analytic geometry
geometry
geometric transformation
rotation

[quote="aidan"]Given 2 sets of 3 coordinates $ (x,y,z)$ where each set represents a triangle in three-dimensional space, how do you determine whether the two triangles intersect (without using graphing methods)?[/quote]
You can find the equations of the planes and find the line of intersection of this plane. And then see if this line is included within the triangle regions. Which is computational awuful.

No, it's worse than that: after we've found the line of intersections of the planes, we first have to decide if it crosses each of the triangles seperately, then (if it does) decide if the places where it crosses the two triangles overlap.  Note that this fails if the two triangles lie on the same plane, in which case you need an entirely new approach (and there are several cases to tackle).

The computations are, indeed, unpleasant.  One way to improve them modestly is to apply some transformations (rotation, translation, skewing) to map one of the triangles to have nice vertices, e.g. $ \{(0, 0, 0), (1, 0, 0), (0, 1, 0)\}$ or $ \{(1, 0, 0), (0, 1, 0), (0, 0, 1)\}$.

You can create a mathematical program that does this. But the program code will be very long.

Have you ever tried to determine if two line segments intersect on a plane? This turns out to be not trivial as well, and doing this should give you an idea of how hard the problem you're asking will be.

Actually, there's a nice criteria for checking whether two segments intersect on a plane. It's useful in computational geometry. It is:

Two lines, $ \overline{AB},\overline{CD}$ intersect if and only if $ A, B$ are on opposite sides of $ \overleftrightarrow{CD}$ and $ C, D$ are on opposite sides of $ \overleftrightarrow{AB}$.

Perhaps a generalization to three dimensions would be:

Triangles $ ABC$, $ DEF$ intersect iff there is at least one of $ A, B,$ and $ C$ on each side of the plane containing $ DEF$ and there is at least one of $ D, E,$ and $ F$ on each side of the plane containing $ ABC$.

I'll leave it up to someone else to prove/disprove it.

Indeed i think this would not be an easy problem, my friend was asking because he wanted to write a computer program to solve this problem.

Hamster1800, wouldn't it be possible for at least one of $ A$, $ B$ and $ C$ to be on each side of the plane containing $ DEF$ and for at least one $ D$, $ E$ and $ F$ on each side of the plane containing $ ABC$ and yet the triangles do not intersect?

Find the intersection of the planes determined by the two triples of points. Then parameterize that line and look at the barycentric coordinates of all the points. If they are all positive, then the point is inside the triangle.

Also I was thinking to project the triangle onto a plane and look at it that way.

Altheman, your initial solution is incomplete: as I noted in my original post on this thread, the line of intersection can pass through both triangles without the triangles intersecting.

Projection will give lots of false positives.  As I noted, though, even the case where the two triangles lie in the same plane is nontrivial.  (One needs to examine both whether any pair of edges of the triangles intersect and whether any vertex of either triangle lies inside the other triangle.)

And, as aidan correctly noted, Hamster1800's simple condition is necessary but not sufficient: for example, the triangles with vertices $ \{(0,0,0), (1,0,0),(0,0,2)\}$ and $ \{(4,0,1),(3,1,1),(3,\minus{}1,1)\}$ satisfy that condition but do not intersect.  In fact, pairs of triangles like this correspond exactly to the case I've pointed out twice now, where both triangles intersect the line of intersection of their containing planes but the two triangles do not intersect each other.


I'm assuming that by the Euler points $ E_{a}$ ,$ E_{b}$, $ E_{c}$ you mean the midpoints of the segments $ HA$, $ HB$, $ HC$, where $ H$ is the orthocenter of $ ABC$. In this case, the construction goes as follows: Note that the parallel $ \gamma$ through $ H_{c}$ to $ E_{a}E_{b}$ coincides with the line $ AB$ (since in triangle $ HAB$, the segment $ E_{a}E_{b}$ is the interception of the $ H$-midline), thus by intersecting the nine-point circle $ H_{c}E_{b}E_{a}$ with $ \gamma$ we obtain the midpoint $ M_{c}$ of $ AB$. Now, consider the parallel $ \beta$ through $ E_{b}$ to the line $ E_{a}M_{c}$ and notice that $ \beta$ coincides with the line $ HB$ (since in triangle $ AHB$, the segment $ E_{a}M_{c}$ is the interception of the $ A$-midline). Thus, we obtain the vertex $ B$ by intersecting $ \gamma$ and $ \beta$. Now, let the parallel through $ M_{c}$ to $ \beta$ meet the nine-point circle at $ M_{b}$. This is the midpoint of $ AC$. So, we can now draw the line $ AC$ by taking the perpendicular from $ M_{b}$ to the line $ BE_{b}$. This line (obviously) intersects $ \gamma$ at $ A$ and so $ C$ is the intersection of $ AM_{b}$ and the perpendicular in $ H_{c}$ to $ \gamma$.


function
limit
induction
algebra unsolved
algebra

Find all $f: R+\to R+$ such that: $f(x+y)-f(x-y)=4\sqrt{f(x)f(y)}$ $\forall x>y>0$

(*) $f(x+y)-f(x-y)=4\sqrt{f(x)f(y)}$

First we can replace $f$ by $cf$ for any $c\in\mathbb{R}_{+}$ and assume $f(1)=1$.

(1) $f$ is strictly increasing:
For $a>b>0$ we have $f(a)-f(b)=4\sqrt{f({a+b\over 2})f({a-b\over 2})}>0$ using $x={a+b\over 2},y={a-b\over 2}$ in (*).

(2) $f(2x)=4f(x)$ for all $x\in\mathbb{R}_{+}$:
There is a natural and easy way to prove that, by showing $\lim_{x\rightarrow 0+}f(x)=0$
and that $f$ is continuous, and then taking $y\rightarrow x$ in (*).
I will give here a direct algebraic approach.
Therefore let $z\in\mathbb{R}_{+}$ and set $c: =\sqrt{f(2z)\over 4f(z)}$.
Then $f(2z)=4c^{2}f(z)$.
$x=2z,y=z$ gives $f(3z)=4\sqrt{f(2z)f(z)}+f(z)=(8c+1)f(z)$
$x=3z,y=z$ gives $f(4z)=4\sqrt{f(3z)f(z)}+f(2z)=(4\sqrt{8c+1}+4c^{2})f(z)$
$x=4z,y=z$ gives $f(5z)=4\sqrt{f(4z)f(z)}+f(3z)=(8\sqrt{\sqrt{8c+1}+c^{2}}+8c+1)f(z)$
$x=3z,y=2z$ gives $f(5z)=4\sqrt{f(3z)f(2z)}+f(z)=(8c\sqrt{8c+1}+1)f(z)$
So $8\sqrt{\sqrt{8c+1}+c^{2}}+8c+1=8c\sqrt{8c+1}+1$
$\Rightarrow \sqrt{8c+1}+c^{2}=(c\sqrt{8c+1}-c)^{2}$
$\Rightarrow 2c^{2}+1=c^{2}\sqrt{8c+1}$
$\Rightarrow 0=8c^{5}-3c^{4}-4c^{2}-1=(c-1)(8c^{4}+5c^{3}+5c^{2}+c+1)$
with $8c^{4}+5c^{3}+5c^{2}+c+1>0$ for $c>0$.
So $c=1$ and $f(2z)=4f(z)$ for all $z\in\mathbb{R}_{+}$.

And so
(3) $f(x)=x^{2}$ for all $x\in\mathbb{R}_{+}$:
We have $f(n+1)=4\sqrt{f(n)}+f(n-1)$ for all $n\in\mathbb{N},n>1$ using $f(1)=1$ and $x=n,y=1$ in (*).
Then we get $f(n)=n^{2}$ for all $n\in\mathbb{N}$ by induction, using $f(1)=1^{2}$ and $f(2)=4f(1)=2^{2}$.
And now (2) gives $f({n\over 2^{r}})={f(n)\over 4^{r}}=({n\over 2^{r}})^{2}$ for all $n,r\in\mathbb{N}$ again by induction.
But now $\{{n\over 2^{r}}: n,r\in\mathbb{N}\}$ is dense in $\mathbb{R}_{+}$. So (1) gives the result.

So the solutions of (*) are $f(x)=cx^{2},x\in\mathbb{R}_{+}$ for any $c\in\mathbb{R}_{+}$.

$\tau: =\inf\{f(x): x\in\mathbb{R}_{+}\}\geq 0$ exists and $\lim_{x\to 0}f(x)=\tau$ because of (1).
Then $x=2y$ in (*) gives $f(3y)-f(y)=4\sqrt{f(2y)f(y)}$ and $y\to 0$ gives $\tau-\tau=4\sqrt{\tau\cdot\tau}$ and $\tau=0$.


calculus
integration
trigonometry
calculus computations

Evaluate $ \int_0^{\frac {\pi}{4}} \frac{dx}{sin^2x\plus{}sinxcosx\plus{}cos^2x}$

[hide]$ I \equal{} \int_0^{\pi/4} \frac{1}{1 \plus{} \sin x \cos x}\,dx \equal{} \int_0^{\pi/2}\frac{1}{2 \plus{} \sin u}\,du \equal{} \int_0^1\frac{1}{t^2\plus{}t\plus{}1}\,dt$

$ \equal{} \frac{2}{\sqrt{3}}\int_0^1\frac{\frac{2}{\sqrt{3}}}{1 \plus{} \left(\frac{2t\plus{}1}{\sqrt{3}}\right)^2}\,dt \equal{} \frac{2}{\sqrt{3}} \left[\arctan \left(\frac{2t\plus{}1}{\sqrt{3}}\right)\right]_0^1 \equal{} \frac{\pi}{3 \sqrt{3}}$.
[/hide]


vector
analytic geometry
abstract algebra
inequalities
algebra
function
domain

I'm stumped on this one: if $ V$ is a vector space over $ F$ and $ v,v' \in V$, why does $ v \otimes v' \equal{} v' \otimes v$ imply that $ v \equal{} av'$ for some $ a \in F$?

It's just more or less the definition of tensor product.  To answer more specifically it is probably good to know the exact definition you use.

Sometimes long calculations can make obvious facts seem more obvious: one way to attack the problem is to express v and v' as linear combination of basis elements ei, then express v ⊗ v' as a linear combination of the basis elements ei ⊗ ej.  For v ⊗ v' it is just vi*vj', but for v' ⊗ v it is vi'*vj.

So you can rephrase your question as a question about real numbers (or field elements): If ai*bj = bi*aj for i,j=1,...,n, then prove ai=c*bi for some fixed real number (or field element) c.

Well, assuming bi and bj are not zero, you get ai/bi = aj/bj for all i,j, and that particular number is "c".  If bi is 0 and ai is 0, then of course any c works in the i'th coordinate, so you can just ignore the i'th coordinate completely.  So you can assume that there is no i such that both ai and bi are zero.  But then if bi is 0, bi*aj is 0, so ai*bj is 0, so bj is 0.  In other words, b=0, and you can choose c=0.

I suppose so, but surely there's a way to do this without choosing a basis. After all, $ V$ could be infinite-dimensional, and I'd rather not have the axiom of choice come into play. Besides, the fact that every vector space has a basis hasn't been introduced at this point in the textbook (Dummit and Foote treat modules in general before vector spaces).

Idea, if you don't like the axiom of choice:

Consider the bilinear map $ (a,b) \mapsto (a\cdot v)(b\cdot v')$.  The universal property of tensor products tells us that $ (v\cdot v)(v' \cdot v') \equal{} (v\cdot v')^2$.  By the Cauchy-Schwarz inequality, this implies that $ v$ and $ v'$ are linearly dependent.

Sorry, what are the domain and range of this bilinear map? And what sort of product does $ \cdot$ denote?

Well, I meant for it to be inner product, but it just occurred to me that choosing an inner product is only slightly less strong than choosing a basis.  But I wonder if it's asking too much to solve this without some structure.  I can't think of any nontrivial bilinear maps from $ V\times V$, if not given an inner product.

Anyway, assuming a given inner product, the map I gave is a bilinear map $ V\times V \to F$.

Cauchy-Schwarz over arbitrary fields?  :wink:

Anyway, I think you can say pretty much NOTHING about tensor products of infinite-dimensional vector spaces without bases. I wouldn't even be surprised if it can be zero in constructive mathematics... Maybe someone familiar with logic can help here.

  darij

Let us assume that $ v,v'$ are linearly independent. Define $ V'\equal{}span\{v,v'\}$ which is a 2-dimensional subspace of $ V$. By flatness of $ F$-vector spaces, the map $ V'\otimes V'\to V\otimes V$ is injective which implies that the relation $ v'\otimes v\equal{}v'\otimes v$ still holds in $ V'$. Thus we can assume WLOG that $ V\equal{}V'$ is 2-dimensional. Then $ V\otimes V$ is 4-dimensional with basis $ v\otimes v$, $ v\otimes v'$, $ v'\otimes v$, $ v'\otimes v'$. In particular $ v\otimes v'\neq v'\otimes v$. Contradiction. So $ \dim span\{v,v'\}\le 1$.

Now prove the flatness of vector spaces without bases or other AC applications in the infinite-dimensional case ;)

  darij

Okokok, so what exactly can we assume to be known here? Especially about modules? What about limits and other nasty things about small categories, e.g.? And how is the tensor product defined - does it fulfil the universal property? Just asking to avoid to many feed-back loops. :) I think there is much more needed from set theory to formally speak about tensor products.

In general for arbitrary commutative (unitary) ring $ R$ and $ R$-modules $ M,N$ if:
$ \sum_{i \equal{} 1}^{n}x_{i}\otimes_{R} y_{i} \equal{} 0$ in $ M\otimes_{R} N$ for some $ x_{i}\in M, \ y_{i}\in N, \ i \equal{} 1,\ldots, n$
then there exist finitely generated submodules $ M_{0} \leq M, N_{0} \leq N$ st
$ x_{i}\in N_{0}, \ y_{i}\in M_{0}, \ i \equal{} 1,\ldots, n,$
and $ \sum_{i \equal{} 1}^{n}x_{i}\otimes_{R} y_{i} \equal{} 0$ in $ M_{0}\otimes_{R}N_{0}.$
Moreover if $ M \equal{} N,$ then we can assume that $ M_{0} \equal{} N_{0}$
[hide="sketch of some proof (I wonder whether/where it uses AC?)"]If we take the tensor product $ M\otimes_{R}N$ constructed in standard way as $ E/K,$
where
$ E \equal{} \bigoplus_{(m,n)\in M\times N}R \equal{} \left\{\sum_{j \equal{} 1}^{l}r_{j}(a_{j},b_{j}) \ : \ l\in \mathbb{N}, \ r_{j}\in R, \ a_{j} \in M, \ b_{j}\in N, \ j \equal{} 1,\ldots, l\right\}$
$ K$ is submodule of $ E$ generated by elements of the forms:
\[ (a, b \plus{} b') \minus{} (a, b) \minus{} (a, b'),\quad (a \plus{} a', b) \minus{} (a, b) \minus{} (a', b), \\
(ra, b) \minus{} r(a,b),\quad (a,rb) \minus{} r(a,b), \quad a,a' \in M, \ b,b'\in N, \ r\in R\]
and $ a\otimes_{R} b : \equal{} \pi((a,b)),$ where $ \pi : E\to E/K$ is canonical epimorphism,
then  $ \sum_{i \equal{} 1}^{n}x_{i}\otimes_{R} y_{i} \equal{} 0$ in $ M\otimes_{R}N$ means that the sum
$ \sum_{i \equal{} 1}^{n}(x_{i},y_{i})\in E$
is equal to finite sum of generators of $ K$ mentioned above.
Define $ M_{0}$ as submodule of $ M$ generated by $ x_{i},\ i \equal{} 1,\ldots, n$ and elements that occur as a first coordinates in that generators, and $ N_{0}$ as submodule of $ N$ generated by $ y_{i}, \ i \equal{} 1,\ldots,n$ and elements that occur as a second coordinates.
Now it is easy to see that similar construction of $ M_{0}\otimes_{R} N_{0}$ yields
$ \sum_{i \equal{} 1}^{n}x_{i}\otimes_{R} y_{i} \equal{} 0$ in $ M_{0}\otimes_{R}N_{0}$.

If $ M \equal{} N,$ then instead of $ M_{0}, N_{0}$ one can take twice $ M_{0} \plus{} N_{0}$[/hide]

Now wlog $ V$ is finite dimensional and in such a case the result follows without AC
[hide]assuming $ v,v'$ are linear independent we can take basis $ e_{1},\ldots, e_{\dim V}$ of $ V$ such that $ e_{1} \equal{} v,\ e_{2} \equal{} v'$ (in finite dimensional case we don't need AC to do it). But then $ \{e_{i}\otimes e_{j}\}_{i,j \equal{} 1,\ldots, \dim V}$ is a basis of $ V\otimes V$ - in particular $ v\otimes v'\neq v'\otimes v.$[/hide]

[quote="darij grinberg"]Now prove the flatness of vector spaces without bases or other AC applications in the infinite-dimensional case ;)[/quote]Well, I will assume the standard construction of the tensor product for arbitrary modules. So we can always reduce to the case that $ V$ is finitely generated over $ F$ since the relations divided out iinvolve finitely many elements of $ V$ only in order to have $ v\otimes v' \minus{} v'\otimes v \equal{} 0$.

Now simply apply the criterion that $ V$ is a flat $ F$-module iff, for every ideal $ \mathfrak{a}$ of $ F$, the natural map $ \mathfrak{a}\otimes V\to \mathfrak{a}V$ is an isomorphism. I cannot see where AC is used in the proof of this lemma (e.g. Lang, Algebra). ;)

[quote="ifk"](I wonder whether/where it uses AC?)]
$ E \equal{} \bigoplus_{(m,n)\in M\times N}R \equal{} \left\{\sum_{j \equal{} 1}^{l}r_{j}(a_{j},b_{j}) \ : \ l\in \mathbb{N}, \ r_{j}\in R, \ a_{j} \in M, \ b_{j}\in N, \ j \equal{} 1,\ldots, l\right\}$
[/quote]

Why is E nonempty?

... because $ E$ can be identified to the set of mappings $ M\times N\to R$ such that it is 0 for almost all arguments. $ E$ is a $ R$-module and contains a 0-element which is the mapping identical to 0.

ifk is right, nice proof!

@ Soarer: Of course, $ E$ is nonempty, since $ \left(a,b\right)\in E$ for every $ a\in M$ and $ b\in N$. And $ 0$, of course.

  darij


[quote="deathwar"]If n<>0 natural and n/(2^n)-2 then 2^n-1 / (2^((2^n)-1))-2.
{ / divides }[/quote]
If I understoud you well then you want us to prove that if $n|2^n-2$=>$(2^n-1)|2^{2^n-1}-2$,but this one is very easy becous if $m|n$<=>$(2^m-1)|(2^n-1)$ :) .


geometry
rectangle
combinatorics unsolved
combinatorics

I can't prove a problem :
The knight (which can move on the diagonal of the rectange 2x3) on the chessboard can move all the places of the chess board
Notice that :
the knight can not move to the place that it has occupied before

Please help me...

(sorry ... my English is not good ... )

Let me get this straight: you want to show that a knight can start from any square of the board and move on the board s.t. it uses each square exactly once?

i guess that is what the meaning is.
anyways it is an old problem (you can find lots of sources for this) that there is a knight's tour on the regular chessboard;for a general nxn board (n even of course)however i  think it is an unsolved problem whether there is a hamiltonian circuit for the knight.

[quote]you want to show that a knight can start from any square of the board and move on the board s.t. it uses each square exactly once?[/quote]
I meant that the knight begin from its origin (at the second cell of the board )

[quote]for a general nxn board (n even of course)however i think it is an unsolved problem whether there is a hamiltonian circuit for the knight.[/quote]
I see that :
For n = 3 or 4 , the knight can't move all of the place in the chess board


superior algebra
superior algebra solved

Let $<S,\cdot>$ be a semigroup, and let $z$ be the "zero" element of $<S,\cdot>$. That is, $xz=zx=z$ for all $x$ in $S$. Define $x$ to be a [i]right zero divisor[/i] of $S$ if there exists an $x'\not= z$ in $S$ such that $xx'=z$. Prove that the set of all right zero divisors in the semigroup $<S,\cdot>$ is form a subsemigroup of $<S,\cdot>$.

They even form a "left-ideal": If $b$ is a right zero divisor, so is $ab$ for all $a$.

Just pick $b' \neq z$ s.t. $bb' = z$, and deduce $(ab)b'=a(bb')=az=z$.

right. but if you were to be more rigorous, how do you prove that the set of all left zero divisors of $S$ inherits associativity from $S$?

ok, i agree. i wasn't sure if my professor would


trigonometry
modular arithmetic
algebra unsolved
algebra

So,the solutions   $ x\equal{}2\pi k x\equal{}\dfrac{\pi}{2}\plus{}2\pi k$
Its no very hard !

_________________________________
Azerbaijan Land of the Fire  :lol:

Who can give me a solution? Thank you very much.  :P

[quote="inversionA007"]$ 2^{\sqrt {\sin{x}}} \plus{} 2^{\sqrt {\cos{x}}} \equal{} 2 \plus{} \sin{x} \plus{} \cos{x}$[/quote]

In order the equation be defined, we need $ x\in[0,\frac {\pi}2]$  $ \pmod{2\pi}$

Let $ f(x) \equal{} 2^{x} \minus{} 1 \minus{} x^2$. And the equation becomes $ f(\sqrt {\sin x}) \plus{} f(\sqrt {\cos x}) \equal{} 0$

It's easy to check that $ f(0) \equal{} f(1) \equal{} 0$ and that $ f(x) > 0$   $ \forall x\in(0,1)$

So $ f(\sqrt {\sin x}) \plus{} f(\sqrt {\cos x}) > 0$ $ \forall x\in(0,\frac {\pi}2)$ and the equation may be solved uniquely for $ x\in\{0,\frac {\pi}2\}$ and it's immediate to see that these two values indeed are solutions.

Hence the answer : $ x \equal{} 2k\pi$ and $ x \equal{} \frac {\pi}2 \plus{} 2k\pi$


number theory
prime factorization
number theory proposed

$ O(n)$ is sum of divisor in the form 4k+1
$ E(n)$ is sum of divisor in the form 4k+3
1.Find all $ n\in N$ such that $ O(n)>E(n)$
2.Find all $ n\in N$ such that $ O(n)\equal{}E(n)$

Jacobi's Two Square Theorem: The number of representations of a positive integer as the
sum of two squares is equal to four times the difference of the numbers of divisors congruent
to 1 and 3 modulo 4.
Let $ r(n)$ be The number of representations of a positive integer as the sum of two squares, so $ r(n)\equal{}4(O(n)\minus{}E(n))\geq 0$.
we have $ O(n)\equal{}E(n)$ iff $ r(n)\equal{}0$ iff $ n$ have a prime factor of the form $ 4m\plus{}3$ with odd exponent in the prime factorization of $ n$.
we have $ O(n)>E(n)$ iff $ r(n)>0$ iff all prime factors of $ n$ of the form $ 4m\plus{}3$ have even exponent in the prime factorization of $ n$.

no, it'sn't the right demonstration, because $ O(n) and E(n)$ aren't the number of divisors but the sum


inequalities
calculus
function
algebra
polynomial
linear equation
inequalities proposed

[quote="xtar"]Let $x,y,z $ be positive real numbers satisfying $x^2+y^2+z^2=3$. Prove that
$\frac{1}{x}+\frac{1}{y}+\frac{1}{z}+\frac{4}{3}(x+y+z)\ge7$[/quote]

Nice problem!

Since $x^2+y^2+z^2=3$, we have $x^2<3$ and thus $x<\sqrt3<6$, so that 6 - x > 0. Hence,

$\left(\frac{1}{x}+\frac43x\right)-\frac16\left(x^2+13\right)=\frac{\left(6-x\right)\left(x-1\right)^2}{6x}\geq 0$,

and thus $\frac{1}{x}+\frac43x\geq\frac16\left(x^2+13\right)$. Similarly, $\frac{1}{y}+\frac43y\geq\frac16\left(y^2+13\right)$ and $\frac{1}{z}+\frac43z\geq\frac16\left(z^2+13\right)$. Summing up, we obtain

$\left(\frac{1}{x}+\frac43x\right)+\left(\frac{1}{y}+\frac43y\right)+\left(\frac{1}{z}+\frac43z\right)\geq\frac16\left(x^2+13\right)+\frac16\left(y^2+13\right)+\frac16\left(z^2+13\right)$;

in other words,

$\frac{1}{x}+\frac{1}{y}+\frac{1}{z}+\frac43\left(x+y+z\right)\geq\frac16\left(\left(x^2+y^2+z^2\right)+39\right)=\frac16\left(3+39\right)=7$,

qed..

  Darij

I have tried it in this manner:
[tex]\displaystyle \frac{1}{x}+\frac{1}{y}+\frac{1}{z}+\frac{4}{3}(x+y+z)\ge7[/tex]
After Multiplying the inequality with [tex] 3xyz[/tex] ,the equivalent inequality turns out to be:

[tex] 4x^2yz +3yz+4y^2xz+3xz+4z^2xy+3xy  \geq \ 21xyz [/tex]

but applying AM-GM inequality to LHS we get:

[tex] 4x^2yz +3yz+4y^2xz+3xz+4z^2xy+3xy \geq 6.4.3.\sqrt{12}\sqrt{x^6y^6z^6} \geq 21xyz [/tex]

Please do comment on my solution for I don't know if it is correct. :?

[quote="darij grinberg"][quote="sebescu"]I have tried it in this manner:
[tex]\displaystyle \frac{1}{x}+\frac{1}{y}+\frac{1}{z}+\frac{4}{3}(x+y+z)\ge7[/tex]
After Multiplying the inequality with [tex] 3xyz[/tex] ,the equivalent inequality turns out to be:

[tex] 4x^2yz +3yz+4y^2xz+3xz+4z^2xy+3xy  \geq \ 21xyz [/tex]

but applying AM-GM inequality to LHS we get:

[tex] 4x^2yz +3yz+4y^2xz+3xz+4z^2xy+3xy \geq 6.4.3.\sqrt{12}\sqrt{x^3y^3z^3} \geq 21xyz [/tex]

Please do comment on my solution for I don't know if it is correct. :?[/quote]

No, it isn't, since you don't always have $6.4.3.\sqrt{12}\sqrt{x^3y^3z^3} \geq 21xyz$.

  darij[/quote]
Ummm.....sorry Darij...It was a mistake on my part :oops: .Now I have edited my post and need ur comments on it.

Thanx.

[quote="sebescu"][tex] 4x^2yz +3yz+4y^2xz+3xz+4z^2xy+3xy \geq 6.4.3.\sqrt{12}\sqrt{x^6y^6z^6} \geq 21xyz [/tex][/quote]

Still this is wrong. Don't forget that xyz can get arbitrarily small.

  Darij

It's immaterial.  If you had correctly applied AM-GM, you would have gotten

$6\sqrt {12}abc$, which is $\leq 21abc$.  Your method is almost guaranteed to not work in problems of this type, because the terms that you are AM-GM-ing are not equal when $a=b=c=1$

[quote="blahblahblah"]Darij, out of curiosity, what method did you use to compute

$\frac {1}{x}+\frac {4}{3}x\geq ax^2+b$?

Did you use some sort of calculus-type method to calculate the minimum, or did you just use the fact that there has to be a double root at $x=1$ and work backwards?[/quote]

That's more or less the same.  :)  I substituted $X=x^2$, $Y=y^2$, $Z=z^2$ to make the condition $x^2+y^2+z^2=3$ become a linear equation, X + Y + Z = 3. Then, I searched for a linear lower bound for the function $\frac{1}{x}+\frac43x=\frac{1}{\sqrt{X}}+\frac43\sqrt{X}$ such that this bound is identical with the function for X = 1 (in order to match the equality case of the inequality in question). Of course, the first thing one tries in such a case is the tangent to the plot of the function $\frac{1}{\sqrt{X}}+\frac43\sqrt{X}$ at the point X = 1. With some calculations, one finds that this tangent is the function $\frac16\left(X+13\right)$. Another simple calculation reveals that this is indeed a lower bound for $\frac{1}{\sqrt{X}}+\frac43\sqrt{X}$ for $0<X\leq 36$, so the method works. Of course, we needed calculus in order to find the solution, but it doesn't appear in the final writeup.

  Darij

[quote="xtar"]Let $x,y,z$ be positive real numbers satisfying $x^{2}+y^{2}+z^{2}=3$. Prove that
$\frac{1}{x}+\frac{1}{y}+\frac{1}{z}+\frac{4}{3}(x+y+z)\ge7$[/quote]
$\frac{1}{x}+\frac{1}{y}+\frac{1}{z}+2(x+y+z)\ge9$ is true too.

[quote="blahblahblah"]$6\sqrt{12}\leq 21$[/quote]
 :rotfl:

In "SI Inequality, A story of Maths", I proposed the following stronger inequality
$\bigstar$ (Hungkhtn's, problem 5.3.5)
Given positive real numbers $a_{1},a_{2},...,a_{k}$ such that $a_{1}^{2}+a_{2}^{2}+...+a_{k}^{2}=k$ and $k$ is a natural number, $k\in\{1,2,...,20\}$, prove that
\[\left(\frac{1}{a_{1}}+\frac{1}{a_{2}}+...+\frac{1}{a_{k}}\right)+2\left(a_{1}+a_{2}+...+a_{k}\right) \ge 3k. \]
Moreover, the following statement is stronger than Xtar's one (also in this book)

Given positive real numbers $a,b,c$ and $a^{2}+b^{2}+c^{2}=3$, prove that
\[\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)+2.9\left(a+b+c+d\right) \ge 11,7. \]

Dear Jichen,

$2.9$ seems to be not so different from $2.9080...$, is'nt it? I know the way to find the best constant (as you show), but it require much heavy computing, so I skip this terrible step and choose the number $2.9$ (as it is so close). Moreover, my proof to my problem (with constant $2.9$) is very easy.

Sorry to say that I hate complicated computing.

In my book, the following inequality of four numbers holds

$\bigstar$
Given positive real numbers $a,b,c,d$ and $a^{2}+b^{2}+c^{2}+d^{2}=4$, prove that
\[\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}+\frac{1}{d}\right)+\sqrt{\frac{23}{3}}\left(a+b+c+d\right) \ge 4+4\sqrt{\frac{23}{3}}.\]

Also this inequality is stronger than yours, Jichen, I like your form. It is nicer than mine.

Moreover, the following is holds for all $n$ (very strong inequality)

$\bigstar$ (hungkhtn's, problem 5.3.5, SII Inequalities, A story of Maths)
Let $a_{1},a_{2},...,a_{n}$ be positive real numbers such that $a_{1}^{2}+a_{2}^{2}+...+a_{n}^{2}=n$. For $k=\sqrt{1+\frac{3\sqrt[3]{n-1}}{\sqrt[3]{n-1}-\sqrt[3]{n-1}+1}}+\frac{2\sqrt{n-1}}{n}$, prove the following inequality
\[\frac{1}{a_{1}}+\frac{1}{a_{2}}+...+\frac{1}{a_{n}}+k(a_{1}+a_{2}+...+a_{n}) \ge n(k+1).\]

[quote="arqady"][quote="Ji Chen"]hungkhtn, you are right. The rational number is nice.
[/quote]
I think, the natural number nicer. :D[/quote]

You are right, too. Arqady. It is the reason why I posed before the inequality with natural number $2$  :lol:

[quote="hungkhtn"]$\bigstar$ (hungkhtn's, SI inequalities, A story of Maths)
Given positive real numbers $a_{1},a_{2},...,a_{k}$ such that $a_{1}^{2}+a_{2}^{2}+...+a_{k}^{2}=k$ ($k$ is a natural number, $k\in\{1,2,...,20\}$), prove that
\[\left(\frac{1}{a_{1}}+\frac{1}{a_{2}}+...+\frac{1}{a_{k}}\right)+2\left(a_{1}+a_{2}+...+a_{k}\right) \ge 3k. \]
[/quote]

Dear Jichen,

Yet I don't know why you deleted your last words. It is hardly you say funny and relaxing word as that  :lol:  Yes I like the number 23 but I like the number 13 more, because I was born in 13-13  :lol: To be honest, I don't check case 21 and 22 because the number 20 is very pretty. What numbers do you like, Jichen?

Dear all others,

My solution to these problems are simple induction. I can attach them right now because I already wrote it in my book, but I wish you can spend more time to solve such nice problems.

[quote="Ji Chen"]hungkhtn, you are right. The rational number is nice.

If $a,b,c,d$ be positive real numbers such that $a^{2}+b^{2}+c^{2}+d^{2}=4$, then
\[\frac{1}{a}+\frac{1}{b}+\frac{1}{c}+\frac{1}{d}+\frac{11}{4}\left(a+b+c+d\right) \ge 15, \]
with equality if and only if $a=b=c=d=1$.[/quote]
http://www.artofproblemsolving.com/Forum/viewtopic.php?f=51&t=590881

[quote="hungkhtn"]
Given positive real numbers $a,b,c,d$ and $a^{2}+b^{2}+c^{2}+d^{2}=4$, prove that
\[\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}+\frac{1}{d}\right)+\sqrt{\frac{23}{3}}\left(a+b+c+d\right) \ge 4+4\sqrt{\frac{23}{3}}\][/quote]
1. The following inequality is also true.
Let $a$, $b$, $c$ and $d$ be positive numbers such that $a^2+b^2+c^2+d^2=4$. Prove that:
\[\frac{1}{a}+\frac{1}{b}+\frac{1}{c}+\frac{1}{d}+\frac{25}{9}\left(a+b+c+d\right) \ge\frac{136}{9}\]

2. A maximal $k$ for which the inequality
\[\frac{1}{a}+\frac{1}{b}+\frac{1}{c}+\frac{1}{d}+k\left(a+b+c+d\right) \ge4+4k\]
is true for all positives $a$, $b$, $c$ and $d$ such that $a^2+b^2+c^2+d^2=4$, is $k=2.7833144...$.
The solutions of these two problems are the same.
[quote="hungkhtn"]

Moreover, the following is holds for all $n$ (very strong inequality)

$\bigstar$ (hungkhtn's, problem 5.3.5, SII Inequalities, A story of Maths)
Let $a_{1},a_{2},...,a_{n}$ be positive real numbers such that $a_{1}^{2}+a_{2}^{2}+...+a_{n}^{2}=n$. For $k=\sqrt{1+\frac{3\sqrt[3]{n-1}}{\sqrt[3]{n-1}-\sqrt[3]{n-1}+1}}+\frac{2\sqrt{n-1}}{n}$, prove the following inequality
\[\frac{1}{a_{1}}+\frac{1}{a_{2}}+...+\frac{1}{a_{n}}+k(a_{1}+a_{2}+...+a_{n}) \ge n(k+1).\][/quote]
Something is wrong here. :maybe:

The following inequality is also true.
Let $a$, $b$, $c$, $d$ and $e$ be positive numbers such that $a^2+b^2+c^2+d^2+e^2=5$. Prove that:
\[ \frac{1}{a}+\frac{1}{b}+\frac{1}{c}+\frac{1}{d}+\frac{1}{e}+\frac{8}{3}(a+b+c+d+e)\ge\frac{23}{3} \]
For six variables it's true for $k=2.5836...$.
For seven variables $k=2.50699...$


A particle P of mass m kg moves vertically in a medium in which resistance to motion has magnitude $ \frac{mv}{10}$ when the speed of the particle is $ vms^{-1}$. The acceleration due to gravity is 10 $ ms^{-1}$.

(i) If the particle falls vertically from rest, show that its acceleration $ ams^{-2}$ is given by $ a=10-\frac{v}{10}$. Hence show that the terminal velocityt of the particle is 100 $ ms^{-1}$.

(ii) The particle is projected vertically upwards with speed 100 $ ms^{-1}$.

(a) show that its acceleration $ ams^{-2}$ is given by $ a=-10-\frac{v}{10}$

(b) Find expressions for the speed $ vms^{-1}$ and the height x metres of the particle after time t seconds

(c) Find the time taken by the particle to reach its max. height and the max. height attained

This is a physics question.  Physics questions should go in the physics forum.  If you really want it in a math forum, it should go in the Calculus Computations and Tutorials section.

Once again all you need are the relatioships: $ a= \frac{dv}{dt}= v. \frac{dv}{dx}$.

And yes I do 4 unit Mathematics.

[quote="bos1234"]how can i transport this to another part of the forum[/quote]

You can't.  Hopefully a moderator will do it.  However, you can in the future post physics questions in the physics subfora  :)


[quote="N.T.TUAN"]You can find here http://www.imo2007.edu.vn/  :wink:[/quote]

Maybe you can give a more precise link. Either the details are not out yet, what I assume, or I did not find the requested information.

'The Official Program will contain, among other things, details of accommodation arrangements (inclusive of food) for Participants and Observers at contracted hotels, the main site and the venues for various official events associated with the IMO 2007, including the opening and closing ceremonies, banquet and examination rooms.' ([url=http://www.imo2007.edu.vn/index.php?module=ViewNewDetailType2&NewId=2]Regulations of the IMO 2007[/url])

[quote="N.T.TUAN"][quote="Jutaro"]Excuse me guys, I don't know whether this is the correct place to ask this, but I would like to know some data about the hotel where we are going to stay at Hanoi. Could someone please get some information about it? :lol:[/quote]
Don't worry about it because http://imo2007.edu.vn/statics/Announcement.htm  :wink:
to orl : I am sorry, I can't find it, too. But please see above link  :)[/quote]

@N.T.Tuan: I have seen the suggested web page among various other IMO 2007 pages out there. But at first this site is targeted at Guides/Volunteers and secondly it does not really tell us more about the accommodation issue.

Hi friends:

I think it depends on your situation. If you are an official members, leader, deputy leaders, observers, the organizing committee will take care of that for you. Do not worry. If you are totally independent, you can ask me for help.

[color=blue]It's important for me to know, in which hotel we will stay.
My teacher need that (all hotels for contestans and leaders, if possible).
She also need the weather forecast.

If someone knows that, tell me please.

(I'm really sorry, if my english is bad)[/color]

[quote]International officials and observers will stay at the La Thanh Hotel while students will stay at the dormitory of the Hanoi Teacher Training University or the Hacinco Student Village.

[/quote]

This is the information I found on [url]http://english.vietnamnet.vn/education/2006/11/634252/[/url]. it's a relatively old article, but the information could be useful...

see that :

http://www1.hut.edu.vn/en/index.php?id=80,71,0,0,1,0


Prove that there are no positive integer d, n that satisfy the following;

$ d | 2n^2$,

$ n^2 \plus{} d$ is a square.

Let $ n^2 \plus{} d \equal{} c^2$ and let $ \gcd(c,n) \equal{} d'$

then we have $ 2n^2 \plus{} 2d \equal{} 2c^2 \Rightarrow d \mid 2\gcd(c,n)^2 \Rightarrow d\mid 2d'^2$ because $ d\mid2n^2$ and $ d \mid 2c^2$

On the other hand we have $ d \equal{} c^2 \minus{} n^2$ which implies that $ d'^2 \mid d$ Hence we have $ d \equal{} kd'^2$ and $ 2d'^2 \equal{} k'd$
it follows that $ kk' \equal{} 2\Rightarrow d \equal{} 2d'^2$ or $ d \equal{} d'^2$

[b][u]Case:1 [/u][/b]: $ d \equal{} d'^2$

we have $ n^2 \plus{} d'^2 \equal{} c^2 \Leftrightarrow a^2d'^2 \plus{} d'^2 \equal{} b^2d'^2 \Leftrightarrow a^2 \plus{} 1 \equal{} b^2$ which is not true !!

[u][b]Case:2 [/b][/u]:$ d \equal{} 2d'^2$

we have $ n^2 \plus{} d'^2 \equal{} c^2 \Leftrightarrow a'^2d'^2 \plus{} 2d'^2 \equal{} b'd'^2 \Leftrightarrow a'^2 \plus{} 2 \equal{} b'^2$ which is not true !!

our problem is proven !!

$ 2n^2  \equal{} kd(k \in N) \Rightarrow n^2  \plus{} d \equal{} n^2 (1 \plus{} \frac{2}{k}) \equal{} m^2  \Rightarrow k \equal{} 1;2.$
But 2n^2;3n^2 isn't a square.

[quote="tronghieu"]$ n^2 (1 \plus{} \frac {2}{k}) \equal{} m^2 \Rightarrow k \equal{} 1;2.$[/quote]

You cant conclude  $ k \equal{} 1;2$ immediately from $ n^2 (1 \plus{} \frac {2}{k}) \equal{} m^2$ :

Take for example $ n^2 (1 \plus{} \frac {5}{k}) \equal{} m^2$ which have solutions for $ k\notin\{1,5\}$ : $ n\equal{}2,k\equal{}4,m\equal{}3$

Picking up from $ n^2\left(1\plus{}\frac2k\right)\equal{}m^2$, we get $ \frac{k\plus{}2}k$ is the square of a rational, which means that in lowest terms, the numerator and denominator are squares.

So if k is odd, we get $ k$,$ k\plus{}2$ squares, no solutions.
And if k is even, we have $ k/2$,$ k/2\plus{}1$ are squares, which gives no solutions. ($ k\equal{}0$ works for this but gives with $ n\equal{}0$)

Sorry. I'm so stupid. we can correct it.
If $ k$ is odd then $ k\plus{}2\equal{}a^2;k\equal{}b^2$.
If $ k$ is even then $ k\equal{}2a^2;k\plus{}2\equal{}2b^2\equal{}>k/2\equal{}a^2;k/2\plus{}1\equal{}b^2.$


limit
integration
trigonometry
logarithms
calculus
real analysis
real analysis unsolved

$\int_{a}^{\frac{\pi}{2}}\sin(n^{4}\cos t)\,dt= \int_{a}^{\frac{\pi}{2}}-n^{4}\sin(t)\sin(n^{4}\cos(t))\cdot \frac{-1}{n^{4}\sin(t)}\,dt$
$=\left[-\cos(n^{4}\cos t)\cdot\frac{-1}{n^{4}\sin t}\right]_{a}^{\frac{\pi}{2}}-\int_{a}^{\frac{\pi}{2}}-\cos(n^{4}\cos t)\cdot-\frac1{n^{4}}\ln\left(\frac{1-\cos t}{\sin t}\right)\,dt$

For $a>0$, both terms obviously go to zero. Taking the limit as $a\to 0$, or equivalently treating a small slice of the integral separately, our desired limit is zero.


So, what are some opinions on the new judge? [url=http://blogs.washingtonpost.com/campaignforthecourt/2006/02/swearing_in_at_.html]Link.[/url]

Personally, I'm sort of scared. The ruling about the right of a woman to an abortion without the husband's permission was sort of draconian.(Completely disregarding the abortion issue! It's the fact that the wife is treated like a kid in need of permission for everything...)
[quote]He's on record in a memo as believing that to shoot an eighth grader, known not to be armed, who was trying to climb over a fence in escape, is a proper use of deadly force by a policeman. In a discussion of immigration cases that have been regularly occasioning inexcusable, vile, un-American heartbreak on people who missed obscure deadlines or violated arcane requirements, all he could say was that the courts get bad transcripts and it was hard to find translators for some of the plaintiffs, but that was a problem for Congress.[/quote]

To me it looks like Alito is probably not a great candidate, but he's not a disaster either. That's because he likes to talk about starry devices or something like that. :D

PenguinIntegral,

You shouldn't be scared.  There was a lot of fear mongering going on in the Senate.  Remember he was selected primarily for being judicially conservative, not politically conservative.  His views will be to stay with settled law, even if he "knows" it is wrong and he knows that the other side would change it at the drop of a hat if they had the votes, as Justice Cardozo said years ago to lead the process leading to legislating from the bench.

The new times are before us.  Many have not been politically aware during any time when Justice O'Connor was on the bench, deciding many close issues in 5/4 decisions.  They also have not been politically aware during any time when Chairman Greenspan hasn't been at the helm of the US monitary system.

It's a brave new world.

If you think about the abortion question rationally (as few do), it is really clear.  If you decide that a human exists at the moment of conception, clearly no one can kill it, just like they can't kill older humans.  If you decide that a human starts at birth, clearly no one can tell the  mother what she can do with something non-human in her body.  It's really that simple.  All the rest (most of) the discussion is smoke and mirrors on both sides.

As for the involvement of the husband's opinion, if one assumes that it is a human at comception and the law is wrong, but established; one doesn't have to go far to see that the financial impact of the child and support obligations are heavy on both parents, not just the mother, who does have a much greater physical involvement in the child's life, particularly before birth.  This is not an irrational restriction placed on a law that someone is not comfortable is correct.  It certainly is better than any other type of restriction that I can think of at the moment.

I hope this helps.

[quote="wizard2378"]
If you think about the abortion question rationally (as few do), it is really clear.  If you decide that a human exists at the moment of conception, clearly no one can kill it, just like they can't kill older humans.  If you decide that a human starts at birth, clearly no one can tell the  mother what she can do with something non-human in her body.  It's really that simple.  All the rest (most of) the discussion is smoke and mirrors on both sides.

As for the involvement of the husband's opinion, if one assumes that it is a human at comception and the law is wrong, but established; one doesn't have to go far to see that the financial impact of the child and support obligations are heavy on both parents, not just the mother, who does have a much greater physical involvement in the child's life, particularly before birth.  This is not an irrational restriction placed on a law that someone is not comfortable is correct.  It certainly is better than any other type of restriction that I can think of at the moment.

I hope this helps.[/quote]

The problem with the requirement that the woman ask her husband before she can get an abortion is that
1) her bodily autonomy is in essence transfered to her husband
2) Why is it the place of the government to force a women to talk to her husband? You are conflating two different things here -- financial arrangements after birth have nothing to do with women needing their husband's consent to get an abortion. In fact, if they got an abortion ... there wouldn't be any financial arrangements after birth.
3) Women in abusive relationships would especially have problems with this law -- afaik, had casey v planned parenthood gone the way alito wanted it to, women would've had to PROVE they were in abusive relationships before they could get an abortion with out their husband's consent. Proving physical, let alone emotional abuse is hazy and very open to abuse.
4) It's not constitutional - see Roe v. Wade

Oh, and the abortion debate is really not that simple...
I think most people agree that fetuses are "human" in the sense that they have 46 chromosomes and human DNA. The difference in agreement occurs on two levels:
1. Whether and when fetuses are sentient/viability standard
2. Whether the woman's autonomy takes precedent over the fetus's life

I'm with PenguinIntegral: Alito is terrifying, even more so than the fact that Anthony Kennedy is now a "moderate" voice on the court.  I'm particularly worried about how civil rights of all sorts will fare under the reconstructed court -- I'm hoping that Roberts will be decent on such issues, but I don't think it's too promising.

[quote] If you decide that a human exists at the moment of conception, [/quote]

And that's rational? Come on.


[quote="Astronum"][quote="SplashD"][quote="Astronum"]Prove by induction that $1^{2}+2^{2}+3^{2}+...+n^{2}= \frac{n(n+1)(2n+1)}{6}$ if $n \geq 1$.

 :roll:[/quote]

[hide]Base case: $n=1$

$1^{2}=1=\frac{1(2)(3)}{6}$

$1^{2}+2^{2}+3^{2}+...+n^{2}+(n+1)^{2}=\frac{n(n+1)(2n+1)}{6}+(n+1)^{2}=\frac{n(n+1)(2n+1)+6(n+1)^{2}}{6}=\frac{(n+1)(2n^{2}+7n+6)}{6}=\frac{(n+1)(n+2)(2n+3)}{6}$

and our induction is complete.[/hide][/quote]

Perfect.  :lol:

How did you know that $2n^{2}+7n+6 = (n+2)(2n+3)$. Was it guess?[/quote]

You don't need to guess.  You could use synthetic division.
http://mathworld.wolfram.com/SyntheticDivision.html

[quote="lingomaniac88"] You don't need to guess.  You could use synthetic division.
http://mathworld.wolfram.com/SyntheticDivision.html[/quote]
You don't need to use synthetic division either.
There are many ways to factor a quadratic.
$1 \qquad 2,3$
$2 \qquad 1,6$
The first 2 numbers are the factors of the leading coefficent. The other 2 pairs of numbers are the pairs of positive integers whose product is the constant. (It gets a little more complicated when the middle term is negative.) Then we have to check only 2 possible choices. First we can look at the pair $1, 6$. One of the numbers of $1, 2$ times one of the numbers of $1, 6$ and the other numbers' product must sum to $7$. We can easily see that's impossible. Then we look at the other pair. $1\cdot3+2\cdot2=7$. So the factorization is $(n+2)(2n+3)$.


Let $ a$ be a natural number which is greater than $ 3$ and consider the sequence $ (u_n)$ $ (n\equal{}1,2,,...)$ defined inductively by
$ u_1 \equal{}a$
$ u_{n\plus{}1}\equal{}u_n\minus{}[\frac{u_n}{2}]\plus{}1$ for all $ n\equal{}1,2,...$.
Prove that there exists $ k\in N^*$ such that $ u_n\equal{}u_k$ for all $ n\geq k$

[quote="sylvect"]Let $ a$ be a natural number which is greater than $ 3$ and consider the sequence $ (u_n)$ $ (n \equal{} 1,2,,...)$ defined inductively by
$ u_1 \equal{} a$
$ u_{n \plus{} 1} \equal{} u_n \minus{} [\frac {u_n}{2}] \plus{} 1$ for all $ n \equal{} 1,2,...$.
Prove that there exists $ k\in N^*$ such that $ u_n \equal{} u_k$ for all $ n\geq k$[/quote]
For any natural $ x\equal{}[\frac{x}{2}]\plus{}[\frac{x\plus{}1}{2}].$ Therefore $ u_{n\plus{}1}\equal{}[\frac{u_n\plus{}3}{2}]$. It give $ U_n\equal{}2$, if $ u_1<3$ and $ u_n\equal{}3$ if $ U_1\ge 3$ for big n.

This is one of the problems from the newest Mathematics and Youth Magazine (7/2008)
And the deadline for submitting solutions is 15 September, 2008.
See [url]http://www.artofproblemsolving.com/Forum/viewtopic.php?t=168836[/url] (post #13 by April).

Moved back (to Pre-Ol, for the easiness of the problem).

  darij


geometry
function
integration
analytic geometry
calculus
calculus computations

Find the total area bounded by the function $y^{2}=x^{2}(1-x^{2})$ and the $x$ axis.

1. That's not a function.
2. The graph of the points satisfying that equation bound an area without the $x$-axis being relevant.
3. It's easy enough to integrate $\int_{0}^{1}x\sqrt{1-x^{2}}dx$ directly, no polar coordinates involved; the answer is exactly 4 times this value.


In the figure, $ \triangle ABC$ is a right triangle with right angle at $ B$. $ \overline {BD}$ is altitude of $ \triangle ABC$, and $ \overline{BE}$ is a median of $ \triangle ABC$. Find the ratio of the area of $ \triangle BEA$ to the area of $ \triangle BDA$. Express your answer as a common fraction.

[asy]pair A,B,C,D,E;
A=(0,0); E=(10,0); C=(20,0); B=(13,9); D=(13,0);
draw(A--C--B--cycle);
label("$A$",A,W);
label("$E$",E,S);
label("$D$",D,S);
label("$C$",C,S);
label("$B$",B,N);
label("8",(6.5,4.5),NW);
label("6",(16.5,4.5),NE);
draw(B--E);
draw(B--D);[/asy]

First, notice that $ \triangle ABE$ and $ \triangle ABD$ share the same height. Therefore, the ratio of the areas is just the ratios of the bases which is $ \frac{AE}{AD} \equal{} \frac{5}{AD}$.

Now to find $ AD$ just look at the similar triangles to get

$ \frac{8}{AD} \equal{} \frac{10}{8}$

$ 64 \equal{} 10 \cdot AD$

$ AD \equal{} \frac{32}{5}$

$ \therefore \frac{\triangle BEA}{ \triangle BDA} \equal{}  \frac{AE}{AD} \equal{} \frac{5}{\frac{32}{5}} \equal{} \boxed{\frac{25}{32}}$


Hello guys,
One of my friends has a small plane (about 20 cm long) and he is wondering if he is able to integrate a remote control system? Theoritacally the plane is fast and powerful and it actually flies. But he wants RC and i decided to ask here :)
thanks
JgFi

Hmm I couldn't help but noticing the "theoretically." If it's a model plane, I would not advise your friend to build an RC, since that would mean "dissecting" the plane to install wires, transmitters, etc. all of which could ruin the aerodynamics.

Yes, but if there is no RC, the plane is not controlled :)

Does the plane already have the machinery to make it fly?  How does it fly (e.g. gas, electricity,...)?  I would suggest that he take it to someone that has experience with this kind of stuff in person.  If it's already got the engine and stuff, I'm pretty sure he would be able to make it remote controllable.  That's not to say it would necessarily be easy.

Even if it already has it's own motor, maybe he could just stick the guts of an old zip-zap in it to steer?
I know you don't have any control over how much it steers, but they do make little cars with good servos...


linear algebra
matrix
vector
probability
linear algebra open

Find all symmetric matrices $A$,with entries in $\{0,1\}$ (and all entries on main diagonal to be zero) such that there exists a non-zero $(0,1)$ vector in the rowspace of $A$ which does not occure as a row of $A$.

Great problem  :wink: !!!
I thought on this problem for hours and nothing happened. Some case of the problem is obvious that there exist two vertices $u,v$ which they don't have any common neighbors. :|

Although I don't know a graph theoretic formulation of the question, these matrices are actually quite common.

In fact, if you choose a symmetric $(0,1)$ matrix at random (for example by flipping a coin independently for every entry above the diagonal, and then picking the other entries to make the matrix symmetric), then the probability that the matrix is singular turns out to approach 0 as the matrix gets larger.  This means that with probability approaching 1 the matrix will be nonsingular, and therefore have plenty of $(0,1)$ vectors in the rowspace.

If we choose elements of the matrix with a Bernouli test it is not clear that rows of the obtained matrix are distinct. So it does not help to this problem.

That's true, but the probability that the rows are distinct approaches 1 for large dimensions. The probability that any particular pair of rows is equal is $2^{-n}$, and there are only $\frac{n(n-1)}2$ pairs of rows.

Thanks for your post.But It would be very nice if you can solve it for any matrices!!!

No answer?????  So I am going to start crying :(

To kevinatcausa:

As I have found if we generate a graph randomly with this model (Every  possible edge has probability $p$ to exist in your graph)that $c ln n/n\leq p\leq 1/2$ which $c$ is a constant greater than 1/2,Then we have:

       $p(rank(G)=n(G)-i(G))=1-O($lnln n^-1/4$)$

    where $n(G)$ and $i(G)$ are number of vertices and number of isolated vertices in the graph.

   Also if $c$ is greater than 1 then the probabilty that your graph has isolated vertex is very low and so with probabilty 1 your graph is fullrank.(as $n$ approaches to infinity)!!!!!!

NO answer :!: Hey guys,come on.Is'nt any body who can say sth intersting about the problem :D


search
number theory proposed
number theory

Prove that $ \exists$ infinitely $ x,y,z\in N$ such that $ 1<x<y<z$ and
$ z| xy\plus{}1$
$ y|xz\plus{}1$
$ x|yz\plus{}1$

we get $ xyz | \sum xy\plus{}1$ or for some number $ k \in N$ we get $ k\equal{}\sum \frac{1}{x}\plus{}\frac{1}{xyz}$.
since $ k$ is a positve integer and we have four fractions on the right side so one of them should be at least $ \frac{1}{4}$ so we get $ x < 4$.
if $ x\equal{}2$ ,since $ z | 2y\plus{}1$  so $ z\equal{}2y\plus{}1$. and the rest is trivial.
if $ x\equal{}3$ ,since $ z |3y\plus{}1$  so either $ 2z\equal{}3y\plus{}1$ or $ z\equal{}3y\plus{}1$.and the rest is trivial.

[quote="shoki"]we get $ xyz | \sum xy \plus{} 1$ or for some number $ k \in N$ we get $ k \equal{} \sum \frac {1}{x} \plus{} \frac {1}{xyz}$.
since $ k$ is a positve integer and we have four fractions on the right side so one of them should be at least $ \frac {1}{4}$ so we get $ x < 4$.
if $ x \equal{} 2$ ,since $ z | 2y \plus{} 1$  so $ z \equal{} 2y \plus{} 1$. and the rest is trivial.
if $ x \equal{} 3$ ,since $ z |3y \plus{} 1$  so either $ 2z \equal{} 3y \plus{} 1$ or $ z \equal{} 3y \plus{} 1$.and the rest is trivial.[/quote]

I dont see how this prove that there is infinitely many solutions  :huh:

According to me, there is just one solution $ (2,3,7)$ and Algadin again posted a [u][b]wrong problem  [/b][/u]:mad:. He really should learn copying texts without errors.

u r right  :P
it doesn't have infinitely many solutions and that's what i wanted to mention. :P
and now i see that the problem was posted [url=http://www.mathlinks.ro/viewtopic.php?search_id=2101753991&t=312142]here[/url] (by shoki!).

[quote="shoki"]
if $ x = 2$ ,since $ z | 2y + 1$  so $ z = 2y + 1$. and the rest is trivial.
[/quote]

How does $ z | 2y + 1$ imply that $ z = 2y + 1$. I may be mistaken but i think it is because of this statement that you are getting finite solution. if anyone has an explanation to the above question or a more complete solution, i'd be grateful.

[color=red]Edit: I understood why. but ill just post it in case someone else has the same doubt.[/color]

$ \& y < z  \\
  \&  \implies y \le z - 1  \\
  \&  \implies 2y + 1 \le 2z - 1 < 2z $

$ z|2y + 1 \implies 2y + 1 = kz < 2z \implies k = 1 \\
\implies z = 2y+1$

Similarly,

$ \& 3y + 1 \le 3z - 2 \le 3z  \\
  \&  \implies k = 1,2  \\
  \&  \implies {\text{Either }}z = 3y + 1{\text{ or }}2z = 3y + 1 $
the rest is simple.
thanks


algorithm
function
inequalities
linear algebra
matrix
algebra
domain

I need a full detail answer for this. I'm still fuzzy on Matrices.

Find matrices x1, x2, x3, for finding the maximization (z) problem in Example 1 can be written as:

Example 1:

Maximize: z = 3x1 + 2x2 + x3

Subject to:

2x1 + x2 + x3 < 150
2x1 + 2x2 + 8x3 < 200
2x1 + 3x2 + x3 < 320

With: x1 > 0, x2 > 0, and x3 > 0

This is a linear programming problem, and appears to be in a standard form. Are you looking for an implementation of the simplex method? Try googling the words "simplex method linear programming"  or "simplex algorithm linear programming" - you'll find several useful links.

The problem is to maximize a linear function on a convex polyhedron. By the nature of linear functions, we can easily show the maximum must occur at a vertex. There are finitely many vertices; in a simple problem like this one, we could simply list them all and test them.
The simplex method is a systematic way of searching among the vertices to find the maximum. For complicated polyhedra in many dimensions, it is usually much faster than simply testing all the vertices.

As for matrices- you could easily rewrite those inequalities in matrix form, but there's no reason to unless you want to do something specific with them later.

If "circuit" should be right, then the superior bound must be reached on an open domain $ D$.
Here there is no solution.
Now we work on $ E\equal{}\bar{D}$ defined by $ 2x_1\plus{}x_2\plus{}x_3\leq{150}\cdots,x_1\geq{0},\cdots$. (we have 6 relations).
The relations 2,4,5,6 imply the relation 3.
The boudary of the set defined by the relations 1,2,4,5,6 has 6 vertices (with 0,0,0). I think that the max is obtained for 75,0,0 but I am in a hurry.

Ah, yes, I missed that because I wasn't looking for it. To be a linear programming problem expressed in a standard way (and to make valid what jmerry said) all of the $ <$ and $ >$ in the original statement should be $ \le$ and $ \ge.$


induction
real analysis
real analysis solved

How do we prove that every sequence of real numbers has a monotonous subsequence?

Let (U(n)) be a sequence of real numbers.
Let E = {U(n) / if k > n then U(k) < U(n)}

- If E is infinite :
Then, the sequence formed with the elements of E gives a decreasing subsequence.

- If E is finite or void :
Let p be the max of the indices n such that U(n) is in E (with p = -1 if E is void).
Let V(1) = U(p+1). Since U(p+1) is not in E then there exists k > p+1 such that U(k) >= U(p+1). Let V(2) = U(k). Note that k > p+1 implies that U(k) is not in E. Thus, we may use the same reasoning as above on U(k). By induction, we construct a non-decreasing subsequence of (U(n)).

Pierre.

Let x_n a real sequence not bounded. Then we can extract from x_n a increasing subsequence wich tends to +oo:

I call a subsequence of x_n a sequence x_f(n) where f:N->N strictely
increasing.

Since x_n is not bounded, by induction we construct a sequence f(n) such
that f(0)=0 and for every n \in N f(n+1)>f(n) ,
x_f(n+1)>sup(n,x_f(n)).

Moubinool

Ok, Moubi.
But how do you construct such a monotone subsequence in the case that (x_n) is bounded?
Moreover, I think you suppose that your sequence is 'not bounded from above'.
At last, I think your construction does not ensure that (x_(f(n))) is increasing. You have to define f(n+1) as one of the elements  of the set {p/ p>f(n) and x_p > x_f(n) }

Pierre.

as proved in a previous question asked again by xxxxtt every sequence contains at least one sub-sequence (in fact two!) that has a limit (either real or \pm \infty).

so the problem reduces to convergent sequences as it is obvious how to extract a strictly increasing / decreasing sequence from a sequence that tends to \pm \infty.

if (xn)n is convergent to x, then we split (xn)n into two sub-sequences (yn)n and (zn)n such that one of them is bounded above by x, say yn \leq  x and the other one is strictly bounded below by x, say zn > x.
Obviously one of these sequences has an infinity of terms, and it is convergent to x (being a sub-sequence of a convergent sequence). It is easy now to extract a monotonous sub-sequence out of it. :)

[quote="xxxxtt"]I found out in an old analysis problem book (author G Siretchi from the university of Bucharest) the following counterexample
Let xn be an enumeration of all the rational numbers from (0,1).Then xn doesnt have any monotonous subsequence.Isnt that wrong?[/quote]

I think it is.


Define g(n) as

g(n) = { 0, if [nx]=[(n-1)x] }          where 0<x<1
g(n) = { 1, otherwise         }

Show that

lim              [Sigma {g(n)/n} ] = x    [Sigma stands for sum from 1 to n]
(n-->infinity)

Sorry for not posting the question with right symbols.

Just to make it look a bit nicer :):

Define $g\left(n\right)$ as

$g\left(n\right)=\begin{cases}0 & \left\lfloor nx\right\rfloor=\left\lfloor\left(n-1\right)x\right\rfloor \\ 1 & \textrm{otherwise}\end{cases}$

for $0<x<1.$

Show that $\lim_{n\rightarrow\infty}\sum_{k=1}^{n}\frac{g\left(k\right)}{k}=x.$


inequalities function parameterization LaTeX integration calculus probability

I would like to compare the two functions in the attached file for the variables B≥1, S≥2, E→∞, and Q∈(-1,1). It is enough if i can do it for integer B, S, and E.

In numerical computation it seems that fr≥fo for all parameters but I can't find a good way to compare them analytically.

$ \text{\LaTeX}$ is enabled on this Forum; just use the standard syntax for math. mode (with dollar signs around the formulae).

As to the question itself, it seems like your functions have some combinatorial (or probabilistic) meaning. It may make sense to share it with us: more often than not, it helps to know the "physical meaning" of the problem if you want to solve it. One can manipulate misterious formulae too (at least, in principle), but it is a much harder and less attractive activity. :)

Also, you stated the recursive relations for $ p$ and $ \varphi$, but not their initial values for $ t=0$. Should I just understand that they are $ p(1,0)=1$, $ \varphi(1,0)=1$ according to the convention that the value of the empty sum is $ 0$?

Yes, the convention fits here well (so initial values for t=o are 1 for both functions). Thanks.

I agree, this looks like a probabilistic computation over a discretization of size $ \frac{1}{E}$ of a continuous space.

If you assume that $ E$ is very large and introduce the new nondimensionalized 'time' parameter $ \tau \equal{} \frac {t}{E}$, then the following approximations are reasonable
\[ p(1,t)\approx \overline{p}(\tau ) \equal{} 1 \minus{} \frac {1 \minus{} Q}{2}\int_0^\tau \overline{\phi} (\xi )^{B \minus{} 1}d\xi \\
\phi (1,t)\approx \overline{\phi}(\tau) \equal{} 1 \minus{} \int_0^\tau \overline{p}(\xi)^{S \minus{} 1}d\xi
\]
Thus we can eliminate $ \overline{\phi}$ entirely to obtain an integral equation for $ \overline{p}$:
\[ \overline{p}(\tau) \equal{} 1 \minus{} \frac {1 \minus{} Q}{2}\int_0^\tau\left[1 \minus{} \int_0^\xi \overline{p}(\eta)^{S \minus{} 1}d\eta\right]^{B \minus{} 1}d\xi
\]
and hence an approximation for $ P$:
\[ P\approx \overline{P} \equal{} \int _0^1 p(\xi )^{S \minus{} 1}d\xi
\]
so we've effectively removed the recursive element.  This is useful for analysis, but it's also useful for computation, since the recursive formulation becomes prohibitively expensive to evaluate even for moderate values of $ E$.  If you take $ B \equal{} 1$ and $ S \equal{} 2$ as a particular case, I believe you can get some exact solutions.

One bit of notation I'm puzzled about: In the definition of $ f_o$, how do we interpret $ {t \minus{} 1 \choose B \minus{} 1}$ when $ t < B$?  Just $ 0$?  It seems rather unconventional.  And for that matter, we could also have $ B > S$, which would make that sum look even less correct. Is there maybe some physical reason to have $ B\leq t\leq S$?

Yes, there is a physical reason to assume that B≤S for all interesting cases. You can also assume that B≤t≤S (this is identical to assuming that t-1 over B-1 is zero if t<B.

Thanks.

Ok, so if my algebra is correct then
\[ \overline{p}(\tau) \equal{} \left(\frac {d}{d\tau}u(\tau)\right)^{\frac {1}{S \minus{} 1}}
\]
where on $ u: (0,1)\to \mathbb{R}$ satisfies the nonlinear ODE
\[ \frac {(Q \minus{} 1)(S \minus{} 1)}{2}(1 \minus{} u)^{B \minus{} 1}\left(u'\right)^\frac {S \minus{} 2}{S \minus{} 1} \equal{} u''
\]
If $ B \equal{} 1$ or $ S \equal{} 2$ the solution has a simple form, and I think the inequality you want will follow (at least in the limit $ E\to \infty$).  Otherwise, it's a mess.  Like fedja said, if it's a probabilistic computation someone here may have better insight if you share the original problem rather than just the discretization.

Thanks.

btw, for B=1 the two equations are equal (I can calculate the discrete version), so the inequality hols weakly.

The problem comes from a probabalistic process over a network, so it is discrete to start with.
Think of a bipartite graph, lets call one group of vertices X and the other group of vertices Y (edges can be established only between a vertices of different groups). Each vertex from group X is connected to B vertices of group Y and each vertex of group Y is connected to S vertices of group Y. A vertex of group X is active with probability (1-Q)/2 and a vertex of Y group is active with probability 1. There are E edges in the graph in total.
Edges in the graph are chosen sequentially. If an edge is chosen and both vertices at its endpoints are active they both become not active. If one of them is not active before the edge is chosen, then no change occur. The process continues until all edges were chosen.
Both functions fr and fo represent (for different network structures) the probability that a vertex of group X that is active at the begining of the process will become not active during this process with its current network position, but would have stayed active if it wasn't connected to a given vertex of group Y that it is now connected to.
The fr function corresponds to a network in which two vetices of group X cannot be connected to the same two vertices of group Y, and the fo function corresponds to a network in which two vertices of group X that have a connected vertex of group Y in common also have all of their other connected vertices in group Y in common.
I can specify a bit more how i got to the formulation of fr and fo, but it is discrete at each step of the way as it is a description of the process over a graph.
Thanks again.

After spending some time trying to prove the inequality, I decided to look at some simple case, namely, at the case $ S\equal{}B\equal{}2$, $ Q\to \minus{}1$, $ E\to\infty$. Let $ \alpha\equal{}\frac{1\minus{}Q}2$, $ \beta\equal{}\frac{1\plus{}Q}2$. Then the recursive relations turn into the system of ODE: $ p'\equal{}\minus{}\alpha \varphi$, $ \varphi'\equal{}\minus{}p$, $ p(0)\equal{}\varphi(0)\equal{}1$, $ t\in[0,1]$. If $ \alpha\approx 1$, we have $ p(t)\approx \varphi(t)\approx e^{\minus{}t}$. Hence $ P\approx 1\minus{}e^{\minus{}1}$ and $ f_r\approx e^{\minus{}1}(1\minus{}e^{\minus{}1})$. Now $ f_0\equal{}\frac 14(\alpha/\beta)^2\beta^2\approx \frac 14$. Thus, in this case, $ f_0>f_r$. Am I missing something? :? (I find it hard to believe that fi_bos has not considered this case, which is one of the very few analytically solvable instances of the problem).


A set partition of a set $ S$ is a way of writing $ S$ as a (unordered) disjoint union of nonempty sets.  For example, there are 5 set partitions of the set $ \{a, b, c\}$: $ abc$, $ ab\cup c$, $ ac\cup b$, $ a\cup bc$, and $ a\cup b\cup c$, where we suppress set brackets for notational simplicity.  the Let the Bell number $ B_n$ denote the number of set partitions of a set with $ n$ elements.

1) Show that $ B_n \leq n!$ for all $ n \geq 0$.

2) Are there constants $ c > 1$ and $ d, e > 0$ such that $ B_n < d\cdot c^n \plus{} e$ for all $ n$?

3) Reprove 1 in the following way: a permutation $ a_1a_2\ldots a_n$ is said to be "bad" if there exists $ 1 \leq i < j \leq n\minus{} 1$ such that $ a_i < a_{j \plus{} 1} < a_j$.  Show that the number of non-bad permutations of $ \{1, \ldots, n\}$ is $ B_n$.

[hide="1"] The number of partitions of a set with $ n$ elements is at most $ n$ times the number of partitions of a set with $ n \minus{} 1$ elements; one can add in the new $ n^{th}$ element to a given partition in at most $ n$ ways (since a partition consists of at most $ n \minus{} 1$ parts and we can add a new partition).  The result follows by induction and the observations that $ B_0 \equal{} 1, B_1 \equal{} 1$. [/hide]
[hide="2"] No.  This is sort of cheating, but $ B_n$ has the exponential generating function $ e^{e^x \minus{} 1}$ which is not bounded by a function of the form $ ae^{cx}$ for any $ a, c$. [/hide]

Ah, the simple proof of 1 hadn't even occurred to me.  (I was thinking of a natural surjection from permutations to set partitions.)  As for number 2, [quote="t0rajir0u"]This is sort of cheating[/quote]  Yes :P -- can you find a non-generating function proof?  (Also, I added part 3 after you posted, if you're interested in something a bit trickier (I think).)


prime numbers
number theory
game
invariant
IMO Shortlist

Four integers are marked on a circle. On each step we simultaneously replace each number by the difference between this number and next number on the circle, moving in a clockwise direction; that is, the numbers $ a,b,c,d$ are replaced by $ a\minus{}b,b\minus{}c,c\minus{}d,d\minus{}a.$ Is it possible after 1996 such to have numbers $ a,b,c,d$ such the numbers $ |bc\minus{}ad|, |ac \minus{} bd|, |ab \minus{} cd|$ are primes?

[color=darkblue]Answer. [/color][color=red]No.[/color]
[color=darkblue]Solution[/color]. Obviously, after the first step the sum of $ a,b,c$ and $ d$ is zero.So after $ 1996$ steps we will have  that $ a\plus{} b \plus{} c \plus{} d \equal{} 0$.
So we get the following relations:
$ ab \minus{} cd \equal{} ab \plus{} c(a \plus{} b \plus{} c) \equal{} (c \plus{} a)(c \plus{} b)$
$ ac \minus{} bd \equal{} ac \plus{} b(a \plus{} b \plus{} c) \equal{} (b \plus{} a)(b \plus{} c)$
$ bc \minus{} ad \equal{} bc \plus{} a(a \plus{} b \plus{} c) \equal{} \minus{} (a \plus{} b)(a \plus{} c)$
So we get that:
$ |ab \minus{} cd|\cdot|ac \minus{} bd|\cdot|ad \minus{} bc| \equal{} (a \plus{} b)^2(b \plus{} c)^2(c \plus{} a)^2$
But the product of three primes can't be a perfect square, so the answer is no.

[b][hide="Opinion about the following problem"]A too easy problem, I can't see how could it get into a shortlist.[/hide][/b]

Indeed, the problem is quite simple. If I am not missing something, it's clear that after at most four moves all the numbers must be even (simple casework), and the numbers will thereafter remain even; then each of the differences in question is divisible by 4, so they are obviously not prime. QED

Really.....when you see this kind of problems you feel that the quality of ISL is decreasing,but when you see certain other problems you really want to have a go at the ISL problems

Trivial. Suppose that at the $1995^{\text{th}}$ step the numbers on the circle are $w,x,y,z$ (in that order). Then we have $$|bc-ad|=|(x-y)(y-z)-(w-x)(z-w)|=|(w-y)(w+y-x-z)|$$
$$|ac-bd|=|(w-x)(y-z)-(x-y)(z-w)|=|(w-y)(x-z)|$$
$$|ab-cd|=|(w-x)(x-y)-(y-z)(z-w)|=|(x-z)(w+y-x-z)|$$
For these three numbers to be prime, we must have either $|w+y-x-z|=|w-y|=1$ or $|w+y-x-z|=|x-z|=1$ or $|x-z|=|w-y|=1$, all of which contradict the fact that $1$ is not a prime. Done.


A point $(x,y)$ is randomly picked from inside the rectangle with vertices $(0,0), (4,0), (4,1),$ and $(0,1)$. What is the probability that $x<y$?

$\left(A\right) \frac{1}{8}$
$\left(B\right) \frac{1}{4}$
$\left(C\right) \frac{3}{8}$
$\left(D\right) \frac{1}{2}$
$\left(E\right) \frac{3}{4}$

[hide="solution"]Graphing the points and the line $y=x$, we find the area of the rectangle is 4 and the set of points is a triangle with area $\frac{1}{2}$, so our answer is $\frac{1}{8}=\boxed{A}$[/hide]


The following inequality appeared in one of Titu Andreescu's inequality lectures at MOP 2004.  The problem was created by Gabriel Dospinescu and has had me stumped for some time.

Show that [tex]\forall x, y, z, a, b, c \in \mathbb{R}^{+} \,| \,xy + yz + zx = 3[/tex], we have the inequality

[tex]\displaymode{
\left(\frac{a}{b+c}\right)\left(y+z\right) + \left(\frac{b}{c+a}\right)\left(z+x\right) + \left(\frac{c}{a+b}\right)\left(x+y\right) \ge 3
}[/tex]

I don't even know where to begin. Help?

I created that one from a nice inequality from Crux. Try to homogenise with respect with x,y,z.THen dehomogenise by taking x+y+z=1 ( I just adore this trick with homogenuous-dehomogenuous) and then Cauchy-Schwarz will he very helpful. Anyway, the homogenuous problem was already solved, but long time ago and I don't know where. So, if these hints don't suffice, I will write the solution. Also Vasc has found an even stronger inequality, but it's really nice that it is quite hard to show that stronger inequality is really stronger than this one.

You can find the ASTONISHING, SURPRISING and FANTASTIC solution of Harazi in the topic : "After a famous CRUX inequality" in Solved Problems Section :lol: .

Harazi, can you please post the Vasc strong version of this inequality?

I am very interested to see it. :maybe:

Thank you very much.

It says that the LHS is at least $\sqrt{(x+y)(x+z)}+\sqrt{(y+x)(y+z)}+\sqrt{(z+x)(z+y)}-(x+y+z) $. In fact what Namdung did with the nice inequality from Crux is the stronger version. But I like my version of the problem!  :)  :)


If $ a,b,c   \ge 0$  then prove:

$ \frac{(a\plus{}b)^5\minus{}a^5\minus{}b^5}{5ab(a\plus{}b)}\plus{}\frac{(b\plus{}c)^5\minus{}b^5\minus{}c^5}{5bc(b\plus{}c)}\plus{} \frac{(c\plus{}a)^5\minus{}c^5\minus{}a^5}{5ca(c\plus{}a)} \ge 3(ab\plus{}bc\plus{}ca)$.


A new user by the name of hellobello has been spamming random forums over this period of time.

Can any admins take care of this issue and ban his account? Thanks.

[url]http://www.artofproblemsolving.com/Forum/search.php?search_author=hellobello[/url]

I would like to bring to attention the fact that hellobello and smilesalltheway had e-mail addresses that were remarkably similar, which could indicate something.

[u]EDIT[/u]
@PowerOfPi: hellobello was already banned early on today

Something similar happened with the user Farmstate. Just ignore them and they will eventually be deleted.


Prove that if $ x,y,z$ are real numbers and $ x\plus{}y\plus{}z\equal{}2$ and $ xy\plus{}yz\plus{}zx\equal{}1$ then $ x,y,z \in [0,\frac{4}{3}]$

[hide]$ z\equal{}2\minus{}x\minus{}y\implies x^2\plus{}xy\plus{}y^2\minus{}2x\minus{}2y\plus{}1\equal{}0$

Write this as $ \left(x\plus{}{y\over 2}\minus{}1\right)^2\plus{}\left({y\sqrt{3}\over 2}\minus{}{1\over\sqrt{3}}\right)^2\equal{}{1\over 3}$

From there $ \minus{}{1\over\sqrt{3}}\leqslant{y\sqrt{3}\over 2}\minus{}{1\over\sqrt{3}}\leqslant{1\over\sqrt{3}}\iff 0\leqslant{y\sqrt{3}\over 2}\leqslant{2\over\sqrt{3}}\iff 0\leqslant y\leqslant{4\over 3}$

Similar symmetric transformations for all variables, hence the conclusion follows.[/hide]


If $ a \equal{} 2$, $ b \equal{} 3$ and $ c \equal{} 4$, what is the numerical value of the
expression $ (b\minus{}c)^2 \plus{} a(b\plus{}c)$ ?

Plugging in, we have $ (b\minus{}c)^2\plus{}a(b\plus{}c)\equal{}(3\minus{}4)^2\plus{}2(3\plus{}4)\equal{}1\plus{}14\equal{}\boxed{15}$.


geometry
circumcircle
incenter
geometry solved

Let O be the circumcenter of triangle ABC, I be the incenter of triangle ABC. If the points of contact of the mixitinear incircles of triangle ABC the circumcircle X,Y, and Z,(A mixtilinear incircle of a triangle ABC is a circle tangent to two sides of the triangle and also internally tangent to the circumcircle of that triangle.) Prove that the lines AX,BY,CZ, and IO are concurrent.

According to http://www.mathlinks.ro/Forum/viewtopic.php?t=5364 post #2, the lines AX, BY, CZ concur at the external center of similitude of the circumcircle and the incircle of triangle ABC. Of course, this center must lie on the line joining the centers of the circumcircle and the incircle, i. e. on the line IO.

  darij


Thoughts?

Any predictions on cutoff?
 The message is too small. Please make the message longer before submitting.

[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=257310]Duplicate topic[/url].

Merged and locked.


Personally [i]Eragon[/i] was horrendous . The special effects -ew Actor-meh, okay, but another BRITISH dude? Why can't he be from some other origin? There are always either British/ Australian ppl playing parts for any fantasy movie. It's so ANNOYING! Another thing: the movie was completely incorrect when it came to the book. Um, I doubt that Eragon was 18?!. Hello? He was 15-16-ish if I might remember( although I may have to reread Eragon again). But what do you think?

* Can't wait till Glaedr comes out in September 2008! It's not a trilogy anymore;; there are 4 books now!

Fourth book is unnamed. And the third book is [i]Brisingr[/i], according to Wikipedia... :roll:

@ZhangPeijin

...The third book, [i]Brisingr[/i] is coming out in September 20th, 2008.

The movie was O.K. for the people who hadn't read the book. I had read the book. I kept cringing. No further commentary....:wink:

[quote="ZhangPeijin"]Wait, how do you know there is going to be a fourth book though?
And yes, I knew the third book was coming out and all, it was on shur'tugal awhile back.[/quote]
[url]http://alagaesia.com/[/url]
Proudly presenting [i]Brisingr[/i] there, apparently.

[url]http://alagaesia.com/christopherpaolini.htm[/url]
gives you
[quote]
Early in 2007 he realized that the plot and characters demanded more space than could fit in one volume and that a fourth book would be necessary to give each story element the attention it deserved. What began as the Inheritance trilogy became the Inheritance cycle. Book Four will complete the story that Christopher envisioned years ago when he first outlined the adventure.[/quote]

I'm fine with Eldest, but Eragon stunk both in book and theater. That movie never should of made it to the cinema hall what with all the fake acting, C list crew, and cheap special effects. The book was the one with the crummy plotline, copying Lord of the Rings/ Star Wars in a lesser way. So yah.... Anyway, Paolini made up for it in Eldest, for the most part...

I personally didn't think the first book was that bad. Sure it had elements from LotR but what fantasy novel doesn't? There were plenty of original elements in [i]Eragon[/i]...  And I'm not sure what you mean by the copy of Star Wars thing (wasn't a big Star Wars fan during the day, so not sure), so could you please elaborate on that?

I didn't see the movie, but I predicted that it would suck after seeing the previews.

[quote="alanchou"]um, dude, how long ago did the movie come out?  like, at least a year I believe[/quote]
No duh. I saw this w/ my class last year. I just wanted to see people's opinions concerning the Inheritance series after this appalling piece of crap that they called a "movie". Our teachers thought it was the best movie; everyone else was shaking their heads in a sad, " oh no, they need a life." sort of way. ;)

Yeah, sort of. (at lifeisapain)
It was way to short for my tastes, and the movie seemed totally unrelated to the book.
The acting was really cheesy too, and the massiveness they create in LoTR isn't there.
Horrible movie.
They call an "army" 50 or so actors poorly dressed in "medieval" plastic "armor"...

I hope they have a midnight party; i'd be going to that! ( like harry potter).

[quote="lifesapain"]The majority of the plot is a copy of Star Wars IV, V, and VI. Just think about it, Eragon is Luke Skywalker...[/quote]

and magic = the Force? I could buy that, I guess.

Actually, Star Wars was a rip-off of another, so I don't think Star Wars itself is original  :wink: .

Modern producers are ripping ideas off of so many sources it doesn't really have an affect on audiences. Usually, identical rip-offs are meant to be parodies.

[quote="dingzhou"]its not that the author can't make up original ideas...its that the readers wont accept some...

i mean....a giant smiley going around eating chickens and go inot a hotdog eating contest with a ninja is a perfectly good original story plot, but with the exceptions of few ppl such as me...no one would buy that book

and actuually...i thougth eldest sucked....

and the author is still a teenager....

and you shouldn't expect too much from a movie based on a book anyways...the books always better....but maybe with the exception of lord of the ring...which i have not read... :P

but i liked the movie[/quote]
Eh the books are really good too.
I think that's the only series that actually has a good book and movie.

[quote="ZhangPeijin"]
Eh the books are really good too.
I think that's the only series that actually has a good book and movie.[/quote]
True that

Though the movies make LotR a lot more accessible, also most people I talk to have not actually read the entire LotR (including all the random songs that they sing in the middle)


A circle passing through vertices $ A$ and $ B$ of triangle $ ABC$ intersects side $ BC$ again at $ D$; A circle passing through vertices $ B$ and $ C$ intersects side $ AB$ again at $ E$, and intersects the first circle again at $ F$.
Suppose that the points $ A,B,C,D$ lie on a circle centered at $ O$.

Show that $ \angle BFO \equal{}\frac{\pi}{2}$


Find a six-digit number, all of whose digits are different.
When you multiply your number by any of the counting numbers 1, 2, 3, 4, 5, or 6,
the resulting product must yield a six-digit number that contains the same set of digits
as the six-digit number you originally selected. What is your six-digit number?

[hide="insight"]Since the multiple of the six digit number is another six digit number, we need to find the greatest to know what boundaries we have.

999999/6=166666.5, so the greatest number that we can try is 166666. That doesn't work, though. So we need to find a number less than 66666 such that when you add 100000 to the number you selected, and you multiply by either 2 though 6, you get a six digit number that is a permutation of the six digits you had originally.[/hide]

I have noticed that any question asking about 6 digits, same digits when multiplying, is usually going to be the first six decimals of 1/7: 142857


This thread is for the beginners who first enter this forum to learn about the basic math skill.

----> Linear equation

Linear equation is a equation with one or more variable with degree of one.

All linear equation is in form of:

$ax+b = y$

For example:

2x+2 = 0

All the basic math you'll see on the algebra I would deal highly with the linear equation so it's important for you people to learn it.

[b]Now, how do you solve it?[/b]

All linear equation can be solved when it's isolated in one side without any other numbers, meaning only variable itself.

2x+2 = 0

We don't want 2 to be with 2x so let's move that out.

2x+2-2 = 0-2

+2-2 = 0 and 0-2 = -2

Therefore, we get 2x = -2

Since we need to find x, we can divide both sides by 2.

2x/2 = -2/2

x = -1

So, x = -1.

Let's check.

2x+2 = 0
2(-1)+2 = 0
-2 + 2 = 0
0 = 0

Correct.

One more important thing is that every linear equation only has ONE solution. It is because it has degree of one. You'll learn more about degrees on quadratic and cubic, quartic later on.

Any suggestions or ideas that add to this post are welcome. Also, ask any questions.

Silverfalcon

you had better design a story for 2x+2=0
e.g. silverfalcon has 2 apples, ....

Once upon a time..

there was a bear and an alien

the bear had two times X apples ,

but the alien has 2 apples that can kill normal apples

when they met each other , alien's apples killed the bear's apples

so the bear got mad

and killed the alien's apples

so there were none left :D

lol

I should turn into a story teller like Brom

the 'alien' and 'kill apple' concepts are good, but the whole story doesn't make sense! How do you get x=-1?


Θα μπορούσατε να δώσετε μια λύση του 3ο Θέματος Προκριματικού Μικρών, γιατί εχω κολλήσει.

Δίνονται οι μη μηδενικοί φυσικοί αριθμοί α,β,γ,δ για τους οποίους ισχύει οτι ο αριθμός
\[ \frac {{a\sqrt 2 \plus{} \beta \sqrt 3 }} {{\beta \sqrt 2 \plus{} \gamma \sqrt 3 }}
\]
είναι ρητός.
Να αποδείξετε ότι ο αριθμός
\[ \frac {{{\alpha ^2} \plus{} {\beta ^2} \plus{} {\gamma ^2}}} {{\alpha \plus{} \beta \plus{} \gamma }}
\]
είναι ακέραιος

[quote="Asynartisies"]Δεν ξέρω ακριβώς ούτε εγώ, δεν έχω προσπαθήσει τη λύση του, αλλά έχω βρει πάνω-κάτω τι να κάνεις. Στο τέλος βγαίνει νομίζω το δεύτερο κλάσμα α, που είναι ακέραιος.

Βασικά πρέπει να βγαίνει α = β = γ, γιατί στο πρώτο κλάσμα ο α(ρίζα2) είναι άρρητος, ο β(ρίζα3) είναι άρρητος κτλ (δεν ξέρω αν άρρητος + άρρητος = ρητός ή άρρητος) και πρέπει να εξουδετερώνονται, δηλ. α = β, όμως τότε και β = γ και ούτω καθ' εξής.

Μάλλον είναι λάθος, μια προσέγγιση έκανα, ας σου δώσει την κανονική λύση κάποιος που ξέρει πραγματικά.  :ninja:[/quote]
και εγω έτσι το σκεφτηκα.Λέω για να είναι ρητός πρέπει α=β=γ για να φεύγουν οι ρίζες

το κλασμα στον διαγωνισμο  ηταν $ \frac{a\sqrt{2}\plus{}b\sqrt{3}}{b\sqrt{2}\plus{}c\sqrt{3}}$
και παρατηρησε οτι δεν ειναι αναγκαστικο a=b=c .. πχ λυσεις αποτελουν και οι (1,3,9) ή (2,4,8)
σαν hint [hide]θεσε το κλασμα ως $ \frac{m}{n}$ και δες τι συμβαινει με τους ρητους και αρρητους αριθμους κτλ.[/hide]

Αν το θεσεις $ k$ το κλασμα τοτε μετα τις απαλοιφες εχεις $ \sqrt{2}(a \minus{} kb) \equal{} \sqrt{3}(kc \minus{} b)$ (η κατι παρομοιο, δεν εχω χαρτι και μολυβι). Οποτε αν $ kc \neq b$ τοτε ο ρητος $ \frac{a \minus{} kb}{kc \minus{} b} \equal{} \sqrt{\frac{3}{2}}$ που ειναι αρρητος, ατοπο. Αρα $ a \equal{} kb \equal{} k^2c \Rightarrow \frac{a}{k} \equal{} b \equal{} kc$, με $ k$ ρητο... Τα υπολοιπα ειναι απλη αλγεβρα :)
[hide]Γραψε το $ a^2 \plus{} b^2 \plus{} c^2 \equal{} (a \plus{} b \plus{} c)^2 \minus{} 2(ab \plus{} bc \plus{} ca)$ και παρατηρησε οτι με τις παραπανω υποθεσεις εχεις $ \frac{ab \plus{} bc \plus{} ca}{a \plus{} b \plus{} c} \equal{} b$[/hide]


LaTeX
inequalities unsolved
inequalities

For $a,b,c$ positive real numbers prove that

$ \sqrt[3]{(\frac{x}{y+z})^2}+\sqrt[3]{(\frac{y}{z+x})^2} + \sqrt[3]{(\frac{z}{x+y})^2} \geq \sqrt[3]{\frac{27}{4}}$

nice one...write

$\displaystyle{\sqrt[3]{\frac{x}{y+z}}=\frac{\sqrt[3]{2}}{\sqrt[3]{2(\frac{y}{x}+\frac{z}{x})^2}}}$

and use AM-GM.

Can you please post your detailed solution after Valentin has solved the Latex problem in the server.

Thank you very much.

you just have to prove that:

$\sqrt[3]{(\frac{x}{y+z})^{2}} \geq \sqrt[3]{\frac{27}{4}} \cdot \frac{x}{x+y+z}$. which is very easy

then you sum


if $ 3000!$ is divisible by $ 30^n$, what is the greatest positive integral value of $ n$ ?

[hide]
$ 30 \equal{} 2(3)(5)$
In $ 3000!$ there will be less factors of $ 5$ than $ 2$ or $ 3$ and $ 5^5 \equal{} 3125 > 3000$ so..
$ \left \lfloor \dfrac{3000}{5} \right \rfloor \plus{} \left \lfloor \dfrac{3000}{25} \right \rfloor \plus{} \left \lfloor \dfrac{3000}{125} \right \rfloor \plus{} \left \lfloor \dfrac{3000}{625} \right \rfloor \equal{} 753$
So the greatest value of $ n$ is $ 753$ [/hide]

[hide]i believe you have incorrectly evaluated the last line
$ \left \lfloor \dfrac{3000}{5} \right \rfloor \plus{} \left \lfloor \dfrac{3000}{25} \right \rfloor \plus{} \left \lfloor \dfrac{3000}{125} \right \rfloor \plus{} \left \lfloor \dfrac{3000}{625} \right \rfloor \equal{} 600\plus{}120\plus{}24\plus{}4\equal{}\boxed{748}$[/hide]
and just a minor addition, to prove that there are at least these many factors of 3, you know every third number will be divisible by 3 and every 2nd number will be divisible by 2, so there are at least 1000 factors of 3 and at least 1500 factors of 2


algebra
polynomial
number theory
greatest common divisor
Miscellaneous Problems

Suppose that both $x^{3}-x$ and $x^{4}-x$ are integers for some real number $x$. Show that $x$ is an integer.

Suppose $ x\neq0$.
We have $ \dfrac{x^4\minus{}x}{x^3\minus{}x}\in\mathbb{Q}$, from where $ \dfrac{x^2}{x\plus{}1}\equal{}q\in\mathbb{Q}$. So $ x^2\equal{}xq\plus{}q$. Then $ A\equal{}x^3\minus{}x\equal{}x(xq\plus{}q)\plus{}q\equal{}x^2q\plus{}xq\plus{}q\equal{}(xq\plus{}q)q\plus{}xq\plus{}q$. Since $ A\equal{}x^3\minus{}x\in\mathbb{Z}$, we have that $ 2xq\plus{}q^2\plus{}q\in\mathbb{Z}$, from where $ x$ is a rational number. Now, if $ x\equal{}\dfrac uv$, with $ \gcd(u,v)\equal{}1$. Suppose $ x\notin\mathbb{Z}$. Then there exists a prime number $ p$ dividing $ v$. To have $ x^3\minus{}x\in\mathbb{Z}$, we need $ p^3|u^3\minus{}uv^2$, which is impossible because $ (p,u)\equal{}1$ but $ p|v$.

I'm not sure wheater the proof is correct. How did you concluded the following equality?

$ A \equal{} x^3 \minus{} x \equal{} x(xq\plus{}q) \plus{} q$

[quote="azo"]I'm not sure wheater the proof is correct. How did you concluded the following equality?

$ A \equal{} x^3 \minus{} x \equal{} x(xq \plus{} q) \plus{} q$[/quote]
Since $ xq \plus{} q \equal{} x^2$ we have $ x^3 \minus{} x \equal{} x(xq \plus{} q) \plus{} q \equal{} x^3 \plus{} q \iff q \equal{} \minus{} x$ which is clearly wrong, so I think there is a mistake.

We can solve it with minimal polynomial - $ x$ is clearly an algebraic integer. Let $ p$ be its minimal polynomial - the monic polynomial with integer coefficients of smallest degree such that $ x$ is its zero (its existence is easy to prove, and it divides any other polynomial with these properties (except for the minimality)). By the given data, it is of degree at most $ 3$. We want to show that it is of degree $ 1$.
Let $ a_1 \equal{} x^3 \minus{} x, a_2 \equal{} x^4 \minus{} x$, and define $ p_1(t) \equal{} t^3 \minus{} t \minus{} a_1, p_2(t) \equal{} t^4 \minus{} t \minus{} a_2$.
If $ a_1 \equal{} 0$ or $ a_2 \equal{} 0$, we get that $ x \equal{} 0$ or $ x \equal{} \pm 1$ and we're done. Otherwise-
Notice that $ p_2(t) \minus{} tp_1(t) \equal{} t^2 \plus{} t(a_1 \minus{} 1) \minus{} a_2$. This polynomial must be divisible by the minimal polynomial. If, for contradiction's sake, we assume that its degree is $ 2$, then it is exactly $ t^2 \plus{} t(a_1 \minus{} 1) \minus{} a_2$, and it follows that $ t^2 \plus{} t(a_1 \minus{} 1) \minus{} a_2|t^3 \minus{} t \minus{} a_1, t^4 \minus{} t \minus{} a_2$.
Putting $ t \equal{} 0,1$ yields $ a_1 \minus{} a_2, a_2|(a_1,a_2)$, and it follows that $ a_2|a_1$ and then $ a_1 \minus{} a_2 \equal{} \pm a_2 \implies a_1 \equal{} 2a_2$.
Now we'll calculate the gcd:
$ (x^3 \minus{} x \minus{} a_1,x^4 \minus{} x \minus{} a_2) \equal{} (x^2 \plus{} x(a_1 \minus{} 1) \minus{} a_2,x^3 \minus{} x \minus{} 2a_2) \equal{} (x^2 \plus{} x(a_1 \minus{} 1) \minus{} a_2,x^2(1 \minus{} a_1) \plus{} x(a_2 \minus{} 1) \minus{} 2a_2)$
and unless $ (1 \minus{} a_1,a_2 \minus{} 1, \minus{} 2a_2) \equal{} C(1,a_1 \minus{} 1, \minus{} a_2)$ for some constant $ C$, the gcd is of degree 1 and we're done. But $ C \equal{} 2$ (because of the coefficient of $ x^0$) so if follows that $ a_1 \equal{} \minus{} 1,a_2 \equal{} \minus{} 3$. But it contradicts $ a_1 \equal{} 2a_2$, so the gcd is indeed linear.

[quote="freemind"][quote="azo"]I'm not sure wheater the proof is correct. How did you concluded the following equality?

$ A \equal{} x^3 \minus{} x \equal{} x(xq \plus{} q) \plus{} q$[/quote]

Thanks for pointing out. My proof is certainly wrong. No idea how could I have concluded that  :D

Anyway, let me try once more and hope I don't mess it up again :)

So $ x^3 \minus{} x \equal{} x\cdot x^2 \minus{} x \equal{} x(xq \plus{} q) \minus{} x$ $ \equal{} x^2q \plus{} qx \minus{} x \equal{} (xq \plus{} q)q \plus{} qx \minus{} x \equal{} (q^2 \plus{} q \minus{} 1)x \plus{} q^2$, yielding $ x$ is rational and I think we can continue in the same way.

This seems way too simple, so sorry if I am wrong again...[/quote]
Your last solution is good!
[quote]Notice that $ p_2(t) \minus{} tp_1(t) \equal{} t^2 \plus{} t(a_1 \minus{} 1) \minus{} a_2$. This polynomial must be divisible by the minimal polynomial.[/quote]
Why is that?

The minimal polynomial divides $ p_2$ and $ p_1$, thus it divides any linear combination of them (and also their GCD).
Proof that the minimal polynomial of $ a$, $ p$, divides any polynomial $ q$ in $ Z[X]$ that $ a$ is its zero:
$ g(t)\equal{}(p,q)$ is of degree less than\equal to $ p$'s degree, and $ g(a)\equal{}0$ (by Bizut Lemma: g is a linear combination of $ p$ and $ q$. put x=a).
Thus, from the minimality of $ p$, $ g$ and $ p$ are the same up to a constant (if they are of the same degree but not the same up to a constant, you can find a linear combination of them of degree less than their degree, and it is 0 at $ x\equal{}a$), which means that $ p|q$, QED.


Non-zero numbers are arranged in $n \times n$ square ($n>2$). Every number is exactly $k$ times less than the sum of all the other numbers in the same cross (i.e., $2n-2$ numbers written in the same row or column with this number).
Find all possible $k$.

[i]Proposed by D. Rostovsky, A. Khrabrov, S. Berlov [/i]

Here is my solution. I hope it's correct:

1. All odd k are possible
Proof: Fill in the square only 1's. Then Every number is 2n-3 times less than the sum of all the other numbers in the same cross.

2. Only odd k are possible
Proof: Let a(i,j) be the the number in the i-th row and the j-th column. Then we have $ (k\plus{}3)a(i,j)\equal{}\sum_{m\equal{}1}^{n}a(i,m)\plus{}\sum_{m\equal{}1}^{n}a(m,j)$
Therefore $ \sum_{i\equal{}1}^{n}(k\plus{}3)a(i,j)\equal{}\sum_{i\equal{}1}^{n}\sum_{m\equal{}1}^{n}a(i,m)\plus{}n\sum_{m\equal{}1}^{n}a(m,j)$

And so $ \sum_{i\equal{}1}^{n}\sum_{j\equal{}1}^{n}(k\plus{}3)a(i,j)\equal{}n\sum_{i\equal{}1}^{n}\sum_{m\equal{}1}^{n}a(i,m)\plus{}n\sum_{j\equal{}1}^{n}\sum_{m\equal{}1}^{n}a(m,j)$

$ \minus{}> k \equal{}2n\minus{}3$

[quote="Max(i)"]Here is my solution. I hope it's correct:

1. All odd k are possible
Proof: Fill in the square only 1's. Then Every number is 2n-3 times less than the sum of all the other numbers in the same cross.
[/quote]
Sorry to disappoint you, but it is not corect. If you fill in the squares only with $1$'s you can't get any odd k, you will get exactly $2n-2$. In fact, the answer is $k=2n-2$ and $k=n-2$.

Proof:Let $a_{ij}$ be the number in the cell placed in the i-th row and j-th column, $R_{i}$ the sum of the elements in the i-th row,  $C_{j}$ the sum of the elements in the j-th column, and $S$ the total sum.
By hypothesis, we have:
$(k+2)a_{ij}=R_{i}+C_{j}$.
Summing up after $i$, we get:
$(k+2)C_{j} = k\sum_{i=1}^{n}a_{ij} = \sum_{i=1}^{n}(R_{i}+C{j}) =  \sum_{i=1}^{n}R_{i} +  \sum_{i=1}^{n}C_{j} = S + nC_{j}$
and $(k-n+2)C_{j} = S$
We split it into 2 cases:

$1. k-n+2=0$ Then $k=n-2$ and $S=0$.
For an example, take all the rows to have identical components on each of them, and their sum to be 0. Then $C_{i} = 0$ and $a_{ij} = \frac{R_{i}}{n}$.

$2. k-n+2$ nonzero. We also get $S$ nonzero.
Then we have $C_{j} = \frac{S}{k-n+2}$ and, summing up after j, we get $\sum_{j=1}^{n} C_{j} = n\frac{S}{k-n+2}$.
So
$S = \frac{nS}{k-n+2}$ and, from here:

$n=k-n+2 , k=2n-2$
We can use the example stated at the beginning, the table filled up only with $1$'s.

Answer is $2n-2,n-2,-2$. Construction are:
[list]
[*] $2n-2$: All numbers of the table equal.
[*] $n-2$: It is not hard to get that it suffices to ensure the sum in any perfect matching is $0$. Now take all numbers of the first row to be $n-1$ and all other numbers of the table to be $-1$.
[*] $-2$: It is easy to get that it suffices to ensure the sum in each line is $0$. Just fill the inner $(n-1) \times (n-1)$ table randomly, and other numbers get determined uniquely (it is easy to ensure all numbers are non-zero by choosing the original $(n-1)^2$ numbers to be positive).
[/list]
Now FTSOC assume $k \ne 2n-2,n-2,-2$. Let $S$ be the sum of all numbers in the table. Using the hypothesis for all numbers of the table and using $k \ne 2n-2$ gives $S = 0$. Now consider any $a_1,\ldots,a_n$ forming a perfect matching $b_1,\ldots,b_n$ in a straight line. Using $k \ne -2,n-2$ respectively yields $\sum a_i = \sum b_i = 0$. But $\sum b_i = 0$ is enough to force $k=-2$, contradiction. $\blacksquare$


Determine all positive integers which are equal to the square of their number of positive divisors (including 1 and the number itself).

In other words : For which n is n = d(n) ?

The forum collapsed twice while I was typing my answer  :evil:  :evil:

I'm a bit tired now !

[hide]

I think n = 1 and n = 9 are the only solutions

n is a square : n = prod pi^2ai

d(n) = prod(1+2ai)

All the pi are odd.

p^2a > (1+2a) for p >= 5 so one of the pi is 3.

3^2a * p^2b > (1+2a) (1+2b)

So we have to solve 3^a = (1+2a). a=0 or a=1 are the only solutions.[/hide]

Indeed. Essentially the same solution can be found at

[url]http://www.kalva.demon.co.uk/canada/casoln/csol993.html[/url]


integration
limit
real analysis
real analysis solved

Here are some problems from the Romanian MO (earlier years):

P1 f:R->R continuous and periodic of period T.
a)  \int {a to a+T} f(x)dx= \int {0 to T} f(x)dx
b) limit n->infinity of  \int {a to b} f(nx)dx = (b-a)/T*  \int {0 to T} f(x)dx

Maybe they won't be too hard, but they are interesting.

Another version

f:R--->R continous,

g:R--->R continous T-periodic then

 \lim_{n-->+\infty} \int_[a;b] f(t)g(nt) dt = 1/T. \int_[a;b]f .\int_[a;b]g

what grobber forgot to say is that the period T is >0. :D :D :D (at least that was how the problem ws given at that contest) :D

Solution:

a) we put g:R-->R, g(x)= \int {x,x+T} f(t)dt and we get that g'(x)=0, which means that g is constant on R and so g(x)=g(0)=\int {0,T} f(t)dt.

b) We haver that:

Int_n= \int {a,b} f(nx)dx = 1/n \int {na,nb} f(t)dt=1/n ( \int {na,na+T} f(t)dt+ \int {na+T,na+2T} f(t)dt+...+ \int {na+(k_n-1)T,na+k_n*T} f(t)dt+ \int {na+k_n*T,b} f(t) dt)= 1/n (\int {na+k_n*T,nb} f(t)dt+ k_n* \int {0,T} f(t)dt) where we have that k_n=[n(b-a)/T].
from the fact that: na+(k_n+1)T>nb we get that:

| \int {na+k_nT,nb} f(t)dt| <= \int {na+k_nT,nb} |f(t)| dt <= \int {na+k_nT,na+(k_n+1)T} f(t)dt= \int {0,T} f(t)dt.
 that's all

cheers! :D :D :D