Coloring
Ramsey Theory
graph theory
IMO
combinatorics
Extremal combinatorics
IMO 1986

Given a finite set of points in the plane, each with integer coordinates, is it always possible to color the points red or white so that for any straight line $L$ parallel to one of the coordinate axes the difference (in absolute value) between the numbers of white and red points on $L$ is not greater than $1$?

I'll use a well-known result: if a connected graph has $2k>0$ vertices of odd degree, then its edge set can be partitioned into $k$ paths, and if all its vertices have even degree, then it has an eulerian circuit.

We have a bipartite graph with bipartition $A,B$ here, constructed as follows: the horizontal lines which contain points from our set represent the vertices in $A$, the vertical lines represent the vertices in $B$, and the point represent the edges between the vertices corresponding to the rows and columns which contain them. What we must prove is that we can color the edge set of a bipartite graph in two colors s.t. the difference between the number of red edges and white edges adjacent to each vertex is $\le 1$ in absolute value. It suffices to prove this for connected bipartite graphs, since then we can apply the result to each connected component of the graph.

If all the vertices have even degree, then we can find an eulerian circuit. The graph is bipartite, so this circuit has an even number of edges, and we can thus color the edges in the circuit alternatively red and white so that two edges which are consecutive in the circuit have different colors. It's clear that this coloration satisfies the requirements.

If, on the other hand, the graph has $2k>0$ vertices with odd degre, then we partition the edge set into $k$ paths, and in each path we color the edges alternatively red and white. Again, it's easy to verify the required properties of the coloration.

P.S.

I distinctly remember solving this before, but as far as I can remember, the argument was different.

A different proof:
Let S be the smallest such set of points that cannot be colored according to the rules.
S cannot contain a  subset of the form {(a,b),(c,b),(c,d),(a,d)}: If S\{(a,b),(c,b),(c,d),(a,d)} can be colored then clearly S can be colored, and the former set is the smallest.
S cannot contain a subset of the form {(a,b),(c,b),(c,d)}: If {(a,d)} U S\{(a,b),(c,b),(c,d)} can be colored, then S can be colored, and the former set is the smallest.

Thus every point is either the only point in the column or the only point in the row, and it is easy to see that S can be colored. Contradiction.

Problem :[color=#4000FF]Given a finite set  $S$ , of points in the plane, each with integer coordinates, show that there is a function $f : P \to \left \{ -1, 1 \right \}  $,  so that for any straight line $L$ parallel to one of the coordinate axes, $\sum_{P \in L} f(P) \in   \left \{ -1,0,  1 \right \}$
[/color]

Solution : We induct on number of points. The small cases are easily checked. Let there exist such a function for $n$ points. We will show there is a function for $n+1$ points.

If there exists a line $\ell$, parallel to any of the coordinate axes (from the next time, any line will be parallel  to either of the coordinate axes, unless otherwise mentioned ), containing [b]odd[/b] number of points, then choose a point $P_x \in \ell$, and consider $S \setminus P_x$. By inductive hypothesis there exists such a function $f : P \to \left \{ -1, 1 \right \}$ for $S \setminus P_x$. Since $P \in \ell, P \ne P_x$ contain even number of points, we have  $\sum_{\substack{P \in \ell  \\ P \ne P_x}} f(P) = 0$. Let $\ell '$ be the line $\perp$ to $\ell$, passing through $P_x$.  Let $\sum _{\substack{P \in \ell ' \\  P \ne P_x}} f(P) = t $, where $t \in \left \{ -1,0 , 1 \right \} $.  Now for $S$ (with $n+1 $ points) define $g$ as $g(P) = f(P)$ for $P \in S \setminus P_x$, and $g(P_x) = -t$, if $t \ne 0$, or $1$ or $-1$ if $t=0$.  It indeed works as such a function for $n+1$ points.

If the above is not the case, i.e.  if all the lines  contain [b]even[/b] number (greater than zero) of points, pick up an arbitrary point $P_y \in S$. Let $\ell$ and $\ell '$ be the two lines containing $P_y$. Also $\ell \perp \ell '$. Consider $S \setminus P_y$, for this set , by the inductive hypothesis, there exists such a function $f$.  In  $S \setminus P_y$ , $\ell$ and $\ell '$ contains odd number of points and if $L$ is a line different from $\ell$ and $\ell '$, then it contains even number of points. So $\sum_{\substack{P \in L \\ L \ne \ell, \ell '}} f(P) = 0 $. Therefore it is seen that

$\sum_{\substack{P \in \ell \\ P \ne P_y}} f(P)  $ $=  - \sum _{\substack{P \in S \\ P \notin \ell , \ell ' }} f(P)$ $ = \sum_{\substack{P \in \ell ' \\ P \ne P_y}}  f(P) = t$.

Since $\ell \setminus P_y$ contain odd number of points, $t$ is either $-1$ or $1$. Wlog $t = 1$. Now for $S$, (containing $n+1$ points) define $g$ as : $g(P) = f(P)$, for $P \in S \setminus P_y$ and $g(P_y) = -t$.

So the induction is complete, and the statement is established.

Take the set with the smallest size that is not colorable like that.

Any line must have en even number of points. Otherwise, take a line with an odd number of points, and forget about one of its points. Color the rest of the points, and you will find you can color the forgotten point accordingly.

So, forget about one point, and color the other points. Suppose that in the vertical line of the point, color white "wins" (more whites than blacks). Then because the other vertical lines have an even number of points, color white wins in the whole set. So color white wins in the horizontal line containing the forgotten point. So we can color the forgotten point black and win.

Done

 So we consider the bipartite graph $R \cup C$, where $R$ denotes the row and $C$ denotes the columns. We want to show that for each vertex $V$, that the number of red edges incident to $V$ minus the number of white edges incident to $V$ is in the set $(-1, 0, 1)$  Now, connect two nodes with an edge if and only if there is a vertex lying at their intersection. The graph is bipartite, so all cycles are even. Now, if we color the edges in the even cycle alternately, we can preserve the difference. This means that we can remove even cycles arbitrarily, until the graph has no even cycles left. Now the graph may be disconnected, so we look at each connected component at a time. Because the connected components are each connected but have no even nor any odd cycles, they are all trees, so look at one tree at a time. I make the following claim. Now call red $1$ and white $-1$. We want to show that the sum of the edges incident to any vertex is always in $(-1, 0, 1)$. Call a vertex balanced if it satisfies this.

We claim the following result:
If there is a tree, and we weight at most one vertex negative one or one (in contrast to normally weighting it zero), we can weight each edge such that the result is satisfied.

Proof: Take an arbitrary root to the tree, and it is clear that there is some way to balance the top vertex. Now, go down the graph and do the same weighting on all the subtrees, which are all disjoint, so we may do the problem inductively Hence, the result is proven.

Now the desired result just comes from zero unbalanced vertices

[i]A different proof:[/i]
[rule]
We consider the graph-theoretical formulation of this problem. First, we get rid of all cycles by coloring them alternately (this can be done as each cycle has even length). We consider the maximal path $\mathcal P$ in the graph. We color it alternately. Then we don't have to worry about vertices of $\mathcal P$ which are not its endpoints. Now endpoints of $\mathcal P$ must degree $1$ (otherwise, we could either find a longer path or a cycle). So we don't have to worry about the endpoints of $\mathcal P$ also.  Doing this repeatedly, we are done. $\blacksquare$


Let $ m,n$ be positive integers such that $ \gcd(m,n)\equal{}1$
Express $ \gcd(5^{m}\plus{}7^{m},5^{n}\plus{}7^{n})$ in terms of m and n.

If m and n are odd, then 12, else 2.
Prove.  Let $ qm\minus{}rn \equal{} 1$, because for odd k $ (5^{a}\plus{}7^{a})|(5^{ka|}\plus{}7^{ka}$ one of (q,r) odd.
If q odd, then $ (a(m),a(n))|(a(qm),a(n)) \equal{} 2\ or 12$ were $ a(m)\equal{}5^{m}\plus{}7^{m}$.

Sorry, Rust, I didn't understand what do you mean that
$ (a(m),a(n))|(a(qm),a(n)) \equal{} 2\ or 12$
More detail, please, thanks.

Let $ a(m)=5^{m}+7^{m}$, if q is odd, then $ a(m)|a(qm)$. I denote $ (m,n)=gcd(m,n)$ (use by many mathematics).
If $ (m,n)=1$, then exist $ q,r$, suth that $ qm-rn=1$. Obviosly one of them is odd.
If $ q$ odd and m even, then r and n odd. Therefore we have only two situations
1. $ q$ and $ m$ are odd or
2. $ r$ and $ n$ are odd.
Obviosly if r is even, then $ a(rn)=7^{rn}-5^{rn}+2*5^{rn}=2*5^{rn)\mod a(n)}$, therefore $ (a(rn),a(n))=2$.
1. $ (a(m),a(n))|(a(qm),a(n))=(7^{rn+1}+5^{rn+1},7^{n}+5^{n})$ and
$ 7^{rn+1}+5^{rn+1}=(7^{n}+5^{n})(7^{(r-1)n+1}-7^{(r-2)n+1}5^{n}+...+7(-5^{n})^{r-1})+(5+(-1)^{r}7)5^{rn}.$
Because $ (5^{rn},a(n))=1$ we get $ (a(m),a(n))|7+(-1)^{r}5$.


Problem: $ e^{x^{e^{x^{...}}}}\equal{}2$ find x such that $ x>0$

Question: I was reading up on counting infinite sets. The book said that you could prove that two infinite sets are of the same size by making a one-to-one correspondence with their elements. For example, to prove the infinite set of positive integers is the same size as the set of all integers, you would do
{1,2,3,4...}
{0,1,-1,2,-3...}
Thus, you are done? My question is, how does this show that two sets are of the same size? I see no rigor in the correspondence given above. All I see is two infinite sets continuing on to, well, infinity.

Any thoughts would be appreciated!

[quote="mihail911"]Problem: $ e^{x^{e^{x^{...}}}} \equal{} 2$ find x such that $ x > 0$

Question: I was reading up on counting infinite sets. The book said that you could prove that two infinite sets are of the same size by making a one-to-one correspondence with their elements. For example, to prove the infinite set of positive integers is the same size as the set of all integers, you would do
{1,2,3,4...}
{0,1,-1,2,-3...}
Thus, you are done? My question is, how does this show that two sets are of the same size? I see no rigor in the correspondence given above. All I see is two infinite sets continuing on to, well, infinity.

Any thoughts would be appreciated![/quote]

Dunno much about the infinite sets, but to answer the question, [hide]since $ e^{x^{e^{x^{...}}}} \equal{} 2$, we can substitute that in:

$ e^{x^{e^{x^{...}}}} \equal{} e^{x^2} \equal{} 2$

$ \log_e e^{x^2} \equal{} \log_e 2$

$ x^2 \equal{} \log_e 2$

$ x \equal{} \pm \sqrt {\log_e 2}$

I think only one answer works though, but which one :([/hide]

good solution, identity! on the contrary, the positive solution is the valid one (*cough*says my book *cough*)

Yes I thought that a positive solution would make the LHS rise to infinity, whereas a negative would stabilize it... oh well, it's tricky stuff  :roll:


ratio
geometry
circumcircle
angle bisector
geometry solved

[quote="silouan"]$ABC$ inscribed triangle in circle $(O,R)$. At $AB$ we take point $C'$ such that $AC=AC'$ and at $AC$ we take point $B'$ such that $AB'=AB$. The segment $B'C'$ intersects the circle at $E,D$ respectively and and it intersects $BC$ at $M$.
Prove that when the point $A$ moves on the arc $BAC$ the $AM$ pass from a standard point.[/quote]

Strange way to formulate a problem (e. g., I don't see what the points E and D are for). By the way, your "standard point" is actually called "fixed point" in English.

Here is a solution to your problem:

Since AB' = AB and AC = AC', the triangles BAB' and C'AC are isosceles. Since the angles at their apices are equal (< BAB' = < C'AC), they must be similar, i. e. we have < ABB' = < AC'C. Hence, BB' || CC'. Thus, by Thales, $\frac{BM}{CM}=\frac{BB^{\prime}}{CC^{\prime}}$ and $\frac{BB^{\prime}}{CC^{\prime}}=\frac{AB^{\prime}}{AC}$. Hence, $\frac{BM}{CM}=\frac{AB^{\prime}}{AC}$. Since AB' = AB, this becomes $\frac{BM}{CM}=\frac{AB}{AC}$. Now, it is well-known that the angle bisector of an angle of a triangle - and only the angle bisector - divides the opposite side in the ratio of the adjacent sides. Thus, from the equation $\frac{BM}{CM}=\frac{AB}{AC}$, it follows that the line AM is the angle bisector of the angle CAB. Now, since the angle bisector of an angle of a triangle bisects the arc intercepted on the circumcircle by the opposite side, we conclude that the line AM, being the angle bisector of the angle CAB, passes through the midpoint of the arc BC (not containing the point A) on the circumcircle of triangle ABC. This midpoint is obviously a fixed point. Thus, the problem is solved.

  Darij

dear siluan i think to answer to this problem is easy.
because as u know line AM bisects the angleBAC.
and it means that by moving A on arc BAC ,AM should go through the mid point of the arcBC.
and this point is the fix point. ;)


linear algebra
matrix
induction
abstract algebra
group theory
Ring Theory
superior algebra

Suppose that $a_1,a_2,\dots ,a_n$ are $n$ integers which $gcd(a_1,a_2,\dots, a_n)=1$. Prove that there exist an $n\times n$ matrix like $A_n$ which its first row $(a_1,a_2,\dots,a_n)$, and $det(A)=1$

?? I don't understand..
matrix must be integers? in this case the theorem is wrong!
Otherwise (a real matrix) the theorem is trivial!

@Uber: Keep cool. The assertion is true ...
@Omid:

We use induction by $n$. If $n=2$, there are $r,s \in \mathbb{Z}$ s.t. $ra_1 + s a_2 = 1$. Then we can choose
\[ \begin{pmatrix}a_1 & a_2 \\ -s & r \end{pmatrix}  \]
Now let $n>2$. Define $d : = gcd(a_1 , ... , a_{n-1})$, $b_i : = a_i/d$ for $i=1,...,n-1$. Then $gcd(b_1 , ... ,b_{n-1})=1$. So by induction hypothesis, there's a matrix of the form
\[ B=\begin{pmatrix} b_1 & \cdots & b_{n-1} \\ & B' & \end{pmatrix} \in \mathbb{Z}^{(n-1) \times (n-1)}  \]
with $\det(B)=1$. Because of $gcd(a_n,d)=1$, there are $s,t \in \mathbb{Z}$ s.t. $s a_n + t d = 1$. Now, define
\[ A=\begin{pmatrix} d b_1 & \cdots & d b_{n-1} & a_n \\ & & & 0 \\ & B' & & \vdots \\ & & & 0 \\ -s b_1 & \cdots & -s b_{n-1} & t \end{pmatrix} \in \mathbb{Z}^{n \times n}  \]
The first row of $A$ is $a_1 , ... , a_n$, and its determinant is by Laplace
\[ det(A) = (-1)^{n+1} a_n (-s) (-1)^{n-2} \det(B) + t d \det(B) = s a_n + t d = 1  \]
Of course, this can be generalized to arbitrary principal ideal domains.

Great Solution -oo-.
There is also a beautiful proof with the Fundamental Theorem of Finitely Generated Abelian Groups. Think on it.

Ok! The idea for this proof arose from [url=http://www.mathlinks.ro/Forum/viewtopic.php?t=112335]this[/url] problem.

Let's consider the finitely-generated abelian group

$A : = \mathbb{Z}^{n}/ \langle (a_{1},...,a_{n}) \rangle$

If $\overline{(b_{1},...,b_{n})}$ is a torsion element of $A$, there are integers $u,v$ s.t.

$u(b_{1},...,b_{n}) = v (a_{1},...,a_{n}) , u \neq 0$

We may assume that $u,v$ are coprime, otherwise divide by $gcd(u,v)$. But then $u | v a_{i}$ implies $u | a_{i}$ for all $a_{i}$. Since the $a_{i}$ are coprime, it follows $u = \pm 1$ and hence $\overline{(b1,...,bn)}=0$. Thus, $A$ is torsionfree, and finitely-generated, hence free of rank $n-1$ by the structure theorem. This implies that $\langle (a_{1},...,a_{n}) \rangle$ is a direct summand of $\mathbb{Z}^{n}$. The other summand is (as a subgroup of a free abelian group) also free, say with basis $s_{2}, ... , s_{n}$. Finally, if we put $s_{1}: = (a_{1},...,a_{n})$, the matrix $(s_{1}| ... | s_{n})$ is invertible, i.e. has determinant $\pm 1$. If necessary, change the sign of $s_{2}$, so that the determinant equals $1$.


trigonometry
vector
function
analytic geometry
search
calculus
derivative

So a friend gave me a problem a while back to make an animation of rain falling on water. I have a decent representation of static waves given by $ z\equal{}h(\sin{(x\minus{}aq)}\plus{}\sin{(y\minus{}bq)}$ or the produce of the two sines times h (constant, as are a,b). $ q$ is a dummied variable that represents water movement by augmenting the center of the waves, while a & b give the general vector motion of the ripples.
Anyhow, that's not the point. Lets say a raindrop hits the surface of the water (let's assume its flat for now i can take care of the merger just fine) with mass $ m$ and velocity $ v$. I'm having trouble with the resulting waves. I don't know what their amplitude is proportional to (exactly, but i have a guessed up demo), nor their speed which is more important. is the speed dependent on the drops momentum or its energy? How fast does the harmonic oscillation of the impact center dissipate?

I know this is a nutty question, but any insight on this would be appreciated. I'll be sure to a link to the final mathematica demonstration once its finished.

I think the speed of the wave will depend on the medium of travel... in this case liquid water.

Here is a little sketch on the basics of surface waves , and then i try to apply them for circular surface waves. I may ommit some parts of the calculations, since they are quite lengthy.

First of all, we need our fluid equations. Lets consider that we are working with an ideal hence, let it be incompressible. Furthermore lets assume that no vorticity developes during the motion of the fluid, which means that we can take it to be irrotational.

Mathematicaly we have, two equations for the velocity field:
\[ \text{curl}\, \textbf{v} \equal{} 0\]
This is the irrotational condition. And:
\[ \text{div}\, \textbf{v} \equal{} 0\]
This is the incrompessibility condition.
From the first equation we see that the velocity field can be expressed as the gradient of some scalar function, $ \Phi$ called the velocity potential. So $ \textbf{v} \equal{} \nabla \Phi$ Plugging this into the other equation we obtain:
\[ \bigtriangleup \Phi \equal{} 0\]
Which means that the velocity potential obeys the laplace equation.
Next we have to set up the boundary conditions. Let $ z$ be the vertical coordinate, and $ \zeta(\textbf{r},t)$ describe the momentary vertical position of the free surface i.e. the wave "amplitude".
The first boundary condition, is that on the free surface the vertical component of the velocity of the fluid is equal to the rate of change of the vertical disposition of the free surface:
\[ \frac {\partial \Phi}{\partial z} \minus{} \dot\zeta \bigg|_{z \equal{} \zeta} \equal{} 0\]
Consider the fluid depth to be $ h$. Then the next boundary condition arises from the ideal nature of the fluid. i.e. the velocity's normal component vanishes at the boundary:
\[ \frac {\partial \Phi}{\partial z}\bigg|_{z \equal{} \minus{} h} \equal{} 0\]
The last boundary condition can be given using the bernoulli equation for non-stationary potential flow. We consider the pressure to be constant above the boundary of the fluid, so we have:
\[ \left(\frac {\partial \Phi}{\partial t} \plus{} gz\right)\bigg|_{z \equal{} \zeta} \equal{} 0\]
Next we solve the laplace equation with these boundary conditions. Lets search for a wave solution wich and travels in the xy plane:
\[ \Phi(x,y,z) \equal{} \psi(z)e^{i(k_x\cdot x \plus{} k_y\cdot y \minus{} \omega t)}\]
$ k_x$ and $ k_y$ are the wave numbers in their respective directions. plugging the above form into the laplace equation we get the following ordinary differential equation for the $ z$-dependence:
\[ \psi''(z) \equal{} k^2\psi(z)\]
Where we introduced $ k^2 \equal{} k_x^2 \plus{} k_y^2$.
Searching for $ \psi(z)$ in the following, seems to be of value:
\[ \psi(z) \equal{} A\cosh(k(z \plus{} h))\]
$ A$ is some constant. As we see this form automatically satisfies the second boundary condition on $ \Phi$ since its derivative wrt $ z$ vanishes at $ z \equal{} \minus{} h$.
To obtain the dispersion relation, differentiate the second boundary condition wrt $ t$, and combine it with the first boundary condition, for the free surface:
\[ \left(\frac {\partial^2\Phi}{\partial t^2} \plus{} g\frac {\partial \Phi}{\partial z}\right)\bigg |_{z \equal{} \zeta} \equal{} 0\]
Plugging our $ \Phi$ into this:
\[ \left[ \minus{} \omega^2\cosh(k(z \plus{} h))e^{i(k_x x \plus{} k_y y \minus{} \omega t)} \plus{} gk\sinh(k(z \plus{} h))e^{i(k_x x \plus{} k_y y \minus{} \omega t)}\right]\bigg|_{z \equal{} \zeta} \equal{} 0\]
assuming that the amplitude of the surface waves are much smaller than the depth of the "pool" ($ \zeta < < h$), we get:
\[ \omega \equal{} \sqrt {gk\tanh(kh)}\]
This is a very interesting dispersion relation. We can get the phase velocity of the wave by dividing the above with $ k$:
\[ c \equal{} \sqrt {\frac {g}{k}\tanh(kh)}\]
Lets analyze this a little. say the wavelength of the surface waves are quite big ($ k$ is small), and the depth of the "pool" is not so big, then we can approximate the hyperbolic tangent function: $ \tanh(kh)\approx kh$:
\[ c_s \equal{} \sqrt {gh}\]
So the phase velocity is proportional to the depth. This is basically(well almost) the reason for the "wave-breaking" at the beach. The water gets shallower as we get nearer to the beach, hence the waves travelling closer to the ground will have smaller velocity than the ones travelling higher, so the ones above break over the lower ones.
But i just mentioned this as a side fact.
The important case in our analysis, is when the wavelength is short, and the pool is deep (a drop of rain doesnt cause so much "disturbance" in a pool, which is sufficiently deep). In this case $ \tanh(kh) \approx 1$, so:
\[ c_d \equal{} \sqrt {\frac {g}{k}} \equal{} \frac {1}{\sqrt {2\pi}}\sqrt {g\lambda}\]
This will be the phase velocity of the surface waves.
So we see that the velocity potential becomes:
\[ \Phi \equal{} \frac {Ae^{k(z \plus{} h)}}{2}e^{i(k_x x \plus{} k_y y \minus{} \omega t)}\]
(since $ e^{ \minus{} k(z \plus{} h)} \to 0$ as $ kh\to \infty$ is "big").

Next we want some sort of equation for the $ \zeta$ describing the free surface.
Differentiating the above $ \Phi$ wrt $ z$:
\[ \frac {\partial\Phi}{\partial z} \equal{} k\Phi\]
Now lets get back to the laplace equation:
\[ \frac {\partial^2\Phi}{\partial x^2} \plus{} \frac {\partial^2\Phi}{\partial y^2} \plus{} \frac {\partial^2\Phi}{\partial z^2} \equal{} \frac {\partial^2\Phi}{\partial x^2} \plus{} \frac {\partial^2\Phi}{\partial y^2} \plus{} k\frac {\partial \Phi}{\partial z} \equal{} 0\]
Differentiating wrt $ t$ (and using Young's theorem):
\[ \frac {\partial^3\Phi}{\partial x^2\partial t} \plus{} \frac {\partial^3\Phi}{\partial y^2\partial t} \plus{} k\frac {\partial^2\Phi}{\partial t \partial z} \equal{} 0\]
At $ z \equal{} \zeta$ we use the boundary condition from the beginning, to get:
\[ \frac {\partial^2\zeta}{\partial x^2} \plus{} \frac {\partial^2\zeta}{\partial y^2} \minus{} \frac {k}{g}\frac {\partial^2\zeta}{\partial t^2} \equal{} 0\]
This is a 2 dimensional wave equation for $ \zeta$.
Now when we see a rain drop falling into a pool, we usually see circular waves, i.e. waves with some sort of cylindrical symmetry. So it is best to transform the space variables into a more suitable coordinate system. We write the space derivatives in polar coordinates $ (r,\varphi)$ instead of cartesian. So the above equation transforms into:
\[ \frac {\partial^2\zeta}{\partial r^2} \plus{} \frac {1}{r}\frac {\partial \zeta}{\partial r} \minus{} \frac {k}{g}\frac {\partial^2\zeta}{\partial t^2} \equal{} 0\]
Where symmetry in the $ \varphi$ variable is assumed.

The time dependence of $ \zeta$ is oscillatory.. so:
\[ \frac {\partial^2\zeta}{\partial r^2} \plus{} \frac {1}{r}\frac {\partial \zeta}{\partial r} \plus{} k^2\zeta \equal{} 0\]
dividing by $ k^2$:
\[ \frac {\partial^2\zeta}{\partial (kr)^2} \plus{} \frac {1}{kr}\frac {\partial \zeta}{\partial kr} \plus{} \zeta \equal{} 0\]
We recognize that this is the Bessel differential equation of order $ 0$ in the variable $ kr$. The non-singular solutions are Bessels functions of the first kind. So our "wavelets" have the form:
\[ J_0(kr)e^{ \minus{} i\sqrt {gk} t}\]
But as we know not only one wave number occurs, hence the general $ \zeta$ will be a superposition (our equation is linear) of these monochromatic components:
\[ \zeta(r,t) \equal{} \int_0^{\infty}dk\,\,k\cdot W(k) J_0(kr)e^{ \minus{} i\sqrt {gk} t}\]
where $ W(k)$ is the weight of the different components. (The reason i wrote the $ k$ in the integrand, is because as shall be seen the integral is easier to be dealt with)
To proceed we have to assume some initial conditions(when the surface is steady). I say that this be the following.
We have a small "fluid cylinder" of height $ a$ and radius $ r_0$, "pulled out" above the surface. i.e. $ \zeta(r,0) \equal{} a$ if $ r < r_0$ and $ \zeta(r,0) \equal{} 0$ elsewhere. With this:
\[ \zeta(r,0) \equal{} \int_0^{\infty}dk\,\,k\cdot W(k) J_0(kr)\]
The weight function is the Hankel-transform pair of this:
\[ W(k) \equal{} \int_0^{\infty}dr\,\,r\cdot \zeta(r,0) J_0(kr) \equal{} a\int_0^{r_0}dr\, r J_0(kr)\]
Using the connection between the first and zero order Bessel functions: $ \frac {d}{dx}(xJ_1(x)) \equal{} xJ_0(x)$, we can perform the integral to get:
\[ W(k) \equal{} \frac {ar_0}{k}J_1(kr_0)\]
And with this the displacement of the surface is:
\[ \zeta(r,t) \equal{} ar_0 \int_0^{\infty} dk \, J_0(kr) J_1(kr_0)e^{ \minus{} i\sqrt {gk} t}\]
There is not much we can do with this integral, but with certain approximating conditions, we can perform the integration. lets assume that the $ r_0 < < r$ . Then using the integral representation of $ J_0$, and the below conditions:
\[ \frac {r}{r_0} > > \frac {gt^2}{4r} > > 1\]
After applying the saddle-point approximation method twice (transforming the integrals into gaussian integrals), I get the following:
\[ \zeta(r,t) \equal{} \frac {agr_0^2}{2\sqrt {2}}\frac {t^2}{r^3}e^{ \minus{} i\frac {gt^2}{4r}}\]
Now i checked my calculation a few times, so I hope i didnt make any errors, or "lose" anything during the calculation.

So finally we are done, and got the shape of the surface.. :trampoline:

I think you can use this for your model, and if you are done, then please send me a link, because i am curious if this really works.. :D

If you have any questions, then please ask

Thank you. I just saw this post a long time afterward because it was dry for a while. Must have taken a bit to write that. I'll link you as soon as I brush up on curl and div. (I'm self taught so... yeah). Should be within a fortnight. What you've got going on there will take a bit to digest.
edit: furthermore, I've used a very small amount of physics in my model (decent) thus far.


Prove that the area of a regular dodecagon (12-gon) is $ 3R^2$, where $ R$ is the circumradius.

EDITED TYPOS

I have a feeling that you mean a dodecagon, which is actually a 12-sided polygon.  :wink:
(As for a proof, simply refresh this page several times, and look on the left.)

Ah, why am I keep making those stupid errors?  Yeah, as you pointed out, it was that animation that inspired me to post this topic; that geometric proof of dividing the dodecagon into 3/4 of the whole square with side length 2R is very elegant, but are there other ways, too?

EDIT: By the way, I'm using mathlinks but not AoPS, but I discovered that animation while reading AoPS wiki, but not the mathlinks forum.  Just curious, is there any significant difference between mathlinks and Art of Problem Solving?  :maybe:

You can trig bash it. Split the 12-gon into 12 isosceles triangles having the base as one side of the 12-gon, and the vertex as the center of the 12-gon. You can find this area by using the formula for interior angles to find the angles, and so on...

Consider one of the 30-75-75 triangles and let $ \angle A \equal{} 30, \angle B \equal{} 75$ and $ \angle C \equal{} 75$.  Note that we can stack all 12 of these triangles into a parallelogram with base $ R$.  To find the height, drop a perpendicular from $ B$ to $ AC$.  $ \angle BAC \equal{} 30$ so the height is $ \frac {R}{2}$.  Thus, the height of the whole parallelogram is $ 6\frac {R}{2} \equal{} 3R$ so the area is $ \boxed{3R^2}$.

[asy]size(5cm);
pair A=(0,0), B=(sqrt(3)/2,1/2), C=(1,0), D=(sqrt(3)/2,0);
D(B--A--C--B);
D(B--D);
label("$A$",(0,0),W);
label("$B$",(sqrt(3)/2,1/2),N);
label("$C$",(1,0),E);[/asy]


is that supposed to say $a_0,a_1,a_2,\cdots a_{2002}\in \mathbb{Z}$?

[hide]$q(x)=(p(x)+5)(p(x)-5)$, so any root of $q(x)$ must be a root of $p(x)+5$ or $p(x)-5$.

Suppose $q(x)$ has at least 2004 distinct integer solutions. Since each of $p(x)+5$ and $p(x)-5$ has at most 2003 roots total, each polynomial must have at least one root (and no root could be shared by the other polynomial). Since the 2004 roots are being divided between two polynomials, at least one must have at least 1002 distinct integer roots.

Let $c$ be an element of $\{\pm 5\}$ such that $p(x)+c$ has at least 1002 distinct integer roots, and label 1002 such roots $r_k$ where $k$ ranges from 1 to 1002. Let $s$ be an integer root of $p(x)-c$.

If $f$ is a polynomial with integer coefficients and $a,b\in \mathbb{Z}$, then $(a-b)|(f(a)-f(b))$, which follows from difference of powers factorization. Then $(r_k-s)|2c$ for each $k$. Since $2c=\pm 10$, its divisors are in the set $\{ \pm 1,\pm 2,\pm 5,\pm 10\}$, for a total of 8 divisors. But as $k$ ranges from 1 to 1002, $r_k-s$ takes on 1002 different values. This is a contradiction, so there are at most 2003 distinct integer roots of $q(x)$.[/hide]


LaTeX
number theory unsolved
number theory

there's no one online give a solution for this exercice  :rotfl:   :rotfl:

Could you please use $\LaTeX$ instead of posting such things?!¿


linear algebra
matrix
induction
function
linear algebra unsolved

If $ n\equal{}2$ then $ det(M)\equal{}a_1^{x_1} a_2^{x_2}\minus{}a_1^{x_2} a_2^{x_1} > 0$. Indeed,
$ a_1^{x_1} a_2^{x_2} > a_1^{x_2} a_2^{x_1}$
which is equivalent to
$ a_1^{x_1\minus{}x_2} > a_2^{x_1\minus{}x_2}$
which is true.
So at least for $ n\equal{}2$ is valid, with the condition $ 0 < a_1$.

[quote="QuyBac"]Consider the real numbers $ 1 \leq a_1 < a_2 < a_3 < ... < a_n$ and $ x_1 < x_2 < x_3 < ... < x_n$ and the matrix
           $ M \equal{} (a_{i}^{x_j})_{1 \leq i,j \leq n}$
Prove that : $ \det(M) > 0$[/quote]
I conjecture that $ M$ is positive definite matrix.

It is not a symmetric matrix.How to say it  is positive define matrix?

But It still a nice problem.First we will prove that:
$ \det(M)\neq 0$,for proving this we should only to show the linear system following equation only have solution $ y_1 \equal{} y_2 \equal{} \cdots \equal{} y_n\equal{}0$
\[ a_i^{x_1}y_1 \plus{} a_i^{x_2}y_2 \plus{} \cdots \plus{} a_i^{x_n}y_n \equal{} 0,i \equal{} 1,2,\cdots,n\]
That is equal to say if $ f(t) \equal{} t^{x_1}y_1 \plus{} t^{x_2}y_2 \plus{} \cdots \plus{} t^{x_n}y_n$ have roots $ 0 < a_1 < a_2 < \cdots < a_n$ we must have
$ y_1 \equal{} y_2 \equal{} \cdots \equal{} y_n \equal{} 0$.We use induction to prove the last proposition.Because $ f(t)$ has $ n$ roots $ a_1,a_2,\cdots,a_n$ such that
$ 0 < a_1 < a_2 < \cdots < a_n$,then $ a_1,a_2,\cdots,a_n$ are still the roots of  $ g(t) \equal{} y_1 \plus{} t^{x_2 \minus{} x_1}y_2 \plus{} \cdots \plus{} t^{x_n \minus{} x_1}y_n$.
Then by the Roll's Theorem there must exist $ b_i$ such that $ 0 < a_1 < b_1 < a_2 < b_2 < \cdots < b_{n \minus{} 1} < a_n$
such that $ g'(b_1) \equal{} g'(b_2) \equal{} \cdots \equal{} g'(b_{n \minus{} 1})$,But $ g'(t) \equal{} y_2(x_2 \minus{} x_1)t ^{x_2 \minus{} x_1 \minus{} 1} \plus{} \cdots \plus{} y_n(x_n \minus{} x_1)t ^{x_n \minus{} x_1 \minus{} 1}$,So by induction
we have $ y_2 \equal{} \cdots \equal{} y_n \equal{} 0$,that is obviously that $ y_1 \equal{} 0$.Hence we proved that $ \det(M)\neq 0$.

The rest  is not  hard,you may assume
$ h(x_1,x_2,\cdots,x_n) \equal{} \det (M)$,then for $ x_i \equal{} i,i \equal{} 1,2,\cdots,n$ we have $ \det(M) > 0$,hence by the intermediate property of the contionuous
function we have $ \det(M) > 0$ for every $ x_1 < x_2 < \cdots < x_n$.


inequalities
real analysis
real analysis unsolved

Let $ A \subset \mathbb{R}^n$ be closed. Proof that $ d_A: \mathbb{R}^n \to \mathbb{R}, x \mapsto \inf\{|x \minus{} a| \mid a \in A\}$ is continuous.

$ |d_A(x)\minus{}d_A(x')|\le |x\minus{}x'|$ by the triangle inequality. It does not matter if $ A$ is closed or not.

[quote="mlok"]$ |d_A(x) \minus{} d_A(x')|\le |x \minus{} x'|$ by the triangle inequality. [/quote]Can you elaborate on this, please?

[quote="berti"].. Can you elaborate on this, please?[/quote]
Remember : if $ \forall a , P(a) \geq k$ then $ inf \{P(a)\} \geq k$
Triangle's inequality : $ \forall a\in A, \parallel{}x\minus{}x'\parallel{} \plus{}\parallel{}a\minus{}x'\parallel{} \geq \parallel{}x\minus{}a\parallel{} \geq d_A(x)$ , so $ \parallel{}x\minus{}x'\parallel{} \plus{}d_A(x') \geq d_A(x)$
In the same way $ \parallel{}x'\minus{}x\parallel{} \plus{}d_A(x) \geq d_A(x')$ , .. and so on ..
 :cool:


inequalities
geometry
inradius
geometry proposed

[color=purple] Let triangle ABC in plane inscribed circle $ (O)$.The circle $ (I,r)$ inscribed triangle. $ AI,BI,CI$ cut $ (O)$ at $ M,N,P$. Denote $ r_{I}$ is inscribed radius of triangle MNP. Prove that :$ \frac{r_{I}}{r}\ge 1\plus{}\frac{(a\minus{}b)^2\plus{}(b\minus{}c)^2\plus{}(c\minus{}a)^2}{8R^2}$
   :D  :) [/color]

If a=b=c, then according the inequality rI=>r, which is not possible because r is the circum radius of triangle MNP and rI is the inradius of triangle MNP

[quote="Siva Nagi Reddy"]I think there is something wrong with the question.pls read the above comment[/quote]
 [color=purple] Hey please read my problem carefully. $ (I,r)$ is the inscribed circle of triangle ABC, and $ (I_{r},r_{I})$ is inscribed circle of triangle $ MNP$[/color] :)


Im dong a problem and ive forgotten a property:

in quadrilateral ABCD, let there be a point P in the interior.

I remember there's a property of PA, PB, PC, and PD.

like PA^2+PB^2+PC^2+PD^2 is a constant or something like that?

Close but not quite. The formula I think you're referring to is PA^2+PC^2 = PB^2+PD^2.

Right, forgot to mention it only works for rectangles, sorry for any confusion.

[quote="Gealgebra"]Close but not quite. The formula I think you're referring to is PA^2+PC^2 = PB^2+PD^2.[/quote]
That "formula" is certainly not true...here's a counterexample:
a quadrilateral with perpendicularly bisecting (at point $ P$) diagonals of lengths 80 and 2. Then, by your "formula," $ 1^2 \plus{} 1^2 \equal{} 40^2 \plus{} 40^2$, which is clearly not true.

It might be you are thinking about the [url=http://www.artofproblemsolving.com/Wiki/index.php/British_Flag_Theorem]British Flag Theorem[/url].


function
modular arithmetic
induction
national olympiad

[quote="doctormath"]2. Prove that $ 7$ | $ 12^n$ + $ 2\times5^{n \minus{} 1}$ .[/quote]

Two integers a and b are said to be congruent modulo n.
It is expressed as:$ a \equiv b \pmod{n}$

Let $ 12^n \plus{} 2\times5^{n \minus{} 1} \equiv 5^n \minus{} 5\times5^{n \minus{} 1} \equal{} 5^n \minus{} 5^n\equal{}0\pmod{7}$

Then the question was done.

[quote="Rosnovisky"][quote="doctormath"]2. Prove that $ 7$ | $ 12^n$ + $ 2\times5^{n \minus{} 1}$ .[/quote]

Two integers a and b are said to be congruent modulo n.
It is expressed as:$ a \equiv b \pmod{n}$

Let $ 12^n \plus{} 2\times5^{n \minus{} 1} \equal{} 5^n \minus{} 5\times5^{n \minus{} 1} \equal{} 5^n \minus{} 5^n \equiv 0\pmod{7}$

Then the question was done.[/quote]
Induction is easy way
2) is very old

3) let $ m\equal{}[\frac{n\plus{}1}{2}]$ then $ |P(a_i)|\equal{}1<2^{\minus{}m}m!$ where i={1,..n} so by Polya theorem P(x) cannot be factorized

I don't understand with number 3 solution. :D please help me, I don't know about Polya's Theorem.. :D

Polya theorem: let f polynom of n degree with integer coef. and $ m\equal{}[\frac{n\plus{}1}{2}]$ if there are n distinct integer numbers a1,a2...an (they are not roots of polynom): $ |f(a_i)<2^{\minus{}m}m!|$ than f(x) cannot be factorized

Guys, each problem should go in its own topic. Thanks!


Given real numbers $ x_1,x_2,\dots,x_n,y_1,y_2,\dots,y_n,a_1,a_2,\dots,a_n$ such that $ a_i\geq0$ for all $ 1\leq i\leq n$ and $ (x_1,x_2,\dots,x_n) \succ (y_1,y_2,\dots,y_n)$.
Let $ X$ be the decreasing sequence of length $ n!$ consisting of all $ n!$ numbers of the form $ a_1x_{\pi(1)}\plus{}a_2x_{\pi(2)}\plus{}\dots\plus{}a_nx_{\pi(n)}$ where $ \pi$ is a permutation of $ (1,2,\dots,n)$. Let $ Y$ be defined similarly.

Prove that $ X\succ Y$

This problem is really not hard...but with it you have a very short proof for Muirhead (like three lines). Is that enough incentive?


*kicks tumbleweed*  It's too quiet here!

Can we get some returning campers to share some past experiences?

You know, stuff you did, things you learned, what you liked or disliked, your favorite part of Mathcamp, etc...

I'll put a question out there: How's the 'problem solving' environment? I know we're not geared towards contest training like some other camps, but I'd still like to know if there are classes for contest math. Also, how often do we have competitions?

Oh, and just for excitement, 19 days until camp! :w00t:

 :D

Last year Bogdan taught a few problem problem solving classes (Intro, Olympiad, Calculus).  I believe he always had two classes per day (Intro lasted throughout camp.  The focus of his other class varied from week to week.).  I didn't take any of his classes, so I can't say what they were like.  But, I don't think he's returning this year.  I'd assume that Mathcamp hired a replacement or that one of the other staff will fill in.  Additionally, there are two weekly problem solving contests.

Team Problem Solving, TPS, is Wednesday nights.  You're assigned a team of 8 (if you chose to participate) that you will be on for the duration of camp.  Each round is an hour and consists of 8 proof-based questions.  The two different types of teams, Hardcore and Mellow, get the same problems but Mellow is generally a bit more relaxed and often a lot of team members won't show up.  The Hardcore teams are generally more interested in winning but aren't overly competitive.

Relays are Saturday afternoons.  Each week you make a team of 8.  Typically, your team is given two numerical answer problems at a time.  Basically, you try to answer as many questions as you can.  I don't want to get too specific on rules because they change every week for variety.  Again, there are Mellow and Hardcore teams, but the Mellow questions are easier than the Hardcore questions.

To answer your other questions, my typical weekday at Mathcamp went something like wake-up, eat breakfast, finish homework/independently working on math, class, another class, lunch, another class, TAU (Time, Academic, Unscheduled.  For me, this usually was working on homework with friends.), dinner, friends/merrymaking, homework/independently working on math, sleep.  It's quite amazing how much math and socializing you can fit into a single day, all of it fun (that is, I never got burnt out despite doing on average 7-8 hours of math a day).  But, their never seemed to be enough time to do both.  So, the second worst thing about Mathcamp is its inability to increase the number of hours in a day.  The best thing about Mathcamp is definitely the number of choices given to campers combined with the freedom to choose as many or as few things as you'd like (with the unfortunate restriction of scheduling conflicts due to time).  No camper ever gets the same experience and alumni say that each year is different.  But everyone says Mathcamp was the best (or tied for the best for alumni) five weeks of their life because everyone is given the freedom to do what they'd like.  So, no one really wants to leave making leaving the worst part of camp.

I'm just wondering, do any of the staff have (at least a rough idea of) what courses will be offered this year? Thanks.

When registering, I assumed that the Hardcore teams would be filled with the people who get perfect scores on GPML Power Rounds solo and never feel drowsy</hyperbole>, and thus decided to choose Mellow despite my competitive streak. How mistaken am I?

The main difference between Mellow and Hardcore is that the Hardcore people are more competitive than the Mellow people. I don't think the typical Hardcore people are any "better" at math than the typical Mellow people.

So if you're a competitive person (i.e. you want to be on a team that's at least trying to win), than you should probably be on a Hardcore team. Don't worry about how good or bad at math you think you are.

I do agree that it's too quiet here.

To be honest, I don't completely agree with some of the opinions here. The people who are crazy with math tend to participate in the Hardcore teams, but there are "normal" people in them, too, so no one should worry about that. I'm not particularly good at math (relative to others), but I'm going to go the Hardcore path (I was mellow last year, and really... it is rather mellow, which is good depending on individual preferences).

There are way too many things to do at Mathcamp. I've only been to one other summer math program (HCSSiM), but there seems to be this feeling that if I were to compare Mathcamp with other programs, Mathcamp offers you soooooooooooooo much more freedom, whereas other programs gives everyone more or less a core curriculum. Thus, it's easy to get carried away with social activities. I should've spent more time studying math last year, but I'll make that a priority this year. =D

[quote="Eternica"]
To be honest, I don't completely agree with some of the opinions here. The people who are crazy with math tend to participate in the Hardcore teams, but there are "normal" people in them, too, so no one should worry about that. I'm not particularly good at math (relative to others), but I'm going to go the Hardcore path (I was mellow last year, and really... it is rather mellow, which is good depending on individual preferences).
[/quote]

Yeah, that's why I said the "typical" Hardcore people are no better than the "typical" Mellow people. It is true that most of the top students (here I mean like MOPpers and such) do tend to choose Hardcore (probably because to do as well as they've done, they have to be at least a bit competitive). But remember that around half of the camp is going to be doing Hardcore. Most of the Hardcore teams are going to end up with one or two MOPpers (who probably end up stuck with much more than their fair share of the work :dry: ), but beyond that "most" of the Hardcore people are "normal".

My main point with that last post was that you don't have to be really awesome at math to do Hardcore.

Hehe. Sorry for being nitpicky, but you emphasized that point well. The safety of MOPpers is why I'm not afraid of going Hardcore. ^_^;;

Hmm, just one note about Mellow, I found it to be [i]really[/i] mellow, as in most of my team never showed up, and I had to join with another team.

By the way, who won mellow last year? (I left early)

I was on Sergei's (who was in black MOP - and decided to go mellow ..) team [and hahafaha's], and surprisingly enough all but one of our members consistently showed up. I'm pretty sure we were the only such team, judging by the scores, and for at least the first couple of rounds, we were just barely trailing the hardcore teams.

I'm pretty sure you guys won Mellow last year (by a lot).  From what I've heard, most Mellow teams are more like calc rulz's.

JAM is certainly right about MOPpers getting stuck with most of the work in Hardcore - our team tried to get away with having Adam Hesterberg do everything.  This worked fine until Adam left to go to the International Linguistics Olympiad.  The last two weeks were interesting . . .  :|

Yeah, we won mellow last year (my name is Leonid, BTW :P). I was rather happy with it -- enough people showed up that we could actually do it, but there was no nagging feeling of obligation. I remember one week, we just sort of hung around and discussed the differences between British and American culture (one of our team members was from Britain). Apparently, many other mellow teams were far more relaxed.

My personal recommendation is to go hardcore on the relays. You form a new team every week (though, of course, you can have the same team as a previous week; the point is that you're not locked into the same group), so if you're feeling enthusiastic for the hour or two, you can have fun, and then just not go the next week. With TPS, you have a responsibility for your team.

As for the average day, for me it was something like: Wake up, shower, eat breakfast/socialize, go to class/socialize (this depended on week, as some weeks I had every class slot filled, and others, I had only a couple of classes -- again, you have full control over this), lunch, TAU, socialization. I actually ended up not doing homework almost ever, and it was fine :D

[quote="aguoman"]A quick question:

I just realized that camp starts on July 5th, which is a Saturday, while our classes go from Tuesday to Saturday (a usual camp week). What do we usually do on the Sunday and Monday at the beginning of camp?[/quote]

So, camp usually starts on Sundays. This year is an exception, meant to deal with 4 July being Friday (I am not certain what the logic was, but my guess is that people would be flying on Friday and then flying back on Sunday). Anyway, usually, people arrive on Sunday, sort of settle in and enjoy themselves, and then Monday, a lot of administrative stuff is taken care of, including a large assembly, meeting various mentors and other things. I don't know what they're planning on this year with the extra day.

Camp usually starts on a Sunday - I think they switched it to Saturday because Sunday travel on the July 4th weekend is expensive.

Usually, Sunday is arrival/settle in/have fun, and Monday is orientation (when they tell you how everything works at camp, you choose your academic adviser, find out about Week 1 classes, etc.).  I'm guessing that they will shift these days back to Saturday and Sunday, and then have a 6 day academic Week 1.  But I could be wrong.

EDIT: Too late . . .

[quote="E^(pi*i)=-1"]EDIT: Too late . . .[/quote]

Owned :P.

Again, completely guessing here, but I doubt they'd go for a six-day academic week. I'm guessing they'll just have a longer weekend.

I doubt they would have a longer weekend - they probably wouldn't have time to pull together a field trip (people signing up for it and such), and a day of nothing happening, while nice later in camp when everyone is acquainted, might be somewhat awkward for new campers so early.

Well, a weekend need not be "a day of nothing happening". It could be a bunch of pre-arranged activities designed to make first-years feel more comfortable and meet each other and alumni

you mean new campers that don't already know other people going :P

I have a question to keep the tumbleweeds out of the way... What do the academic advisors actually do? How much control do they have over the classes you attend? Would they say, tell you that you should take/not take a certain course? Would they yell at you for not taking enough or for taking too many classes?

thanks dudes :)

The academic advisors are just there to advise you about academics  :P . You have complete control over what classes you take/how many classes you take etc. Basically as long as [i]you're[/i] happy with your course schedule, everything's fine. Your academic advisor (or any staff member for that matter) will never [i]yell[/i] at you.

So, basically the function of an academic advisor: at the end of every week, on like Saturday, you'll have a 10 minute meeting with your AA about your classes for the next week. A lot of times this meeting will end up being something like:

"I'm planning to take this class, this class, this class and this class"
"OK. That sounds cool."

However, a lot of times it won't be this easy. If you're having some trouble deciding what classes to take, this is an excellent time to ask for some advise. Your AA will explain a bit more about the classes your trying to choose between. He/she might ask you want sort of things you're interested in, and give some recommendations based on that. Basically their goal is to help you come up with the class schedule that [i]you[/i] will be most happy with. Of course, this 10 minute meeting isn't the only time you get talk to them/get advise. If you got a question some other time, feel free to ask them (or any staff member for that matter, you don't have to consult your AA exclusively. In particular, if you want to know more about a specific class, the best person to ask is, of course, the person teaching that class).

Right, I think this point should be stressed. You have COMPLETE control over what you take, each week -- ANY subset of the classes offered that week, including the empty set or the whole set, is fair game. The advisor is there to give suggestions, and may suggest you take or not take a certain class, but the choice remains yours.

Unfortunately, your ability to take the whole set of classes offered is limited by the difficulty of being in 5 places at once.  :(

The AA meetings, as JAM said, are nothing to worry about.  They are there to help you.  Sometimes they will try to introduce some sanity into your schedule (i.e. if you want to take four 4-star classes, they might encourage you to seek a little more balance (I know this from experience  :) )), but they will NEVER force you to take any particular class.  Indeed, on at least one occasion I told my AA I was going to take one class and then changed my mind over the weekend and went to a different one.

I would add that the AA's are also the (only) ones who read the "log tables" ( :wink: ) you submit at night evaluating the classes you took that day.

Right, when I said "all classes" I meant "all class slots". Though, I suppose you can hang out for 15 minutes of each lecture...

A corollary of this whole AA not forcing you into classes bit is that they do have your best interests in mind. Although if you really want to take a class you are more than welcome to, the suggestions of the AA tend to be good ones. I have had precisely one disagreement with my AA, went to the class he recommended for two days and then switched into another one. It was absolutely fine.

It should also probably be noted that star level (one to four) indicates difficulty of the class, but it is entirely possible for easy classes to be very interesting. I actually made the mistake last year of pretty much only taking four star classes. While the classes were almost all good, in retrospect there were some one and two-star classes that would've been really cool as well that I missed out on, and I wasn't able to keep up with all of the work from all of the classes, so I probably would've understood everything better as well if I had taken some of those classes.

Note: For those not familiar with the system, every class is given a difficulty from * to **** to indicate its expected difficulty. Of course, different staff have different notions of difficulty, so this isn't perfect, but it gives you an idea of how much work you'll have to do if you want to understand the material well.


A coin of radius r is stretched in an x direction by a factor of X.

At what degree from a vector normal to the center of the coin must one look at the coin for it to appear circular?

Assuming $ X \geq 1$...

[hide="Answer"]$ \arccos 1/X$[/hide]
[hide="Solution that is probably five times as long as it needs to be"]Let $ O$ be the center of the coin. Consider the plane formed by the vector normal to the coin and the vector pointing along the coin's longest radius (the direction it was stretched in). In this plane the coin is projected on to a line segment of length $ 2Xr$. Let the endpoints of this segment be $ A$ and $ B$, so $ OA \equal{} OB \equal{} Xr$. Draw a circle in the plane around $ O$ with radius $ Xr$ so $ A$ and $ B$ lie on it.

Let $ C$ be a point on the circle such that $ AC \equal{} r$. Going back into space, to view the coin as having radius $ r$ again, we want to view it from the vector normal to $ AC$. The angle between that vector and the one normal to $ AB$ is $ \angle CAB$. Since $ \triangle ACB$ is right, $ \angle CAB \equal{} \arccos r/Xr \equal{} \boxed{\arccos 1/X}$.[/hide]


Nebraska, the home of the AMC, changed its license plate scheme. Each old license plate consisted of a letter followed by four digits. Each new license plate consists of three letters followed by three digits. By how many times is the number of possible license plates increased?

$ \textbf{(A)}\ \frac{26}{10} \qquad
\textbf{(B)}\ \frac{26^2}{10^2} \qquad
\textbf{(C)}\ \frac{26^2}{10} \qquad
\textbf{(D)}\ \frac{26^3}{10^3} \qquad
\textbf{(E)}\ \frac{26^3}{10^2}$

Nebraska, the home of the AMC, changed its license plate scheme. Each old license plate consisted of a letter followed by four digits. Each new license plate consists of three letters followed by three digits. By how many times is the number of possible license plates increased?

$ \textbf{(A)}\ \frac {26}{10} \qquad \textbf{(B)}\ \frac {26^2}{10^2} \qquad \textbf{(C)}\ \frac {26^2}{10} \qquad \textbf{(D)}\ \frac {26^3}{10^3} \qquad \textbf{(E)}\ \frac {26^3}{10^2}$
[hide]
The number of combinations for the old license plates:

$ 26 \cdot 10 \cdot 10 \cdot 10 \cdot 10$
$ \equal{} 26 \cdot 10^4$

The number of combinations for the new license plates:

$ 26 \cdot 26 \cdot 26 \cdot 10 \cdot 10 \cdot 10$
$ \equal{} 26^3 \cdot 10^3$

To find the number of times the number of possible license plates was increased, divide the number of possible new license plates by the number of possible old license plates.

$ \dfrac{26^3 \cdot 10^3}{26 \cdot 10^4}$
$ \equal{} \boxed{\dfrac{26^2}{10}, (C)}$[/hide]

hi

[hide]C) 26^2/10 because there are 2 more letters in the new license plates, therefore, x26^2 and one less number, therefore /10.[/hide]

[hide=Solution]

The old amount of possible license plates used to be $26\cdot 10^4$.
Now, it's $26^3\cdot10^3$
The total increase would be $\frac{26^3\cdot10^3}{26\cdot 10^4}=\boxed{\frac{26^2}{10}}$


symmetry
geometry
geometric transformation
reflection

[quote="GameBot"]How many lines of symmetry does a square have?[/quote]

Unless I'm missing something, 4.

[quote="alex31415"]Can't a square have infinitely many lines of symmetry?[/quote]

How would you get any other lines of symmetry other than the 4 ernie showed?

[quote="alex31415"]Can't a square have infinitely many lines of symmetry?[/quote]

A line of symmetry must divide the figure into two halves which are [b]reflections of each other[/b].

In general, a polygon with $n$ sides has $n$ lines of symmetry.

[quote="LiBoy"]Good advice: If you have doubts about symmetry for figures, just start drawing the lines, making sure they make perfect reflections.

[quote="AwesomeToad"]

In general, a polygon with $n$ sides has $n$ lines of symmetry.[/quote]

Do you mean [b]regular[/b] polygons? I can draw a triangle with no lines of symmetry.[/quote]

Yes, I meant regular polygons.


How would I make a c or c++ program that prints out all the permutations of 1,2,3,4,5?

There is something interesting on this topic in Paul Yiu's notes "Recreational Mathematics", chapter 38. You can find these notes at http://www.math.fau.edu/yiu/Geometry.html .

  Darij

the easiest way is this:
[code]
#include<algorithm>
#include<iostream>
using namespace std;

int main(){
   int i,arr[5];
   for (i=0;i<5;i++)
      arr[i]=i+1;
   do{
     for (i=0;i<5;i++) cout<<arr[i]<<" ";
     cout<<endl;
   } while(next_permutation(arr,arr+5));
    return 0;
}
[/code]
there are tons of other ways to do this...  but for starters this should be sufficient


geometry
incenter
function
analytic geometry
ratio
geometric transformation
homothety

A triangle $\bigtriangleup ABC$ is given and let $I,$ $N,$ its incenter and Nagel point respectively.

We denote as $A',$ $B',$ $C'$ be, the intersection points of $BC,$ $CA,$ $AB,$ from the lines through $N$ and parallel to the angle bisectors
$AI,$ $BI,$ $CI,$ respectively.

Prove that the segment lines $AA',$ $BB',$ $CC',$ are concurrent at one point, so be it $P.$

[b][size=100]Conjecture:[/size][/b] (By the drawing) Prove that the point $P,$ lies on the segment line $NGe,$ where $Ge$ is the Gergone point of $\bigtriangleup ABC.$

Kostas Vittas.

PS. I don't know, dear all my friends, if they are well known results and if somebody has a reference, please inform me.

For the concurrency:

Let D the intersection of the angle bissector of A and BC, X the intersection of AD and CN and K the intersection of AB and CN.

We have that $CA' = \frac{CN.CD}{XC}$.

By the Menelaus theorem in ABD (with the transversal CXK), we find $CX = \frac{CK . b}{p}$.

So, $CA' = \frac{CN.CD.p}{CK.b}$.

By the Menelaus theorem in ABM (with transversal CNK, where M is the intersection of AN and BC),
we find $\frac{CN}{CK}= \frac{c}{p-c}$.

We know that $CD = \frac{ab}{b+c}$, so, $CA' = \frac{pac}{(p-c)(b+c)}$.

In the same way, we find the values of BA', CB', AB', AC' and BC', and the problem follows easily by Ceva!

_

To solve the collinearity part, let K the intersection of NP and BC.
We can use Menalaus to find $KX_{a}$ and $KA'$ and after we can use Menelaus again to find the razon $\frac{AG}{GD}$ as a function of this values.
(D is the intersection of BC and $AG_{e}$, G is the intersection of AD and NP and $X_{a}$ is the intersection of BC and AN).
So, we need only to prove that the obtained value for AG/GD is the same of $AG_{e}$/$G_{e}D$.

 :)

Now for the collinearity we can use barycentric coordinates, as from Van Aubel we have that $\frac{PA}{PA'}=\frac{C'A}{C'B}+\frac{B'A}{B'C}=\frac{ca(p-b)+ba(p-c)}{bc(p-a)}$.

Thus $\overrightarrow{r_{P}}=\frac{ca(p-b)+ab(p-c)}{\sum{ab(p-c)}}\cdot \overrightarrow{r_{A}}+\frac{bc(p-a)}{\sum{ab(p-c)}}\cdot \overrightarrow{r_{A'}}$ $(1)$.

Now because $\frac{A'B}{A'C}=\frac{b(p-c)}{c(p-b)}$, we have that:

$r_{A'}=\frac{c(p-b)}{b(p-c)+c(p-b)}\cdot \overrightarrow{r_{B}}+\frac{b(p-c)}{b(p-c)+c(p-b)}\cdot \overrightarrow{r_{C}}$.

Thus by replacing it in $(1)$, we obtain the barycentric coordinates of $P$, let them $P(\alpha,\beta,\gamma)$.

Now because $G_{e}(\frac{(p-b)(p-c)}{r(4R+r)}, ...)$, and $N(\frac{p-a}{p}, ...)$,
Thus $P \in G_{e}N \iff \left|\begin{array}{ccc}\frac{p-a}{p}& \frac{p-b}{p}& \frac{p-c}{p}\\\ \frac{(p-b)(p-c)}{r(4R+r)}& \frac{(p-c)(p-a)}{r(4R+r)}& \frac{(p-a)(p-c)}{r(4R+r)}\\\ \alpha& \beta& \gamma\end{array}\right|=0$, which can be proved easily.

Thank you dear friends [b][size=100]e.lopes[/size][/b] and [b][size=100]pohoatza[/size][/b], for your interest and solutions.

Because of I am not good in calculations and barycentrics, I try to solve the problems based on simple ideas, as applications of well known Theorems and Lemmas.

So, the solution I have in mind for the concurrency, is based on a Lemma, helping us to prove that the points $A',$ $A''\equiv BC\cap AI,$ are isotomic conjugates with respect to $BC$ ( symmetric each other with respect to the midpoint of $BC$ ) and similarly for the pairs of points $B',$ $B''\equiv AC\cap BI$ and $C',$ $C''\equiv AB\cap CI.$

Hence, it is easy to show that the segment lines $AA',$ $BB',$ $CC',$ are concurrent at one point, so be it $P,$ as the isotomic conjugate of $I,$ with respect to the triangle $\bigtriangleup ABC.$

About the collinearity of $N,$ $P,$ $Ge,$ I thought that it was possible to me to prove the congruence of the cross ratios $(B,D,A',D') = (A,E,B',E'),$ but I still have not succeed in it.

Probably this is not a good idea.

Best regards, Kostas Vittas.

$ \bullet$ Let $ \triangle A_0B_0C_0$ be the anticomplementary triangle of $ \triangle ABC.$ $ A_0,B_0,C_0$ against $ A,B,C.$ Incircle $ (N')$ of $ \triangle A_0B_0C_0$ is tangent to $ B_0C_0$ at $ X.$ Let $ X'$ be the antipode of $ X$ WRT $ (N')$ and $ Y \equiv A_0X' \cap B_0C_0$ the $ A_0$-Nagel ray of $ \triangle A_0B_0C_0.$ Since $ A$ and $ N'$ are the midpoints of $ XY$ and $ XX',$ then $ AN'$ is the X-midline of $ \triangle XYX',$ hence $ AN' \parallel A_0Y$ $ \Longrightarrow$ $ AN'$ is identical with the homologous of $ A_0Y$ through the homothety centered at the centroid $ G$ of $ \triangle ABC$ with factor $ \minus{} \frac {_1}{^2}$ $ \Longrightarrow$ $ N'$ is the Nagel point of $ \triangle ABC$ $ \Longrightarrow$ $ N \equiv N'.$ Then $ A' \in A_0N.$ Let $ V_a,V_b,V_c$ be the feet of the internal angle bisectors of $ \triangle ABC.$ Since $ A_0V_aAA'$ is a parallelogram, its diagonals $ AA_0$ and $ A'V_a \equiv BC$ bisect each other $ \Longrightarrow$ $ A'$ and $ V_a$ are symmetric about the midpoint of $ BC.$ Similarly, $ B',V_b$ and $ C',V_c$ are symmetric about the midpoints of $ BC,CA$ $ \Longrightarrow$ $ AA',BB',CC'$ concur at the isotomic conjugate of the incenter $ I$ WRT $ \triangle ABC.$

$ \bullet$ Let $ O,G_e$ be circumcenter and Gergonne point of $ \triangle ABC.$ Isogonal conjugation $ \mathcal{I}$ WRT $ \triangle ABC$ maps $ G_e$ and $ N$ into the similitude centers of $ (I) \sim(O)$ and  $ I$ is a double point, thus $ G_e,N,I,A,B,C$ lie on a circumconic $ \mathcal{F},$  the isogonal conjugate of the diacentral line $ IO,$ namely the Feuerbach hyperbola. Isotomic conjugation $ \mathcal{L}$ WRT $ \triangle ABC$ takes $ G_e,N$ into each other and $ I$ into $ P.$  $ \mathcal{L} : \mathcal{F} \mapsto NG_e$ $ \Longrightarrow$ $ P \in NG_e.$


Theoretically, if we beam enough electrons to the moon, we can make the moon accelerating to Earth. The attraction force between the positive charged Earth and the negative charged moon causes the acceleration.

As the moon pulled closer, it will circle the Earth faster in smaller orbit. At some point maybe we can make the moon a geostationary satellite, then we can build a moon bridge to the moon.

Could someone please calculate this idea, appreciate!

The bridge is new.

Can you tell me how long it should be?  :D

Kind close to another result. Does the Earh spin faster at this new orbit? How fast?

"yeah, it's not a bad idea, but remember, the moon has to be slowed down, by reducing the EMF, and it will have to enter the new orbit at the right angle, and for it to be geosychronous, it'll be at about 42,000 km from the center of earth. That's a pretty long bridge, but perhaps with this proposed new nano (something) space elevator it'll happen. "

This was posted at another forum.

Suggest, build electron gun on the moon, use solar energy to shoot electrons to Earth.

When Earth negative charged to a point, it may help the weather or cure cancer.


What is the value of the least base ten number which requires six digits for its binary representation?

It must be at least $ 2^{6\minus{}1}\equal{}2^5\equal{}\boxed{32}$.

Why does it have to be 2 to the 6-1?
Thanks :wacko:


analytic geometry
probability
rotation
symmetry

[quote="gauss1181"]You'd probably get partial credit...so maybe a 2 or 3?[/quote]

I think he would get a 4. It is a trivial thing that can be fixed very very easily.

[quote="BOGTRO"][quote="barkashdanbar"]Even though I used wrong coordinates, my solution was rigorous and perfect....
Do I get any point for that?[/quote]

Uh.... isn't this already on the x-axis?[/quote]

It is, but I also managed to do this at first.  I mean, the particle must leave the x-axis on its first move, so I found the probability that it would hit y=6 before returning to the x-axis.  But then I reread the problem and realized I had found the wrong probability.  But it was pretty quick, because I just used the same method.

I dont get what you are trying to ask here?

You say your solution was perfect?  How, if you started at (0,3.5), your answer would be 1.

Unless you misread it two times and found the chances that it hits y-axis before hitting x=6...then your answer would be correct and you would get 5 (maybe a 4), since you can just rotate it around.

Actually, if the particle is on (0,3.5), it goes down to (-1,3.5) and so on, never hitting a lattice point?

I see even you have conflated the coordinates as well. If you start at (0, 3.5), you go down to (0, 3). Barkashdanbar said he started at (3.5, 0). He most likely just rotated everything by 90 degrees and got the same answer, otherwise, like Brut3Forc3 said, the particle will never hit a lattice point.

^oops i meant if he started at (3.5,0)...yea which would be 1.

The question said moving towards the origin, not moving south...so he probably just rotated everything by 90 degrees...since this question has a lot of symmetry, i doubt they would count off.


This problem is [url=http://web.mit.edu/hmmt/www/datafiles/problems/2006.shtml]HMMT '06, Combinatorics #2[/url].

[hide="Thought Process"]1. If the innermost term is simply 1, what is your summation really doing?
2. See if you can express the rules for the above in some way other than a summation.
3. Find an effective way of counting this[/hide]
[hide="Solution cont from above"]
1. It counts the number of terms you add [i](since the value of each term is 1)[/i]

2. You want the number of non-decreasing sequences of length 61, where each element is from the set {0,1,2} [i](if you don't see this, work from outside to inside, observing how each summation's bounds affects what happens with the summation to its right)[/i]

3. Suppose you have 61 markers. Place two dividers in there so that the objects left of both dividers represent 0, those between the two dividers represent 1, and those right of both represent 2. Since there are 61 markers and 2 dividers, we have a total of 63 objects, of which we must chose 2 to be dividers. The answer is $ \binom{63}{2} \equal{} \boxed{1953}$.

The [url=http://web.mit.edu/hmmt/www/datafiles/solutions/2006/scomb06.pdf]official HMMT solution (PDF)[/url] creates a bijection to a different set of sequences.[/hide]

Thank you for all the help Zuton Force.
That makes perfect sense.


calculus
real analysis
real analysis solved

A fossilized leaf contains 14% of its normal amount of carbon 14.  How old is the fossil (to the nearest year)?  Use 5,600 years as a half-life of carbon 14.

Can you show me how you solve it?

Half-life is a first order reaction, meaning that the rate of decay is proportional to the quantity at that time or in mathematical terms

$\frac{dQ}{dt} = kQ$

Solving this yields

$Q=Q_0e^{kt}$, where $Q_0$ represents the initial quantity.

So for your problem $1/2 = e^{k(5600)}$, solve for $k$, then use it to solve for $t$ in this equation $.14=e^{kt}$.

Hope this helps.

The value of $k$ is negative value, isn't it?

Yes, because the model I gave assumes growth, however the final result is the same.

[quote="probability1.01"]Can also be solved in a non-calculus way...[/quote]

Well ya, if you just memorize the equations, which are derived from calculus.


inequalities
function
inequalities unsolved

OKay.. cause today is my birthday and I promissed myself that at least on my birthday I will stop solving math problems.. I will just ask you if you used the oldie but goldie trick a=x/y, b=z/x and c=y/z because their product will be 1.

hope that this is how is made! :D

(I am not solving anything today :) )

1st of all happy birthday Lagrangia!

sencond of all - this inequality was given at a Romanian Test for 8-th graders (last school class before entering high-school in Romania), and obviously no-one did it. I've tried it then, using all elementary (and nice) ideas I could possible think of. Didn't work. Last year I found out that it was also used in a Bulgarian TST, and I succeded in solving it using the "brute force" (which they actually have in their original solution).

Besides that, like that problem from Iran 1996, it is very unlikey that one can find a nice solution ... :)

For the trick I prefer a=x^2/yz, b=y^2/zx, c=z^2/xy instead since it doesn't cause the loss of symmetry..

Well,we can use a well-known method here.Take $f(a,b,c)=RHS-LHS$.
Assume that $c=max \lbrace{a,b,c}\rbrace$ then we have:
[tex] f(a,b,c)-f( \frac{a+b}{2}, \frac{a+b}{2},c)= [/tex]
[tex] (a-b)^2(\frac{1}{(2+a)(2+b)(4+a+b)}-\frac{1}{(1+a+c)(1+b+c)(2+2c+a+b)}) \geq 0 [/tex]
since  [tex] c \geq 1 [/tex]
Now,prove it for [tex] f(a,a,b) [/tex].I have expanded it and this is really disgusting :( .
However,the last result is [tex] (a-1)^2(2a^4+9a^3+9a^2+4a+3) \geq 0 [/tex]. :)

I just wanted to mention the appearance of the inequality with solution in two sources:

[b]1.[/b] Titu Andreescu, Vasile Cîrtoaje, Gabriel Dospinescu, Mircea Lascu, [i]Old and New Inequalities[/i], Zalau: GIL 2004, problem 99.

[b]2.[/b] American Mathematics Competitions: [i]Mathematical Olympiads 1997-1998: Olympiad Problems from Around the World[/i], Bulgaria 21, p. 23.

Both solutions are almost the same: Brute force. The inequality doesn't seem to have a better proof. The second source can be freely downloaded from

http://www.unl.edu/amc/a-activities/a4-for-students/problemtext/s-index.html

[b]EDIT:[/b] Since this website does not work properly, here are direct links to the files at the site:

http://www.unl.edu/amc/a-activities/a4-for-students/problemtext/
http://www.unl.edu/amc/a-activities/a4-for-students/problemtext/geom-080399.ps
http://www.unl.edu/amc/a-activities/a4-for-students/problemtext/geom-080399.tex
http://www.unl.edu/amc/a-activities/a4-for-students/problemtext/ineqs-080299.ps
http://www.unl.edu/amc/a-activities/a4-for-students/problemtext/ineqs-080299.tex
http://www.unl.edu/amc/a-activities/a4-for-students/problemtext/mc96-97-01feb.dvi
http://www.unl.edu/amc/a-activities/a4-for-students/problemtext/mc96-97-01feb.pdf
http://www.unl.edu/amc/a-activities/a4-for-students/problemtext/mc96-97-01feb.ps
http://www.unl.edu/amc/a-activities/a4-for-students/problemtext/mc97-98-01feb.dvi
http://www.unl.edu/amc/a-activities/a4-for-students/problemtext/mc97-98-01feb.pdf
http://www.unl.edu/amc/a-activities/a4-for-students/problemtext/mc97-98-01feb.ps

  Darij

FOR several months I am trying to prove the [url=http://www.mathlinks.ro/Forum/viewtopic.php?t=25935]general case of this damn inequality[/url], but I did not succeded. Did you?

It seem be post some times in mathlinks.
And when you Multi two hand side,it's very clearly,not difficult.
I had solve it before by this way.

Other way,It can be prove immediate by the generel:

$\sum\frac{1}{n-1+x_i}\le 1$ if $\prod{x_i}=1$

And then:

$\sum\frac{1}{n-1+x_i} \le 1 \le \sum\frac{1}{1+(n-1)x_i}$

[quote="hungkhtn"]It seem be post some times in mathlinks.

$\sum\frac{1}{n-1+x_i}\le 1$ if $\prod{x_i}=1$

And then:

$\sum\frac{1}{n-1+x_i} \le 1 \le \sum\frac{1}{1+(n-1)x_i}$[/quote]
Here is a proof for the first one.
$f(x_{1},x_{2},...,x_{n})=\sum\frac{1}{n-1+x_i}$.As  $\prod{x_i}=1$ we may assume $x_{1}\ge{1}, x_{2}\le{1}$. We shall prove that  $f(x_{1},x_{2},...,x_{n})\le{f(1,x_{1}x_{2},...,x_{n})}$. And this is true because after a little computation we obtain $(1-x_{1})(x_{2}-1)(x_{1}x_{2}+(n-1)^{2})\ge{0}$ which is obviously true.
So we have $f(x_{1},x_{2},...,x_{n})\le{f(1,x_{1}x_{2},...,x_{n})}\le ...\le{f(1,...,1)}=1$.
QED

i have just solved by clearing denominators + muirhead symmetric inequality.
but is there a short solution???????

i'm very sure that mixing variables works, but i'm not sure if it's short.

Here is solution :asumme [tex]a\geq b\geq c[/tex]
[tex]R.H.S-L.H.S=\sum \frac{b-1}{(2+a)(1+a+b)}\geq \frac{a+b+c-3}{(2+a)(1+a+b)}\geq 0[/tex]

this is from bulgaria ninety something i guess the best solution is bruteforce put :

x=a+b+c and y=ab+ac+bc and express each side in terms of these variables and then pure calculation till you get this very ugly inequality wich is true using AM-GM

[quote="Bojan Basic"][quote="keira_khtn"]$\sum \frac{b-1}{(2+a)(1+a+b)}\geq \frac{a+b+c-3}{(2+a)(1+a+b)}$[/quote]
This inequality is not trivial at all, because $\frac{c-1}{(2+b)(1+b+c)}$ is negative and can't simply change the denominator.[/quote]
Oh,i'm false..

i m not sure whether i m right, i found a very easy solution though :?
look, use the classical technic , put $ a = \frac xy , b = \frac yz , c = \frac zx $ and assume that $ x \geq y \geq z $
then we have
$ \sum \frac {1}{1+a+b} = \sum \frac {yz}{yz + zx + y^2} \leq \sum \frac {yz}{2yz + zx} $ since $ y^2 \geq yz $
then
$ \sum \frac {yz}{2yz + zx} =  \sum \frac {1}{2 + \frac xy} = \sum \frac {1}{2 + a} $
hence
$ \sum \frac {1}{1+a+b} \leq \sum \frac {1}{2 + a} $

any mistake??

[quote="Al-Khawarizmi"]$\sum \frac {yz}{yz + zx + y^2} \leq \sum \frac {yz}{2yz + zx} $ since $ y^2 \geq yz $

any mistake??[/quote]

Yes. We do have $y^2 \geq yz$ but the other, similar inequalities do not necessarily hold.

:oops_sign:  my solution is wrong, sorry!

thank you Arne, i did not see your post,  :blush:  you are so fast here people :lol:

[hide="Here is a more direct form. Hope it is correct"]

We have to prove that the inequality $\frac{1}{1+a+b}+\frac{1}{1+b+c}+\frac{1}{1+c+a}\le\frac{1}{2+a}+\frac{1}{2+b}+\frac{1}{2+c}$ holds for all $a,b,c>0$ that satisfy $abc=1$.

Lets consider the function $f(x)=\frac1{x}$, which is convex. So, Jensen's inequality yields:

$\frac{1}{1+a+b}+\frac{1}{1+b+c}+\frac{1}{1+c+a}\leq \frac{3}{\frac1{3}(1+a+b+1+b+c+1+c+a)}$$=\frac9{3+2a+2b+2c}\leq 1$

The last one is true since $\frac{3+2a+2b+2c}{9}\geq \sqrt[9]{a^2b^2c^2}=1$.

So, we need to prove that $1\leq \frac{1}{2+a}+\frac{1}{2+b}+\frac{1}{2+c}$. After clearing denominators and simplifying it a bit, it remains $ab+bc+ca\geq 3$, which is true by AM-GM.[/hide]

[b][EDIT][/b] I applyed Jensen's incorrectly. It is funny how you get the desired result with the stupid sign mistake :(

Guys, you know this is a hard problem and the only known solution is brute-force, so if you think you have a short simple proof, why can't you bother check it twice before posting it?

For instance, Silouan's inequality

[quote="silouan"]$\frac{1}{2a+1}+\frac{1}{2b+1}\geq\frac{2}{a+b+1}$[/quote]

is true, but doesn't help solving the initial problem. Yassin's

[quote="Ezbakhe Yassin"]$1\leq \frac{1}{2+a}+\frac{1}{2+b}+\frac{1}{2+c}$[/quote]

is true, but with $\geq$ instead of $\leq$, so it isn't too useful either. Sorry, but I just wanted to tell the truth... :(

  Darij

[quote="Ezbakhe Yassin"]Here is a more direct form. Hope it is correct.

$\frac{1}{1+a+b}+\frac{1}{1+b+c}+\frac{1}{1+c+a}\leq \frac{3}{\frac1{3}(1+a+b+1+b+c+1+c+a)}$$=\frac9{3+2a+2b+2c}\leq 1$

[/quote]

$\frac{1}{1+a+b}+\frac{1}{1+b+c}+\frac{1}{1+c+a}\geq \frac9{3+2a+2b+2c}$ by Cauchy

[quote="silouan"]I think the best solution is to use the fact that
$\frac{1}{2a+1}+\frac{1}{2b+1}\geq\frac{2}{a+b+1}$[/quote]

Then We must prove

$\frac{1}{a+2}+\frac{1}{b+2}+\frac{1}{c+2}\geq \frac{1}{2a+1}+\frac{1}{2b+1}+\frac{1}{2c+1}$

But it was wrong  :o

[quote="alekk"]Let $ a$, $ b$, $ c$ be positive real numbers such that $ abc \equal{} 1$.
Prove that
$ \frac {1}{1 \plus{} b \plus{} c} \plus{} \frac {1}{1 \plus{} c \plus{} a} \plus{} \frac {1}{1 \plus{} a \plus{} b}\leq\frac {1}{2 \plus{} a} \plus{} \frac {1}{2 \plus{} b} \plus{} \frac {1}{2 \plus{} c}$.[/quote]
Let $ x \equal{} a \plus{} b \plus{} c;y \equal{} ab \plus{} bc \plus{} ca$
Inequality becomes:
$ \frac {x^2 \plus{} 4x \plus{} y \plus{} 3}{x^2 \plus{} 2x \plus{} y \plus{} xy} \leq \frac {12 \plus{} 4x \plus{} y}{9 \plus{} 4x \plus{} 2y}$
<=>$ \frac {2x \plus{} 3 \minus{} xy}{x^2 \plus{} 2x \plus{} y \plus{} xy} \leq \frac {3 \minus{} y}{9 \plus{} 4x \plus{} 2y}$
Note that $ x\geq3;y \geq 3;x^2 \geq 3y$
so we have $ \frac {5}{3}x^2y\geq 5x^2;\frac {x^2y}{3}\geq y^2;xy^2 \geq 9x;5xy \geq 15x;xy \geq 3y;x^2y \geq 27$
Done!

[quote=alekk]Let $ a$, $ b$, $ c$ be positive real numbers such that $ abc=1$.
Prove that
$ \frac{1}{1+b+c}+\frac{1}{1+c+a}+\frac{1}{1+a+b}\leq\frac{1}{2+a}+\frac{1}{2+b}+\frac{1}{2+c}$.[/quote]
Let $ a$, $ b$, $ c$, $ d$ be positive real numbers such that $ abcd=1$.Prove that
$$  \frac{1}{1+b+c+d}+\frac{1}{1+c+d+a}+\frac{1}{1+d+a+b}+\frac{1}{1+a+b+c} \le\frac{1}{3+a+} +\frac{1}{3+b} +  \frac{1}{3+c} +  \frac{1}{3+d} $$

[quote=alekk]Let $ a$, $ b$, $ c$ be positive real numbers such that $ abc=1$. Prove that
$$ \frac{1}{1+b+c}+\frac{1}{1+c+a}+\frac{1}{1+a+b}\leq\frac{1}{2+a}+\frac{1}{2+b}+\frac{1}{2+c}$$[/quote]
Let $ a$, $ b$, $ c$ be positive real numbers such that $ abc=1$.[url=https://artofproblemsolving.com/community/c6h2098041p15184819]Prove that[/url]
$$\frac{1}{3}\left(\frac{1}{\sqrt{2+a}} +\frac{1}{\sqrt{2+b}}+\frac{1}{\sqrt{2+c}}\right)^2\leq \frac{1}{2+a}+\frac{1}{2+b}+\frac{1}{2+c} \leq 1$$
$$\frac{1}{3+a}+\frac{1}{3+b}+\frac{1}{3+c}\le \frac{3}{4}$$
[url=https://artofproblemsolving.com/community/c6h1315544p7064916]here[/url] [url=https://artofproblemsolving.com/community/c6h1235118p6815861]here
[/url]

[url]https://artofproblemsolving.com/community/c6h1655438p10498091:[/url]
Let $a,b,c>0$ and $abc=1$. Show that
$$\frac{a}{b+c+1}+\frac{b}{c+a+1}+\frac{c}{a+b+1} \geqslant \frac{a}{a+2}+\frac{b}{b+2}+\frac{c}{c+2}\ge 1$$Let $a,b,c>0$ and $abc=1$. Show that
$$\frac{1}{b+c+1}+ \frac{1}{c+a+1}+ \frac{1}{a+b+1}\leq\frac{1}{a+2}+\frac{1}{b+2}+\frac{1}{c+2}\leq 1.$$

[quote=sqing]Let $ a$, $ b$, $ c$, $ d$ be positive real numbers such that $ abcd=1$.Prove that
$$  \frac{1}{1+b+c+d}+\frac{1}{1+c+d+a}+\frac{1}{1+d+a+b}+\frac{1}{1+a+b+c} \le\frac{1}{3+a} +\frac{1}{3+b} +  \frac{1}{3+c} +  \frac{1}{3+d} $$[/quote]
[url=https://math.stackexchange.com/questions/907336/how-prove-this-inequality-sum-limits-cyc-frac1a3-sum-limits-cyc-frac?rq=1]here[/url]


linear algebra
matrix
complex numbers
superior algebra
superior algebra solved

We have a 2nd ordered matrix X with real inputs and we also know that Tr(X)=1. Now prove that
a) det(X^2+I2)>=0;
b) det(X^2+X+I2)>=3;
c) is 3 the best constant for b); ?

I2=
1 0
0 1

a) det(X^2 + I) = |det(X+iI)|^2  \geq 0
b)
First case if X is diagonalisable in M_2(R)

X is similar to
(a 0)
(0 b) with b=1-a

X^2+X+I is similar to

((a^2+a+1) 0)
(0 (b^2+b+1))

det(X^2+X+I)
= (a^2+a+1)(b^2+b+1)
= ( (a+1/2)^2 + 3/4)((b+1/2)^2 + 3/4)
= ( (a+1/2)^2+ 3/4)( (a-3/2)^2 + 3/4)
= a^4-2a^3+a^2+3

Second case if X is trigonalisable in M_2(R) then X is similar to
(a c)
(0 b)

det(X^2+X+I)
=((a+1/2)^2+ 3/4)( (a-3/2)^2 + 3/4)
=a^4 -2a^3 + a^2 +3

f(a) = a^4 -2a^3 + a^2 +3 study the variations of  f we have

min f = 3 = f(0) = f(1)

For
X=
(0 0)
(0 1)
det(X^2+X+I_2) = 3

Third case X is not diagonalisable and not trigonalisable in M_2(R) but
trigonalisable in M_2(C)

X is similar to
(a  c)
(0 a*) where a \in C, a* is the complex conjugate of a

det(X^2 + X+I_2)
= |a^2 + a +1|^2  square of the modulus a = x+iy, trX =a+a* = 1
= |a|^4 + |a|^2 + a^2 + (a*)^(-2) +a+a* +1
=(x^2+y^2)^2 + x^2 + y^2 + 1-2(x^2+y^2) + 2x+1
=(x^2 + y^2)(x^2+y^2-1) + 2(x+1)   but x=1/2trX=1/2

det(X^2 + X+I_2)
= g(y) = (1/4 + y^2)^2.(y^2-3/4) + 3

det(X^2 + X+ I)  \geq min_(y\in R) g(y)

If I am not mistaken min_(y\in R) g(y) = g( \sqrt (5/12) ) ~ 2,8519

It seems 3 is not the best contant for b
maybe g(  \sqrt  (5/12) ) = 77/27 ~ 2,8519


Solve for all x: (x - 3)^3 + (x - 7)^3 = (2x - 10)^3

Damn!
Emoticoncs don't carry over in copy-paste

Sorry guys


MATHCOUNTS
AMC
USA(J)MO
USAMO
AIME
AMC 10
AIME I

[quote="kstan013"]126 AMC 10B
60 AIME

so 186 index, and I barely qualified![/quote]

oh man that's pretty weak
wait so [i]that[/i] index is for amc 10? so what 150+10(7)=220? that's still somewhat weak, eh, should be a 9 or so in there, whatever at least I made it

but my [b]real[/b] index is 120+70=190, I miss it there.

[quote="Aryth"]144 10A (oops) + 7(0) AIME.
Yay![/quote]

144 10A+7(10)AIME, is what you mean, just telling you.

139.5 + 70 = 209.5

I wasn't too happy with that, but I made it.  :)

[quote]
144 10A+7(10)AIME, is what you mean, just telling you.[/quote]

Right. Or else that would be bad.

W00t, I'm one of 5 7th graders in the nation?!?
Anyways...
136.5 AMC10
7 AIME (stupid careless errors =D)

I really don't think that it matters who made USAMO in 8th grade and under this year, because MathCounters had a huge advantage on this AIME compared to normal years. I'm definitely not proud of my score considering that I've had tons of MathCounts + AIME practice.

Yes, I didn't like the fact that there were 39 middle schoolers, I'm sure more than half are in the USAMO because they were taught to brute force problems by looking for patterns blindly and not really understand the concepts, that happened to be very doable on the AIME I this year. Unfortuately, the only problems I got right were either the easy ones, or the ones in which that was doable, 1-6 and 11. I answered 4 others (7,8,10,12), but made silly mistakes on all (some were more subtle than others).

[quote="erdogankerem123"]magixter made USAMO with 7...I only got a 5, im so mad...[/quote]

Thanks for posting up my score, erdogankerem123.

144(1) + 7(10) = 214... Not that good.

Oh, erdogankerem123, I didn't know you took the AIME officially.

I didn't... I took it for practice at the same time as you, remember?? I completely bombed the AMCs.
What do you mean your index isn't that good?! I think you were lucky

What do you do to the AMC 12?
108 12A
111 12B

3 AIME

[quote="daine"]136.5 (12b) + 10(10)(AIME II) = 236.5
I actually qualified by index![/quote]

Well you can't qualify through the floor if you don't take the AMC 10, and they included the "AMC 10 index" in the AMC 10 qualification thing. I think that was a terrible idea, because now 121.5/6 will get you in, and any MathCounts pattern guesser/brute forcer can get that pretty easily. The sucky thing is that I made stupid mistakes on the problems that actually required thinking (7, 8, 10, [12]) and now I look like one of those people.

ok. Here's my score:

141. :(

I took both AMC 12's, and my highest score was 111.

[quote="jhredsox"]just wondering...but has a person not in high school ever made mop?

jorian[/quote]
Yes, several times.
The most recent is Eric Larson.
Before him Reid Barton made it in 7th and 8th grade.

MOP is for the people who do well on the USAMO.

I think around 28? Maybe less, not exactly sure.

I looked it up, and I think it was 30... But I may have misunderstood it.

I think its for USAMO/IMO.
But anyways, diophantient, you can hardly call two people "several times"...

There have probably been a lot more people, but those were the only ones he remembered at the time.

Really? Do you know what they got on the USAMO?

[quote="ra5249"]no those are the only ones i checked the website[/quote]
Those were the only ones I knew of, I assumed there would be more.

Could you please ask more about MOP and qualifiers and stuff on the AMC forum?  Thanks, I don't want stuff to get too off topic.

k, to finish...did neal come close last year (he had a 14)?

jorian

on usamo? wow thats probably like really really close
like maybe 2-3 or at most 5-6 points

ok, cause i remember *a lot* of kids said he powned!

jorian


I don't know whether this has been done already.......

What are your favourite rock bands?And don't forget guitarists ;)

This topic belongs in Fun and Games, by announced policy of the AoPS administrators.

Pink Floyd is by far my favorite
Next up is Yes
and then the Beatles

Muse, U2, Kansas, Switchfoot, Death Cab . . . the list goes on!

Beatles, Nirvana, CCR: I'm not that into rock.

I love therasmus,though they aren't too famous...please listen to in the shadows by the rasmus,it's an wesome song...
Linkinpark is cool...evanescence,[b]CREED[/b],alterbridge,bonjovi are other favorites

I heard this band called "Flipsyde",anyone heard about them??

[/quote]

all the stuff said above is awesome, yeah.. and i realy couldnt list ALL my favorite bands, but here are some of mine that are consistently really good:

3 Doors Down, Coldplay, Creed, Dave Matthews, Green Day, Guns N Roses, Linkin Park, Nickelback... and... the red hot chilli peppers

this topic is really old and worn out :rotfl:  , but i 'll reply again

yo ,tobe finally a fellow GNR lover!

my favourite rock bands...

1. GUNS N ROSES ( the greatest !, sweet child o mine, paradise city, november rain)
2. Deep Purple ( smoke on the water)
3. Eagles( hotel california - an epic)
4. metallica( enter sandman, coolest riff)
5. Nirvana( man who sold the world)

The Mars Volta, Avenged Sevenfold, Franz Ferdinand, Finch, At The Drive-In, Coldplay, The Cure, System Of A Down...

[quote="shadysaysurspammed"]Oh sorry...i forgot to mention that  Mr.bring a.k.a emonem ain't allowed to bore us with his *yawn* bo :| ring choices[/quote]

dont use aint , or the big,bad iversonfan2005 ( just kidding) will pounce on you and u will meet the same fate as me in the india forum!( these guys roasted my pants off not to use slang)

Aren't we in the fun forum,so aren't we a bit free here with our language?Anyeay eminem,you post like you live on the streets or something,i'm not kidding.Good you learnt a lesson :rotfl:  :rotfl:  :rotfl:  :rotfl:

I heard simpleplan's "untitles",though the tune was nice but if they extended the solo,the song would have been a big hit...does anyone know where to download "CoheedandCambria" mp3s from...i want hteir song "Welcome home",hell of a song!!!

Does this topic only pertain to Rock, or can Rap "bands" also be included?

[quote="eminem"][quote="shadysaysurspammed"]Aren't we in the fun forum,so aren't we a bit free here with our language?Anyeay eminem,you post like you live on the streets or something,i'm not kidding.Good you learnt a lesson :rotfl:  :rotfl:  :rotfl:  :rotfl:

I heard simpleplan's "untitles",though the tune was nice but if they extended the solo,the song would have been a big hit...does anyone know where to download "CoheedandCambria" mp3s from...i want hteir song "Welcome home",- of a song!!![/quote]

well seriously you better watch what you say or i'll really show you what street talk is , dont act like my brains are half mashed ok, i 'm getin tired of your put-downs. Remember the signature episode in the physics forum ,hehehhehehe, you didnt like that " humilation" did ya?well you'll recieve something like that again , so watch it[/quote]

Okay i do not want to fight....i'll punch you at school....but really ur posts were having quite some slangs and stuff and my mistake i said you live on the streets.....but it wasn't too harsh that it made you cry,was it?And please show me ur street talk (P.M)

Anyway people...i'll delete this stuff soon,so sorry mods,i'll delete it as sure sas eminem reads it,he just wants to humiliate me infront of everyone,isn't very satble u know :roll:

I find it amusing that a ton of nerdy math types have large black rappers as their usernames/avatars. Also, to add to my list, Green Day and Leningrad.

Eminem isn't black.  :) ... or for that matter, large either.

Updated list of my favorites:

Pink Floyd, Led Zeppelin, Yes, Genesis, Beatles, King Crimson, Electric Light Orchestra, Eagles

Pearl Jam
Soundgarden
Temple of the Dog
Pixies
Breeders
Neil Young
The Who
Pink FLoyd
Syd Barrett
Beatles
Black Sabbath
Vaselines
REM
Nirvana
Alice in Chains
Smashing Pumpkins
White Stripes
Queens of the Stone Age
Kelley Deal 6000
Led Zeppelin
King Crimson
Ministry
NIN
Tool
Zombies
Green Day
early Chili Peppers
The Clash
Smashmouth
Mahavishnu ORchestra
Jane's Addiction
Hendrix
SOnic Youth
Radiohead
Misfits
Moby
Screaming Trees
Velvet Underground
Blind Melon
Stone Temple Pilots
Avalanches
Uncle Tupelo
BOB Dylan
The Distillers
The Doors

Beatles, Stones, Billy Joel, Green Day, Don McClain (sp?)...


geometry
geometric transformation
reflection
analytic geometry
geometry proposed

There is a straightforward proof by contradiction. But I suppose the easiest
(and in my view most elegant) proof is the following:

By successively reflecting the billiard by one of its sides we get a tilling of the plain. Identifying all the arising
triangles in the right manner the laws of reflection imply that the path of the ball in the billiard can be
represented as a straight line in the plain. Now we introduce coordinates in the plain such that the
original billiard has vertices $(0,0)$, $(0,1)$ and $(1,1)$, where the initial position of the ball is the origin.
By the above described identifaction this vertex corresponds to every point in the plain with even coordinates.
Now the clain follows from the simple fact that any straight path from the origin to a point with
even coordinates passes through another lattice point first.

(I hope the idea is not obstructed by my inability to properly express myself in English... )

We can prove easily that the line formed by the $4k+i$ move is parallel to the line formed by the $4(k-1)+i$ move, and that is impossible go to our initial vertice with lines parallels to our first 4 lines!

[quote="solyaris"][quote="e.lopes"] ... the line formed by the $4k+i$ move is parallel to the line formed by the $4(k-1)+i$ move ...[/quote]

Are you sure about that?  Denote the sides of the billiard by $s_{1}$, $s_{2}$ and $l$, where $l$ is the longest
side and the ball starts at the vertex between $l$ and $s_{1}$. Then your statement is true,
if the ball first hits $s_{2}, l, s_{1}, l$ in that particular order. However we also could have $s_{2},l,s_{1},s_{2}$
or $s_{2},l,s_{2}, s_{1}$, and after that we don't seem to get something parallel.  :|[/quote]

You're right, my friend! But in this case, we will have basically the same thing. After many 'moves', our lines will be parallel!  :)

thank you for this beautiful solution, solyaris.

[hide=Solution][quote=Iran TST 2007 Day 2 P1]In an isosceles right-angled triangle shaped billiards table , a ball starts moving from one of the vertices adjacent to hypotenuse. When it reaches to one side then it will reflect its path. Prove that if we reach to a vertex then it is not the vertex at initial position
[i]By Sam Nariman[/i][/quote]
[b][color=#000]Solution:[/color][/b] Let the given triangle be $\Delta ABC$ right angled at $B$ & Assume, the ball starts from $A$, then it must hit & bounce off $\overline{BC}$ in the end, in order to reach $A$ back. Say, the ball hits the sidelines of $\Delta ABC$, $n$ times, at $X_1,X_2, \ldots , X_n$. We know, $X_1,X_n $ $\in$ $\overline{BC}$. Also, $AX_1,AX_n > AB$.
-----------------------------------------------
When, $BX_1 < CX_1$, keep reflecting $\Delta ABC$ over $\overline{BC}$ such that the ball traces a straight line. Build a coordinate system with $AB$ as $X-$axis & $BC$ as $Y-$axis. Let, $A$ $\equiv (-1,0)$, $B$ $\equiv$ $(0,0)$ & $C$ $\equiv$ $(0,1)$. Suppose, the straightline- trace endpoint is $T$, Then, due to $BX_1 < CX_1$ $\implies$ $T \equiv (t,1)$. Join the line segments, $(2x-1,0),(2x-2,1)$ and $(2x-1,0),(2x,1)$ and $(x-1,0),(x-1,1)$ for $x$ $\in$ $\mathbb{N}$. Then, due to reflection argument, the number of times $AT$ hits these line segments is $n$. Let these new hit-points be $Y_1,$ $Y_2,$ $\ldots$ $,Y_n$. Also, it's clear, $X_1 \equiv Y_1$. Now Due to Congruence,
$$Y_iY_{i+1}=X_iX_{i+1} \implies X_nA=Y_nT > AB$$
Clearly, when $t$ is odd $\implies$ $Y_nT$ $<$ $AB$, because $Y_n$ $\in$ $\overline{(2x-1,0),(2x-2,1)}$. When $t$ is even $\implies$ $Y_nT$ $=$ $X_nA$ $=$ $AX_1$ $>$ $AB$ $\implies$ $X_1$ $\equiv$ $X_n$. This fact implies, $X_2 \equiv X_{n-1}$ $\implies$ $\ldots$ $\implies$ $X_k $ $\equiv$ $ X_{(n+1)-k}$. Regardless of parity of $n$, this coincidence of point eventually lead that there exists a perpendicular bounce $\implies$ Contradiction!
----------------------------------------------------------
When, $BX_1 > CX_1$, the only difference is that $T \equiv (t,u)$ for $t > u > 1$. Now, Erase all the existing lines, Draw the lines only in the form of $x=k$, $y=\ell$, $x+y=2r+1$ and $x-y$ $=$ $2r+1$ for $k, \ell ,r \in \mathbb{Z}$ and only focus in the region enclosed by $y-x=1$ and $y=0$. When, $u$ is odd, the method demonstrated in $(t,1)$ works similarly too. When $u$ is even & $t$ is odd, we can use the method demonstrated in "$(t,1)$ when $t$ is even" and When $u,t$ both are even, we can use the method in "$(t,1)$ when $t$ is odd". Summing up, all cases lead contradiction & hence, the ball cannot bounce back in point $A$ $\qquad \blacksquare$
[/hide]

[asy]  /* Geogebra to Asymptote conversion, documentation at artofproblemsolving.com/Wiki go to User:Azjps/geogebra */
import graph; size(8.963615163720652cm);
real labelscalefactor = 0.5; /* changes label-to-point distance */
pen dps = linewidth(0.7) + fontsize(12); defaultpen(dps); /* default pen style */ real xmin = -2.7955935436028976, xmax = 9.155893341357972, ymin = -4.706972242258498, ymax = 3.485349585955354;  /* image dimensions */
pen ffxfqq = rgb(1.,0.4980392156862745,0.); pen ttffcc = rgb(0.2,1.,0.8); pen ffzzff = rgb(1.,0.6,1.); pen ffqqff = rgb(1.,0.,1.);
pair A = (0.,0.), B = (1.,1.), C = (1.,0.), D = (2.,0.), F = (3.,1.), G = (3.,0.), H = (4.,0.), I = (4.,1.), J = (5.,1.), K = (5.,0.), L = (6.,0.), M = (6.,1.);
 /* draw figures */
draw(B--A, linewidth(0.8));
draw(A--L, linewidth(0.8));
draw(L--M, linewidth(0.8));
draw(M--B, linewidth(0.8));
draw(B--C, linewidth(0.8));
draw(D--B, linewidth(0.8));
draw(D--(2.,1.), linewidth(0.8));
draw(D--F, linewidth(0.8));
draw(F--G, linewidth(0.8));
draw(F--H, linewidth(0.8));
draw(H--J, linewidth(0.8));
draw(I--H, linewidth(0.8));
draw(J--K, linewidth(0.8));
draw(L--J, linewidth(0.8));
draw(M--A, linewidth(0.8) + linetype("4 4") + ffzzff);
draw(J--A, linewidth(0.8) + linetype("4 4") + ffqqff);
 /* dots and labels */
dot(A,linewidth(7.pt) + ffxfqq);
dot(B,linewidth(7.pt) + ttffcc);
dot(C,linewidth(7.pt) + blue);
dot(D,linewidth(7.pt) + ffxfqq);
dot((2.,1.),linewidth(7.pt) + blue);
dot(F,linewidth(7.pt) + ttffcc);
dot(G,linewidth(7.pt) + blue);
dot(H,linewidth(7.pt) + ffxfqq);
dot(I,linewidth(7.pt) + blue);
dot(J,linewidth(7.pt) + ttffcc);
dot(K,linewidth(7.pt) + blue);
dot(L,linewidth(7.pt) + ffxfqq);
dot(M,linewidth(7.pt) + blue);
clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle);
 /* end of picture */[/asy]
 [hide=solution]
consider this shape we reflect the shape every time by the hypotenuse or the perpendicular side to get this shape.
it is obvious that if the ball starts its journey through any point from the initial orange point on the lower left. it doesn't reach any orange point. note that the orange points are the initial vertex where the ball started its movements.
[/hide]


Hello !

I wonder if anybody could help me with this problem.
I just cannot solve it and I hope you can give me a hand with it.

f : R to R

F(x)^2 + f(x)f(y) = x^2 + xy

A sort of obvious solution if you think about is

[hide]$\ f(x)=x$[/hide]

But you'd have to prove or disprove the fact that this is the only solution.

[quote="nickolas"]I  think  f  must  be continous  on R.[/quote]
Why?
Can you please explain the usages of continuity in functional equations?

The  function  $f(x)=\begin{cases}x, x \ rational \\-x, x\ not\ rational \end{cases}$  is also  satisfing the condition $(f(x))^2=x^2$  but  is continous only  for $x=0$.

Also note that : $(f(x))^2=x^2$  for  every  $x\in R$  gives $(f(x)=x$  or $f(x)=-x)$ for every  $x\in R$,  but not  $f(x)=x$  for  every  $x\in R$  or  $f(x)=-x$  for every $x\in R$.


A circle has diameter $AB$ and $X$ is a fixed point of $AB$ lying in between $A$ and $B$. A point $P$, distinct from $A$ and $B$, lies on the circumference of the circle. Prove that, for all possible positions of $P$,

$\frac{\tan \angle APX}{\tan \angle PAX}$ remains constant.

[hide="Hints"]$\angle APB$ is right; Sine Law; given ratio is equal to ${AX\over BX}=\text{const.}$[/hide]


A freight train one mile long is travelling at a steady speed of 2o miles per hour. It enters a tunnel one mile long at 1 p.m. At what time does the rear of the train emerge from the tunnel?

[hide]
Total distance the front of the train travels $ 2 \ \text{miles}$.
Then $ \frac{2}{20}= 0.1 \equiv 6 \ \text{minutes}\Rightarrow 1: 06 \ \text{pm}$.
[/hide]


The standard deviation of a set of numbers is the average distance the set of numbers is from mean of the set of numbers. If the standard deviation of a set of numbers is 6, which of the following would be the standard deviation of the new set of numbers formed by adding three to each number in the original set of numbers?

A. 3
B. 6
C. 9
D. 12
E. 15

[hide="hope its right"]lets say the set is $S_{1}=[a,b,c,d]$  let $m=\frac{a+b+c+d}{4}$

from the information given the standard whatever is the average of the diferences between the elements of the set and $m$

something like $D_{1}=\frac{1}{4}(|a-m|+|b-m|+|c-m|+|d-m|)$ and that equals 6.

Now, the new set should be $S_{2}=[a+3,b+3,c+3,d+3]$ let $n=\frac{a+3+b+3+c+3+d+3}{4}=m+3$

$D_{2}=\frac{1}{4}(|a+3-(m+3)|+|b+3-(m-3)|+|c+3-(m+3)|+|d+3-(m+3)|)=$

This should hold no matter how many elements the sets contain.

We can see that the see the $3s$ cancel out so that $D_{1}=D_{2}=6$ so the correct answer should be $\boxed B$[/hide]

Picture this relation as a line. All points on this line pushed the exact same distance, in the exact same direction.
Does the line's length change?[hide="ANSWER"][b][i][u]NO[/u][/i][/b]
$ B$; $ 6$[/hide]
 :spam: