if $ a_n\minus{}a_{n\minus{}2}\rightarrow 0$ prove that $ \frac{a_n}{n}\rightarrow 0$

From Cesaro theorem we have $ \frac{a_{n\plus{}2}\plus{}a_{n}\minus{}a_{1}\minus{}a_{2}}{n}\minus{}>0$ .plus, $ \frac{a_{n\plus{}2}\plus{}a_{n}\minus{}a_{1}\minus{}a_{2}}{n}\equal{}\frac{a_{n\plus{}2}\minus{}a_{n}\minus{}a_{1}\minus{}a_{2}}{n}\plus{}2\frac{a_{n}}{n}$ the first one converge to $ 0$ (cuz $ a_{n\plus{}2}\minus{}a_{n}\minus{}>0$) so...!  :)

It suffices to show that if $ a_{n \plus{} 1} \minus{} a_n\to 0$, then $ a_n/n\to 0$ (for fixed $ a_n$ satisfying the condition, take $ b_{2n \plus{} 1} \equal{} 0$ and $ b_{2n} \equal{} a_{2n}$ and $ c_{2n} \equal{} 0$ and $ b_{2n} \equal{} a_{2n \minus{} 1}$).  Say that $ |a_{n \plus{} 1} \minus{} a_n| < \varepsilon$ for $ n > N$.

For given $ j\gg n$, $ a_j \equal{} (a_j \minus{} a_{j \minus{} 1}) \plus{} (a_{j \minus{} 1} \minus{} a_{j \minus{} 2}) \plus{} \ldots \plus{} a_N$.  Hence $ |a_j| < |a_N| \plus{} (j \minus{} N)\varepsilon$; if we pick $ j$ large enough that $ |a_N|/j < \varepsilon$, then $ |a_j/j| < |a_N|/j \plus{} (j \minus{} N)/j\varepsilon < 2\varepsilon$.  Hence $ a_n/n\to 0$ as $ n\to\infty$.

as noted above, the proof is very similar to the proof that 'normal' convergence implies cesaro convergence.


modular arithmetic
combinatorics unsolved
combinatorics

There is a game at which $ 26$ persons take place. They all sit at a round table. At one point the game master enters the room and writes one symbol on each person's head so no one can see his own symbol but can see all of the other symbols on the head of everyone else. There are as many symbols as persons and it is not a rule that all symbols are used. After this is done everyone writes down a symbol on piece of paper that only he/she sees. The goal of the game is that at least one persons writes down  the symbol on his head.
Prove that:
a) there is a strategy for the "persons" to win without any form of communication after the game master enters;
b) if one persons can see only $ 24$ symbols there is no strategy;

a) Assign each symbol a different residue mod 26. Have the $ k$th player assume that the sum of all $ 26$ symbols (including his) will be $ k \pmod{26}$, and have him write down the symbol on his head that makes this equation true. The players cover all residues in this way. Thus, since the actual sum of the $ 26$ symbols has to be some residue $ m$ modulo 26, player $ m$ will be the one who gets his symbol right.

b) Before we consider the variation, take the game in (a). Fix a player $ A$ and consider all $ 26^{25}$ combinations of symbols he can see on everyone else's head. $ A$ must compute a guess from this based on the strategy, and after constraining everyone else there is only one configuration in which $ A$ will be right. This means that of all $ 26^{26}$ configurations, $ A$ can only guess right in $ 26^{25}$ of them. The same applies for all players. But in a perfect strategy the players must win in all of the $ 26 \cdot 26^{25}$ configurations, and there are $ 26$ players. This means in any given configuration, exactly one player must guess right, or the overlap means there is a configuration that will defeat them.

This result, that only one player can get the correct symbol at a time, carries through in the variation too. Suppose the variation is implemented so that player 1 cannot see player 2's symbol. Now fix the symbols on the heads of players 3 through 26. Player 1 must make some guess based on this and the strategy. Have the game master write that guess on his head. Now player 2 must make a guess based on everyone else's symbols. Have the game master make him get it right too. Thus, we have a configuration where two players are correct, so some other configuration must have none of them be correct.

please past problems to SEEMOUS shortlist last years


function
geometry
3D geometry
sphere
calculus
real analysis
real analysis solved

let $f:R^n -> R$ be a differentiable function such that:
1/$f(O)<0$
2/$f(x)>0$ if $x \in S$ where S is the unit sphere of $R^n$
3/$lim_{|x|->+\infty} f(x) =0$

what is the minimal number of point $x$ where $df(x)=0$

I'd say a minimum of three critical points. The function must have a minumum near the origin. It must have a maximum, where its value is positive. Finally, consider all paths from the minumum near to origin to "infinity". Since each such path much cross the unit sphere, $f$ becomes positive somewhere on the path but tends back down to zero. Hence, $f$ restricted to the path must have a maximum. Consider the infinum of all such pathwise maxima. We need an argument (which I don't have at the moment) to show that there is a path that realizes this least maximum. The point at which that special path has its maximum is a minimax point - a "saddle point" to use the usual third-semester calculus jargon. And the differential is zero at such a point.

The only way for this minimax point to be the same as the maximum would be for the function to take on its maximum at every point on a set that looks like a sphere - and that would be infinitely many critical points.

As usual, Kent is right (if $n>1$, but the case $n=1$ obviously gives the same answer $3$). The only thing missing in his post is an example of a function having exactly 3 critical points. Here it is:
Let $x=r\omega$ ($r\geq 0$, $\omega\in S$) be the polar representation of $x\in R^n$. Let $e$ be any unit vector. Let $\psi(x)$ be any smooth function on $[0,+\infty)$ such that $\psi(0)=0$, $\psi$ increases on $[0,1]$ to $1$ and then decreases to $0$ on $[1,+\infty)$. Now just put
\[
f(x)=\psi(r^2)(2+\langle e,\omega\rangle)
\]
The critical points are $0$ (minimum), $e$ (maximum), and $-e$ (saddle).
To be honest, this function is $0$ at $0$, but it is easy to push the minimum down a bit...

[quote="Kent Merryfield"]Consider the infinum of all such pathwise maxima. We need an argument (which I don't have at the moment) to show that there is a path that realizes this least maximum. The point at which that special path has its maximum is a minimax point - a "saddle point" to use the usual third-semester calculus jargon. And the differential is zero at such a point.[/quote]
We don't need these arguments, since Mountain Pass theorem states that minmax is a critical value for $f$, i.e. there is $x$ s.t. $f(x)=minmax$ and $f'(x)=0$.  :?


Hi,

I'm new to Latex, and I wonder if there is a method to include c++ code with typical layout (and colors : comments green,....) in a latex text ??

Any help is greatly appreciated

plinius

This sort of thing is usually done with the verbatim environment:

\begin{verbatim}
code
\end{verbatim}

This command prints your code literally.  Don't know about the colours.

I think you'll have to tell the compiler what colors you want. Put the line \usepackage{color} at the top. Then use \textcolor{red}{whatever you want in red} to get stuff in red. I think you can use red, blue, green, yellow, black, and white. You can also define them by using shadings of RGB.

thanks jhaussmann5, this works indeed... However, the tabs disappear....
(it's not that bad, since VC allows to replace the tabs by whitespaces, but if you know a solution, it would be nice).
ComplexZeta, thanks. This works also.

I still have one more problem:
when I press  \textcolor{green}{\\the comment...}  in the "verbatim environment" it gets printed (off course), so , How can I combine colors and verbatim ????

Thanks again for your helps.

Hm tough problem. You can put the textcolor command before the verbatim, but then you have to put a new verbatim every time you want to change colors.

that's the problem indeed (I have more than 10 pages of code, and:
->comments are green
->"int" "char" "long" etc etc are blue
->strings are red
->...
)
I wonder if someone has had the same problem before and made a "plugin" or something like that for it ???
(I'm new to Latex and don't know what is possible and what isn't)

Is there any good place where I can find plugins (If they exist, at least...) ?
I googled, but didn't found anything I could use
 :?

Thanks.

There is a listings.sty package which perhaps does what you want.

[url]http://www.atscire.de/index.php?nav=products/listings[/url]

You may be able to find alternatives or more info by Googling for "latex listing package" or something along those lines.

Disclaimer: I've never used it.  I have no idea how it works, how easy it is to use, or how robust it is.

thanks, DPatrick, but it doesn't allows color and it hasn't a predefined file for c++ (only for assembler , vb and basic   :( ).

plinius.

I found c++2latex, but that's only for linux ...

You can find a list at http://www.tex.ac.uk/cgi-bin/texfaq2html?label=codelist
I haven't tried it but highlight listed there looks interesting and has a version for Windows - see [url=http://www.tex.ac.uk/tex-archive/support/highlight/README]Readme[/url].
[quote]Is there any good place where I can find plugins[/quote]
There are 2 places to look for packages:
[list=1][*][url=http://www.tex.ac.uk/cgi-bin/texfaq2html?introduction=yes]TeX FAQ[/url] which also offers help for common problems
[*]Many (most?) LaTeX programs are available from CTAN and you can search it in a number of places, for example, at http://www.ctan.org/search either by searching the archive or the catalogue.[/list]

thanks, stevem, that "highlight " works really wel!!!
It does exact what I needed.

Thanks.
plinius.


[quote="mr.affable"]So, when I first started flaying FTW, I thought that everybody I was against had considerable mathematical skill.  Now I dont think so.

Example 1:

For the problem which states

Find the sum of

1/2 + 1/2 sqrd + 1/2 cubed ....... 1/2 tenth

my opponent gave the answer within 1 or 2 seconds.  I believe that this is only possible if the problem had been seen before, or if it had been previously examined.  I couldn't even read the exponent on the end of the series before the question had been answered.  Doesn't a previous knowledge of the questions defeat the "challenge" inherent in playing the game?[/quote]

The answer is quite obviously 2^10-1/2^10, or 1023/1024.  THey just know their powers of 10.

Even if the operation is that simple.. I see no way that my opponent could have answered in a time span too short to comprehend all of the aspects of the question..

Yes memming is a part of the game of FTW (for the win). Some people (myself included) have memorized a few of the questions. But one thing FTW does is helps you look for shortcuts in solving the problems! For that one, it is the sum $ 1\plus{}2\plus{}4....\plus{}512$ over $ 1024$ After a while playing you are much more familiar with typing the numbers and can do 1023/1024 in less than 3 seconds.

Trust me, alot of us can get that <2 seconds. (just need to see who types the fastest)

alright.. now that I remain convinced that there are mathematicians on this board who can intuitively solve problems of that calibre in a very short time span.  I still dont quite understand "sums" of that type... I know about one method of finding a common summend and then multiplying by half the amount in the series...

Well, here is the sum of the first powers of 2
\[ \displaystyle\Sum^{n}_ {k = 1} = \frac {1}{2} + \frac {1}{4} + \cdots + \frac {1}{2^n} = \frac {\frac {1}{2}(1 - \frac {1}{2^n})}{1 - \frac {1}{2}} = 2\frac {1}{2}\frac {2^n - 1}{2^n} = \frac {2^n - 1}{2^n}
\]
And we all know that, so, if we plug in 10
\[ \frac {2^{10} - 1}{2^{10}} = \frac {1023}{1024}
\]

@Isabella below me: Yeah I fixed it :D. Thanks for catching that! (I've noticed I've done that on my challenge set too.... gah more editing)

To make exponents, do

[code]
$2^{10}$
[/code]

Some questions have been seen before, but there are only a few thousand questions. I have played 2700 games, so I have seen quite a few.

Pretty much. And also, generally, when you post a question about a specific problem you encountered on FTW, you should post it in the FTW Problems Discussion Forum. Here's the link: http://www.artofproblemsolving.com/Forum/index.php?f=503

[quote="(^_^)"]he definitely memed the problem, this question is very common,(and i'm not talking about ftw) and if you do enough problems, you will see this multiple times, he might not have seen this on ftw, but he must have encountered this problem before, theres no way you can answer it in less than 2 sec if its his first time seeing this type of problem for the first time, and typing 1023/1024 takes a while.[/quote]
unless you have a fast brain ;)

[quote="mr.affable"]so.. basically.. because we know that the sum of the 1/2^n series as n approaches infiniti, becomes 1, and from there you derive the equation?  Thanks alot for the help, btw.[/quote]

Its the general equation for geometric sums

Let $ S\equal{}a\plus{}ar\plus{}ar^2\plus{}\cdots\plus{}ar^{n\minus{}1}$

Then $ rS\equal{}ar\plus{}ar^2\plus{}\cdots\plus{}ar^{n}$

Subtracting gives

\[ (1\minus{}r)S\equal{}a(1\minus{}r^n)\]
\[ S\equal{}\frac{a(1\minus{}r^n)}{1\minus{}r}\]

Another way to think about it, is thinking about a slot 1 unit long. We are filling it with 1/2, 1/4, ... . We know each new block we put in, we half the remaining space. In this question, the last block put in is 1/1024, so there is 1/1024 left, and subtracting from one gives the answer.


geometry
geometric transformation
rotation
analytic geometry
combinatorial geometry
combinatorics
Russia

Clearly, you have to add some hypothesis (think about the case where you only use one or two colors). Is each color used an infinite number of times or something like that?

Pierre.

Ok, all we have to assume is that each color is used at least once.

If M(x,y) is an integer point, then i will say that the points (x+1,y), (x-1,y), (x,y+1) and (x,y-1) are its neighbours.
Note that if for each point M all its neighbours have the same color as M then all the grid is colored by only one color, which contradicts our hypothesis. Thus, we may consider two neighbours with different colors. Wlog, we may assume that these points are A(0,0) which is green and B(1,0) which is blue.

- If there is a red point belonging to the axis y = 0.
Let C(a,0) be such a red point. Then a > 1 or a < 0.

If some point M(0,y) is blue (resp. red) then the triangle MAC (resp. MAB) is tricolored and right-angled, so we are done. Thus we assume that each of the points M(0,y) is green.
In the same manner, if the point D(a,1) is not red we are done, so we assume that it is red.
But D is the image of M(0,a-1) (which is not A) by the rotation with center B and angle $- \frac {\pi} {2}$, so that MBD is tricolored and right-angled.

- If there is no red point on the axis y = 0.
Let C(a,b) be a red point, with b =/= 0.
Let P(a,0). From our hypothesis P is not red. If P is blue, then P is not A, and the triangle PAC is tricolored and right-angled. If P is green then we use PBC, and we are done.

Pierre.

Call the colours A,B,C. Assume that there is no line or column which contains all three colours. Then we select a column or a line l in which exactly two colours are used, say A and B. Then Consider a point P which is not on l. If P has colour C, then the projection of P onto l has colour A or B, say it is P' and has colour A. There is K on l such that K has colour B by the definition of l. In this case KP'P has the desired property.

Now, if there is a line l (if there is a column, rotate the plane by 90 degrees) which contains A,B,C then take a point P on l. Select Q,R on l such that P,Q,R are pairwisely different coloured. We can assume WLOG that P has colour A, Q has colour B and R has colour C. Take P' such that the projection on l of P' is P. Then if P' has colour B, triangle P'PR does the jobe. If it has colour C, triangle P'PQ does the job. Therefore we can assume that for a certain point, all points on its "column" have the same colour.  Now, take the points P,Q,R from above again. If Q is between P and R (we judge all other cases in a similar manner) , we draw from Q two lines which make angles of 45 degrees with l, to the left and to the right. These lines intersect the "columns" of P and R in points P' and R'. Triangle P'QR' does the job and we are done.

Hello,
I'm just wondering whether the original solution is like the ones offered by iandrei and pbornsztein. I mean the task is meant to be solved by students of the 9th form. Could they have applied the same techniques?

1) It seems staightforward that if we take three points on the plane with coordinates (x_1;y_1), (x_2;y_2) and (x_3;y_3) (x and y are intergers), then there always exists a right-angled triangle with respective vertices, since we can always find a triple of integers, the lengths of the sides of the triangle, which can be calculated using the coordinates of the respective vertices. In other words, there is always a triple of numbers, intergers, satisfying p^2+q^2=r^2.

2) With the vertices the colours are also given. Each vertex can be of any of the three colours, for example a, b and c. Thus there are three elements which can be variated 3^3=27 times. Among the 27 possibilities to variate 3 elements over 3 places there [u]must[/u] be the special cases abc, bca, cba.

Is there the original solution to this problem I can look up?

Actually it's taken from Saint Peterburg's math olympiad 1996 for the 7th class. Hm, it's strange that this problem appeared in a Russian olympiad once again since it may had been known to some participants

How about an inductive solution?

Assume that the three colors used were Red (R), White (W), and Blue (B). We induct on R. The base case R=1 holds. To prove it, consider a random point being R. Call the row R is in R1, and call its column C1. If some point on R1 is W, and some point on C1 is B, we are done. So assume that they are not. This implies that all the points on R1, and C1 (except for R) is either all B or all W. WLOG, let them be all B. If we were to avoid having a right triangle, then note that R combined with the B's in all of R1 and C1 accounts for all the other points in the plane. Thus we would have no W, but that's a contradiction. Thus, we must have a W somewhere, and so we have a right triangle. For the inductive step, assume that for some R=k, we get he desired result. We now prove it for R=k+1. Since we are assured that we have a right triangle with the desired property for R=k, we can just choose some point not affecting the right triangle and color that R. That completes the induction.

I can't help but think there's something missing there...

[hide="Solution"]
Case 1: There's a line with all three colors.
Then pick three points on that line with different colors, and draw the perpendiculars to the line passing through those points.  Then clearly those three lines can only have one color.  WLOG color 2 is between color 1 and 3 on the original line.  Then draw the circle with diameter being the line segment determined by color 1 and color 3 on the original line.  The point where this circle hits the perpendicular at color 2, with the points at color 1 and color 3 give us our desired right triangle.

Case 2: There's no line with all three colors.
Pick a line going through color 1 and color 2.  Then project a point with color 3 onto this line, and we are done!
[/hide]

Elegant! :)
[quote=orl]Each grid point of a cartesian plane is colored with one of three colors, whereby all three colors are used. Show that one can always find a right-angled triangle, whose three vertices have pairwise different colors.[/quote]

Suppose no such triangle exists. Fix a red point $A$, a green point $B$ and a blue point $C$. Consider the circle $\omega$ with diameter $\overline{AB}$. Note $C \not \in \omega$. Let $\overline{CA}, \overline{CB}$ meet $\omega$ again at $B', A'$ respectively. If $B'$ is red then $\triangle CBB'$ contradicts the hypothesis. Since $B'$ is not blue, it must be green. Now if any point $P$ on line $\overline{BB'}$ is red then $\triangle CB'P$ contradicts the hypothesis; if it is blue then $\triangle AB'P$ contradicts the hypothesis. Thus, the entire line is green. Similarly line $\overline{AA'}$ is red. However $\overline{AA'}, \overline{BB'}$ intersect at the orthocenter $H$ of $\triangle ABC$. Thus, $H$ is both green and red; and we don't buy that logic. Contradiction!

Edit: v_E points out that an edge case is when $C$ lies on line $\overline{AB}$. Consider lines through $A, B, C$ perpendicular to line $ABC$. Then it is clear that all points on each of these lines have the same color. Now it is easy to find a tri-colored right triangle.

Is this statement true for acute-triangle? :ewpu:

Cute!!

Let $\triangle ABC$ be a triangle with three different colored vertices. Say that $A$ is colored red, $B$ is colored yellow, and $C$ green. WLOG say that the orthocenter of this triangle, $H$, is colored green. Let the perpendicular foot to the sides of the triangle $AB, BC, CA$ be $D$, $E$, and $F$. First, notice that $F$ should be green. Then, $E$ should be red and $D$ should be yellow! But then we have a contradiction in the right triangle $ADC$.

[hide=Solution][quote=All-Russian MO 2004 Grade 9 P1]Each grid point of a cartesian plane is colored with one of three colors, whereby all three colors are used. Show that one can always find a right-angled triangle, whose three vertices have pairwise different colors.[/quote]
[b][color=#000]Solution:[/color][/b] Suppose there exists a line $\ell$ with point $X_1,X_2,X_3$ in the same order and have the colour $C_1,C_2,C_3$. Let $\ell_i$ be the perpendicular line to $\ell$ at $X_i$. Note that, every point on line $\ell_i$ must have the colour $C_i$, otherwise we readily have the triangle. Name $X_2$ as $A$. Choose point $B$ on $\ell_1$ and $C$ on $\ell_3$. Let $x,y$ be distances $X_1X_2$ and $X_2X_3$. Let $B'$ be foot from $B$ to $\ell_3$ and Let $k,m$ be distances $BX_1$ and $B'C$ (WLOG, assume, $X_3C$ $>$ $X_1B$). Choose $k,m$ such that $xy$ $=$ $k(k+m)$. We can obviously find such $k,m$ when $xy>1$ and the construction is obvious when $xy=1$, thus by Pythagoras' Theorem, $AB^2$ $+$ $AC^2$ $=$ $BC^2$.

Now, We'll show that such a configuration, where $3$ points with different colour on the same line exists. Take a line $\ell$ with all points on it having colour $C_1$. Let $A,B$ be a points having colour $C_2$, $C_3$. Consider the intersection of perpendicular from $B$ to $\ell$ and parallel to $\ell$ through $A$. Clearly this point must have the colour $C_2$ otherwise we're done and thus we arrive at the configuration. Suppose there exists a line $\ell$ with all points having colour $C_1$ and a point $A$ with colour $C_2$. Take point $B$ with colour $C_3$ and notice that the intersection of perpendicular from $B$ to $\ell$ must have colour $C_3$ otherwise we're done and thus we again arrive at our required configuration $\qquad \blacksquare$[/hide]

[url=https://artofproblemsolving.com/community/c6h195933]Indian National Mathematics Olympiad 2008 - Problem 4[/url]

btw, some of the solutions above solve a different (and easy) problem - when all points of the plane are colored, not just the lattice ones.

Lets assume the contrary - there is no different-colored triangle.\\
[color=#f00][b]Lemma[/b][/color]: There are no three different-colored points on the same line.\\
	 Assume there are three different-colored points on the same line - say A, B, C. Let the color of A be a, of B be b, of C be c. Wlog let them lie on the line in this order. Let $X$ be the intersection of the circle with diameter AC and the line perpendicular to AC at point B(one of them).\\

	Now:\\
	-$X$ can't be in b because of $\triangle AXC$\\
	-$X$ can't be in a because of $\triangle BXC$\\
	-$X$ can't be in c because of $\triangle ABX$\\
	$\implies$ contradiction\\
	 We know that there exist three points A, B, C with different colors say a, b, c. From the lemma they are not collinear so they form a triangle.\\
	Let $k$ be the circle with diameter AB. CB is not tangent to $k$ or $ABC$ is different-colored triangle. Let $\{Y,B\} = k\cap CB$.\\
	-$Y$ is not in a because of the lemma\\
	-$Y$ is not in b because of $\triangle ACY$\\
	-$Y$ is not in c because of $\triangle ABY$\\
	$\implies$ contradiction.\\

As said in #13, many of the solutions (including the above) solve a different problem. "Each grid point of a cartesian plane..." means each point with [b]integer[/b] coordinates.

Okey so i will poste the solution of this problem with some motivation(is same as anantmudgal09)
So we have a triangle that no matter how the vertices are colored,it will be there.

So for the sake of contradiction we suppose that such a triangle does not exist....(1)

Let's fixe 3 vertixe of different color,$A$-red,$B$-blue and $C$-yellow.
we have two cases ,those 3 points being in a line and none colinear.

$Case1$, $A$,$B$,$C$ are not collinear.

Lets take a circle with diameter $AB$ ,if $C$ is on that circle then we get a contradiction with ....(1)

So $C$ is not on the circle with diameter $AB$
The reason why we took a circle with diameter $AB$ is that we always want to create perpendicular angles,and we know that each point in that arc will gives us $\angle XAB=90$,

Let the intersections of the lines $CA$ and $CB$ with the circle be $A'$ and $B'$ respectively.If $A'$ is yellow than we get contradiction with ...(1)

Also if $A'$ is red than the triangle $CA'B$ will get us a contradiction with ...(1)
so this means that $A'$ is blue,,so if any point on line $BA'$ is Differen color of these 2 points we get a contradiction with ...(1)
so the whole line is colored with Blue.
Similarly the line $AB'$ is colored all red
But we see that their intersection is the orthocenter of the triangle $ABC$ so $H$ can we blue(which means that $CB'H$ GIVES CONTRADICTION)
                                                                                or  $H$ can be red  (which gives that $CA'H$ GIVES CONTRADICTION)
so for this case we are done

Now the second case
$A$-$B$-$C$ are collinear
We take a line $l$ and points $A$,$B$,$C$ on it,
Now late take $l_A$ ,  $l_B$ and $l_C$ lines perpendicular to $l$ and passind through $A$ , $B$ , $C$ respectively, we see that the colors of these lines should be the same as their point passing throught,otherwise we get a contradiction.

So now just Take points in these 3 lines that form a right angeled triangle .And we are done    $\blacksquare$

@above: Only the lattice points are colored, i.e. the points with both coordinates integer numbers. It was emphasized at least 2 times in this thread, e.g. just above your post. So, some of intersection points you construct may not be colored and your argument fails.

Let the colors be red, yellow, and blue. Assume for the sake of contradiction no such triangle exists. Let $A$ be a red point and $B$ be a yellow point and let $\ell_A$ be the point perpendicular to $AB$ at $A$.

[b]Claim 1:[/b] $\ell_A$ only contains red points

If there is some blue point $C \in \ell_A$ then $\angle ABC = 90^{\circ}$. If there is some yellow point on $\ell_A$ then let us choose a blue point, not on $\ell_A$, and call it $X$ (such a blue point exists since there are no blue points on $\ell$). Choose some $Y \in \ell_A$ such that $XY \perp \ell_A$. If $Y$ is red then we can choose a yellow point $Z \in \ell_A$ which gives $\angle XYZ = 90^{\circ}$. If $Y$ is yellow then $\angle ZYA = 90^{\circ}$. Thus, $\ell_A$ contains no blue or yellow points, proving the claim.

By the same logic, $\ell_B$ (Defined similarly) only contains yellow points. Thus, we have a pair of parallel lines, one all red and the other all yellow. If we take $P$ to be a blue point we can choose a sufficiently far away $Q \in \ell_A$ such that the circle with radius $PQ$ intersects $\ell_B$, say... at $R$. This gives $\angle PQR = 90^{\circ}$. Contradiction. $\blacksquare$

[quote=dgrozev][quote=blackbluecar]... Choose some $Y \in \ell_A$ such that $XY \perp \ell_A$....[/quote]
Apparently you also misunderstood the problem (see #13, #15 and #18). Only the lattice points are colored! Only the points with both coordinates integers.[/quote]

Oops yeah, your right. This was part of a worksheet I was working on which doesn't mention the Cartesian plane  :P

Suppose the proposition in false. Let $A$, $B$ and $C$ be the 3 colours. We must have a column where, w.l.o.g, $A$ appears infinitely. In this column, choose 3 points which are coloured $A$, points $X$, $Y$ and $Z$. On the $X$'s row, no point can be coloured $A$, or else we would have an $A$ triangle, and therefore there should be infinite points coloured, w.l.o.g, $B$ on this row. On $Y$'s row, every point that lies in a column where, over the $X$'s row the point is coloured $B$, must be painted $C$ or else we would have a $B$ triangle. Now, take the intersection from a column which there's a $B$ on the $X$'s row and a $C$ on the $Y$'s row, and $Z$'s row. This point can't be coloured with any of the 3 colours and we get a contradiction. $\blacksquare$

@above I dont see why having an A - triangle is not good, we want to prove a triangle with A, B, C colours exist. A,A,A is probably fine ig ?

If no row or column contains at least two colors, then the entire grid must only contain one color, so assume WLOG that the $x$ axis contains points $A,B$ with colors $a,b$.

If any point $C$ with color $c$ is on the $x$ axis, then for an arbitrary point $P$ not sharing a column with $A,B$ or $C$, one of triangles $AHP,BHP,CHP$ works where $H$ is the projection of $p$ onto the $x$ axis. Otherwise, select an arbitrary $C$ with color $c$; $C$'s projection $F$ onto the $x$ axis is one of colors $a,b$, hence one of the triangles $AFC,BFC$ works.

Not totally sure if there are any cases left out :( . Also, sorry for the horrible write-up, I couldn't come up with a better one :(
$\textbf{Case 1:}$ All of the vertical lines in the grid are of same colour. In this case we choose any point (say $A$) on one of the vertical lines and draw $2$ diagonal lines through that point, each one of which will intersect each vertical lines at exactly one lattice points and $2$ of these vertical lines must have different colours than $A$, giving us the desired triangle.
 $\textbf{Case 2:}$ There exist at least one vertical line with two distinct colours. Say $A,B$ are two points of different colours on a vertical line. Now if we have a point on the perpendiculars through $A,B$ which are of different colours than $A,B$ then we are done, if not, then pick a point $C$ of different colour than  $A,B$ and drop a perpendicular on the line containing $A,B$ at $D$. Then one of triangle $ACD$ or $BCD$ must have vertex of different colours.
$\textbf{Case 3:}$ There exist at least one vertical line with $3$ different colours. In this case, say those 3 points are $A,B,C$. We draw the perpendicular through each of $A,B,C$. If any of the perpendiculars have a point of different colour then we are done. Suppose not. Then the perpendicular lines must be all monochromatic. In fact every perpendicular on the vertical line must be monochromatic, which makes it similar to case 1 (this time we have horizontal monochromatic lines instead of vertical ones) hence we draw the diagonals through a point as given in case 1, and we will get our desired triangle.

[hide=Solution]
Name the colors red, blue, green. Assume the contrary. Note that at least one line is non-monochromatic, since if otherwise, the line $y = 0$ is all one color, WLOG let it be red. Then, the line $x = c$ for any integer $c$ intersects $y = 0$ at $(c, 0)$, so since the line $x = c$ is monochromatic, it must also be all red. This implies that $\mathbb{Z}^2$ is all red since any $(a, b) \in \mathbb{Z}^2$ lies on $x = a$, which is all red, contradicting the fact that all colors are used at least once.

Thus, there exists a line with at least two colors. WLOG let those colors be red and blue, and WLOG
let the line be $y = 0$. Let the coordinate of the red point be $P_1 = (a, 0)$ and let the coordinate of the blue point be $P_2 = (b, 0)$. Then, there exists a green point $Q = (c, d)\in \mathbb{Z}^2$. Then, let $P_3 = (a, d)$ and $P_4 = (b, d)$. Then, $P_3$ is red, since if $P_3$ was blue, then $\triangle P_1P_3Q$ would be right and its vertices would all be pairwise different colors,  and if $P_3$ was green, then $\triangle P_3P_1P_2$ would be right and its vertices would all be pairwise different colors. Similarly, we can show that $P_2$ is blue. Then, it follows that $Q\notin \{P_3, P_4\}$, and the line $y = d$ contains three points $P_3, P_4, Q$ of pairwise distinct colors. Now, this means that the whole line $x = a$ is red, since if a point $A$ on it was blue, then $\triangle AP_3Q$ is a right triangle with vertices all pairwise different colors, and if $A$ was green, then $\triangle AP_3P_4$ is a right triangle with all vertices pairwise different colors. Similarly, the whole line $x = b$ is blue and the whole line $x = c$ is is green. Then, if $Q_3 = (b, d+b-a)$ and $R_3 = (c, d+a-c)$, then the slope of $P_3Q_3$ is $$\frac{(d+b-a) - d}{b-a} = 1,$$ and the slope of $P_3R_3$ is $$\frac{(d+a-c) - d}{c-a} = -1,$$ and these multiply to $-1$, so it follows that $\angle Q_3P_3R_3 = 90^{\circ}$, so $\triangle P_3Q_3R_3$ is right. However, $P_3$ is red since it lies on $x = a$, $Q_3$ is blue since it lies on $x = b$, and $R_3$ is green since it lies on $x = c$, so $\triangle P_3Q_3R_3$ is right and all of its vertices are pairwise distinct colors, a contradiction.

Thus, there exists such a right triangle with pairwise distinct colored vertices, as desired. $\blacksquare$

Say the colors are red, green, and blue. If there does not exist two points in the same row or column of different colors, then the entire grid would be the same color, contradiction.\\

Suppose FTSOC that such a triangle does not exist. Suppose WLOG that we have two points, $R_0$ and $G_0$, in the same row or column such that $R_0$ is red and $G_0$ is green. Now, consider a blue point $B.$ The foot from $B$ to $R_0G_0$, which we call $B_0$, must also be a blue point, since otherwise we would have a right triangle with different colors.\\

Let $\ell_R$ be the line through $R_0$ perpendicular to line $R_0G_0B_0$, and define $\ell_G$ and $\ell_B$ similarly. By our earlier logic, for any color $C$, all points on $\ell_C$ must be of color $C$.\\

WLOG the order of the lines from left to right are $\ell_R,\ell_G,\ell_B.$

Then, let $O=(0,0)$ be some point on $\ell_G$. Let $R_0G_0=r$ and $B_0G_0=b$. Then, consider $$R_1=(-r,br),O=(0,0)=B_1=(b,1).$$ The slope of $OB_1$ is $\frac{1}{b}$, and the slope of $OR_1$ is $-\frac{1}{b}$, so this is a right triangle, and its vertices are all different colors, so we are done.

Nice and easy.

Assume for the sake of contradiction that the result is false. Pick two points that are colored red and green respectively; suppose they lie on a line $\ell$. Let $\ell_r$ be the perpendicular line to $\ell$ through the red point and $\ell_g$ be the perpendicular line to $\ell$ through the green point.

[b][color=#f00]Claim.[/b][/color] Every point on $\ell_r$ is red, and every point on $\ell_g$ is green.

[i]Proof.[/i] Obviously, there cannot exist a blue point on $\ell_r$. Then, fix some blue point $B$, and consider the projection of $B$ onto $\ell_r$, say $P$. This point must be red. But taking the right triangle formed by $B$, $P$, and any other point $A \in \ell_r$, that point $A$ must also be red. Hence every point on $\ell_r$ is red, and a similar result follows from $\ell_g$. $\blacksquare$

Now, take any blue point $B$ in the plane, and draw two lines $a, b$ through $B$ that are perpendicular to each other. Then $a$ intersects $\ell_r$ at some red point $R$, and $b$ intersects $\ell_g$ at some green point $G$. It follows $RGB$ satisfies the desired condition.


Prove that:

(n!m!)/(m + n)! = (nC1)/(m + 1) - 2.(nC2)/(m + 2) + 3.(nC3)/(m + 3) - ... + (-1)^(n+1)n.(nCn)/(m + n)

NOTE: nC1 means n choose 1. I struggled with latex so I had to eventually use this notation.

You'll want to take a look at [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?search_id=689790046&t=242870]this thread[/url].

$ \text{\LaTeX}$:  $ \frac{n!m!}{(n+m)!} = \sum_{k=1}^{n} \frac{(-1)^{k-1}}{m+k} {n \choose k}$.

I meant:
\LaTeX : \frac{n!m!}{(n+m)!} = \sum_{k=1}^{n} \frac{(-1)^{k + 1}}{m+k}{k}{n \choose k}
What changes would that lead to?


If you want an easy book, "Art Of Problem Solving".
I think it is good since it has problems and solutions, too. Therefore you can learn a lot of strategies.

davron

You might try:

http://mathcircle.berkeley.edu/cnt.pdf
http://www.maths.gla.ac.uk/~ajb/dvi-ps/3q-notes.pdf
http://modular.fas.harvard.edu/ent/ent.pdf

For more, I suggest using this Google query (no quotes):

number theory filetype:pdf

Another good book is "Number Theory" by George Andrews.

A great journal I found today:

http://www.integers-ejcnt.org/

[quote="dieuhuynh"]i want to find book about number  theory  but it is hard .someone can help me?[/quote]

site http://www.animath.fr/ has book:


linear algebra
matrix
probability and stats

May anyone suggest an introductory text for Random Matrix Theory?

[url=http://www.amazon.com/Random-Matrices-Third-Applied-Mathematics/dp/0120884097]This[/url] is probably the most complete reference.  [url=http://www.umpa.ens-lyon.fr/%7Eaguionne/cours.pdf]Here[/url] are some very recent, and reasonably complete, lecture notes on the topic, covering many of the more recent developments.


calculus
integration
calculus computations

How do you find the integral of arcsin(x) dx ?

$ \int\arcsin xdx\equal{}x\arcsin x\minus{}\int\frac{x}{\sqrt{1\minus{}x^{2}}}dx$ and now its quite easy  :wink:

To get it, we require the "Integration by parts".

So take $ u\equal{}\arcsin x$ and $ dv\equal{}dx$

Integration by parts:

$ P \equal{}\int\arcsin x\,dx \equal{}\int1\cdot\arcsin x\,dx \equal{} x\arcsin x\minus{}\int x\cdot (\arcsin x)'\,dx$

$ P \equal{} x\arcsin x\minus{}\int\frac{x}{\sqrt{1\minus{}x^{2}}}\,dx \equal{} x\arcsin x\plus{}\frac{1}{2}\int(\minus{}2x)(1\minus{}x^{2})^{\minus{}1/2}\,dx$

$ P \equal{} x\arcsin x\plus{}\sqrt{1\minus{}x^{2}}\plus{}C$.


linear algebra
matrix
symmetry
analytic geometry
graphing lines
slope
trigonometry

Let $ A \equal{} \left(\begin{array}{cc}a & b \\
b & \minus{} a \end{array}\right)$ and $ f,\ g_1,\ g_2$ be the linear transeformations expressed by matrices $ A,\ I \plus{} A,\ I \minus{} A$ respectively, where $ I$ is identy matrix.

Soppose that $ g_1$ maps the whole plane to a line $ l_1$, prove the following contexts.

(1) $ a^2 \plus{} b^2 \equal{} 1$.

(2) $ g_2$ maps the whole plane to $ l_2$ which is perpendicular to $ l_1$.

(3) $ f$ is a symmetric transformation across the line $ l_1$.

$ 1.$ Since $ g_1$ maps the plane to a line we have $ \det(g_1)\equal{}0\equal{}1\minus{}a^2\minus{}b^2$, whence the part one follows.
$ 2.$ Since $ a^2\plus{}b^2\equal{}1$ we have $ \det(g_2)\equal{}0$, so it it maps the plane to $ l_2$. But $ l_1$ and $ l_2$ correspond to nonzero eigenvalues of $ g_1$ and $ g_2$, respectively, ie $ l_1\equal{}<(a\plus{}1, b)>, l_2\equal{}<(a\minus{}1, b)>$, so $ l_1$ and $ l_2$ are perpendicular.

PS.I had misunderstood the problem.

$ 3.$ Let $ B$ be the matrix of the symmetry with respect to the line $ l_1$, whose slope is $ m\equal{}\tan(\theta)\equal{}\frac{b}{a\plus{}1}$. Then $ B\equal{}\left(\begin{array}{cc} \cos(2\theta) & sin(2\theta)\\\sin(2\theta) & \minus{}\cos(2\theta)\end{array}\right)$. But $ \cos(2\theta)\equal{}a,\ \sin(2\theta)\equal{}b$, so $ A\equal{}B$, and we are done.

How about $ a\equal{}\minus{}1,\ b\equal{}0$?

In that case $ l_1\equal{}<(0,1)>$, and $ A\equal{}\left(\begin{array}{cc}\minus{}1 & 0 \\
0 & 1 \end{array}\right)\equal{}B$. (I hope I haven't left anything else;) )

That's right. Consequently, $ B$ is the symmetric transformation across the line $ y \equal{} (\tan \theta)x$.

Prove that :
If $ S$ adn $ T$ are both linear transformations and $ I\minus{}SoT$ is injective, then $ I\minus{}ToS$ is also injective.
$ I$ is the identity transformations

[quote="Qruni"]Prove that :
If $ S$ adn $ T$ are both linear transformations and $ I \minus{} SoT$ is injective, then $ I \minus{} ToS$ is also injective.
$ I$ is the identity transformations[/quote]
[url=http://www.mathlinks.ro/Forum/viewtopic.php?t=86370]http://www.mathlinks.ro/Forum/viewtopic.php?t=86370[/url]

[quote="Qruni"]Can you prove that : The amount of nonzero eigenvalues of $ A$ is less than rank $ A$ for any matrix $ A$ ?? Thank you for your explanation.[/quote]
It is better to ask such questions in a new topic, for example in [url=http://www.mathlinks.ro/index.php?f=353]this forum[/url].


Below I found an interesting story and fun for you guys to rate your 'mindless reading'. Have fun :lol:
------------------------------------------
By BEN FELLER, AP Education Writer
Mon Jul 3, 1:44 PM ET

WASHINGTON - Better pay attention, reader. This whole story may be a blur.

For the first time, researchers have demonstrated the ill effects of mindless reading — a phenomenon in which people take in sentence after sentence without really paying attention.

Ever read the same paragraph three times? Or get to the end of a page and realize you don't know what you just read?

That's mindless reading. It is the literary equivalent of driving for miles without remembering how you got there — something so common many people don't even notice it.

In a new study of college students, researchers from the University of Pittsburgh and the University of British Columbia established a way to study mindless reading in a lab.

Their findings showed that daydreaming has its costs.

The readers who zoned out most tended to do the worst on tests of reading comprehension — a significant, if not surprising, result. The study also suggested that zoning out caused the poor test results, as opposed to other possible factors, such as the complexity of the text or the task.

The researchers hope their work inspires more research into why zoning out happens, and what can be done to stop it.

For now, they want the problem to be taken seriously.

"When you talk about this work at conferences, it does lend itself to a lot of jokes," acknowledges University of Pittsburgh professor Erik Reichle, co-leader of the study.

"It's so ubiquitous. Everybody does it," he said. "I think that's one of the main reasons it's been overlooked. And there's been a view that it is tough to study experimentally. Hopefully, now, there will be more interest in the topic."

The federal government is showing some.

Reichle and fellow psychology professor Jonathan Schooler did the study on a $691,000 grant from the Institute of Education Sciences, an arm of the Education Department. It is one of 178 federally backed projects aimed at giving schools a scientific basis for sound policies.

Over three experiments, students used computers to read the first five chapters of Leo Tolstoy's "War and Peace." (Reichle wanted some boring reading — better for zoning out.)

Reichle said the dry text itself did not skew the results toward mindless wandering. After all, the students were on alert, unlike the typical reader.

Participants were told to monitor and report instances of zoning out as they read text on a computer. Half of them got computer reminders, too: "Were you zoning out?"

Despite all that, many still reported zoning out at a regular pace.

"That's the amazing thing," Reichle said. "It shows how often this can happen even under conditions that are designed to keep it from happening."

The students said as their eyes scanned the words, their minds often were elsewhere.

They were hungry, or thirsty, or tired. They were thinking about plans, worries or memories. Some drifted into fantasies. Others stuck with the book, but their minds wandered into tangents about the plot.

Karen Wixson, a nationally recognized reading expert and professor of education at the University of Michigan, cautioned not to read too much into all this.

"This is a long ways away from having implications for reading instruction," Wixson said. "It could, eventually, down the line. But to draw inferences about this as a contributing factor toward reading comprehension would be a huge, huge leap."

To apply to younger kids — the target audience of reading classes — the findings would have to be replicated among school-age children, Wixson said.

She said participants may have zoned more often because they were reading off computer screens, and because they had no real incentive to pay attention, as they would in school.

But at the International Reading Association, Cathy Roller sees some direct payoff. She directs research and policy for the association, which represents literacy professionals.

By recognizing zoning out as problem, she said, teachers can do something.

Like asking students to put a checkmark next to paragraphs as they finish them and then summarize what they just read.

Or having students scan all the pictures and bold type before reading the text of a story, so that they have a general understanding of what's to come.

Zoning out may simply mean that the prose isn't interesting, Roller said. But it could also be a clear signal that students don't understand the work.

"You don't want to generalize narrow studies into large implications," Roller said. "But zoning out is probably not a whole lot different than not comprehending. And telling people to start using some good comprehension strategies is not likely to do any harm."

Roller knows. She had just been zoning out while reading a literature review.

By the way, last sentence here. If you missed anything, there's no shame in rereading.
___

I only zone out when the math problems are really long and complicated.

But I sometimes read mindlessy when I read long or boring (to me) books.  But that hasn't hapenned to me in a while.

sometimes, i read the question, and i think "oh, that's mildly interesting." then i move on. then i remember, "oh maybe i was supposed to solve the problem."

I only zone out when reading about probability, combinatorics, or inequalities. In fact, I turn into this guy---> :wacko:

I don't really zone out when reading math, but on stuff like history  :oops:

It happens all the time.

Shouldn't "sometimes" be an option?
I zone out way more on literature and history, but sometimes I won't focus on math when I read over it (derivations, dirty/lengthy algebra...).

Maths at times is so boring that it is difficult not to zone out. Though I did not know that the thing was called zoning out.


geometry
geometric transformation
rotation
function
linear algebra
matrix
trigonometry

Find the formula of function $y=x^2$ after rotation round the point$(0; 0)$ by angle $60$

As far as I know, the use of matrix is not prohibited in IMO. :)
I assume the $60^\circ$ here means a anti-clockwise rotation.
$\begin{pmatrix} x' \\ y' \end{pmatrix}=\begin{pmatrix} \cos\theta & -\sin\theta \\ \sin\theta & \cos\theta \end{pmatrix}\begin{pmatrix} x \\ y \end{pmatrix}=\begin{pmatrix} \frac{x}{2}-\frac{\sqrt{3}y}{2} \\ \frac{\sqrt{3}x}{2}+\frac{y}{2} \end{pmatrix}$
Hmm, does this count as a formula? :huh:  :P

Yes.

You then can look at the 'function' $y'=x'^2$, which is a curve of the form $ax^2+bxy+c^2+dx+ey=0$

I have a similar method.  Here it goes.
Let $x$ and $x_n$ be the old and new x-values, respectively; $y$ and $y_n$ be the old and new y-values, respectively.

$\\\begin{pmatrix}x_n & y_n\end{pmatrix}=\begin{pmatrix}x & x^2\end{pmatrix}\begin{pmatrix}\cos\frac{\pi}{3} & \sin\frac{\pi}{3}\\-\sin\frac{\pi}{3} & \cos\frac{\pi}{3}\end{pmatrix}\\=\begin{pmatrix}x\cos\frac{\pi}{3}-x^2\sin\frac{\pi}{3} & x\sin\frac{\pi}{3}+x^2\cos\frac{\pi}{3}\end{pmatrix}\\ y_n=x\sin\frac{\pi}{3}+x^2\cos\frac{\pi}{3}\\ =\frac{\sqrt{3}}{2}x+\frac{1}{2}x^2\\\\ x_n=x\cos\frac{\pi}{3}-x^2\sin\frac{\pi}{3}\\ =\frac{1}{2}x-\frac{\sqrt{3}}{2}x^2\\ \sqrt{3}x_n=\frac{\sqrt{3}}{2}x-\frac{3}{2}x^2\\\\ \sqrt{3}x_n+y_n=\frac{\sqrt{3}}{2}x-\frac{3}{2}x^2+\frac{\sqrt{3}}{2}x+\frac{1}{2}x^2\\ =\sqrt{3}x-x^2\\\\ x=\frac{-\sqrt{3}\pm\sqrt{3-4(x_n\sqrt{3}+y_n)}}{-2}\implies\ell=-\sqrt{3}\pm\sqrt{3-4(x_n\sqrt{3}+y_n)}$
Placing $x=\frac{-\sqrt{3}\pm\sqrt{3-4(x_n\sqrt{3}+y_n)}}{-2}$ into $x_n=x\cos\frac{\pi}{3}-x^2\sin\frac{\pi}{3}$ gives us the new equation $y_n$ as a function of new $x_n$ values.
$\therefore$ our formula becomes
\[ 2\ell+\sqrt{3}\ell^2+8x_n=0  \]

Ahh man.  This took me so long to type...

Is this correct (I just learnt this type of rotation right after the problem)?

[quote="leepakhin"]As far as I know, the use of matrix is not prohibited in IMO. :) [/quote]
This is just using $e^{i\pi/3}$ in matrix notation, so...yeah...complex numbers.
Masoud Zargar

how did this get in the olympiad section? this is a straightforward theorem application


Alicia earns $ \$20$ per hour, of which $ 1.45\%$ is deducted to pay local taxes.  How many cents per hour of Alicia's wages are used to pay local taxes?

$ \textbf{(A)}\  0.0029 \qquad \textbf{(B)}\  0.029 \qquad \textbf{(C)}\  0.29 \qquad \textbf{(D)}\  2.9 \qquad \textbf{(E)}\ 29$

[hide="Solution"]C. Multiply $(0.0145)(20)$.
[/hide]

4everwise! How could you get it wrong! :mad:

[hide]E[/hide]

Jeez if I get this wrong...
[hide]20*0.0145 = 0.29, so $%Error. "block" is a bad command.
{C}$[/hide]

Wrong, Apprentice :D Read the question more carefully ...

Your answer IS wrong, Apprentice. The amount should be in cents, not in dollars (read carefully).

Lol this problem is so confusing (I got it wrong when I took the test too :blush:)

[hide]$(20)(.0145)(100)=$$\$ 29 \Leftrightarrow E$[/hide]

[quote="chess64"]Lol this problem is so confusing (I got it wrong when I took the test too :blush:)

[hide]$(20)(.0145)(100)=$$\$ 29 \Leftrightarrow C$[/hide][/quote]

It should've been E.

[hide]The tax is $0.0145\cdot\$20=\$0.29$, which is 29 cents, so $\boxed E$[/hide]

Wow, so the first problem is the hardest problem on the test... I will take note of that  :jump:

[hide]E.[/hide]

By the way, the first 5 or so problems of the AMC are all basically about reading carefully.

[hide]A cheapshot: you can't earn a fraction of a cent.  So the answer is E.[/hide].


abstract algebra
parameterization
induction
Ring Theory
advanced fields
advanced fields unsolved

Let $ (A, \mathfrak{m})$ be a local Noetherian ring of dimension $ d$. Let $ x_1,\ldots,x_d$ be a system of parameters and $ M$ an $ A$-module. Then $ H_{\mathfrak{m}}^i(M) \equal{} 0$ for all $ i < d$ and $ H_{\mathfrak{m}}^d(M) \neq 0$ implies that $ x_1,\ldots,x_d$ is an $ M$-quasi-regular sequence.

Basically this proposition tells you that the depth of $ M$ is maximal, i.e. the dimension $ d$ of the regular Ring $ A$. Note that in general the depth of $ M$ is bounded by the dimension of the ring and it is the smallest integer $ n$ such that $ H^n_\mathfrak{m}(M)\neq 0$. All this invokes dimension theory and the machinery of cohomological algebra of modules over commutative rings.

Get your cohomological engine started and prove the claim by induction on $ d$:

The case $ d \equal{} 0$ is trivial ($ A$ is a field) and $ d \equal{} 1$ is easy too, since then $ A$ is a discrete valuation ring.

Assume $ M\neq 0$ and  $ d > 1$. For the induction step, pick  one of the $ x_i$'s, say $ x_d$, which is not a zerodivisor of $ M$. If it should not exist, $ \mathfrak{m}$ consists of non-zerodivisors only so that $ \mathfrak{m}$ is an associated prime ideal and $ M$ is zero-dimensional, i.e. $ \dim A/Ann(M) \equal{} 0$. Thus $ M$ sits at the closed point of $ Spec(A)$, but there it is zero ($ H^0_\mathfrak{m}(M) \equal{} 0$) which implies that $ M \equal{} 0$ - a contradiction. Use the short exact sequence $ 0\rightarrow M\stackrel{x_d}{\rightarrow} M\rightarrow M/x_dM\rightarrow 0$ to apply the technique of a cohomological shift. Apply dimension theory to establish the situation such that you can use the induction hypothesis.


induction
inequalities
inequalities unsolved

(2^0/0!) + (2^1/1!)+...+ (2^n/n!) <= 8-(2^n/n!) for n>=3

I am having a hard time proving the induction step.  I think I am doing some algebra error. Because in the end I get n-3>=0 for when I try to show that 8-((2^n/n!) - (2^(n+1)/(n+1)!)) <= 8-((2^(n+1)/(n+1)!)) and while this is true, the next question (B) Prove that in fact the inequality holds for all integers n>=0. And if we extend our basis this is easy to prove, but when i try and prove the same inequality for n>=0, with my current algebra, I get n-3>=0 and given the updated range of N, you can see it does not work. Any help would be enjoyed.

[quote="Juliantg"]
$\frac{2^0}{0!} + \frac{2^1}{1!}+...+ \frac{2^n}{n!} \leq 8-\frac{2^n}{n!}$ for $n\geq 3$
[/quote]
$S_n = \sum_{k=0}^{n}{\frac{2^k}{k!}}$

$S_n\leq 8-\frac{2^n}{n!} \Leftrightarrow \boxed{8-S_n\geq \frac{2^n}{n!}} \ \ (1)$

for $n=3$ it is true

$S_3 = \frac{2^0}{0!}+\frac{2^1}{1!}+\frac{2^2}{2!}+\frac{2^3}{3!} = 1+2+2+\frac{8}{9}=5+\frac{8}{9}<8-\frac{8}{9}$

Asume that $(1)$ holds for $n$

$8-S_n\geq \frac{2^n}{n!} \Rightarrow$

$8-S_n-\frac{2^{n+1}}{(n+1)!}\geq \frac{2^n}{n!}-\frac{2^{n+1}}{(n+1)!} \Rightarrow$

$\boxed{8-S_{n+1} \geq \frac{2^n}{n!}-\frac{2^{n+1}}{(n+1)!}}$

$\boxed{\frac{2^n}{n!}-\frac{2^{n+1}}{(n+1)!}} = \frac{(n+1)2^n}{(n+1)!}-\frac{2\cdot 2^n}{(n+1)!} = \frac{(n-1)2^n}{(n+1)!}$

Since $n-1\geq 2^{(*)}$ we get $\frac{(n-1)2^n}{(n+1)!}\geq \frac{2 \cdot 2^n}{(n+1)!} = \boxed{\frac{ 2^{n+1}}{(n+1)!}}$

$\Rightarrow 8-S_{n+1}\geq \frac{2^{n+1}}{(n+1)!}$

So, $(1)$ holds for $n+1$

[b][u]Note[/u][/b] $(*)$

We got a not-strong inequality ($\geq$). This doesn't mean that we have equality for some $n$.
For example, if we start from $n=4$ we have $n-1>2$, so the symbol $\geq$ is replaced by $>$
and we saw that for $n=3$ we have [u]strong[/u] inequality.

Also, it's easy too see that $S_n<8-\frac{2^n}{n!}$ for $n=0,1,2$

So, for every natural number $n$ (including zero of course) we have $S_n<8-\frac{2^n}{n!}$

:oops_sign: you are right! sorry  :oops:

but it's still $5+\frac{8}{6}<8-\frac{8}{6}$

I'll check it this time  :D

$5+\frac{8}{6} = \frac{38}{6}$

$8-\frac{8}{6} = \frac{40}{6}$


There exists a polynomial $P$ of degree $5$ with the following property: if $z$ is a complex number such that $z^{5}+2004z=1$, then $P(z^{2})=0$. Find all such polynomials $P$


linear algebra
matrix
algebra
polynomial
superior algebra
superior algebra solved

Let A,B belong to M_3(R) so that det(AB-BA)=0. Prove that for every real numbers x,y with x+y<>0 we have the realation:

det(I_3+(xAB+yBA)/(x+y))=det(I_3+(yAB+xBA)/(x+y))

so let's prove that
det(I+kAB+(1-k)BA)=det(I+(1-k)AB+kBA)

Q(k)=det(I+kAB+(1-k)BA)=det(I+BA+k(AB-BA))
polynomial deg Q  \leq 3
with det(AB-BA)=0
Q(k)=ak+bk+c
Q(0)=Q(1) => Q(k)=ak-k+c = Q(1-k)

Omarjee


[quote="AIME 1984 #14"]
What is the largest even integer that cannot be written as the sum of two odd composite positive integers?
[/quote]

I know the answer but not the solution. How does one approach this problem?

I would go off from 1000 down, since

1. AIME answers from 0-999 in integer,
2. Since it is #14, brute forcing works just great.

However, there has got to be some kind of solution. I am assuming u looked at Kalva's solution, which i do not get either. Some sort of I-E method and McNugget principle, but to do that we have too many odd primes.....

Explanation of Kalva solution:

6n+18 = 9 + 6n+9,
6n+44 = 35 + 6n+9,
6n+34 = 25 + 6n+9.

We have 38 working, and 39 to 44 not working.  If x > 44 works, then consider it modulo 6.  Since it is even, it is either 0 2 or 4.

But now from the above, is a valid decomposition of x into the sum of two odd composite numbers.

Avoiding the Kalva solution and going to the problem book solution (which I think is the most elegant), find a composite ending in each (odd) digit:

21
33
15
27
9

Add 10n+5 (i.e., a number greater than 5 ending in 5) to each of these and you will find that you can make all even numbers greater than these ending in the same digit:

34
48
30
42
24

The first number not accounted for is 38, which we test to see that it is, in fact, the answer.

Or you could do it my way (which I still like even though it's not as fast): The first three consecutive composites are 91, 93, 95. If you add something to 91, that minus two to 93, and that minus 4 to 95, one set has to be two composites. Therefore, no numbers greater than 98 work. Then brute force it (which is surprisingly easy).

Thanks guys. I think I actually understand now :idea: How about #13, which I think is even harder?

[quote="Kalva's version of AIME 1984 #13"]
Find $10 \cot( \cot^{-1} 3 + \cot^{-1} 7 + \cot^{-1}13 + \cot^{-1}21)$.
[/quote]

A slightly quicker derivation for $\cot{(a+b)}$ is

$\cot{(a+b)} = \frac{1}{\tan{(a+b)}} = \frac{1-\tan{a}\tan{b}}{\tan{a}+\tan{b}}$

if you know the $\tan{(a+b)}$ formula, which you should. Then $\tan{(\cot^{-1}{x})} = \frac{1}{x}$.


AMC
AIME
MATHCOUNTS
USA(J)MO
USAMO
geometry
geometric transformation

[quote="bpms"]Just wondering how Floridians have done so far on AMC and AIME.[/quote]

i got bored on the 10B and randomlly started guessing and got a 108, woulda had a 117 if i hadnt

was sick during the A

ahhh i failed the aime

could've had a 7, pretty sure i could've made the floor with that.. -_-

What are the AMC/AIME's like?  I didn't know I was eligible for math contests other than mathcounts, but I've only prepared for MC.  How should I get ready for next year's AMC/AIME?

[quote="indianamath"][quote="Pakman2012"]owned the balls out of everything. don't worry bout it.

i think we're getting fewer and fewer qualifiers each year. which is just ridiculous

Guaranteed qualifiers this year:
Saiem Gilani
Chenyu Lin
Gaku Liu
Carolyn Kim
Preet Patel
Rui Yang
Jason LaRue
David Jia (we think)
Vincent Dang (we think)
Marcel Nunez
Wendy Hou

Then we have our maybes
Cheran "Nao Nao" Wu
Louie Wu
Jonathan Mei
Michael Duckett
Karlanna Lewis

I'm sure a couple other people are somewhere on the list.
Its just not looking very good. We really ought to be getting like.. 20-25 qualifiers a year.[/quote]
didn't karlanna lewis make the Countdown at the 2005 Nats?(and no, i am not supposed to be here.)[/quote]
yes, she was a semifinalist

shes only in 10th.  she doesnt do much math anymore either.  only that ballet crap.  she didnt come to 1 competition all year.

Wait, why aren't Yannik, Karlanna, and Vincent listed under USAMOers?

[quote="solafidefarms"]I don't see any Yannik, Karlanna, or Vincent on the AMC website as USAMOers...[/quote]
Yes, that's what I was asking about.

Hey this is Yannik.

Yeah I didn't get to officially take the AIME this year so my scores didn't count.

Long story short...my school has this weird rule that I couldn't do math contests because I'm not taking any math classes.  So I was disqualified.  *I was also out of state for both AIME dates...which didn't help matters much.*

But if you cared...I had a 132 AMC 12B and a 9 AIME I (unofficial).  Kind of lame that I didn't get to try and make USAMO again, I guess.  Whatever.

"being good at mathcounts has NOTHING to do with the AIME"

Good- I didn't make national mathcounts and want to make USAMO.

Well, I'll go take the 2007 AMC/AIME (or try to- does anyone have a link?) soon and report my scores.

[quote="diophantient"][quote="junggi"][quote="indianamath"]how is she a "maybe?"[/quote]
being good at mathcounts has NOTHING to do with the AIME[/quote]
Then why did 5 of the top 10 written at nationals last year make USAMO?[/quote]
whats your point

[quote="junggi"][/quote][quote="diophantient"][quote="junggi"][quote="indianamath"]how is she a "maybe?"[/quote]
being good at mathcounts has NOTHING to do with the AIME[/quote]
Then why did 5 of the top 10 written at nationals last year make USAMO?[/quote][quote="junggi"]
whats your point[/quote]

Don't start guys.
The point here is, there are many people who will be good at math, no matter what competition it is.

Indianamath is just pointing out that you are not going to do really well on the AIME just because you did well on MathCounts.

And dioph. has a good point as well.

And junggi is junggi.

And wait, isnt this the Florida forum.  No outta staters fighting on here.  :P

Just leave it there.

[quote="Walk Around The River"][/quote][quote="junggi"][/quote][quote="diophantient"][quote="junggi"][quote="indianamath"]how is she a "maybe?"[/quote]
being good at mathcounts has NOTHING to do with the AIME[/quote]
Then why did 5 of the top 10 written at nationals last year make USAMO?[/quote][quote="junggi"]
whats your point[/quote][quote="Walk Around The River"]

Don't start guys.
The point here is, there are many people who will be good at math, no matter what competition it is.

Indianamath is just pointing out that you are not going to do really well on the AIME just because you did well on MathCounts.

And dioph. has a good point as well.

And junggi is junggi.

And wait, isnt this the Florida forum.  No outta staters fighting on here.  :P

Just leave it there.[/quote]
eh what?

let them fight

its funny

got a 4 on the 2007 AIME II, will do amc 10 soon

[quote="junggi"][quote="Walk Around The River"]
Indianamath is just pointing out that you are not going to do really well on the AIME just because you did well on MathCounts.
[/quote]
um no that was me
indianamath expressed shock and disgust at the fact that a good mcer is labeled maybe
i relieved him of that shock and disgust by making a statement
but seriously i think it has been quite some time since i made that statement
why are we talking about it
*cough* dioph *cough*
ok i will stfu and get out now[/quote]

lol. you need to be a stand-up comedian. :rotfl:

I think it should be the other way around:

your AIME level usually reflects how well you do at MC. Unless, of course you're super-slow at everything and you suck at easy problems.

[quote="indianamath"]
your AIME level usually reflects how well you do at MC. Unless, of course you're super-slow at everything and you suck at easy problems.[/quote]
No.
THat's complete bull.
Just because you pwn the AIME doesn't mean jack about MC.
Even if you have speed.
I got a 0 on the AIME last year and 27th at nationals.
This year I got an 8 on the AIME and not top 57 at nationals.
[quote="junggi"]*cough* dioph *cough* [/quote]
I agree that just because you pwned mathcounts doesn't mean jack about AIME, but you can't say that they have NOTHING to do with each other.

[quote="diophantient"]
[quote="junggi"]*cough* dioph *cough* [/quote]
I agree that just because you pwned mathcounts doesn't mean jack about AIME, but you can't say that they have NOTHING to do with each other.[/quote]
ok fine
i was particularly emo that day probably(seeing as it was less than a month after the aime) so i exaggerated a bit
oh well

Hmmm...I can't edit that post.

2007 AMC 10A - 119.5
2007 AIME II   -    4

I just need to get that AIME up to 6-8 to qualify for USAMO, right?

Does AMC 12 use knowledge of Algebra II and/or Precalculus?

If so, how much?  (Is it a waste of time to do AMC 12 if you're only in algebra II?)

[quote="The Herald of Doom"]hello floridians

are you actually from florida

florida is very much like a tomato

they both are kinda red and squishy

yay for florida[/quote]
???? are you just random, or are you just insulting us?

he did something similar in the MA forum

Lol, you guys don't even know Karlanna so why are you expressing "shock and disgust" about how she's considered a "maybe"? I've been good friends with Karlanna for a few years now and we were on the same MathCounts team in middle school. I have no doubt that she's capable of making the USAMO, but there are circumstances that must be taken into account, which I will not go into too much detail on a public forum. I'm sure there is a correlation between MC and AMC, but it is not a guarantee. Saiem, since he goes to Rickards and is a Floridian, clearly understands the situation better than the rest of you and the predictions were largely on track.

Vincent Dang didn't make it because he got a 4 on the AIME, but that was unforeseeable since he aced the AMC 12B. Everybody else in the guaranteed qualifiers indeed made it to the USAMO. The rest of the maybes did not, except for Naonao, but he is also a special case. Due to the obscenely low cutoff score and after the 20 people from the 5% club were disqualifed from the USAMO qualifiers list, Naonao was one of the 19 or so people who became late qualifiers, and even then he only made it by about 0.5 points. Overall, I'd say the predictions were pretty good. Some people who were left out include Brent and Jimmy and Chen Ye, the third of whom only recently came from China and was again unforeseeable.

[quote="hunter34"]I think the admins did something to his/her account, now there is "Herald of Doom" instead of "The Herald of Doom".[/quote]

HE WAS ARNAV SPAMMING

[quote="dgreenb801"]I am 9th grader from Hollywood,Florida (near Miami) and missed the USAMO with 6 on the AIME (two careless mistakes) and a 127.5 on the AMC10  :( . Only Rui Yang, a 12th grader from Miami Palmetto made it from south Florida this year. I go to Nova High School, one other kid from my school(12th grade) made the AIME with an AMC 12 score of 103. I was just wondering, is there anyone else in South Florida that is into the AMC, AIME, etc?[/quote]

Well, I don't know what is really considered S. Florida, since i'm from Tally, but isn't Vero Beach down there?

[hide=" :ninja:  :ninja: "]Huh What is this?? WHat?[/hide]

[hide]Awesome avatars. How do people find those?? Everything i find is too big[/hide]

sob :( I dont even knoew What the AIME nd AMC are!!!!!!!!!!!!!!  :huh:


Is there any way to disable the spell checker? BTW, is this a new feature?

There is no spell checker on the forum. You must have installed one in your browser.

If you have Firefox 2.0.0.3, then go to Tools >> Options >> Advanced, and you should find the "Check my spelling as I type" option there.

Thanks! I tried to type a sequence of letters, but the annoying red line kept on coming.


linear algebra
matrix
linear algebra unsolved

Find the number of matrix $ A\in M_{n}$with properties:
1) all element are 1 or -1
2) $ Tr(A)\equal{}0$
Obvious n is odd

You must see in how many posibilities you can put $ \frac n2$ of  $ \minus{}1$s on the principal diagonal and all the other elements can be $ \pm 1$. So you get $ C_n^{n/2}\cdot 2^{n^2\minus{}n}$


Does there exist a sequence $ a_{1}, a_{2}, ... ,a_{2010}$ of positive reals, such that for all integer $ k$, $ 1\le$ $ k$ $ \le 2010$ all the roots $ z$ of the polynomial $ p_{k}(x) \equal{} a_{k \plus{} 2009}x^{2009} \plus{} a_{k \plus{} 2008}x^{2008} \plus{} \cdots \plus{} a_{k \plus{} 1}x \plus{} a_k$ satisfy $ |Im(z)| \le |Re(z)|$? Here $ a_{i \plus{} 2010} \equal{} a_{i}$, for $ i \ge 1$.

The answer is NO!

Let $ n \equal{} 2010$. Assume there exists such polynomials.
Fix $ k$ and let $ x_i \equal{} \alpha_i \plus{} \beta_i\sqrt { \minus{} 1},1\le i\le n \minus{} 1$ be all the roots to $ p_k(x)$.
Notice $ \sum_ix_i^2$ is a real number.
Hence, by assumption, we have
$ \sum_ix_i^2 \equal{} \sum_i(\alpha_i^2 \minus{} \beta_i^2)\ge0$, or
$ \sum_ix_i^2 \equal{} (\sum_ix_i)^2 \minus{} 2\sum_{i < j}x_ix_j \equal{} (\frac {a_{k \plus{} n \minus{} 2}}{a_{k \plus{} n \minus{} 1}})^2 \minus{} 2\frac {a_{k \plus{} n \minus{} 3}}{a_{k \plus{} n \minus{} 1}}\ge0$, or
$ 2\frac {a_{k \plus{} n \minus{} 1}}{a_{k \plus{} n \minus{} 2}}\le \frac {a_{k \plus{} n \minus{} 2}}{a_{k \plus{} n \minus{} 3}}$.

Let $ k$ go through $ 1$ to $ n$. We will have
$ 2^{n \minus{} 1}\frac {a_2}{a_1}\le \frac {a_2}{a_1}$. Contradiction!


I have not seen any yet. I think I got an 11, but not sure.

Team or Individual? I have figured them all out. Which ones do you want to talk about?

I think my school did regional. Which problem?

are you allowed to start discussing mandelbrot?
I think schools had til today to take the test(our school took it today).

Some schools are still taking the Mandelbrot this Saturday (ICAE, for instance). Don't discuss the questions until next week!


how much do u enjoy doing math? do u actually like it?

Judging by the results so far a 10 must mean that you hate math  :D  ;)   :P

Ooookay, if everyone's interpreting it that way, I'll say 10.

[quote="EFuzzy"]lol, why would someone come here if they don't love math?[/quote]

what are u talking about?......i don't LOVE math....... :?

[quote="h_s_potter2002"][quote="KBabe"]btw, 10 means u hate math and 1 means u love math.[/quote]

I'm pretty sure you should change that.  I think everyone thought it was the other way around.  It would be pretty weird if over 2/3 of the members of a math forum hate math.[/quote]

i did too.....at first....but anyways, 10 means u love math and 1 means u hate math. :blush:  :)

This is pretty rediculous I mean all the active members here post in the math forums and apparently love math so why even post this poll?

What if somebody's girlfriend or boyfriend broke up with them because of math?

[quote="LynnelleYe"]Could be their parents make them come or something. I know that's what's up with some people in my school's MathCounts club.[/quote]

...my friends' parents make them come, some of them seriously HATES math.....

Having your parents make ou do math always makes it a whole lot less fun.

i'm not as smart as most of you but i would assume that the members of a math forum (based on a math book) would enjoy math a whole lot so this topic should not be necessary. but what do i know? :D

I voted w/o finding out whether or not 10 is u luv math or u hate it... should make it clear.....

I like math okay... it's not like I can't live without doing math problems.... but it's more like I like how u feel when you find the answer to a really hard question, and besides, after you start competing, you can't really stop b/c ppl except you to be good at it.....

BRUH IM THE ONLY ONE WHO VOTED ON THIS POLL LIKE ITS BEEN 15 YEARS WHAAAAAT.
BUMPITY BUMP BUMP :rotfl:


chứng minh rằng tích phân suy rộng sau không phụ thuộc $ \alpha$
$ \int \limits_{0}{+\infty}\frac{1}{(1+x^\alpha)(1+x^2)}$

[quote="vanky"]chứng minh rằng tích phân suy rộng sau không phụ thuộc $ \alpha$
$ \int \limits_{0}{ + \infty}\frac {1}{(1 + x^\alpha)(1 + x^2)}$[/quote]
Đây là 1 bài trong kỳ thi olympic toán sv thì phải.Bài này bạn thử đặt t=1/x đc 1 tích phân mới bằng tich phân ban đầu cộng tích phân này với tích phân ban đầu ta đc đpcm.Mình ko ranh gõ công thức toán trên diễn đàn này nên ko thể trình bày lời giải cụ thể đc.Thông cảm.Có ji ko hiểu thì đối đáp sau nhé.


Of the 60 students in the drama club, 36 take mathematics, 27 take physics and 20 students take both mathematics and physics. How many drama club students take neither mathematics nor physics?

There is a total of 36+27-20=43 people who actually take math and/or science. Thus, there are 60-43=17 people who dont take math ( :(  ) and/or science at all.

The reason for the above numbers is clear if you imagine a Venn diagram.  Think of two circles, overlapping a bit, one for math and one for physics.  The overlapping part has 20 people, so if I add 36+27, I counted those 20 people twice.  So, subtract off 20 to fix that problem.  That's how dragon arrived at 36+27-20 people taking math and/or science.

[asy]size(200);
draw(circle((4,0), 8));
draw(circle((-4,0), 8));
label("16", (-6,0), N);
label("7", (6,0), N);
label("20", origin, N);
label("Math", (-4,8), N);
label("Physics", (4,8), N);[/asy]

Maybe I wasn't that clear, but heres a Venn diagram I made just for practice on Asy.


A 28 on a Nats Sprint? I think I got 22 on the 90-91 Nats Sprint :( :( .

Ok, these are the "kind of" rules for posting on out-dated topics (>3 weeks):

a:you must have a question about that particular question
b:if it's like more than a year old, just redo the question on a new topic
c:don't bring it back if you're gonna spam, and even more if it doesn't add at all to anything (aka: you have a test that old?)

thanks,
jorian

[hide="Answer - 29"]
Subtracting, you know that $50\%$ is 1200. So, 100% of the tank is 2400. You subtract that from 3700, which gives you $\boxed{1300}$[/hide]

[quote="yhh3002"]You have Nat. that old? I have ones that are from 1987...  :D[/quote]

i have every single one.(access software)

[quote="yhh3002"]You have Nat. that old? I have ones that are from 1987...  :D[/quote]
Dude, DO NOT AWAKEN OLD THREADS TO SPAM IN.
This thread is soooo old, it's even older than Tare.
It's pointless to awaken old threads, you really shouldn't.


absolute value
algebra
linear equation
graph theory
combinatorics unsolved
combinatorics

A finite graph is connected. A positive real number is assigned to each point. Each point is colored red or blue and at least one point is colored red. Show that, given a knowledge of (1) the points and edges of the graph, (2) the number assigned to each red point, and (3) for each blue point the average of the numbers for the points joined to it, one can find the number assigned to each point.

What about the following gragh? Zombies A, B, C, and D (in that order) make a square and hold hands with the zombies to their left and right. A and C are red (because they are covered in blood), while B and D are blue (because they are dead). The number of people eaten by A is three, and the number eaten by C is seven. The average of the number of eaten people for zombies holding hands with B is 5, same for D.

Now go ahead and tell me how many people B and D have eaten.

Edit: ok, just in case I misinterpreted your question, let's say this time that the zombies take themselves into account with the average. Then since this is such a ridiculous possibility, I'm going to let B and D grow extra arms and hold hands with each other as well.

[hide]
Well, we get a lot of linear equations. For example, we get like $ a_1 \equal{} \frac{a_2 \plus{} a_3 \plus{} 3}{3}$ if we have a graph where 3 points are colored blue and a_1 is the number of the first vertice colored blue, and then he is connected to vertice 2, vertice 3, and a red vertice with number 3. We want to show that these linear equations have at most 1 solution. Assume they have 2 solutions. Then, if b_1, b_2 ,b_3 are the 3 solutions, our equation turns into $ b_1 \equal{} \frac{b_2 \plus{} b_3 \plus{} 3}{3}.$ We subtract these equations to get the equation $ c_1 \equal{} \frac{c_2 \plus{} c_3}{3}$ where c_1 = a_1-b_1. Note that the "+3" disappeared.  We now consider the c_k with greatest absolute value. For example, lets say c_1 has the greatest absolute value and we have the equation $ c_1 \equal{} \frac{c_2 \plus{} c_3\plus{}c_4\plus{}c_5}{4}.$ This would imply that c_2 and c_3 and c_4 and c_5 have the same absolute value. Since the graph is connected, we have that all the c_ks have the same absolute value. However, we know that there is some linear equation where the right side is like $ \frac{c_2 \plus{} c_3\plus{}c_4\plus{}c_5}{5}$ instead of $ \frac{c_2 \plus{} c_3\plus{}c_4\plus{}c_5}{4}$ because there is some red vertice. This implies that if k is the absolute value of all the c_i,  then k = 0.

[/hide]

I am not very good at explaining this...

pythag011, the problem never said that the value at any blue point is the average of its neighbors' values.  It just said that you know the average of the neighbors' values.

I'm as confused as zeb.  If the averages around the blue points don't include the points themselves, a graph consisting of a red point connected to a blue point is a counterexample.  If the averages include the points themselves, a complete graph consisting of a red point connected to two blue points is a counterexample, and these are the two most trivial cases.

I'll just leave the link to the previous discussion on the same problem.

http://www.mathlinks.ro/viewtopic.php?t=57327


trigonometry
logarithms
calculus
derivative
calculus computations

Answer the following question.

[1] Let $f(x)=\sqrt{\sqrt{x}+\frac{1}{\sqrt{x}}}\ (x>0)$, find the value of $f'(4)$

[2] $\frac{d}{dx} \frac{\sin (2x+1)}{1-x}$

[3] $\frac{d}{dx} x^{e^x}\ (x>0)$

[4] Sinsyu University Faculty of Textile and Sience technology

(1) $\frac{d}{dx} \ln (\cos ax)$

(2) $\frac{d}{dx} \frac{1}{\cos ^ 2 (ax)}$

(3) Given $\tan y=x$, find the value of  $\frac{d^2y}{dx^2}$  at  $y=\frac{\pi}{4}$

[4]

(1) $\displaystyle \frac{1}{\cos ax}(-\sin ax) = -\tan ax$

(2) $\displaystyle \frac{d}{dx} \frac{1}{\cos ^ 2 (ax)} = \frac{d}{dx} \cos ^ {-2} (ax) = -2\cos ^ {-3} (ax)(-\sin ax) = \frac{2\sin ax}{\cos ^ 3 (ax)} = \frac{2\tan ax}{\cos ^ 2 (ax)}$

[quote="ambierona"][4]

(1) $\displaystyle \frac{1}{\cos ax}(-\sin ax) = -\tan ax$ :(

(2) $\displaystyle \frac{d}{dx} \frac{1}{\cos ^ 2 (ax)} = \frac{d}{dx} \cos ^ {-2} (ax) = -2\cos ^ {-3} (ax)(-\sin ax) = \frac{2\sin ax}{\cos ^ 3 (ax)} = \frac{2\tan ax}{\cos ^ 2 (ax)}$ :) [/quote]

[quote="adidasty"](1) $-a\tan ax$[/quote] :)

Whoops, I always forget the inside derivative  :(

[quote="adidasty"]One more try at 3.

[3] $x^{e^{x}}\left(\frac{e^{x}}{x}+e^{x}\ln x\right)$[/quote] :)


$ a,b,c$ are positive real numbers.

Prove that

$ (a\plus{}b\plus{}c)^{3}\ge a^{3}\plus{}b^{3}\plus{}c^{3}\plus{}24abc$

 :D

[hide="Solution"]
Expand $ (a+b+c)^3$: $ a^3+b^3+c^3+3ab^2+3ac^2 +3ba^2+3bc^2+3ca^2+3cb^2 + 6abc$.

Now we have to prove: $ a^3+b^3+c^3+3ab^2+3ac^2 +3ba^2+3bc^2+3ca^2+3cb^2 + 6abc\ge a^3+b^3+c^3 + 24abc$

Subtract the cubes.

\begin{eqnarray*}
3ab^2+3ac^2 +3ba^2+3bc^2+3ca^2+3cb^2 + 6abc&\ge&24abc\\
3(ab^2+ac^2 +ba^2+bc^2+ca^2+cb^2)+6abc&\ge&24abc\\
3(ab^2+ac^2 +ba^2+bc^2+ca^2+cb^2)&\ge& 18abc\\
ab^2+ac^2 +ba^2+bc^2+ca^2+cb^2&\ge& 6abc
\end{eqnarray*}

Which is true by applying AM-GM on the LHS.

[/hide]

[quote="PKMathew"]$ a,b,c$ are positive real numbers.

Prove that

$ (a \plus{} b \plus{} c)^{3}\ge a^{3} \plus{} b^{3} \plus{} c^{3} \plus{} 24abc$

 :D[/quote]

[hide="Another sol."]

$ (a \plus{} b \plus{} c)^{3}\equal{}a^3\plus{}b^3\plus{}c^3\plus{}3(a\plus{}b)(b\plus{}c)(c\plus{}a)$, and

$ (a\plus{}b)(b\plus{}c)(c\plus{}a)\ge 8abc$  by using AM_GM.

So it is obviously true.
 :P
[/hide]

$ LHS\equal{}(a\plus{}b\plus{}c)^3\equal{}a^3\plus{}b^3\plus{}c^2\plus{}3(a\plus{}b)(b\plus{}c)(c\plus{}a)\ge$ $ a^3\plus{}b^3\plus{}c^3\plus{}3\cdot 2\sqrt{ab}\cdot 2\sqrt{bc}\cdot 2\sqrt{ca}\equal{}RHS$


limit
trigonometry
integration
calculus
real analysis
calculus computations

[quote="Lovasz"]Find:
\[ \lim_{n \to \infty}\frac{1}{n}\left( \sin{\frac{\pi}{n}}+\sin{\frac{2\pi}{n}}+\cdots+\sin{\frac{(n-1)\pi}{n}} \right).  \][/quote]
\[ \lim_{n \to \infty}\frac{1}{n}\left( \sin{\frac{\pi}{n}}+\sin{\frac{2\pi}{n}}+\cdots+\sin{\frac{(n-1)\pi}{n}} \right)=\int_{0}^{1}sin(\pi x)dx =\frac{2}{\pi} \]

how did you know that the integral goes from 0 to 1?

Write it in series form. What are you using as $\Delta x$ in the Riemann sum? How does it vary? That's how you find the bounds.

Suppose $\Delta x$ is a small positive number, $x_0 \in \mathbb{R}$, and $x_1=x_0 + \Delta x, x_2 = x_0 + 2 \Delta x, \ldots , x_{n}= x_0 + n \Delta x$.  Also suppose $f: \mathbb{R} \to \mathbb{R}$ is Riemann integrable on $[x_0,x_n]$.

Then

$f(x_0)\Delta x + f(x_1) \Delta x + f(x_2)\Delta x + \cdots + f(x_{n-1})\Delta x$ is a Riemann sum that approximates $\int_{x_0}^{x_n} f(x) dx$.

$(1/n) \left ( \sin{\frac{0\pi}{n}} +\sin{\frac{1\pi}{n}} + \sin{\frac{2\pi}{n}} + \cdots + \sin{\frac{(n-1)\pi}{n}} \right)$ is a Riemann sum with $\Delta x = \frac{1}{n}, x_0=0$, and $x_n = 0 + n \Delta x = 1$.  (And $f(x) = \sin{\pi x}$.)

So this Riemann sum approximates $\int_0^1 \sin{\pi x} dx$.


In the circle below, $m\angle M=37^{\circ}$ and $m\angle P=89^{\circ}$. If $MP=NQ$, find the number of degrees in arc $NQ$.

[quote="rnwang2"][hide]
$\frac{2\cdot (180-37-89)}{2}=\boxed{54^{\circ}}$.[/hide][/quote]
I am not sure if this is the same method as you gave but here is my method:
[hide] Say that the two lines intersect at Point A.  This means Angle QAN=180-37-89=54 degrees.  This means that Angle PAQ=180-54=126 degrees.  Now, since Angle M=Angle N, Angle P=Angle Q, and MP=NQ, we have that Triangle MAP is congruent to Triangle NAQ.  Hence, AP=AQ, so Angle APQ=Angle AQP.  Thus, angle APQ=(180-126)/2=27.  Therefore, Arc NQ makes an angle of 2(27)=$54$ degrees, which is our answer.  [/hide]

[quote="The QuattoMaster 6000"]Say that the two lines intersect at Point A.  This means Angle QAN=180-37-89=54 degrees.  This means that Angle PAQ=180-54=126 degrees.  Now, since Angle M=Angle N, Angle P=Angle Q, and MP=NQ, we have that Triangle MAP is congruent to Triangle NAQ.  Hence, AP=AQ, so Angle APQ=Angle AQP.  Thus, angle APQ=(180-126)/2=27.  Therefore, Arc NQ makes an angle of 2(27)=$54$ degrees, which is our answer. [/quote]
That is equivalent to what I did.

I'm pretty sure you use the following theorem without proof:

Given chords $AB$ and $CD$ of a circle such that $AB=CD$, minor arcs $AB$ and $CD$ are equal.

Note that the converse of this statement is also true.


Two intersecting circles $C_{1}$ and $C_{2}$ have a common tangent which touches $C_{1}$ at $P$ and $C_{2}$ at $Q$. The two circles intersect at $M$ and $N$, where $N$ is nearer to $PQ$ than $M$ is. Prove that the triangles $MNP$ and $MNQ$ have equal areas.

[color=darkblue][b]Remark:[/b] Compare this question 1 with the [url=http://www.mathlinks.ro/Forum/viewtopic.php?highlight=touches+tangent+common&t=14547]question 1 on the first round of this olympiad[/url][/color]

$X$ is the intersection of $MN$ and $PQ$.

Then: $(XP)^{2}=(XN)(XM)$ and $(XQ)^{2}=(XN)(XM)$  $\Rightarrow$  $XP=XQ$

So we have:$[NXP]=[NXQ]$ and $[MXP]=[MXQ]$

$\Rightarrow$ $\boxed{ [MNP]=[MNQ]}$ $\mathbb{QED}$

The radical axis of two circles passes through the midpoint of any common tangent.


[quote="sinajackson"]$a_{1}, a_{2}, \cdots , a_{n}$ are positive reals ( $n \geq 2$ ). Prove that:
$\sum^{n}_{i=1}( a^{3}_{i}+1 ) \geq \sum^{n}_{i=1}( a^{2}_{i}a_{i+1}+1)$

$( a_{n+1}= a_{1})$[/quote]
If you really meant this, it's very easy. We can subtract $n$ from both sides. Then it becomes $\sum^{n}_{i=1}a^{3}_{i}\geqslant \sum^{n}_{i=1}a^{2}_{i}a_{i+1}$ which is just plain rearrangement.  :D
I'm sure that there's a typo. Maybe it should be $\prod^{n}_{i=1}( a^{3}_{i}+1 ) \geq \prod^{n}_{i=1}( a^{2}_{i}a_{i+1}+1)$ (I saw this inequality somewhere, but I don't remember the solution).  :maybe:

Consider $a,b,c$ such that $b\geq a\leftrightarrow b\geq c$.

Then  it holds $(b^{3}+1)\geq \frac{(a^{2}b+1)(b^{2}c+1)}{a^{2}c+1}$, since it is equivalent to

$(b^{3}+1)(a^{2}c+1)\geq (a^{2}b+1)(b^{2}c+1)$
$a^{2}b^{3}c+b^{3}+a^{2}c+1\geq a^{2}b^{3}c+a^{2}b+b^{2}c+1$
$b^{3}+a^{2}c\geq b^{2}c+ba^{2}$
Which follows from rearranging $[b^{2}, \;a^{2}]$ and $[b, \;c]$.
Now we will prove the first inequality by induction. Consider it proven for $n-1$ and suppose without loss of generality that between the $a_{i}$,  $1\leq i\leq n$, $a_{n}$ is a maximal or minimal element.
Then $(a_{n}^{3}+1)\geq \frac{(a_{n-1}^{2}a_{n}+1)(a_{n}^{2}a_{1}+1)}{a_{n-1}^{2}a_{1}+1}$  and:

$\prod_{i=1}^{n-1}(a_{i}^{3}+1)\geq \prod_{i=1}^{n-1}(a_{i}^{2}a_{i+1}+1) \Rightarrow$
$(a_{n}^{3}+1)\prod_{i=1}^{n-1}(a_{i}^{3}+1)\geq \frac{(a_{n-1}^{2}a_{n}+1)(a_{n}^{2}a_{1}+1)}{a_{n-1}^{2}a_{1}+1}\prod_{i=1}^{n-1}(a_{i}^{2}a_{i+1}+1) \Leftrightarrow$

$\prod_{i=1}^{n}(a_{i}^{3}+1)\geq \prod_{i=1}^{n}(a_{i}^{2}a_{i+1}+1)$

Where we take the indices in each case modulo the number on top of the productorial.
The base case $n=2$ is just rearrangement.

[quote]I saw this inequality somewhere,[/quote]

I think this one is from Czech Slovak M.C. and solution is same as ThAzN1's one


Why is the number of users in the memberlist different from the number of registered users (at the bottom of the main page)?

Just wondering...


I don't know if this questions belongs started.

The digital root of a number is the sum of the integers until it becomes one digit.

Example:

Number      [color=blue] [i][b]Digital Root[/b][/i][/color]

18 -               1+8 = [color=blue][i][b]9[/b][/i][/color]
345 -             3+4+5 = 12, 1 +2 = [color=blue][i][b]3[/b][/i][/color]
1456 -           1 + 4 + 5 + 6 = 16, 1 + 6 =[color=blue][i] [b]7[/b][/i][/color]

How can you prove that the digital root of perfect squares only ends in 1,4,7 and 9? (Of course excluding zero)

[quote="Altheman"][hide="hint"]mod 9[/hide][/quote]

Yup.  I know that, but what Im looking for is the proof for all numbers.

If you aren't sure if a problem should be in Getting Started, please put it in intermediate (which is where I'm moving this).

[quote="Altheman"]that is a hint for the solution of the problem...the rule for divisibility by 9 is tells us that summing the digits has the same remainder mod 9 as the initial number, all that remains to show is that all perfect squares mod 9 have residues of 1,4,7, or 9, that is easy case work since you can just go through 0 to 8 squared and you are done[/quote]

Sorry if I'm so dumb, but is there an algebraic proof that would prove that all digital roots of all perfect squares only ends in 1, 4, 7 and 9?

Yeah Altheman has the right idea. That's exactly what I thought about.


[quote="ToBe"]$-2 \frac{4}{5}$  X  $(3 \frac{5}{7})$[/quote]
here is my solution:
$-2 \frac{4}{5}$  X  $(3 \frac{5}{7})=-\frac{14}{5} \times \frac{26}{7}= -\frac{364}{35}$
ashegh did i make a mistake or you made a mistake. :?:

No offence, but I think this is a kindof pointless problem. It involves almost NO thinking.

[hide]first, make them improper fractions.

$-2\frac{4}{5} = -\frac{14}{5}$
$3\frac{5}{7} = \frac{26}{7}$

$-\frac{14}{5} \cdot \frac{26}{7}$

Cancel 7, and put 2 in place of 14

$\frac{52}{5} = 10 \frac{2}{5}$[/hide]

The correct answer is $-\frac{52}5$ (or $-10\frac25$).
[i]jensen hasn't simplified the fraction, and diophantine27 has omitted the negative.[/i]

$\frac{-52}{5}$ is the answer. so simple


geometry
conics
hyperbola
Asymptote
geometric transformation
homothety
geometry unsolved

Lb  side
Lc  side
r   line conecting I with G

i want the triangle

 :(

This is very similar to the problem [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?f=46&t=224690]Lb Lc IH[/url].

Construct incircle $(I)$ tangent to $\ell_b,\ell_c$ and denote by $M,N,L$ the unknown midpoints of $BC,CA,AB.$ As $BC \equiv \ell_a$ envelopes $(I),$ $B \mapsto C$ is a homography between $\ell_b$ and $\ell_c,$ thus $L \mapsto N$ is clearly a homography. Since the directions $LM \parallel \ell_b$ and $NM \parallel \ell_c$ remain fixed, then  $M$ describes a hyperbola $\mathcal{M}$ with asymptotes parallel to $\ell_b,\ell_c$ $\Longrightarrow$ $G$ describes the hyperbola $\mathcal{G}$ homothetic to $\mathcal{M}$ under homothety with center $A$ and coefficient $\tfrac{2}{3}.$ Hence, $G \equiv \mathcal{G}\cap r$ $\Longrightarrow$ $G$ are the double points of a homography that $\mathcal{G}$ induces on $r.$ Once $G$ is determined the construction of $\triangle ABC$ is straightforward.


Fin all function f(x) \in R[x] has property :
 exist n \in N* such that
 \sum {i=1->2n+1} (-1)  i [i/2]f(x+i)=0 ,for all x \in R

Do you mean:
  $\sum_{i=1}^{2n+1}(-1)^i [\frac{i}{2}] f(x+i)=0\forall x \in R$
Here is my soln:
 $ A=\sum_{i=1}^{2n+1}(-1)^i [\frac{i}{2}] f(x+i)
    =\sum_{i=1}^{n}i [f(x+2i) - f(x+2i+1)] $
If $deg f \geq 1 \Rightarrow \exists N $such that f(x) is strictly increasing func. or strictly decreasing func. in$(N;\infty)\Rightarrow A>0 $ or $A<0$,a contradiction.
Hence, f(x)=constant .Thus ,we are done.Or at leats I hope so
 ;)  :P


Pythagorean Theorem
geometry
geometry proposed

AB and AC are tangent to circle O at B and C, respectively, and D is a point on the major arc BC. Lines BD and AC meet at E, and lines DC and BA meet at F. If BF = a, CE = b, and EF = x, prove that $ x^{2}\equal{}a^{2}\plus{}b^{2}\minus{}ab$

The proof is analogous to the one given in the other topic [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?f=47&t=291972]resembling pythagorean theorem[/url] apart from some sign changes. Tangent of $(O)$ through $D$ cuts $BC$ at $P.$ Then $P,E,F$ are collinear on the Lemoine axis of $\triangle BDC.$ Let $AB=AC=L.$ By Menelaus' theorem for $\triangle ABC$ cut by $\overline{PFE},$ we have

$\frac{a}{L+a} \cdot \frac{L+b}{b}=\frac{PB}{PC}=\frac{DB^2}{DC^2}= \frac{a^2}{b^2} \cdot \frac{EB^2}{FC^2} \Longrightarrow \frac{EB^2}{FC^2}=\frac{b(L+b)}{a(L+a)} \ \ (1)$

On the other hand, by Stewart theorem for the cevians $EB,FC$ in $\triangle AFE,$ we have

$EB^2=\frac{x^2L+a(L+b)^2-aL(L+a)}{L+a} \ , \ FC^2=\frac{x^2L+b(L+a)^2-bL(L+b)}{L+b}$

$\frac{EB^2}{FC^2}=\frac{L+b}{L+a} \cdot \frac{x^2L+a(L+b)^2-aL(L+a)}{x^2L+b(L+a)^2-bL(L+b)} \ \ (2)$

From $(1)$ and $(2)$ we have then

$\frac{x^2L+a(L+b)^2-aL(L+a)}{x^2L+b(L+a)^2-bL(L+b)}=\frac{b}{a} \Longrightarrow \ x^2=a^2+b^2-ab$

[b]Nice problem and nice proof ![/b] Remark that the relation $\boxed{x^2=a^2+b^2-ab}$ is truly (analogous
proof) for any point $D$ which belongs to the circle $(O)$ , on the minor arc $BC$  or on the major arc $BC$ .


linear algebra
matrix
induction
algebra proposed
algebra

$\alpha_1=1$, $\alpha_2=6$, $\beta_1=0$, $\beta_2=1$
Use product of matrices $A_{n+1}(A_nA_{n-1}...A_2A_1)$ we get
$z_{n+2}=(4n+6)z_{n+1}+z_n$ where $z_n$ is $\alpha_n$ or $\beta_n$

With $a_n=3\alpha_n+\beta_n$ we have

(1) $a_{n+2}=(4n+6)a_{n+1}+a_n$, $a_1=3$,$a_2=19$

Now for $n=1$, $n=2$ in $\sum\limits_{k=0}^n\frac{(n+k)\;!}{k\;!(n-k)\;!}$
we find also $1+2=3$,  $1+6+12=19$

It seems induction and relation (1) will give the answer of question (i)

This problem is related to continous fraction  :cool:


Let $ ABC$ be a triangle with altitude $ AH$ and median $ AM$, let $ A'$ be the point such that $ \vec{MA'}\equal{}\frac{1}{3}\vec{MH}$ construct $ B',C'$, similarly, for all arbitrary point $ P$ prove that
\[ \vec{PA'}\cdot\vec{BC}\plus{}\vec{PB'}\cdot\vec{CA}\plus{}\vec{PC'}\cdot\vec{AB}\equal{}0\].

$ \triangle A'B'C'$ is pedal triangle of the centroid $ G.$ Let $ \triangle XYZ$ be pedal triangle of $ P.$ The proposition is equivalent to $ \overline {XA'} \cdot \overline {BC} \plus{} \overline {YB'} \cdot \overline {CA} \plus{} \overline {ZC'} \cdot \overline {AB} \equal{} 0.$ But this holds for pedal triangles of arbitrary points $ P, Q,$ not just for pedal triangles of arbitrary $ P$ and centroid $ G.$ The reason is that any $ P$ can be reached from arbitrary $ Q$ by traveling on the line $ QA' \perp BC$ for a while and then on the line $ QB' \perp CA$ for a while.

Nice! Thanks, there is other way to prove the perpendicular lines w.r.t side of triangle concurrent as following

Let $ d_a\perp BC,d_b\perp Ca,d_c\perp AB$ and $ d_a\cap BC\equal{}A',d_b\cap CA\equal{}B',d_c\cap AB\equal{}C'$ then $ d_a,d_b,d_c$ are concurrent iff for all point $ P: \vec{PA'}\cdot \vec{BC}\plus{}\vec{PB'}\cdot\vec{CA}\plus{}\vec{PC'}\cdot\vec{AB}\equal{}0$.


inequalities
trigonometry
inequalities proposed

hi,
Prove that for every x in real numbers we have the inequality
\[ (1\plus{} \sin{(x)})(1\plus{} \cos{(x)}) \leq \frac{3}{2} \plus{} \sqrt{2}.\]

$ (1\plus{} \sin{(x)})(1\plus{} \cos{(x)}) \leq \frac{3}{2} \plus{} \sqrt{2} \Leftrightarrow \frac{sin 2x}{2}\plus{}(sin x\plus{}cos x) \le \frac{1}{2}\plus{}\sqrt{2}$
Obviously $ \frac{sin 2x}{2}\plus{}(sin x\plus{}cos x) \le \frac{sin 2x}{2}\plus{}\sqrt{1\plus{}sin 2x} \le \frac{1}{2}\plus{}\sqrt{2}$ and we have when $ x\equal{}\frac{\pi}{4} (mod \;\; \pi)$, so done.


A particle can travel at speeds up to $ \frac{2m}{s}$ along the $ x$-axis, and up to $ \frac{1m}{s}$ elsewhere in the plane. Provide a labelled sketch of the region which can be reached within one second by the particle starting at the origin.


Our national language is hindi...and we have about 18 constitutional languages :)

Hindi is not exactly our national language....

But Khadi Boli dialect(which we call Hindi) of Devnagri is the most popular language in India owing to its use in Bollywood, The Indian film industry....

Other forms like Bhojpuri and Awadhi,Maithili  are popular in UP and Bihar only...

I speak Tamil, my mothertounge, English, the most used language in India(not Khari Boli!! :P) and Hindi, (Khari Boli) with specification..
Though I can understand parts of Malayalam, Telugu, Marathi, and many other languages..:D

I speak tamil (MT), english, a bit of hindi and telugu.  :lol:

i speak a bit of malayalam,hindi and telugu...

my mother tongue is tamil and i speak that very well

Well... I speak Tamil and English very well and can manage Hindi pretty good :D Apart from that, I understand Punjabi, Bengali and Spanish... Of course, Telugu and Malayalam take a lot from Tamil, almost similar, so I could say its comprehendible too

i speak tamil,hindi,oriya,english and a bit of sanskrit...can understandt a bit of malayalm :)

Mine is predominantly Tamil but apart from that obviously English and I understand and can read both Hindi and Sanskrit(not spoken in today's world). :D

I know xhosa, afrikaans, finnish, hawaiian, mandarin, and ofcourse, very evidently gibberish :D  :lol:

Yeah i forgot all those  :D  As a matter of fact aint i the one who taught you all that?  :huh: And since when did you forget tamil my boy? I thought i taught you that too  :P

adhum enakku theriyum.. aana adhu avallo importantana baashai kadayathu :D

side note: for all of those who imagined me sayin that pls dont laugh :D  :rotfl:

[quote="hash_include"]side note: for all of those who imagined me sayin that pls dont laugh :D  :rotfl:[/quote]
 :rotfl:  :rotfl:   :P

Shrids, seriously, if you had left it without the sidenote, I might not have laughed so hard you know :D  :rotfl:

edhan kettu appe vangikiruthu :P  :P  :P  :D

Mainly I know and speak - Oriya, Bihari, Hindi (All dialects), Spanish, French, German (yay!!), Tulu, Urdu, Braille, Sign Language, Luganda, Lithuanian, Slovak, Romanian, Ewe and some which are not even listed here. http://en.wikipedia.org/wiki/Languages_by_speakers

Just for time pass, now and then I speak these English, Tamil, Kannada.

Fluent in English,Tamil,Hindi. Can speak and understand what i would like to call "SLOW" french

[quote="Cyber-Shreyas"][quote="verizonwires"][quote="Cyber-Shreyas"Just for time pass, now and then I speak these English, Tamil, Kannada.[/quote]

me 2, bye the way, i am indian. i just wanted to know what to speak in these forums.

ANY PUNJABIS or GUJJUS HERE???[/quote]

Dunno about that but english and tamil dominate this forum mainly cause of the dominating population of tamilians and the mod himself being a tamil .:D[/quote]

hope that was a complement :D

Dont think there are any punjabis, not sure about the Gujjus :D

Can understand Punjabi and Bengali to a certain extent as two of my friends were Punjabi and Bengali resp :)

[quote="hell_ever"]I speak tamil as my mt but it not so good i speak onlt with my mom and dad but otherwise bambaiyya hindi[/quote]
visit this forum more regularly da...and I guarantee u ur tamil will improve :P  :P  :D  :D  :rotfl:  :rotfl:

Lol. I second pardesi :D You will learn tamil very very fast if you stay here for some more time :)

Can speak english, hindi, urdu, punjabi, (readonly) arabic and (understandonly) bengali.. :wink: .

[quote="verizonwires"][quote="Cyber-Shreyas"Just for time pass, now and then I speak these English, Tamil, Kannada.[/quote]

me 2, bye the way, i am indian. i just wanted to know what to speak in these forums.

ANY PUNJABIS or GUJJUS HERE???[/quote]

Ok, ami bangali........doesn't anyone else here know bengali ?


geometry
combinatorics unsolved
combinatorics

Let ABCD be a square, by a triangulation of ABCD we understand a division of ABCD in triangles by lines, such that every two triangles are disjoint or they share a side, or they share a vertex...

make a triangulation of ABCD where every triangle is acuteangled

determine the samallest number of triangles n such that it is possible to triangulate ABCD into n acuteangled triangles

The answer is $8$.

We need segments to divide the right angles, $3$ of which will necessarily meet inside the square, so there's at least an inner vertex. If there's only one, at least $5$ segments would start from it, and at least one of them would meet a side of the square (not in a vertex), producing a non-acuteangled triangle that cannot be eliminated. If the inner vertices are $2$, the number of inner segments is at least $9$, and the total sides are at least $9 \cdot 2+4=22$, so there are at least $8$ triangles. With more than $2$ internal vertices, we have at least $26$ sides, and $9$ triangles.

Finally, an acuteangled triangulation of a square with $8$ triangles is easily found: for example, let $ABCD$ be the square, $E$ the midpoint of $AB$ and $F$ the midpoint of $CD$. Trace the $4$ circles with diameters $AE$, $EB$, $BC$ and $DA$ respectively, and choose $G$ and $H$ in the small area among the circles, so that $GH \parallel AB$. The triangulation can be obtained joining $G$ with $A$, $E$, $H$, $F$, $D$, and joining $H$ with $E$, $B$, $C$ and $F$.

With similar arguments, one can prove that the minimal acuteangled triangulation of a non-acuteangled triangle has $7$ triangles.


How might I go about solving a system of equations,

x^2006 + y^2006 + z^2006 =2 -----(1)
x^2007 +y^2007 + z^2007 = 2 -----(2)
x^2008 + y^2008 + z^2008 =2 -----(2)

in the real number system? The only obvious solutions I can see are the permutations of the set {0,1,1}
Are there others? How do I formally find them without the wild guessing phase?

My current interest in cyclic expressions lead me to a proof that the given set is the only possible solutions of this system,
we subtract, say (1) from (2), now treating this in x,

x^2007 - x^2006 + y^2006 .(y-1) + z^2006.(z-1) = 0

the only rational solution for this is x=1
Similarly for other pairs of eq's in y and z,taking the set on the entire system we have our set {1,1,0}
A set of irrational solutions may exist for the given equation and its cyclic equivalents, however in the system given any presented non rational solution set will not satisfy all given equations.[/hide]


i know modulo is one of the best 'weapon' to solve int equations but i often have to guess. Can someone show me the way to find the modulo in some situation?For exam:
x^3 - 4xy + y^3=-1
(5^x)(7^y) + 4 = 3^z
4xy - y -x = z^2
I have not solved any of them but i think we can use modulo.  :lol: Help me!

For the second one if you try modulo 7(or 5) you can get that z is even $ z \equal{} 2k$. From here $ 5^{x}7^{y}\equal{}(3^{k}\minus{}2)(3^{k}\plus{}2)$. Since numbers $ 3^{k}\minus{}2$ and $ 3^{k}\plus{}2$ are coprime we get that $ 3^{k}\plus{}2\equal{}5^{l}$ and $ 3^{k}\minus{}2\equal{}7^{n}$ or vice versa.
These equations should be easier than the original.