inequalities
function
inequalities proposed

Show that $\sum_{cyc}x(4x^{2}+5yz)^{3}\geq (x+y+z)^{7}$
given that x,y,z are pozitive reals :)

it's easier than you think... :lol:  :wink:

[quote="Nameless"]
I don't understand why $\sum x(4x^{2}+5yz)^{3}\geq (4(x^{3}+y^{3}+z^{3})+15xyz)^{3}$[/quote]
$f(x)=x^{3}$ is convex function when $x\geq0.$
Hence, by jensen's inequality $xf(a)+yf(b)+zf(c)\geq f(ax+by+cz)$
 for all $a,$ $b,$ $c,$ $x,$ $y$ and $z$ are non-negative numbers such that $x+y+z=1.$
And use $a=4x^{2}+5yz,$ $b=4y^{2}+5xz,$ $c=4z^{2}+5xy.$


How would you make your document in the format like that of the MathCounts sprint round?

Please, someone help?
I mean the format where you have a question, then a column of numbered blanks to the side.

I have a template for you.  This is incredibly useful, I got it from another AoPS member...I can't remember who...

Yes, but would you know how to put it in TeXnicCenter? I doesn't have to be exactly like the mathcounts format, I'm just trying to figure out how to put a column of numbered blanks to the side...


Cover a 23*23 square by 1*1,2*2,3*3 squares without gaps and overlaps.
Then how many 1*1 squares are required at least?

Not hard, but it takes some time until you analyse the cases. Or is there a shorter and simpler solution?
The configuration with 4 $1x1$ squares is the following:  form a $20x20$ square. Now add horizontally a band of width $3$ and length  $18$, and another one vertically. Add a $3x3$ in the corner and another two $2x2$ squares, and there are 4 squares left.
Now we will get some relations. Color the plane in 4 colors(by translating a $2x2$ 4-colored square). Let's look after one of the colors (red). Every $2x2$ has exactly one red square. Divide the $3x3$'s in the ones that have $1,2,4$ red squares, and denote the number of them as $n_1,n_2,n_4$, respectively. Suppose we have x $1x1$ squares. We want to show that cases x=0,x=1,x=2,x=3 don't work.
Now color in 2 colors. We get that $n_1+n_4-n_2=y$(counting the squares from one of the colors), where $|y|$ can be :
1)for x=0, 0
2)for x=1, 1
3)for x=2, 0 and 2
4)for x=3, 1 and 3.
Counting for the 4-coloring we get $2n_2+4n_4+n_1+a+z=144$, where $z\leq x$, where a is the number of $2x2$ squares, $z$ is the number of single $1x1$ squares which are red. And adding all the squares $9(n_1+n_2+n_4)+4a+x=23^2   (1)$.Multiply the first one by 4 and subtract the second one. We get $7n_4-n_2-5n_1+4(z-x)=7\Rightarrow 6(n_4-n_1-1)=1-n_1-n_4+n_2+4(x-z)$. So $n_4+n_1-n_2$ is odd. That means cases 1) and 3) fall. In case 2), the relation becomes $M_6=1\pm 1+2\pm 2$, so it follows that $y=1$ and $x=z$. But substituting $n_1+n_4=n_2+1$ in $(1)$, gives a contradiction $mod 4$.
For case 4) the relation is a) $M_6=1\pm 1+4(x-z)$. So we can have $y=1, x=z+3\Rightarrow z=0$, or $y=-1, and z=2$. But the red points are also counted in the first coloring so if $z=2$, then $|y|\geq 2$, so the second case falls. The first case because of a $mod 2$ contradiction in $(1)$.
b)$M_6=1\pm 3+4(3-z)\Rightarrow y=3,z=1, or y=-3,z=1$, and again contradiction in $(1)$ $mod 2$.It seems I can get a contradiction in $(1)$ for any odd $x$. Sorry, didn't see in time. It would have shortened my solution.
Anyway, I like the idea.

:blush: , u're right.
oops, that means I should reconsider all the cases. I won't do that, I'm not in the mood for re-analyzing the cases. maybe someone else will ?


[quote="N.T.TUAN"]Prove that $\frac{2}{9}\leq a^{3}+b^{3}+c^{3}+3abc<\frac{1}{4}$, where $a,b,c$ are sides of a triangle with perimeter $1$.[/quote]

We have many rezults relating the one. For example:
$\frac{52}{27}\leq a^{2}+b^{2}+c^{2}+2abc<2$
$\frac{13}{27}\leq a^{2}+b^{2}+c^{2}+4abc<\frac{1}{2}$

I think all of them can solve by the same way but now i have no idea


Calculate the empirical formula of a compound containing $ 22.7$ percent sodium, $ 21.6$ percent boron, and $ 55.7$ percent oxygen by mass.

I figured out that the answer is $ Na_{2}B_{4}O_{7}$ using guess and check.  Is there another faster way?

Thanks in advance!

Guess and check is a pretty fast way, assuming you understand the structure of boron oxides and some of their chemistry.

If I had no idea how to guess at the problem, I'd do the standard "assume 100g of the compound" approach, albeit tedious.

 guess and check is the last resort you should do. In any of the test or chemistry exam, you have to solve a problem in a logical way.
Here's how you solve it:
Assume 100 g of the original compound, so you have 22.7g Na; 21.6g boron, and 55.7 g oxygen.
Mol of sodium = 22.7g/ 22.99g/mol = 0.987 mol Na
Mol of boron: 21.6g / 10.8 g/mol  = 2 mol B
Mol of oxygen: 55.7g / 16 g/mol = 3.48 mol O
Take three numbers 0.987; 2; and 3.48 divided by the smallest number which is 0.987:
0.987 / 0.987 = 1
2 / 0.987 ~ 2
3.48 / 0.987 ~ 3.5
because the subscript have to be a whole number, thus you have to multiply 1, 2, and 3.5 by 2
the result is 2, 4 and 7.
so, you have 2 Na, 4 Boron and 7 oxygen.
Thus the emperical formula is Na2B4O7


geometry
rectangle
complex analysis
function
limit
complex analysis unsolved

Suppose $ \Omega$ is a bounded region.  Let $ L$ be a (two-way infinite) line that intersects $ \Omega$.  Assume that $ \Omega \cap L$ is an interval $ I$.  Choosing an orientation for $ L$, we can define $ \Omega_l$ and $ \Omega_r$ to be the subregions of $ \Omega$ lying strictly to the left or right of $ L$, with $ \Omega \equal{} \Omega_l \cup I \cup \Omega_r$ a disjoint union.  If $ \Omega_l$ and $ \Omega_r$ are simply connected, then prove $ \Omega$ is simply connected.

my idea is to find a simply connected region around $ I$ ( I use a rectangle) and try to use that as a "bridge" to transform curves across the two regions, but I tried for a while and kept on finding myself proving bunch of specific sub-cases...

Is there a perhaps more rigorous, general way of going about this?? (can't find too many literature on proving simply connected set in a simple way, so I guess I'm quite inexperienced...)

Here is something sketchy if your line is meant to be straight: Apply the criterion that a connected open subset $ D$ of $ \mathbb C$ is simply connected iff every holomorphic function on $ D$ without any zero has a holomorphic square root.

So let $ f: \Omega\to \mathbb C$ be a holomorphic function without any zero. Since each of the $ \Omega_i$'s is simply connected, there are holomorphic square roots $ f_i: \Omega_i\to\mathbb C$ of $ f\mid\Omega_i$, i.e. $ f_i^2 \equal{} (f\mid\Omega_i)^2$. Now we will try to patch the $ f_i$'s together to get a global square root of $ f$: For every $ z\in I$ choose an open disk $ U_z\subset\Omega$ containing $ z$ and a holomorphic square root $ f_z: U_z\to\mathbb C$ of $ f$ on $ U_z$. Using $ f_z$ we see that the limits $ \lim_{z'\to z, z\in\Omega_i}f_i(z')\in\{\pm f_z(z)\}$ exist and that for each $ z\in I$ there is a unique $ \alpha(z)\in\{\pm 1\}$ with
\[ \lim_{z'\to z, z\in\Omega_1}f_1(z') \equal{} \alpha(z) \cdot\lim_{z'\to z, z\in\Omega_2}f_2(z')\,.
\]
Using $ f_z$ again, the function $ \alpha: I\to\{\pm 1\}$ is constant at $ U_z$, hence locally constant. Since $ I$ is connected, $ \alpha$ is constant. WLOG assume that $ \alpha\equiv 1$ (otherwise replace $ f_2$ by $ \minus{} f_2$). Now define $ F: \Omega\to\mathbb C$ by $ F\mid\Omega_i \equal{} f_i$ and $ F(z) \equal{} \lim_{z'\to z, z\in\Omega_{1}}f_{1}(z')$ for $ z\in I$. Show that $ F$ is holomorphic (e.g., using the $ f_z$'s one more time) with $ F^2 \equal{} f$.

I believe  that this method generalizes to the case when $ L$ is a curve which is not too crazy.


The disscociation constant of water at 18 degree celcius is $ \alpha$ find the $ K_{a}$of water at this temperature and the ionic product of water?

i have a conceptual doubt someone solve this first and then i will ask the doubt :D

u ve said the dissociation constant is $ \alpha$ no?

HEY do you want dissociation constant at room temperature ah as you have already said alpha at 18 celsius!!! :D

[quote="pardesi"]The disscociation constant of water at 18 degree celcius is $ \alpha$ find the dissociation constant of water at this temperature and the ionic product of water?

i have a conceptual doubt someone solve this first and then i will ask the doubt :D[/quote]

I think pardesi wants the equilibrium constant????Nice typo.

$ K_{eq} \equal{} \frac{\alpha ^2}{2(1\minus{} \alpha )}$

[quote="rituraj007"]how do we write fractions and other codes???????[/quote]

learn it [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=188533]here[/url]  :)


algebra
polynomial
number theory open
number theory

Do there exist two non-constant polynomials $ P$ and $ Q$ with integer coefficients such that for all integers $ x$ and $ y$, $ P(x)$ and $ Q(y)$ are coprime ?

[quote="Thomas B"]Do there exist two polynomials $ P$ and $ Q$ such that for all integers $ x$ and $ y$, $ P(x)$ and $ Q(y)$ are coprime ?[/quote]

$ P(x)\equal{}2$ $ \forall x$
$ Q(y)\equal{}3$ $ \forall y$


The digits 1, 3, 7 and 9 each occur at least once in the decimal notation of the positive integer n.
Show that some permutation of digits of n gives a number which is a multiple of 7.

Note that the following congruences are true:
\[1379\equiv 0\mod{7}\]
\[1793\equiv 1\mod{7}\]
\[3719\equiv 2\mod{7}\]
\[1739\equiv 3\mod{7}\]
\[1397\equiv 4\mod{7}\]
\[1937\equiv 5\mod{7}\]
\[1973\equiv 6\mod{7}\]

Now note that any number containing these four digits has a permutation of the form
\[10^4A+X\]
where X is one of the 7 above numbers.  Then whatever residue $10^4A$ has $\mod{7}$ there is a number above that when added to it results in $0\mod{7}$ since all residues are possible.


Has anyone heard about UIL mathmatics, is doing UIL mathmatics or has experience with UIL mathmatics? My school has decided to do UIL mathmatics instead of mc this year and I am confused about what I need to study and which calulator should I buy for the exam. Any help is apreciated.

Yes, yes and yes!  :D When you say UIL Mathematics, do you mean the entire Math contest program of UIL(Number Sense, Mathematics, and Calculator) or do you mean just simply the UIL Math test itself? Either way, this link will help: http://www.texasmath.org/. Go to the Downloads section. There's a few middle school tests there so you can get a general idea of the test format (since this is in the middle school forum, I assume you are in middle school). Also, if your school isn't a member of TMSCA (Texas Math and Science Coaches Association), I encourage you to try and get your school to join (my school's not actually a member  :( but it is still worthwhile to do.) You can get more tests and if you qualify you can go to a state competition in March.

As for calculators: UIL boils down to speed and recognition, so go with the calculator that is most comfortable for you (I'd suggest either the TI-83 or TI-84, but that's just me.) Just by looking at past tests you should get a good feel for what the tests ask for and whatnot.

Just a question: Why isn't your school doing Mathcounts? (Lack of funding, interest,etc.) I always thought Mathcounts was really fun and awesome. Note: you don't have to answer it, I'm just curious :lol:


geometry
function
algebra
domain
modular arithmetic
trigonometry
limit

Notation used in this series of problems:

Let $N$ be a fixed positive integer (large enough to avoid trivialities - say $N\ge 3.$)

Let $\omega=e^{2\pi i/N}.$

Suppose $z$ is a function $z: \mathbb{Z}/(N\mathbb{Z})\mapsto\mathbb{C}.$

Then the points $z(0),z(1),\dots,z(N-1),$ in that order, form the vertices of an $N$-gon in the plane. (It wouldn't necessarily be simple - that is, no self-intersections - unless we imposed that as an extra hypothesis.) In what is said below, any arithmetic manipulation of the indices in the domain of such a function $z$ (for instance, sending $n$ to $n+1$) is presumed to be done $\pmod{N}.$

Definition of Fourier transform:

If $k\in \mathbb{Z}/(N\mathbb{Z}),$ then $\widehat{z}(k)=\frac1N\sum_{n=0}^{N-1}z(n)\omega^{-kn}.$

We can recover $z$ by $z(n)=\sum_{k=0}^{N-1}\widehat{z}(k)\omega^{kn}.$

---------------------

Now for my questions:

1. Suppose $z$ defines a simple polygon traced out in a counterclockwise direction. Prove that the area of the polygon is given by the following formula:

(Area of polygon) $= \frac N2\sum_{k=1}^{N-1}\sin\left(\frac{2\pi k}N\right)| \widehat{z}(k)|^2.$

2. Define an operation $V$ on the polygon $z$ by finding the midpoint of each edge of the original polygon and making those midpoints (in the same order) the vertices of a new polygon.

For any $N$-gon $w$, does there exist an $N$-gon $z$ such that $w=V(z)?$

If the answer is false, then can you characterize those $w$ which are in the range of $V?$

(The $N=4$ case of this is fairly well known and can be dealt with by elementary means.)

3. Consider iterating the operator $V.$ What can you say about

$\lim_{m\to\infty}\frac{\text{area}(V^{m+1}(z))}{\text{area}(V^m(z))}?$

4. Continuing with #3, if we choose an appropriate scale factor (which may depend on $z$), we should be able to make the iterates of $V$ applied to $z$ tend to a certain limiting shape. Describe that shape more fully.

Your inversion formula has a typo- \widewhat is not a recognized code.

[color=green][The fix is obvious, and has been done.][/color]

I'm going to indulge myself and bump this topic. Hey, guys - don't be scared off, this is do-able, and there's some nice stuff here. You do part of it. #1 and #2 are completely independent of each other. #3 does need both #1 and #2, but you can just assume that #1 is true and attack it anyway.

A hint for #1: do you know a formula - a nice, more or less quadratic formula - that computes the area of the polygon directly from the coordinates of the vertices?

[quote="Kent Merryfield"]1. Suppose $z$ defines a simple polygon traced out in a counterclockwise direction. Prove that the area of the polygon is given by the following formula:

(Area of polygon) $= \frac12\sum_{k=1}^{N-1}\sin\left(\frac{2\pi k}N\right)| \widehat{z}(k)|^2.$[/quote]

Indeed, this can be done  :D  Even I almost got it.

First, I'll also include $k=0$ for simple-calculation reasons. Let $T =$ right hand side.

Using the fact that $|z|^2 = z \cdot \overline z$, we get that \[ T = \frac{N}2 \sum_{k=0}^{N-1} \sin\left(\frac{2 \pi k} N \right) \frac{1}{N^2} \sum_{s=0}^{N-1} \sum_{t=0}^{N-1} z(s) \overline{z(t)} \omega^{(t-s)k}  \]

Changing the order of the sums,
\[ = \frac{N}{2N^2} \sum_{s=0}^{N-1} \sum_{t=0}^{N-1} z(s) \overline{z(t)} \sum_{k=0}^{N-1} \sin\left(\frac{2 \pi k} N \right) \omega^{(t-s)k}  \]

We know that $\sin\left( \frac{2 \pi k}{N} \right) = \frac{\omega^k - \omega^{-k}}{2i}$, so
\[ = \frac1{2N} \sum_{s=0}^{N-1} \sum_{t=0}^{N-1} \frac{1}{2i} z(s) \overline{z(t)} \sum_{k=0}^{N-1} \left( \omega^k - \omega^{-k} \right) \omega^{(t-s)k}  \]

Using geometric series, we see that for a fixed $s$, only $t=s-1$ and $t=s+1$ count. (the first is $N$, whilst the second is $-N$)
\[ = \frac1{2} \sum_{s=0}^{N-1} \frac{1}{2i} \left( z(s) \overline{z(s-1)} - z(s) \overline{z(s+1)} \right)  \]
\[ = \frac1{2} \textrm{Im} \left( \sum_{s=0}^{N-1} z(s) \overline{z(s+1)} \right)  \]
\[ = \frac1{2} \sum \left( x_k y_{k+1} - y_k x_{k+1} \right)  \]
(all indices are modulo $n$; also, $z_k = x_k + i y_k$)

Now, I used the determinant formula for the area of a triangle,
\begin{eqnarray*} 2 K_{\textrm{poly}} &=& \sum_{k=1}^{N-2} \left| \begin{matrix} x_0 & y_0 & 1 \\ x_k & y_k & 1 \\ x_{k+1} & y_{k+1} & 1 \end{matrix} \right| \\ \ &=& \sum_{k=1}^{N-2} \left( x_k y_{k+1} - y_k x_{k+1} - x_k y_0 - x_0 y_{k+1} + y_0 x_{k+1} +x_0 y_k \right) \\ \ &=& \sum_{k=0}^{N-1} \left( x_k y_{k+1} - y_k x_{k+1} \right) \end{eqnarray*}

I probably did a computation mistake; where did I missed that $N$?  :(   [After Kent's edit, all is OK]

By the way, does this formula have any interesting application? (besides 3 and 4)

Oops! About that factor of $N:$ perfect_radio is correct. I somehow misplaced that factor in Parseval's Identity in this context. I have edited the original.

Here's a particular case, useful for checking. Let $z(n)=\omega^n.$ This defines a regular $N$-gon incribed in the unit circle. The Fourier transform is $\widehat{z}(k)=1$ for $k=1$ and is zero for all other $k.$

Applying the formula to this gives an area of $\frac{N}2\sin\left(\frac{2\pi}N\right).$ This is the correct area for the regular $N$-gon, and tends to $\pi$ as $N\to\infty,$ as it should.

That's problem #1. Now, what about problem #2? [hide="Slight hint."]The answer differs between the cases of $N$ even and $N$ odd.[/hide]

On problem #3, the limit is only needed for $N\ge 5.$ The answer is exact for $N=3$ and $N=4.$

$\# 2$:

[i]Solution 1.[/i] Let $B_{1}(u_{1},v_{1}),\ldots,B_{N}(u_{n},v_{n})$ be the polygon we want to obtain. Let $A_{1}(x_{1},y_{1})$ be a random point in the plane. Then, $A_{2k}\left( 2 \left( u_{2k-1}-u_{2k-2}+\ldots \right)-x_{1}, 2 \left( v_{2k-1}-v_{2k-2}+\ldots \right)-y_{1}\right)$ and $A_{2k+1}\left( 2 \left( u_{2k}-u_{2k-1}+\ldots \right)+x_{1}, \left( v_{2k}-v_{2k-1}+\ldots \right)+y_{1}\right)$. (so that $B_{k}$ is the midpoint of $A_{k}A_{k+1}$)

We wish to have $A_{N+1}=A_{1}$.

If $N$ is odd, then $N+1$ is even, so we must have $u_{N}-u_{N-1}+\ldots = x_{1}$ and $v_{N}-v_{N-1}+\ldots = y_{1}$, which is of course doable :D

Otherwise we must have $u_{N}-u_{N-1}+\ldots = 0$ and $v_{N}-v_{N-1}+\ldots = 0$.

[i]Solution 2.[/i]I also tried it with DFT:

Let $A_{0}A_{1}\ldots A_{N-1}$ be a given polygon and let $B_{k}$ be the midpoint of $A_{k}A_{k+1}$.

We have $\widehat{z_{A}}(k) = \frac{1}{N}\sum_{n=0}^{N-1}z(n) \omega^{-kn}$ and
\begin{eqnarray*}\widehat{z_{B}}(k) &=& \frac{1}{N}\sum_{n=0}^{N-1}\frac{z(n)+z(n+1)}{2}\omega^{-kn}\\ \ &=& \frac{1}{2N}\sum_{n=0}^{N-1}z(n) \omega^{-kn}+\omega^{k}\frac{1}{2N}\sum_{n=0}^{N-1}z(n+1) \omega^{-k(n+1)}\\ \ &=& \frac{1+\omega^{k}}{2}\cdot \frac1{N}\sum_{n=0}^{N-1}z(n) \omega^{-kn}\\ \ &=& \frac{1+\omega^{k}}{2}\widehat{z_{A}}(k) .\end{eqnarray*}

Using Inverse DFT, for $N$ odd we can always find a good polygon, $A_{0}A_{1}\ldots A_{N-1}$, since $\frac{1+\omega^{k}}{2}\neq 0$, for all $k \in \mathbb Z$; otherwise, let $N=2t$, so that $\widehat{z_{B}}(t) = 0$, which yields the same condition as above, i.e. $z_{1}-z_{2}+\ldots+z_{2t-1}-z_{2t}= 0$.

The first solution is valid, I think. What about the second?

$\textrm{Edit}_{1}$: Ok, I will post a more detailed/clear solution. But you'll have to wait a little.

$\textrm{Edit}_{2}$: I actually did what I said in the previous edit.

The second solution is valid (although I'd appreciate the notation cleaned up a little bit, distinguishing $n$ from $N$ and actually naming the midpoint.)

The key thing there: the operation $V$ that sends $z$ to $V(z)$ is a convolution operation. It's convolution with the vector $u=\left(\frac12,\frac12,0,\dots,0\right)$ (or its reflection, depending on how you enumerate the points of $w.$

Convolution on one side is pointwise multiplication on the other, and the Fourier transform of $u$ is a scalar multiple of $\frac{1+\omega^{-k}}2,$ which is essentially what you had. (You may have been working with the reflection of $u.$)

I was reading something on Green's Theorem (more of an informal presentation) and I saw there a proof of the area of a polygon formula (which seems kind of strange to me). That made me remember this topic.

So, I tried to solve $\# 3$ using the area formula from $\# 1$, but I got stuck in the computations, maybe I messed them up. Anyway, can you please tell me at least the answer? Thanks.

Here's an interesting form of Green's Theorem:

Suppose $\gamma$ is a closed curve that simply enclosed the region $E$ (this is as in the usual formulations of the theorem). Suppose $f$ is a sufficiently regular function on $E\cup\gamma.$

Then $\int_{\gamma}f(z)\,dz=2i\iint_{E}\frac{\partial f}{\partial\overline{z}}\,dA$

If $f$ happens to be analytic, this becomes Cauchy's theorem: the integral of an analytic function around a closed curve is zero. For our purpose, we want to apply this with $f(z)=\overline{z}.$

The hint on #3 is that the limit only depends on the $k=1$ and $k=-1$ terms of the sum. ($k=-1$ can also be written $k=N-1.$)

Am I supposed to get \[ \textrm{area}\left( V^{m}(z) \right) = \sum_{k=1}^{N-1}\sin \left( \frac{2 k \pi}N \right) \cdot \left| \hat z (k) \right|^{2}\cdot \left| \cos \left( \frac{k \pi}N \right) \right|^{2m}\, ?\] $\left( \textrm{using some trig to transform}\, \, \left| \frac{1+\omega^{-k}}2 \right| \, \, \textrm{into}\, \, \left| \cos \left( \frac{k \pi}{N}\right) \right| \right)$

I'm asking this because if it is like this, then some complications appear when $\left| \hat z (1) \right|^{2}= \left| \hat z (n-1) \right|^{2}$. But maybe that can't happen since our polygon is convex. I'll think more tomorrow, now it's just too late.

As for the result itself, I find it pretty interesting (I'm assuming that it's $\left| \cos \left( \frac{\pi}{N}\right) \right|^{2}$, which is true for $n=3,4$ at least) as I don't see any other way to pull that $\cos$ from "the hat".

You're on the right track. And it's fair to assume that $\left| \hat z (1) \right|^{2}\ne \left| \hat z (n-1) \right|^{2}$ because otherwise we wouldn't have a simple curve.

Hmm sorry to bump but I was reading through this old topic (that might have been recently linked to) and since both of you are still active users, do you know how to finish the problem now?


function
algebra
domain
polynomial
rational function
advanced fields
advanced fields unsolved

I'm stuck with this exercise which shouldn't be difficult:

Let $ V$ be an (irreducible) affine variety, let $ U$ be an open non-empty subset of $ V$ and let $ f : U \to k$ - where $ k$ is the base field, which is assumed to be infinite but not necessarily algebraically closed. Suppose that $ f$ is locally given by rational functions, i.e. for every $ P \in U$ there exists an open $ P \in U_P \subseteq U$ and a rational function $ f_P \in k(V)$ such that the domain of $ f_P$ contains $ U_P$ and $ f_P(Q) \equal{} f(Q)$ for every $ Q \in U_P$.

Show that $ f$ is the restriction to $ U$ of a rational function on $ V$.

Can anyone help me out?

What about the following example: $ k$ algebraically closed (to make it very easy), remove from $ V\equal{}\mathbb A^1_k$ the origin to get $ U$ and take $ f(x)\equal{}1/x$? Is your claim true for this example?

I think it is, why wouldn't it be? It is true that $ f(x) \equal{} 1/x$ has a singularity at $ x \equal{} 0$ but it is still a rational function.

A rational function is by definition just a rational function on a nonempty open subset. Thus it's true by definition.

What are your points $ P$ of the variety? Are they rational over $ k$ or over an algebraic closure of $ k$? Or, in other terms, do you consider $ V$ to be a classical variety like a closed subset of $ k^n$ or as scheme?

This might be important to understand how you evaluate the locally existent rational representations of $ f$. Note that $ f$ is given as an ordinary function on $ U$ taking values in $ k$ which is compared to the function induced by an element of the rational function field $ K(V)$.

I consider $ V$ as a classical variety, not as a scheme.

[quote="KOFO"]I do not understand it... it is not given that $ f$ is a rational function itself![/quote]
Apart from what ThirdEdition asked:
A rational function is a function on a nonempty open subset that is locally a quotient of polynomials. And $ f$ fulfills that property.

This is what ZetaX pointed out already from another perspective using Hartshorne's notation (Chap. I.3):

Your $ f$ is a regular function on $ U$, i.e. $ f\in\mathcal O(U)$. Now the ring $ \mathcal O(U)$ sits in the field $ K(V)$ of rational functions on $ V$, by the definition of $ K(V)$ - apparently without using the assumption on $ k$.  :o

My remarks on the distinction between polynomials and polynomial functions is obsolete in the theory of classical algebraic varieties, since ordinary functions on the varieties with values in the base field are considered here. This becomes a bit nasty when switching to schemes, but the Hilbert Nullstellensatz builds the bridge between the bridge if $ k$ is algebraically closed.


logarithms
function
ratio
combinatorics proposed
combinatorics

Find the number of matchings in a n-cycle, i.e simple cycle consisting of $ n$ vertices and $ n$ edges.

(Parental advisory: symbolic language. Read with care!)

A (possibly empty) matching of $ n$-cycle can be described as a non-labelled cycle of singletons and 2-sequences.
This symbolically written as
\[ \mathcal M \equal{} \mathrm{Cyc} (\mathcal Z \plus{} \mathcal Z^2).\]
Thus the OGF of matchings is the Polya logarithm of $ z \plus{} z^2$, that is
\[ M(z) \equal{} \sum_{n\ge 1}\frac {\varphi(n)}{n}\log\frac1{1 \minus{} z^n \minus{} z^{2n}}.\]
It will take me some time to figure out the answer.

So I got
\[ m_n \equal{} \sum_{d|n} \frac1n \varphi(d) L(n/d),\]
where $ L(n) \equal{} \varphi^{n} \plus{} ( \minus{} \varphi)^{ \minus{} n}$ are Lukas numbers (Fibonacci numbers starting from $ 1,3$), $ \varphi(n)$ is Euler's function and $ \varphi$ is the golden ratio (sorry for using the same letter: the traditions are beyond me).

Deduct $ 1$ from $ m_n$ if you don't like empty matchings.

(It is highly likely from the formula that this can be done with the help of Burnside's lemma rather than the symbolic language .)

(And I forgot to say that I consider cyclically different matchings the same. Otherwise this number is just some Fibonacci number.)

An alternative way is to let $ A(n)$ denote the number of matchings in an $ n\minus{}$cycle and let $ B(n)$ denote the number of matchings in a path with $ n$ edges.  By considering the inclusion or non-inclusion of a given edge in a cycle, you get a formula for $ A$ in terms of $ B$, and by considering the inclusion or non-inclusion of the last edge in a path, you get a recursion for $ B(n)$.

[quote="kevinatcausa"]An alternative way is to let $ A(n)$ denote the number of matchings in an $ n \minus{}$cycle and let $ B(n)$ denote the number of matchings in a path with $ n$ edges.  By considering the inclusion or non-inclusion of a given edge in a cycle, you get a formula for $ A$ in terms of $ B$, and by considering the inclusion or non-inclusion of the last edge in a path, you get a recursion for $ B(n)$.[/quote]
That's what I wrote:
 [quote](And I forgot to say that I consider cyclically different matchings the same. Otherwise this number is just some Fibonacci number.)[/quote]
But it's a bit different problem.


geometry
trapezoid
ARML
number theory
relatively prime

In trapezoid $ ABCD$, with $ AB<DC$, the sum of the areas of regions I and II is 1996. If the lengths of bases $ AB$ and $ CD$ are integers and the distance between them is an integer, compute the minimum area of $ ABCD$  (See attached picture)

I read the ARML contest book's solution to the following problem and I don't understand the motivation for the steps of the solution. Some steps that I did not understand included:
(a) Temporarily assuming that lengths AB and DC were relatively prime
(b) Solving for the height of the trapezoid in terms of other lengths.

It's asking for minimum area, so having side lengths that aren't relatively prime would give you a multiple of the answer you wanted.  Also, having the height of the trapezoid in terms of other lengths lets you use stuff associated with it (perhaps the triangles formed?).

Oh! I get it..

What are some other ways to minimize lengths or areas in geometry problems?

[hide]
Let $ AB\equal{}m$, $ DC\equal{}n$. Since triangles $ APB$ and $ CPD$ are similar, we introduce another variable $ k$ such that the distance from point $ P$ to $ AB$ and $ DC$ are $ km$ and $ kn$ respectively.

It follows that the height $ h\equal{}km\plus{}kn$ so $ k\equal{}\frac{h}{m\plus{}n}$.

Thus $ [APB]\equal{}0.5km(m)\equal{}\frac{hm^2}{2(m\plus{}n)}$.

Similarly, $ [DPC]\equal{}0.5kn(n)\equal{}\frac{hn^2}{2(m\plus{}n)}$.

Plugging our known values of the area of the entire trapezoid and setting it equal to the general formula for the area of a trapezoid, we find that
$ 1996\plus{}\frac{hm^2}{2(m\plus{}n)}\plus{}\frac{hn^2}{2(m\plus{}n)}\equal{}\frac{1}{2}h(m\plus{}n)$

Isolating h, we find that $ h\equal{}\frac{1996(m\plus{}n)}{mn}$.

The area of $ ABCD$ is then $ \frac{1998(m\plus{}n)^2}{mn}\equal{}998(\frac{m}{n}\plus{}\frac{n}{m}\plus{}2)$

To minimize this experssion, we assume that $ m$ and $ n$ are relatively prime. Thus, m+n and mn are relatively prime so both m and n must be divisors of 1996*

Choosing $ m\equal{}1$ and $ n\equal{}2$, $ \frac{m}{n}\plus{}\frac{n}{m}\equal{}5/2$, so the area of ABCD is $ 4491$.

[/hide]

Actually, why must m and n must be divisors of 1996? Why are m+n and mn are relatively prime?

They must be divisors of $ 1996$ because you had $ \displaystyle h\equal{}\frac{1996 \left(m\plus{}n\right)}{mn}$, and the problem says that $ m, n, h \in \mathbb{Z}$. Also, we can assume they are relatively prime, because even if they are not, the common factors will cancel in the fractions. For example, let's consider the numbers $ km$ and $ kn$. Then:
\[ \frac{km}{kn}\plus{}\frac{kn}{km}\equal{}\frac{m}{n}\plus{}\frac{n}{m}\]In order words, every fraction can be written in simplest form, so why bother considering multiples of numbers?


limit
logarithms
real analysis
real analysis unsolved

Study for the the sequence $x_{0}>0,t>0$  $x_{n+1}=x_{n}+\frac{1}{e^{tx_{n}}}$ the limit $\lim_{n\rightarrow \infty}\frac{x_{n}}{\ln n}$.

The limit is 1/t.
To see this we apply Cesaro stolz lemma to get that:

$\lim_{n\rightarrow \infty}\frac{x_{n}}{\ln(n)}=\lim \frac{ne^{-tx_{n}}}{n\ln \frac{n+1}{n}}=\lim \frac{n}{e^{tx_{n}}}=\lim \frac{1}{e^{tx_{n+1}}-e^{tx_{n}}}$

$=\lim \frac{1}{e^{tx_{n}}\left(e^{t(x_{n+1}-x_{n})}-1\right)}=\lim \frac{e^{-tx_{n}}}{e^{te^{-tx_{n}}}-1}=1$,
since $\lim_{x\rightarrow 0}\frac{x}{e^{tx}-1}=\frac{1}{t}$.

It is not hard to see that $x_{n}\rightarrow \infty$ hence $e^{-tx_{n}}\rightarrow 0$.

A connection:

Find the same limit for the case when  the sequence $(x_{n})_{n\in N}$ is defined by the following recurrence relation:

$x_{n+1}=x_{n}+\frac{1}{e^{t(x_{1}+x_{2}+\cdots+x_{n})}}$, $x_{1}>0$ and $t>0$, $n\geq 1$.

Observe that the limit is finite.


Prove that: the binomial coefficient
                        $ C_n^k\equal{}\frac{n(n\minus{}1)...(n\minus{}k\plus{}1)}{1.2...k}$ with $ n\geq 2k$
always a prime factor $ p>k$
I read it in a book but I can not prove it
Can any one hel me?

for a nice proof of Bertrand's postulate with $ \binom{2n}{n}$ see here:
http://en.wikipedia.org/wiki/Proof_of_Bertrand%27s_postulate


probability
integration
inequalities
real analysis
real analysis unsolved

[hide]$\left(\int \sum_{i}\chi_{A_{i}}\right)^{2}\le\int \left(\sum_{i}\chi_{A_{i}}\right)^{2}=\int \sum_{i,j}\chi_{A_{i}\cap A_{j}}$[/hide]

And that inequality is precisely the Schwarz inequality on a probability space :)

What is the improvement of Borel-Cantelli, then?

Let $(A_{n})_{n}$ be a sequence of events satisfying $\sum_{i}P(A_{i})=\infty$.  Then $P(\limsup A_{i})=1$ if $A_{j}$ and $A_{k}$ are independent whenever $|j-k|>N$ for some nonnegative integer $N$.

The hard part of the proof is showing that $P(\limsup A_{i})$ is at least as big as the quantity in my first post in this thread.  I think there's a proof in (Frank?) Spitzer's book on Random Walks.


In ABC , AC=BC, the bisectrix CC' and AA' are in the following relation : AA'=CC'. Find the measure of C.

[hide][b]Lemma[/b]: If two internal bisectors in a triangle are equal, then the corresponding sides are equal as well.

[b]Solution[/b]: By the given conditions, $AC=BC$ and $BC=AB$, hence the triangle is equilateral, giving $\angle C=60^\circ$.[/hide]


geometry
rectangle
analytic geometry
graphing lines
slope

A string is wound symmetrically around a right circular cylinder which has height 12 cm and circumference 6 cm, starting at one end of the cylinder and finishing at the other end, having gone exactly 6 times around the cylinder. How long is the string?

A. 48 cm
B. 12$ \sqrt{10}$ cm
C. 20$ \sqrt{3}$ cm
D. 18$ \sqrt{5}$ cm
E. 36 cm

[hide] Okay, so "unwind" the cylinder to get a rectangle with height 12 cm, and a width of 6 cm. Now, we know that the string starts on the bottom left of the rectangle, and follows a slope up until it hits the right edge. After hitting the right edge (since this is a 3D cylinder) it reappears on the left at the same height it hit the right edge on and continues up at the same slope. Now, since it winds around 6 times, it does this pattern 6 times, forming 6 right triangles with legs of length 6 cm and $ \frac{12}{6}\equal{}2$cm. Now, its a simple job to find the length of the string:

$ 6^2\plus{}2^2\equal{}c^2$
$ 40\equal{}c^2$
$ 2\sqrt{10}\equal{}c$

Now, since there are 6 of these triangles, we multiply:

$ 6 \cdot 2\sqrt{10}\equal{}12\sqrt{10}$

Which is our answer.

B[/hide]


trigonometry
trig identities
Law of Cosines

i was arguing with a friend about if you are given two sides of a triangle and the angle in between, you can figure out the perimeter. Is it really possible, i mean that is what SAS is all about, right?, and there is such thing as trigonometry and sines, secants, etc.?

Yes it is possible. Use sines and cosines.

Yeah, if you have two sides and the angle in-between, you can use Law of Cosines: $c^2=a^2+b^2-2ab\cos C$, where $C$ is the angle opposite $c$. However, that doesn't really belong in this forum, since it includes trig.


$ a,b,c > 0$
\[ 17\sum a^2 \plus{} 13\sum ab\ge\sum(8c \plus{} a \plus{} b)\sqrt {(2c \plus{} a)(2c \plus{} b)}\ge16\sum{a^2} \plus{} 14\sum{ab}
\]

\[ \sum\sqrt {\frac {(4c \plus{} a)(4c \plus{} b)}{c^2}}\ge\frac {8(\sum{ab})^2}{\sum{a}\cdot abc} \minus{} 9
\]

[quote="shfdfzhjj"]$ a,b,c>0$
\[ \sum_{cyc}\frac{ab}{c}\plus{}\sum{b^2}\ge3\plus{}\frac{3(\sum{a^2b^2\plus{}\sum{a^3bc}\minus{}1\minus{}abc})}{(abc\plus{}1)^2}\][/quote]

This one is easy.

[quote="shfdfzhjj"]\[ \sum\sqrt {\frac {(4c \plus{} a)(4c \plus{} b)}{c^2}}\ge\frac {8(\sum{ab})^2}{\sum{a}\cdot abc} \minus{} 9
\][/quote]
This inequality is wrong. Try $a=2,$ $b=c=1.$
[quote="shfdfzhjj"]$ a,b,c>0$
\[ \sum_{cyc}\frac{ab}{c}\plus{}\sum{b^2}\ge3\plus{}\frac{3(\sum{a^2b^2\plus{}\sum{a^3bc}\minus{}1\minus{}abc})}{(abc\plus{}1)^2}\][/quote]
This can be proved using the AM-GM Inequality.

[quote="shfdfzhjj"]$ a,b,c > 0$
\[ 17\sum a^2 \plus{} 13\sum ab\ge\sum(8c \plus{} a \plus{} b)\sqrt {(2c \plus{} a)(2c \plus{} b)}\ge16\sum{a^2} \plus{} 14\sum{ab}
\]
[/quote]
The left inequality follows from the inequality
$(8a+b+c)[(2a+b)+(2a+c)] \ge 2(8a+b+c)\sqrt{(2a+b)(2a+c)}.$
The right inequality follows from the inequality
$(8a+b+c)\sqrt{(2a+b)(2a+c)} \ge 16a^2+6ab+6ac+2bc.$


Prove that

$ \tan(\frac{1}{4}\pi \minus{} \frac{1}{2}x) \equiv \sec x\minus{} \tan x$. $ \, (*)$

(i) Use (*) to find the value of $ \tan \frac{1}{8} \pi$. Hence show that

$ \tan \frac{11}{24}\pi \equal{} \frac {\sqrt{3} \plus{} \sqrt{2} \minus{}1}{\sqrt{3}\minus{}\sqrt{6} \plus{} 1}.$

(ii) Show that

$ \frac{\sqrt{3}\plus{}\sqrt{2}\minus{}1}{\sqrt{3}\minus{}\sqrt{6}\plus{}1} \equal{} 2 \plus{} \sqrt{2}\plus{}\sqrt{3} \plus{} \sqrt{6}$.

(iii) Use (*) to show that

$ \tan \frac{1}{48}\pi \equal{} \sqrt{16 \plus{} 10\sqrt{2} \plus{} 8\sqrt{3} \plus{}6\sqrt{6}}\minus{}2\minus{}\sqrt{2} \minus{} \sqrt{3} \minus{}\sqrt{6}$.

It just want some manipulation.
\[ \sec x \minus{} \tan x \equal{} \frac 1{\cos x} \minus{} \frac {\sin x}{\cos x} \equal{} \frac {1 \minus{} \sin x}{\cos x}
\]

\[ \Rightarrow \sec x \minus{} \tan x \equal{} \frac {(\sin \frac x 2 \minus{} \cos \frac x 2)^2}{\cos^2 \frac x 2 \minus{} \sin^2 \frac x 2} \equal{} \frac {(\sin \frac x 2 \minus{} \cos \frac x 2)^2}{(\cos \frac x 2 \minus{} \sin \frac x 2)(\cos \frac x 2 \plus{} \sin \frac x 2)}
\]

\[ \Rightarrow \sec x \minus{} \tan x \equal{} \frac {\cos \frac x 2 \minus{} \sin \frac x 2}{\cos \frac x 2 \plus{} \sin \frac x 2} \equal{} \frac {\sqrt 2 \sin( \frac {\pi}{4} \minus{} \frac x 2)}{\sqrt 2 \cos( \frac {\pi}{4} \minus{} \frac x 2)} \equal{} \tan (\frac {\pi}{4} \minus{} \frac x 2)
\]
(i) From above; we have:
\[ \tan \frac {\pi}8 \equal{} \sec \frac {\pi}4 \minus{} \tan \frac {\pi}4 \equal{} \sqrt 2 \minus{} 1
\]

\[ \tan \frac {11} {24} \pi \equal{} \tan (\frac {12}{24}\pi \minus{} \frac {1}{24}\pi) \equal{} \cot \frac {\pi}{24} \equal{} \frac {1}{\tan \frac {\pi}{24}}
\]

\[ \tan \frac {\pi}{12} \equal{} \sec \frac {\pi}{3} \minus{} \tan \frac {\pi}{3} \equal{} 2 \minus{} \sqrt 3
\]
let $ \tan \frac {\pi}{24} \equal{} a$ then:
\[ \tan \frac {\pi}{12} \equal{} \frac {2a}{1 \minus{} a^2}
\]

\[ \Rightarrow 2 \minus{} \sqrt 3 \equal{} \frac {2a}{1 \minus{} a^2} \Rightarrow 2 \minus{} \sqrt 3 \minus{} (2 \minus{} \sqrt 3)(a^2) \equal{} 2a
\]

\[ \Rightarrow (2 \minus{} \sqrt 3)(a^2) \plus{} 2a \minus{} (2 \minus{} \sqrt 3) \equal{} 0 \Rightarrow a \equal{} \frac { \minus{} 2\pm \sqrt {4 \minus{} 4( \minus{} 2 \plus{} \sqrt 3)(2 \minus{} \sqrt 3)}}{2(2 \minus{} \sqrt 3)}
\]
Since $ a > 0$ so
\[ \Rightarrow a \equal{} \frac { \minus{} 1 \plus{} 2 \sqrt {2 \minus{} \sqrt 3}}{2 \minus{} \sqrt 3}
\]
now with some manipulation the desired result will be obtain.
for other sections do like this.

Thanks madadpoor. Unfortunately, I cannot get to see hidden text on my computer.

For (i) it seems much easier to do

$ \tan \frac{11\pi}{24} \equal{} \tan ( \frac{\pi}{8} \plus{} \frac{\pi}{3})$

$ \frac{\tan \frac{\pi}{8} \plus{} \tan \frac{\pi}{3}}{1\minus{}\tan \frac{\pi}{8} \tan \frac{\pi}{3}}$

$ \equal{}\frac{\sqrt{2} \minus{} 1 \plus{} \sqrt{3}}{1\minus{}(\sqrt{2} \minus{} 1)\sqrt{3}}$

$ \equal{} \frac{\sqrt{3} \plus{} \sqrt{2} \minus{}1}{\sqrt{3} \minus{} \sqrt{6} \plus{} 1}$.

For part (ii), multiplying above and below by $ \sqrt{3} \minus{} \sqrt{6} \minus{}1$ gives $ \frac{2 \minus{} 2\sqrt{2} \minus{}2\sqrt{3} \plus{} \sqrt{6}}{4 \minus{} 3\sqrt{2}}$.

Multiplying above and below by $ 4\plus{}3\sqrt{2}$ then gives the required result.

Any offers for part (iii)?

[quote="AndrewTom"]For (i) it seems much easier to do

$ \tan \frac {11\pi}{24} \equal{} \tan ( \frac {\pi}{8} \plus{} \frac {\pi}{3})$

$ \frac {\tan \frac {\pi}{8} \plus{} \tan \frac {\pi}{3}}{1 \minus{} \tan \frac {\pi}{8} \tan \frac {\pi}{3}}$

$ \equal{} \frac {\sqrt {2} \minus{} 1 \plus{} \sqrt {3}}{1 \minus{} (\sqrt {2} \minus{} 1)\sqrt {3}}$

$ \equal{} \frac {\sqrt {3} \plus{} \sqrt {2} \minus{} 1}{\sqrt {3} \minus{} \sqrt {6} \plus{} 1}$.

For part (ii), multiplying above and below by $ \sqrt {3} \minus{} \sqrt {6} \minus{} 1$ gives $ \frac {2 \minus{} 2\sqrt {2} \minus{} 2\sqrt {3} \plus{} \sqrt {6}}{4 \minus{} 3\sqrt {2}}$.

Multiplying above and below by $ 4 \plus{} 3\sqrt {2}$ then gives the required result.

Any offers for part (iii)?[/quote]

$ \tan \frac {\pi}{48}\equal{}\tan\frac {12\pi\minus{}11\pi}{48}\equal{}\tan\left(\frac {\pi}{4}\minus{}\frac {11\pi}{48}\right)\equal{}\sec \frac {11\pi}{48}\minus{}\tan \frac {11\pi}{48}$

$ \implies\tan \frac {\pi}{48}\equal{}\sqrt{1\plus{}\tan^2\frac {11\pi}{48}}\minus{}\tan \frac {11\pi}{48}$

$ \implies\tan \frac {\pi}{48}\equal{}\sqrt {1\plus{}\left(2\plus{}\sqrt{2}\plus{}\sqrt{3}\plus{}\sqrt{6}\right)^2}\minus{}2\minus{}\sqrt{2}\minus{}\sqrt{3}\minus{}\sqrt{6}$

$ \implies\tan\frac{\pi}{48}\equal{}\sqrt{16\plus{}10\sqrt{2}\plus{}8\sqrt{3}\plus{}6\sqrt{6}}\minus{}2\minus{}\sqrt{2}\minus{}\sqrt{3}\minus{}\sqrt{6}$

Thanks makar. (The $ 48$s at the end of your first two lines are typos for $ 24$s.) Much appreciated.


[quote="mathson"]where must I put packages on mac?[/quote]

I downloaded "olympiad.asy", if i put it in

[quote]Application/asymptote-1.43/base[/quote]

It gives the error

[quote] error: could not load module 'olympiad'[/quote]

[quote="fedja"]I use PC, so I cannot tell you the exact directory name for mac, but it should be in the same directory as, say, graph.asy or math.asy. Just search for those and you'll know the answer :).[/quote]

I put it in the same directory:
Application/asymptote-1.43/base

And I get:
[quote]error: could not load module 'olympiad' [/quote]

This is strange. Do the following:

a) check if import graph; works fine.

If not, you have a path problem and we'll have to play with settings a bit. If yes,

b) create some essentially empty file named mypackage.asy (the content should be something like pair A=(0,0); ) and put it to that base directory.

c) include the line import mypackage; into some working .asy file in your working directory and try to execute it. See if it works.

If it doesn't, I'll be very surprised and will have to think more. If it does,

d) put the olympiad package into your working directory, i.e., the same directory where the file calling it is. See if it can be called without an error now.

If it works, this will solve the problem temporarily but we'll still have to see what the problem is. If it doesn't, put mypackage.asy there and try to call it instead. If the problem happens even with it, reinstall asymptote. If not, try to execute olympiad.asy that sits in your working directory and see if you have some error messages.

Post your observations and we'll proceed from there :)

Import graph is working, put mypackage doesn't work !! It give me the same error:
[quote]error: could not load module 'mypackage' [/quote]
 :?:

[quote="fedja"]That's crazy. Try to put mypackage.asy in the working directory and see if it can be called from there (and double check that it can be executed without an error by itself).[/quote]

Yes !!!  :lol:  That's working. I tried for olympiad.asy and ... also it is working.
But do I need to put this packages when I need them? I'm usually work in different directories.

Thanks again fedja  :blush:

OK, create some harmless mypackage2.asy and put it in the base directory you mentioned.
Now, create the testpath.asy file in your working directory with the following content

access settings;
settings.dir="Application/asymptote-1.43/base";
import mypackage2;
size(200);
draw((0,0)--(1,1));

(I assume that this string is a full path, i.e., if you write [b]cd Application/asymptote-1.43/base[/b] (or whatever the change directory command is on Mac), it'll bring you to this directory from wherever you are. On Windows, you should start full path with the drive name, but I'm completely ignorant about Mac conventions in this respect. If it is not a full path, you'll have to figure out what the full path is first and put it instead).

Does this file give you an error?

Yes  :!:
[quote]error: could not load module 'mypackage2'[/quote]

Yeah, that's fine. Apparently an attempt to reset settings.dir from the executable file doesn't work :(. What you really need to do then is to create the file named config.asy with the lines

import settings;
settings.dir="Application/asymptote-1.43/base";

and put it in the same directory where asy.exe is.

After you've done it, run Asymptote in the interactive mode and execute two lines

access settings;
settings.dir;

if it shows Application/asymptote-1.43/base as the result, you should be fine. Try some file with import mypackage2; now and see if it works. In all other cases, we'll need to place that config.asy file somewhere else (i.e., the problem reduced to itself). Can you tell me the whole asymptote related directory tree?

Sorry, but I can't understand this step. Because I don't have asy.exe. Also, I don't know how to run Assympotate !!!

[quote="fedja"]OK, what is the name of the Asymptote executable program file on Mac then? There should be one, right? Use it instead of asy.exe. To run Asymptote in the interactive mode, just run that file.[/quote]

OK, I did that and I got:

[quote]1.8: no matching variable 'setting.dir'[/quote]

[quote="fedja"]setting[b]s[/b].dir, not setting.dir![/quote]

Ops  :blush: !
Sorry. Ok, I did that but I did not get anything.

[quote="fedja"]This means that the file config.asy should be placed somewhere else, i.e., the problem reduced to itself. Rather than trying the directories one by one, place one copy of it into every Asymptote related directory and see if it works after that. Also, read [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=223490]this thread[/url] and the threads mentioned in it and see if you can find anything relevant there.[/quote]

I tried to copy it in every directory, but it still not working.


Did anyone here participate? Did anyone here go to NYC yesterday?

For those of you who haven't heard of it, look it up. It's applied math/math modeling, it's fun, and the prizes are sick. Winning team gets 20,000 USD

It's limited to the North-East US (check the website) now, but it's getting bigger every year.

Yeah, I did that this year.  Didn't win anything though... :(

I did too
won nothing.
I thought we were amazing.


Let ABC be a triangle with acute angle A and AC= 2 AB . The angle bisector AD cuts the altitude BH at K ( D lies on BC , H on AC) . The line CK cuts AB at E . Prove that ABC is right at B when and only when the areas of the triangles BDE and HDE are equal

You've also posted this in intermediate -- please don't post the same questions in different places.


If $ y\triangleleft$ is defined as $ y^2 \minus{} y$, find $ ((3\triangleleft)\triangleleft)\triangleleft$.

Just like the other problem, simply plug and chug. $ 3^2 \minus{} 3 \equal{} 6 \implies 6^2 \minus{} 6 \equal{} 30 \implies 30^2 \minus{} 30 \equal{} \boxed{870}$.


[b][u]Dirichlet's Theorem[/u][/b]

If $(a,b)=1$ then there exist infinite numbers of prime in the form of $an+b$ for positive integers of $n$.

Divisors of $n^{2}+1$. And very likely to be posted before.

[quote="Bictor717"]Suppose there were a finite number of such primes. Take their product, add 4 and get a new 4k+1 prime?[/quote]
it's not true for example $5.13+4=69=3.23$ ! and $3,23$ are $4k+3$ . you must prove that if $p|n^{2}+1$ then $p=2,4k+1$
and then you put $n=2p_{1}p_{2}...p_{k}$ where $p_{1},...,p_{k}$ are the all prime numbers in the form of $4k+1$ .

[quote="durmstrang"]How can we prove that there are infinite number of primes that take the form of [b]4k + 1[/b] ?[/quote]

First, I like to mention that is really upsets me when someone answers this question "...use Dirichlet's theorem..." Because it is a useless post. I think the original poster knows this. And doesnt want to use this theorem.
---

Assume there is a finite amount of primes of this form $p_{1},p_{2},...,p_{k}$. For the integer $n=(2p_{1}p_{2}...p_{k})^{2}+1$. Cleary $n>1$ is odd, so it is divisible by some odd prime $p$. That is,
$(2p_{1}p_{2}...p_{k})^{2}\equiv-1 (\bmod p)$
But that means the the Lengendre symbol $(-1/p)=1$.
And that only happens when $p\equiv 1(\bmod 4)$.
So $p$ has to be among those primes, an impossibilitiy for it would imply that $p|1$ which is clearly false.


MATHCOUNTS
Columbia
geometry
analytic geometry
trigonometry
Support
AMC

Was anyone here at the Seattle MATHCOUNTS? If Adam Hesterberg's reading, you gave me this logic problem there about invisible flies. (Mr.Howell told you to give the Eckstein team a logic problem apparently.) So what's the answer?

Adam probably wont give you the answer...although i have solved it, i will not give it to you either... :D

it is a very beautifull problem, its worth the time to think on it... (took me almost 3 months)...

also about the spider problem, look here(the answer is not there, but you can find some hints):

http://www.artofproblemsolving.com/Forum/viewtopic.php?highlight=invisible+fly&t=64295

1: No discussion of the problems or scores on the official competition, please (see announcement in MATHCOUNTS forum).

2: I was there as a grader, and I got down to see the Countdown round- Kelsey was amazing.

Yeah, Kelsey was crazy! I had to skim all the problems to even know what the question was asking before she answered. It was a bit sad I finished the way I did because I got a really good score. I was in a tie for the highest score in the competition.

I volunteered- I'm currently a grad student at the university, so I got a message asking for volunteers, and I went for it. I care about Mathcounts because it's how I got started in contests- and I made it to nationals on the California team in 1996.

Do you any advice then? Any formulas or useful strategies, especially for the sprint/target rounds?

Hailing from the Lake Washington Chapter!

Who's moving on to state from Seattle?

Team wise, Washington MS is, Lakeside MS, Eckstein MS and Seattle Country Day. Washington totally creamed everyone and was close to a perfect team score. All 4 members on the team (plus another student not on the official team) were in the top 10.

Individually, Kelsey James (Eastside Homeschool) went from 10th to 1st, Andy Morgan (WMS) was 2nd, Jack Coughlin was 3rd (WMS), I was 4th (Eckstein), and that's all I remember well.

What happened at your chapter?

Teams: Northshore, Redmond (me), Odle, and Islander I think.

Northshore had a perfect score, but he wasn't on their official team  :o

Isn't Sam Keller on your team? All the math coaches I know keep telling me how great he is. Gary Pounder, the coach of WMS, calls him "the goods." Who wasn't on the Northshore official team? How many perfect scores were there? My coach heard there were 3, but there's probably more. Nobody in Seattle had a perfect score, but 3 (including me) were 1 away.

Sam Keller....oh, yeah, I know him.  He's like my brother's brother or something  ;)

3 perfect...me, some seventh grader from odle, and Nick Defiesta of Northshore.

So Northshore couldn't have had a perfect score if all the members didn't. Do you remember the individual places? I think your chapter is the hardest in the state. Who's on your team?

My team....nobody you've heard of.

And I meant that there was somebody with a perfect score from Northshore, not that Northshore had a perfect score  :blush:  :lol:  :rotfl: .

Individual places that I can remember

1. Kerry Xing (Northshore) (He moved from 4th to 1st in the countdown, he got a 45 and was on the Washington team with me last year)
2. Me
3. Nick Defiesta (Northshore)
4. Derek something (Odle 7th grade)
9. Puneet Dewan (7th Grader from Redmond)
10. Evan Sheih (Pine Lake)

I can't remember any others. Everyone I listed is going to state.

What's nationals like? What score did you have to get to be in the top 4?
It seems like this year's chapter was a lot easier than other years. Last year I got a very bad 33 because I really didn't care for math competitions last year. At least I got 12 more points this year.
For the last 2 competitions, I've been sick. It sucks. At Math is Cool Masters I had like 2 hours of sleep and lots of coffee, even though I hate coffee. I was 9th and they gave awards up to 8th.

Yes.  Why?  You sound like you think it's a bad thing.

well, I was just wondering if they would put the girl in the same room with a guy. nothing bad would happen, but it'd feel kinda awkward.

No, they put them with people of the same gender from other states.

do you know a  fast way to do this problem w/o a calculator:
what is the greatest common divisor of 10961 and 11413?

Ooh, ooh, it was called Euler's watchamacallit or something...

Anyway, if 10961 and 11413 are both divisible by x, then isn't (11413-10961) also divisible by x?

So the problem is really asking for the GCF of 10961 and 452
Which is the GCF of 10509 and 452
Which is the GCF of 10157 and 452
.
.
.
Which is the GCF of 113 and 452
Which is 113.

Basically you keep on subtraction the smaller of the two until it's manageable enough to find the greatest common divisor.

I knew it had something to do with subtrating them!

And I never figured out how to do these kinds of problems quickly: if you can order boxes containing 6, 9 or 20 pieces, what is the greatest number of pieces that you can't order if you have to buy whole boxes?

Use modular arithmetic (do you know what that it is?).  We can get any number that is equivalent to 1 (mod 6) with the equation $20+20+9+6x=49+6x$, where x is a whole number.  Similarily, you can get any number that is equivalent to:
2 (mod 6): $20+6x$
3 (mod 6): $9+6x$
4 (mod 6): $20+20+6x=40+6x$
5 (mod 6): $20+9+6x=29+6x$
0 (mod 6): $6x$

Obviously, every number equal to or greater than 49 can be had.  The greatest numnber of boxes possible, then, is equal to the greatest possible number that fits no e of the equations when x is not positive but is an integer: $49+(-1)x=43$

Is there a chicken McNugget theorem for 3 or more cases?

what scores did you all get at state mathcounts?

There's already a thread with this topic in the Mathcounts section of the forum.

I know, but there are pages and pages of other people's scores and conversations.

Jonathan and cowpi, does Joy have an AoPS account?   If not, I think we should get her to get one.

hey ppl
who are on your state team?
tell me the names & score


function
real analysis
real analysis unsolved

How to construct a function f:

1)f is a continuous function which defined on [0,1]
2)f(0)=f(1)
3)∀x∈[0,1], f( x+(2/5) ) ≠  f(x)

or f does not exist?
because I don't have any idea to construct a function satisfied the above conditions.

thanks~

Consider the points $ P_1(0,0), P_1(1/5, \minus{}2), P_2(2/5, 1), P_3(3/5, \minus{}1), P_4(4/5, 2), P_5(1,0)$, and connect them by a broken line to get such a function $ f$, since $ f(x\plus{}2/5) \minus{} f(x) \equal{} 1$ over all $ [0, 3/5]$ (you cannot consider points in $ (3/5, 1]$, since $ f(x\plus{}2/5)$ is not defined there).


In a photographer's studio, a 4-inch wide frame is placed around a
1-foot square photo of a champion dog. The photo is then enlarged
by multiplying each dimension by $ y$. Given that the dimensions of the
enlargement are now the same as the outside of the original frame, find
the value of $ y$. Express your answer as a common fraction.

The width of the original photo is $ 1\text{ foot}\left(\frac{12\text{ inches}}{\text{foot}}\right)\equal{}12\text{ inches}$. After putting the frame around, we enlarge its dimension by $ 2(4\text{ inches})\equal{}8\text{ inches}$. That means the new width is $ 20\text{ inches}$.
Now we have to find a number $ n$ such that $ 12n\equal{}20$. Dividing both sides by $ 12$ gives $ n\equal{}\frac{20}{12}\equal{}\boxed{\frac{5}{3}}$


The scores are not based on the comments.  The graders might deduct for something that they didn't necessarily comment upon, and as it says at the top of the grading rubric there is a 1-point discretionary factor in all cases.

To have a valid protest, you must specifically show how your solution was graded incorrectly based on the grading criteria on the USAMTS website.  What may or may not be written in the comments box is not necessarily relevant.

[quote="mathtyro"]Should I protest if there is a error(as far as I understand) in the comments itself? I don't know how much that affected my score or anything, but is there any way I can ask/comment on the comments - if I don't understand the comments/why the comments are true, for example.

Any advice on this situation would be helpful - but I am reluctant to post the comment until I receive some advice.[/quote]
We graders write our comments in haste and in ink, so sometimes our phrasing is unclear and our emphasis is misplaced. We usually try to explain the error in the solution, but we also give suggestions on how to avoid similar errors in future solutions. And sometimes we lack the space to mention all the errors.

However, on rare occasions a grader misunderstands a solution, especially one using an innovative technique, and misinterprets it as an incomplete solution based on another technique. In that case, an erroneous comment is a clue that the grader misunderstood the solution and the solution should be graded anew. To determine this, check your solution against the grading rubric and see if it deserves its score according to the rubric.

I graded Problem 1 this round, so if it was a comment on that problem, you could repeat it here and I might be able to explain it in detail.

Erin Schram
USAMTS grader


Melanie ran at $ 5$ m/s for $ 400$ meters, then ran at $ 4$ m/s for the next $ 800$ meters, and finally ran at $ 8$ m/s for the final $ 400$ meters. What was her average speed, in meters per second, for the $ 1600$ meter run?  Express your answer as an improper fraction.

Her average rate is the distance travelled divided by the time taken. We already know the distance is 1600 meters, so now we calculate the time.

5 m/s for 400 meters takes 80 seconds.
4 m/s for 800 meters takes 200 seconds.
8 m/s for 400 meters takes 50 seconds.
This totals 330 seconds.

$ \frac {1600}{330} \equal{} \frac {160}{33}$ m/s.

i agree with your solution, though i think it's much faster to just do 80+200+50 and then divide by 400+800+400.
words are useles.


inequalities
rearrangement inequality
inequalities unsolved

Ten people queue up before a tap to fill their buckets. Each bucket requires a different time to fill. In what order should the people queue up so as to minimize their combined waiting time?

In my solution, I assume that the time spent by a person while their bucket is being filled is included in their waiting time. If not, then the solution is almost identical.

[hide="Solution"]Let $ a_i$ be the amount of time required to fill the bucket held by person $ i$.

Person 1's waiting time: $ a_1$.
Person 2's waiting time: $ a_1\plus{}a_2$.
Person 3's waiting time: $ a_1\plus{}a_2\plus{}a_3$.
...
Person 10's waiting time: $ a_1\plus{}a_2\plus{}a_3\plus{}a_4...\plus{}a_{10}$

Summing these, we get that the total combined waiting time is:

$ a_{10}\plus{}2a_9\plus{}3a_8\plus{}4a_7...\plus{}10a_1$.

And by the rearrangement inequality, this sum is minimal when $ [1,2,3,4...,10]$ and $ [a_{10},a_9,a_8,a_7...,a_1]$ are oppositely sorted.

Thus, $ a_1 < a_2 < a_3 < a_4 ... < a_{10}$.

So the people should queue with the shortest fill time at the front, then the second shortest, and so on in increasing order of fill time.[/hide]


I have a drawer with 6 forks, 6 spoons, and 6 knives in it. If I reach in and randomly remove three pieces of silverware, what is the probability that I get one fork, one spoon, and one knife?

There are $ 6 \times 6 \times 6 \equal{} 216$ ways to select one knife, one spoon, and one fork. There are $ \binom{18}{3}$ ways in total to select three objects. Thus our answer is $ \frac{216}{816}$ or $ \boxed{\frac{9}{34}}$.

We first find the numerator which is 6*6*6=216 then we need the denominator which is 18C3=816

our answer is 9/34.


[quote="moldlee"]There is an $11 \times 6\times 5$ solid made up of unit cubes.

A termite chewed throught it by one of its diagonals. How many unit

cubes did the termite chew through?[/quote]
[hide]
Include exclude
20[/hide]

all of them are relatively coprime so it's fairly easy to see that it's 5 + 6 + 11 - 2 or 20.


Να υπολογιστεί το  $\int^{\frac{\pi}{2}}_{-\frac{\pi}{2}}(-x^4+8x^2+2x)sinxdx$

kai to  $\int^{\frac{\pi}{2}}_{-\frac{\pi}{2}}\frac{cosx}{1+e^x}dx$

Σιλουανέ  σόρυ  που  θα  στο  πώ  αλλά  τα  ολοκλήρωματα  αυτά  είναι  αρκετά  εύκολα. Να  φανταστείς  για  το πρωτο :  την  μέθοδο  για  τον  υπολογισμό  τέτοιων  ολοκληρωμάτων  την  έχει  μέχρι  και  το  σχολικό  βιβλίο................ :o  :o  .  Το  δε  δέυτερό  αρκεί  να  το  σπάσεις  σε  δύο  ολοκληρωματα  και  κάνοντας  τον  μετασχηματισμό  x = -u  και  ύστερα  από  λίγες  πράξεις  καταλήγεις  σε  ένα  στοιχειώδες  ολοκλήρωμα . :lol:


Let:$ a\neq b\neq c;a;b;c>0$.Prove that:
$ \frac {a^4}{(a \plus{} b)(a^2 \plus{} b^2)} \plus{} \frac {b^4}{(b \plus{} c)(b^2 \plus{} c^2)} \plus{} \frac {c^4}{(c \plus{} a)(c^2 \plus{} a^2)} > a \plus{} b \plus{} c \minus{} \frac {1}{2}(\frac {a^3 \minus{} b^3}{a^2 \minus{} b^2} \plus{} \frac {b^3 \minus{} c^3}{b^2 \minus{} c^2} \plus{} \frac {c^3 \minus{} a^3}{c^2 \minus{} a^2})$

I'll take $ a,b,c > 0$ :)

Since $ LHS\ge \frac {a \plus{} b \plus{} c}{4}$ it is enough to prove that :

$ \frac{1}{2}\sum_{cyc}\frac {a^2 \plus{} ab \plus{} b^2}{a \plus{} b}\ge \frac {3}{4}(a \plus{} b \plus{} c)$ , but this inequality results from :

$ \sqrt {x^2 \plus{} xy \plus{} y^2}\ge \frac {\sqrt {3}}{2}(x \plus{} y)$ , which is a trivial inequality .

Equality occurs only when $ a \equal{} b \equal{} c$ , so there's no equality .

[quote="alex2008"]I'll take $ a,b,c > 0$ :)

Since $ LHS\ge \frac {a \plus{} b \plus{} c}{4}$ it is enough to prove that :

$ \sum_{cyc}\frac {a^2 \plus{} ab \plus{} b^2}{a \plus{} b}\ge \frac {3}{4}(a \plus{} b \plus{} c)$ , but this inequality results from :

$ \sqrt {x^2 \plus{} xy \plus{} y^2}\ge \frac {\sqrt {3}}{2}(x \plus{} y)$ , which is a trivial inequality .

Equality occurs only when $ a \equal{} b \equal{} c$ , so there's no equality .[/quote]

Why $ LHS \ge \frac {a \plus{} b \plus{} c}{4}$
My souliton:It's only Am-Gm:
We need prove:
$ \sum \frac {a^4}{(a \plus{} b)(a^2 \plus{} b^2)} \plus{} \sum \frac {a^2 \plus{} ab \plus{} b^2}{2(a \plus{} b)} > a \plus{} b \plus{} c$(*)
By Am-Gm We have: :P
$ \sum \frac {a^4}{(a \plus{} b)(a^2 \plus{} b^2)} \plus{} \sum \frac {a^2 \plus{} b^2}{4(a \plus{} b \plus{} c)} \ge \sum \frac {a^2}{a \plus{} b} \ge \frac {a \plus{} b \plus{} c}{2}$
and:$ \sum \frac {a^2 \plus{} b^2}{4(a \plus{} b)} \plus{} \sum \frac {ab}{2(a \plus{} b)} \equal{} \sum \frac {a \plus{} b}{4} \equal{} \frac {a \plus{} b \plus{} c}{2}$
Done!  :P

[quote="caubetoanhoc94"]

Why $ LHS \ge \frac {a \plus{} b \plus{} c}{4}$
[/quote]

First let  observe that :

$ \sum_{cyc}\frac{a^4}{(a\plus{}b)(a^2\plus{}b^2)}\equal{}\sum_{cyc}\frac{b^4}{(a\plus{}b)(a^2\plus{}b^2)}$

Then it is enough to prove that :

$ \sum {\frac{{a^4  \plus{} b^4 }}{{\left( {a \plus{} b} \right)\left( {a^2  \plus{} b^2 } \right)}}}  \ge \frac{{a \plus{} b \plus{} c}}{2}$

But :

$ \sum {\frac{{a^4  \plus{} b^4 }}{{\left( {a \plus{} b} \right)\left( {a^2  \plus{} b^2 } \right)}}}  \ge \sum {\frac{{\left( {a^2  \plus{} b^2 } \right)}}{{2\left( {a \plus{} b} \right)}} \ge } \frac{{a \plus{} b \plus{} c}}{2}\left( {Cauchy} \right)$ , done .


Find a solution to the equation if possible.

$ 1 \equal{} 15tan(9x \plus{} 1) \minus{} 6$

Re-arrange this to yield
$ 7\equal{}15\tan(9x\plus{}1)$
Divide both sides by 15:
$ \frac{7}{15}\equal{}\tan(9x\plus{}1)$
Take the arctangent of both sides, subtract 1 from both sides, and divide both sides by 9 to get:
$ \frac{\tan^{\minus{}1}\left(\frac{7}{15}\right)\minus{}1}{9}$
This has multiple values since trig functions are periodic.

hello, $ x\equal{}\minus{}\frac{1}{9}\plus{}\frac{1}{9}(\arctan(\frac{7}{15})\plus{}k\pi)$ with $ k \in \mathbb{Z}$.
Sonnhard.

[b]@ JRav:[/b] The arctan function does not have multiple values; it is a true function. It is the inverse of the branch of tangent function which contains the origin.

How about $ 1 \equal{} 35 tan(4x\plus{}6) \minus{}6$

Same thing...
$ 35\tan(4x \plus{} 6) \equal{} 7$
$ \tan(4x \plus{} 6) \equal{} \frac {1}{5}$
$ 4x \plus{} 6 \equal{} \arctan\frac {1}{5}$
$ x \equal{} \frac {\arctan\frac {1}{5}}{4} \minus{} \frac{3}{2}\approx \minus{}1.451$

Edit: Sorry, stupid mistake, fixed it.

[quote="TZF"]@ JRav: The arctan function does not have multiple values; it is a true function. It is the inverse of the branch of tangent function which contains the origin.[/quote]
See what Dr. Graubner wrote.  I meant that depending on which quadrant you're in, you can get a different angle.

I am such an idiot. :oops: Never mind.  Of course arctangent is one-one or else it couldn't be used as a bijective mapping.

For some stupid reason, I was associating the graph of arctangent with the graph of some other function.  So ignore what I previously claimed.


logarithms
real analysis
real analysis unsolved

Show that the equation $ x(1\plus{}\ln \frac{1}{\epsilon \sqrt{x}})\equal{}1,   x>0,  \epsilon >0$ has, for each sufficiently small $ \epsilon >0$ ,exactly two solutions. Let $ x(\epsilon)$ be the smaller one. Show that

1)$ x(\epsilon) \longrightarrow 0$ as $ \epsilon \longrightarrow 0\plus{}$.

yet for any $ s>0$

2) $ \epsilon ^{\minus{}s} x(\epsilon) \longrightarrow \infty$ as $ \epsilon \longrightarrow 0\plus{}$.

[hide="solution part 0"]
Let $ x (1 \plus{} \ln{1 \over \epsilon \sqrt {x}}) \minus{} 1 \equal{} y$. Since $ y$ is obviously continuous for positive $ x$ and $ \epsilon$, we have $ y' \equal{} \minus{} \ln \epsilon \sqrt {x}$ and $ y'' \equal{} \minus{} {1 \over 2x}$, which give $ y$ a maximum at $ x \equal{} {1 \over \epsilon^2}$, no minimum, and no inflection points. Therefore, if $ 0<\epsilon<1$, $ y$ must cross the x-axis exactly twice.[/hide]


[quote="ch1n353ch3s54a1l"]

[hide]C. just cube root and fourth power it[/hide][/quote]

don't forget your reciprocals

don't bother calculating very much.  take the negative sign on the exponent and flip the fraction.  Then just look at the 27 which when cubed drops to 3 then taken to the 4th power goes to 81 so the numerator will be 81 so the answer is C since that is the only answer with 81 in the numerator.

Hope that helps you do it quickly in your head.


$f(x)$ is a monic $2007$th degree polynomial such that the coefficient of $x$ is $2007$ and the constant term is $-1$.  Given that all the roots of this polynomial are positive, prove that they are all equal.


ratio
IMO Shortlist
geometry proposed
geometry

Let a triangle  $ABC$   and the cevians  $AL, BN , CM$ such that  $AL$  is the  bisector of angle  $A$. If  $\angle ALB = \angle ANM$, prove that  $\angle MNL = 90$.

Let $K$ be the foot of the external bisector of $\angle{A}$.
Then form the Menelaos' theorem we have that the points $N,M,K$ are collinear.
Moreover, the quadrilateral $KANL$ is cyclic, thus $\angle{MNL}=90^\cdot$.

A teriffic solution !
 This momment I read for first time the official solution. It is too composite , but it could be better !!!!. Any way , to help our  younger visitors we must say that first we apply Ceva for the three Cevians .
 Probably this was a problem from IMO shortlist 2005. We' ll see!

babis

It is actually no terrific at all if you have learnt harmonic conjugate, (internal and external) angle bisectors before ;)

Mmm!

 [b]The problem is BALTIC WAY 2002 !!![/b]
 Mecrazywong:

 Yes , we must have strong and wide backround in geometry. I have not it , so I have never been an olympiad contestant in the ...  long past ! For all younger students , the world is better. Believe me that.

 [u]babis[/u]

let $MN$ intersects $AC$ at $T$,then $TALC$ are a harmonic sequence.
hence $\angle TBL=90$
since $\angle ALB=\angle MNB$ we get $BNLT$ cyclic
so $\angle TNL=\angle TBL=90$.


trigonometry
inequalities
inequalities proposed

[quote="hxy09"]Given that $ A,B,C>0$ and $ A + B + C = \pi$
$ x,y,z \ge 0$
Prove or disprove:$ (\frac {x}{\sin A} + \frac {y}{\sin B} + \frac {z}{\sin C})^2\ge 4(xy + yz + zx)$[/quote]
Dear friend, your inequality is true and nice. Here is my work for it, this may have some errors in calculating since I did it in mind and didn't have any paper to write down... Hope you all forgive me if there are any mistakes.

Without loss of generality, we may assume that $ A=\max\{A,B,C\}.$ Then, from the AM-GM Inequality, we get
$ 4(xy+yz+zx)=4\cdot (y+z) \sin A \cdot [ \frac{x}{\sin A}+\frac{yz}{(y+z)\sin A}] \le [ (y+z) \sin A + \frac{yz}{(y+z)\sin A}+\frac{x}{\sin A}]^2,$
and so it suffices to prove that
$ \frac{y}{\sin B}+\frac{z}{\sin C} \ge (y+z)\sin A+\frac{yz}{(y+z) \sin A},$
which is equivalent to
$ (z+y)[y( \frac{1}{\sin B}-\sin A) +z(\frac{1}{\sin C}-\sin A) ] \ge \frac{yz}{\sin A}.$
Now, we use the AM-GM Inequality again to reduce the problem into proving that
${ 4yz \sqrt{( \frac{1}{\sin B}-\sin A})(\frac{1}{\sin C}-\sin A)} \ge \frac{yz}{\sin A},$
which is equivalent to
$ 4\sqrt{( \frac{1}{\sin B}-\sin A)( \frac{1}{\sin C}-\sin A)} \ge \frac{1}{\sin A}.$
Note that for any $ m,n,p$ such that $ m(n+p) \le 1,$ we have
$ ( \frac{1}{n}-m) (\frac{1}{p}-m) \ge ( \frac{2}{n+p}-m)^2.$
This inequality is very easy to check. Now, applying it with noting that $ \sin A (\sin B+\sin C) <1,$ we infer that
$ \sqrt{( \frac{1}{\sin B}-\sin A)( \frac{1}{\sin C}-\sin A)}  \ge \frac{2}{\sin B+\sin C} -\sin A \ge \frac{1}{\sin \frac{B+C}{2}}-\sin A,$
where the last inequality holds since $ \sin B+\sin C =2\sin \frac{B+C}{2}\cos \frac{B-C}{2} \le 2\sin \frac{B+C}{2}.$ So, it is enough to check that
$ 4( \frac{1}{\sin \frac{B+C}{2}} -\sin A) \ge \frac{1}{\sin A}.$
Setting $ t=\cos \frac{B+C}{2},$ and using the double formula $ \sin A =\sin (B+C)=2\sin \frac{B+C}{2}\cos \frac{B+C}{2},$ it becomes
$ 4(\frac{1}{\sqrt{1-t^2}} -2t\sqrt{1-t^2}) \ge \frac{1}{2t\sqrt{1-t^2}},$
which is equivalent to
$ 8t(2t^3-2t+1) \ge 1.$
Now one can easily factorize $ 8t(2t^3-2t+1) -1=(2t-1)^2(4t^2+4t-1),$ so that the last inequality is equivalent to
$ (2t-1)^2 (4t^2+4t-1) \ge 0.$
This is true since $ t=\cos \frac{B+C}{2} \ge \cos \frac{\pi }{3} =\frac{1}{2}.$ The proof is now completed.

Hi,
thank you for your nice thought :lol:
But I don't think $ \sin A (\sin B\plus{}\sin C) <1,$ holds.Try $ A\equal{}\frac{\pi}{2},b\equal{}\frac{\pi}{4},c\equal{}\frac{\pi}{4}$
And I think $ 4\sqrt{\left(\frac{1}{\sin B}\minus{}\sin A\right)\left(\frac{1}{\sin C}\minus{}\sin A\right)}\ge\frac{1}{\sin A}.$ is not true
try $ A\equal{}89,b\equal{}89,c\equal{}2$ :maybe:
Am I right?

P.S.This ineq is my conjecture(I havn't proved it yet)
So the topic said "Prove or disprove" :oops:

I am sorry for the mistakes. I haven't written it out and that's the reason why I haven't checked it carefully. So sorry.

Can anyone help me to accomplish my conjecture? :)

[quote="hxy09"]Given that $ \alpha, \beta , \gamma \ge 0$ and $ \alpha \plus{} \beta \plus{} \gamma \equal{} \pi$

$ u,w,v \ge 0$

Prove : $ (\frac {u}{\sin \alpha} \plus{} \frac {v}{\sin \beta} \plus{} \frac {w}{\sin \gamma})^2\ge 4(uv \plus{} vw \plus{} wu);$[/quote]

Let : $ f(\alpha,\beta,\gamma)\equal{}\frac {u}{\sin \alpha} \plus{} \frac {v}{\sin \beta} \plus{} \frac {w}{\sin \gamma}$

By Lagrange multipliers, $ min(f(\alpha,\beta,\gamma)$ holdes for  :  $ \frac{u\cos \alpha}{\sin^2 \alpha} \equal{} \frac{v\cos \beta}{\sin^2 \beta} \equal{} \frac{w\cos \gamma}{\sin^2 \gamma} \equal{} \lambda$ $ (*)$

$ \Leftrightarrow \cos \alpha \equal{}\frac{1}{2\lambda}(\sqrt{u^2\plus{}4\lambda^2}\minus{}u)$ , $ \cos \beta \equal{}\frac{1}{2\lambda}(\sqrt{v^2\plus{}4\lambda^2}\minus{}v)$ , $ \cos \gamma \equal{}\frac{1}{2\lambda}(\sqrt{w^2\plus{}4\lambda^2}\minus{}w)$

For any triangle we have :$ \sum cos^2A \plus{} 2cosAcosBcosC \equal{} 1 \Rightarrow$ for $ \lambda$ we obtain the equation :

$ 2 \plus{} \frac{u^2\plus{}v^2\plus{}w^2}{2\lambda^2}\minus{}\frac{1}{2\lambda^2}(u\sqrt{u^2\plus{}4\lambda^2}\plus{}v\sqrt{v^2\plus{}4\lambda^2}\plus{}w\sqrt{w^2\plus{}4\lambda^2})\plus{}$ $ \frac{1}{4\lambda^3}(\sqrt{u^2\plus{}4\lambda^2}\minus{}u)(\sqrt{v^2\plus{}4\lambda^2}\minus{}v)(\sqrt{w^2\plus{}4\lambda^2}\minus{}w)\equal{}0$

$ \lambda$ - single root of this equation $ (\lambda > 0)$.

So we only need to prove inequality for conditions $ (*)$

$ \Leftrightarrow \boxed{(tan(\alpha)\plus{}tan(\beta) \plus{} tan(\gamma))^2 \ge 4(\frac{\sin^2\alpha}{\cos \alpha}\plus{}\frac{\sin^2\beta}{\cos \beta}\plus{}\frac{\sin^2\gamma}{\cos \gamma})}$ - which is true for any triangle ($ \alpha , \beta , \gamma \ne \frac{\pi}{2}$).

$ tan(\alpha)\plus{}tan(\beta) \plus{} tan(\gamma)\equal{}tan(\alpha)tan(\beta)tan(\gamma)$

$ \Leftrightarrow \cos \alpha \cos\beta \cos \gamma (\frac{\cos\alpha}{\sin^2\alpha}\plus{}\frac{\cos\beta}{\sin^2\beta}\plus{}\frac{\cos\gamma}{\sin^2\gamma}) \le \frac{1}{4}$

$ \cos \alpha \cos\beta \cos \gamma\equal{} \frac{s^2\minus{}(2R\plus{}r)^2}{4R^2}$

$ \frac{\cos\alpha}{\sin^2\alpha}\plus{}\frac{\cos\beta}{\sin^2\beta}\plus{}\frac{\cos\gamma}{\sin^2\gamma} \equal{} \frac{1}{4}(\frac{s^2}{r^2}\minus{}8\frac{R}{r}\minus{}\frac{(4R\plus{}r)^2}{s^2})$

$ \Leftrightarrow \boxed{\frac{s^2}{r^2}\minus{}8\frac{R}{r}\minus{}\frac{(4R\plus{}r)^2}{s^2} \le \frac{4R^2}{s^2\minus{}(2R\plus{}r)^2}}$ , $ s \ne 2R\plus{}r$ $ (**)$

$ LHS(**)$ - increase by $ s^2$ and $ RHS(**)$ - decrease by $ s^2$ , $ s^2 \le 2R^2\plus{}10Rr\minus{}r^2\plus{}2\sqrt{R}(R\minus{}2r)^{3/2}$

$ (**) \Leftrightarrow 2t^2\plus{}2t\minus{}1\plus{}2\sqrt{t}(t\minus{}2)^{3/2}\minus{}\frac{(4t\plus{}1)^2}{2t^2\plus{}10t\minus{}1\plus{}2\sqrt{t}(t\minus{}2)^{3/2}}\plus{}$ $ \frac{2t^2}{t^2\minus{}3t\plus{}1\minus{}\sqrt{t}(t\minus{}2)^{3/2}} \le 0$ , ($ t\equal{} \frac{R}{r} \ge 2$) - which is true .

Thank you very much Sergic Primazon :lol:
I tried lots of methods but failed just without realizing Langrange multipliers kills it :!:
But I think you have a typo $ \Leftrightarrow\boxed{(tan(\alpha)\plus{}tan(\beta)\plus{}tan(\gamma))^{2}\ge 4(\frac{\sin^{2}\alpha}{\cos\alpha}\plus{}\frac{\sin^{2}\beta}{\cos\beta}\plus{}\frac{\sin^{2}\gamma}{\cos\gamma})}$ should be replaced by $ \Leftrightarrow\boxed{(tan(\alpha)\plus{}tan(\beta)\plus{}tan(\gamma))^{2}\ge 4(\sum \frac{\sin^{2}\alpha \sin^{2}\beta}{\cos\alpha \cos\beta})}$

And I post a different proof for it:

[quote="Sergic Primazon"][quote="hxy09"]Given that $ \alpha, \beta , \gamma \ge 0$ and $ \alpha \plus{} \beta \plus{} \gamma \equal{} \pi$

$ u,w,v \ge 0$

Prove : $ (\frac {u}{\sin \alpha} \plus{} \frac {v}{\sin \beta} \plus{} \frac {w}{\sin \gamma})^2\ge 4(uv \plus{} vw \plus{} wu);$[/quote]

Let : $ f(\alpha,\beta,\gamma) \equal{} \frac {u}{\sin \alpha} \plus{} \frac {v}{\sin \beta} \plus{} \frac {w}{\sin \gamma}$

By Lagrange multipliers, $ min(f(\alpha,\beta,\gamma)$ holdes for  :  $ \frac {u\cos \alpha}{\sin^2 \alpha} \equal{} \frac {v\cos \beta}{\sin^2 \beta} \equal{} \frac {w\cos \gamma}{\sin^2 \gamma} \equal{} \lambda$ $ (*)$

$ \Leftrightarrow \cos \alpha \equal{} \frac {1}{2\lambda}(\sqrt {u^2 \plus{} 4\lambda^2} \minus{} u)$ , $ \cos \beta \equal{} \frac {1}{2\lambda}(\sqrt {v^2 \plus{} 4\lambda^2} \minus{} v)$ , $ \cos \gamma \equal{} \frac {1}{2\lambda}(\sqrt {w^2 \plus{} 4\lambda^2} \minus{} w)$

For any triangle we have :$ \sum cos^2A \plus{} 2cosAcosBcosC \equal{} 1 \Rightarrow$ for $ \lambda$ we obtain the equation :

$ 2 \plus{} \frac {u^2 \plus{} v^2 \plus{} w^2}{2\lambda^2} \minus{} \frac {1}{2\lambda^2}(u\sqrt {u^2 \plus{} 4\lambda^2} \plus{} v\sqrt {v^2 \plus{} 4\lambda^2} \plus{} w\sqrt {w^2 \plus{} 4\lambda^2}) \plus{}$ $ \frac {1}{4\lambda^3}(\sqrt {u^2 \plus{} 4\lambda^2} \minus{} u)(\sqrt {v^2 \plus{} 4\lambda^2} \minus{} v)(\sqrt {w^2 \plus{} 4\lambda^2} \minus{} w) \equal{} 0$

$ \lambda$ - single root of this equation $ (\lambda > 0)$.

So we only need to prove inequality for conditions $ (*)$

$ \Leftrightarrow \boxed{(tan(\alpha) \plus{} tan(\beta) \plus{} tan(\gamma))^2 \ge 4(\frac {\sin^2\alpha}{\cos \alpha}\frac {\sin^2\beta}{\cos \beta} \plus{} \frac {\sin^2\beta}{\cos \beta}\frac {\sin^2\gamma}{\cos \gamma} \plus{} \frac {\sin^2\gamma}{\cos \gamma}\frac {\sin^2\alpha}{\cos \alpha})}$

- which is true for any triangle  ($ \alpha , \beta , \gamma \ne \frac {\pi}{2}$).

$ tan(\alpha) \plus{} tan(\beta) \plus{} tan(\gamma) \equal{} tan(\alpha)tan(\beta)tan(\gamma)$

$ \Leftrightarrow \cos \alpha \cos\beta \cos \gamma (\frac {\cos\alpha}{\sin^2\alpha} \plus{} \frac {\cos\beta}{\sin^2\beta} \plus{} \frac {\cos\gamma}{\sin^2\gamma}) \le \frac {1}{4}$

$ \cos \alpha \cos\beta \cos \gamma \equal{} \frac {s^2 \minus{} (2R \plus{} r)^2}{4R^2}$

$ \frac {\cos\alpha}{\sin^2\alpha} \plus{} \frac {\cos\beta}{\sin^2\beta} \plus{} \frac {\cos\gamma}{\sin^2\gamma} \equal{} \frac {1}{4}(\frac {s^2}{r^2} \minus{} 8\frac {R}{r} \minus{} \frac {(4R \plus{} r)^2}{s^2})$

$ \Leftrightarrow \boxed{\frac {s^2}{r^2} \minus{} 8\frac {R}{r} \minus{} \frac {(4R \plus{} r)^2}{s^2} \le \frac {4R^2}{s^2 \minus{} (2R \plus{} r)^2}}$ , $ s \ne 2R \plus{} r$ $ (**)$

$ LHS(**)$ - increase by $ s^2$ and $ RHS(**)$ - decrease by $ s^2$ , $ s^2 \le 2R^2 \plus{} 10Rr \minus{} r^2 \plus{} 2\sqrt {R}(R \minus{} 2r)^{3/2}$

$ (**) \Leftrightarrow 2t^2 \plus{} 2t \minus{} 1 \plus{} 2\sqrt {t}(t \minus{} 2)^{3/2} \minus{} \frac {(4t \plus{} 1)^2}{2t^2 \plus{} 10t \minus{} 1 \plus{} 2\sqrt {t}(t \minus{} 2)^{3/2}} \plus{}$ $ \frac {2t^2}{t^2 \minus{} 3t \plus{} 1 \minus{} \sqrt {t}(t \minus{} 2)^{3/2}} \le 0$ , ($ t \equal{} \frac {R}{r} \ge 2$) - which is true .[/quote]

[quote="hxy09"]Given that $ A,B,C\ge 0$ and $ A \plus{} B \plus{} C \equal{} \pi$
$ x,y,z \ge 0$
Prove or disprove:$ (\frac {x}{\sin A} \plus{} \frac {y}{\sin B} \plus{} \frac {z}{\sin C})^2\ge 4(xy \plus{} yz \plus{} zx)$[/quote]
My method did not fail, I just failed because I have used the "so trong" estimation. The following is my new proof:

Very very [b]spectacular[/b] solution can_hang2007 :!:
Thank you so much,and I finally knew how god does inequalities :lol:

[quote="hxy09"]...[b]how god does inequalities [/b]:lol:[/quote]
[size=200][b]Yeah, he loves inequalities[/b]  :lol: [/size]


geometry
3D geometry
sphere
parallelogram
geometry proposed

Find the maximal number of spheres with radius $\frac{1}{2}$ fitting in a rectangular parallelepiped with edge lengths $10,$ $10$ and $1.$

if the 1-side isn't perpedicular to the 10&10 plane, it's 0
elseway it is reduced to x circles with radius 1/2 fitting in a 10&10 parallellogram

am I wrong? :? seems a weird question...

peter, the parallelepiped is rectangular, implying that the 1-side is perpendicular to the 10x10-plane. it also follows that the parallelogramm is a rectangle.
it is well-known that one has to put the circles such that three neighboring always form an equilateral triangle. if i got time and don't forget about this problem, i'll try to find a nice proof.

Peter

Please post your solutions. This is just a solution template to write up your solutions in a nice way and formatted in LaTeX. But maybe your solution is so well written that this is not required finally. For more information and instructions regarding the ISL/ILL problems please look here: [url=http://www.mathlinks.ro/Forum/viewtopic.php?t=15623]introduction for the IMO ShortList/LongList project[/url] and regarding[url=http://www.mathlinks.ro/Forum/viewtopic.php?t=15623]solutions[/url]  :)


In how many different ways can the letters in the word PEOPLE be scrambled?

[quote="mrsaturns"]The answer when I reveiwed was 128. How do you get this?[/quote]

For me, I reviewed and it said 180.

[quote="icecapmeltdown"]wait do you mean [b]180?[/b]

That's what i got, but the answer is actually 126... I think there might be an error  :([/quote]

Like I said, it says 180.


Let $ a,b>0$ and $ a\plus{}b\equal{}1$ . Show :

$ \left(a\plus{}\frac{1}{a}\right)^2\plus{}\left(b\plus{}\frac{1}{b}\right)^2\ge \frac{47}{4}\plus{}a\sqrt{b}\plus{}b\sqrt{a}$

That's wrong. Take for example $ a\equal{}b\equal{}1/2$.

Are you sure ? It gives $ LHS\equal{}12.5$ and $ RHS \sim 12.457106...$

[quote="Pain rinnegan"]Let $ a,b > 0$ and $ a \plus{} b \equal{} 1$ . Show :

$ \left(a \plus{} \frac {1}{a}\right)^2 \plus{} \left(b \plus{} \frac {1}{b}\right)^2\ge \frac {47}{4} \plus{} a\sqrt {b} \plus{} b\sqrt {a}$[/quote]

[hide]

It is easy to prove by Cauchy Schwartz that

$ \left(a \plus{} \frac {1}{a}\right)^2 \plus{} \left(b \plus{} \frac {1}{b}\right)^2\ge \frac {25}{2}$

So we have to prove that $ \frac {3}{4} \geq a\sqrt {b} \plus{} b\sqrt {a}$

Define the function $ f(x) \equal{} \sqrt {x}$ which is concave since $ f''(x) \equal{} \minus{} \frac {x^{ \minus{} \frac {3}{2}}}{4} < 0$

Since $ a \plus{} b \equal{} 1$ we apply Jensen's inequality

$ a\sqrt {b} \plus{} b\sqrt {a} \leq f\left(\frac {ab \plus{} ba}{1}\right) \equal{} \sqrt {2ab} \leq \frac {3}{4} \iff 9 \geq 32ab$

This is true since $ 9 \equal{} 9(a \plus{} b)^2 \geq 9 \cdot 4ab \equal{} 32ab$ [/hide]

there is nothing wrong with your proof in this case but the last line seems troublesome. Anyway, it is just a mistake in typo. However, your solution is right until the last statement  :lol:

I edited , can you check it again great math  :)

Thanks

[hide="by Am-Gm"]

$ a^2 \plus{}ab \plus{}\frac{1}{4}a \plus{}\frac{1}{4} b \geq  4 \sqrt[4]{\frac{a^4 b^4}{4^2}}\equal{}2a\sqrt b$

and

$ b^2 \plus{}ab \plus{} \frac{1}{4}a \plus{}\frac{1}{4}b \geq 2 b \sqrt a$

$ \frac{3}{4}\equal{}\frac{1}{2}(a\plus{}b)^2 \plus{}\frac{1}{4}(a\plus{}b) \geq a\sqrt b \plus{}b\sqrt a  \iff (a\plus{}b)^2 \plus{}\frac{1}{2} (a\plus{}b) \geq 2a\sqrt b \plus{}2b\sqrt a$

We have that $ \frac{3}{4} \geq a\sqrt b \plus{} b\sqrt a$
[/hide]


Let $a,$ $b$ and $c$ are non-negative real numbers.
Prove that $5(a+b+c)\geq9\sqrt[3]{abc}+6\cdot\sqrt{\frac{a^2+b^2+c^2}{3}}.$

[quote="amcavoy"]$5\cdot\textrm{AM}\geq 3\cdot\textrm{GM}+2\cdot\textrm{QM}$

Then maybe somehow use the fact that $\textrm{QM}\geq\textrm{AM}\geq\textrm{GM}$?[/quote]
I am not sure that you are right.

[quote="Melissa"]It should be $15 AM \ge 9 GM + 6 QM$. But I don't think you can then use $QM \ge AM \ge GM$ to prove it.[/quote]
I divided by $3$; it is equivalent (even if you can't use QM-AM-GM).

[quote="amcavoy"]$\textrm{QM}=\textrm{Quadratic Mean}=\sqrt{\frac{1}{n}\prod_{k=1}^{n}{x_{k}}^{2}}$[/quote]
Are you sure? Shouldn't it be $\sqrt{\frac{1}{n}\sum_{k=1}^{n}{x_{k}}^{2}}$?

arqady, besides sos and expansion, do you have any other method to handle it?

[quote="Rzeszut"]For 2 variables we have $2\mathrm{AM}\geq \mathrm{QM}+\mathrm{GM}$ and it allows to prove the inequality immadietly. But is it true for 3 variables?[/quote]

how do u prove that for 2 variables?

[quote]how do u prove that for 2 variables?[/quote]

http://www.mathlinks.ro/Forum/viewtopic.php?t=107442


[quote="MrWise"]A,B,C are angles of a triangle and $2\sum Sin^3A=\sum SinA(Sin^2B+Sin^2C)$
Find A,B,C[/quote]

$2\sum\sin^3A=\sum_{sym}\sin^3A$ and $\sum\sin A(\sin^2B+\sin^2C)=\sum_{sym}\sin^2A\sin B$.

$\sum_{sym}\sin^3A\geq\sum_{sym}\sin^2A\sin B$ by Muirhead or Rearrangement and equality holds when all term are equal, so $\sin A=\sin B=\sin C$ and since $A+B+C=\pi$, $A=B=C=\frac{\pi}{3}$


Hi guys, again in the tradition of this topic [url=http://www.mathlinks.ro/Forum/viewtopic.php?t=5608]forum members at IMO 2004 Athens, Greece[/url], it would be great if we'd get to know each other before the IMO takes place. So if you are a contestant that is the the IMO Team for 2005, a deputy, observer or a team leader, or just a visitator just leave a word here :)

My country selection Test is like 2 weeks before IMO, so I guess we dont know...However, I am pretty confident I will be there, and I hope to meet many mathlinkers there! :D  BTW for those who are going if they want to know something in Spanish (useful things, like what to tell a girl, and that) you can asks us in the spanish comunity forum, or in another treath.

Best!

Hello everyone. I am going, too, for the second time.

[quote="Pascual2005"]BTW for those who are going if they want to know something in Spanish (useful things, like what to tell a girl, and that) you can asks us in the spanish comunity forum, or in another treath.

Best![/quote]You can make yourself a semi-guide there :P :D

I won't be there, but I can announce the Turkish team
Fatih Hafalir
Metehan Ozsoy
Hasan Eruslu
Murat Sevim
Samet Oymak
Said Tunc ( for second time)

Hong Kong Team members:
Hui Jack (jackhui), grade 11
Siu Ho Chung :), grade 12
Tsoi Yun Pui (yptsoi), grade 12
Wong Chiu Wai (mecrazywong), grade 9
Chung On Yip, grade 10
Poon Ming Fung, grade 13

Only Poon Ming Fung went to IMO 2004, others will be going for the first time.

Congratulations mecrazywong, siuhochung, jackhui and yptsoi!! :)

[quote="mecrazywong"]I'm quite delighted with this :)
and hopefully I will go to IMO till 2009 :D[/quote]I have no doubt about that :)

Australian team members:
Sam Chow
Steve Muirhead
Vinoth Nandakumar   :)
Konrad Pilch
Kim Ramchen
Graham White
and Reserve is Vinh Pham

[quote="142857"][quote="mecrazywong"]I'm quite delighted with this :)
and hopefully I will go to IMO till 2009 :D[/quote]
I think you will be in university before 2009. :P[/quote]
You mean the early admission scheme?
I think I won't be good enough to apply for it...it's too tough for me to get >=6A

Kyrgyzstan team is:
Bekjan Jumabaev  :) ,grade 11(IMO 2004)
Ernist Tillenbaev,grade 11(IMO 2003)
Bektur Mambetov,grade 11
Mamat Ishmatov,grade 10
Ulanbek Omurkanov,grade 11(IMO 2004)
Said Mavlanov,grade 10 (he was chosen)
Interesting to note that me,Mamat and Said are from the same school,Bektur,Almir and Ulanbek are also from the same school.

[quote="tywebb"]i sympathise with rustam and am pleased valentin has tried to help.

but [i]pleeeease[/i] tell me the names of the china team (in english).

thx.[/quote]

Why are you so curious to know that?
Anyway, here's my poor translation.
Yen, XinChun
Diao, HanSheng
Luo, Ye
Kang, JiaYin
Shao, ??Cheng
Zhao, TongYuan

well, they have been polled as the favourites in valentin vornicu's poll as to which country will be the best and i therefore would like to know who they are (in english).

in 1998, china did not compete in the imo because it was in taiwan and taiwan called itself the republic of china. china seems to have withdrawn from the icho this year because again it's in taiwan, who are again calling themselves the republic of china.

anyway, thanks.

[quote="shobber"]I only recognize three of them when their names are translated into English, even though I already know their Chinese name.[/quote]

I'm terribly sorry for my poor pinyin....

Who can tell me is there any oportunities to study in abroad universities after participating in the IMO?
for example In Australiya , USA , Romaniya , Canada................ and being 16 years old ?
being participant but not come to IMO? :?  :wallbash:  :bruce:

[quote="tywebb"]thanks. but i can't read chinese. can you or someone else translate them into english and put them in this thead please?[/quote]

The Chinese Team

Ren Qingchun - 11
Diao Hansheng - 12
Luo Ye - 12
Kang Jiayin - 12
Shao Xuancheng - 12
Zhao Tongyuan - 12

[quote="Orif"]i am ex :( UZB4[/quote]I am sorry but we tried anything that was possible here. Unfortunately your leaders communicated to us that nothing can be done anymore :(

It's really sad ....  :(

We've got a 11th year! Really cool......

Mongolian team :
Dorjnamjil  m
Batzaya m
Bat-erdene m
Jargalsaihkan m
Chandmani f
Baigalmaa f

unfortunately we can't participate IMO because of USA transit visa :mad:

We are going by UlaanBaatar-seoul-los angeles-monterrey-merida.

yesterday we were take the plane UB-Seoul. but we could't transfer cause of transit visa Transit visa needed 3 or more hours. We will be in los angeles only 2 hours 55 minut. What is contradiction?
 Our counsil in korea called Korean air many times , they didn't  allowed on the time. In other way, we can fly 1 day later(Jul12 ) they said. If we do that, we will miss examination. So we are in Mongolia. it's very very bad thing.
all members of team are actually  sad now.

Another team??  :? This is incredible  :(

Here is the list of Georgian Math team for IMO 2005

Irakli Merabishvili
David Tsirekidze
George Mariamidze
Zviad Metreveli
Alexander Lomadze
Otar Chitaia

I wish them good luck in Merida

If any of team members get on this site contact me
Regards


let a, b are natural numbers such that a>b.

prove that (a^2+b^2)/(a^2-b^2) isn't in N .

[u]le 2 ème tour de l'olympiades internationales marocaines. [/u]
this problem is very important to me

can you give a detailled solution .

This is number theory, not calculus. It's also easy.

WLOG, $ a$ and $ b$ are relatively prime. If $ a^{2}\minus{}b^{2}$ divides $ a^{2}\plus{}b^{2}$, it also divides $ 2a^{2}$ and $ 2b^{2}$- which means that $ a^{2}\minus{}b^{2}$ divides $ 2$. This is impossible, and we are done.

[quote="jmerry"]
WLOG, $ a$ and $ b$ are relatively prime.[/quote]

Why?

If they aren't, we just divide each by their common factor.

ok

$ a^{2}\minus{}b^{2}$devides$ 2$ is true for $ a \equal{} 1$ $ b \equal{} 0$

I think $ b$ must be positive

[quote="greatestmaths"]let a, b are [b]natural numbers[/b][/quote]

Based on the problem statement, the convention appears to be that $ 0\not\in\mathbb{N}$.


[quote="easyas3.14159..."][hide]If we simplify each number in parentheses, we get $\frac{3}{2}*\frac{4}{3}*...\frac{2005}{2004}$.Numerators and denominators will cancel out and leave us with $\frac{2005}{2}$.[/hide][/quote]

You forgot about (1+1/1)

[quote="mysmartmouth"][hide]So it is $\frac{2009}{2}$ when you include that first term.[/hide][/quote]
[hide]Note that you are multiplying by 2 not adding 2.[/hide]

So far no one has got the correct answer.

[hide]Take the $\frac{2005}{2}$ that easyaspi got and multiply that by $2$ to get $2005$.[/hide]

[quote="redsmartie"]Find the Value of
(1+1/1) * (1+1/2) * (1+1/3) * (1+1/4) *...... * (1+1/2004)
The symbol, *, means multiplication.[/quote]

[hide]

That is equal to: $\left(1+\frac{1}{1}\right)\times\left(1+\frac{1}{2}\right)\times\left(1+\frac{1}{3}\right)\times\dots\times\left(1+\frac{1}{2004}\right)$

By simplifying that, you get $2\times\frac{3}{2}\times\frac{4}{3}\times\dots\times\frac{2005}{2004}$. From that, by cancelling out, the answer is $\mbox{2005}$?
[/hide]

I got $500\frac{1002}{2003}$ somehow
-jorian

Well obviously your solution was wrong then, post it if you want feedback on how to correct it.


geometry
trigonometry
ratio
geometry unsolved

[b]1. Let ABC be an equilateral triangle with area 7 and let M;N be points on sides AB;AC, respectively,
such that AN = BM. Denote by O the intersection of BN and CM. Assume that triangle BOC has area 2.

(a) Prove that MB/AB equals either 1/3 or 2/3.

(b) Find angle AOB.[/b]

Let $ AO$ meets $ BC$ at $ L$. Then $ \frac{OL}{AL}=\frac{\triangle{BOC}}{\triangle{BAC}}=\frac{2}{7} \Longrightarrow \frac{OL}{AO}=\frac25$.

By Menelaus, we have $ \frac{AM}{MB}\cdot\frac{BC}{LC}\cdot\frac{LO}{OA}=1$ and $ \frac{AN}{NC}\cdot\frac{CB}{BL}\cdot\frac{LO}{OA}=1$. Multiply the two idendities and notice that $ BC=BL+LC$, we have:

$ \frac{4}{25} \cdot \frac{(BL+CL)^2}{BL\cdot CL}=1$

Then we have that $ (4CL-BL)(CL-4BL)=0$. Hence $ \frac{BL}{CL}=1 \text{ or }4$. Thus $ \frac{MB}{AB}=\frac13 \text{ or } \frac23$.

[quote="msg_1989"]I dont understand how u got MB/AB = 1/3 or 2/3. please explain... also please find angle AOB[/quote]
There is already $ \frac{AM}{MB}\cdot\frac{BC}{LC}\cdot\frac{LO}{OA}\equal{}1$, since $ \frac{LO}{OA}$ and $ \frac{BC}{CL}$ are already known, so it is easy to compute $ \frac{AM}{MB}$.

As for the second part, by using the Law of Cosine, it is easy once we get the ratio of the segments.


geometry
circumcircle
geometric transformation
reflection
trigonometry
angle bisector
IMO Shortlist

Let $  H  $ and $ O $ the orthocenter and the circumcenter of a triangle $  ABC $. If the bisector of angle $  A $ is perpendicular to $   HO$  , find angle $  A $.

    [i](taken from an excellent geometry book in Greece , based on problems from (the old) Soviet Union)[/i]

Observe that $\angle BAC$ must be acute.
A simple angle chasing shows that AO and AH are reflections in the angle bisector of $\angle A$. Therefore AH=AO which implies $A=60^\circ$.

BTW, this problem should belong to Intermediate or Pre-Olympiad.

I did not say it is an IMO - level problem but it could be a first round problem for grades 9-10.Could this problem be  preolympiad  ? I think no! If you think that the right place is intermediate , please ask the mod /tors  to do it.Any way , your solution is very good and it is the same as mine.It took me 10 minutes to make it(the solution) but this has no importance .Keep strong and solving good olympiad problems.

           Babis

See also http://www.mathlinks.ro/Forum/viewtopic.php?t=31157 .

  darij

[quote="shobber"]Why AO=AH straightly gives <A=60? I used the fact that if the distance from the circumcentre is $d$ then $AH=2d$ to get $\angle{BOC}=120^o$.[/quote]
I just skipped the steps which I regard as "trivial" as I always do when posting solutions in mathlinks...

Well, megacrazywong, you are probably surprised to hear that the following was a problem in some IMO shortlist in the early 90's?

Let $ABC$ be an acute-angled triangle such that $AO = AH$, where $O$ and $H$ are the circumcentre and orthocentre of $ABC$. Find $\widehat{A}$.

Too easy, you are right :)

Maybe the problem selection committee of that year "overlook" something in the problem :D

BTW, I believe you will be even more surprised that Valentin has also mis-read my name as megacrazywong before(no ga ;) )

The lines AH, AO are "isogonal" w.r.t the $\angle BAC$. Thus,
$AH=AO\Longleftrightarrow 2R\cdot  | \cos A |=R\Longleftrightarrow |\cos A |=\frac 12$.
But the $\angle BAC$ is acute (you can show easy). Therefore, $A=60^{\circ} .$


Determine all functions which satisfy the functional identity \[ ye^{f(x)} = xe^{f(y)}. \]

[hide]From the equation it follows that $xy$ must be positive.

Therefore $x\neq 0$, so we can write

\begin{eqnarray*}{y\over x}=e^{f(y)-f(x)}\iff \ln |y|-\ln |x|=f(y)-f(x)\iff f(x)-\ln |x|=f(y)-\ln |y|.\end{eqnarray*}

The last result must be valid for all $x,y$, hence $f(x)-\ln |x|=\text{const.}=C$, giving $f(x)=\ln |x|+C.$ Plugging this back into the initial equation, with the condition $xy>0$, proves that this indeed is a solution.[/hide]


pigeonhole principle
superior algebra
superior algebra unsolved

Suppose $ R$ is a finite ring. Show that for each $ x\in R$ there is a positive integer $ n$ such that $ nx \equal{} 0$. Here $ nx \equal{} \underbrace{x\plus{}x\plus{}...\plus{}x}_{\text{n times}}$

Nothing, just consider the series $ x, 2x, \dots , (n\plus{}1)x$ for $ x\in R$ by assuming $ |R|\equal{}n$ and use pigeonhole principle.


Euler integration limit trigonometry logarithms calculus derivative

$ A_n = \int_{1/n}^{n}sint/t dt $
Prove that A_n ~ 1- C + ln(n) when n -->+oo

The prooblem is analagous to proove that
${\lim_{n\to \infty }\, \left(\frac{\int_{\frac{1}{n}}^{n}\frac{\sin (t)}{t^{2}}\, dt}{-c_{1}+\log (n)+1}\right)=1}$

L'Hospital Rule
${\lim_{n\to \infty }\left(\, \frac{\left(\frac{\partial \left(\int_{\frac{1}{n}}^{n}\frac{\sin (t)}{t^{2}}\, dt\right)}{\partial n}\right)}{\left(\frac{\partial \left(-c_{1}+\log (n)+1\right)}{\partial n}\right)}\right)=\lim_{n\to \infty }\left(\, \frac{\sin \left(\frac{1}{n}\right)+\frac{\sin (n)}{n^{2}}}{\frac{1}{n}}\right)}$

Since both limits exist, we have
${\lim_{n\to \infty }\, \left(n \left(\sin \left(\frac{1}{n}\right)+\frac{\sin (n)}{n^{2}}\right)\right)=\left(\lim_{n\to \infty }\left(\, \frac{\sin (n)}{n}\right)\right)+\left(\lim_{n\to \infty }\, \left(n \sin \left(\frac{1}{n}\right)\right)\right) =1}$

so the approximation is true

it took me 3 years to solve  :D

Just a remark: why do you use partial derivatives if both the numerator and denominator are only functions of n?

for 3 reasons

1 - I type mathematical expressions in Mathematica 5.2. So, if I use the function D, total derivative, mathematica will think all others constants are variables and take the total derivative, and is pointless tell to mathematica what is constant or not when i can easily take the partial derivative, the result is the same and, does not matter :)

2 - the total derivative of a function of one variable is equal to the partial derivative of this function

3 - i love prime numbers


logarithms
floor function
number theory
greatest common divisor
number theory open

Is it true that for any t>0 we can find an n_t such that for any n>n_t there are integers a_n and b_n such that n<(a_n)^2+(b_n)^2<n+n^t? I suspect the answer is yes, but I don't know how to prove this.

I would guess not.  Let $N(r)$ is the number of lattice points inside a circle of radius $r$, and define $h(r) = N(r)-\pi r^2$.  Hardy and Landau proved that $h(r)$ is [i]not[/i] $o(r^{1/2}(\log r)^{1/4}$.  So it seems unlikely that one can take $t$ much less than 1/2.

[quote="harazi"]Is it true that for any t>0 we can find an n_t such that for any n>n_t there are integers a_n and b_n such that n<(a_n)^2+(b_n)^2<n+n^t?[/quote]
Even by elementary means, it could be proven that the statement is true if $ t=\frac 1 4+\varepsilon$ for any $ \varepsilon>0$. Here is that proof:

Let $ a_n=\lfloor\sqrt n\rfloor$. Define ${ \rho\in\mathbb{R}_0^+}$ with $ \rho=\sqrt{n-a_n^2}$, and let $ b_n=\lfloor\rho\rfloor +1$. Then\[ a_n^2+b_n^2>a_n^2+\rho^2=n\quad(1)\]and\[ a_n^2+b_n^2\leqslant (n-\rho^2)+(\rho+1)^2=n+2\rho+1.\quad(2)\]However, since $ \rho^2=n-a_n^2<n-(\sqrt n-1)^2=2\sqrt n-1<2\sqrt n$, it follows $ \rho<\sqrt 2n^{\frac 1 4}$ and\[ 2\rho+1<2\sqrt 2n^{\frac 1 4}+1<n^{\frac 1 4+\varepsilon}\quad(3)\]for all $ n>n_{\varepsilon}$. Therefore, from $ (1)$, $ (2)$ and $ (3)$, for all $ n>n_{\varepsilon}$ it holds\[ n<a_n^2+b_n^2<n+n^{\frac 1 4+\varepsilon},\]what was to be proven. $ \blacksquare$

Unfortunately, this is the best known result. Tsz Ho Chan managed [1] to reduce the exponent to $ \frac 1 5+\varepsilon$ [b]conditionally[/b], that is, given that the following conjecture is true:\[{ \sum_{\substack{1\leqslant h\leqslant H\\\mathrm{GCD(h,q)=1}}}\exp(2\pi i\lambda h)\sum_{0\leqslant k<K}\exp(2\pi i\mu k)(\frac h q})\exp(2\pi i\frac{a\overline{h}k^2}q)=\\
\phantom{xxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxx}={\cal O}((\sqrt{HK}+\sqrt[4]{H^3}+K+\frac{HK+K^2}{\sqrt q})q^{\varepsilon}),\]where $ a,q\in\mathbb{Z}$ and $ \mathrm{GCD}(a,q)=1$, $ q$ is not a perfect square, $ \lambda$, $ \mu$, $ H\geqslant 1$, $ K\geqslant 1$ are reals, ${ (\frac{\cdot}{\cdot}})$ is the Jacobi symbol, $ \overline{h}$ denotes the multiplicative inverse of $ h$ modulo $ q$, and the implied constant depends only on $ \varepsilon>0$.

Finally, I would like to point to Tsz Ho Chan's site, where you can see the same conjecture harazi asked about: https://umdrive.memphis.edu/tchan/public/problem.html, problem 2.

References:

1. T. H. Chan, Finding Almost Squares, [i]Acta Arith.[/i] [b]121[/b] (2006), 221–232.