geometry
function
inradius
modular arithmetic
rectangle
inequalities
national olympiad

Download [url=http://www.mathlinks.ro/Forum/download.php?id=2007]PDF Version[/url] of the problems.

[hide="Test 1 - 31 March 2005"]
[url=http://www.mathlinks.ro/Forum/viewtopic.php?t=31947]Problem 1[/url]
Solve the equation $3^{x}=2^{x}y+1$ in positive integers.

[url=http://www.mathlinks.ro/Forum/viewtopic.php?t=31948]Problem 2[/url]
Let $n\geq 1$ be an integer and let $X$ be a set of $n^{2}+1$ positive integers such that in any subset of $X$ with $n+1$ elements there exist two elements $x\neq y$ such that $x\mid y$. Prove that there exists a subset $\{x_{1},x_{2},\ldots, x_{n+1}\}\in X$ such that  $x_{i}\mid x_{i+1}$ for all $i=1,2,\ldots, n$.

[url=http://www.mathlinks.ro/Forum/viewtopic.php?t=31950]Problem 3[/url]
Prove that if the distance from a point inside a convex polyhedra with $n$ faces to the vertices of the polyhedra is at most 1, then the sum of the distances from this point to the faces of the polyhedra is smaller than $n-2$.[/hide]
[hide="Test 2 - 1 April 2005"]
[url=http://www.mathlinks.ro/Forum/viewtopic.php?t=32029]Problem 1[/url]
Prove that in any convex polygon with $4n+2$ sides ($n\geq 1$) there exist two consecutive sides which form a triangle of area at most $\frac 1{6n}$ of the area of the polygon.

[url=http://www.mathlinks.ro/Forum/viewtopic.php?t=32028]Problem 2[/url]
Let $m,n$ be co-prime integers, such that $m$ is even and $n$ is odd. Prove that the following expression does not depend on the values of $m$ and $n$: \[\frac 1{2n}+\sum^{n-1}_{k=1}(-1)^{\left[ \frac{mk}n \right]}\left\{ \frac{mk}n \right\}.\] [url=http://www.mathlinks.ro/Forum/viewtopic.php?t=32027]Problem 3[/url]
A sequence of real numbers $\{a_{n}\}_{n}$ is called a [i]bs[/i] sequence if $a_{n}= |a_{n+1}-a_{n+2}|$, for all $n\geq 0$. Prove that a bs sequence is bounded if and only if the function $f$ given by $f(n,k)=a_{n}a_{k}(a_{n}-a_{k})$, for all $n,k\geq 0$  is the null function.[/hide]
[hide="Test 3 - 19 May 2005"]
[url=http://www.mathlinks.ro/Forum/viewtopic.php?t=34119]Problem 1[/url]
Let $A_{0}A_{1}A_{2}A_{3}A_{4}A_{5}$ be a convex hexagon inscribed in a circle. Define the points $A_{0}'$, $A_{2}'$, $A_{4}'$ on the circle, such that \[A_{0}A_{0}' \parallel A_{2}A_{4}, \quad A_{2}A_{2}' \parallel A_{4}A_{0}, \quad A_{4}A_{4}' \parallel A_{2}A_{0}.\] Let the lines $A_{0}'A_{3}$ and $A_{2}A_{4}$ intersect in $A_{3}'$, the lines $A_{2}'A_{5}$ and $A_{0}A_{4}$ intersect in $A_{5}'$ and the lines $A_{4}'A_{1}$ and $A_{0}A_{2}$ intersect in $A_{1}'$.

Prove that if the lines $A_{0}A_{3}$, $A_{1}A_{4}$ and $A_{2}A_{5}$ are concurrent then the lines $A_{0}A_{3}'$, $A_{4}A_{1}'$ and $A_{2}A_{5}'$ are also concurrent.

[url=http://www.mathlinks.ro/Forum/viewtopic.php?t=34122]Problem 2[/url]
Let $ABC$ be a triangle, and let $D$, $E$, $F$ be 3  points on the sides $BC$, $CA$ and $AB$ respectively, such that the inradii of the triangles $AEF$, $BDF$ and $CDE$ are equal with half of the inradius of the triangle $ABC$. Prove that $D$, $E$, $F$ are the midpoints of the sides of the triangle $ABC$.

[url=http://www.mathlinks.ro/Forum/viewtopic.php?t=34123]Problem 3[/url]
Let $P$ be a polygon (not necessarily convex) with $n$ vertices, such that all its sides and diagonals are less or equal with 1 in length. Prove that the area of the polygon is less than $\frac{\sqrt 3}2$.[/hide]
[hide="Test 4 - 23 April 2005"]
[url=http://www.mathlinks.ro/Forum/viewtopic.php?t=34583]Problem 10[/url]
Let $a\in\mathbb{R}-\{0\}$. Find all functions $f: \mathbb{R}\to\mathbb{R}$ such that $f(a+x) = f(x)-x$ for all $x\in\mathbb{R}$.

[url=http://www.mathlinks.ro/Forum/viewtopic.php?t=34584]Problem 11[/url]
On the edges of a convex polyhedra we draw arrows such that from each vertex at least an arrow is pointing in and at least one is pointing out.
Prove that there exists a face of the polyhedra such that the arrows on its edges form a circuit.

[url=http://www.mathlinks.ro/Forum/viewtopic.php?t=34585]Problem 12[/url]
Let $n\geq 0$ be an integer and let $p \equiv 7 \pmod 8$ be a prime number. Prove that \[\sum^{p-1}_{k=1}\left \{ \frac{k^{2^{n}}}p-\frac{1}{2}\right\}= \frac{p-1}2 .\] [url=http://www.mathlinks.ro/Forum/viewtopic.php?t=34586]Problem 13[/url]
a) Prove that there exists a sequence of digits $\{c_{n}\}_{n\geq 1}$ such that or each $n\geq 1$ no matter how we interlace $k_{n}$ digits, $1\leq k_{n}\leq 9$, between $c_{n}$ and $c_{n+1}$, the infinite sequence thus obtained does not represent the fractional part of a rational number.

b) Prove that for $1\leq k_{n}\leq 10$ there is no such sequence $\{c_{n}\}_{n\geq 1}$.[/hide]
[hide="Test 5 - 24 April 2005"]
[url=http://www.mathlinks.ro/Forum/viewtopic.php?t=34679]Problem 14[/url]
On a $2004 \times 2004$ chess table there are 2004 queens such that no two are attacking each other\footnote[1]{two queens attack each other if they lie on the same row, column or direction parallel with on of the main diagonals of the table}.

Prove that there exist two queens such that in the rectangle in which the center of the squares on which the queens lie are two opposite corners, has a semiperimeter of 2004.

[url=http://www.mathlinks.ro/Forum/viewtopic.php?t=34683]Problem 15[/url]
Let $n\geq 2$ be an integer. Find the smallest real value $\rho (n)$ such that for any $x_{i}>0$, $i=1,2,\ldots,n$ with $x_{1}x_{2}\cdots x_{n}= 1$, the inequality \[\sum_{i=1}^{n}\frac 1{x_{i}}\leq \sum_{i=1}^{n}x_{i}^{r}\] is true for all $r\geq \rho (n)$.

[url=http://www.mathlinks.ro/Forum/viewtopic.php?t=34684]Problem 16[/url]
Let $\mathbb{N}=\{1,2,\ldots\}$. Find all functions $f: \mathbb{N}\to\mathbb{N}$ such that for all $m,n\in \mathbb{N}$ the number $f^{2}(m)+f(n)$ is  a divisor of $(m^{2}+n)^{2}$.

[url=http://www.mathlinks.ro/Forum/viewtopic.php?t=34686]Problem 17[/url]
We consider a polyhedra which has exactly two vertices adjacent with an odd number of edges, and these two vertices are lying on the same edge.

Prove that for all integers $n\geq 3$ there exists a face of the polyhedra with a number of sides not divisible by $n$.[/hide]

[quote="chavez"]when will occur the second test?  :)[/quote]Today, and I've just posted the questions in the post above ;)

Hi Valentin,
What do you know about the tests for the juniors?

Thanks.

[quote="RHS"]Sorry Valentin, but what does the phrase "the distance from a point inside a convex polyhedra with n faces" in problem 3 from the first TST mean? I'm not sure I understand the English; what distance does it refer to?[/quote]Sorry I've forgotten a "to the vertices of the polyhedra"  :D

Just in case anyone's wondering, here are the complete results (the first 15 people - I got lazy :P).

Zamorzaev Alexandru (Sasha - R. Moldova) 7 7 7 7 7 7 = 42
Zahariuc Adrian (Dzeta) 7 7 2 7 7 7 = 37
Boreico Iurie (Iura - R. Moldova) 7 7 7 2 7 5 = 35
Kreindler Gabriel (Kappa) 5 7 7 1 7 5 = 32.5
Stefanescu Andrei (Andreis) 7 5.5 7 0 7 5.5 = 32
Michnea Dragos 7 7 0 0 7 7 = 28
Borsenco Ivan (Prowler - R. Moldova) 7 2 0 0 7 7 = 23
Purcaru Alin 7 7 0 0 7 1.5 = 22.5
Ungureanu Andrei 7 7 1 0 0.5 7 = 22.5
Ionescu Dragos 7 0 0 7 7 1 = 22
Banica Dan 7 0.5 0 0 7 7 = 21.5
Dicu Gabriel 7 7 6.5 0 0 1 = 21.5
Dragus Andrei 7 0 0 0 7 7 = 21
Pachitariu Marius (Ikap) 7 0.5 0 0 7 6 = 20.5
Filip Simeon (Dust - R. Moldova) 7 7 0 1 0 5 = 20

Congrats to the Moldovian team, it seems this year we (the Romanians) need a lot of luck to beat the Moldovian team :P :)

[quote="Valentin Vornicu"][quote="valeriano"]PDF and tex versions (from a model de Myth)  ;)[/quote]Sorry, but there is a good reason for which I have no included the .pdf and .tex files. Please stop doing that for now.[/quote]

Will it be inappropiate if I'd ask for this reason ? Just out of curiousity... :?

[quote="Valentin Vornicu"]Just in case anyone's wondering, here are the complete results (the first 15 people - I got lazy :P).

Zamorzaev Alexandru (Sasha - R. Moldova) 7 7 7 7 7 7 = 42
Zahariuc Adrian (Dzeta) 7 7 2 7 7 7 = 37
Boreico Iurie (Iura - R. Moldova) 7 7 7 2 7 5 = 35
...

Congrats to the Moldovian team, it seems this year we (the Romanians) need a lot of luck to beat the Moldovian team :P :)[/quote]

Romania got defeated by his small brother one more time.  :D I see some perfect scores coming up at the IMO horizon...  ;)

[quote="Tales"]when the third test will occur?  :)[/quote]This upcoming Tuesday. It's going to be a Geometry Test (only geo problems in it) :) Some students look forward to this, others are really scared :D :D

Test 3 (a full-geo test) has been posted. Discuss the problems in their appropiate topics please.

[quote="prowler"]Could somebody post the rezults of the third TST.
And when the forth will occure ?[/quote]All the results will be posted at once (the BMO team, after the last two tests). The last two tests will take place this week-end, so stay tuned in :)

[quote="Zubr"]test 3 is from ISL[/quote] :huh:

Test 4 has been posted (given today). Good luck to all contestants for the last test for Balkan MO tomorrow :)

The final test has been posted. Soon the results ;)

The results :D

Kappa 97
Dzeta 91.5
Andreis 90
Ikap 76.5
Micnea Dragos 76
Ungureanu Andrei 56
--------------------------------------------
Ganea Octavian 51.5
Ionescu Dragos (Cirpian) 49.5
Banica Dan 47
Talau Cristian 45
Banda Narcisa 44
Chirila Cezar 42.5
--------------------------------------------

All of the above students will participate in the BMO 2005 at Iasi, Romania, but only the first 6 will receive a Diploma if they receive a medal. The BMO will count as a final TST, but all the points will be divided by 2, so instead of a maximum of 40 points, one student can accumulate $\dfrac {40}2 = 20$ points.

This means that the first 5 are already qualified to the IMO 2005 :D
Most probably places 6-7-8 are in the battle for the final spot. Good luck to all contestants at the BMO 2005! :D

Hi!
The team is great!!! :)
Good luck to everybody and hope to see them all in Iasi. Btw, how did you check that fast the $5^{th}$ TST???
Congratulations to everybody!!!!

Valentin said: "This means that the first 5 are already qualified to the IMO 2005 :D".
 Harazi said: " Great!".  :D  :D  ;)  ;)

[quote="Valentin Vornicu"]The BMO will count as a final TST, but all the points will be divided by 2, so instead of a maximum of 40 points, one student can accumulate $\dfrac {40}2 = 20$ points.

This means that the first 5 are already qualified to the IMO 2005 :D [/quote]

What are the reasons for using BMO as 5th Romanian TST and not allowing non-participating students to retrieve further points? How many TSTs are there in total this year?  :)

Let's say that this year the organizers are experimenting a bit with the organization :D , for example the geometry test.

  I think they put the 6th test as the balkan so that everybody in the team (first 12) remain motivated.  Also for the sake of motivation they said there [i]might [/i]be another test in late may etc.

 orl: what non-participating students do you mean?

  So the total number of tests will be either 6 or 7, totaling 21-24 problems...

does anyone know why there is a geometry test this yr?

(Hong Kong also had one and I flunked it..  :( it seems that not only HK is having it, what's the purpose of having it?)

[quote="kappa"]well, the organizers here decided upon a geo test after the results at problem 5 at the imo last year. The conclusion was that we were not well prepared for geo:(... But the effect, beside motivating us to prepare better, was that some students who loved geometry were advantaged <--- me too :D[/quote]

For HK, the test was held because of the results at problem 1 and 5 of imo last yr...
It didn't motivate me to prepare better for geometry, but it motivated me to prepare better for other tests..

[quote="kappa"]some students who loved geometry were advantaged <--- me too :D[/quote]<--- and me too :D.

Does anyone know if we will receive the detailed scores?

[quote="kappa"]who loved geometry[/quote] and probably they still do :P  ;)

Can you post the geo test here , Valentin ?

I has been posted. A long time ago in fact!  ;)

[quote="Frozen"][quote="Valentin Vornicu"]
Talau Cristian 45
Banda Narcisa 44
Chirila Cezar 42.5
--------------------------------------------
[/quote]
As I know, they didn't even qualify after the first two tests.[/quote]Initially ...

[quote="Frozen"]ONM...Made in Romania.
I'm dissapointed that this happens. :(
What's the purpose of Tests 1 and 2?[/quote]Well it's not my place to make any comments about the choice, but I can say that since these 3 students have proved great ability in the upcoming tests, it was a good decision taking then along to continue the TSTs. A student must do well in all TSTs, not only the first two TSTs.

[quote="harazi"]Ok, let's not make a case on this. Anyway, I have a message to Valentin, he seems to have forgotten some of my previous messages: you're absolutely LAZY!  When will we have the results of mathlinks contest and the junior TST's problems?  :P  :P[/quote]I am absolutely BUSY! :) I am working on numerous things so have patience. It is the greatest virtue :P

Ok, ok, exactly the same thing you said the last time ;) . It seems you like very much this virtue... ;)

[quote="rustam"]hi guys i am fromn Uzbekistan and i was just wondering how can i get the solutions here.
Help me please I am for the just second time to be in such Forum   :([/quote]Welcome on AoPS-MathLinks  :welcomeani: The discussion about the problems is made in a separate topic for each problem. Links to where the topics are for the Romanian IMO TSTs 2005 are presetned in the [url=http://www.mathlinks.ro/Forum/viewtopic.php?p=198305#p198305]first post[/url] in this topic.

Are the junior tests up on the site yet?


rotation
analytic geometry
geometry proposed
geometry

With 1/4 circles we construct a closed path. Prove that the number of the 1/4 circles is divisible by 4. :)

How can this be true? :? Maybe I am misunderstanding it?

There are two equal circle quarters having $A,B$ as their endpoints, and the path could consist of these two alone. Since $4\not|2$.. :? Maybe the path has to be smooth (by this I understand that it should have a unique tangent in each point)?

You are right. For example, if a circle is fixed you are allowed to use  the 4 1/4 circles but you are not allowed to rotate them. An 8 is an example of such a path.

Then if my understanding of the problem is correct, we could do something like this:

To each "step" consisting of one quarter of a circle, associate the path going from its first endpoint to its second endpoint. Like I said, if my understanding of the problem is correct, we have to travel along these straight segments (I'm also assuming the quarters of circles are all equal.. :? you did not mention this and I think it's important) so that we change direction with either $\frac\pi 2$ or $-\frac\pi 2$ at each change of segment, and we have to show that there are $4k$ segments.

The problem would then be equivalent to this one: if we travel along the lines of the usual plane lattice so that we change direction whenever we change segments and we return to the point where we started from, show that we walk along $4k$ segments.

I had this as a problem when I was a junior, at a selection tset for the JBMO. Assume you start from $(0,0)$. At each step, the parity of exactly one of your coordinates changes, so it takes $4$ steps to get back to (even,even). This means that it takes a multiple of $4$ steps to get back to $(0,0)$.

Once again, I must stress that this only works if I understood the problem correctly.


Hi I have a question here involving probability.
I was wondering if this is true:  For r.v.s $ X$ and $ Y$, is $ P(X\plus{}Y\equal{}x\plus{}y)\geq P(X\equal{}x,Y\equal{}y)$
If it is true, how may one proof this?

$ \{(a,b) : (a \equal{} x) \wedge (b \equal{} y)\} \subseteq \{(a,b) : a \plus{} b \equal{} x \plus{} y\}$ and by additivity, if an event A is a subset of another event B, the probability of A must be less than or equal to the probability of event B.


topology
vector
function
Functional Analysis
real analysis
real analysis theorems

What is the significance of the [b]concept of separability in a topological space[/b]?

This is a very indirect response, and it's about functional analysis rather than topology [i]per se.[/i]

The normed vector spaces (and perhaps other topological vector spaces) that are separable (have countable dense sub[b]sets[/b]) are also those with "nice" dense sub[b]spaces[/b].

This leads to a certain style of proving theorems, perhaps especially in functional analysis in the service of differential equations. We'd like to prove some boundedness result concerning a linear operator $ L$ - maybe it's something like the inverse of a differential operator - on some normed space, perhaps a Sobolev space of some kind. It's not even clear that this operator is defined on that space - we have just a heuristic sketch of what $ L$ is supposed to be. What we do is first assume that $ f$ is "nice" - maybe $ C^{\infty}$ compactly supported or something like that. We work out a definition of $ L(f)$ just for those "nice" $ f.$ But then we work to prove a norm estimate for such functions - and once we have a norm estimate, we have that $ L$ is uniformly continuous and can be uniquely extended to a bounded operator on our whole Banach space.

There are Banach spaces that aren't separable: $ L^{\infty},$ or the space of bounded (Borel) measurse, or some of the friends and relatives of those spaces. And not being separable means we can't really use the argument I've outlined above. Some subspace of "nice" functions won't be dense in the whole Banach space, and to extend a bounded operator defined on such a subspace to the whole space would require us to invoke Hahn-Banach - and that's non-constructive and not unique.


I believe you mean...

Given: $g(x)=3x-1$, $[f \circ g](x) = 5x-1, [g \circ h](x) = 2x+4$.  f and h are linear functions.

Find: $[g \circ(h+f)](x)$

[quote="myc"]Given: $g(x)=3x-1$, $[f o g](x) = 5x-1, [g o h](x) = 2x+4$.  f and h are linear functions.

find: [g o(h+f)](x)

this is what I have so far...need help.. [hide] $f(x)=m_{1}x+b_{1}$
$h(x)= m_{2}x+b_{2}$

$f[f o g](x)=m_{1}(3x-1)+b_{1}$

$f[f o h](x)=m_{2}(3x-1)+b_{2}$

so for [g o(h+f)](x) i get a really long equation with multiple variables...

[/hide]  Thanks!  :P[/quote]

Look at the equations you wrote and compare to the ones in the problem.

let m1=a, m2=b, b1=c, b2 = d

3a=5 --> a = 5/3
c-a = -1 --> c= 2/3
So, f(x) = 5/3 x + 2/3

do this for h(x) too and voila!


\[ \sqrt {29 \plus{} 12\sqrt {5}}
\]
[hide="hint"]Try the form $ a \plus{} b\sqrt {c}$.[/hide]

We have that
\[ \left(\sqrt{a}+\sqrt{b}\right)^2=29+12\sqrt{5}.\]
Expanding,
\[ \left(a+b\right)+2\sqrt{ab}=29+12\sqrt{5}.\]
Setting things equal, we have
\begin{align*}
a+b&=29 \\
2\sqrt{ab}&=12\sqrt{5}
\end{align*}
The second equation implies $ ab=180$. Solving, we find that $ a=20$ and $ b=9$, so
\[ \sqrt{9}+\sqrt{20}=\boxed{3+2\sqrt{5}}.\]


calculus
derivative
function
analytic geometry
trigonometry
geometry
circumcircle

I  find very interesting following proposition :
[b]Lemma (A.S. Besicovitch-[1] )[/b][i]
 Given a set F of coplanar circles, the center of no one of them being in the interior of another, and   (C) a circle of G whose radius does note exceed the radius of any other circle of  F, then the number of circle meeting (C) does not exceed $ {\mathbf 18 } $. [/i]

               R E F E R E N C E S

[1] A.S. Besicovitch , [i]A general form of the covering principle and relative  differentiation of   additive functions[/i] , Proc.Cambridge.Philos.Soc., 41 (1945)  103-110 .
[2] E. F. Reifenberg ,[ ??] ,  Math.Gaz., 32 (1948) 290-292.

There are generalization of this result by Besicovitch ?
It's possible to find online a proof of above Lemma ?

Moreover, $18$ is optimal as proved by the $18$ circles with respective center in polar coordinates $(1, \frac {h \pi } 3)$ for $h=0,1, \cdots, 5$ and $(2 \cdot \cos ( \frac { \pi } {12} ) , \frac {(2k+1) \pi } {12} )$ for $k=0, \cdots , 11.$

This has been rediscovered by Bateman and Erds [1] in 1951 and thy have shown the equivalence of that result with the following one :
it is impossible to dispose more than $19$ points in a closed disk with radius $2$, one being at the center of the disk, and with mutual distances at least $1$.

Therefore, it is easy to generalize the result in this area :
Let $n \geq 2$ be an integer.
1) What is the leat radius, say $R(n)$, of a closed disk into which we can choose $n$ points, one being the center, and with mutual distances at least $1$?

2) Among all the sets of $n$ points in the plane with mutual distances at least $1$ (without any other condition on the position of the points) which ones have the minimal diameter? And what is this minimal diameter, say $D(n)$?

Clearly $D(n) \leq 2 \cdot R(n)$.

From abov, we then have $R(20) > 2$. In the same paper, they proved that $D(7) = 2$ and the only one set with this property is the set of the six vertices of a regular hexagon together with its center.

It is knwon that :
$R(2) = R(3) = \cdots = R(7) =1$
$\dsp R(n) = \frac 1 {2 \cdot \sin ( \frac {\pi } {12} ) }$ for $n = 8,9,10,11$
$R(12) = 1,68 \ldots $
$R(13) =\sqrt 3 $
$R(19) \leq 2 \cdot \cos ( \frac {\pi } {12} ) =1,93 \ldots $

and :

$D(2) = D(3 ) = 1$
$D(4) = \sqrt 2 $
$D(5) = 2 \cos ( \frac { \pi } 5 ) =$ the golden mean.
$D(6) =  2 \sin ( \frac { \pi } 5 )$
$D(7) = 2$.

Moreover, for $n=3,4,5$ th minimal diameter is reached only by the vertices of the regular convex $n$-gon with side $1$.
For $n=6$ it is reached only for the vertices of the regular pentagon and the center of its circumcircle with radius $1$.

And, asymptotically ([2],[3]) : $\dsp D(n) \sim 2R(n) \sim ( \frac {12} { \pi ^2 } )^{\frac 1 4} \sqrt n.$

References :
[1] P. Bateman, P. Erds, Geometrical extrema suggested by a lemma of Besicovitch, A.M.M., 58 (1951), p. 306-34.
[2] B. Segre, K. Mahler, On the densest packing of circles, A.M.M., 51 (1944), p.261-270.
[3] L. Fejes Toth, Extremal distributions of points in the plane, on the surface of the sphere and in the space, Acta Sci. Math. Nat., n23 (1944).

Pierre.


function
limit
logarithms
integration
probability
real analysis
real analysis unsolved

Here's a quite neat result: suppose that $ f_n$ is a bounded (in uniform norm) sequence of continuous functions on $ [0,1]$ and that if $ M_n\equal{}(f_1\plus{}...\plus{}f_n)/n$, then the function $ \sum_{n}{\frac{M_n^2}{n}}$ is integrable on $ [0,1]$. Then almost surely $ M_n$ goes to $ 0$.

Am I right that $ F_n$ almost surely goes to 0 means that
\[ \lim_{\varepsilon\to 0}mes \{x: \ |F_n(x)| > \varepsilon\} \equal{} 0?
\]
I really can't remember the definition :(

And in what sense integrable?

Oh, sorry for being imprecise. The integrability is $ L^1$ and the fact that it tends almost surely to $ 0$ means that $ M_n(x)$ tends to $ 0$ for all $ x$ except a set of measure $ 0$. I'm sorry for the vocabulary, I'm used with the french one and I guess there are some differences with the english one.

Suppose the statement is wrong. Thus there is set $ E\subset [0,1]$ of non-zero mesure (that is $ mes^*(E) > 0$) s.t. for any $ x\in E$ there is increasing sequence $ \{n_k(x)\}$ with property $ |M_{n_k(x)}(x)| > \varepsilon(x) > 0$.
Consider now those $ x\in E$ that $ \varepsilon(x) > \frac {1}{n}$, denote the coresponding set as $ E_n$. Since $ E \equal{} \cup E_n$ we conclude there is $ m$ s.t. $ E_m$ is not zero-mesuare set. Denote $ E_m$ as $ E$ and $ d \equal{} mes^*(E_m) > 0$, $ \delta \equal{} \frac {1}{m}$.
Next observation is the following. There is a constant $ s > 1$ s.t. for all sufficiently large $ n$ if $ |M_n(x)| > \delta$ then $ |M_i(x)| > \frac\delta2$ for all $ i \equal{} n,...,ns$. It comes from the fact that $ \{f_n(x)\}$ is bounded in $ C[0,1]$. Also we can conclude that $ |M_i(U(x))| > \frac\delta3$ for some neighbourhood $ U(x)$ for all $ i \equal{} n,...,ns$. It comes from the fact that $ f_n$ are continuous.
Note that $ \frac {1}{n} \plus{} \frac {1}{n \plus{} 1} \plus{} ... \plus{} \frac {1}{ns}\approx \ln s$.
 :arrow: Let $ E^p \equal{} \{x\in E: \ n_1(x)\leq p\}$. Since $ E^1\subset E^2\subset E^3\subset...$ and $ \cup E^p \equal{} E$ we obtain that there is $ p_1$ s.t. $ mes^*(E^{p_1}) > \frac d2$. Construct $ U(x)$ for each $ x\in E^{p_1}$, they cover set of mesuare at least $ \frac {d}{2}$ and for each p. $ y$ in this set we can find summands from $ \sum_{n \equal{} 1}^{p_1s}\frac {M_n^2(y)}{n}$ those sum will be at least $ \left(\frac {\delta}{3}\right)^2\cdot \ln s$. Finally
\[ \int_0^1\sum_{n \equal{} 1}^{p_1s}\frac {M_n^2(t)}{n}dt\geq \frac {d}{2}\left(\frac {\delta}{3}\right)^2\cdot \ln s.
\]
Exclude from each sequence $ n_k(x)$ all members $ \leq p_1s$ and renumber sequence from index 1, repeat process  :arrow: by constructing $ p_2$ with property $ mes^*(E^{p_2}) > \frac d2$, then
\[ \int_0^1\sum_{n \equal{} p_1s \plus{} 1}^{p_2s}\frac {M_n^2(t)}{n}dt\geq \frac {d}{2}\left(\frac {\delta}{3}\right)^2\cdot \ln s.
\]
And so on. Hence $ \int_0^1\sum_{n \equal{} 1}^{\infty}\frac {M_n^2(t)}{n}dt$ is unbounded. A contradiction.

P.S. Mostly technical proof. I doubt someone will read carefully through it.

[quote="harazi"]I'm used with the french one and I guess there are some differences with the english one.[/quote]

what you said is fine in english although people usually use 'almost sure' (a.s.) convergence in a probability context and 'almost everywhere' (a.e.) convergence in an analysis context.  in english what myth referred to is usually called 'convergence in probability' (again in a probability context) or 'convergence in measure'.


geometry
geometric transformation
reflection
conics
hyperbola
perpendicular bisector
geometry unsolved

Let $ P$ be one of the unknown tangency points, say on the left branch of the hyperbola. Its corresponding tangent $ \tau$ bisects $ \angle F_1PF_2$ internally. Let $ X$ be the reflection of  $ F_1$ about $\tau$ $\Longrightarrow$ $ X$ lies on $ PF_2.$ Note that $ F_2X$ is the lenght of the major axis, therefore $\triangle  XF_1F_2$ is constructible, since the sides $ F_1F_2,F_2X \equal{} V_1V_2$ and the angle $ \angle XF_1F_2 \equal{} 90^{\circ} \minus{} \varphi$ are known, where $ \varphi$ is the acute angle formed by the given direction $ r$ of the tangent and the focal axis $ F_1F_2.$ Such construction permits to  find the vertex $ X$ $\Longrightarrow$ $\tau$ is the perpendicular bisector of $ F_1X.$ Reflect $\tau$ about the center of the hyperbola to get the another solution.


inequalities
trigonometry
inequalities unsolved

in triangle ABC,prove
$\frac{\cos^{2}A}{sin^{2}B+sin^{2}C}+\frac{\cos^{2}B}{sin^{2}C+sin^{2}A}+\frac{\cos^{2}C}{sin^{2}A+sin^{2}B}\geq \frac{1}{2}$


algebra unsolved
algebra
system of equations


A solid 6-by-6-by-6 cube is constructed from unit cubes and the entire outer surface is painted. What fraction of the unit cubes has at least two painted faces? Express your answer as a common fraction.

There are 8 cubes with 3 faces and 12*4=48 cubes with 2 faces.

56/216=7/27.

[hide="my carefully explained answer"]ill anser this in a series of steps.

1.1st we know that there are 12 edges to a cub

2.then next we know that each corner on the cube has 3 painted edges.

3.on each edge there are 4 that are painted 2 sides

4.now that we know that there are 4 painted on 2 sides on each edge and there are 12 edges so we get the equation 4*12=48

5.now we also know that there are 8 cubes that are painted on 3 sides so we get a total of 48+8=56

6.then next we know that there are 216 unit cubes so we get the ratio  56/216

7.[hide="ANSWER"]56/216=7/27[/hide][/hide]

[hide="look"][size=200][color=#FF40FF][b][i][u] PLEASE RATE MY POST THANK YOU SO MUCH!!!!!!!!!!![/u][/i][/b][/color][/size][/hide]

:starwars:

may the force of AoPS rest within you !


For some reason, when I press F5 to output a pdf, adobe reader pops up without the pdf itself.
Please look at the powerpoint and tell me how to fix my problem.

I don't have Powerpoint nor wish to, so save to pdf instead. Adobe reader does presentations as well so best to use that than use a system that's not used by many manthematicians.

I suspect the pdf was not created and the log file will tell you that the file wasn't created and what the error was. TexnicCenter will also show the errors in the bottom window.

So we need more info - is it just this file that's a problem or all of them? Does LaTeX=>DVI work or should you be using LaTeX=>PS=>PDF? See [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=157945]Minimal example please![/url]

Attached is the bottom window where "build" is attached. Is that the log file?

Yes, but why not copy and paste from the file itself Week_3.log?

Anyway, as you'll see from the log you'll have to correct the error in the document where it looks like you don't have \end{enumerate} following \begin{enumerate} on line 72.

I fixed the error (now it says 0 errors and 0 warnings) but when I hit the view output button, the pdf didn't show up. I tried going to build, and then define output profile to set adobe reader under the path of executable. This made it so when I pressed F5, adobe reader opened but not the file itself.

[quote="stevem"]So we need more info - is it just this file that's a problem or all of them? Does LaTeX=>DVI work or should you be using LaTeX=>PS=>PDF? See [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=157945]Minimal example please![/url][/quote]
Why won't you answer the questions? It is is impossible to diagnose the problems without this information. So please tell us:
1. Do you have the same problem with other tex files or just this one?
 a) if it's only this file keep stripping out lines until you find the cause or else post a minimal example which doesn't open the pdf file.
2. If you compile to DVI using LaTeX=>DVI does that work?
3. Is the pdf actually created? Look for it, using My Computer, in the same directory as the tex file? If it's there will Adobe open it? Is there an error message?
4. What happens if you save the tex file to a path without any spaces in it such as C:\latex? Is the pdf opened there?

1. Do you have the same problem with other tex files or just this one?

All tex files.

2. If you compile to DVI using LaTeX=>DVI does that work?

No, it didn't work for the file I was working with. The image attached is what I saw. However, after I tried using LaTeX=>DVI for the "First LaTeX Document," it worked. I tried copying and pasting to a new, blank file and doing Latex=>DVI for the original file and I saw the same message from the image attached.

3. Is the pdf actually created? Look for it, using My Computer, in the same directory as the tex file? If it's there will Adobe open it? Is there an error message?

I don't know what the same directory as the tex file is. I looked where all the files I save in TeXnicCenter go. Is that the directory? The pdf is created, I can view it with adobe reader, and I can make edits to it. There is no error message when I open it.

4. What happens if you save the tex file to a path without any spaces in it such as C:\latex? Is the pdf opened there?

I don't know what a path is. The folder I currently save stuff in TeXnicCenter is sent to a folder with no spaces in its name and the pdfs do appear in that folder.

[quote="aufha"]I don't know what the same directory as the tex file is. I looked where all the files I save in TeXnicCenter go. Is that the directory? [/quote]Yes
[quote="aufha"]4. What happens if you save the tex file to a path without any spaces in it such as C:\latex? Is the pdf opened there?

I don't know what a path is. The folder I currently save stuff in TeXnicCenter is sent to a folder with no spaces in its name and the pdfs do appear in that folder.[/quote]That's clearly not true! The error message you show says the path is C:\Documents and Settings\Blues\My Documents\LaTeX which has 3 spaces in it.

It appears that the programs are objecting to folders with spaces in them. So create a new folder C:\latex then open your tex file and save it in C:\latex. When you compile to pdf does that work ok? If so, move all your files to that folder (or something similar, but with no spaces in the name) and then all should be well.

Many LaTeX programs have problems with spaces in file/folder names so best to avoid them always.

When I right click, choose new, and click folder I can't make the name with a colon.
How do make "C:\latex?"

What is a path? This thing: "C:\Documents and Settings\Blues\My Documents\LaTeX which has 3 spaces in it."
When I saw the icon for the folder I was saving stuff in, it wasn't labeled with this. It was labeled: LaTeX.
How can I make sure something has no spaces in its name?

[quote]When I right click, choose new, and click folder I can't make the name with a colon.
How do make "C:\latex?" [/quote]

Where are you right-clicking?

Anyway, go to 'My Computer' from your start menu, then double-click on C drive.
Now right-click, and go to 'New -> Folder'
Then right-click on this "New Folder", and rename it to 'latex'

[quote="aufha"]I made the folder called latex in the c drive.
The pdf is created and I can edit it but when I press F5 after building, the pdf file doesn't appear. Is it supposed to?[/quote]It should do provided the viewer setup in Tools, Define Output Profiles looks is correct and looks something like this:

Thank you very much. I manually typed all of that in and it works now. Did I mess up the installation or something or was I supposed to type that in from the start?

It should have been entered automatically by the wizard that ran when TexnicCenter first started and which asked you questions. You can always re-run the wizard if anything else is missing by pressing the Wizard button at the bottom left of that picture.


calculus
derivative
function
trigonometry
real analysis
calculus computations

Let $ f\in C^\infty(\mathbb R)$. Does there exist a sequence of real numbers $ \{c_n\}_{n\equal{}1}^\infty$ such that for all $ x_0\in \mathbb R$
\[ c_n f^{(n)}(x_0) \to 0, \quad n\to \infty\]
 :?:

Trivially yes. Just take $ c_n\equal{}\frac 1{n (1\plus{}\max_{[\minus{}n,n]}|f^{(n)}|)}$

Indeed, trivially. But is it possible to find a sequence $ c_n$ independent of $ f$ ?

No (unless you allow $ c_n$ to be $ 0$, which is, probably, not what you had in mind). For every sequence $ b_n$ of real numbers, you can construct a $ C^\infty$-function such that $ f^{(n)}(0)\equal{}b_n$.

In [color=red][b]a[/b][/color] textbook, yes. In one of the modern introductory calculus textbooks that are good mainly for making a satisfactory thump when dropped to the floor, no.

Anyway, the trick is simple if you think of it for just a moment. It is especially transparent if you just want $ |f^{(k)}(0)| > b_{k}$ where $ b_k$ is some sequence of positive numbers (which is enough to answer your question). Just choose an increasing sequence $ n_k > b_k$ of positive integers and consider $ f(x) \equal{} \sum_k b_k n_k^{ \minus{} k}[\cos n_k x \plus{} \sin n_k x]$. Once you understand how this works, the theorem I mentioned (which is due to Borel, if I don't mistake) will be not too far away.

A proof by Besicovitch of Borel's theorem is in: Krantz-Parks,  A Primer of Real Analysis, p.55

[quote="nckg"]Ok, and can I find the theorem you use in a textbook?[/quote]


Find all functions $ f: \mathbb {Z}\to\mathbb {Z}$ for which we have $f(0)\equal{}1$ and
\[ f(f(n))\equal{}f(f(n\plus{}2)\plus{}2)\equal{}n,\quad \forall n\in \mathbb {Z}.\]

[quote="mathVNpro"]Find all functions $ f:$ $ \mathbb {Z}\mapsto \mathbb {Z}$ for which we have:
\[ f(0) = 1, f(f(n)) = f(f(n + 2) + 2) = n, \forall n\in \mathbb {Z}\]
[/quote]

$ f(f(n)) = n$ shows that $ f(n)^$ is injective. So $ f(f(n)) = f(f(n + 2) + 2)$ implies $ f(n + 2) = f(n) - 2$ and so :
$ f(2p + 1) = f(1) - 2p$
$ f(2p) = f(0) - 2p$

Then $ f(0) = 1$ implies $ f(1) = f(f(0)) = 0$ and so $ f(n) = 1 - n$ $ \forall n$ and a quick check confirms that this indeed is a solution.

Hence the unique solution : $ \boxed{f(n) = 1 - n}$


geometry
function
algebra
polynomial
complex analysis
complex analysis unsolved

find number of roots of  $ z^7\minus{}2z^5\plus{}6z^3\minus{}z\plus{}1$  in $ \{|z|<1\ \ |\ z\in \mathbb{C}\}$

Let $ g(z)\equal{}z^{7}$ and $ h(z)\equal{}\minus{}2z^{5}\plus{}6z^{3}\minus{}z\plus{}1$.  Then $ \left|h(z)\right|\equal{}\left|z\right|^{7}$ and $ \left|g(z)\right|\equal{}\left|\minus{}2z^{5}\plus{}6z^{3}\minus{}z\plus{}1\right|$.

Use Rouche's theorem.

Are you sure this will work? I can't find the number of roots of this equation $ g(z)\equal{}\minus{}2z^5\plus{}6z^3\minus{}z\plus{}1$ easily in the unit disc, but I think this will work better, consider $ f(z)\equal{}6z^3$ and $ g(z)\equal{}z^7\minus{}2z^5\minus{}z\plus{}1$, then on $ S^1$ we have $ |g(z)|<|f(z)|$ and $ f$ has three roots in unit disc, so according to Rouche's theorem the number of roots of $ f$ and $ f\plus{}g$ are the same.

You could probably use the Principe of argument.

$ N_f \minus{} P_f \equal{} \frac {1}{2\pi} \Delta_\phi arg f(z)$

$ N_f$ ; number of roots
$ P_f$ ; number of pools
Let $ \phi$ be the area defined by $ |z| \equal{} R$ Further more, let R be big enough to encircle all the roots.

Okidoki
The function seems to be analytic. So all we have to do now is to find the argument variation.

$ arg (f(Z)) \equal{} arg( z^7 \plus{} \epsilon)  ; \epsilon$ = a small contribution when R is BIG
Follow the egde of $ \phi$ and you get 2$ \pi$.
$ arg( z^7 ) \equal{} 2\pi * 7$

$ N_f \minus{} P_f \equal{} \frac {1}{2\pi} 14\pi$

Since we have no pools we get:

$ N \equal{} 7$

let $ f \equal{} z^{7} \minus{} 2z^{5} \plus{} 6z^{3} \minus{} z \plus{} 1$
let $ g \equal{} 6z^{3}$ then $ |f \minus{} g| \equal{} |z^{7} \minus{} 2z^{5} \minus{} z \plus{} 1 |\leq |1 \plus{} 2 \plus{} 1 \plus{} 1| \equal{} 5 < 6|z|^{3} \equal{} 6$ for $ |z| \equal{} 1$ , since $ g$ has 3 zeros inside the unit disc so does f

[quote="yberfuzz"]Further more, let R be big enough to encircle all the roots.
...
Since we have no pools we get: $ N \equal{} 7$[/quote]  The correct spelling is "poles."  The question gave that $ R \equal{} 1$, so you don't get to choose it.  If you did, what you would have just demonstrated is that a degree-7 polynomial has 7 roots  :wink:  :P


Δίνεται ο θετικά ορισμένος πίνακας Α και η τετραγωνική μορφή
$ Q(x)\equal{}\frac{1}{2}(x,Ax)\minus{}(x,b)$
Δείξτε ότι η μορφή Q ελαχιστοποιείται στο σημείο $ x_o$,για το οποίο ισχύει
$ Ax_o\equal{}b$ και ότι $ Q(A^{\minus{}1}b)\equal{}\minus{}\frac{1}{2}(b,A^{\minus{}1}b)$

Αν δεν κάνω λάθος, για να ισχύει αυτό που λες χρειάζεται ο $ A$ να είναι συμμετρικός :huh:

Πείτε μου αν κάνω κάπου λάθος:
$ Q(x) \equal{} \frac {1}{2}\sum_{i,j}A_{ij}x_i x_j \minus{} \sum_i b_i x_i$
Εύκολα φαίνεται ότι το σημείο ελαχιστοποίησης θα έχει $ \frac {\partial Q}{\partial x_k} \equal{} 0$ οπότε:
$ \frac {\partial Q}{\partial x_k} \equal{} 0\implies$
$ \frac {1}{2}\sum_j(A_{kj}x_k \plus{} A_{jk}x_j) \minus{} b_k\implies$
$ \frac {1}{2}(A \plus{} A^\text{T})x \equal{} b$
απ'όπου και προκύπτει το ζητούμενο για τον συμμετροποιημένο πίνακα $ A_{\text{sym}} \equal{} \frac {1}{2}(A \plus{} A^\text{T})$ (ο οποίος και είναι ίσος με τον $ A$ όταν ο $ A$ είναι συμμετρικός).

Χάνω κάτι? :?:

Cheerio,

Durandal 1707

Edit: εναλλακτικά, complete the square στην αρχική παράσταση...

[quote="Durandal"]
Χάνω κάτι? :?:
[/quote]

Α θετικα ορισμένος (= positive definite) ==> A συμμετρικός

[quote="Demetres"]Α θετικα ορισμένος (= positive definite) ==> A συμμετρικός[/quote]

Huh? O πίνακας Α = [1 2; 0 1] είναι θετικά ορισμένος ($ x^T A x \equal{} x^2 \plus{} 2xy \plus{} y^2 \geq 0$), αλλά δεν είναι συμμετρικός :huh:

Εκτός αν συμπεριλαμβάνεις τη συμμετρία στον ορισμό των θετικά ορισμένων πινάκων κατ'αναλογία με τη μιγαδική περίπτωση που συνήθως μιλάς για ερμιτιανούς πίνακες... αυτό εννοείς?

Cheerio,

Durandal 1707

[quote="Durandal"]
Εκτός αν συμπεριλαμβάνεις τη συμμετρία στον ορισμό των θετικά ορισμένων πινάκων κατ'αναλογία με τη μιγαδική περίπτωση που συνήθως μιλάς για ερμιτιανούς πίνακες... αυτό εννοείς?
[/quote]

Αυτό ακριβώς εννοούσα.

Eγώ πάλι τον ορισμό τον ήξερα χωρίς τη συνθήκη συμμετρίας, αλλά... big deal ;)

Cheerio,

Durandal 1707


Hello AoPS users. I am starting a new Problem Solving Tournament. This is going to be a lot like the previous one except it will probably be a little bit harder and there will be a corresponding post in Games and Fun Factory where users can try and guess who they think is going to win and guess some other things about the tournament as well.
  For those who don't know how to play, there will be signups starting now, and the tournament will be beginning in December and ending around June, but I can't know for sure until I know how many people are playing.
  Well, these are the rules;

1. Each contestant will be given a list of questions. Getting them all right would result in a score of 50 points.
2. The contestants that become eliminated each round are the ones that have the lowest total scores at the end of each week.
3. At the top 8 there will be a round taking place at FTW.
4. The winner at the first for the win challenge will autimatically make it to the top 6.
5. The person in the top eight that has the highest score will also move on to the top 6.
6. The top 6 will also be a FTW challenge but the winner does not get any advantage.
7. The top 4 will also be a FTW challenge where the top three move on.
8. The top 2 contestants will face off in three countdown rounds and whoever wins 2 out of 3 or more will get 1 Bonus Point. The contestant that had the highest overall score at the beginning of the final round will also get 1 Bonus Point. The last Bonus point will be rewarded to the person that gets the most points in the list of questions given to them in that round.
9. The person with the most Bonus Points will be the champion.

Signups start now, and questions will be sent to them in December;

For those that are interested about how it will be narrowed down, this will be how it gets narrowed down to the top contestant;

One week; More than 100 contestants. (Not that I actually believe there will be more than 100 contestants)
The Next Week; 100 contestants
The Next Week; 75 Contestants
The Next Week; 60 Contestants
The Next Week; 50 Contestants
The Next Week; 40 Contestants
The Next Week; 30 Contestants
The Next Week; 20 Contestants
The Next Week; 19 Contestants
The Next Week; 18 Contestants
The Next Week; 15 Contestants
The Next Week; 12 Contestants
And then from there it narrows down one by one.

Signups start now;

Okay, so far there are two people that have joined the tournament.

Isabella2296
Poincare

I want to join!
Some questions:
1) What is FTW?
2) Can everybody join these tournaments?
Please answer somebody!

[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=179638]FTW[/url] is an interactive, flash math game on AoPS. Under "Classroom" in the Navigation bar, it says "For The Win". Click that, and you will arrive at FTW.

These are the three people that are currently in the competition;

Isabella2296
Poincare
Bugi

I'll join, even though I'll probably get eliminated in the first round.
Oh well. :ninja:

Wickedestjr, you should do something or there won't be even 20 contestants until December!

I really don't know. Try to find a way to promote it to other users of this forum. If I come up with a good :idea: , I'll tell you!

Like offering it to other users on this website that I know?

Ill join (The message is too small. Please make the message longer before submitting.)

Hmmm... it was posted on the FF too, you know.

You can have a multiple, as long as you don't use it on FTW.  Also, I join.

Allright, you are everyone out of topic!!!
If you want, make a topic in FF about that Wickedestjr=alingaling.
Here is the topic for signups for the tournament, so stick to that please.

Thank you.

Isabella2296
Poincare
Bugi
vamathletes
budi713
Ketvoman
ajain
Bachukas
ThetaPi
AIME15

Isabella2296
Poincare
vamathletes
budi713
Ketvoman
ajain
Bachukas
ThetaPi
AIME15

unfortunately, Bugi quit the tournament.

Wicked stated about 3 times that he will start in December. :wink: Please read the first post before asking questions.

[quote="AIME15"]It doesn't seem that many people are joining....so when are we gona start?[/quote]

If you would like more people to join, then that is fine, but you should just tell people about this tournament, to get them to sign up.

I'll join! What level will the problems be?

Isabella2296
Poincare
vamathletes
budi713
Ketvoman
ajain
Bachukas
ThetaPi
AIME15
kookamunga

It will range from MOEMS level to AIME level, if that is okay.

Please feel free to encourage others to join, because as you can see, there are not a lot of people competing right now.

I would love to join!  Please send me a private message when it starts though.

Isabella2296
Poincare
vamathletes
budi713
Ketvoman
ajain
Bachukas
ThetaPi
AIME15
kookamunga
iphailedusamo


calculus
integration
limit
function
calculus computations

Thank you for your reply, xufrog.

Your answer is incorrect.

kunny

[hide="Hint 1. "]$e^{x^2}$ is something of a red herring. Think of this as

$\lim_{n\to\infty}n\int_0^1(1+x)^{-n-1}f(x)\,dx$ where $f$ is some arbitrary smooth well-behaved function on $[0,1].$[/hide]

[hide="Hint 2. "]Integrate by parts.[/hide]


function
search
topology
real analysis
real analysis unsolved

E is a metric space.
f a mapping from E to R , real number set.
A is the set of points a from E suth that
lim as x converge to a and different from a , of the fuction f , exists and is different from f(a).
Show that a is at most countable

It's false. A counterexample:

Let $ S$ be an uncountable set, and let $ E\equal{}S\times [0,1]$, with metric $ d((a,b),(a',b'))\equal{}\begin{cases}1&a\neq a'\\ |b\minus{}b'|&a\equal{}a'\end{cases}$. If $ f(a,b)\equal{}\begin{cases}1&b\equal{}0\\0&b\neq 0\end{cases}$, the set of removable discontinuities is precisely $ S\times\{0\}$.

In fact i forgot an information : the metric space E is separable.So now it should be true

See http://www.mathlinks.ro/viewtopic.php?search_id=764906518&t=31694
I'm a total noob at topology stuff, but I think we can easily adapt grobber's work: Let $ x_{n}$ be the countable dense subset. For each $ x\in A_{n}$, there is a small ball $ B(x,r)$ such that $ y\not\in A_{n}$ for all $ y\in B\setminus\{ x\}$. Now, we build a function that assigns to each $ x$ a certain $ x_{i}$ found inside $ B\left( x,\frac{r}{2}\right)$. The $ \frac{1}{2}$ part is put there to ensure the injectivity of the function, so that $ A_{n}$ is countable.

In other news, you really shouldn't have posted [url=http://www.mathlinks.ro/viewtopic.php?t=164086]the same topic twice[/url].


function
real analysis
real analysis unsolved

Give an example of a function f:R->R, bijective, f(0)=0, f continuous in 0, but its inverse not continuous in 0.

A tryout:

It should be easy to define a bijection g: (-00,0)->(-00,0)\{-1/n|n in N*} \cup {n|n in N*} s.t. g(x)->0 as x->0. Now take f(x)=g(x) if x<0, f(x)=x if x>=0 and x isn't of the form n in N* and f(n)=-1/n for all n in N*.

The problem asks for the condition x doesn't tend to 0 as f(x)->0, and we take the sequence f(1), f(2), .., f(n),.. which tends to 0 but 1, 2, 3, .., n, .. doesn't tend to 0, so we're done. The only thing left is to see if all of this is correct. I know it's not even complete, because I didn't actually define the function g, but is it so hard to define? I didn't take my time to think about that, because it's not really that nice..


Čestitke svima, a posebno ovim nabrojanim ispod  :)

 :winner_first: Stjepan Glavina [HRV]

 :winner_second: Ivan Pilat [HRV], Goran Žužić [HRV], Jakša Markotić [HRV], Luka Milićević [SRB]

 :winner_third: Uglješa Stojanović [SRB], Boris Grubić [SRB]

Cestitke svima, najvise osvajacima!!!

E sad malo da se hvalim, Boris ide sa mnom u razred  :rotfl:


geometry
rectangle
combinatorics proposed
combinatorics

The $2n$ squares of a rectangle $2\times n$ are colored in 3 colors. We say that a certain coloration has a $cut$ if on one of the $n$ collums we have 2 squares with the same color.
a) Determine the number of the colorations that have no $cuts$
b) Determine the number of the colorations that have one $cut$.

[quote="aram_24"][b]"collums"[/b] :?:  excuse me does it mean [b]"columns"[/b]?[/quote]
Yes.  :blush:

Unless I misunderstand, this is very easy:  there are 6 ways to color any column so that the two colors are different, so the answer to the first is $ 6^n$.  For the second, there are 3 ways to color a column so that the colors are the same, so the answer is $ 3 \cdot n \cdot 6^{n \minus{} 1}$.


Math Olympiads is over this year. How did everybody do?

i didn't participate.  It's offered in 5th grade in my school and i'm in 7th

I did. I did okay. Our school got like 12 in the nation or something.

you wrote "i did" twice. And 12 is not okay, its great.


algebra
polynomial
number theory
greatest common divisor
floor function
induction
search

Help me solve these following problems:
1/Prove that there isn't any polynomial P with integer coefficient(except constants) satisfies that
P(n) is a prime number with all n $ \in$ N.
2/Let $ a,b$ are positive integer satisfy that $ p\equal{}\frac{b}{4}\sqrt\frac{2a\minus{}b}{2a\plus{}b}$ is a prime number.Find the max value of p.
3/Let $ a,b$ are positive integer satisfy that $ a<\frac{b^2}{4}$ and all prime divisors of $ a$ are smaller than $ b$.Prove that $ b! \vdots a$.

for the first one...if $ P(0)\equal{}a_0\equal{}p$,a prime number ,and if $ P(n)$ is always prime number,take for example $ P(kp)$ where k is a positive integer.this expression is obviously divisible by p,and because of the fact that $ P(kp)$ is also a prime we should have $ P(kp)\equal{}p$.hence,the polynomial takes the same value infinitely many times which is obviously a contradiction.... :wink:

Can't you just use the fact:

$ x \minus{} y|P(x) \minus{} P(y)$

Edit: Sorry, maybe not. That was stupid.

[quote="Rafikafi"]I also thought of that one first but $ 0\notin \mathbb{N}$[/quote]
well in some countries 0 belongs to $ N$.when $ 0$ doesn't belong to our set of positive integers we write $ N^*$.ok i will post another solution later.. :)

Let $ r_i,q$ be primes
$ P(1)\equal{}q$
$ kq\equal{}(kq\plus{}1)\minus{}1|P(kq\plus{}1)\minus{}P(1)\equal{}r_k\minus{}q \Rightarrow r_k\equal{}q$

Therefore $ P(kq\plus{}1)\equal{}q$ for all $ k\in \mathbb{N}$ contradiction

[quote="FelixD"]What does $ b! \vdots a$ mean?[/quote]

I think it is $ a | b!$ $ \iff$ $ a$ divides $ b(b\minus{}1)(b\minus{}2)...2.1$

Thanks pco. ^^

[quote="SherlockBond"]Help me solve these following problems:
1. Prove that there isn't any non-constant polynomial $ P$ with integer coefficients such that $ P(n)$ is a prime number for all $ n \in \mathbb{Z_ \plus{} }$.
2. Let $ a$, $ b$ be positive integers satisfying $ p \equal{} \frac {b}{4}\sqrt {\frac {2a \minus{} b}{2a \plus{} b}}$ is a prime number. Find the maximum value of $ p$.
3. Let $ a$, $ b$ be positive integers satisfying $ a < \frac {b^2}{4}$ and $ p < b$ for each prime $ p$ dividing $ a$ .Prove that $ b! \vdots a$.[/quote]

1. Let $ P(x) \equal{} \sum_{i \equal{} 0}^{k} c_ix^i$ and let wlog $ c_k > 0$. Then we have $ P(n)\to \infty$ as $ n \to \infty$ and $ P(0) \equal{} c_0$ is prime. Hence, $ P(mc_0)$ is divisible by $ c_0$ and therefore it equals $ c_0$. But as $ m$ gets large, this is a contradiction.

2. Let $ d$ be the gcd of $ a$ and $ b$ and write $ a \equal{} da'$, $ b \equal{} db'$. We have $ \gcd(2a' \minus{} b'$, $ 2a' \plus{} b') \in \{1$, $ 2$, $ 4\}$. Then there are three cases to consider.
Case 1. $ \gcd \equal{} 1$. This is the case iff $ b'$ is odd. Hence $ 2a' \minus{} b' \equal{} s^2$ and $ 2a' \plus{} b' \equal{} t^2$ for some odd integers $ s$, $ t$. Then $ 4a' \equal{} s^2 \plus{} t^2\equiv2 (4)$, a contradiction.
Case 2. $ \gcd \equal{} 2$. Then $ 2a' \minus{} b' \equal{} 2s^2$, $ 2a' \plus{} b' \equal{} 2t^2$, where $ (s$, $ t) \equal{} 1$. Hence $ s^2 \plus{} t^2 \equal{} 2a' \equiv0 (2)$, which implies that $ s$, $ t$ have the same parity. Thus, they are both odd. We get $ b' \equal{} t^2 \minus{} s^2$ and
\[ \frac {b}{4}\sqrt {\frac {2a \minus{} b}{2a \plus{} b}} \equal{} \frac {d(t^2 \minus{} s^2)s}{4t}.\]
Thus $ d \equal{} td'$. Set $ s \equal{} 2s' \plus{} 1$, $ t \equal{} 2t' \plus{} 1$. Then the number $ d'(2s' \plus{} 1)(s'^2 \plus{} s' \plus{} t'^2 \plus{} t')$ is prime. This is the case if and only if $ d' \equal{} 1$, $ s' \equal{} 0$, $ t' \equal{} 1$ and hence $ p \equal{} 2$.
Case 3. $ \gcd \equal{} 4$. Then $ 2a' \minus{} b' \equal{} 4s^2$, $ 2a' \plus{} b' \equal{} 4t^2$ and $ (s$, $ t) \equal{} 1$. Moreover, $ a'$ is odd. Hence $ a' \equal{} s^2 \plus{} t^2 \equiv 1(4)$, which implies that $ s$, $ t$ have different parity. We get $ b' \equal{} 2(t^2 \minus{} s^2)$ and
\[ \frac {b}{4}\sqrt {\frac {2a \minus{} b}{2a \plus{} b}} \equal{} \frac {d(t^2 \minus{} s^2)s}{2t}.\]
Hence $ d \equal{} td'$. If $ s$ is even, then $ t \equal{} s \plus{} 1$. Write $ s \equal{} 2s'$ and the number $ d(4s' \plus{} 1)s'$ has to be prime, which is the case iff $ s' \equal{} 1$ and hence $ p \equal{} 5$. If $ s$ is odd, then $ d' \equal{} 2d''$ and the number $ d''(t \minus{} s)(t \plus{} s)s$ has to be prime, which is the case iff $ s \equal{} 1$, $ t \equal{} 2$, $ d'' \equal{} 1$ and hence $ p \equal{} 3$.

Thus, the maximum is $ p \equal{} 5$.

3. Let $ a \equal{} p_1^{\alpha_1}\dots p_r^{\alpha_r}$. We have to prove that $ v_{p_i}(b!) \ge v_{p_i}(a) \equal{} \alpha_i$. We have $ b > 2p_i^{\frac {\alpha_i}{2}}$ and $ v_{p_i}(b!) \equal{} \sum_{k \equal{} 1}^{\infty} \left\lfloor \frac {b}{p_i^k} \right \rfloor > \sum_{k \equal{} 1}^{\infty} \lfloor 2p_i^{\frac {\alpha_i}{2} \minus{} k} \rfloor$. This can be done by induction, but I'm too lazy to type it now.

Problem 2 is in PEN (A95).go there.always search :wink:

I always wonder how to search. There are thousands of problems (not in PEN of course, but there are also enough^^) and they have most likely names like "very hard" or "easy number theory". So, how to find this problem in PEN?

:huh:  u can type from which country the problem is ,in this way (because PEN says from where it is)  if it is in PEN u will find it(e.g problem 2 is from iran 1998) if u didn't find it then type the problem itself or an important part of the problem
for example if u type "1324+279m+5^n" u will find a problem containing exactly this expression. :wink:

[quote="SherlockBond"]Could you tell me,FelixD,what do $ v_{p_{i}}(b!)$ and $ v_{p_{i}}(a)$  mean?[/quote]

$ v_p(m)$ is the greatest power of $ p$ which divides $ m$.

When $ p$ is prime, $ v_p(n!)\equal{}\sum_{k>0}[\frac n{p^k}]$


[quote="Quevvy"]A square pyramid has base edges each 8 inches long and lateral edges each 12 inches long. To the nearest inch, what is the number of inches in its altitude?

I tried doing this but I couldn't do square roots in my head.[/quote]
[hide]
think of it as a right triangle with leg $4\sqrt{2}$ (half the diagonal of the base) and hypotneuse $12$ (lateral edge)

figure out the last leg. so it is the square root of 144 minus 32, so it is the square root of 112, which is rounded to 11.
[/hide]

[quote="ehehheehee"][hide]The altitude is $\sqrt{12^{2}-4^{2}*2}$ if my own stupidity did not strike again.

So that would be $\sqrt{144-32}=\sqrt{112}$.  Now, to find the altitude to the nearest inch you first notice that $10<\sqrt{112}<11$.  Then, find the square of 10.5.  Since that is less than 112, the answer is 11. [/hide][/quote]
Could you also use the fact that $\sqrt{112}=4\sqrt{7}$?

For MC, would it be good to memorize square roots to the nearest hundreth? or should you always estimate?

112 is between 100 and 121. you know that the square of 10.5 is closer to 100 than to 121, and 112 is closer to 121 than 100, so the square root of 112 is closer to 11
i think thats easier than multiplying 4 and the square root of 7


http://citeseer.ist.psu.edu  很好的免费搜索数学论文的网站(好)
http://www.math.gatech.edu/~cain/textbooks/onlinebooks.html 下数学电子书的好
地方
http://ocw.mit.edu/OcwWeb/Global/all-courses.htm#Mathematics MIT开放课程资源
下载，不过好像分布很乱
http://reflections.awesomemath.org/  一个数学杂志，适合大中学生看，而且还可以
免费下到。
http://www-history.mcs.st-andrews.ac.uk/  在这里可以找到所有数学家的资料
http://www.mathlinks.ro/Forum  一个数学论坛，传说也许应该是世界上最大的了(好)

http://www.cecm.sfu.ca/organics/papers/boyd/paper/html/paper.html  一个研究be
ta展开与Salem数关系的网站
http://www.emis.de/journals/short_index.html 欧洲网上数学电子资源下载
http://www.jmilne.org/math/index.html J.Milne的主页，他非常喜欢杭州，呵呵
http://www.loria.fr/~sustreto/listki/ 练习题下载
http://www.math.ucla.edu/~tao/  陶哲轩的个人主页
http://www.iisc.ernet.in/nias/hrj.htm  Hardy-RAMANUJAN JOURNAL。A  Journal
devote to primes,diophantine equations,transcendental numbers and other
problems.
http://arxiv.org/ 搞数学的应该都知道(好)

This is similary with http://www.mathlinks.ro/Forum/viewtopic.php?t=151289  :P


probability
search
probability and stats

We have an urn of $ n$ balls with exactly one ball white.Two persons A and B have a probability of $ P_{1}$ and $ P_{2}$ of telling the truth.A ball was taken out of the urn it was reported by both of them as white .What is the probability the ball is actually white?

[hide="my solution"]
comsider the events $ E_{1}$:Both speak truth
                               $ E_{2}$:Both lie
                              $ E_{3}$:A lies and B speaks the truth
                               $ E_{4}$ B lies and A speaks the truth
Clearly the $ E_{i}$ are mutually exclusive and exhaustive events
so baye's theorem is applicable
now let $ C$ be the event of reporting the ball as white by both A and B
now we have to find $ P(E_{1}|C)$
now $ P(C|E_{1})P(E_{1}) \equal{} \frac {P_{1}P_{2}}{n}$
       $ P(C|E_{2})P(E_{2}) \equal{} \frac {(1 \minus{} P_{1})(1 \minus{} P_{2})(n \minus{} 1)}{n}$
 clearly $ P(C|E_{3}) \equal{} P(C|E_{4}) \equal{} 0$
so by bayes theorem we must have
$ P(E_{1}|C) \equal{} \frac {P_{1}P_{2}}{P_{1}P_{2} \plus{} (n \minus{} 1)(1 \minus{} P_{1})(1 \minus{} P_{2})}$

but the answer given in the book is $ \frac {(n \minus{} 1)P_{1}P_{2}}{(n \minus{} 1)P_{1}P_{2} \plus{} (1 \minus{} P_{1})(1 \minus{} P_{2})}$
where am i wrong???
[/hide]

EDIT:Due to mlok and fedja :)

Perhaps the book's answer is wrong. If $ n\equal{}1$, the probability of the ball being white is equal to $ 1$.

Try to come up with a more descriptive title next time.

You aren't, the book is. Consider the case when both people speak truth with probability $ 1/2$. Then their information is completely useless and you should get $ 1/n$. Your formula gives it, the book's doesn't.

By the way, I understand that minor spelling errors are inavoidable but try at least to spell the key words by which the people will search your post in the future like "white" and "Bayes" correctly ;)

yes thanks mlok and fedja :)
i will try to put things in a better way next time


Josh writes down the 91 digits in the natural numbers from 1 to 50, inclusive. What is the arithmetic mean of all of the digits Josh writes? Express your answer as a decimal to the nearest tenth.

EDIT: Sorry, I meant to put "digits." In fact, I noticed that earlier, and I thought I had edited it, but apparently I didn't.  :huh:

[hide]
continually, you have the 1-9, or 45 (sum).  you have that 5 times.  then for 10s, there are 10*1, 20s are 10*2, then 10*3, 10*4 + 5.

so 225+105=330

330/91 is approx. 3.6.[/hide]

[quote="13375P34K43V312"]What do you mean numbers. Do you mean digits only, or does "numbers" include the two digit numbers as well?[/quote]
I'm like 99% sure he means the digits only.
[hide]Note that all of the numbers can be written as two digit numbers with the tens digit being 0 for one digit numbers. It cycles through 1-9 in the units digits 5 times giving us 225. For the tens we find that the sum is 45. So the sum of the numbers is 270.
So our answer is $\frac{270}{91}$ which is about 3.0 (too lazy to make sure)...[/hide]

its kind of funny when this is the 2nd time he's done this :P
-jorian

[quote="nebula42"]Yes, I meant to put "digits," and I've changed it in the post above.

I believe that Walk Around the River's answer is correct  :) , but bpms, I think you just miscalculated the sum of the tens digits.[/quote]
You didn't know that 10*10=40?
:P


[hide]When you multiply it out, you can see that the product is equal to
$1+x+x^{2}+x^{3}\dots$
(This can be proved easily with induction if a proof is necessary.)
We have $1+x+x^{2}+x^{3}\dots=\boxed{\frac{1}{1-x}}$ by the formula for an infinite geometric series.[/hide]

[hide]When this is expanded, all the 1s will multiply together to make 1.  Also, every number of $x$s will be multiplied together, because it is like binary for which exponents to add in.  Thus, the sequence is

$\sum_{k=0}^{\infty}x^{k}$

Use the formula for sum of an infinite geometric sequence to get:

$\frac{1}{1-x}$[/hide]

Yup, good job.

Now let me find another easy $F^{3}$ problem to post. Being a mod is hard...

Another way to look at it -

Multiply the Numerator and Denominator of the expression by $(1-x)$.
All the terms in the numerator would be reduced to form $(1-a)(1-b)$.
Thus in the end expression would be -
$\frac{(1-x^{\infty})}{1-x}$
As $\mid x\mid < 1$, hence
$x^{\infty}\to 0$
Thus the answer is - $\frac{1}{1-x}$


The table below shows the distance $ s$ in feet a ball rolls down an inclined plane in $ t$ seconds.
\[ \begin{tabular}{|c|c|c|c|c|c|c|}
\hline
t & 0 & 1   & 2   & 3 & 4 & 5\\
\hline
s & 0 & 10 & 40 & 90 & 160 & 250\\
\hline
\end{tabular}
\]
The distance $ s$ for $ t \equal{} 2.5$ is:

$ \textbf{(A)}\  45\qquad
\textbf{(B)}\ 62.5\qquad
\textbf{(C)}\ 70\qquad
\textbf{(D)}\ 75\qquad
\textbf{(E)}\ 82.5$

For each value, $ t$, the value of $ f(t)\equal{}10t^2$

So plugging this is for $ t\equal{}2.5$, we get $ \boxed{62.5}$ as the output


number theory
greatest common divisor
relatively prime
linear algebra
linear algebra unsolved

p prime number
M element of $SL_n(Z)$ of order p
m is the positiv gcd of all coefficient of $M-I_n$

prove that m divide p

Show that, either $ m \equal{} 1$, or $ m \equal{} p \equal{} 2$.

Let M-I=N;thus (N+I)^P=I or by Newton's formula:
N^2(N^(p-2)+.....+p(p-1)/2.I)+pN=0; therefore m^2 divides
pN(i,j) for all(i,j).  m=gcd(N(i,j)i,j) implies that m divides p.
The two cases m=1 and m=p are possible.

The condidion "p is prime" is useless:
m divides the (pK(i,j))i,j where the (K(i,j)=N(i,j)/m)i,j are relatively prime .
By Bézout m divides p.


assume that f ,g,h are continiuos function f,g,h:R+-->R. find all f,g,h s.t:
f(x+y)+g(xy)=h(x)+h(y)
(x,y are in R+)

Only a little bit hard? Wow! After a struggling battle, I gave up. I don't how what to do with it. Please, anyone post a solution.

i need enough time to post the solution because it's realy too long! :(

now I dare to  copy the sol from the book :lol:
assume that $f,g,h$ satisfy the condition,if $y=1$ and $x>0$ we have
$g(x)=h(x)-f(x+1)+h(1)$ (1) now we conclude from (1) and the condition of the problem $h(x)+h(y)-h(xy)=f(x+y)-f(xy+1)+h(1)$ suppose that:
$H(x,y)=h(x)+h(y)-h(xy)$
if $x,y,z>0$ then we have $H(xy,z)+H(x,y)=H(x,yz)+H(y,z)$(2)
according to (1) we have $H(x,y)=f(x+y)-f(xy+1)+h(1)$ so we can write (2) in this way $f(xy+z)-f(xy+1)+f(1)=f(x)+f(y)-f(x+y)$so since f is continiuos if $z\rightarrow0(z>0)$ we conclude that $f(xy)-f(xy+1)+f(1)=f(x)+f(y)-f(x+y)$ now assume that $f^*(t)=f(t)-f(t+1)+f(1)             (t>0)$
$F(x,y)=f(x)+f(y)-f(x+y)             (x,y>0)$  if x,y,z>0 we have :
$F(x+y,z)+f(x,y)=F(x,y+z)+F(y,z)$ so $F(x,y)=f^*(xy)$according to above we can say $f^*(xz+yz)+f^*(xy)=f^*(xy+xz)+f(yz)     (x,z,y>0)$if $z=\frac{1}{y} ,u=\frac{x}{y},v=xy$ then we have $f^*(u+1)+f^*(v)=f^*(u+v)+f^*(1)$
if you swich the place of $u$ and $v$ then $f^*(u+1)+f^*(v)=f^*(v+1)+f^*(u)$
so if v=1,
$f^*(u+1)=f^*(u)+f^*(2)-f^*(1)$ so we can say according to above
$f^*(u+v)+f^*(1)=f^*(u)+f^*(v)+f^*(2)-f^*(1)$
so since $f^*$ is contiuos we can conclude $f^*(t)=\alpha t+\beta$ so $\alpha xy+\beta=f(x)+f(y)-f(x+y)$ then we can easily conclude that $f(x)=-\frac{\alpha}{2} x^2 +\gamma x+\beta$ and in this way we can get that
$h(x)=-\frac{\alpha}{2} x^2+\gamma x+k\ln(x)-\sigma$ and $g(x)=k\ln(x) +\alpha x -2\sigma +\beta$
huh :) endless typing is realy boring specialy when you are coping from some material :maybe:

Wow! O boy, this is a hell of a problem. It's so nice not to have it in a contest! But the solution is very nice.


modular arithmetic
number theory unsolved
number theory

$0\leq x,y,z$   and  $x,y,z$ are integer . solve that :

 $5^{x}7^{y}+4=3^{z}$

It's still wrong since in the first step you get $2 \not | x$ instead of $2|x$.

Hi! Here is my soln.

Taking $\mod 3$, we find that $2^{x}+1 \equiv 0 \pmod{3}$, thus $x$ must be odd. Taking $\mod 5$, we have $-1 \equiv 3^{z}\pmod{5}$, so $z \equiv 2 \pmod{4}$ and so is even. Taking $\mod 8$, since $5^{x}\equiv 5 \pmod{8}$ for odd $x$ and $3^{z}\equiv 1$ for even $z$, we get $5 \cdot 7^{y}\equiv 5 \pmod{8}$, so $y$ must be even.

Letting $z = 4z_{0}+2$, we have $5^{x}7^{y}= 3^{4z_{0}+2}-4 = (3^{2z_{0}+1}-2)(3^{2z_{0}+1}+2)$. Since $\gcd ( 3^{2z_{0}+1}-2, 3^{2z_{0}+1}+2)$ must be a divisor of $4$, it can only be $1$.

We thus have 2 cases:

Case 1:
$3^{2z_{0}+1}-2 = 5^{x}, 3^{2z_{0}+1}+2 = 7^{y}$.
Since $y$ is even, by letting $y = 2y_{0}$, we get $7^{2y_{0}}-4 = 5^{x}$, or equivalently, $(7^{y_{0}}-2)(7^{y_{0}}+2) = 5^{x}$, which is a contradiction, since $\gcd (7^{y_{0}}-2, 7^{y_{0}}+2)$ can only be $1$.

Case 2:
$3^{2z_{0}+1}-2 = 7^{y}, 3^{2z_{0}+1}+2 = 5^{x}$.
We have $5^{x}-7^{y}= 4$. Note that since $y$ is even, we have $7^{y}\equiv 1 \pmod{25}$.
If $x>1$, then taking $\mod 25$, we get $-1 \equiv 4 \pmod{25}$, a contradiction. Thus since $x$ is odd, we can only have $x =1$.

Therefore, the only solution is $(x,y,z) = (1,0,2)$.

dblues thats like to my solution. and  after the solution we can look that question is true

haha..  :P

[i]TomciO[/i]'s post was in response to [i]Megus[/i]'s wrong solution, which he had already deleted.


geometry
geometric transformation
reflection
geometry unsolved

Draw the incircle $ (I)$ passing though $ F$ and tangent to the sidelines $ \mathcal{L}_b$ and $ \mathcal{L}_c.$ This is the Apollonius problem (line, line, point). However, it is equivalent to the easier case (point, point, line) by reflecting $ F$ on the internal bisector of $ A.$ This yields two solutions but consider the one according to the sketch. Let $ Y,Z$ be tangency points of $ (I)$ with $ AC,AB.$ It is well-known that the reflections of the diacentral line $ IO$ about the sidelines of the intouch triangle concur at its orthopole, i.e. the Feuerbach point $ F.$ Hence, construct the reflection $ F'$ of  $ F$ on $ ZY$ $\Longrightarrow$ $ F'I$  is the diacentral line $ d \equiv IO.$ Since  $ F$ is the orthopole of $ d,$  construct the projection $ A'$ of $ A$ on $ d$ $\Longrightarrow$ $ FA'$ is perpendicular to $ BC.$ Construct the tangent to the incircle (at the opposite arc of ZFY) perpendicular to $ FA'$ to complete the $ \triangle ABC.$


function
trigonometry
Trigonometric inequality
Inequality
IMO
IMO 1977

Let $a,b,A,B$ be given reals. We consider the function defined by \[ f(x) = 1 - a \cdot \cos(x) - b \cdot \sin(x) - A \cdot \cos(2x) - B \cdot \sin(2x). \] Prove that if for any real number $x$ we have $f(x) \geq 0$ then $a^2 + b^2 \leq 2$ and $A^2 + B^2 \leq 1.$

$ f(x) \equal{} 1\minus{}\sqrt{a^2\plus{}b^2}\sin (x\plus{}arctan\frac{a}{b}) \minus{} \sqrt{A^2\plus{}B^2}\sin (2x\plus{}arctan\frac{A}{B}) \geq 0$.
$ f(x\plus{}\pi) \equal{} 1\plus{}\sqrt{a^2\plus{}b^2}\sin (x\plus{}arctan\frac{a}{b}) \minus{} \sqrt{A^2\plus{}B^2}\sin (2x\plus{}arctan\frac{A}{B}) \geq 0$

Therefore, $ \sqrt{A^2\plus{}B^2}\sin (2x\plus{}arctan\frac{A}{B}) \leq 1$. Since this identity is true for any real $ x$, let the sine term be one, $ \longrightarrow A^2\plus{}B^2 \leq 1$.

To get cancellation on the rightmost terms, note $ \sin (x\plus{}\pi/2) \equal{} \cos x, \sin (x\minus{}\pi/2) \equal{} \minus{}\cos x$.

$ f(x\plus{}\pi/4) \equal{} 1\minus{}\sqrt{a^2\plus{}b^2}\sin (x\plus{}\pi/4\plus{}arctan\frac{a}{b}) \minus{} \sqrt{A^2\plus{}B^2}\cos2x\plus{}arctan\frac{A}{B}) \geq 0$.
$ f(x\minus{}\pi/4) \equal{} 1\minus{}\sqrt{a^2\plus{}b^2}\sin (x\minus{}\pi/4\plus{}arctan\frac{a}{b}) \plus{} \sqrt{A^2\plus{}B^2}\cos2x\plus{}arctan\frac{A}{B}) \geq 0$.

Let $ x\plus{}arctan\frac{a}{b} \equal{} y$.
Then $ \sqrt{a^2\plus{}b^2}(\sin (y\plus{}\pi/4) \plus{} \sin (y\minus{}\pi/4)) \leq 2$
$ \sqrt{a^2\plus{}b^2} \leq \dfrac{2}{\sqrt{2}(\sin y)}$. Since it's valid for all real $ x$ let $ \sin y \equal{} 1$, and we are done.

[quote=arqady][quote="orl"]Let $ a,b,A,B$ be given reals. We consider the function defined by
\[ f(x) \equal{} 1 \minus{} a \cdot cos(x) \minus{} b \cdot sin(x) \minus{} A \cdot cos(2x) \minus{} B \cdot sin(2x).
\]
Prove that if for any real number $ x$ we have $ f(x) \geq 0$ then $ a^2 \plus{} b^2 \leq 2$ and $ A^2 \plus{} B^2 \leq 1.$[/quote]
By the way, if $ f(x)>0$ for all value of $ x$ then, $ f(x)<3$ for all value of $ x.$ :wink:[/quote]

What's the reason?

If we have $f(x)=1-asin(x)-bcos(x)-csin(2x)-dcos(2x)>3$
Then $f(x+2\pi/3)+f(x-2\pi/3)=2+asin(x)+bcos(x)+csin(2x)+dcos(2x)<0$
So $f(x) \leq 3$


How do you find the number of arrangements for a 3x3 Rubik's Cube? 4x4, 5x5, and so on?

For the $3\times 3\times 3$ cube, consider dismantling the cube and then putting it back together.

You have pieces in the center of each face; they'll stay put. Then you have 8 corner pieces and 12 edge pieces. Each edge piece has two possible orientations, and each corner piece has 3 possible orientations.

We can permute the edges in $12!$ ways.
We can orient the edges in $2^{12}$ ways.
We can permute the corners in $8!$ ways.
We can orient the corners in $3^8$ ways.

That gives us $12!\cdot2^{12}\cdot8!\cdot3^8.$

But you can't do all that with an intact cube; there are restrictions. One of the restrictions is that with all of the corners in "home" position, the sum of the clockwise twists of corners must be divisible by 3, and with all of the edges in "home" position, the number of flipped edges must be even. There are also restrictions on which permutations you can use.

The resulting number (which can be interpreted as the order of a finite group) is $\frac{12!\cdot2^{12}\cdot8!\cdot3^8}{12}.$

[quote="gopherhole112"]Unfortunately its harder than that. Not all of those are possible solutions. For example, a solveable rubik's cube will never, no matter how hard you try, be in a position where it is completely  solved, except one of the edge (not corner) pieces is flipped the wrong way. I don't know how to solve the problem asked, but I would assume that you start using the dismantling idea and then subtract impossible configurations.[/quote]

I thought Mr. Merryfield explained pretty well how to count which cases were impossible and that's why he divided by 12.


find all posible integer A which is a solution of 6A^2+1=B^2, 12A^2+1=C^2 (B,C also integer)

- I tried a proof using pell's equation, but it's something hard  :(  plz help

Write your problems with $ L$A$ T$E$ X$ Please:

find all posible integer $ A$ which is a solution of
$ 6A^2\plus{}1\equal{}B^2$,
$ 12A^2\plus{}1\equal{}C^2$
($ B,C \in Z$)

This is a very nice question.
My Solution:

We show that the solutions for $ A$ are $ 0,2 or -2$

$ \{\begin{array}{cc}6A^2+1=B^2\\12A^2+1=C^2\end{array}$ $ \Longrightarrow \{\begin{array}{cc}2B^2-C^2=1     (1)\\B^2+C^2=18A^2+2     (2)\end{array}$

$ B^2+C^2=18A^2+2 \Rightarrow  B^2+C^2 \equiv 2 (mod 18)$

We know that $ X^2 \equiv 0,1,4,7,9,10, 13 or 16$.

$ (2) \Rightarrow$ $ \{\begin{array}{cc} B^2 \equiv 7 or 1\\ C^2 \equiv 13 or 1(mod 18)\end{array}$

So,We have two parts:

[size=150][color=darkred]Part one:[/color][/size]

$ \{\begin{array}{cc} B^2 \equiv 7 \\ C^2 \equiv 13(mod 18)\end{array}$

Then we can write $ B^2=18k+7$ and $ C^2=18k_1+13$ where$ k \in Z$ .since $ (1)$ we have:

$ 2B^2-C^2=1 \Longrightarrow 36k+14-(18k_1+13)=1 \Longrightarrow k_1=2k$

So $ B^2=18k+7$ and $ C^2=36k+13$.

 $ (2) \Rightarrow 18A^2+2=54k+20 \Rightarrow A^2=3k+1$

Now, we know that $ B^2=18k+7 \geq 0 \Longrightarrow k \geq 0$ .

Assume that $ k >1$ so $ 18k+7>25$  and $ 36k+13>49$ that means $ B^2>25$ and $ C^2>49$.So $ A^2=2B^2-C^2>1$ that is clearly wrong.So $ k=1$ and  we have: $ A=2 or -2, B=5 or -5,C=7 or -7$

[size=150][color=darkred]Part two:[/color][/size]

$ \{\begin{array}{cc} B^2 \equiv 1 \\ C^2 \equiv 1(mod 18)\end{array}$
Which gives us $ B=C=1$ and $ A=0$

So,the proof is complete.

where did you get a $ A^{2}\equal{}2B^{2}\minus{}C^{2}$?

[quote="drst123"]where did you get a $ A^{2} \equal{} 2B^{2} \minus{} C^{2}$?[/quote]

Oh I'm sorry,a very bad mistake.
I think the solution is right. but $ 1 \equal{} 2B^{2} \minus{} C^{2}$ and not equalls to $ A^2$.
ok?

amparvardi


What is the number of cubic feet in the volume of a box that measurse $ 6''\times8''\times4'$? Express your answer as a mixed number.

Converting all units to feet, one arrives at $ \frac12\cdot\frac23\cdot4\equal{}\boxed{1\frac13}$.


I have Fedora 10 and use Kile as my LaTeX Editor. It was working fine yesterday, but this morning it wouldn't launch, after the latest Fedora patches were applied. Though it showed the blue rectangle thing, whereas it would normally take about ten seconds to start, I was waiting for over five minutes and nothing happened. Please help. Thanks in advance!

You could try reinstalling Kile or see if there's a later version than the one you are using. Apart from that, probably best to ask in a Kile forum such as http://sourceforge.net/forum/forum.php?forum_id=292014


[quote="mdk"]Prove that the number of partitions of n in which all the even parts are distinct is the same
as the number of partitions of n where each part is repeated at most three times.[/quote]

This should be done with generating functions.
The generating function for expression [b]even[/b] parts is:
$ f(x)=(1+x^{2}+x^{4}+x^{6}+...)(1+x^{4}+x^{8}+x^{12}+...)...$
So,
$ f(x) = \frac{1}{1-x^{2}}\cdot \frac{1}{1-x^{4}}\cdot ...$

Now find the generating function for expression [b]at most 3[/b]
$ g(x)$
(I didnt find that yet).

And so that,
$ f(x)=g(x)$.

In fact, the first partition states that the even parts are distinct but there can be odd parts.

So $ f(x)\equal{}\prod_{n\equal{}1}^{\infty} \frac{1\plus{}x^{2n}}{1\minus{}x^{2n\plus{}1}}\equal{}\prod_{n\equal{}1}^{\infty} \frac{1\minus{}x^{4n}}{(1\minus{}x^{2n})(1\minus{}x^{2n\plus{}1})}\prod_{n\equal{}1}^{\infty} \frac{1\minus{}x^{4n}}{1\minus{}x^{n}}$

And $ g(x)\equal{}\prod_{n\equal{}1}^{\infty}(1\plus{}x^n\plus{}x^{2n}\plus{}x^{3n})\equal{}\prod_{n\equal{}1}^{\infty}\frac{1\minus{}x^{4n}}{1\minus{}x^{n}}$


[quote="tmrfea"]$ \det{A} \equal{} \left|\begin{array}{cccccc} 1 & 2 & 3 & ... & n \minus{} 1 & n \\
2 & 3 & 4 & ... & n & 1 \\
3 & 4 & 5 & ... & 1 & 2 \\
... & ... & ... & ... & ... & ... \\
n \minus{} 1 & n & 1 & ... & n \minus{} 3 & n \minus{} 2 \\
n & 1 & 2 & ... & n \minus{} 2 & n \minus{} 1\end{array}\right|$[/quote]
[color=blue]You can find answer since click here :[/color]http://www.mathlinks.ro/viewtopic.php?p=1457691#1457691
 [hide]$ \det(A) \equal{}\minus{}( \minus{} 1)^{\frac {n(n \plus{} 1)}{2}}\frac {n(n \plus{} 1)}{2}n^{n \minus{} 1}\prod_{k \equal{} 1}^{n \minus{} 1}{\frac {1}{\lambda_k \minus{} 1}}$
 Here $ \lambda_k \equal{} cos(\frac {2k\pi}{n}) \plus{} isin(\frac {2k\pi}{n}) ,\forall k \in$ {$ 1,2,..,n \minus{} 1$}[/hide]


geometry
3D geometry
sphere
analytic geometry
number theory unsolved
number theory

FInd all $a,b,c,d \in \mathbb{N}$ such that $a^2+b^2+c^2=d^2$.

Use the same methode as for any number of variables:
- reduce it to the rational solutions of $x_1^2 + ... + x_n^2 = 1$ (a sphere)
- take the solution $s=(1,0,0,...)$
- take any line through $s$ that intersects the hyperplane with first coordinate $0$ in a rational point
- look at the second intersection of this line with the sphere

:?:
If that isn't elementary, what then¿

[quote]If that isn't elementary, what then¿[/quote]

some divisibility for example. Came on, I'm sure that exist something like that. If you can to do it on "elem." can you give me forms of $a,b,c,d$.


A sequence of length seven has only 1 or -1 as terms. Two consecutive terms in the sequence cannot have a negative sum. For example, 1, -1, 1, 1, -1, 1, -1 is a valid sequence, while 1, 1, -1, -1, 1, -1, 1 is not. How many such sequences exist?

[hide]So basically, no two $ -1$s can be adjacent.  Then you can just use balls and urns, right?[/hide]

[hide="Maybe?"]
We count the amount of sequences that exist: there are two options for each place, for 7 places: $ 2^{7}$ unique sequences.
Then we subtract the invalid parts:
so basically, we can't have a few things:
(-1, -1) in any of 6 places
(-1, -1, -1) in any of 5 places
(-1, -1, -1, -1) in any of 4 places...
(-1, -1, -1, -1, -1, -1, -1) in any of 1 place

So there are $ 2^{7}-(6+5+4+3+2+1) = 107$ such sequences?

[/hide]

Hmm you're right.

I under count those types, but I'll think it through.

Archemides hint is pretty big.

Its another fancy way to say "How many 7 sequence terms exist consisting of 1s and -1s without two consecutive -1s."

Key is fibonacci (or combinatorics but a little longer that way).

It should be $ 34$

Recursion approach:
[hide]
Let $ A_{n}$ be a valid sequence with length $ n$, ending with a positive $ 1$.
Let $ B_{n}$ be a valid sequence with length $ n$, ending with $ -1$.

Trivially, $ A_{1}=1; B_{1}=1$.

It is easy to see that $ B_{n+1}=A_{n}$ because you just use the $ A_{n}$ list and attach a $ -1$ at the end.

$ A_{n+1}=A_{n}+B_{n}$ because it combines all possible valid sequence of length $ n$, and adds the positive $ 1$ at the end of the sequence to acheive the length $ n+1$.

Using a table, we can see that $ A_{7}= 21$ and $ B_{7}= 13$, so all possible sequences of length 7 are $ 21+13 = \boxed{34}$[/hide]

Recursion as well, though a glimpse at your solution told me that yours is different from mine.

[b]Solution.[/b] The problem is equivalent to finding the number of n-tuples in $ \{-1,1\}^{n}$ with no two consecutive $ -1$s.  Let $ a_{n}$ be the number of such n-tuples.  If our n-tuple begins with a $ 1$, then we can add an (n-1)-tuple in $ a_{n-1}$ ways.  If our n-tuple begins with a $ -1$, then we can add an (n-1)-tuple in $ a_{n-2}$ ways.  Therefore, $ a_{n}=a_{n-1}+a_{n-2}$.  Since $ a_{1}=2$, we have that $ a_{n}=F_{n+2}$, where $ F_{n+2}$ is the (n+2)-th Fibonacci number. The desired answer to your question is $ a_{7}=F_{9}=34$. $ \Box$

If there were a set amount of -1s you have, you can visulize this by using balls and urns.  for this, there can be up to 4 -1s.


We have a circle of radius R.Find the triangle of minimum area which is circumscribed to the circle.

1.Area=s*r where s is semiperimeter.So for area minimum s must be minimum.We have 2(s-b)+2(s-c) must be minimum.So a must be minimum.So triangle ABC is isosceles with AB=AC
2.Area=ah/2 and h=2*a^2*r/(a^2-4*r^2)
3.For  minimum area a=2*3^.5*r and h=3*r
4.Area=3*3^.5*r^2


topology
advanced fields
advanced fields unsolved

Is it true that on any surface homeomorphic to an open disc there exist two congruent curves homeomorphic to a circle?

I know that it is not very rigorous solution but let's try to intersect this surface with itself in such a way that intersection curcve will be different on first and second copies of surfaces ;)


[b]Solve:

sin3xcosx = cos3xsinx[/b]

Where I am so far:
sin3xcosx - cos3xsinx = 0

Then, I'm not sure if I should either do:
sin3x - sinx = 0 (because of angle subtraction identity), or

sin(2x+x)cosx - cos(2x+x)sinx = 0.

Any help is appreciated!

With $ \cos x , \cos 3x \neq 0$ we have $ \tan3x \equal{} \tan x\implies3x \equal{} x \plus{} \pi k\implies x \equal{} \pi k/2$
edit; if k is odd both cos x and cos 3x will be 0 so we're fine...

[quote="Zhero"]Alternatively,

[hide="Solution"]
$ \sin 3x \cos x \minus{} \cos 3x \sin x \equal{} 0$
$ \sin(3x \minus{} x) \equal{} 0$
$ \sin(2x) \equal{} 0$
$ x \equal{} \frac {k\pi}{2}$, with k any integer.
[/hide]

I hope I'm right. :/[/quote]

Ok, so you can just subtract the x from the 3x, without making the 3x into (2x + x), then doing angle addition?
It looks right but just seems too easy...

He used the identity $ \sin(a \minus{} b) \equal{} \sin a\cos b \minus{} \cos a\sin b$ backwards.

By the way, this is much too easy for Intermediate Topics, so post this kind of thing in HSB next time.

If someone doesn't know how to solve a problem, they probably can't know how difficult it is until they see a solution.

hello, one can prove that $ \sin(3x)\cos(x)\minus{}\cos(3x)\sin(x)\equal{}2\sin(x)\cos(x)$.
Sonnhard.


geometry
geometric transformation
reflection

Is anyone else here a Josh Groban fan? I am!!! He is my absolute favorite singer and I love him to death!  :blush:  lol. He has an AMAZING voice, especially for someone his age. I first heard him on Oprah (of course :) ) and I was like WOW! I have his self titled CD and his CD Closer, and I want his Live at the Greek CD/DVD. I just got tickets to his concert in Fort Wayne, IN for March 8, 2005 and am ecstatic!!! So...is anyone else a big fan?

heh, you can tell this is a great topic cause there are so many posts :P

You know I was kidding miss moderator ma'am......  :lol:  heheh, right?..... hey your awesome Ems

There are lots of Grobanites on a forum I used to be pretty active on. Never heard him though. Therefore, last option.

[quote="Valentin Vornicu"][quote="MathFiend"]naters!!! how rude...ur gonna pay for tha... ;)  ;)[/quote]Now really, who the heck is he?  :cool:[/quote]

he's a goode friend of mine, who i recently got to join and he was just being annoying. it's natertater89...lol

[quote="Valentin Vornicu"][quote="MathFiend"]naters!!! how rude...ur gonna pay for tha... ;)  ;)[/quote]Now really, who the heck is he?  :cool:[/quote]

I believe it's the person in Mathfiend's avatar

[quote="MathFiend"]he knows...i talk about josh ALL the time...he wasjust givin me a hard time :)[/quote]In fact I was expecting an answer of the sort:

[quote="MathFiend"]naters!!! how rude...ur gonna pay for tha... ;)  ;)[/quote] :D ...

Behold the new Orlando Bloom  ;)

[edit]srry! i meant to hit the 'quote' button but hit edit instead...this is how the post was b4 i messed it up ~MathFiend[/edit]

[quote="MathFiend"][quote="A+MATH"]Behold the new Orlando Bloom  ;)

oh goodness...i was hoping i was past that. it really was embarrassing...the saga reflects it the most. were u around then? or have u just read it? i don't rmr u in the saga...but i may be mistaken...

and, no, i will not make this the new orlando bloom. josh is simply a great singer who looks goode to. Wink[/quote][/quote]

Look on the bright side. There were probably over a million people just like that. Including about twenty in my town.

arg. you GUYS! http://www.joshgroban.com GO THERE. FEEL THE LOVE! Josh rocks my socks!

[quote="BHorseMath"]mathfiend your never gonna live the saga down you were an airhead in it. lol

I still don't know who this guy is...[/quote]

i kno i kno...i was young and naive... :)


Daca numerele $\ a,b,c$ sunt strict pozitive si mai mici ca $\ 1$, atunci avem $\ 5-(2a+3c+8b)+ab+bc+ca+2(a^2+b^2+c^2)>0$.

Pai considera ca o functie de gradul 2 in a.

poti da o solutie completa??ca eu nu sunt prea familiarizat cu functiile de gradul 2.Apoi iti arat si solutia mea.

Sunt destul de multe calcule, dar ideea este ca daca pentru $f(x)=ax^{2}+bx+c$ avem $a>0$ si $\Delta=b^{2}-4ac\leq 0$ atunci $f(x)\geq 0$.
Considerand $b,c$ fixate si expresia ca o functie in $a$ ar trebui sa iasa daca faci calculele.

p-asta o stiam, se numeste exploatarea trinomului de gradul 2.Uite solutia mea, care e greu de ghicit:
$\ 1-a-b+ab>0$
$\ 1-b-c+bc>0$
$\ 1-c-a+ca>0$
$\ b^{2}-3b+\frac{3}{4}>0$
$\ c^{2}-c+\frac{1}{4}>0$
$\ b^{2}+b>0$,$\ c^{2}+c>0$,$\ a^{2}+a>0$
$\left(a-\frac{1}{2}\right)^{2}>0$
$\left(b-\frac{1}{2}\right)^{2}>0$
$\left(c-\frac{1}{2}\right)^{2}>0$
   Adunand inegalitatile evidente de mai sus obtinem inegalitatea mea. :wink:

si totusi
cand $a$ si $c$ tind la $0$, si $b$ tinde la $1$, ajungem la
$-1\geq 0$...
de exemplu
$a=c=0,0000000000000000000001$.
$b=0,9999$.
termenii care contin $a$ sau $c$ sunt neglijabili, pot fi considerati $0$. ramane sa aratam ca $5-8b+2b^{2}>0$, care nu prea e adevarat, deoarece $b$ e apropiat de $1$, si $5-8+2=-1$.
inegalitatea nu prea e corecta....

cred ca n-am calculat eu prea corect.. :oops:


I want to find a complete sample of either a fall startup or the team scramble.  Please show links.

Please, somebody reply!!!!!!!  I really want one.


I came across two questions that I simply can't seem to find the answer on my own.

(1) (r^2 + 2rs + s^2)/(rs - r) divided by (r^2 - s^2)/(s^2 - s)

I can factor every expression given except for the numerator of the left side fraction: (r^2 + 2rs + s^2).

(2) (a^2 - 5)/(a - b) + (b^2 - 5)/(b - a)

NOTE:

a) Can (a^2 - 5) and (b^2 - 5) be factored?

b) Can I rewrite (b - a) as ( a + b) after multiplying (b - a) by (-1)?

Thank you

[quote]I can factor every expression given except for the numerator of the left side fraction: (r^2 + 2rs + s^2). [/quote]

isn't that a perfect square binomial?

a) Can (a^2 - 5) and (b^2 - 5) be factored?

[hide="Hint"] a^2 - 5 = a^2 - 4 - 1[/hide]

[quote="sharkman"]

I can factor every expression given except for the numerator of the left side fraction: (r^2 + 2rs + s^2).

[/quote]
That is $(r+s)^{2}$

[hide="1"]
$\begin{eqnarray*}\frac{\frac{r^{2}+2rs+s^{2}}{rs-r}}{\frac{r^{2}-s^{2}}{s^{2}-s}}&=&\\ \frac{(r+s)^{2}}{r(s+1)}\cdot\frac{s(s-1)}{(r+s)(r-s)}&=& \end{eqarray*}$
[/hide]
You should be able to handle it from there.

[quote="sharkman"]b) Can I rewrite (b - a) as ( a + b) after multiplying (b - a) by (-1)?[/quote]

If you meant $a-b$, then yes.


Solve in integers the system:

$ 3a^{4}\plus{}2b^{3}\equal{}c^{2}$

$ 3a^{6}\plus{}b^{5}\equal{}d^{2}$

[hide="hint"]what is b mod 3, infinit descent[/hide]

a=b=c=d=0 satisfies.
If $ a,b,c,d\neq 0$,
first equation tells us that $ 2b^3 \equiv c^2\equiv0,1(mod 3)$
If $ 2b^3\equiv1(mod 3)\Rightarrow b^5\equal{}b^3b^2\equiv 2 (mod 3)\Rightarrow d^2\equiv2(mod 3)$, contradiction.
Also $ 2b^3\equiv0(mod 3)$ is not possible as can be proved easily by infinite descent.

Hi, delegat.  :)

From where can you find the Bulgaria 2007 problems?

[quote]Hi, delegat. Smile

From where can you find the Bulgaria 2007 problems?[/quote]

Sorry for late reply

Deputy leader of Bosnia at last IMO ( H . Jamak) gave me that problem He has complete book with problems and solutions from Bulgarian M. C. 2007 ( I think that he got it in Vietnam  from Bulgarian team leader)

Ah, I see.

Can you post some of these competition problems?

I will  try to do so Also  yesterday I posted some problems from Russian regional comp. 2007

But you can find some problems at [url]http://www.imo.org.yu[/url] (there is a topic in forum about Bulgarian TSE 2007)


Let's consider the inequality $ a^2 \plus{} b^2 \plus{} c^2 \ge ab \plus{} bc \plus{} ac$. We can prove this by application of the Trivial Inequality:

$ (a \minus{} b)^2 \ge 0 \implies a^2 \plus{} b^2 \ge 2ab \\
(b \minus{} c)^2 \ge 0 \implies b^2 \plus{} c^2 \ge 2bc \\
(a \minus{} c)^2 \ge 0 \implies a^2 \plus{} c^2 \ge 2ac.$

Summing them all up we obtain the desired inequality (after division by $ 2$). Now the equality condition for each of those inequalities combined turns out to be $ a \equal{} b \equal{} c$ by consequence of the trivial inequality. So this implies the equality condition for $ a^2 \plus{} b^2 \plus{} c^2 \ge ab \plus{} bc \plus{} ac$ is also $ a \equal{} b \equal{} c$ right? So in general, if one proves a given inequality that has an equality condition, can one find this condition by consequence of how it was proved?

Yes -- if $ f_1(x) \geq g_1(x)$ with equality if and only if $ x \in S$ and if $ f_2(x) \geq g_2(x)$ with equality if and only if $ x \in T$ then $ f_1(x) \plus{} f_2(x) \geq g_1(x) \plus{} g_2(x)$ with equality if and only if $ x \in S \cap T$.  (Here $ x$ represents some collection of variables.)

In the case you present, we have three equality conditions; represented as sets, they are $ \{(x, x, y) \mid x, y \in \textbf{R}\}$, $ \{(x, y,x)\mid x,y \in \textbf{R}\}$ and $ \{(x,y,y)\mid x,y \in \textbf{R}\}$, and their intersection is $ \{(x,x,x) \mid x \in \textbf{R}\}$.

Note that it's quite possible that we could have this intersection empty: for example, from $ x^2 \geq 0$ and $ (x \minus{} 1)^2 \geq 0$ we can certainly deduce $ x^2 \plus{} (x \minus{} 1)^2 \geq 0 \plus{} 0 \equal{} 0$, and equality will hold in this later inequality if and only if $ x \in \{0\} \cap \{1\} \equal{} \varnothing$, i.e., never.


Hello all! I'm a 2nd year math PhD student looking to become a quant. It's my understanding from talking to other students and my potential advisor that quants generally don't spend their entire careers as quants. They do it for 5-10 years, burn out, and then do something more "relaxing".

For those that do leave their jobs as quants, what do they do? Are there positions outside of NYC? Do these new jobs use mathematics, and is the mathematics the same or as interesting? Finally, how well are ex-quants paid?

On a completely different topic, when should I start looking for internships? I just finished my 1st year and am hoping to complete my qualifiers this year.

[quote="Xevarion"]I'm not even a quant, let alone an ex-quant, but I thought I should at least point out that part of the point of working in a high-powered quantitative finance job for just 5-10 years and then doing to do something more "relaxing" (or whatever you're seeking) is to earn a lot of money in phase I so that you don't have to worry about the salary you're getting in phase II.[/quote]

I will note that very few people actually choose this path, though many intend to.

The two most common paths are:

1) Go in, stay in, even if you intended to leave after 5 years.

2) Go in, decide you don't like it, leave within 2 years.

The next most common path is:

3) Go in for 4-8 years, quit, then come back in 1-3 years.

The least common is:

4) Go in for 4-8 years, leave.

I'm the so-called exception that proves the rule.  Many people at Shaw thought I'd be back at it at some point.  Had AoPS not worked out so well, they might have been right.

[quote="rrusczyk"]I'm the so-called exception that proves the rule.  Many people at Shaw thought I'd be back at it at some point.  Had AoPS not worked out so well, they might have been right.[/quote]

I think you guys did an amazing job!  :)

Sure start-ups are attractive but you are very dependent on the money of venture capitalists. Thus many people as Richard first wanted to have a solid financial fundament before getting started on a more risky endeavour. Another good example is [url=http://www.google.com/search?hl=en&q=Jeffrey+Bezos+amazon+hedge+fund&btnG=Google+Search]Jeff Bezos[/url] who is the founder of Amazon:

"Bezos graduated summa cum laude and Phi Beta Kappa in Electrical Engineering and Computer Science from Princeton University in 1986. After college, Bezos joined FITEL, a high-tech start-up company in New York building a network to conduct international trade. Two years later, Bezos began working for Bankers Trust Company in New York, where he led the development of their computer systems and became the company's youngest vice president in 1990. From 1990 to 1994, Bezos worked for D.E. Shaw & Co where he helped build one of the most technically sophisticated quantitative hedge funds on Wall Street." [url=http://mitworld.mit.edu/video/417/]Source[/url]

[quote="Xevarion"][quote="rrusczyk"]I will note that very few people actually choose this path, though many intend to. [/quote]
In fact, almost all of my friends who are working for hedge funds or big banks think that they're going to do some variation of this. Go back to grad school, start a biotech firm, retire by 30 -- everyone seems to think that working in finance for a few years provides financial security and a more secure platform from which to launch into what they really want to do with their lives.

I'm interested to see how many of us actually do what we say we will...[/quote]

I'm betting the number is quite small, but it'll be great if it isn't.

As for the Bezos example, I'll note that he was working on building internet businesses at Shaw, so his leaving to found Amazon wasn't so much a change of career as it looks.

[quote]everyone seems to think that working in finance for a few years provides financial security and a more secure platform from which to launch into what they really want to do with their lives. [/quote]

There is a similar phenomenon in academia, if you substitute "a well-known area" for "finance" and "tenure" for "financial security."

The areas and approaches which tend to produce interesting new results tend to be the ones which are obscure or have been overlooked by other researchers.  Of course, these are exactly same the areas where it is hard to:

-- get hired (because departments would rather hire people in a secure area than take a chance on something new)

-- get funding (because you have to convince the grant agency that a project has a reasonable chance of success, and how do you know that if it hasn't been tried before?)

-- get tenure (because the scientific community may not recognize the importance of your results in time).

And so, many of my friends who want to go into academics to work on something new and interesting are postponing their plans until they get tenure.  They do a senior thesis in a typical area so they can get into grad school; in grad school, they will do a dissertation in an area where they can get hired; once they get hired, they will work in an area in which they can get tenure.

The problem is, by that point, they have invested so much of their time, energy, and reputation in one area that it is hard to switch to another.  By then, too, many people find they would rather work on something which is familiar and comfortable to them than taking a leap into the great unknown.

Some people do switch.  But, as Mr. Rusczyk said about finance, it would be great if more people did, but it is the exceptions which prove the rule.

Very interesting point -- I'd never heard this explanation before.  Very sad point, too.

It's not quite as bad as that, and there are some ways around the obstacles I described:

Getting hired--  Working in an unpopular area makes you less likely to be hired.  But, you're more likely to be hired if you have graduated from a top graduate program (which helps the prestige of the department), have published prolifically (which shows that you already have an active research life) and are qualified ot teach many different things (which helps more than anything, because most departments don't have the resources to hire a specialist to teach in each area).

Getting recognition--  People tend to write off the work done in unpopular areas because they don't seem important to the field.  So, it's important to understand the broader significance of your work and articulate that clearly (because most people won't see the significance of anything unless you tell them).

Getting grants--  It's hard to get grants for projects which have no evidence of success.  So, wait until the project is nearly done (so that there are results to show) to apply for a grant, and then use that grant to fund the next project.

Still, all other things being equal, someone working in a "popular" area has a much better chance of getting hired.  How much better (or luckier) do you have to be to compensate for that fact?  There's always room at the top, but the uncertainty of the outcome drives many people away, and those who do try and don't succeed are forced out of the field.  Others hedge their bets, working partly in a conventional area, and party in the area they want to work in.  On the bright side, I've gone straight for what I want to work on, and that has worked very well for me so far.


inequalities
trigonometry
geometry unsolved
geometry

In any triangle $ ABC$ prove that :

\[ \cos^4 \frac{A}{2}\plus{}\cos^4 \frac{B}{2}\plus{}\cos^4 \frac{C}{2}\le \frac{(a\plus{}b\plus{}c)^3}{16abc}\]

hello, using the well-known formulas
$ \cos(\alpha/2)=\sqrt{\frac{s(s-a)}{bc}}$
$ \cos(\beta/2)=\sqrt{\frac{s(s-b)}{ac}}$
$ \cos(\gamma/2)=\sqrt{\frac{s(s-c)}{ab}}$
and inserting these in your inequality we get after an ugly computation
the following equivalent inequality
$ -1/16\,\left( a+b+c \right) ^{2}\cdot$
$ {\frac{\left(a^2bc+ab^2c+abc^2+a^4-2a^3b-2a^3c+2a^2b^2+2a^2c^2-2b^3a+b^4-2b^3c+2b^2c^2-2c^3a-2c^3b+c^4\right)}{a^2b^2c^2}\geq0}$
setting
$ a=y+z,b=x+z,c=x+y$
in the inequality above we get
$ {1/2\,{\frac { \left( y+z+x \right) ^{2} \left( -2\,{y}^{2}{x}^{2}+{y}^
{3}x+{y}^{3}z-2\,{y}^{2}{z}^{2}-2\,{z}^{2}{x}^{2}+{z}^{3}x+{z}^{3}y+y{
x}^{3}+z{x}^{3} \right) }{ \left( y+z \right) ^{2} \left( x+z \right)
^{2} \left( x+y \right) ^{2}}}\geq0}$ and this is true because
the second factor can written as
$ xy(y-x)^2+yz(y-z)^2+zx(z-x)^2$
Sonnhard.


superior algebra
superior algebra solved

show An is genearated by 3 cycle where n>2.

Every element in $A_{n}$ is a product of an even number of transpositions. For two adjacent transpositions we have the following possibilities:
1). The are distinct of the form $(a,b)$ and $(c,d)$. Then $(a,b)(c,d) = (a,c,b)(a,c,d)$.

2). They have 1 elemnt in common, thus they are of the form $(a,b)$,$(a,c)$.Then $(a,b)(a,c) = (a,c,b)$.

So we can pair our transpositions and the product of these transpositions in each pair is generated by $3$-cycles.