b)=p(a)=0, so q(x) | f(x)=p(r-x) since q is the irreducible polynomial of b. Similar p(x) | q(r-x). This implies that  f(x)=p(r-x) | q(r - (r-x)) = q(x). Therefore q(x) =  :pm: f(x) =  :pm: p(r-x) since the leading coefficient of f(x) is  :pm: 1. So p^(r/2) - q^2(r/2) = 0.

 [/hide]

By the way , the ComplexZeta function is much more interesting in mathematics

[quote="JBL"][quote="fuzzylogic"]so q(x) | f(x)=p(r-x) since q is the irreducible polynomial of b.[/quote]

Can you explain this better, please?[/quote]

It can be proved using Euclidean Divison Algorithm that if [tex]g(\alpha)=0[/tex] for some nonzero [tex]g\in\mathbb{Q}[x][/tex], then the set [tex]\{f(x) | f(\alpha)=0, f\in\mathbb{Q}[x]\}=\{h(x)p(x) | h\in\mathbb{Q}[x]\}[/tex], where [tex]p(x)\in\mathbb{Q}[x][/tex] is a unique monic irreducible polynomial over [tex]\mathbb{Q}[/tex]. Furthermore, [tex]p(x)[/tex] is a polynomial of minimal degree >=1 having [tex]\alpha[/tex] as a zero.

( I am trying to avoid the standard language of abstract algebra, algebraic over a field, principle ideal domain, etc.)

I don't see why using abstract algebra should be a problem on the advanced board.


no the results will be out only next year  :maybe:

[quote="karthik0112358"]Have any of recieved a letter tellin u that u have got thru physics or chem olympiad 2007-08.
I wamted to know if the results are out??[/quote]

You will see the results online first.

when??? :D and whats the selection cut off??? :blush:

Official Announcement. Results will be out TOMORROW! All the best everybody.

http://www.hbcse.tifr.res.in/olympiads/Data/ObjectType/e/exam_dates/viewObjectType

You have posted the wrong link. The real link is one link away  :P
Btw we have school today you know. Marks out today  :maybe:

AaaaH!!!! so who's so sure of clearing the olympiad ???? :maybe:  Pardesi i am sure you will clear even the second round with ease as you are a god in physics to get marks like that in AITS :o  so you are one sure shot what abt others??? :ninja:

@srinitrs
thanks but i have not done well in NSEP.Expecting around 128 in the objective ,My subjective has been very very miserable just one is perfectly right, I couldn't  adjust myself with the time for the subjective, so lots of silly mistakes.
NSEC god knows how i have done ...really fussed up the Biological  part didn't have time even to look at them :( .Forgot how i answered..no rough work
NSEA was simple but quite information based so lots of mistakes in those parts and silly mistakes in maths related portion...expecting around 134,not sure if that would be enough though.

Anyone else wants to come out with last minute review  :)

[quote="pardesi"]@srinitrs
thanks but i have not done well in NSEP.Expecting around 128 in the objective ,My subjective has been very very miserable just one is perfectly right, I couldn't  adjust myself with the time for the subjective, so lots of silly mistakes.
NSEC god knows how i have done ...really fussed up the Biological  part didn't have time even to look at them :( .Forgot how i answered..no rough work
NSEA was simple but quite information based so lots of mistakes in those parts and silly mistakes in maths related portion...expecting around 134,not sure if that would be enough though.
[/quote]

You wrote NSEA?? Our stupid school never told us. Anyway, I'm keeping my fingers crossed.

@Pardesi: We have a lot in common. I'm getting exactly your mark in objective give or take 4. Subj, am getting around 2 but am not really sure about my clarity of expression and also about how steps are awarded.

shreyas i attended 4 subjectives and 3 of them got miserably wrong somwhere or the other.I just had 15 mins for the subjective  :(
I was too tired while giving NSEA after some 4 and half hours and the questions were even more terrible

The results would be announced day after tommorow that is 4th of January,2008(Friday) by 12 noon

Results out only on 7th January ( Monday ) 12 noon

Maybe they are changing some answers after all  :P  :D

How so inconvenient, I am sitting online instead of studying for CSC and results aren't cropping up...:(

and i am sitting here and not studying English(Shreyas,madness..i will go to school today  :D )  :P
and the results aren't out

big deal not studing english
we have to nv about 50 pages of notes in computers for our exam today  :mad:

what the hell its just past 12 and the results are not yet out  :mad:

When will the bloody results come out?? Useless people..
Its well past 12..

Dei what happened after you and Shreyas were expelled from the exam  :rotfl:
Is Shreyas with you now?
What did you guys do?

We were studyin english!! :P  Gotta rite compsci on wed...

Nothing much really CSC on wednesday. it was quite fun actually...

WTH I rushed home with fear and trepidation for this:((((

When will the results come?

When did you go home?
Waiting for the results
Even Ranga rushed home  :rotfl:

And by the way do u think it is sufficient to refresh that page every time?
Or should we start afresh?

I am highly irritated. I am in no mood to wait any longer. Idiots postpopned it and now not posting. :furious:  :furious:  :bomb:

LOL The whole bunch is online. Nice Smiley da rock! :bomb:  :oops_sign:

I am not sure if refreshing alone is enough. Depends on browsers and their caching... It never works for me, so I always Clear the full cache before checking again.

And can we click on a link that has been posted before?
The fools :fool:

Fool is too good a word to use..  :wallbash_red:

The curiosity is killing me..:D  :play_ball:

Good die  :rotfl:   :D
One competition less for IIT JEE  :D

Since results are out i am locking this  :D


function
algebra
domain
trigonometry
functional equation
algebra unsolved

Find the all $f$:
$f(x+y)=\frac{f(x)+f(y)}{1-f(x)f(y)}$

*(My NOTE: This function can be seen as $f(x)=tanx$ but showing the formal proof i could not)

[hide="My usual rantings again"]
1) When someone gives a a functional equation to solve, it would be useful to say the domain of the wanted equations. For instance: Find all functions $f: \mathbb{R}\to\mathbb{R}$ with ... (It was just an example. In this case, $f(x)=tan(x)$ is not a $f: \mathbb{R}\to\mathbb{R}$, because it's not defined for every real.)
Why am I asking this? Notice that you can't take special cases $x=0$, or $x=1$ if you don't know if $0$ or $1$ belong to the domain of $f$. Where can we begin, then?

2)It would be useful too to say if you only want continuous or differentiable solutions. Certain functional equations, like Cauchy's equation, can be easily solved for continuous functions, but more sophisticated math is needed to solve them generally (more than olympiad level, anyway)[/hide]
Anyway, following your hint, we can write $f(x)=tan(g(x))$. We can do it because the range of the function $\tan{x}$ is $\mathbb{R}$.
Substituting, $g(x+y)=g(x)+g(y)$, so $g$ must be any solution (continuous or not) to Cauchy's equation (the only continuous ones are $g(x)=Cx$).

[quote="abdurashidjon"]So its enough to say  that without proving $f(x)=tanx$[/quote]
No!
But thanks to your idea, the "change of variable" $f(x)=\tan (g(x))$ helps us finding all the solutions:
$f(x)$ is a solution iff $g(x)$ is a solution to Cauchy's equation.
All the continuous solutions (where the $\tan$ is continuous) are $f(x)=\tan (Cx)$.


[quote="AIME15"]Bribed Mr. Rusczyk.[/quote]

It all makes sense now.

[quote="pianoforte"]AIME15: That was rude.[/quote]

But true.

Wait a minute truffles you are a rl missourian?

:rotfl: Are you being sarcastic mewto?

And I'm a girl and yeah my location is correct.

Dude what really that is pretty uber I did not know that; where in Missouri?

Dude mewto are you serious?  You know what my previous username was right?

Failz also I remember you telling me you changed your username to truffles lol i failz.

suma_milli. Wait, I'm from Washington!! :D


geometry
circumcircle
projective geometry
geometry unsolved

Let $ABC$ be an acute triangle with $AB\neq AC,$ and let $D$ be the foot of perpendicular from $A$ to line $BC.$ Point $P$ is on altitude $AD.$ Rays $BP$ and $CP$ meet sides $AC$ and $AB$ at $E$ and $F,$ respectively. If $BFEC$ is cyclic, prove that $P$ is the orthocenter of triangle $ABC.$

Follows Ceva's theorem:
$\frac{FA}{FB}. \frac{EC}{EA}. \frac{DB}{DC}= 1 \\ \Rightarrow \frac{FA.EC}{FB.EA}= \frac{DC}{DB}\\ \Rightarrow \frac{FA.BA.EC}{FB.EA.CA}= \frac{DC}{DB}.\frac{AB}{AC}\\ \Rightarrow \frac{EC}{FB}= \frac{tanB}{tanC}. \frac{sinC}{sin B}= \frac{cosC}{cosB}\ \ \text{(Note that }FA.BA = EA.CA) \\ \text{But we have: }\frac{EC}{FB}= \frac{sin \angle EBC}{sin \angle BCF}= \frac{sin \angle BEC. cos C+sin C. cos \angle BEC}{sin \angle BFC. cos B+sin B. cos \angle BFC}\\ \text{So }\frac{sin \angle BEC. cos C+sin C. cos \angle BEC}{sin \angle BFC. cos B+sin B. cos \angle BFC}= \frac{cosC}{cosB}= \frac{sin \angle BEC. cos C}{sin \angle BFC. cos B}=\frac{sin C. cos \angle BEC}{sin B. cos \angle BFC}$
If $cos \angle BEC = cos \angle BFC \neq 0$ we have $\frac{sinB}{sinC}= \frac{cosB}{cosC}\Rightarrow tan B = tan C \Rightarrow B=C$, which is impossible. So $cos \angle BEC = cos \angle BFC= 0$, it means that $BE \perp AC, BF \perp AB$, so $P$ is the orthocenter.

Because of the point $P$ lies on $AD,$ we conclude that the segment line $EF,$ passes through the point $D',$ as the harmonic conjugate of $D,$ with respect to $B,$ $C$ $($ From the complete quadrilateral $AEPFBC$ $).$

Because of the quadrilateral $BFEC$ is cyclic, we conclude that $EF$ is antiparallel of $BC,$ with respect to the angle $\angle A$ and so, we have that $OA\perp EF,$ where $O,$ is the circumcenter of $\bigtriangleup ABC.$

We denote as $E',$ $F',$ the traces on $AC,$ $AB$ respectively, of the altitudes through $B,$ $C$ and as $P',$ the orthocenter of $\bigtriangleup ABC,$ lies on $AD.$

It is easy to show that $E'F'\equiv EF,$ because of the segment line $E'F'$ passes through the point $D'$ and also is perpendicular to $OA.$

Hence we conclude that $P'\equiv P$ and the proof is completed.

Kostas vittas.


linear algebra
matrix
algebra
polynomial
superior algebra
superior algebra unsolved

what is the interior in Mn(C) of the diagonalizable matrices ? and in Mn(R)

In the first case (when we deal with complex matrices), I believe the interior consists of those diagonalizable matrices which have distinct eigenvalues. Let $\mathcal D$ be the set of complex diagonalizable matrices.

We may as well assume that we're working with diagonal matrices, and if there are two equal eigenvalues, then the first two diagonal entries are equal.

If the matrix has distinct eigenvalues, then all the matrices in a sufficiently small ball around our matrix have distinct eigenvalues, because the characteristic polynomial, which is $\det(xI_n-A)$, depends continuously on $A$.

On the other hand, if the matrix $A$ has the first two diagonal entries equal, then we can take $A_k\to A$, where $A_k$ is exactly the same as $A$, except that it has a $\frac 1k$ in position $(2,1)$. It's easy to see that this sort of matrix can be brought through a similarity to a non-diagonal Jordan form (which is the same as $A_k$, except that instead of that $\frac 1k$ it has a $1$), so it's not diagonalizable. This means that $A$ is a limit point of the complement of $\mathcal D$.

nice.  :)
and for $Mn_(R)$ ? It is a little harder.

Should the answer be the same? That's what I'm getting.

For diagonaliuzable matrices with two equal eigenvalues, we can do the exact same thing as above to prove that they are not interior points. As for the ones with distinct eigenvalues, the matrices in a ball of small enough radius centered in such a matrix will have distinct eigenvalues. The problem is that they might be complex. However, they are also roots of a real polynomial, so if there are any complex eigenvalues, they come in pairs of conjugates. If the ball is small enough, the real parts of the eigenvalues of the matrices inside the ball will also be distinct, because this is the case for our matrix, so these eigenvalues must be real.

Is it Ok? :?


a,b and c are three positive real numbers , prove this following inequalities :
$ (\frac{b}{a\plus{}b})^4\plus{}(\frac{c}{b\plus{}c})^4\plus{}(\frac{a}{a\plus{}c})^4 \geq \frac{3}{16}$

[hide="Hidden by mod"]Let a/b=x,b/c=y and c/a=z so the inequality turns out to be: 1/(1+x)^4+1/(1+y)^4+1/(1+z)^4>=3/16(xyz=1;x,y,z>0)
It is just the easier case of the Chinese inequality 2004 :)[/hide]


trigonometry
limit
real analysis
real analysis unsolved

Define $ f(z) \equal{} \frac {\cot(z)\, \coth(z)}{z^3}$ and the contour path $ \gamma_n$

as the square with the corners $ (\pm 1\pm i) \left(n \plus{} \frac12\right) \, \pi$ .

$ f(z) \equal{} \frac {1}{z^3} \, \frac {\cos z}{\sin z} \, \frac {\cosh z}{\sinh z} \equal{} \frac {1}{z^3}\,\, \frac {1 \minus{} \frac {z^2}{2} \plus{} \frac {z^4}{24} \minus{} ...}{z \minus{} \frac {z^3}{6} \plus{} \frac {z^5}{120} \minus{} ...}\,\,\, \frac {1 \plus{} \frac {z^2}{2} \plus{} \frac {z^4}{24} \plus{} ...}{z \plus{} \frac {z^3}{6} \plus{} \frac {z^5}{120} \plus{} ...}$

$ \equal{} \frac {1}{z^5}\, \frac {1 \minus{} \frac {z^4}{6} \plus{} O(z^8)}{1 \minus{} \frac {z^4}{90} \plus{} O(z^8)} \equal{} \frac {1}{z^5}\, \left(1 \minus{} \frac {z^4}{6} \plus{} O(z^8)\right)\, \left(1 \plus{} \frac {z^4}{90} \plus{} O(z^8)\right)$

$ \equal{} \frac {1}{z^5}\, \left(1 \minus{} \frac {7}{45}z^4 \plus{} O(z^8)\right) \, \Rightarrow \, \text{Res} f\big|_{z \equal{} 0} \equal{} \minus{} \frac {7}{45}$ .

$ \text{Res} f\big|_{z \equal{} k \pi} \equal{} \lim_{z\to k \pi} \frac {z \minus{} k\pi}{\sin z}\, \frac {\cos z\, \coth z}{z^3} \equal{} \frac {\coth(k\pi)}{(k\pi)^3}$

$ \text{Res} f\big|_{z \equal{} i k \pi} \equal{} \lim_{z\to i k \pi} \frac {z \minus{} i k\pi}{\sinh z}\, \frac {\cot z\, \cosh z}{z^3} \equal{} \frac {\coth(k\pi)}{(k\pi)^3}$

$ \sin z \equal{} \sin(x \plus{} iy) \equal{} \sin(x)\, \cosh(y) \plus{} i\, \cos(x)\, \sinh(y)$

$ \Rightarrow |\sin z|^2 \equal{} \sin^2(x)\, \underbrace{\cosh^2(y)}_{1\plus{}\sinh^2(y)} \plus{} \cos^2(x)\, \sinh^2(y) \equal{} \sin^2(x) \plus{} \sinh^2(y)$

which is at least 1 if $ x$ or $ y$ equals $ \pm \left(n \plus{} \frac12\right) \, \pi$ .

For all values $ z$ on the contour path $ \left|\cot z\right|^2\le 1 \plus{} 1$ , since $ \cot^2 z \equal{} \frac {1}{\sin^2 z} \minus{} 1$ .

Because of $ iz\in \gamma_n$ for all $ z\in \gamma_n \quad \left|\coth z\right|^2 \equal{} \left|\cot (iz)\right|^2$ is also $ \le 2$ .

If follows $ \left|f(z)\right|\le \frac {2}{|z|^3}$ and therefore $ \left|\oint_{\gamma_n} f\, dz\right|\le 4 \, (2n \plus{} 1)\, \frac {2}{\left(n \plus{} \frac12\right)^3}\to 0$ .

So the sum of all residues must be 0.

$ 4\sum_{k \equal{} 1}^\infty \frac {\coth(k\pi)}{(k\pi)^3} \minus{} \frac {7}{45} \equal{} 0 \, \Rightarrow \, \sum_{k \equal{} 1}^\infty \frac {\coth(k\pi)}{k^3} \equal{} \frac {7\pi^3}{180}$

$ \frac {2}{e^{2\pi k} \minus{} 1} \equal{} \frac {2 \, e^{ \minus{} k\pi}}{e^{k\pi} \minus{} e^{ \minus{} k\pi}} \equal{} \frac {e^{k\pi} \plus{} e^{ \minus{} k\pi}}{e^{k\pi} \minus{} e^{ \minus{} k\pi}} \minus{} 1$

$ \Rightarrow \, \sum_{k \equal{} 1}^\infty \frac {2}{k^3\, \left(e^{2\pi k} \minus{} 1\right)} \equal{} \sum_{k \equal{} 1}^\infty \frac {\cosh(k\pi)}{k^3} \minus{} \sum_{k \equal{} 1}^\infty \frac {1}{k^3}$


geometry
incenter
geometric transformation
reflection
circumcircle
symmetry
perpendicular bisector

[quote="shoki"]$ (a)$ Let $ \Delta ABC$ be a triangle and $ I,$ $ O$ be its incenter and circumcenter. Let $ l_a$ be a line which is perpendicular to $ AI$ at $ I$ and denote its intersection with $ BC$ by $ A_1$. The same definition holds for $ B_1,$ $ C_1$. Let $ A_2$ be the reflection of $ A_1$ wrt line $ OI$. Defining similarly $ B_2,$ $ C_2$. Prove that $ AA_2,$ $ BB_2,$ $ CC_2$ are concurrent.[/quote]
Here is my solution.

First, let us prove that $ A_1,$ $ B_1$ and $ C_1$ are collinear $ (*)$. In fact, this collinearity can be generalized, which is for any point $ I'$ on the plane of $ \triangle ABC$, let $ A',$ $ B'$ and $ C'$ respectively be the intersections of the  perpendicular lines with $ AI',$ $ BI',$ $ CI'$ passing through $ I'$ and $ BC,$ $ CA,$ $ AB$ respectively. Then, $ A',$ $ B'$ and $ C'$ are collinear. Indeed, consider the inversion through pole $ I'$, with any power $ k'$. Let $ A'',$ $ B'',$ $ C''$ and $ A'_1,B'_1,C'_1$ respectively be the image of $ A,B,C$ and $ A',$ $ B',$ $ C'$ through $ \mathcal {I}(I',k')$. Therefore, we have $ \mathcal {I} (I',k')$ maps $ \overline {ABC'}$ into $ (A''B''C'_1I')$, $ \overline {ACB'}$ is taken into $ (A''C''B'_1I')$ and $ \overline {BCA'}$ is mapped into $ (B''C''A'_1I')$. Then $ I'C''$ $ \perp$ $ I'C'_1$, $ I'B''$ $ \perp$ $ I'B'_1$ and $ I'A''$ $ \perp$ $ I'A'_1$. Let $ O_1,$ $ O_2,$ $ O_3$ respectively be the circumcenters of $ (A''C''B'_1I')$, $ (B''C''A'_1I')$ and $ (A''B''C'_1I')$. Since $ O_2O_3\perp I'B''$, $ I'B'_1\perp I'B''$, thus $ O_2O_3\| I'B'_1$. Therefore, the perpendicular bisector of $ I'B'_1$ will be the line $ d_1$ passes $ O_1$ and $ d_1$ $ \perp$ $ O_2O_3$. With the same argument holds for $ O_1O_2$ and $ O_1O_3$. As the consequence, $ d_1,d_2,d_3$ repsectively will be the altitudes from $ O_1,$ $ O_2,$ $ O_3$ wrt $ \triangle O_1O_2O_3$. Obviously, they are concurrent at the orthocenter of $ \triangle O_1O_2O_3$, which implies that $ I',$ $ B'_1,$ $ A'_1,$ $ C'_1$ are concyclic. Therefore, $ A',$ $ B',$ $ C'$ are collinear. Just let $ I$ $ \equiv$ $ I'$, we shall get the result $ (*)$.

Now, since $ A_2,$ $ B_2,$ $ C_2$ respectively are the reflections of $ A_1,$ $ B_1,$ $ C_1$ with respect to $ OI$, thus, $ A_2,$ $ B_2,$ $ C_2$ are also collinear. Let us denote $ f_{OI}$ by the the reflection through $ OI$. Then if $ f_{OI}(A) \equal{} A'$, $ f_{OI}(B) \equal{} B'$ and $ f_{OI}(C) \equal{} C'$, hence, $ f_{OI}(\triangle ABC) \equal{} \triangle A'B'C'$ and $ (I)$, $ (O)$ respectively will be the incircle and circumcircle of $ \triangle A'B'C'$. Therefore, $ A_2,$ $ B_2$ and $ C_2$ now do the same positions with respect to $ \triangle A'B'C'$ as $ A_1,$ $ B_1,$ $ C_1$ did with their triangle $ \triangle ABC$. Let $ D,$ $ E,$ $ F$ respectively be the tangency points of $ (I)$ with $ BC$, $ CA,$ $ AB$ and $ D',$ $ E',$ $ F'$ are the image of $ D,$ $ E,$ $ F$, respectively under $ f_{OI}$, then, $ D',$ $ E',$ $ F'$ will be the tangency points of $ B'C',$ $ C'A',$ $ A'B'$ with $ (I)$. Consider the inversion through pole $ I$, power $ k \equal{} r^2$, where $ r$ is the radii of $ (I)$. Then $ \mathcal {I}(I,k)$ maps $ A_2$ into $ I_1$, then $ I_1$ can be redefined as the intersection of the altitude $ D'D'_1$ of $ \triangle D'E'F'$ and the line passes through $ I$ and parallel with $ E'F'$. The same argument holds for $ I_2,$ $ I_3$. Moreover, $ \mathcal {I}(I,k)$ also maps $ A$ into $ M$, which is the midpoint of $ EF$, and hence by the property of $ f_{OI}$, then, $ M$ is also the image of the midpoint $ M'$ of $ E'F'$ wrt $ f_{OI}$. Argue the same for $ N$ and $ P$. Therefore, $ (IMI_1)$ $ \equiv$ $ (\mathcal {C}_1)$ is the image of $ AA_2$. Define the same for $ (\mathcal {C}_2)$ and $ (\mathcal {C}_3)$. We claim that these circle go through the [i]Anti-steiner[/i] point $ W$ of $ \triangle M'N'P'$ $ (**)$.

Indeed, let $ I_e$ be the circumcenter of $ \triangle M'N'P'$ and $ O_e$ is the center of the $ 9 \minus{}$ point circle with respect to $ \triangle M'N'P'$. Let $ O_1$, $ O'_1$ respectively are the reflections of $ I,$ $ I_e$ with respect to $ N'P'$. Note that $ I$ is the orthocenter of $ \triangle M'N'P'$, then, $ O_1$ $ \in$ $ (I_e)$ and $ W$ $ \in$ $ O_1O'_1$. Let $ M''$ be the symmetry point of $ M'$ with respect to $ I_e$. Then it is known that $ (I_eM''O'_1)$ goes through $ W$ and $ O'_1,$ $ M$ and $ M''$ are collinear. Then $ \angle (MW,MM')$ $ \equal{}$ $ \angle (M''W,M''I_e)$ $ \equal{} \angle (O'_1O_1,O'_1I_e)$ $ \equal{} \angle (IM',II_e)$, which implies $ MW$ $ \perp$ $ M'I$, yielding that $ MW$ $ \|$ $ N'P'$ $ \equiv$ $ II_1$. But it is easy to note that $ I_1$ is the reflection of $ I$ with respect to $ I_eO'_1$ and obviously, $ W$ is the reflection of $ M$ with respect to $ I_eO'_1$, which yields that $ I,$ $ W,$ $ M,$ $ I_1$ are concyclic, or we can say $ W$ $ \in$ $ (\mathcal {C}_1)$. Argue the same for $ (\mathcal {C}_2)$ and $ (\mathcal {C}_3)$. We obtain the result $ (**)$.

As the consequence, $ AA_2$, $ BB_2,$ $ CC_2$ will go through a common point $ W'$ $ \in$ $ (O)$, where $ W'$ is the image of $ W$ under the $ \mathcal {I}(I,k)$. Our proof is completed then. $ \square$

[hide="For Shoki"]May I ask, where did you find such this nice problem? Did you come it out with yourself? And I am also very curious with an "easy" solution that you offered for your nice problem.

Best regard,
mathVNpro
[/hide]

[quote="shoki"]a)let $ \Delta ABC$ be a triangle and $ I,O$ be its incenter and circumcenter.
denote by $ D,E,F$ the touch points of the incircle with the sides $ BC,CA,AB$ respectively.
let $ l_a$ be a line which is perpendicular to $ AI$ at $ I$ and denote its intersection with $ BC$ by $ A'$.
define similarly $ B',C'$.
let $ A_1$ be the reflection of $ A'$ wrt line $ OI$.
define similarly $ B_1,C_1$.
prove the concurrency of $ AA_1,BB_1,CC_1$.
b)let $ \Delta ABC$ be a triangle.line $ l$ intersects $ BC,CA,AB$ at $ A',B',C'$.
let $ H$ be a point on the line $ l$.
denote by $ A_1,B_1,C_1$ the reflection of $ A',B',C'$ wrt point $ H$ respectively.
prove the concurrency of $ AA_1,BB_1,CC_1$.[/quote]

I will  first prove b , then use b to prove a  :oops:

[b]b/ [/b]
[img]http://img42.imageshack.us/img42/9053/hinh2.png[/img]

Let the intersection of $ BB_2$ and $ CC_2$ is $ K$ ; $ AK \cap \overline{A_1C_1B_1} \equiv A'_2$

We will prove that  $ \boxed{A'_2 \equiv A_2}$

Use the [b]Menelaus's theorem[/b] for the secant $ \overline{AKA'_2}$ :

+ And $ \bigtriangleup BC_1B_2$ : $ \frac {AB}{AC_1}.\frac {A'_2C_1}{A'_2B_2}.\frac {KB_2}{KB} \equal{} 1$

+ And $ \bigtriangleup CB_1C_2$ :  $ \frac {AB_1}{AC}.\frac {A'_2C_2}{A'_2B_1}.\frac {KC}{KC_2} \equal{} 1$

[b]Multiply[/b] two equations we got : $ \frac {A'_2C_1}{A'_2B_1}.\frac {A'_2C_2}{A'_2B_2}.\frac {A'B}{A'C}.\frac {A'B_1}{A'C_1}.\frac {KC}{KB}.\frac {KB_2}{KC_2} \equal{} 1$

Similarly with secant $ \overline{BKB_2}$ and secant $ \overline{CKC_2}$ . [b]Multiply[/b] three equations with note that :

$ \prod_{cyc \ A,B,C} \frac {A'_2C_2}{A'_2B_2}.\frac {AB}{AC}.\frac {AB_1}{AC_1}.\frac {KC}{KB}.\frac {KB_2}{KC_2} \equal{} 1$

$ \Longrightarrow \frac {A'_2C_1}{A'_2B_1}.\frac {B_2A_1}{B_2C_1}.\frac {C_2B_1}{C_2A_1} \equal{} 1$

Now we use the problem's data :  $ B_2C_1 \equal{} C_2B_1$

$ \Longrightarrow \frac {A'_2C_1}{A'_2B_1} \equal{} \frac {A_1C_2}{A_1B_2}$

$ \Longrightarrow \boxed{A'_2 \equiv A_2}$
[color=blue]
Our proof completed[/color]

[b]a/[/b]
[img]http://img145.imageshack.us/img145/243/hinh11.png[/img]

It's easy to see that :$ \widehat{A_1IB} \equal{} \frac {\widehat{C}}{2}$ and $ \widehat{A_1IC} \equal{} 180^o \minus{} \frac {\widehat{B}}{2}$

It mean's : $ \frac {A_1B}{A_1C} \equal{} \frac {BI}{CI}.\frac {sin \frac {\widehat{C}}{2}}{sin \frac {\widehat{B}}{2}}$

Use the [b]Menelaus's theorem[/b] we implie that $ A_1$ ; $ B_1$ and $ C_1$ are collinear .

Now, we shall prove that $ \boxed{OI \perp \overline{A_1B_1C_1}}$

Note that the polar line of $ A_1$ wrt $ (I)$ is perpendicular to $ AI$ and go through $ D \Longrightarrow$ this line is paralell to $ EF$ and go through $ D$

$ \Longrightarrow$  the polar line of $ A_1$ wrt $ (I)$ is the altitude from $ D$ of $ \bigtriangleup DEF$

It's means the POLE of $ \overline{A_1B_1C_1}$ wrt $ (I)$is orthorcenter of $ \bigtriangleup DEF$ which is collinear with $ I$, $ O$

$ \Longrightarrow \boxed{OI \perp \overline{A_1B_1C_1}}$

[color=blue]Use [b]b[/b] with $ H$ is the intersection of $ \overline{A_1B_1C_1}$ and $ OI$ we end the proof  :P[/color]

[quote="shoki"]@mathVNpro:
well, i don't know the source i just found it in an olympiad book.and it didn't mention the source unfortunately.
but to prove it we first prove by pole and polar that $ OI$ is in fact perpendicular to $ A'B'C'$
(as hophinhan did) also we can prove the collinearity by Menelaus theorem easily like [url=http://www.mathlinks.ro/viewtopic.php?t=313257]here[/url]
then the problem is just a special case of part b) which can be proven just using céva's theorem(sinus form).
i think it was easy because it uses just a collinearity (and a well-known one $ O,H_{\Delta DEF},I$) and some other lemmas which are well-known.
Best regards,
shoki.[/quote]
Dear Shoki,

I think this problem can be more harder than it is in its present version. I found out that, as in the process of proving this nice problem, the concurrent point is also a point lies on the circumcircle $ (O)$ of $ \triangle ABC$ (or you can see that directly in my proof above), and such proof like Menelaus alone is not enough to kill it. I think it would be nice if you please try to prove the property I had told you above and see whether I am right.  :)

Best regard,
mathVNpro

[hide="P/s"]Sorry if it is still easy for you  :D  ![/hide]

the second part of problem is Balakie lemma and its interesting that if $ H$ be the projectoin of $ O$ on the line $ L$ then the point of conurrency lies on the circle $ (O)$.

Nice MJ GEO . Could you plz give me some links to find more knowledge about this lemma

I searched on google but ...  :(

im sorry because i dont know any good link to give you.

I' ll prove the lemma mentioned.
[color=red] Lemma:
$ \Delta ABC$ is a triangle with circumcircle $ (O)$ and $ l$ is a line which intersects $ AB,AC$ at $ L,G$. Let $ H$ be the projection of $ O$ on $ l$. Let $ J,K$ be the reflection of $ G,L$ wrt point $ H$ respectively.
prove that $ BJ,CK$ intersect each other at a point $ N$ which lies on $ (O)$.[/color]
Let $ BH \cap (O)\equal{}B'$.similarly for $ C'$.
because of butterfly theorem and the fact that $ H$ is the midpoint of $ QQ'$ where $ l \cap (O)\equal{}\{Q,Q'\}$ and also the fact that $ HJ\equal{}HG$ we get that $ AH,C'J$ intersect each other on $ (O)$.
let this point of intersection be $ A'$.
we can prove similarly $ B'K,AH$ intersect each other on $ (O)$ which implies that $ A'\equal{}AH \cap C'J \cap B'K \cap (O)$.
now consider the hexagon $ C'A'B'BNC$.
also consider this lemma which can easily be proven (using contradiction and pascal's therem):
[color=red]consider a hexagon $ ABCDEF$ such that the intersections of its opposite sides are collinear.and also there exists a circle which passes through $ A,B,C,D,F$($ 5$ vertices) .Prove that this circle  passes through $ E$(the remaining vertice),also.[/color]
so we r done !


Since I just competed in my last USACO competition (I'm a senior), I'm thinking about trying out TopCoder.  I was an average, or slightly below average competitor in the silver division.

How difficult is TopCoder Division II compared to Silver or Bronze, and what type of problems appear on TopCoder?  Should I wait until I get better at programming before I enter?

the problems in TC div2 are easier than USACO silver, but you have less time.  Go ahead and start now, there's no reason not to go ahead and get experience.

You can't really compare USACO and TopCoder tasks.

They are... well... different!

USACO is more focused on thinking and planning, while TopCoder is "think, but think fast! type, but type fast!". :)

Topcoder is good practice for USACO, as you become very familiar with the standard set of advanced algorithms, especially on D1 1000-pointers. Also you get practice coding accurately, meaning you are more consistent on USACO.


abstract algebra
topology
group theory
vector
advanced fields
advanced fields unsolved

What are the compact connected abelian Lie groups?

I think they would be (up to isomorphism) the $n$-tori $T^n$.. ALthough I cant prove it.

One of the problems on my manifolds final was to show that any connected abelian Lie group $G$ was isomorphic to $\mathbb{R}^k\times\mathbb{T}^l$.
An outline of the proof:

Show that the Lie algebra $\mathfrak{g}$ is abelian, and that $\exp$ is a homomorphism from $\mathfrak{g}$ to $G$. Using connectedness, show that $\exp$ is surjective. An abelian Lie algebra is isomorphic to $\mathbb{R}^n$. Since $\exp$ is a diffeomorphism on some neighborhood of 0, its kernel is discrete. Show that every discrete additive subgroup of $\mathbb{R}^n$ is the lattice of integer combinations of some linearly independent vectors $v_1,v_2,\dots,v_l$. Then by choosing a basis for $\mathfrak{g}$ that includes $v_1,v_2,\dots,v_l$, we can construct a homomorphism from $\mathfrak{g}$ to $\mathbb{R}^{n-l}\times\mathbb{T}^l$ with the same kernel as $\exp$. This induces an isomorphism between $G$ and $\mathbb{R}^{n-l}\times\mathbb{T}^l$, and we are done.

The compact connected abelian Lie groups are of course the tori.


logarithms
trigonometry
ratio
limit
calculus
calculus computations

use the ratio comparison test with the harmonic series. I.e. show that $\lim_{x \to 0}\frac{1}{x}\ln \left( \frac{1+\sin x}{1-\sin x}\right) = 2$...

[quote="Carcul"]Ascertain the nature of the following series:

$\sum_{n=1}^{\infty}\ln\left(\frac{1+\sin(\frac{1}{n})}{1-\sin(\frac{1}{n})}\right)$

Carcul[/quote]
Sos440 is right. :)
Your sum is equivalent $2\sum_{1}^{+oo}\frac{1}{n}$ wich is equivalent $2\ln{n}+2\gamma+o(1)$

[quote="Sos440"]use the ratio comparison test with the harmonic series.[/quote]

That's right, that's also what I did.

[quote="Kent Merryfield"]Just a vocabulary note: the textbooks I use would call what Sos440 did the "limit comparison test." We would rather not use the word "ratio" in this context because of the potential confusion with a completely different test.[/quote]

That's right Professor Merryfield, we also use that distinction: "limit comparison test" or "comparison through a limit" for the test used in this problem, and "ratio test" for the D'Alembert test.

Carcul


geometry
parallelogram
trigonometry
analytic geometry
graphing lines
slope
complex numbers

Let $ ABCDE$ be a convex pentagon. If $ P$, $ Q$, $ R$ and $ S$ are the respective centroids of the triangles $ ABE$, $ BCE$, $ CDE$ and $ DAE$, show that $ PQRS$ is a parallelogram and its area is $ 2/9$ of that of $ ABCD$.

Let $ P'$, $ Q'$, $ R'$, $ S'$ be the midpoints of $ AB$, $ BC$, $ CD$, $ DA$, respectively.
To prove that $ P'Q'R'S'$ is a parallelogram, draw $ AC$ and $ BD$ and note that $ P'S'\parallel BD\parallel Q'R'$ and $ P'Q'\parallel AC\parallel S'R'$.
$ \frac{EP}{EP'}=\frac{EQ}{EQ'}=\frac{ER}{ER'}=\frac{ES}{ES'}=\frac{2}{3}\implies$ $ PQ\parallel P'Q', QR\parallel Q'R', RS\parallel R'S', SP\parallel S'P'\implies$ $ PQRS$ is a parallelogram
We know $ [P'Q'R'S']=\frac{1}{2}[ABCD]$, and since $ \frac{PQ}{P'Q'}=\frac{QR}{Q'R'}=\frac{RS}{R'S'}=\frac{SP}{S'P'}=\frac{2}{3}$, $ \frac{[PQRS]}{[P'Q'R'S]}=\frac{4}{9}\implies [PQRS]=\frac{2}{9}[ABCD]$.

[quote="Jutaro"][color=darkred]Let $ ABCDE$ be a convex pentagon.
Denote the centroids $ P$, $ Q$, $ R$, $ S$ of the triangles $ ABE$, $ BCE$, $ CDE$, $ DAE$ respectively.
Show that the quadrilateral $ PQRS$ is a parallelogram and $ 9\cdot [PQRS]=2\cdot [ABCD]$.[/color][/quote]
[color=darkblue][b]PROOF.[/b] Denote the midpoints $ X,Y,Z,U,V$ of the sides $ [AB]$, $ [BC]$, $ [CD]$, $ [DE]$, $ [EA]$ respectively. Prove easily that

$ \left\{\begin{array}{c}PQ\parallel XY\parallel AC\ ,\ PQ=\frac{2}{3}\cdot XY=\frac{1}{3}\cdot AC\\\\ QR\parallel YZ\parallel BD\ ,\ QR=\frac{2}{3}\cdot YZ=\frac{1}{3}\cdot BD\\\\ RS\parallel AC\ ,\ RS=\frac{1}{3}\cdot AC\\\\ SP\parallel BD\ ,\ SP=\frac{1}{3}\cdot BD\end{array}\right\|$ $ \implies$ $ \left\{\begin{array}{c}PQ\parallel RS\parallel AC\ ,\ PQ=RS=\frac{1}{3}\cdot AC\\\\ QR\parallel SP\parallel BD\ ,\ QR=SP=\frac{1}{3}\cdot BD\end{array}\right\|$ $ \implies$

the quadrilateral $ PQRS$ is parallelogram and $ m\left(\widehat{AC,BD}\right)=m(\widehat{PQR})=\phi$ $ \implies$ $ [PQRS]=QP\cdot QR\cdot\sin \phi=$

$ \left(\frac{1}{3}\cdot AC\right)\cdot\left(\frac{1}{3}\cdot BD\right)\cdot\sin\phi =$ $ \frac{2}{9}\cdot\left(\frac{1}{2}\cdot AC\cdot BD\cdot\sin\phi\right)=$ $ \frac{2}{9}\cdot [ABCD]$ $ \implies$ $ 9\cdot [PQRS]=2\cdot [ABCD]$.

[b][u]Remark I.[/u][/b] I"ll use the [b]complex numbers[/b]. Denote $ X(x)$- the point $ X$ with the affix $ x\in \mathcal C$ Therefore,

$ \left\{\begin{array}{c}\left\{\begin{array}{c}3p=a+b+e\\\ 3r=c+d+e\end{array}\right\|\implies 3(p+r)=a+b+c+d+2e\\\\ \left\{\begin{array}{c}3q=b+c+e\\\ 3s=a+d+e\end{array}\right\|\implies 3(q+s)=a+b+c+d+2e\end{array}\right\|$ $ \implies$ $ p+r=q+s$ $ \implies$ $ PQRS$ is parallelogram.

[b][u]Remark II.[/u][/b] Consider the hexagon $ ABCDEE$, i.e. suppose that we have identical six balls such that only one ball in each from $ A,B,C,D$ and two balls in $ E$.

$ P$ is centroid of $ \triangle ABE$ and $ R$ is centroid of $ \triangle CDE$ $ \implies$ the centroid of the hexagon $ ABCDEE$ is the midpoint of $ [PR]$.

$ Q$ is centroid of $ \triangle BCE$ and $ S$ is centroid of $ \triangle ADE$ $ \implies$ the centroid of the hexagon $ ABCDEE$ is the midpoint of $ [QS]$.

Thus, the centroid of the hexagon $ ABCDEE$ is [b]both the midpoint of the segment $ [PR]$ and the midpoint of the segment $ [QS]$ ![/b]

Therefore, the quadrilateral $ PQRS$ is parallelogram.[/color]

[quote="Altheman"]Interesting. It is a nice problem, but I don't think that this problem is appropriately placed on an olympiad...It takes a total of 'no thinking' to coordinate bruteforce with the shoelace formula. The parallelogram thing can be shown by vectors.[/quote]

Sure, Altheman! :) But you should take into account that the problem belongs to one of the first Olympiads ever organized in our region (by the way, my country hosted this one, in 2000 :D). Also, the competition is aimed to high school students under 16, so the problems are not supposed to be really difficult. :)

[hide]AnalyticsSolution

Let $ E(0,0), D(a,0), C(b,c), B(d,e), A(f,g) $ be the coordinates of the vertex of the convex pentagon $ABCD$.

Taking the triangles $AEB$, $BEC$, $ECD$, $AED$ their respective barycenters have coordinates $P(\frac{d+f}{3} , \frac{e+g}{3})$ $Q(\frac{b+d}{3}, \frac{c+e}{3})$ $R(\frac{a+b}{3}, \frac{c}{3})$ $S(\frac {a+f}{3}, \frac {g}{3})$

Getting the equations of $PQ$, $QR$, $RS$, $SP$;

Equation of $PQ$

$\frac {y - \frac {c+e}{3}}{x-\frac{b+d}{3}} = \frac{\frac{c+e}{3} - \frac {e+g}{3}}{\frac{b+d}{3} - \frac {d+f}{3}}$

Thus $ y = (\frac{c-g}{b-f})x + \frac{(c+e)(b-f)-(b+d)(c-g)}{3(b-f)}$

Equation of $SR$

$\frac {y - \frac {g}{3}}{x-\frac{a+f}{3}} = \frac{\frac{g}{3} - \frac {c}{3}}{\frac{a+f}{3} - \frac {a+b}{3}}$

Thus $ y = (\frac{g-c}{f-b})x + \frac{g(f-b)-(a+f)(g-c)}{3(f-b)}$

Equation of $RQ$

$\frac {y - \frac {c+e}{3}}{x-\frac{b+d}{3}} = \frac{\frac{c+e}{3} - \frac {c}{3}}{\frac{b+d}{3} - \frac {a+b}{3}}$

Thus $ y = (\frac{e}{d-a})x + \frac{(c+e)(d-a)-e(b+d)}{3(d-a)}$

Equation of $PS$

$\frac {y - \frac {g+e}{3}}{x-\frac{d+f}{3}} = \frac{\frac{e+g}{3} - \frac {g}{3}}{\frac{f+d}{3} - \frac {a+f}{3}}$

Thus $ y = (\frac{e}{d-a})x + \frac{(g+e)(d-a)-c(f+d)}{3(d-a)}$

Let $m_{IJ}$ be the slope of the line $IJ$. From the equations above we have that $m_{PQ}=m_{SR}$ and $m_{PS}=m_{QR}$ therefore $PQRS$ is parallelogram.

Now let´s work with areas. For any convex cuadrilateral with vertex $(x_{1}, y_{1}), (x_{2}, y_{2}), (x_{3}, y_{3}),(x_{4}, y_{4})$ we can determine its area as $Area=\frac{(x_{1}-x_{3})(y_{2}-y_{4}) + (x_{2}-x_{4})(y_{3}-y_{1})}{2}$ (you can deduce this formula by solving the matrix). Substituting with $P, Q, R, S$

$ [PQRS]=\frac{(\frac{a+f}{3}-\frac{b+d}{3})(\frac{c}{3}-\frac{e+g}{3}) + (\frac{a+b}{3}-\frac{f+d}{3})(\frac{c+e}{3}-\frac{g}{3})}{2}$

$=\frac{ac+bc+dg-dc-fe-ag}{9}$

Doing the same with $ABCD$.

 $[ABCD]=\frac{(f-b)(e-0) + (d-a)(c-g)}{2}=-(\frac{ac+bc+dg-dc-fe-ag}{2})$

Since the area when working with analytics must be given in absolut value we finally get $[PQRS]=\frac{2}{9}[ABCD]$

[/hide]

[hide = Solution]
The midpoints form a parallelogram with area half of $[ABCD]$. Perform a homothety centered at $E$ with scale factor $2/3$ and we prove both parts!

[/hide]


For $x,y \in R$ show that $3(x+y+1)^2 +1 \geq 3xy$

[hide]This is probably a bad solution, but it works I think...

Expand and factor into

$3(x-y)^2+(3x+2)(3y+2) \ge 0$

Note that if $x, y \ge -\frac{2}{3}$ or $x,y \le -\frac{2}{3}$ we're done. Suppose neither is true. Then WLOG assume $x > -\frac{2}{3}$ and $y < -\frac{2}{3}$.

Let $x = -\frac{2}{3}+a$ and $y = -\frac{2}{3}-b$ where $a$ and $b$ are positive.

Then our inequality becomes

$3(a+b)^2-9ab \ge 0 \Rightarrow a^2-ab+b^2 \ge 0$ which is clearly true.[/hide]

That's how I did it except I substituted $w=x+y$ and $z=x-y$.


linear algebra
matrix
vector
linear algebra unsolved

Hey,

I've been assigned the following problem, but made no progress on it:

let V be the vector space of 2x2 matrices over a field F.  I want to find a basis
{A,B,C,D} such that each basis element satisfies the following property:

A^2=A, B^2=B, etc.

I can come up with a couple of idempotent matrices,

Identity matrix

1 1
0 0

0 0
1 1

1 0
1 0

etc.  But I cannot construct a basis out of them, any hints or thoughts would be appreciated.

Think a bit simpler ;)
$ \left(\begin{array}{cc} 1 & 0 \\
0 & 0\end{array}\right)$ and $ \left(\begin{array}{cc}0 & 0 \\
0 & 1\end{array}\right)$ are also idempotent.

More generally: For any $ n\in\mathbb{N}$, there exists a basis of the $ K$-vector space $ K^{n\times n}$ consisting of idempotent matrices. Indeed, for every $ u$ and $ v$ from $ \left\{1,2,...,n\right\}$, the matrix $ \left(a_{i,j}\right)_{1\leq i\leq n,\ 1\leq j\leq n}$ with $ a_{i,j}=\begin{cases} 1, & \text{ if }i=j=u\text{ or }\left(i=u\text{ and }j=v\right);\\ 0, & \text{ otherwise}\end{cases}$ is idempotent.

  darij

:blush:  Heh. Thanks,  I guess I should work with matrix multiplication a little more!

Thanks again!

[quote="littlefermat23"]Hey,

I've been assigned the following problem, but made no progress on it:

let V be the vector space of 2x2 matrices over a field F.  I want to find a basis
{A,B,C,D} such that each basis element satisfies the following property:

A^2=A, B^2=B, etc.

I can come up with a couple of idempotent matrices ...[/quote]
You already seem to have four such matrices. Now show that they are linearly independent.


geometry
geometric transformation
reflection
geometry proposed

let $ ABCDE$ a pentagon such that
[$ ABC$] $ \equal{}$ [$ ABD$] $ \equal{}$ [$ ACD$ ] = [$ ADE$ ]  $ \equal{} a$
what is the area of the triangle  $ BCE$
[    ]  is design the area

$ [ABC] \equal{} [ABD] \longrightarrow CD$ parallels $ AB$.
$ [ADB] \equal{} [ADC] \longrightarrow AD$ parallels $ BC$.
$ ABCD$ is a parallelogram.
$ [ADE] \equal{} [ADC] \longrightarrow$ The possible location of $ E$ is the image of reflection of BC (line, not segment).
The altitute of E on BC is double the height of A on BC, thus we have $ [BCE] \equal{} 2 [BCD]$


If: $ x \plus{} y \plus{} z \equal{} 6$, $ x,y,z \ge 0$ (Thanks can_hang2007, I forgot to add this condition).
Prove: $ \sum_{cyc}\frac {(x \plus{} y)}{(x \minus{} y)^2} \ge 3/2$
Using: Awesome Cauchy Schwartz (cooler than regular old cauchy schwartz)

[quote="Rofler"]If: $ x \plus{} y \plus{} z \equal{} 6$
Prove: $ \sum_{cyc}\frac {(x \plus{} y)}{(x \minus{} y)^2} \ge 3/2$
Using: Awesome Cauchy Schwartz (cooler than regular old cauchy schwartz)[/quote]
I think we need the condition $ x,$ $ y,$ $ z \ge 0.$

Write the inequality as
$ \frac {x \plus{} y}{(x \minus{} y)^2} \plus{} \frac {y \plus{} z}{(y \minus{} z)^2} \plus{} \frac {z \plus{} x}{(z \minus{} x)^2} \ge \frac {9}{x \plus{} y \plus{} z}.$
Without loss of generality, assume that $ z \equal{} \min\{x,\; y,\; z\}.$ We have
$ \frac {y \plus{} z}{(y \minus{} z)^2} \ge \frac {1}{y \minus{} z} \ge \frac {1}{y},$
$ \frac {x \plus{} z}{(x \minus{} z)^2} \ge \frac {1}{x \minus{} z} \ge \frac {1}{x},$
and
$ \frac {9}{x \plus{} y \plus{} z} \le \frac {9}{x \plus{} y}.$
Therefore, it suffices to show that
$ \frac {x \plus{} y}{(x \minus{} y)^2} \plus{} \frac {1}{x} \plus{} \frac {1}{y} \ge \frac {9}{x \plus{} y},$
or
$ \frac {1}{xy} \plus{} \frac {1} {(x \minus{} y)^2} \ge \frac {9}{(x \plus{} y)^2}.$
Using the Cauchy-Schwarz Inequality, we have
$ \frac {1}{xy} \plus{} \frac {1}{(x \minus{} y)^2} \equal{} \frac {4}{4xy} \plus{} \frac {1}{(x \minus{} y)^2} \ge \frac {(2 \plus{} 1)^2}{4xy \plus{} (x \minus{} y)^2} \equal{} \frac {9}{(x \plus{} y)^2}.$
The proof is completed.

Ah, well then, it took me a while to find this solution (which I thought was quite original), and it took you a grand total of 5 minutes :o. You have an inequality sign switched when you say $ \frac {9}{x \plus{} y \plus{} z} \ge \frac {9}{x \plus{} y}$, but it is clear what you meant from context.

Anyways, [url]http://www.artofproblemsolving.com/Forum/viewtopic.php?t=325108[/url] should therefore be trivial for you :P

I have just edited that typo. I am very sorry. By the way, nice inequality. :-)

Its not mine, I just figured out that Cauchy Schwartz worked.

My actual solution was just changing z to be 0, and replacing $ x \plus{} y \plus{} z \equal{} 6$ with $ x \plus{} y \le 6$, which surprisingly works!

BTW, this can create quite unusual equality cases because of the way cauchy-schwartz is applied.

I thought

$ \Leftrightarrow\sum{yz(y^4 \plus{} z^4)} \minus{} 8\sum{y^2z^2(y^2 \plus{} z^2)} \plus{} 18\sum{y^3z^3} \plus{} 4xyz(3\sum{x^3} \minus{} 5\sum{yz(y \plus{} z)} \plus{} 18xyz) \ge 0$.

[quote="Rofler"][quote="frankvista"]I thought

$ \Leftrightarrow\sum{yz(y^4 \plus{} z^4)} \minus{} 8\sum{y^2z^2(y^2 \plus{} z^2)} \plus{} 18\sum{y^3z^3} \plus{} 4xyz(3\sum{x^3} \minus{} 5\sum{yz(y \plus{} z)} \plus{} 18xyz)$.[/quote]

How does that prove the inequality? Especially since the inequality has odd equality cases...[/quote]

Yes,I had no idea yesterday. I only wanted to show that it might be okay in this way.


Errr, is that right? 2002! isnt a power of two...

Absurd.  Should that be some kind of inequality?

the obvious way is to calculate it in a long list on your calculator...

I don't think there is any other way.

Yup, ok then! New question, which is bigger?

Though thats really obvious...

[hide="There is always another way!"] We recourse to Stirling's Approximation,

$\sum_{k=1}^{n}\ln n \approx n \ln n-n$

Which gives us

$\sum_{k=1}^{2006}\ln n \approx 13247$

Note that this is the natural log.  Using the base change formula,

$\sum_{k=1}^{2006}\log_{2}n \approx 19112 > 17984$ [/hide]

Oops.  Misread the question.  :oops:

Well, $\log_{2}2003+...+\log_{2}2006 \approx 4 \log_{2}2048 = 44$ so the error is small (smaller than the error from Stirling's).


Al, Bob, Cam, and Dan play one game of rock, paper, scissors at the same time. What is the probability that exactly two show one symbol and the other two show the same symbol, but a different symbol from the first two?

If you think that's easy...

Players $P_{1},P_{2},\ldots,P_{2n}$ for some positive integer $n$ play one game of rock, paper, scissors at the same time. What is the probability that exactly $n$ show one symbol and the other $n$ people show the same symbol, but a different symbol from the first two?

Then if you're a real stud...

Players $P_{1},P_{2},\ldots,P_{n},\ldots,P_{kn}$ for some positive integers $k$ and $n$ play one game of $S_{1},S_{2},\ldots,S_{z}$ at the same time, where $z\ge k$. What is the probability that we can group the players into $k$ groups of exactly $n$ such that each member within a group shows the same symbol, and people in different groups show different symbols?

(For the third one, it's a little beyond Mathcounts level, so don't discuss that to too much extent).

I think the first one is
[hide]$\frac{2}{27}$[/hide]
but i'm pretty sure i'm wrong
My work:
[hide]The first person could do any sign so it is $1$
The second person has to do the same sign as the first person so the probability is $\frac{1}{3}$
The third person could do any of the 2 remaining signs so the probability is $\frac{2}{3}$
The fourth person has to do the same sign as the third person so the probability is $\frac{1}{3}$
$1\times\frac{1}{3}\times\frac{2}{3}\times\frac{1}{3}=\frac{2}{27}$
But, I still think thats wrong[/hide]

EDIT: I made a list and got the same result

The second person doesn't necessarily have to do the same thing as the first, we can have Bob-Cam and Al-Dan. Your solution assumes Al-Bob and Cam-Dan. You're close, but you're just missing that.

Here's a safer solution, it's usually more effective in other problems, it doesn't really make a difference here.

[hide]We first group our four people. There are ${4\choose2}=6$ ways to do this. Consider one group. There are 3 signs they can throw. Consider the other group. There are 2 signs they can throw. Therefore there are 36 possible ways to have 2 and 2.
Clearly there are $3^{4}$ possible combinations of four signs so our answer is $\frac{36}{81}=\frac{4}{9}$[/hide]

[hide="My solution"]
For the first part, there are $3 \times 1 \times 2 \times 1 \times \dbinom{4}{2}$ ways to get what we want, because of the 4 people, 2 must be chosen to show the same symbol and the other 2 will show another. The total number of cases is $3^{4}$ so our answer is $\frac{36}{81}$ $=$ $\boxed{\frac{4}{9}}$

The second one would be $6 \times \dbinom{2n}{n}$ for what we want divided by $3^{2n}$, so we have an answer of $\boxed{\frac{6 \times \dbinom{2n}{n}}{3^{2n}}}$
(yikes! I bet that can be simplified though...)
Eh, I'll do the third one later...
EDIT: Oh darnit, you already posted the solution to the first one!
[/hide]

No, that's not it.

[hide="2"]There are $\binom{2n}{n}$ ways to group them. Then, there are $3\times2$ ways to assign RPS, so this yields $6$. There are $3^{2n}$ total possibilities. So, the prob. is $\frac{6\binom{2n}{n}}{3^{2n}}=\frac{2\times\binom{2n}{n}}{3^{2n-1}}$.[/hide]

[hide="1"]Substituting in $n=2$ from #2, we get $\frac{2\times\binom{4}{2}}{27}=\frac49$.[/hide]

[hide="3"]Using similar reasoning, there are $\binom{kn}{n}\binom{k(n-1)}{n}...\binom{k}{n}$ ways to group the $kn$ people. The number of ways to assign aan $S$ to each group is $\frac{z!}{(z-k)!}$. The total possible number of outcomes is ${(kn)}^{z}$, so the prob. is $\frac{\binom{kn}{n}\binom{k(n-1)}{n}...\binom{k}{n}z!}{(z-k)!{(kn)}^{z}}$, [/hide]

Eh, might be wrong.

Oh right, why am I including P in my answer...
oh and perfect, your answer to 1 is definitely wrong because 1337 already posted the solution, and it was
[hide]$\frac{4}{9}$[/hide]
also, combinations, ($\dbinom{a}{b}$), is when order does NOT count.


College app season is coming 'round, and I was wondering which part of applications--say, the Common App--would be most appropriate to list AoPS classes. Thanks for any input.

Your AoPS classes are one of your extracurricular activities, except in the extremely rare instance that you are getting school credit for an AoPS course.


induction
college
summer program
Mathcamp
geometry
AMC
USA(J)MO
\/closed

We're having a Math Jam this Friday night, at 7:30 PM ET (4:30 PM PT), to introduce induction to younger problem solvers who have not yet learned about this powerful proof technique.  The instructor will be Ari Nieh, who has a PhD in math from Berkeley and is a very popular Mathcamp instructor.

Would it be for someone in 7th/8th grade or younger? I know what induction is, but I haven't done much with it.

Then maybe you should listen to this talk. :)

Come check it out.  This Math Jam is appropriate for students who have some algebra background and either haven't heard of induction before, or have heard of induction but haven't really used it much.  The ideal grade range is 7th-10th grade (of AoPSer), though those who already have used induction a lot would find this Math Jam to be review.

[quote="Walk Around The River"]Will there be a transcript of it posted afterward, like usual?[/quote]
For all math jams, there will probably a transcript. :)
[quote="hasan4444"]I can't attend this jam because the time is very bad for me even I need it very much  :(
So if there is a transcript when it usually get published? After the Jam finish immediately?[/quote]
It probably won't be [b]right[/b] after the mathjam, but usually they come up within the day, I believe. (Don't quote me though, I can't be sure).

This Math Jam does not apply to me, but I would like to thank the AoPS staff for offering it.  It is very considerate of you to offer some free instruction to anyone who visits the site.  If this Jam were next fall, I would point some of my high school senior friends to it.

As with all our mathjams, a transcript should be available within 24 hours after the mathjam has ended. We generally post them within an hour after the jam session.

[quote="1=2"]Fact: Induction is part of Discrete Mathematics[/quote]

obviously...

I have a question.
Even though I am a high school junior, can I still attend this?
I learned induction/used it to prove some stuff, like the hardest being something to the extent of : prove that n! > 2^n for all n>=4 or something to that extent)
But farther than that, I haven't done much problems.
 :maybe:

Oh. Alright. Thank you!
Im going to set my alarm so I remember. =)

Wait, math jams take place in the classroom, right?

[quote="PowerOfPi"]It should take about an hour, but sometimes it takes longer. You can enter or exit the classroom at any time.[/quote]

What I said in a previous post.

I'll probably go just because I'm bored, even though I [b]think[/b] I can use induction.

Will there be math jams that teach concepts like this more often (there was an inversion one last year or the year before, I believe, but I was too stupid to understand it back then)? Then again, I understand how everyone at AoPS is extremely busy and it would be annoying to arrange many of these It would be great if these occurred more often, however.

We probably won't do these too often.  They're very time-consuming to prepare.

in how long will the join jam button appear?

He does teach them, he's just not teaching this one.

Not really... he mainly teaches the informational ones, not the actual math-y ones

Well, I was in an AIME one with him a couple months ago. And the math jam has started btw

[quote="tobedotty"]I think I will stay up for this one (its at 12:30 am here...).[/quote]

just asking where do u live?

Yes, that was an INFORMATIONAL one... you can tell because he was advertising a class

asking me? Im possibly dead. No jk. I live in seattle, washington

[quote="connaissance"]Why don't you teach any of these subject Math Jams, Mr. Rusczyk? You could talk about why you started AoPS, advertise it some more, advertise USAMTS, advertise a whole bunch of other stuff, and talk about AoPS some more. I'm sure that's an incentive...[/quote]

He's probably not familiar enough with induction to teach something like that.  He probably attended it as a student.  (Secretly, of course.)

Or it's probably because he's too busy trying to finish the Precalculus book. I'm pretty sure that all the instructors at AoPS are extremely busy and it's too much work creating a math jam all the time. Looking through the old math jams, they used to have math jams like this in the beginning, but AoPS has become much larger and busier since then.


group theory
abstract algebra
superior algebra
superior algebra unsolved

Let $ m>2$ be a natural number and $ G$ a finite group with $ ord(x)>m$ for all $ x$ in $ G$. prove that $ G$ cannot be writen as a union of $ m$ proper subgroups.

the order of the group is product of primes greater than m. so any proper subgroup will have index greater than m. obviously their union has less elements than the sum of their orders, which in turn can not exceed m*(|G|/m)=|G|.


I read a topic by toadoncart called 'an easy riddle'.

I thought it was kind of fun, so I thought

'Gee, that's fun why don't I give it a try?' :lol:

So I did.
Here's the riddle.(It's from a book called Gregor and the Code of Claw)

[b]There are 7 creatures: [/b]

2 humans(Overlander & Underlander),
1 bat, 1 mouse, 1 cockroach, 1 spider, and 1 rat.

[u][b]They had a lunch and they have each eaten their favorite food.
No two of them ate the same favorite foods. [/b][/u]

[b]Lists of favorite food:[/b]

Fish, Cheese, Cake, Cookies, Bread, Mushrooms, and shrimp in cream sauce.

[b]Here are the clues:[/b]

The bat's favorite food is either mushrooms or cake.

Cookies are not the cockroach's favorite food.

The mouse will eat cheese, but she didn't today.

The Underlander's favorite food is either cookies or shrimp in cream sauce.

The mushrooms and cookies were not eaten by mammals.

The Overlander's favorite food is either cake or bread.

[b]Question: Who ate the cheese?[/b]

Answer: [hide]Rat[/hide]

i think you just make a chart that has 7*7 one side has different ppl/animals names and the other side has the different foods. and cross things out as you read the clues.

Thank you for your thought.

I solved it.

It really isn't hard at all! I pretty much did the same thing abacadaea did.

Huh, interesting, ANOTHER mario fan, just like me.......

I like Donkey Kong, he's pretty cool.

I got my username from thinking about Mario Kart Double Dash, and about Toad. So I thought of TOAD-ON-CART! get it? .........  :maybe:

That riddle of yours is pretty interesting, and hard. Did you make it up?

its kinda funny...you can pretty much guess that the rat would eat the cheese, but whatever...

[quote="toadoncart"]That riddle of yours is pretty interesting, did you make it up?[/quote]
It's from a book (as mentioned earlier). If your interested in reading it, it's by Suzanne Collin. Underland Chronicles FTW!

[quote="mz94"]you can pretty much guess that the rat would eat the cheese, but whatever[/quote]
funny as it sounds, in the book the rat actually likes to eat shrimp and cream sauce.

Although the Underland Chronicles is one of my favorite series, too, the Mathcounts forum isn't really the place to talk about it or Mario. Since the problem itself has been solved, this topic will be locked.


I just finished unrolling the master theorem to prove it's correctness.  But I'm a bit stumped on the following...

T(n) = aT(n-c) + dn^b

n = problem size
a = # of recursive calls
c = size of sub-problems
d = represents O-notation
b = is used in the cases of the master theorem

what's the pattern the above recurrence has?  It's been a while since i've done this...It's not supposed to work out though, right?

[ (a^k) * T(n-c) ] + summation( dn^b ) ^ k ......with lower bound of k=1 and upper bound of n-1?

Sorry for not knowing propper summation notation for the forum...

What do you mean by "the pattern the above recurrence has"?  The Master theorem is concerned with asmptotic bounds; are you asking for a closed form for $ T(n)$?

Ya the closed form of T(n).  Sorry,  I thought that would have been clear with my statement about what I thought the recurrence pattern was near the bottom of my last post.  Was what I posted the closed form?  Do you happen to know what the closed form is?

To explain further my original post...

Before I was working with a general recurrence of the form T(n) = aT(n/c) + dn^b.  By unrolling that recurrence a pattern is revealed that simplifies to (a^k)*T(n/c^k) + (dn^b) * summation(( a/c^b)^i)..where i goes from 0 to k-1.  The recurrence T(n) = aT(n/c) + dn^b is the form a recurrence needs to be in, in order for the master theorem to apply.  That was what I had to prove, the correctness of the master theorem.  But now I have to apply the exact same steps on my new recurrence of the form aT(n-c) + dn^b, but I wasn't 100% sure on the closed form because it's been a while since i've done these.  Does what i'm saying make sense?

A summation usually isn't considered a "closed form."  $ T(n \minus{} c)$ shouldn't be there; you probably want $ T(n \minus{} ck)$.  And you're treating $ d$ as a constant instead of as O-notation.

i'm having a tough time communicating what I want to do.  I appologise but you said closed form, I just went along with you...like I said it's been a while since I've done any of this.  I want the summation there, and I want it to look the way it does.  I need to do some more math and simplifying afterwards anyways, so I can't get rid of any of that yet.  Do you happen to know the answer i'm looking for?  Can you please re-read my posts from earlier and see when I took T(n) = aT(n/c) + dn^b and got an answer for that general recurrence.  I'm just doing the exact same thing for the new recurrence....A pattern emerges if you "unroll" it, than you need to put the pattern into a general form (which is what I want), which can than be further simplified in the future...as for dn^b it's just notation...

[quote="frankiecostello"]as for dn^b it's just notation...[/quote]

The summation you wrote contains the term $ (dn^b)^k$.  This isn't meaningful if $ d$ is interpreted as O-notation.

What you probably want to do is just treat $ d$ as a constant instead; this'll give you the correct upper bound (since that's all O-notation really means).  The summation you wrote for the other problem includes a factor of $ a$ in the summation; this factor is still going to be there (unless you're actually using O-notation).  The correct summation is

$ T(n) \equal{} a^k (T \minus{} ck) \plus{} \sum_{i \equal{} 0}^{k\minus{}1} a^{i} d (n \minus{} ic)^b$.

Note that you mixed up $ n$ and $ k$.  (If you also did the first summation interpreting $ d$ as a constant, that's not quite right either.)

ok, i'm not going to be able to work on it anymore tonight but i'll definately take a look at it tomorrow.  Thanks for all your help, sorry for not being clear, it's been far to long since i've had to do any of this.  If I get stuck again I guess i'll just end up right back here :S, thanks again.


I have no clue where I should ask, so I chose here. How do people get the "Spinning Fractal Balls" as their avatars? A lot of people have them, and they are on all the AoPS books.
Thanks

Dojo made them for couple of us. You can ask him to make it for you too.

The software used to generate the spinning balls may be found [url=http://www.geometrygames.org/KaleidoTile/]here[/url].

However, I believe that you need to use special (a.k.a. costly) software to make the animated GIF file.

you don't  Just put multiple files in one gif, then it display them one by one!

Well, I don't know how to make a GIF animation, maybe someone else will be able to tell us how?

How do you put multiple files into a GIF? :maybe:

dunno, cause the two times that worked, someone helped me.
but check out http://www.gifs.net/gif/

[quote="PowerOfPi"]I have seen in people's signatures that they got it from Dojo. Ask him/her.[/quote]

or MCrawford

[quote="PowerOfPi"]I have seen in people's signatures that they got it from Dojo. Ask him/her.[/quote]
Him. The software I used needed money. As to how much, I really don't know.

Ah, so you just don't have it anymore?

And what's that about MCrawford? Can he make one for me?

If you have the images you want converted into a GIF animation, download GIMP (if you don't have something like photoshop) and move the images into their own layers and save as a GIF (animation)

I have no idea as to what you just said, but pink, I still have the software.

As you can now see with my avatar, I've now got it. Thanks so much to Dojo, once again, as well as everybody else for the advice.


inequalities
topology
real analysis
real analysis unsolved

I am working on the following question  :

let $ A\subset\mathbb{R}$ st $ m_{*}(A) > 0$ where $ m_{*}$ is the outer measure on $ \mathbb{R}$. Let $ 0 < \alpha < 1$ then prove that there exists an open interval $ I$ st : $ m_{*}(A\cap I)\geq \alpha m_{*}(I)$

the following is "my proof" :

Given $ \epsilon > 0$ then we can find an open subset $ O = \cup_{i\geq 1}I_{i}$ of $ \mathbb{R}$ st :
- $ \{I_{i}\}$ is mutually disjoint
- $ A\subset O$
- $ m_{*}(O) < m_{*}(A) + \epsilon$

Since  $ 0 < \alpha < 1$,  I have $ \alpha m_{*}(O) < m_{*}(O) < m_{*}(A) + \epsilon$ then :  $ \alpha m_{*}(O)\leq m_{*}(A)$   (?)    ( I am not sure about this)

$ A\subset O$ so $ A = A\cap O = \cup_{i\geq 1}(A\cap I_{i})$ therefore $ m_{*}(A) = \sum_{i\geq 1}m_{*}(A\cap I_{i})\geq \alpha m_{*}(O) = \alpha \sum_{i\geq 1}m_{*}(I_{i})$

if $ \alpha m_{*}(I_{i}) > m_{*}(A\cap I_{i})$ then take summation over $ i$ I have $ m_{*}(A) < \alpha m_{*}(O)$ thus I have a contradiction, so there exists $ I_{i}$ st : $ \alpha m_{*}(I_{i})\leq m_{*}(A\cap I_{i})$

Could you please give me some comments? is my proof ok? Many thanks

You need to clearly state your assumptions. Here's how I would organize it:

Let $ 0<\alpha<1$ and let $ A\subset R$ (*1) be a set such that for each interval $ I$, $ m_*(A\cap I)< \alpha m_*(I)$. We want to show that the outer measure $ m_*(A)\equal{}0$. (*2)
Let $ \epsilon>0$. By the definition of outer measure, there is some open set $ U$ such that $ A\subset U$ and $ m_*(U)<m_*(A)\plus{}\epsilon$. Decomposing $ U$ into its component intervals $ I_j$, we apply the hypothesis to each of these to find that $ m_*(A\cap I_j)<\alpha m_*(I_j)$. Summing over $ j$ (outer measure is countably additive over disjoint intervals), $ m_*(A)\equal{}m_*(A\cap U)<\alpha m_*(U)<\alpha\left(m_*(A)\plus{}\epsilon\right).$ (*3) Solving this inequality, $ m_*(A)<\frac{\alpha\epsilon}{1\minus{}\alpha}.$
Since $ \epsilon$ was arbitrary, we can make it as small as we want. This makes the outer measure of $ A$ smaller than any positive number, hence $ m_*(A)\equal{}0$.
Taking the contrapositive, if $ m_*(A)$ is positive, we have that for each $ \alpha\in (0,1)$, there is some interval $ I$ with $ m_*(A\cap I)\ge \alpha m_*(I)$.

(*1) Your original has a typo here, with E instead of A as elsewhere. Not a very big deal, but it does add to the general confusion.
(*2) This is the big change. I find that most of the time arguments by contradiction are better handled as proofs by contrapositive. This sentence isn't essential in my version, but yours makes no sense without announcing the intent to contradict and the relevant assumptions in advance.
(*3) The spot you marked with a question mark is a complete waste of space. Having multiples of $ \alpha$ on the small side of an inequality can't help. The next line is the one that matters, and you need the $ \epsilon$ there to cover the slack in the approximation of $ A$ by $ U$ (your $ O$).


calculus
derivative
function
real analysis
real analysis unsolved

Let $f$ be a continuous function on $[0,1]$. The upper right Dini
derivative $D^{+}f$ is bounded below on $(0,1)$. Show that $f$ is differentiable almost everywhere on $(0,1)$.

Try this:

It is known that if $f$ is a continuous function on {a,b] and one of its derivatives , say $D^{+}f$ is everywhere nonnegative on (a,b) then f is nondecreasing on [a,b] hence everywhere differentiable.

now let assume that $D^{+}(f)(x)\geq a\geq 0$ then you are done.

If $a$ is negative then use this technique:

it is known that $D^{+}(f+g)\leq D^{+}f+D^{+}g$
so assume $D^{+}f\geq a$. Consider the function $g=f-ax$.

Then $f=g+ax$ hence we have that $a\leq D^{+}(f)=D^{+}(g+ax)\leq D^{+}(g)+D^{+}(ax)=D^{+}g+a$ it follows that $D^{+}g \geq 0$ hence $g$ is differentiable almost everywhere from which it follows that f is.


real analysis
function
limit
real analysis unsolved

Prove that if $f$ is a Lebesgue measurable complex function on $\mathbb{R}$ then there is a Borel function $g$ on $\mathbb{R}$ such that $f=g$ a.e.

Let's take a $f: \mathbb{R}^n \to \mathbb{R}$ function here. We can then write $f = f^+ - f^-$. Because $f^+ \ge 0$, there is a sequence of simple measurable
functions $0 \le f_1 \le f_2 \le \cdots$ so that $f_j \nearrow f$ and $f_j = \sum_{i=1}^k a_i\chi_{A_i}$. Let's choose such Borel sets $B_i \subset A_i$ that
$m(A_i \setminus B_i) = 0$. We then set $\varphi_j = \sum_{i=1}^k a_i\chi_{B_i}$. This is clearly a sequence of Borel functions with the property that
$0 \le \varphi_j \le f_j$ and $\varphi_j = f_j$ almost everywhere. Now $\varphi^+ = \liminf_{j \to \infty} \varphi_j$ is a Borel-function so that
$\varphi^+ = f^+$ almost everywhere. Note here that $\lim_{j \to \infty} \varphi_j$ doesn't necessarily exist. We do the same with $f^-$ to get $\varphi^-$ and
set $g = \varphi^+ - \varphi^-$. This does the job. For a complex function we write $f = u + iv$ and the rest should be clear.


Indeed easy:
Since $(a^2+d^2)(b^2+d^2) = (ab+cd)^2 + (ac-bd)^2 = 1517 = 37 \cdot 41$ we know that $a^2+d^2 \in\{ 37 , 41 \}$. This gives the solutions $a=6,d=1,c=5,b=4$ and some permutations, which easily can be checked if they fulfill the equations or not.

There also an another solution. a=-1, b=-4, c=-5, d=-6

We search for integers not only positive integers

[quote="jhaussmann5"]The problem: Find the nonzero integers a, b, c, d with a > b > c > d.  We know that ab + cd = 34, ac - bd = 19.

This could have easily been posted as plain text.[/quote]

I answered to the above, when it's not the same as in your document then you could:
a) have told this before
b) written your statement to the forum by yourself

and it's also easy to conclude from positive to all solutions since only changing some signs will do.

Actually, it was the same as his document...nowhere, neither in jhaussman's post or the document, was 'nonnegative' mentioned.


[quote="chien than"]Prove that:$x+y+z \leq \frac{1}{2}xyz+\sqrt{8}$
where $x;y;z$ are positive real numbers satisfying $x^{2}+y^{2}+z^{2}=4$[/quote]
See here:
http://www.artofproblemsolving.com/Forum/viewtopic.php?t=32399

nice ! It is Mr.Damhuydong's ineq  :wink:
@chien than :Bai nay chua het han gui bai  :mad:

@mathboy:Het thang roi ma,ma o do da co giai dau,em se tu giai!


What percent of the people in the survey like drama, romance or science fiction as their favorite type of movie?
\begin{tabular}{cc}&Fraction of Those Surveyed\\Type of Movie&Who List It as Favorite\\\hline Drama&$ \frac7{16}$\\Science Fiction&$ \frac3{16}$\\Historical&$ \frac1{16}$\\Romance&$ \frac18$\\Comedies&$ \frac3{16}$\\\end{tabular}

[quote="GameBot"]What percent of the people in the survey like drama, romance or science fiction as their favorite type of movie?
\[ \begin{array}{cc}
&\text{Fraction of Those Surveyed}\\
\text{Type of Movie}&\text{Who List It as Favorite}\\ \hline
\text{Drama}&\frac{7}{16}\\
\text{Science Fiction}&\frac{3}{16}\\
\text{Historical}&\frac{1}{16}\\
\text{Romance}&\frac{1}{8}\\
\text{Comedies}&\frac{3}{16}\\
\end{array}\][/quote]

$ \left(\frac{7}{16}\plus{}\frac{1}{8}\plus{}\frac{3}{16}\right)\times100\equal{}\frac{12}{16}\times100\equal{}\frac{3}{4}\times100\equal{}\boxed{75\%}$


I don't problem in creating Explicit and Recursive Formulas that's can appear easily.
But is there a way to know what is the formula even it was so complex  :?:  :)

OK what the formula of $ 1,3,5,7$ it's very easy to know that it's $ a_n \equal{} 2n \minus{} 1$ right?
But how you face a hard sequence that you can't know it's formula in an easy way like:$ a_n \equal{} (n)^ {\minus{} 1} 2n \plus{} 4 7^{n \minus{} 2}$ just for example.

Well, you could find the differences and the differences of those differences, and etc. so that you know the degree of the final function. Other than that, I don't know :)

Just observation I think , and more practice !

Thanks for your replies but I was just wondering if there was a way or rule  :D


logarithms
algebra
function
domain
complex analysis
complex analysis unsolved

Question:  find all the Z complex that satisfied the following equation :
e^(1/Z)=-1 ,  I know the answer, but I don´t know how to solve it...     Please help-me ! :oops:
Cheers
J.Ladeira

Take the $\log$ of both sides.

So you get $\frac{1}{z}= i \pi (\mod 2i \pi)$.

Thus, $\frac{1}{z}= (2k+1)i \pi$.

Therefore, $z = \frac{1}{(2k+1)i \pi}$.

Thanks,  I forgot that the domain of the log is C\{0}  and (-1) is under complex number group.


Suppose that $ \varphi$ is a homomorphism from $ S_4$ onto $ \mathbb{Z}_2$.  Determine $ \text{ker} \varphi$.  Determine all homomorphisms from $ S_4$ to $ \mathbb{Z}_2$.

Here, $ S_4$ is the symmetric group of degree $ 4$ under composition, and $ \mathbb{Z}_2 \equal{} \{0,1\}$ under addition $ \mod 2$.

you have to determine the subgroups of $ S_4$ (or as easy: $ S_n$) of index 1 or 2.

The result being alluded to there is usually called the "First Isomorphism Theorem," and it's the natural way to attack this problem.

An alternate technique would be to look how $ \varphi$ acts on generators of $ S_4$.  You can write $ S_4\equal{}<(1\ 2), (1\ 2\ 3\ 4)>$, for example.  Of the four ways to assign these elements to members of $ Z_2$, only two generate legitimate homomorphisms.

EDIT: Nevermind this post, got it! Thanks guys!


calculus integration derivative MATHCOUNTS

i want to buy a graphing calculator, but i don't know which kind to get, i need one that i could use through high school, and maybe even college. Any recommendations? i'm trying to decide between 83 sliver plus and 89, 83 sliver have most of the things 89 have and it is prettier, but 89 is more powerful, but less memory , please HELP

my suggestion is to wait till school ends and get TI-89 Titanium or at least TI-84

in ur current condition, the TI 83 silver would probly be the best choice unless u wanna do some heavy calculus and whatnot...

most high school cirriculums are written with reference to the 83+'s...also, it's by far the most popular calculator and almost everyone could help u if u ever need anything...

89 has excatly the same internal memory as TI-83 Silver Edition however the OS for 89 is about five times larger than the TI-83. plus more of the TI-89's is also about four times larger than the TI-83 Plus.

er...I have a 83 Silver and I'm fairly sure it has a LOT more memory (Archive space) than the 89...

On a minor technicality, you are wrong about the memory size being the most. The 89 has a 2MB flash ROM chip, as well as 250KB of RAM. That totals to 2353152 bytes. The OS for the 83+(se) is somewhere between 128KB and 384KB, making it slightly less than that of the 89. If TI rewrote the 89's OS in hand-hacked assembly like the z80 OSes are, it would be an even more profound difference, as the 89 would have more of its massive flash ROM free. Actually, I wouldn't be surprised if it's the same flash ROM in the 89 and 83+se, just with sharp differences in the OS sizes. However, RAM is definitely a victory for the 89, as is the fact that it can handle numbers much longer than the z80 calcs (even the 85/86): it can handle integers up to 614 digits long, with ALL their digits, and is accurate to up to 14 decimal places on floating point numbers, much better than the 12 and 10, respectively, for the 82/83(+(se)) series.

I agree with you that it is pretty stupid to buy the 83+se when the 89 is much more powerful for almost the same price. imacs may be some of the ugliest computers we've ever seen, but look at Apple's main market: graphics people (both professional and personal), and Joe Average. To avoid going into a rant, I'll stop here on Apple. However, TI is targeting Joe Average Jr. with the 83+se, hoping that junior will be terrified of the half-GUI/half-CLI interface of the 89, and will spend the same amount of money on a technologically inferior calculator, giving TI a much bigger profit margin. The ONLY time the 83+se has an advantage over the 89 is on the ACT, since the latter is banned from it (does anyone know if the HP equivalent(s) of the 89 are banned too?).

As far as casing colors go, I hate it when companies tease you with what's inside by using a translucent plastic case. Either make it opaque or transparent, but not translucent. Despite my dislike of the frosted appearance, it makes sense, since most electronic devices generate potentially disruptive radio waves, and must be shielded. This is generally done with either some sort of steel (often steel mesh) or copper (NSA headquarters uses copper mesh built into the exterior walls--basically a giant Faraday cage). If TI made it the normal black/gray color (the 82/83 line is actually gray--the 85/86 and 68k lines are black), it would defeat the purpose of the name, although it does look more like it should be called 'platinum edition' instead of 'silver edition'.

I personally just detest the whole idea of yet *another* 82/83 series calculator. TI needs to realize that their target audience for the 83+se (at least right now...) uses the 86 and/or 89, not the 83 or 83+, so why would they want an 83+se? They wouldn't; they would much rather see an 86+, but TI has repeatedly turned down requests and inquiries from several people of various statures in the TI community.

wow...okay...did not know that stuff...i just use the calculators...lol

but i still do hold to the theory that 83's are alot more popular than 89s...although i rather like 89's games...

That is so true, the SE costs the same as the 89, but it is severely inferior! They should make a super duper version of the 89 with more memory, more flash, apps, and other stuff. But wait, they have, it is called THE TI-VOYAGER!!!

DANG straight, man! This 83+ SE shiz disgusts me. Does TI think we're stupid or something? I have a TI-89, and i dont care if the 83+ SE has 10 MEGs of rom, the 89 still whips it's sorry butt!

Quoted from a TI-89 Fan!

yeah...i need some more calculators myself...

Why don't you get Mathematica for the price of a TI-89?

or Maple which is used by the universities in my state.

[url]http://www.maplesoft.com/[/url]

that's pretty nice...but it's a computer thing, no?

I don't know what level of math you're in, but many teachers will not allow you to use a TI-89 in Precalculus or calculus classes/tests.

most teachers let you the ti-89 because its legal on the AP Exam and you need practice.

Well a calculator is only legal on about half the AP exam and it's illegal on parts where you can quickly plug and chug in derivates and integrals.  I think TI-89's are very handy and I would recommend them.  But if you're planning to use this on calculus AP exams you need to be able to do all integrals and derivatives by hand.  it's kind of like calculator on target MATHCOUNTS, but not on a sprint; thats how they judge if you get to use a calc or not.

welll i'm not taking the ap calc exam right now so i sort of cheat on classroom ap calc tests.
i copy the answers and teh thing to do. then when the calc section comes i get my calc and i do it and have the right answer.

[quote="churchilljrhigh"]or the free CAS by the germans Mupad. (maple is canadian, and mathematica is american).[/quote]

I downloaded MuPAD and started using it.  It's very good, especially for being free.  Thanks for suggesting it.

so i should get 89, i wish they make clear 89, it will be soo cool

download an emulator of TI-89 or use the similator that TI has in the programmer software.

rainynight, tell spikkes to bring his link cable on mon.
he'll know why i asked.

If you want to talk to only one person, private message them, don't use the forum.  Or call them on the telephone, for crying out loud!