Given $ a, b, c \geq\ 0.$
Prove that: $ \sqrt[]{a^2\plus{}4bc}  \plus{} \sqrt[]{b^2\plus{}4ca}   \plus{} \sqrt[]{c^2\plus{}4ab}  \geq\ \sqrt[]{15(ab\plus{}bc\plus{}ca)}$

[quote="nguoivn"]Given $ a, b, c \geq\ 0.$
Prove that: $ \sqrt []{a^2 \plus{} 4bc} \plus{} \sqrt []{b^2 \plus{} 4ca} \plus{} \sqrt []{c^2 \plus{} 4ab} \geq\ \sqrt []{15(ab \plus{} bc \plus{} ca)}$[/quote]
<=>$ (\sqrt []{a^2 \plus{} 4bc} \plus{} \sqrt []{b^2 \plus{} 4ca} \plus{} \sqrt []{c^2 \plus{} 4ab})^2 \geq\ {15(ab \plus{} bc \plus{} ca)}$
Use CS we have $ \sqrt []{a^2 \plus{} 4bc}\sqrt []{b^2 \plus{} 4ca} \geq a\sqrt {ac} \plus{} b\sqrt {bc}$
Last let $ \sqrt a \equal{} x$ and Use $ \sum xy(x^2 \plus{} y^2)\geq 2({xy}^2 \plus{} {yz}^2 \plus{} {zx}^2),\sum x^4\geq \sum {xy}^2$
We have done

[quote="Allnames"] Use $ \sum xy(x^2 \plus{} y^2)\geq 2(xy \plus{} yz \plus{} zx),\sum x^4\geq \sum xy$
We have done[/quote]
But it's wrong. Try $ a\equal{}b\equal{}c\equal{}\frac{1}{2}.$ :wink:

[quote="nguoivn"]Given $ a, b, c \geq\ 0.$
Prove that: $ \sqrt []{a^2 \plus{} 4bc} \plus{} \sqrt []{b^2 \plus{} 4ca} \plus{} \sqrt []{c^2 \plus{} 4ab} \geq\ \sqrt []{15(ab \plus{} bc \plus{} ca)}$[/quote]
by holder we have
$ \sqrt []{a^2 \plus{} 4bc} \plus{} \sqrt []{b^2 \plus{} 4ca} \plus{} \sqrt []{c^2 \plus{} 4ab} \geq \sqrt{\frac{(\sum a^2\plus{}4\sum ab)^3}{\sum(a^2\plus{}4bc)^2}}$
we need prove that $ \sqrt{\frac{(\sum a^2\plus{}4\sum ab)^3}{\sum(a^2\plus{}4bc)^2}}\ge \sqrt{15(ab \plus{} bc \plus{} ca)}$
$ \Leftrightarrow \sqrt{\frac{(p^2\plus{}2q)^3}{p^4\minus{}4p^2q\plus{}18q^2\minus{}20pr}}\ge \sqrt{15q}$
$ \Leftrightarrow p^6\minus{}9p^4q\plus{}72p^2q^2\minus{}262q^3\plus{}300qpr\ge 0$
assume $ p\equal{}a\plus{}b\plus{}c\equal{}1, q\equal{}\sum ab,r\equal{}abc$
$ \Leftrightarrow f(r)\equal{}1\minus{}9q\plus{}72q^2\minus{}262q^3\plus{}300qr\ge 0$
$ f'(r)\ge 0 \Rightarrow f(r)\ge f(\frac{4q\minus{}1}{9})\equal{}\frac{1}{3}(1\minus{}3q)(262q^2\minus{}118q\plus{}3)\ge 0$

Allnames was wrong at use BCS:
$ \sqrt[]{(a^2\plus{}4bc)(b^2\plus{}4ca)} \geq\ \sqrt[]{2}(a\sqrt[]{ac} \plus{} b\sqrt[]{bc})$
Are you sure equality can occurs ? Because equality in BCS is only occurs when $ \frac{a}{x} \equal{} \frac{b}{y}$
 :)

[quote="mitdac123"]Bunhiacovski's proof is also wrong :P[/quote]
where..? ex? :P

[quote="bunhiacovski"]
we need prove that $ \sqrt {\frac {(\sum a^2 \plus{} 4\sum ab)^3}{\sum(a^2 \plus{} 4bc)^2}}\ge \sqrt {15(ab \plus{} bc \plus{} ca)}$
[/quote]
Try $ a\equal{}b\equal{}1$ and $ c\equal{}0.$ :wink:

[quote="Sunjee"]we can use Mean Inequality and AM-GM ![/quote]
Show it. Thank you!

[quote="Sunjee"]we can use Mean Inequality and AM-GM ![/quote]
And what is Mean ineq ,detail it please

Hi, this problem is easy. But I don't like my solution. (I used AM-GM and pqr)
Used AM-GM, we only need to prve that:
$ 216(a^2\plus{}4bc)(b^2\plus{}4ca)(c^2\plus{}4ab) \geq\ (11\sum\ ab\minus{} \sum\ a^2)^3$
With $ a\plus{}b\plus{}c\equal{}3$, it's becames:
$ 216(125r^2\plus{}4q^3\minus{}180qr\plus{}432r) \geq\ (13q\minus{}9)^3$
$ f'(r)\equal{}216(432\plus{}250r\minus{}180q) \geq\ 0$
$ \Rightarrow  f(r) \geq\ f(0) \geq\ 0$ always trues with $ q \leq\ 2.59$
With $ q \geq\ 2.59$, we have: $ f(r) \geq\ f(\frac{4q\minus{}9}{3}) \equal{}(3\minus{}q)(q\plus{}4.48)(q\minus{}2.02) \geq\ 0$
$ \Rightarrow$ q.e.d
@Sunjee: Can u post your solution, please  :)

[quote="nguoivn"]
Used AM-GM, we only need to prve that:
$ 216(a^2 \plus{} 4bc)(b^2 \plus{} 4ca)(c^2 \plus{} 4ab) \geq\ (11\sum\ ab \minus{} \sum\ a^2)^3$
[/quote]
Beautiful idea! :lol:

$ M^{\frac12}\geq M^{\minus{}1}$ and $ M^{\minus{}1}\leq M^{1}$ and AM-GM

To allnames :
First, squaring  .Next, use am-gm :P

[quote="nguoivn"]
$ f'(r) \equal{} 216(432 \plus{} 250r \minus{} 180q) \geq\ 0$[/quote]
why? :wink:

Because of Schur: $ 9 \plus{} 3abc \geq\ 4(ab \plus{} bc \plus{} ca)$
PS: If $ k \equal{} 5$, ineq is not true.If $ k < 3.59$, we can prove easily by:
$ (a^2\plus{}kbc)(b^2\plus{}kca)(c^2\plus{}kab) \geq\ \frac{(1\plus{}k)^3}{27}(ab\plus{}bc\plus{}ca)^3$
So, $ k \equal{} 4$ is nice  :)

[quote="nguoivn"]Because of Schur: $ 9 \plus{} 3abc \geq\ 4(ab \plus{} bc \plus{} ca)$
PS: If $ k \equal{} 5$, ineq is not true.If $ k < 3.59$, we can prove easily by:
$ (a^2 \plus{} kbc)(b^2 \plus{} kca)(c^2 \plus{} kab) \geq\ \frac {(1 \plus{} k)^3}{27}(ab \plus{} bc \plus{} ca)^3$
So, $ k \equal{} 4$ is nice  :)[/quote]
$ \Rightarrow f'(r)\ge \minus{}68688\plus{}33120q$
if $ f'(r)\ge0 \Leftrightarrow p\ge 2.0739130434782608696$  why?  :P

I don't know what you say  :maybe:
$ f'(r) \equal{} 432 \plus{} 250r \minus{} 180q \equal{} 48(9 \plus{} 3r \minus{} 4q) \plus{} 12q \plus{} 106r > 0$
If you want to ask about the ineq:
$ (a^2 \plus{} kbc)(b^2 \plus{} kca)(c^2 \plus{} kab) \geq\ \frac {(1 \plus{} k)^3}{27}(ab \plus{} bc \plus{} ca)^3$
with $ k < 3.59$
I think you can kill it by yourself because it's too easy. (It has been mentioned on toanthpt.net)
I think you understood  :)

If using my old way, we'll have to prove:
$ 13824(a^2\plus{}4bc)(b^2\plus{}4ca)(c^2\plus{}4ab) \geq\ (41\sum\ ab \minus{} \sum\ a^2)^3$
Is this ineq true? :maybe:

No, it's wrong.  :(  $ a\equal{}0$ and $ b\equal{}c\equal{}1.$ :wink:

See also here:
http://www.artofproblemsolving.com/community/c6h1242915p6360695

[quote=Allnames][quote="bunhiacovski"][quote="nguoivn"]Hi, this problem is easy. But I don't like my solution. (I used AM-GM and pqr)
Used AM-GM, we only need to prve that:
$ 216(a^2 \plus{} 4bc)(b^2 \plus{} 4ca)(c^2 \plus{} 4ab) \geq\ (11\sum\ ab \minus{} \sum\ a^2)^3$
[/quote]
please.., detail :wink:[/quote]
I dont't think it is easy,nguoivn
To bunhiacovski:since $ (\sqrt []{a^2 \plus{} 4bc} \plus{} \sqrt []{b^2 \plus{} 4ca} \plus{} \sqrt []{c^2 \plus{} 4ab})^2 \geq\ {15(ab \plus{} bc \plus{} ca)}$
then$ 2 \sum \sqrt []{a^2 \plus{} 4bc}\geq 11\sum\ ab \minus{} \sum\ a^2$ then use AMGM
[hide="to nguoivn"]em nghĩ anh nên post mấy bài này ở mục http://www.mathlinks.ro/index.php?f=52 này thì hợp hơn đó ạ :lol: [/hide][/quote]

well the link you used seems to be going to the artofproblemsolving's home. not the link related to this problem

maybe the link could have some problems

[quote=arqady]It seems that the following inequality is true too.
Let $ a,$ $ b$ and $ c$ are non-negative numbers. Prove that:
\[ \sqrt[3]{a^2\plus{}4bc} \plus{} \sqrt[3]{b^2\plus{}4ca} \plus{} \sqrt[3]{c^2\plus{}4ab} \geq\ \sqrt[3]{45(ab\plus{}bc\plus{}ca)}\][/quote]
I just proved this inequality. It's true! :-D


induction
combinatorics solved
combinatorics

Given a graph with $2n+1$ vertices, such that for any $n$ vertices, there exists another one connected to all these $n$ vertices. Show that there is a vertex connected to all the other $2n$ vertices.

Hey, it seems that I might have a solution.

We construct an $(n+1)$-clique: It's very easy to prove that there is a $3$-clique in the graph. Now, if we have constructed a $k$-clique , with $k\leq n$, choose a vertex not in the clique who is connected to all the vertices in the clique. Thus the former vertices and the new one form a $(k+1)$-clique. We can go on like this as long as $k\leq n$.

Ok, so now we have an $(n+1)$-clique. Choose the vertices in the graph that are not in the clique ($n$ of them). There is a vertex connected to all those $n$, who is also in the clique, so that this vertex is connected with all other vertices,  QED.

Nice solution, very easy to explain (unlike mine)

Keep up the good work

hey.. yes iandrei's solution is wonderful.. andrei negut mind if you post your solution... I would also be interested in another solution... if you don't mind, do you?  :D

So, pick out of the $2n+1$ vertices several pairs of points that are not connected between them (So we will have the pairs $\left(A_1,A_2\right),\ \left(A_3,A_4\right),\ ...,\ \left(A_{2k},A_{2k+1}\right)$, such that $A_{2i}$ is not connected to $A_{2i+1}$, and the $A_i$'s are all distinct). We extract these pairs points until we either have just one point left, or all the remaining points are all connected between them.

Well, suppose that we extracted $k$ such pairs of points ($k$ can be any number between $0$ and $n$) and the remaining $2n+1-2k$ points (which we will call FREE points) are all connected between them. One element of each pair we will call a "left point", and the other will be "a right point".

Now let's look at the free points. Let's show that there are $n-k+1$ of the free points that are connected to all the $k$ left points. Suppose we have $L\leq n-k$ such free points. Then we can find another, because of the following argument: Select the $L$ free points, the $k$ left points, and $n-L-k$ other free points. There must be another point connected to these $n$ points, and that point can't be a right point (because any right point is not connected to at least one left point). So that point is a free point and connected to all the left points, and so we found $L+1$ free points that are connected to all the left points.

In the same way we must have $n-k+1$ free points connected to all the right points. Because there are only $2n-2k+1$ free points, there must exist a free point connected to all the left points AND all the right points. And that point will be connected to all the other points.

I don't know why, but I suddenly feel like posting my solution...

The problem asks to prove the following statement:

[b](A)[/b] Given a graph with $2n + 1$ vertices, if for any $n$ vertices there exists another vertex connected to all these $n$ vertices, then there is a vertex connected to all other $2n$ vertices of the graph.

Proving this statement [b](A)[/b] is equivalent to proving the inverse statement:

[b](B)[/b] Given a graph $G$ with $2n + 1$ vertices, if there is no vertex of $G$ connected to all other $2n$ vertices of the graph $G$, then there exist $n$ vertices of $G$ such that there doesn't exist another vertex of $G$ connected to all these $n$ vertices.

This can be slightly rewritten:

[b](B')[/b] Given a graph $G$ with $2n + 1$ vertices, if there is no vertex of $G$ connected to all other $2n$ vertices of the graph $G$, then there exist $n$ vertices of $G$ such that every other vertex of $G$ is connected not to all of these $n$ vertices.

Now we consider the complementary graph $G'$ of the graph $G$. This is the graph whose vertices are the $2n + 1$ vertices of $G$, but two vertices are connected to each other in $G'$ if and only if they are not connected to each other in $G$. Now, we can rewrite the statement [b](B')[/b] from the viewpoint of the graph $G'$ (instead of $G$), replacing "connected" by "not connected" and vice versa:

[b](C)[/b] Given a graph $G'$ with $2n + 1$ vertices, if there is no vertex of $G'$ which is not connected to any of the other $2n$ vertices of the graph $G'$, then there exist $n$ vertices of $G'$ such that every other vertex of $G'$ is connected to at least one of these $n$ vertices.

This statement can be slightly simplified:

[b](C')[/b] Given a graph $G'$ with $2n + 1$ vertices, if each vertex of $G'$ is connected to at least one other vertex, then there exist $n$ vertices of $G'$ such that every other vertex of $G'$ is connected to at least one of these $n$ vertices.

Now this is easy to prove: First, remove as many edges of the graph $G'$ as you can without breaking the condition that each vertex of $G'$ is connected to at least one other vertex. After this removal of edges, the graph $G'$ becomes a forest (because as long as our graph is not a forest, it has a cycle, and thus we can remove a further edge (namely, any arbitrary edge of this cycle) from this graph without breaking the condition that each vertex of the graph is connected to at least one other vertex); denote this forest by $H$. It is clear that, in order to prove the assertion [b](C')[/b], it is enough to show that there exist $n$ vertices of this forest $H$ such that every other vertex of $H$ is connected to at least one of these $n$ vertices (this is enough because if two vertices are connected in $H$, they are surely connected in $G'$). But $H$ is a forest and therefore bipartite; thus, we can color the vertices of the forest $H$ black and white such that no two adjacent vertices have the same color. Either the number of black vertices or the number of white vertices is $\leq n$ (in fact, if both of these numbers were $> n$, then they would be $\geq n+1$, so their sum would be $\geq 2\left(n+1\right)$, but this impossible because their sum is the number of all vertices of $H$ and thus equal to $2n + 1$). We can WLOG assume that the number of white vertices of $H$ is $\leq n$. Thus, by taking all white vertices of $H$ and (maybe) a few black vertices, we can obtain a set $S$ of $n$ vertices of $H$. Every other vertex of $H$ is connected to at least one of the $n$ vertices from this set $S$ (in fact, every other vertex $q$ of the graph $H$ must be black (because all white vertices are in the set $S$), and it must be connected to at least one other vertex of $H$, but this other vertex must be white since $H$ is bipartite, and thus this other vertex must lie in $S$ (since all white vertices are in the set $S$), so our vertex $q$ is connected to at least one vertex from the set $S$). This completes the proof of [b](C')[/b], and thus the problem is solved.

  Darij

I find andrei negut's solution kind of neat.  And darij's is really nice.

This problem can also be solved with a combination of the well-ordering priniciple and induction.  Let $v$ be the vertex of largest degree, and assume $d(v)<2n$. We will show some vertex has a larger degree, which gives the desired contradiction.  Consider a vertex $u$ such that $vu\not\in E$ and denote $G\backslash\{v,u\}$ as $G'$.  Then any $n$-vertex subset containing $u$ must be adjacent to some vertex in $G'$.  This implies that any $n-1$-vertex subset of $G'$ must be adjacent to some vertex in $G'$.  Since $|G'|=2(n-1)+1$, it follows by induction that some vertex in $G'$, say $w$, is adjacent to all other vertices in $G'$, as well as $u$.  It can be checked that $d(w)>d(v)$ which completes the proof.

Could this be proven using induction?

Say n = 1. That's easy to prove.

n = k would be the induction hypothesis. And when n = k + 1, you'd just add two more vertices. Work one of them into the graph, and the remaining vertex could be added in as well.

Sorry if this doesn't make any sense or if it's just plain wrong. I'm kinda new at this... :blush:

Sorry to tell this after all the years, but there is something I don't understand in Negative_3's posting:

[quote="Negative_3"]it follows by induction that some vertex in $G'$, say $w$, is adjacent to all other vertices in $G'$, as well as $u$.[/quote]

Why "as well as $u$"?

PS. I have edited all the above posts (including mine) to fix some typos, TeX the formulas, and sometimes (I believe) slightly improve the writing ("Then we can find another, because of the following argument" used to be "Then we can find another and"). While I am here, let me add that both Andrei Negut and iandrei have beautiful proofs.


geometry
circumcircle
geometric transformation
homothety
geometry unsolved

Let $ABC$ a triangle and P an interior point, the lines $AP, BP$ and $CP$ intersects the circumcircle of the triangle $ABC$ in $A_{1}$, $B_{1}$ and $C_{1}$, respectively, $M_{a}$, $M_{b}$ and $M_{c}$ are the midpoints of $BC$, $AC$ and $AB$, respectively, finally let $A_{2}$ the simetric of $A_{1}$ respect $M_{a}$, define similarly to $B_{2}$ and $C_{2}$. Prove that the circumcircle of the triangle $A_{2}B_{2}C_{2}$ pass throught a fixed point.

Note: It works for P: exterior point.  :wink:

O: circuncentro, G: baricentro; H: ortocentro; A': pto. medio de AA1; B',C': analogamente.
G es baricentro de AA1A2 y, en particular, divide al segmento A2A' en la relacion de 2:1. Lo mismo ocurre con B2B' y C2C'.
A', B' , C'  y O estan sobre una circunferencia (de diametro OP) . Asi, por la homotecia de radio 2 y centro G tenemos que A2, B2, C2 y H tambien estan sobre una circunferencia.

Really nice solution!  :o

The theorem also holds when, instead of taking the symmetrics of A,B,C respect $M_{A},M_{B},M_{C}$ you take the symmetrics respect the lines AB,BC,CA.
This is proved by taking the isogonal conjugate of P... but is there a more direct proof to this? (i thought this case were simplier... maybe it's not)

This picture refers to the original problem:

[quote]G es baricentro de AA1A2[/quote]
¿¿Como afirmas esto??
Muy buena solucion... :lol:

Ma es pto. medio de A1A2 y G divide al segmento AMa en la relacion de 2:1.

Can someone please translate the solution by jonfv?

I don't know Portuguese but it's very similar to italian, so I can translate  :D

Let O be the circumcentre of ABC, H the orthocentre of ABC, G the centroid of ABC.
Let A',B',C' be respectively the midpoints of $ AA_{1},BB_{1},CC_{1}$.
Note that G is also the centroid of $ AA_{1}A_{2}$, therefore we have $ \frac{A_{2}G}{GA'}= 2$. The same for $ B_{2}B'$ and $ C_{2}C'$.
Since $ \angle OA'P = \angle OB'P = \angle OC'P = 90$, $ A',B',C',O$ lie on the same circle.
Using an homotety with centre G and factor -2, we discover that $ A_{2},B_{2},C_{2},H$ are on the same circle, too.  :)

To jonfv and conejita: Please use English in this forum!  :mad:


inequalities
quadratics
function
inequalities unsolved
Vietnam

Prove that for any real numbers a,b,c such that $a^{2}$ + $b^{2}$ + $c^{2}$=9
2(a+b+c) - abc \leq 10

[quote="Thao"]Prove that for any real numbers $a,b,c$ such that $a^{2}+b^{2}+c^{2}=9$
$2(a+b+c)-abc \leq 10$[/quote]
Let $a=\min\{a,b,c\}.$
If $a\leq0$ then $2(a+b+c)-abc \leq2a+2\sqrt{2(b^{2}+c^{2})}-a\cdot\frac{b^{2}+c^{2}}{2}=$
$=2a+2\sqrt{2(9-a^{2})}-\frac{a(9-a^{2})}{2}.$
But for $a\leq0$ holds $2a+2\sqrt{2(9-a^{2})}-\frac{a(9-a^{2})}{2}\leq10\Leftrightarrow$
$\Leftrightarrow4\sqrt{2(9-a^{2})}\leq20+5a-a^{3}\Leftrightarrow(a+1)^{2}(a^{4}-2a^{3}-7a^{2}-24a+112)\geq0.$
If $a>0$ then it's true but I found ugly proof only.

For $a>0$ we have $a,b,c>0$
Take $x=a+b+c,\ y=ab+bc+ca,\ z=abc$ with Schur Inequality we have
$a(a-b)(a-c)+b(b-a)(b-c)+c(c-a)(c-b)\geqslant0\\ \Longleftrightarrow x^{3}+9z\geqslant4xy$
The inequality (what we need to prove)
$2x-z\leqslant10$
comes to the stronger one
$2x-\frac{4xy-x^{3}}{9}\leqslant10$
Notice that $x^{2}-2y=9$, so the one becomes
$2x-\frac{2x(x^{2}-9)-x^{3}}{9}\leqslant10\\ \Longleftrightarrow x^{3}+90\geqslant 36x$
Actually with AM-GM we have
$x^{3}+90=x^{3}+45+45\geqslant3\sqrt[3]{x^{3}\cdot45\cdot45}>36x$
Done! :D

[hide]
If $a,b,c$ all nonnegative, it's clearly true when $a+b+c<5$. On the other hand, $a+b+c\geq 5$ requires $a,b,c\geq 1$ (to show $a\geq 1$, use $(b+c)^{2}\leq 2(b^{2}+c^{2})$ to get quadratic in $a$). We then have $abc\geq 1$ and $a+b+c\leq 3\sqrt{3}<\frac{11}{2}$, which completes this case.

Otherwise, at least one variable is negative. WLOG let $c=\min \{ a,b,c\}$, and let $r=\sqrt{ \frac{a^{2}+b^{2}}{2}}$. Then we have
\[2(a+b+c)-abc\leq 2(2r+c)-r^{2}c \]
by RMS-AM and AM-GM ($-c$ is positive). Now let $z=-c$ so that $2r^{2}+z^{2}=9$ and we must prove
\[2(2r-z)+r^{2}z\leq 10 \]
which is equivalent to
\[4\sqrt{18-2z^{2}}+z(5-z^{2})\leq 20 \]
But we have $2\sqrt{18-2z^{2}}\leq 9-z$; since $z\leq 3<9$, this can be verified by squaring both sides, and it reduces to $(z-1)^{2}\geq 0$. Then it remains to prove
\[2(9-z)+z(5-z^{2})\leq 20 \]
which follows from $(z+2)(z-1)^{2}\geq 0$.

[/hide]

[quote="Thao"]Prove that for any real numbers a,b,c such that $ a^{2}$ + $ b^{2}$ + $ c^{2}$=9
2(a+b+c) - abc \leq 10[/quote]
We have:
$ 2(a\plus{}b\plus{}c)\minus{}abc \equal{}2(a\plus{}b)\plus{}c(2\minus{}ab) \leq \sqrt{(a^2\plus{}b^2\plus{}2ab\plus{}c^2)(8\minus{}4ab\plus{}a^2b^2)} \equal{}\sqrt{(9\plus{}2ab)(a^2b^2\minus{}4ab\plus{}8)}$
Put $ ab\equal{}t$,we can prove that $ (9\plus{}2t)(t^2\minus{}4t\plus{}8) \leq 100$(very easy) :wink:

[quote="Thao"]Prove that for any real numbers a,b,c such that $ a^{2}$ + $ b^{2}$ + $ c^{2}$=9
2(a+b+c) - abc \leq 10[/quote]
Let $ a \plus{} b \plus{} c \equal{} 3u,$ $ ab \plus{} ac \plus{} bc \equal{} 3v^2$ ( $ v^2\in\mathbb R$ ) and $ abc \equal{} w^3.$
Hence, $ 2(a \plus{} b \plus{} c) \minus{} abc$ is linear of $ w^3.$
Thus, $ 2(a \plus{} b \plus{} c) \minus{} abc$ gets a maximal value when $ w^3$ gets an extremal value,
which happens when two numbers from $ \{a,b,c\}$ are equal.
Id est, we need to check only one case: $ b \equal{} c.$

[quote="earldbest"][quote="arqady"]

Hence, $ 2(a \plus{} b \plus{} c) \minus{} abc$ is linear of $ w^3.$
[/quote]

What do you mean? :huuh:[/quote]
Let $ f(w^3) \equal{} 2(a \plus{} b \plus{} c) \minus{} abc\equal{}6u\minus{}w^3.$ Hence, $ f$ is a linear function since, $ a^2 \plus{} b^2 \plus{} c^2$ does not depend of $ w^3.$

[quote="Thao"]Prove that for any real numbers $a,b,c$ such that $a^{2}+b^{2}+c^{2}=9$
$2(a+b+c)-abc \leq 10$[/quote]
http://www.artofproblemsolving.com/community/c6h380556p2105124

Let $a,b,c$ be reals such that $a^2+b^2+c^2\le 2.$ Prove that$$a+b+c-abc\le 2.$$
[size=50]p/6230228238[/size]
[url=https://artofproblemsolving.com/community/c4h1114715p5096739]h[/url]

[quote=Thao]Prove that for any real numbers $a,b,c$ such that $a^{2} + b^{2}+ c^{2}=9,$
$$2(a+b+c) - abc \leq 10$$[/quote]
[quote=sqing]Let $a,b,c$ be reals such that $a^2+b^2+c^2\le 2.$ Prove that$$a+b+c-abc\le 2.$$[/quote]
Let $a,b,c$ be real numbers such that $a^2+b^2+c^2 \leq 9$. Prove that
$$-10 \leq 2(a+b+c)-abc \leq 10$$
Let $a,b,c$ be real numbers such that $a^2+b^2+c^2 \leq 2$. Prove that
$$-2 \leq a+b+c-abc \leq 2$$
Let $a,b,c$ be real numbers such that $a^2+b^2+c^2 \leq 3$. [url=https://artofproblemsolving.com/community/c6h2663717p23076435]Prove that[/url]
$$-8 \leq 3(a+b+c)-abc \leq 8$$

[hide=Beautiful.Thank ytChen. ][quote=ytChen][quote=sqing]Let $a,b,c$ be real numbers such that $a^2+b^2+c^2 \leq 2$. Prove that
$$-2 \leq a+b+c-abc \leq 2.$$[/quote]

[b]Solution[/b]. The assumption $a^2+b^2+c^2 \leq 2$ gives $2bc\le a^2+b^2+c^2 \leq 2$, thereby $bc\le1$. It suffices to show
$$(a+b+c-abc)^2\le4.$$From the Cauchy-Schwarz's Inequality, it follows that
\begin{align*}(a+b+c-abc)^2&=[a(1-bc)+(b+c)]^2\\
&\le[a^2+(b+c)^2]\cdot[(1-bc)^2+1]\\
\left(\because a^2+b^2+c^2 \leq 2\right)\,\,&\le(2+2bc)\cdot\left[(bc)^2-2bc+2\right]\\
&=2\left[(bc)^2-2bc+2+(bc)^3-2(bc)^2+2bc\right]\\
&=2\left[2+(bc)^2(bc-1)\right]\\
\left(\because bc\leq1\right)\,\,&\le2\times2=4,
\end{align*}where the equality can be attained when $(a,b,c)=(0,1,1)$ or $(0,-1,-1)$. $\blacksquare$[/quote]

[/hide]


function
algebra
domain
calculus
derivative
integration
algebra proposed

[b]Problem [/b]: Let $ f$ be a continuous function
\[ f: [0;1]\to [0;1]
\]
satisfying: For all $ x\in [0;1]$ there exists $ f'(x)$ (which is also a continuous function)
and [b]$ f'(x)| < 1 \forall x\in [0,1]$[/b].
The sequence $ x_n$ such that $ x_1 \equal{} \frac {1}{2}$
\[ x_{n \plus{} 1} \equal{} f(x_n)
\]
Prove that Exist a real number $ a\in [0;1]$ satisfying $ lim_{n\to \infty } x_n \equal{} a$
[b]My opinion [/b]
I think this problem is wrong
[b]Old result [/b]
If $ f'(x)|\le c < 1 \forall x\in [0,1]$ there $ c$ is a constant then $ lim_{n\to \infty } x_n\equal{} a$ where $ a$ is the unique solution of the equation $ x \equal{} f(x)$
The Lagrangre theorem shows it easily
[b]My requirements [/b]
If you think the problem  is true ,please give a complete solution
else give me a counter example !
Thank you for you attention.
My poor English makes you annoyed :blush:
Help me!.
Edited after pco's post:
So the problem is true  :( . Can you check your solution ( or demo ) carefully. Thank you!.
The old result is a basic knowledge and well-known, So I think my problem is wrong, dear.

be l£[0;1] f(l)=l (it is easy to show  existence of l since f is continuou)

f'(x)<1 so for all n  $ \left|(f(x_n) \minus{} f(l) \right| < \left| x_n \minus{} l \right|$

so $ \left| (x_{n \plus{} 1} \minus{} l \right| < \left| x_n \minus{} l \right|$

so $ \left| x_n \minus{} l \right| < \left| x_1 \minus{} l)\right|^n$

so $ lim_{n\to\infty}(x_n\minus{}l)\equal{}0$

[quote="Allnames"][b]Problem [/b]: Let $ f$ be a continuous function
\[ f: [0;1]\to [0;1]
\]
satisfying: For all $ x\in [0;1]$ there exists $ f'(x)$ (which is also a continuous function)
and [b]$ f'(x)| < 1 \forall x\in [0,1]$[/b].
The sequence $ x_n$ such that $ x_1 \equal{} \frac {1}{2}$
\[ x_{n \plus{} 1} \equal{} f(x_n)
\]
Prove that Exist a real number $ a\in [0;1]$ satisfying $ lim_{n\to \infty } \equal{} a$[/quote]

$ f(x)$ has a fixed point $ "a"$ and this fixed point is unique (else, if $ f(a)\equal{}a$ and $ f(b)\equal{}b$, $ \exists c\in(a,b)$ such that $ f'(c)\equal{}\frac{f(a)\minus{}f(b)}{a\minus{}b}\equal{}1$, which is impossible).
Since $ f^{[n]}(x)$ ($ n^{th}$ composition) respects obviously the same properties (domain and derivative bounds), all $ f^{[n]}(x)$ have the same unique fixed point $ "a"$.

Then, either cardinal of $ A\equal{}\{x_i\}$ is finite and so $ x_p\equal{}x_q\equal{}u$ for some $ p,q$ and then this $ u$ is a fixed point of $ f^{[q\minus{}p]}$, so is $ "a"$ and so $ x_k\equal{}a$ $ \forall k\geq p$,

Either cardinal of $ A\equal{}\{x_i\}$ is infinite and so $ A$ has some accumulation points. But each accumulation point is a fixed point of some  $ f^{[n]}(x)$. So we have one unique accumulation point, so the sequence is convergent towards this point.

<<to be full honest, I'm quite sure that "each accumulation point is a fixed point of some  $ f^{[n]}(x)$" is true but have no simple demo >>

[quote="Allnames"][quote="mohamed-01-01"]
so $ \left| (x_{n \plus{} 1} \minus{} l \right| < \left| x_n \minus{} l \right|$

so $ \left| x_n \minus{} l \right| < \left| x_1 \minus{} l)\right|^n$

[/quote]
But why?[/quote]

because f'(x)<1 ==> for all x;y£[0,1] $ \left| (f(x)\minus{}f(y)\right| < \left|x \minus{} y \right|$

so take x=x_n et y=l ==>$ \left| (f(x_n) \minus{} f(l) \right| < \left| x_n \minus{} l \right|$

so $ \left| (x_{n \plus{} 1} \minus{} l \right| < \left| x_n \minus{} l \right|$ for all n

                    $ \left| (x_{n } \minus{} l \right| < \left| x_{n\minus{}1} \minus{} l \right|$

                    $ \left| (x_{n \minus{}1} \minus{} l \right| < \left| x_{n\minus{}2} \minus{} l \right|$
                      .                                      .                  .

                      .                                        .                 .

                      $ \left| (x_2 \minus{} l \right| < \left| x_1 \minus{} l \right|$

and to make the multiplication ==> $ \left| x_n \minus{} l \right| < \left| x_1 \minus{} l)\right|^n$

[quote="pco"][quote="mohamed-01-01"] ...
so $ \left| (x_{n \plus{} 1} \minus{} l \right| < \left| x_n \minus{} l \right|$ for all n

                    $ \left| (x_{n } \minus{} l \right| < \left| x_{n \minus{} 1} \minus{} l \right|$

                    $ \left| (x_{n \minus{} 1} \minus{} l \right| < \left| x_{n \minus{} 2} \minus{} l \right|$
                      .                                      .                  .

                      .                                        .                 .

                      $ \left| (x_2 \minus{} l \right| < \left| x_1 \minus{} l \right|$

and to make the multiplication ==> $ \left| x_n \minus{} l \right| < \left| x_1 \minus{} l)\right|^n$[/quote]

Unfortunately not.

the "multiplication" ==> $ \left| x_n \minus{} l \right| < \left| x_1 \minus{} l\right|$[/quote]

oh yes ; sorry

you have f' continuous f'[0,1] is a segment ( f'[0,1]=[a,b])

since to$ \left| f'(x)\right|<1$ so $ \left|b\right|<1$

==>$ \left| (x_{n } \minus{} l \right| < \left|b\right| \left| x_{n \minus{} 1} \minus{} l \right|$

by multiplication $ \left| x_n \minus{} l \right| < \left|b\right|^n\left| x_1 \minus{} l)\right|$

...

You're both right.

And this quite quick demo is much more simple than mine. Congrats!  :)

[quote="Allnames"][b]Lemma[/b]
If $ f(x)$ is a both continuous and bounded   function  $ |f(x)| < 1 \forall x\in [0;1]$ then there exists $ c$ is a constant satisfying $ |f(x)|\le c < 1$
Indeed, we can choose $ c \equal{} max_{x\in [0;1]}$
Now Lagrangre theorem shows that $ |x_n \minus{} x_0| \equal{} c^{n \minus{} 1}|x_1 \minus{} x_0|$
where $ x_0$ is the unique solution of eqution $ x \equal{} f(x)$ (quite obviously). When two hands tend the infinite we get $ lim x_n \equal{} x_0$
We have done!  :([/quote]

This fact is very famous in Japanese University entrance examination.

or, you can use $ \boxed{f(b)\minus{}f(a)\equal{}\int_a^ b f'(x)\ dx}$


calculus
derivative
logarithms
analytic geometry
limit
calculus computations

[quote="cal137"]A: 1)lim x(lnx)^2 as x approach to 0+
    2)lim x(lnx)^2 as x approach to +infinite

B:first and second derivative of y=x(lnx)^2[/quote]

B: $ f' \equal{} \frac{x*2*\ln{x}}{x}\plus{}\ln^2x\equal{}2\ln{x}\plus{}\ln^2x$
$ f'' \equal{} \frac{2}{x}\plus{}2\frac{\ln{x}}{x}$

what are the critical points for y'
and what are the inflection points for y"

critical points: $ y' = \lnx(2 + lnx) = 0$ so $ x = 1$ and $ x = e^{ - 2}$

Inflection points: $ 2/x(1 + lnx) = 0$ gives $ x = e^{ - 1}$

plug all these values back into the original equation to get their $ y$ coordinates

a) $ \lim_{x \to 0^ + } \frac {\ln^2 x}{1/x}$ - indeterminate at $ 0 +$, so l'hopital it!
$ \lim_{x \to 0^ + } \frac {2/x lnx}{ - 1/x^2} = \lim_{x \to 0^ + } - 2\frac {lnx}{1/x}$ still indeterminate, l'hopital again
$ \lim_{x \to 0^ + } \frac { - 2/x}{ - 1/x^2} = \lim_{x \to 0^ + } 2x = 0$

$ \lim_{x \to \infty} x l\n^2 x = \lim_{x \to \infty} x (\lim_{x \to \infty} \ln x)^2 = \infty \ln^2 \infty = \infty \cdot \infty^2 = \infty$ since natural log is strictly increasing and unbounded, the natural log of an arbitrarily large number can be made arbitrarily large (i.e. $ \ln \infty = \infty$) and then we are just multiplying a bunch of infinite limits.


$ABCD$ is a square . $E$ is a point in $\triangle ABD$ such that $\angle ABE=\angle EDB=19^{\circ}$ . Find $\angle ECD$

 :)

[quote="delta"][hide]Let $\angle ABE=\delta$ so $\angle EBD=45-\delta$ and $\angle BED=135$. Consider the circumcircle around $\Delta BED$ its center lies on line $AC$ and if $M$ is an arbitrary point of it from the semiplane $BDC$ then $\angle BMD=45$ as $BEDM$ is inscribed quadrilateral. So as $\angle BMD$ is inscribed this means that the central angle is $2*45=90=BCD$ and this proves $C$ is the circumcenter.
Then $\angle ECD=2*\angle EBD=90-2*\delta$ in this problem $\angle EBD=52$ [/hide][/quote]

nice solution  :) .Mine is more less the same by proving $C$ is the circumcenter , but by noticng that the major angle $\angle BCD=270=2\times 135=2\times \angle BED$ . :)


There exist $ r$ unique nonnegative integers $ n_1 > n_2 > \cdots > n_r$ and $ r$ unique integers $ a_k$ ($ 1\le k\le r$) with each $ a_k$ either $ 1$ or $ \minus{} 1$ such that
\[ a_13^{n_1} \plus{} a_23^{n_2} \plus{} \cdots \plus{} a_r3^{n_r} \equal{} 2008.
\]Find $ n_1 \plus{} n_2 \plus{} \cdots \plus{} n_r$.

while you might call the solution "guess and check", it is probably a little bit more than simply guessing and checking.
that is,
you can convince yourself that you need to be subtracting some powers of three, so you know that you need at least one of the $ n_{i} \geq 7$
However, you see that $ 3^{8}$ is far too big (just estimating, around 6500?) and you'd be subtracting too many powers of three to get down to something near 2008.

So you start by saying the largest $ n_{i} \equal{} 7$ and from there you proceed as someperson01

[hide="Here is how I did it..."]
First we proceed to write 2008 in base 3, that is:
$ 2008 \equal{} 2202101_3$

Then we write out the representation that this produces, namely:
$ 2008 \equal{} 2 \cdot 3^6 \plus{} 2 \cdot 3^5 \plus{} 2 \cdot 3 ^ 3 \plus{} 1 \cdot 3 ^ 2 \plus{} 3^0$

And now we just replace all the multiples of 2 above to fit the specifications of the problem, namely that $ 2 \cdot 3 ^ k \equal{} 3 ^ {k \plus{} 1} \minus{} 3^k$:
$ 2008 \equal{} (3^7 \minus{} 3^6) \plus{} (3 ^ 6 \minus{} 3 ^ 5) \plus{} (3 ^ 4 \minus{} 3 ^ 3) \plus{} 3 ^ 2 \plus{} 3 ^ 0$
$ 2008 \equal{} 3 ^ 7 \minus{} 3 ^ 5 \plus{} 3 ^ 4 \minus{} 3 ^ 3 \plus{} 3 ^ 2 \plus{} 3 ^ 0$

Summing up the above exponents gives $ 7 \plus{} 5 \plus{} 4 \plus{} 3 \plus{} 2 \plus{} 0 \equal{} \boxed{21}$
[/hide]

realize that 2008 = multiple of 3 + 1. so ar must equal 1 and nr = 0
Factor 2007 after all the factors of 3's are gone and repeat the cycle.

well i guess it was kinda lucky for me....like about a week before the contest, i did the "parting shots" problems in the aops vol2, and one of the question is about how to write numbers in the way this problems is asking for....so i got this one like in a second or so...

This is basically just balanced ternary, something which everyone on our math team knows  :D

[hide="Easy with Brute Force :D "]
Basically we can see that the first term must be $ 3^7$, because if it is $ 3^8$, then even if we subtract each previous term, it wouldn't be less than 2008. Starting with $ 3^7$, we use terms less than that to get at 2008. We finally get $ 3^7\minus{}3^5\plus{}3^4\minus{}3^3\plus{}3^2\plus{}3^0\equal{}2008$. Adding up teh powers gives us $ \boxed{021}$.[/hide]

Convert 2008 to base 3, then manipulate to get the powers all different (which I forgot to do)

[hide=Solution]
We know that $3^{\color{red}{7}}=2187$ is very close to $2008$.

We want to subtract $179$.

We know that $3^{\color{red}{5}}=243$ is close to $179$.

We want to subtract $64$.

We know that $3^{\color{red}{4}}=81$ is close to $64$.

We want to subtract $17$.

We know that $3^{\color{red}{3}}=27$ is close to $17$.

We want to subtract $10$.

We know that $3^{\color{red}{2}}+3^{\color{red}{0}}=9+1=10$.

We know that our answer is $7+5+4+3+2+0=\boxed{021}$
[/hide]


cac ban nghi the nao ve no
                http://www.mathlinks.ro/Forum/viewtopic.php?t=82276

em nghĩ là bác dieuhuynh spam bài thì nhìu hơn là kiu người ta học hỏi đấy ạ!!! :huh:


i think it is not true.because
$ b=a\sqrt{8}$
$ a=\frac{b}{\sqrt{8}}$

Assuming this is what you meant:

[quote="bos1234"]$ \frac{a}{b}$=$ \frac{1}{\sqrt8}$

$ a=1$
$ b=\sqrt8$[/quote]

(Spaces don't show up in LaTeX so you'll have to start a new line for the variable.)

That's one possible solution, but there are several others. Cross-multiply and you'll get $ a\sqrt{8}=b$ which gives you all possible solutions.


logarithms
calculus
calculus computations

Solve the differential equation y^y = e^(dy/dx) with the initial condition y = e when x = 1.

$ \frac{dx}{dy}=\frac{1}{y\ln y}\Longrightarrow x=\ln |\ln y|+C$, by $ (x,\ y)=(1, e)$, yielding $ x=\ln |\ln y|+1$.


Fourteen friends met at a party. One of them, Fredek, wanted to go to bed early. He said goodbye to 10 of his friends, forgot about the remaining 3, and went to bed. After a while he returned to the party, said goodbye to 10 of his friends (not necessarily the same as before), and went to bed. Later Fredek came back a number of times, each time saying goodbye to exactly 10 of his friends, and then went back to bed. As soon as he had said goodbye to each of his friends at least once, he did not come back again. In the morning Fredek realized that he had said goodbye a different number of times to each of his thirteen friends! What is the smallest possible number of times that Fredek returned to the party?

Define $ g_i$ as the amount of goodbye received by friend $ i$,
WLOG $ g_1 < g_2 < \cdots < g_{13}$

And $ k$ the number of times of Frederiks salutes.
$ 10k\equal{}\sum_{i\equal{}1}^{13}{g_i}$

$ g_{13} \le k$ then
$ 10k \le (g_{13}\minus{}12)\plus{}(g_{13}\minus{}11)\plus{}\cdots\plus{}(g_{13}\minus{}1)\plus{}g_{13}$
$ \le (k\minus{}12)\plus{}(k\minus{}11)\plus{}\cdots\plus{}(k\minus{}1)\plus{}k$
$ 0 \le 13k\minus{}78\minus{}10k$
$ 26 \le k$ and for the case $ k\equal{}26$ it is not difficult to find an example.

[quote=mszew]...and for the case $ k\equal{}26$ it is not difficult to find an example.[/quote]
That's what happens when you don't sit down and actually write down an example.
Your argument until here is correct but not sharp. In fact only $k \ge 33$ is possible (and one can actually construct an example for $k=33$!).
The point is that you forgot about the special rôle of the last friend to whom he only said goodbye in the last night i.e. only once!
So you actually get $g_1=1$ and hence
\[10k \le k+(k-1)+(k-2)+\dots+(k-11)+1=12k-65\]
so that  $k \ge 33$.


Firstly, this problem belongs to Inequalities, not to Algebra. Secondly, it is wrong: try $a=\frac12,\; b=4,\; c=0$.


Let $ ABCDE$ be a given pentagon.Find the necessary condition for the sides and dioganals of this pentagon,such that this pentagon has an inscribed circle.In the answer you can use only the lengths of the sides and dioganals of the $ ABCDE$.
P.S For example,for the quadrilateral the necessary condition is the equality between the sums of the opposite sides.
Thank you.


Among the integers 1 to 300, how many are divisible by 3 or 5?

[hide] The number of positive integers up to $ 300$ divisible by $ 3$ is $ \frac{300}{3}\equal{}100$. The number of positive integers divisible by $ 5$ up to $ 300$ is $ \frac{300}{5}\equal{}60$. However, we must subtract those numbers which are divisible by $ 15$ (since they have been counted twice). $ \frac{300}{15}\equal{}20$. So, we have $ 100\plus{}60\minus{}20\equal{}\boxed{140}$.[/hide]


hallo,
let $(x_{n})_{n \ge 1}$ be a sequence with property \[x_{n+2}=\alpha x_{n+1}+\beta x_{n}\] where $\alpha , \beta \in N$ and $x_{1}, x_{2}\in N$.
If $\alpha^{2}+4 \beta$ isnt a square then prove that there doesnt exist a natural $n_{0}$ with \[x_{n_{0}}^{2}= x_{n_{0}+1}x_{n_{0}-1}.\] :)
this problem isnt very hard but i think its interesting  :lol:

$x_{n}=c_{1}\cdot \lambda_{1}^{n}+c_{2}\cdot\lambda_{2}^{n}$, where $\lambda_{j}$ are (different) roots of $\lambda^{2}=\alpha\lambda+\beta$.
Since $x_{1},x_{2}\in\mathbb{N}$ $c_{1}c_{2}\ne0$. Substitute in $x_{n}^{2}=x_{n-1}x_{n+1}$ and divide by $c_{1}c_{2}\lambda_{1}^{n}\lambda_{2}^{n}\ne0$ ($\lambda_{j}$ are irrational). We get $\frac{\lambda_{1}}{\lambda_{2}}+\frac{\lambda_{2}}{\lambda_{1}}=2$... Am I right?


superior algebra
superior algebra unsolved

Prove that class number of $\mathbb{Q}(\zeta_{23})$ equal $3$.

The Minkowski bound gives that we just have to check the ideals with norm $\leq 6$ to find all classes. This is now just case bash (but still a lot of work).


linear algebra
matrix
abstract algebra
group theory
vector
superior algebra
superior algebra solved

Let $M=\left\{A\in M_2(\mathbb{R})\left| \right.\exists x\in\mathbb{R},A=\begin{pmatrix}1&x\\0&1\end{pmatrix}\right\}$ and for $a>0$, $M_a=\left\{\begin{pmatrix}1&x\\0&1\end{pmatrix}\left|\right.x\in(-a,a)\right\}$.
a)Prove that $(M,\cdot)$ is an abelian group, isomorphic with $(\mathbb R,+)$.
b)If $G$ is a nonempty subgroup of $(M,\cdot)$ with the property that  $G\cap M_a$ is finite for every $a>0$, prove that $(G,\cdot)$ is isomorphic with $(\mathbb Z,+)$.

a) is rather tivial. I will write the matrices in the vector form $(1,x,0,1)$  for matrices of $M$. The multiplication yields $(1,x,0,1)(1,y,0,1)=(1,x+y,0,1)$. Now it is both clearly that $M$ is a group and $x\rightarrow (1,x,0,1)$ is an isomorphism.

Now I will think about b).

BTW: What is this contest you are talking about?

Misha

It's a Romanian contest held among participants from several counties (not the entire country, but a large part of it). Dan Barbilian was a brilliant Romanian mathematician.

Is it high school or university contest ?

High school, but I think there are also students from the $7$'th and $8$'th grade (high school starts from $9$'th grade).

Wow,   it is like math-sup or math-sp problem in France
after baccalaureat 12th grade in France.

b) is not very hard also. Consider $G$ generated by $(x_i)_{i\in I}$. If $G$ is the trivial group the statement is clear. If not consider $x_0:=min(|x_i|,i\in I)$. This minimum exists because $G\cap M_a$ is finite. Then for each $x_i$ we have a $k\in\Bbb Z$ with $x_i\in]kx_0,(k+1)x_0[$. But then
$x_i-kx_0\in G$ which is a contradiction to the minimality of $x_0$. So $G$ is generated by $x_0$. The map $Z\rightarrow G$, $k\rightarrow kx_0$ is clearly an isomorphism, qed.

PS: In my solution I identified $(1,x,0,1)$ with $x$ as we can do because of a).

Misha

Because $G\cap M_a$ is finite results that $G$ is at least countable (I also identified the matrix $\begin{pmatrix}1&x\\0&1\end{pmatrix}$ with $x$ because of a) ).
Now we have that $G$ is isomorphic to $\mathbb Q$ and $\mathbb Z$ is also isomorphic with $\mathbb Q$, results that $G$ isomorphic with $Z$.
is it correct?

Same cardinality and isomorphy is not the same! $\Bbb Q$ and $\Bbb Z$ are both countable but the groups $(\Bbb Q,+)$ and $(\Bbb Z,+)$ are not isomorphic because $(\Bbb Z,+)$ is generated by $1$ and there is no element in $\Bbb Q$ which generates $(\Bbb Q,+)$. Same cardinality means that there exists a bijective mapping from one set onto another. Isomorphy means that there exists a bijective homomorphism between the groups (in this case) which is a much stronger condition.

Misha


Suppose we have a triangle $ABC$. We choose two points like $D$ and $E$ out of $\triangle ABC$ so that $DA=DB,EA=EC$ and $\angle ADB=\angle AEC=90$. Suppose that $F$ is in the middle of $BC$. Prove that $EF=DF$ and $\angle DFE=90$.

It,s easy. Let $X$ be a point s.t. $X$ is out of $ABC$, $BX=CX$ and $\angle BXC=90^\circ$. We have$\triangle BDA\sim \triangle BFX$ and $\triangle CEA\sim \triangle CFX$, so $\triangle BDF\sim \triangle BAX$ and $\triangle CEF\sim \triangle CAX$. Thus $EF=\frac{AX}{\sqrt{2}}=DF$ and $\angle DFE=180^\circ-\angle BFD-\angle CFE=180^\circ-\angle BXA-\angle CXA=90^\circ$. :rotfl:  :ninja:  :D


advanced fields
advanced fields unsolved

Is there exists a measure space $ (X, \emph{M}, \mu)$ with non-trivial $ \sigma$-algebra, that for every measurable $ A\subset X$ is $ \mu(A)\equal{}0$ or $ \mu(A)\equal{}\plus{}\infty$?

Take $ X$ uncountable, $ M: \equal{}\{A\subseteq X | |A|\leq|\mathbb{N}| \vee |X\setminus A|\leq|\mathbb{N}|\}$ and $ \mu(A): \equal{}\begin{cases}0 & |A|\leq|\mathbb{N}| \\ \infty & |X\setminus A|\leq|\mathbb{N}|\end{cases}$


partition
combinatorics
Additive combinatorics
Set systems
IMO
IMO 1989

Prove that in the set $ \{1,2, \ldots, 1989\}$ can be expressed as the disjoint union of subsets $ A_i, \{i \equal{} 1,2, \ldots, 117\}$ such that

[b]i.)[/b] each $ A_i$ contains 17 elements

[b]ii.)[/b] the sum of all the elements in each $ A_i$ is the same.

Other ways are possible:

Take $S_i=\{39i+1,\cdots, 39i+39\}$
and take $S_{ij}=\{39i+13j+1,\cdots, 39i+13j+13\}$
And let $S_{ijk}=39i+13j+k$
Then we see see we can take for $13:$

$(4,11,6),(1,12,8),(13,3,5),(9,10,2)$ and one time $7,7,7$ as $k's$ in:
for $S_{00},S_{11},S_{22}$
$S_{01},S_{12},S_{20}$ and the other $3.$

Then it was also easy we can divide each subset of $\{117k+1\cdots 117k+233\}$ symmetrically in two equal sums of sets.

****
Yet an other way.
Find for $S_1$ until $S_3$ $21$ subsets with equal sum and of each $S_i$ one element.
Then we can make the other $96$ sets symmetrically in the easy way the maximum mistake we have is less than $108$ now.

In $S_kS_{k+1}$ where $k$ is even, we search $7$ couples of points with a difference of a "mistake" and change the symmetrically points and other we let in their correctsumtransformation.
So without calculating a solution, we have proved at this way also.

Extension: Find all $(a,b) \in \mathbb{Z}^2$ such that the set $\{1, 2, ..., ab\}$ can be expressed as the disjoint union of subsets $A_i$, $\{i=1, 2, ..., b\}$ such that

i.) each $A_i$ contains $a$ elements
ii.) the sum of all the elements in each $A_i$ is the same.

Extension solution:
Throughout the solution congruences will be taken $mod 2$.
We claim that $(a,b) \not\equiv  (1,0)$,with $a \neq 1$ is the entire family of solutions for our problem.

By a trivial argument on the integerness of the $A_i$ sums,$(a,b) \equiv (1,0)$ is impossible.Also we can immediately disprove the $a=1$ case.
By a similarly trivial induction with step $2$ on $a$,$(a,b) \equiv (0,1),(0,0)$ works.

It remains to prove that $(a,b) \equiv (1,1)$ with $a \neq 1$ is valid.Once again we resort to step-2 induction on $a$.It is trivial to go from $a$ to $a+2$,so now we only need to do the base case $a=3$ to be done($a=1$ is not valid,thus it isn't our base case).For this we have to exhibit a solution family.Since $b$ is $1 \mod 2$,we can write $b=2l+1$.Now consider the set family
$A_i=\{i,4l+3-2i,5l+3+i\}$ for $i=1,2,...,l$
$A_i=\{i,6l+4-2i,3l+2+i\}$ for $i=l+1,...,2l+1$.
It's easily checkable by hand that all the $A_i$ are disjoint,their union is $\{1,2,...,6l+3\}$ and the sum in each $A_i$ is $9l+6$,thus we have found a solution set family for the case $a=3$ and we are done.

Let us start pairing numbers in the following fashion, where each pair sums to $1990$:
$(1, 1990-1),
(2, 1990-2),
\ldots
(936, 1054)$

There are a total of $117*8$ pairs above. Let us start putting these pairs into each of the $117$ subsets starting with the first pair $(1, 1990-1)$ going into $A_1$, $(2, 1990-2)$ into $A_2$ and so on $\ldots$ with $(k, 1990-k)$ going into $A_{k(mod 117)}$.

Now we have $16$ numbers present in each of the $117$ subsets all of which have the same total sum. We need $1$ more number to be filled in each subset from the remaining $117$ numbers $(937, 938, ... 1053)$.

Let us arrange these $117$ numbers in the following manner:
$995-58, 995-57,  \ldots, 995-1, 995, 995+1, 995+2, \ldots, 995+57, 995+58$  $==> (1)$

Now notice that for each number that's off by $x$ from $995$, there's a counter number off by $x$ from $995$ in the opposite direction.

So we need to create similar imbalances in the Subsets $A_1$ to $A_{117}$, so that we could add the above $117$ numbers to those imbalances to keep the total sum unchanged.

Now start swapping the $1st$ number of each pair that we added to subsets $A_1$ to $A_{117}$ to create the above imbalances:
Swap $1$ from $A_1$ with $59$ from $A_{59}$ - this creates an imbalance of $+58$ and $-58$
Swap $2$ from $A_2$ with $58$ from $A_{58}$ - this creates an imbalance of $+56$ and $-56$
$\ldots$
Swap $29$ from $A_{29}$ with $31$ from $A_{31}$ - creates an imbalance of $+2$ and $-2$
Leave $30$ in $A_{30}$ as it is - $0$ imbalance.

Similarly,
Swap $60$ from $A_{60}$ with $117$ from $A_{117}$ - this creates an imbalance of $+57$ and $-57$
Swap $61$ from $A_{61}$ with $116$ from $A_{116}$ - this creates an imbalance of $+55$ and $-55$
$\ldots$
Swap $88$ from $A_{88}$ with $89$ from $A_{89}$ - this creates an imbalance of $+1$ and $-1$

Now start adding the $117$ numbers from $(1)$ to the subsets $A_1$ to $A_{117}$ to counter the imbalances $-58$ to $+58$ so that the sum remains unchanged. Notice that each subset now has $17$ elements with total sum = $8*1990 + 995 = 16915$.


Find all numbers $a,b \in \mathbb N^\ast$ such that for all $\displaystyle x,y \in \left[ a,b \right]$ we have \[ \frac{1}{x} + \frac{1}{y} \in \left[ a,b \right] .  \]

[i]Mircea Fianu[/i]

Does $\mathbb N^\ast$ mean all natural numbers including 0? (all whole numbers?)

Letting $x=y=b$, we get $\frac{2}{b}\ge a$, and hence $2\ge ab$. Since $a,b\in\mathbf{N}$, we must have $a=b=1$ (which obviously does not satisfy the given property) or $a=1$, $b=2$, which does.


trigonometry
geometry
angle bisector
trig identities
Law of Cosines
geometry unsolved

[color=darkblue]Let $ \triangle ABC$, known that $ \angle C < \angle B < 90^0$. Let $ AH \perp BC$, $ M$ is midponit of $ BC$, $ AD$ is bisector of $ \angle BAC$. Let $ S_{\triangle ABC} \equal{} 14S_{\triangle ADM}$, $ S_{\triangle ABC} \equal{} \frac {50}{7}S_{\triangle AHM}$. Find $ \angle BAC$.[/color]

[quote="Ahiles"][b][color=darkblue]We have:

$ \left\| \begin{array}{ccc} \left\|\begin{array}{cc} S_{ABC} = \frac {1}{2}AH \cdot BC \\
 \\
S _{ADM} = \frac {1}{2} AH \cdot DM \end{array} \right\| \Rightarrow \frac { BC}{DM} = 14 \\
 \\
\left\|\begin{array}{cc} S_{ABC} = \frac {1}{2}AH \cdot BC \\
 \\
S _{AHM} = \frac {1}{2} AH \cdot HM \end{array} \right\| \Rightarrow \frac { BC}{HM} = \frac {50}{7} \end{array} \right\|$

$ \begin{array}{cc} BC = 14DM \Leftrightarrow BC + \7\cdot \frac {1}{2}BC = 14DM + 14MC \Leftrightarrow 4 BC = 7 CD \Leftrightarrow 4CD + 4 DB = 7 CD \Leftrightarrow \boxed{\frac {AB}{AC} = \frac {BD}{CD} = \frac {3}{4}} \\
 \\
7 BC = 50 HM \Leftrightarrow 7 BC = 50 (\frac {1}{2}BC - BH) \Leftrightarrow 9BC = 25BH \Leftrightarrow \boxed{BH = \frac {9}{25}BC, \ CH = \frac {16}{25}BC} \end{array}$

Note that:

$ \begin{array}{cc} AH^2 = AB^2 - BH^2 = AC^2 - HC^2 \\
 \\
AC^2 - AB^2 = CH^2 - BH^2 \\
 \\
AC^2 - \frac {9}{16}AC^2 = \frac {16}{25}BC^2 - \frac {9}{25}BC^2 \\
 \\
\frac {7}{16}AC^2 = \frac {7}{25}BC^2 \\
 \\
BC = \frac {5}{4}AC \end{array}$

Observe that $ AB^2 + AC^2 = \frac {25}{16}AC^2 = BC^2$, so $ \boxed{\angle{A} = 90^\circ}$[/color][/b][/quote]
It's true. Thank you very much.

WLOG, let BC=14, then DM=1, DH=24/25, BH=126/25, MC=7. By converse of angle bisector, we can let AB=3n, AC=4n. Using Pythag to solve for AH two ways, we have $ 9n^2 \minus{} (126/25)^2 \equal{} 16n^2 \minus{} (224/25)^2$. Solving gives n=14/5. Using the Law of Cosines to find A, we find A is 90.

Basically the same as the previous proof, except we assume one side, and we use Law of Cosines instead of guessing it's right.


AMC
AIME
calculus
integration
algebra
system of equations
SFFT

For how many values of a < 1000 do the following system of equations have integral solutions?

x + xy + y = -a
x2 + y2 = 2a-1

Warning: This question may be slightly hard (as in, take you longer to solve than a typical AIME problem). Consider it two AIME questions in one.

Wow, that's a great question -- is it your own, or is it from an AIME exam?  (It looks a little familiar, but only very very vaguely.)

I adapted it from a problem in a local contest.

[hide]We add 1 to both sides of the first equation, and add twice the first equation to the second to get (x+1)(y+1)=(1-a) and (x+y)(x+y+2)=-1. The second one looks more approachable, so we start with that. The second factor is greater than the first, and as -1 factors into integers only as 1*-1, we see x+y=-1. This makes us happy. We actually didn't need Simon's favorite factoring trick, ha! Anyway, x+y=-1 => sub into first equation => xy=1-a. Whoo, this means xy=(x+1)(y+1). Cool. Not that it matters. Anyway, what matters is that we have a system of equations for x and y: x+y=-1 and xy=1-a. => x=-(y+1) => y(y+1)+1=a. The first few values of y(y+1)+1 are 1, 3, 7, 13, 21, blah. The largest such value less than a thousand is 993 which occurs for y=31, while the smallest occurs for y=0. (Actually all values occur twice, for y and -1-y, so we're just considering the positive values of y). So anyway this gives us 32 possible values for a. Bubbled in, this would be 032. [/hide]was I right?

You could also [hide]add twice the first equation to the second, to get:

x2 + 2xy + y2 + 2x + 2y = -1

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

(x + y + 1)2 = 0

Ends up the same. [/hide]And I really liked this problem too.

Syntax Error wrote:On the AIME do you have to put a 0 in the hundreds place if it's a 2 digit answer?

Just to bubble it in, yes.

I think the answer is[hide] 32[/hide], here is how I did it(in spoiler):[hide]call x+y=N. So N+xy=-a, x^2+y^2=2a-1.  Multiply 2 to

to the first equation then add it to the second equation we have N^2+2N=-1, or (N+1)^2=0, so N=-1. x+y=-1, x^2+y^2=2a-1. x=-(1+y), substitute it in to the second equation and the second equation simplifies to y^2+y+(1-a)=0. the discriminant has to be perfect square, or 1-4(1-a)=P^2. That is (P+1)(P-1)=4(a-1). P has to be an odd positive integer < or =63. If it =65, then( P^2-1)/4 -1 is greater than 1000. so the answer is 32.[/hide]

Thanks :) Good job you guys.

For others who come in, try working on it before you look at the others' solutions.


function
topology
real analysis
real analysis unsolved

Let $ X,Y$ topology spaces, $ f: X\rightarrow Y$ function,$ \mathcal{U}$ set of all open sets in $ Y$,$ \mathcal{B}$ some base for topology $ \mathcal{U}$,$ \mathcal{F}$ set of all closes sets in $ Y$.Let $ D\subseteq X$ set of all $ x\in X$ such that $ f$ isn't continuous in $ x$.
Prove that $  D\equal{}\cup_{U\in\mathcal{U}}(f^{\leftarrow}(U)\backslash Int f^{\leftarrow}(U))\equal{}\cup_{V\in\mathcal{B}}(f^{\leftarrow}(V)\backslash Int f^{\leftarrow}(V))\equal{}\cup_{F\in\mathcal{F}}(Cl f^{\leftarrow}(F)\backslash f^{\leftarrow}(F))$.

I avoided using $ f^{\minus{}1}$.

There are no ideas in this exercise, only reformulations of the definition of continuity.
If $ f$ is not continuous at $ x$, then there is a neighborhood $ U$ of $ f(x)$ such that $ f^{\minus{}1}(U)$ fails to contain any neighborhood of $ x$. Hence $ x\in f^{\minus{}1}(U)\setminus\mathrm{Int}f^{\minus{}1}(U)$. And so on... for closed sets you should also use the fact that preimages commute with taking complements.


Try this one. But I don't know this inequality is shape or not.
If $a,b,c$ is real number such that $\frac{1}{ab}+\frac{1}{bc}+\frac{1}{ca}=9$ then:
$\Sigma \frac{c^2+a^2}{b^2}+2 \Sigma \frac{1}{a^2}+2 (\frac{a-c}{b^2}+\frac{b-a}{c^2}+\frac{c-b}{a^2} )\geq 4 \Sigma \frac{c+a}{b}$
  I hope that it is not easy! :huh:


[quote="MCrawford"]Suppose the time is now 1 o'clock on a twelve-hour clock which runs continuously.  What time will it show 1,000 hours from now?[/quote]
[hide]
1000 mod 12 is 4
4 hours from 1 o'clock is 5 o'clock
[/hide]

whoa what just happened to my latex?
Let me redo it---[hide="da answer"]
$\frac{1000}{12}=83$ with a remainder of 4.
1 plus 4 is 5.  [b]5:00[/b] is the answer.[/hide]

[hide]1000/12 = 83 remainder 4. The extra 4 hours added on to 1:00 = would be 5:00[/hide]


a) Why does the strength of dispersion forces increase with increasing polarizability?
b) Explain how the observation that helium can be liquefied provides evidence for the existence of dispersion forces.
c) How do the strengths of dispersion forces vary with molecular size?

a) What is the nature of dispersion forces? In a given instant, the electron cloud (whose electrons are not static, remember) of a molecule may assume an assymetric distribution: this means that we have, in that moment, an instantaneous dipole which will induce in nearby molecules opposite charge distributions. The net result is an atraction between molecules. This type of force is present in any molecule, but its strenght is, in general, comparably weak. What are the factors that influence London (or dispersion) forces? They are three:

1) Polarizability: the greater or less capacity for the electrons of an atom to move as a result of an applied electric field. As we might expect, the bigger the atom, less atracted are their valence electrons, and more polarizable he is.

2) Molecular weight: more mass means stronger London forces.

3) Molecular shape: what pair can have more surface area of contact, two sausages or two eggs?


There exists a group $ G$ of order $ 8$ having generators denoted by $ i,j,k$ s.t. $ ij\equal{}k, jk\equal{}i, ki\equal{}j, i^2\equal{}j^2\equal{}k^2$.
Denote $ i^2$ by $ m$.
Show that every element of $ G$ can be written in the form $ e,i,j,k,m,mi,mj,mk$ and also show that these are precisely the distinct elements of $ G$.

That's the quaternion group.
So, what have you done so far¿


#1 [hide]

3-2 :rt2:  = (a+b) :^2:

3-2 :rt2:  = a :^2:+2a+b :^2:

3=a :^2:+b:^2

-2 :rt2: =2ab

- :rt2: =ab

2=a :^2: b :^2:

So, a :^2:  and b :^2:  are roots of x :^2: -3x+2, which factors to (x-1)(x-2). so a :^2: =1 and b :^2: =2. So, the answer is 1- :rt2: . Notice how the other combinations of -1 and  positive  :rt2:  don't work. [/hide][/hide]

3.[hide]

The sqrts of a+b*sqrt(c) and a-b*sqrt(c) are conjugates of each other, so from question 1 we see that the sqrt of 3+2:rt2: is 1+:rt2:, which is the squareroot part of our expression.

5*1+:rt2: is 5+5:rt2:. Adding this to 2 gives 7+5:rt2:.

Expanding (a+b):^3: gives a:^3: + 3a:^2:b + 3ab:^2: + b:^3:.

If b is in the form x*sqrt(y), then 3a:^2:b and b:^3: are the irrational parts of the expression.

So, a:^3:+3ab:^2:=7 and

3a:^2:b+b:b^3:=5:rt2:

We notice that we can factor the bottom equation into (3a:^2:+b:^2:)b=5:rt2:

We can use guess that b=:rt2: and plug that into the bottom equation to find out that a=1.

Putting those values into the top equation works, so our answer is 1+:rt2:

[/hide]

nice to see that you could use my answer to help you out...

umm, when I try to do #2  I end up with an expression with no real roots...help!

for #2, rationalize the denominator and consider the square of the numerator; lots of stuff cancels

you can also do #3 this way:
[tex]\sqrt{3+2\sqrt{2}}=1+\sqrt2[/tex]
the expression is [tex]\displaystyle\sqrt[3]{7+5\sqrt2}[/tex]
Let [tex]x=\displaystyle\sqrt[3]{7+5\sqrt2}[/tex]
      [tex]y=\displaystyle\sqrt[3]{7-5\sqrt2}[/tex]
let [tex]x+y=k[/tex]
[tex]k^3=x^3+y^3+3xy(k)[/tex]
sub in values of x^3, y^3, 3xy we have
[tex]k^3+3k-14=0 [/tex]which factors into
[tex](k-2)(k^2+2k+7)=0[/tex], k must=2.
If [tex]x=a+b\sqrt2[/tex], then [tex] y=a-b\sqrt2[/tex], x+y=k=2a, a=1.
[tex](1+b\sqrt2)^3=7+5\sqrt2[/tex]
so 1+6b^2=7, b=1

1. [hide]Actually mlee06, if you look at the other combinations of positive and negative 1's and :rt2:'s, :rt2:-1  is the answer. Notice that :rt2: is bigger than one, so :rt2:-1 is positive, but 1-:rt2: is not.[/hide]

2.[hide]Start by squaring the top, as Rep123Max suggested :

Then factor out the two.

We then take the sqrt, since we squared it in the beginning.

We cancel  out of the top and bottom of the fraction:

As we saw in #1,

Using that fact,  is our answer.[/hide]


Can I graph in Cartesian planes without Asymptote? If it is possible, could someone please explain how?

I need to be able to graph lines and points in Cartesian planes.

Is there a specific version of Asymptote that is made especially for Mac OS X's?

I couldn't find one

Thanks for the help! :lol:

There are many threads about asymptote on mac, and this is the best one:

http://www.artofproblemsolving.com/Forum/viewtopic.php?t=223490

I'm not exactly sure as to what you're asking otherwise. How can you graph Cartesian planes without asymptote? Do you mean some other program?

Anyway, [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=213540]this[/url] thread talks about graphing in asymptote. (For lines, just use the standard [b]draw()[/b] command.)

what I mean with my question is, can i graph in cartesian planes without any kind of software, just with texshop?

sorry dojo, i was unclear


abstract algebra
group theory
induction
invariant
superior algebra
superior algebra unsolved

i think this is kind of simple but i'm getting stuck. if G is a finite abelian group and if m divides order of G then G contains a subgroup of order m. i tried using induction and cauchy's theorem but i'm getting stuck somwhere, could anyone help me out? any help appreciated

Well, it seems like somehow you will have to use Sylow's theorem and some kind of result about the strutucture of a finite abelian $p$-group. But if you do that, the problem becomes trivial. I don't have much time to think about a more elementary way now ...

As Julien_Santini said, if $n=p_1^{\alpha_1}p_2^{\alpha_2}...p_k^{\alpha_k}$ and $m=p_1^{\beta_1}p_2^{\beta_2}...p_k^{\beta_k}$ you may use Primary Decomposition Theorem for abelian group. The group $G=A_1 \times A_2 \times... \times A_k$, where $A_i$-is $p_i^{\alpha_i}$-Sylo subgroup(note that since $G$ is abelian every Sylow is nornmal hence there just one such Sylow subgroup). Now it is only left to deal with $A_i=\mathbb{Z}_{p_i^{\alpha_i-\beta_i}} \times \mathbb{Z}_{\beta_i}$(read up invariant decomposition and note that $(\alpha_i-\beta_i,\beta_i$ make up a partition of $\alpha_i$). So, you have a subgroup of $G$ of order $p_i^{\beta_i}$. Do the same for all the primes $p_i$ and their direct product will have the order $m$.

Edit: I merged two threads, the other starting with:

[quote="barra"]Show that a finite abelian group has a subgroup of every order dividing the order of the group.[/quote]

By the Sylow theorems (extended ones ;) ), we have that there is one subgroup for every prime power $p^{k}$ dividing $|G|$ (=order of the group). But if $A,B$ are subgroups with $m,n$ elements such that $m,n$ are coprime, then $A,B$ have trivial intersection and $AB$ is a subgroup of order $mn$.
So we can generate these subgroups by taking "multiplying" with $p$-groups.

[quote="ZetaX"]By the Sylow theorems (extended ones ;) ), we have that there is one subgroup for every prime power $p^{k}$ dividing $|G|$ (=order of the group). But if $A,B$ are subgroups with $m,n$ elements such that $m,n$ are coprime, then $A,B$ have trivial intersection and $AB$ is a subgroup of order $mn$.
So we can generate these subgroups by taking "multiplying" with $p$-groups.[/quote]

Where do we use the fact that the group is abelian?

When $AB$ is supposed to be a subgroup again.

How to prove your 'extended' sylow subgroup?
Is this proof OK?

Let $|G|=p^{a}m$ where $p\nmid m$. By Sylow's there is a subgroup $P$ of order $p^{a}$. But this kind of group is solvable. If \[1=P_{0}%Error. "normal" is a bad command.
< P_{1}<\cdots <P_{k}=P\] is a composition series of $P$,  then then $k=a$ and $|P_{i}|=p^{i}$ by using the fact that the center of a $p$-group is nontrivial.

Yes thats ok.
Th extended version I meant says more: the number of subgroups with $p^{a}$ elements is $\equiv 1 \mod p$. But yours i more direct I think.

Alternative proof: By the structure theorem of finite abelian groups, it suffices to show the assertion for finite abelian $p$-groups. But it's already true for finite $p$-groups, see for example [url=http://www.mathlinks.ro/Forum/viewtopic.php?t=104268]this thread.[/url]


ratio
geometric series
absolute value
geometric sequence

how come the formula for the sum of an infinite geometric series doesn't work when the absolute value of the common ratio is equal to or greater than one?

If the absolute value of the common ratio is equal to or greater than one, then the terms increase indefinitely and the sum approaches $ \infty$.

And if the ratio is $ \minus{}1$, the partial sums oscillate between 0 and 1, therefore no limit exists.

And if the ratio is 1, then you don't have a geometric series; you have an arithmetic series with no difference between terms.

the intro to algebra book has a bunch of proofs....u can go check it out.  :lol:

Although it might seem like overkill, if you know your limits, apply that to geometric series to create an infinite one.  You should find that if $ |r|\geq 1$, the series diverges.  If $ |r|<1$, then you get $ \frac{a}{1\minus{}r}$ as the limit.

and I don't believe the intro to algebra book mentions convergence and stuff.  At least not in a rigorous way.

well as a proof of what alanchou said:

the normal equation for a non-geometric sequence is

$ \frac{a(1 \minus{} r^k )}{1 \minus{} r}$ and if k( number of terms is $ \infty$, then you get:

$ \frac{a(1 \minus{} r^{\infty})}{1 \minus{} r}$ = $ \frac{a}{1 \minus{} r}$, so ya.


geometry
3D geometry
analytic geometry
ratio
geometric transformation
reflection

[quote="Tenoreoz"]Given that 3 vertices of a cube in space have coordinates (3,4,1), (5,2,9), and (1,6,5), find the coordinates of the center of the cube.[/quote]

[hide]
i are bad at 3d geo
but heres an idea
how many types of triangles can you make with the points of a cube?
how is each one different?

[/hide]

[hide="Its fairly simple actually,"]
The distances amongst the points respectively are $ \sqrt {4 \plus{} 4 \plus{} 64} \equal{} \sqrt {72}$, $ \sqrt {16 \plus{} 16 \plus{} 16} \equal{} \sqrt {48}$, and $ \sqrt {4 \plus{} 4 \plus{} 16} \equal{} \sqrt {24}$. The sides of this triangle are in the ratio $ 1: \sqrt {2}: \sqrt {3}$, so the segment of length $ \sqrt {72}$ is a long diagonal of the cube. Hence we want the midpoint of $ (3,4,1),(5,2,9)$, which is $ (4,3,5)$.
[/hide]

[hide="Comment"]I think that if you're given three different vertices $ A$, $ B$, and $ C$ of a cube of side length $ s$, the only possible lengths of $ AB$, $ BC$, and $ AC$ are $ s,s\sqrt{2},s\sqrt{3}$, $ s,s,s\sqrt{2}$, and $ s\sqrt{2},s\sqrt{2},s\sqrt{2}$, not necessarily in that order.  If you get the second one, then $ A,B,C$ lie on one face of the cube, so it doesn't determine one cube (you can reflect over that face).  If you get the third one, it doesn't determine the cube either, because you can reflect over the plane that contains $ \triangle ABC$.  Therefore if you can determine the cube, the sides must be $ s,s\sqrt{2},s\sqrt{3}$.  At least, I think so.[/hide]


function
integration
LaTeX
algebra
floor function
real analysis
real analysis unsolved

Prove that { if [] denote Greatest Integer Function, f is continous, x>1 }

\int_{1}^{x} [u]([u] + 1)f(u)du= 2 \sum_{i=1}^{[x]} i \int_{i}^{x}f(u)du

Please help me with this.


Let a,b,c > 0. Prove that \[ \frac{a^2}{b}\plus{}\frac{b^3}{c^2}\plus{}\frac{c^4}{a^3} \geq 2b\plus{}2c\minus{}a\]

[quote="apollo"]Let a,b,c > 0. Prove that
\[ \frac {a^2}{b} \plus{} \frac {b^3}{c^2} \plus{} \frac {c^4}{a^3} \geq 2b \plus{} 2c \minus{} a
\]
[/quote]

$ \frac {a^2}{b} \plus{} \frac {b^3}{c^2} \plus{} \frac {c^4}{a^3} \geq 2b \plus{} 2c \minus{} a \iff \frac {a^2}{b} \plus{} b \plus{} \frac {b^3}{c^2} \plus{} c \plus{} c \plus{} \frac {c^4}{a^3} \plus{} a \plus{} a \plus{} a \geq 3b \plus{} 4c \plus{} 2a$, which becomes obvious, because from AM-GM we have that:
$ \frac {a^2}{b} \plus{} b\geq 2a$, $ \frac {b^3}{c^2} \plus{} c \plus{} c\geq 3b$ and $ \frac {c^4}{a^3} \plus{} a \plus{} a \plus{} a\geq 4c$.


limit
ratio
calculus
calculus computations

Which of the following series converge (justify the answer)?

$ \sum_{n\equal{}1}^\infty \frac{1}{1\plus{}n^{2}}$

$ \sum_{n\equal{}1}^\infty \frac{n!}{n^{n}}$

$ \sum_{n\equal{}1}^\infty \frac{1}{\sqrt{n^{2}\plus{}n}}$

$ \frac {1}{1 \plus{} n^2} < \frac {1}{n^2}, \forall n \in \mathbb{N}$, and the series with general term $ \frac {1}{n^2}$ is convergent, since it is a Dirichlet series with p > 1. Therefore, the given series converges.

$ \lim \frac {a_{n \plus{} 1}}{a_n} \equal{} \lim \frac {(n \plus{} 1)!}{(n \plus{} 1)^{n \plus{} 1}} \cdot \frac {n^n}{n!} \equal{} \lim \left(\frac {n}{n \plus{} 1}\right)^n \equal{} \frac1e < 1$,

so, according to the ratio test, the given series converges.

$ \lim n a_n \equal{} \lim \frac{n}{\sqrt{n^2\plus{}n}} \equal{} \lim \frac{1}{\sqrt{1 \plus{} \frac1n}} \equal{} 1$, and so the given series diverges (why?).

[quote="Carcul"]

$ \lim n a_n \equal{} \lim \frac {n}{\sqrt {n^2 \plus{} n}} \equal{} \lim \frac {1}{\sqrt {1 \plus{} \frac1n}} \equal{} 1$, and so the given series diverges (why?).[/quote]

Maybe the last series diverges due to the limit comparison test?

hello, the last series disverges because
$ \frac{1}{\sqrt{n^2\plus{}n}}\geq \frac{1}{2n}$ holds for all $ n\geq 1$.
Sonnhard.

[quote="Pythagorean12"]Maybe the last series diverges due to the limit comparison test?[/quote]

Yes, but my point was:

Theorem. Let $ a_n \geq 0$. If $ \sum a_n$ is a convergent series then $ \lim n a_n \equal{} 0$.

Dr Sonnhard Graubner's comparision is perhaps a simpler alternative way.

[quote="Pythagorean12"]$ \sum_{n \equal{} 1}^\infty \frac {1}{1 \plus{} n^{2}}$[/quote]
[quote="Carcul"]$ \frac {1}{1 \plus{} n^2} < \frac {1}{n^2}, \forall n \in \mathbb{N}$, and the series with general term $ \frac {1}{n^2}$ is convergent, since it is a Dirichlet series with $ p > 1$. Therefore, the given series converges. [/quote]

It is easier (you don't need to know any theory) to use $ \frac {1}{1 \plus{} n^2} < \frac {1}{n^2 \minus{} n} \equal{} \frac {1}{n \minus{} 1} \minus{} \frac {1}{n}, \forall n > 1$.
So we get $ \sum_{n \equal{} 2}^s \frac {1}{1 \plus{} n^{2}} < \sum_{n \equal{} 2}^s \frac {1}{n^{2} \minus{} n} \equal{} 1 \minus{} \frac {1}{s}$.

[quote="Carcul"]

Yes, but my point was:

Theorem. Let $ a_n \geq 0$. If $ \sum a_n$ is a convergent series then $ \lim n a_n \equal{} 0$.
[/quote]

That's false. For example, let $ a_{2^k} \equal{} 1/k^2$, $ k \equal{} 1, 2, ...$, $ a_n \equal{} 0$ if $ n$ is not a power of $ 2$.

But the correction is easy: we just add "provided this limit exists" at the end of the condition.  In any event, it's just a simple special case of the limit comparison test.


This is a very clever problem, but do NOT brute force it, because you will not get anywhere. Actually, I'm not sure how you can brute force it. So yeah...

Okay, say you have a matrix $\underline{A} = $\[{ \left( \begin{array}{ccc}4 & 9 \\16 & 25\endarray} \right)\]

Now let's say you have another matrix $\underline{B}$ such that $\underline{AB}^8 = \underline{A}$.  And also, everytime you multiply $\underline{A}$ by another power of  $\underline{B}$ less than 8, you get a new matrix. So $\underline{B}$ is [i]not [/i]the identity matrix, if you're wondering.  What is the value of $\underline{B}?$

[hide="Hint"]

Try solving this problem first: The $90^\circ$ rotational matrix is \[{ \left( \begin{array}{ccc}0 & -1 \\1 & 0\endarray} \right)\].  Find $\underline{A}^4$ without writing anything down.
[/hide]

Remind me again how to multiply matrices?  We learned it, but I've forgotten :blush:  Don't you go row by column and add like if we have
${{
\left( \begin{array}{ccc}4 & 9 \\16 & 25\endarray} \right) \left( \begin{array}{ccc}2 & 4 \\3 & 1\endarray} \right)
$
You end up with a matrix like
${
\left( \begin{array}{ccc}4(2)+9(3) & 4(4)+9 \\16(2)+25(3) & 16(4)+25\endarray} \right)
$

right?

Yep, that's matrix multiplication for you. :)
Although multiplying it out may not be the best way to go...
I think you probably don't want to go around guessing/checking with $\underline{A}^8$.

Nice problem, pkothari13. I wouldn't have noticed the rotational effect without your hint. :)

any takers??

[hide="Hint"]
Use the rotational matrix do rotate the matrix an eighth of the way back to its original spot.[/hide]


induction
function
number theory
prime numbers
algebra
binomial theorem
number theory unsolved

Can anyone prove Chebychev's Theorem with induction??

Can you say what is the Chebichev's theorem???? (I don't know this theorem)

Chebyshev's theorem (as I know it) is a weaker version of the Prime number theorem, it gives bounds on $\Pi (x)$.

The proof I know revolves around estimating the sum for n<x of the Van Magnoldt function of n.

Yes, but there is not [u]one[/u] theorem by Chebyshev...
Please give us the upper and lower constant you want.
But I think there is no 'nice' proof by induction.
(there are some proofs estimating some binomial coeefficients)

I mean bertrand's postulate. You know between $n$ & $2n$ there always exists a prime.

I believe that is possible somehow because we can define natural numbers by recursively. However I have never seen a proof of Bertrand postulate based on induction.

isn't anyone interested in answering this question???????? :mad:

I think nobody is answering you because it's even when possible not that easy to show it this way (I think it's as good as impossible when really doing it with induction).
But be more patient also and wait, there are a lot more problems here that are unsolved...  ;)

[quote="armanf"]I know one and only one prove for this nice theorem which is done by induction itself.

the true theorem : for any n>5 we have at least 2 prime numbers between n and 2n.

this prove have 12 lemma and 10 of these lems are solved by induction. I will tell u the prove later. :D[/quote]

i can't wait to see this  :D

[quote="armanf"]The first lemma is (2n)!/(n!)^2 > (4^n)/(2sqrt(n))

Second lemma: The product of all primes less or equal than n is less that 4^n

Third lemma :if p is  a prime divisor of (2n)!/(n!)^2 and p >= 2sqrt(n) then the power of p in number (2n)!/(n!)^2  is equal 1.

I will more lems later ...

_____
Good luck :D[/quote]

Is there any better bound for $\prod\limits_{p=\text{prime}} p$, i.e. $< k^n$, with $k<4$?

I dont know because the prove of 4^n is combinatorical and it is especially for 4. we have (2n)!/n!^2 < 4^n by binomial theorem . now what do you think about the lower bound? :?:

[quote="DCo"]On second thought, that's the way it's supposed to go, so induction does work...

Edit: Perfect Radio, there is a very well known estimate for the sum of ln(p) over all p<x, but I don't have it on the tip of my tongue, though you could easily get bounds by estimating the sum of the Van Magnoldt function of m for m<x.[/quote]

this looks really too advanced for me  :(

[quote="perfect_radio"]
Is there any better bound for $\prod_{p=\text{prime}}p$, i.e. $< k^{n}$, with $k<4$?[/quote]
Rosser proved that $\prod_{p\leq n}p<(2.83)^{n}$.
*I read this in this book by Leo Moser:
[url]http://www.trillia.com/moser-number.html[/url]

At a math camp this summer for one of the special lectures a counselor proved it in the way outlined above with lemmas... it is a really simple proof but long and ingenious...

you can find the proof in the [b]W.Sierpinski [/b]'s book. :)


OK, Asians were completely pwning in the game before, so let's change the rules:

Asians = +1
Non-Asians = -3.

Same general rules, no posting twice in an hour, no double-posting.

+1

[quote="perfect628"]1=2 only once per hour :wink:[/quote]

oops. sorry.

*pushes back to 13*

$ 2$

What are you talking about? They have always been pointless.

-1

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The tug of war things are seriously pointless...

They should all be locked/deleted..

Show me a thread in GFF that is not pointless.

0

oh man i was just about to say that

wait hamster i didn't know you were asian?

+1

Half, but I associate more with the asians, so I felt like going asian.

[b]-6[/b] just so it takes longer and please lock this already


topology
advanced fields
advanced fields unsolved

Prove that every $ k$-dimensional manifold $ X$ may be immersed in $ \mathbb{R}^{2k}$

You really should have specified a smooth manifold; "immersion" doesn't make sense for a topological manifold.

As for the proof? Look it up in a textbook. It takes about three pages in mine.

The one I've got is Lee (Intro. to Topological Manifolds, Intro. to Smooth Manifolds), and I like it. I haven't looked at any others to compare.
The general answer of "look it up in a textbook" applies to several of your other questions as well. Proving standard theorems isn't usually very interesting.

Also, as a new member, your five posts within five minutes gives the appearance that you're just trying to get people to do your homework for you (or worse).


Find all positive integer solutions to abc - 2 = a+b+c

Is there any solution for $a \neq b \neq c$ :?: .

[hide]
WOLOG $a \ge b \ge c$

first observe that if we increase a variable (let's say $a$) by 1, then we increase the left by a value of $bc$ and increase the right by $1$.  Since all variables are positive integers, $bc \ge 1$ with equality iff $b = c = 1$.  Therefore, in general, increasing a variable affects the left more than the right, so There's an upperbound for $a$ (and therefore an upperbound for $b$ and $c$).

Lemma: $c \le 2$.

If $c = 3$, then $a \ge b \ge 3$.  The minimal solution (here minimal means the minimal value of the left subtract the right) we can have is $a = b = c = 3$ (because from above increasing $a$ or $b$ increase the gap between the left and the right).  So plug in this minimal solution we have $25 > 9$, so $c=3$ or above have no solutions.

case 1:$c=2$.
plug in the minimal solution in this case (which is $a = b= c = 2$), we have $6=6$, a solution.  Since increasing any variable increase the gap between the left and the right, this is the only solution in this case.

finally, $c = 1$.  we have $ab-2 = a+b$, using Simon's favorite trick we obtain $(a-1)(b-1) = 3$, so another solution is $(4, 2, 1)$.

Now, removing the ordering imposed upon the variables, we have solutions $(2, 2, 2)$ and all cyclic permutations of $(4,2,1)$.  QED.

[/hide]


geometry
circumcircle
ratio
power of a point
radical axis
geometry proposed

[color=darkred]The radical axis between the circumcircle and the incircle of the triangle $ ABC$ meets the sideline $ BC$ in the point $ X$.

The radical axis between the circumcircle and the $ A$- exincircle of the triangle $ ABC$ meets the sideline $ BC$ in the point $ Y$.

Denote the middlepoint $ M$ of the segment $ [BC]$. Prove that $ MX=MY$ and $ \frac{XB}{XC}=\left(\frac{p-b}{p-c}\right)^{2}$.[/color]

Actually there is a more general result in [i]"Some radical axis associated with the circumcircle", D.Moody Bailey, Mathematics Magazine, Vol. 38, No. 5. (Nov., 1965), pp. 276-283.[/i]

Let $ ABC$ be a triangle and $ P$ a point inside for which consider $ \triangle{DEF}$ its cevian triangle for which $ D \in (BC)$, $ E \in (CA)$, $ F \in (AB)$ and denote by $ X$ the point where the radical axis of the circles $ DEF$ and $ ABC$ cuts $ BC$. Then
\[ \frac{XB}{XC}=\frac{a^{2}(\frac{BF}{FA}+1)(\frac{AE}{EC}+1)-b^{2}(\frac{BD}{DC}+1)(\frac{BF}{FA}+1)+c^{2}(\frac{BD}{DC}+1)(\frac{CE}{EA}+1)}{a^{2}(\frac{AF}{FB}+1)(\frac{CE}{EA}+1)+b^{2}(\frac{CD}{DB}+1)(\frac{BF}{FA}+1)-c^{2}(\frac{CD}{DB}+1)(\frac{CE}{EA}+1)}\]
For $ P$ as the Gergonne point, obtain the proposed problem.

[quote="Virgil Nicula"][color=darkred]The radical axis between the circumcircle and the incircle of $ \triangle ABC$ meets the side $ [BC]$ in the point $ X$.

The radical axis between the circumcircle and the $ A$- exincircle of the triangle $ ABC$ meets the sideline $ BC$ in the point $ Y$.

Denote the middlepoint $ M$ of the segment $ [BC]$. Prove that $ MX=MY$ and $ \frac{XB}{XC}=\left(\frac{p-b}{p-c}\right)^{2}$.[/color][/quote]
[color=darkblue][b][u]A synthetical proof.[/u][/b] Denote the circumcircle $ \mathcal C$ of $ \triangle ABC$ and the point $ D$ where the incircle $ \omega$ of $ \triangle ABC$ touches the side $ [BC]$.
Let $ \Omega$ be the circle which passes through the points $ B$, $ C$ and is tangent in the point $ T$ to the circle $ \omega$.
Let $ X$ be the point where the tangent in the point $ T$ to the circle $ \Omega$ meets the line $ BC$. The ray $ [TD$ is the bisector of the angle $ \widehat{BTC}$.
Thus, $ \frac{XB}{XC}=\left(\frac{TB}{TC}\right)^{2}=\left(\frac{DB}{DC}\right)^{2}=\left(\frac{p-b}{p-c}\right)^{2}$. Denote the radical axis $ r$ between the circles $ \mathcal C$ and $ \omega$.
The circles $ \mathcal C$, $ \omega$, $ \Omega$ by twos have the radical axis $ d$, $ TX$ and $ BC$ which is concurrently. Thus, $ X\in d$.
Analogously prove and the second part of this problem and you"ll obtain $ \frac{YB}{YC}=\left(\frac{p-c}{p-b}\right)^{2}$ a.s.o.[/color]

Instead a solution for Virgil Nicula's proposed problem follows nicely with some computations.

Denote $ X$ the point where the radical axis cuts $ BC$ and the ratio $ \frac{XB}{XC}=m$.

Since the power of $ X$ w.r.t. the circumcircle is $ XB \cdot XC$ and the power of $ X$ w.r.t the incircle is $ XD^{2}$, where $ D$ is the point of tangency of the incircle with $ BC$, the condition that $ X$ lieing on the radical axis, writes as $ XB \cdot XC = XD^{2}$. $ (1)$

But $ \frac{XB}{XC}=m$, implies $ XB = \frac{am}{m+1}$ and $ XC = \frac{a}{m+1}$, so $ XB \cdot XC = \frac{a^{2}m}{(m+1)^{2}}$

Assume, WLOG, $ X \in (DC)$, thus $ XD=XB-BD=\frac{am-(p-c)(m+1)}{m+1}=\frac{m(p-b)-(p-c)}{m+1}$

Therefore $ (1)$, rewrites as $ \left[m(p-b)-(p-c)\right]^{2}=a^{2}m$.

Now denote $ t=\sqrt{m}$, so the above relation is equivalent with $ t^{2}(p-b)-ta-(p-c)=0$.

The discriminant for this quadraic equation is $ \Delta_{t}=a^{2}+4(p-b)(p-c)$ and after some computations, using $ x=p-b$, $ y=p-c$, $ x+y=a$, we obtain
$ \Delta_{t}=\frac{4(p-b)^{4}+a^{2}(p-b)^{2}-4(p-b)^{2}a(p-c)}{(p-c)^{2}}=a^{2}+4(p-b)(p-c)$, i.e

$ \Delta_{t}=\frac{\left[2(p-b)^{2}-a(p-c)\right]^{2}}{(p-c)^{2}}$, thus $ t=\frac{a+\frac{2(p-b)^{2}-a(p-c)}{p-c}}{2(p-b)}=\frac{p-b}{p-c}$, i.e. $ m=\left(\frac{p-b}{p-c}\right)^{2}$


How many positive integers can be represented as the
difference of two distinct members of the set
\{1, 3, 5, 7, 9, 11, 13\}?

The smallest difference is 2 consecutive odd numbers , which is 2. the largest difference is 13-1=12. Since this set only includes odd numbers, we cannot get an odd difference so we must divide by 2. There are 6-1+1=6 differences.


Usually, it is the set of primes.

Pierre.

I can prove that a) doesn't have the solution.But with b) i can't   find all the solution.Can anyone help me? :(

I think both a) and b) have no solutions . If the equations have possitive integer solutions , they must be divisors of 2


[quote="shobber"]$ O$ is the circumcenter of an acute triangle $ ABC$. $ BO$ and $ CO$ meets $ CA$ and $ AB$ at $ E$ and $ F$ respectively.

Show that if $ \angle{A} \equal{} 45^o$, then $ OB \equal{} 2OE \Longleftrightarrow OB \equal{} 3OF$.[/quote]

Let's $ \angle OCE\equal{} \alpha$
Then we have that $ \angle OBF\equal{}45^o\minus{}\alpha$
we have:
$ tan(45^o\minus{}\alpha)\equal{}\frac{1\minus{}tan(\alpha)}{1\plus{}tan(\alpha)}$
and
$ tan(\alpha)\equal{}\frac{1\minus{}tan(45^o\minus{}\alpha)}{1\plus{}tan(45^o\minus{}\alpha)}$

I) we have that $ OB\equal{}2OE$ so $ tan(\alpha)\equal{}\frac{OE}{OB}\equal{}\frac{1}{2}$
therefore:
$ \frac{OF}{OB}\equal{}tan(45^o\minus{}\alpha)\equal{}\frac{1\minus{}\frac{1}{2}}{1\plus{}\frac{1}{2}}\equal{}\frac{1}{3}$

II) we have that $ OB\equal{}3OF$ so $ tan(45^o\minus{}\alpha)\equal{}\frac{OF}{OB}\equal{}\frac{1}{3}$
therefore:
$ \frac{OE}{OB}\equal{}tan(\alpha)\equal{}\frac{1\minus{}\frac{1}{3}}{1\plus{}\frac{1}{3}}\equal{}\frac{1}{2}$


Find all functions $ f$ from non-negative integers to non-negative integers that satisfy:

i) $ f(2)\equal{}2$;
ii) $ f(mn)\equal{}f(m)f(n)$ for $ gcd(m,n)\equal{}1$;
iii)$ f(m)>f(n)$ whenever $ m>n$;

That problem is slightly different; the $ (m,n)\equal{}1$ restriction is not part of the second condition.

MellowMelon's inductive step only uses (ii) on the prime factorizations of $ p\minus{}1,p\plus{}1$ so it still works.

How do you know that $ p\minus{}1$ and $ p\plus{}1$ aren't prime powers themselves? Matt's right; the method doesn't generalize. I did miss that detail.

Now that I see the problem for what it is, I am fairly sure that Erdos proved (in a nontrivial manner) that any function from $ \mathbb N$ to $ \mathbb R$ satisfying (ii) and (iii) has to be of the form $ n^\alpha$ for some $ \alpha$, which implies that here the identity is the only one that works.

Aha, here we go: http://www.artofproblemsolving.com/Forum/viewtopic.php?p=346551#346551


function
real analysis
real analysis unsolved

is there a function  from $\mathbb{R}$ to $\mathbb{R}$ such that the image of any segment is a segment and which is not continuous on any?

$x$ a real number and $x=[x].x_1x_2x_3..x_n..$ its decimal expansion.
$u_n$ the number of $i$ such that: $1 \leq i \leq n$ and $x_i=1$.
Then define $F(x)=liminf(\frac{u_n}{n})$. Then clearly, the image of set with non void interior is $[0;1]$.

How about this: Let $\mathbb{R}/\mathbb{Q}$ be the set of equivalence classes under the relation $x\sim y\text{ if }x-y\in\mathbb{Q}.$ $\mathbb{R}/\mathbb{Q}$ has the same cardinality as $\mathbb{R}$, so we can find a surjective function $f$ from $\mathbb{R}/\mathbb{Q}$ to $\mathbb{R}.$ $f$ extends naturally to a function $g$ from $\mathbb{R}$ to itself which takes any interval to the whole real line, and for obvious reasons, $g$ must be discontinuous everywhere.


Weighted AM-GM gives $\frac{x}{y} + \sqrt{\frac{y}{z}} + \sqrt[3]{\frac{z}{x}} > 2 \cdot \left( \frac{1}{6} \cdot \frac{x}{y} + \frac{1}{3} \cdot \sqrt{\frac{y}{z}} + \frac{1}{2} \cdot \sqrt[3]{\frac{z}{x}} \right) \geq 2 \cdot \sqrt[6]{\frac{x}{y}} \cdot \sqrt[6]{\frac{y}{z}} \cdot \sqrt[6]{\frac{z}{x}} = 2$.

[quote=hardsoul]Hello,everybody here're some problems that seems to be difficult.
If x,y,z>0 then show that:
x/y+(y/z)^1/2+(z/x)^1/3>3/2[/quote]
If $x,y,z>0.$ Then show that:$$\frac{x}{y} + \sqrt{\frac{y}{z}} + \sqrt[3]{\frac{z}{x}} > \frac{3}{2}.$$

[quote=sqing][quote=szl6208]The following inequality is also true!

If $x,y,z>0$,Prove that $\frac{x}{y}+\sqrt{\frac{y}{z}}+\sqrt[3]{\frac{z}{x}}\ge{2+\sqrt[6]{\frac{x^4}{y^3z}}}$[/quote]
Nice.
Simple proof?[/quote]

Let $\sqrt{\frac{y}{z}}=u,\sqrt[3]{\frac{z}{x}}=v=>x=\frac{z}{v^3},y=u^2z$,then,the inequality becomes:
\[\frac{1}{u^2v^3}+u+v\ge{2+\frac{1}{uv^2}}\]
\[<=>\frac{1}{u^2v^3}+u+v-2-\frac{1}{uv^2}=\frac{(v-1)^2}{v}+\frac{(uv+1)(uv-1)^2}{u^2v^3}\ge{0}\]


modular arithmetic
function
number theory
greatest common divisor
least common multiple
divisibility tests

Find the last three digits of $ 2^{8731}$
(Hide your answers please)

you can use the carmicheal function. Tell me if i'm usin it wrong..

$ 2^{8731} \equiv 2^{\lambda{(1000)}\ \cdot100 \plus{} 731} \equiv 2^{731} \equiv 648 \mod{1000}$

None of the three congruences you've listed appear to have any obvious justification.  Also, use \cdot for a multiplication dot -- the asterisk looks ugly in Latex.

Edit: Ah, I see, you don't know how to spell lamb[b]d[/b]a  :wink:

[hide="Clarification"]
It's well-known that $ \lambda(1000)\equal{}200$; we hence have $ 2^{8731}\equiv 2^{200\cdot 43 \plus{} 131}\equiv 1^{43}\cdot 2^{131}\equiv 2^{131}\pmod{1000}$. We could be lazy and simply calculate out this last quantity, or note that $ 2^{\phi(125)}\equiv 2^{100}\equiv 1 \pmod{125}$ and hence $ 2^{131}\equiv 2^{31}\equiv 23 \pmod{125}$ and then solve the system of linear congruences to get that $ 2^{8731}\equiv 648 \pmod{1000}$.
[/hide]

[quote="Temperal"][hide="Clarification"]
It's well-known that $ \lambda(1000) \equal{} 200$; we hence have $ 2^{8731}\equiv 2^{200\cdot 43 \plus{} 131}\equiv 1^{43}\cdot 2^{131}\equiv 2^{131}\pmod{1000}$. We could be lazy and simply calculate out this last quantity, or note that $ 2^{\phi(125)}\equiv 2^{100}\equiv 1 \pmod{125}$ and hence $ 2^{131}\equiv 2^{31}\equiv 23 \pmod{125}$ and then solve the system of linear congruences to get that $ 2^{8731}\equiv 648 \pmod{1000}$.
[/hide][/quote]

Please check if you use  Carmicheal function $ \lambda(n)$  for  $ a^{\lambda(n)}\equiv 1 \pmod n$ , you must make sure $ n$ is coprime to $ a$.

Oh; you're right. Apologies for the oversight.  :oops:

These solutions so far have been very poor in that
a.) One of them was not hidden
b.) Important calculations were not shown, so that someone who didn't already know how to solve the problem could understand how they were made
and c.) They had an inappropriate use of Carmichael's Theorem (as vinskman
 pointed out)

[hide="Perhaps better?"]
Firstly, $ \lambda(125) \equal{} 100$ (admittedly I just looked this up; I don't know how to actually calculate the Carmichael function), and furthermore $ (125,2) \equal{} 1$. By the definition of the Carmichael function, $ 2^{8731}\equiv 2^{87\cdot 100}\cdot2^{31}\equiv 1^{87}\cdot 2^{31}\equiv 2^{31} \pmod{125}$, and clearly $ 2^{8731}\equiv 0 \pmod{8}$. Note that because $ 125|1000$, we need only consider the last three digits of $ 2^{16}$ and $ 2^{15}$ (which should be easy to calculate) in the first congruence, so we have that it's $ 736\cdot 536 \equal{} 411648 \equiv 23 \pmod{125}$. There are only a few integers $ \pmod{1000}$ which are of the form $ 23 \plus{} 125k$, so we simply check all of these for divisibility by 8 and find that $ 648$ is the only viable remainder. (There exists a unique residue class $ \pmod{1000}$ which is a solution to the two congruences by CRT)
[/hide]

[hide="Solution"]
I will try to make this as easy to follow as possible.

We want the last three digits of $ 2^{8731}$, which is equivalent to finding $ 2^{8731}\bmod 1000$.

Let $ 2^{8731}\equiv 2^3x\bmod 1000$ (because $ 2^3$ is the highest power of $ 2$ that divides $ 1000$) so that $ 2^{8728}\equiv x\bmod 125$ after division. We're essentially using the Chinese Remainder Theorem here.

(Recall that division in modular arithmetic mandates the following: $ a\equiv b\bmod m\implies \frac {a}{d}\equiv \frac {b}{d}\bmod\frac {m}{\text{gcd}(m,d)}$)

$ \lambda(125) = \lambda(5^3) = (5 - 1)5^2 = 100$, so $ 2^{100}\equiv 1\bmod 125$. Obviously, $ 8728 = 87\cdot 100 + 28$, so $ x\equiv 2^{28}\bmod 125$.

$ 2^{28}\equiv 16^7\equiv 16\cdot 256^3\equiv 16\cdot 6^3\equiv 81\equiv x \bmod 125$.

$ 8x = \boxed{648}$.
[/hide]

@Temperal: ${ \lambda(p_1^{k_1}p_2^{k_2}\cdots p_n}^{k_n}) = \text{lcm}(\lambda(p_1^{k_1}),...,\lambda(p_n^{k_n}))$, and $ \lambda(p^q) = \varphi(p^q)$ if $ p$ is an odd prime, or $ \lambda(2^n) = 2^{n - 2}$.

The Carmichael function works fine for non-coprime bases; you just can't reduce all the way to $ 1$.  More precisely, the following is true.

[b]Proposition:[/b]  Let $ v_p(n)$ denote the greatest power of $ p$ dividing $ n$ and let $ k$ be the smallest positive integer such that, for every prime $ p$ dividing both $ a$ and $ n$, it is true that $ v_p(a^k) \ge v_p(n)$.  Then

$ a^{\lambda(n) \plus{} k} \equiv a^k \bmod n$.

When $ n$ is prime, this reduces to the usual statement of Fermat's little theorem for arbitrary bases.


inequalities
conics
hyperbola
vector
triangle inequality
complex numbers

If $ z\in \mathbb{C}$ and $ 1\plus{}|z|\equal{}|1\plus{}z|$ show that $ z$ is real and positive. I don't know if my proof is correct, i'm searching for new ones.

[hide]Using the triangle inequality, we see that equality occurs when the complex numbers are aligned, or equivalently they arguments differ from a multiple of $ 2\pi$. Then $ z$ is aligned with 1. That is $ z$ is a positive real.[/hide]

[hide="Algebraic proof"]\begin{eqnarray*} 1+|z| &=& |1+z|\qquad/^2\\ 1+2|z|+|z|^2 &=& |1+z|^2\\ 1+2|z|+z\bar z &=& (1+z)(1+\bar z)\\ 1+2|z|+z\bar z &=& 1+z+\bar z+z\bar z\\ 2|z| &=& 2\Re\{z\}\qquad/: 2, ^2\\ \Re^2\{z\}+\Im^2\{z\} &=& \Re^2\{z\}\\ \Im\{z\} &=& 0\end{eqnarray*}

Therefore $ z$ is real, so $ z=\Re\{z\}$, but we got up there that $ \Re\{z\}=|z|$, hence $ z=|z|$ which means that $ z$ is non-negative.[/hide]

[quote="math_music"]If $ z\in \mathbb{C}$ and $ 1 \plus{} |z| \equal{} |1 \plus{} z|$ show that $ z$ is real and positive. I don't know if my proof is correct, i'm searching for new ones.

[hide]Using the triangle inequality, we see that equality occurs when the complex numbers are aligned, or equivalently they arguments differ from a multiple of $ 2\pi$. Then $ z$ is aligned with 1. That is $ z$ is a positive real.[/hide][/quote]

Could you show me a diagram of this statement?

[quote="t0rajir0u"]Why bother?  This is the equality case of the triangle inequality.[/quote]

1. If someone asks for an alternative proof, it's usually not helpful to say "why bother?" :)

2. Working with conjugates, real and imaginary parts can prove very useful in other, more complicated proofs, so it doesn't hurt to introduce the idea.

Thanks Farenhajt, nice use of properties! Thanks for the replies.
bos1234: Think at the complex numbers as vectors, i'll try to explain in words (sorry for the bad english). Draw a triangle a make some height $ h$ tend to zero, then you will see that the base side tend to the sum of the others sides. The two vectors (non-base sides of the triangle) become parallel to the third vector (base). Then in complex plane we will have theese two vectors with equal argument (or differing by $ 2\pi$)


Flood in your votes. I'll start ToW again soon!

u certainly should coz any new game is to be welcomed....... but bfr' doing so please revise the rules once again could u?

 :blush:

plz no more spam lol. even though those increased my post count so much  ¬_¬

ok, ok. How about one ToW where anyone can play? You don't have to sign up or anything like that. That's what I really dislike. Signing up for things.

[quote="madatmath"]How abt Tug of War 3 with improvized rules?//

 signing up in advance,
 madatmath[/quote]

With improvised rules?! If the players do the improvisation, :rotfl: that's going to be reallllllly interesting. (And somehow I don't think it'd last very long either. :D)

No more tugs of war, chains, or anything else intended to increase post count. That's not what this forum is here for. Games are just fine, as long as you make them meaningful in some way.

:-) And no, spam is not meaningful.


This problem is from ACoPS (pg.96, if you have it)

[color=blue]Let $n$ be a positive integer. Show that if you have $n$ integers, then either one of them is a multiple of $n$ or a sum of several of them is a multiple of $n$. [/color]

The thing im having trouble with understanding is the second part of the casework they present.

[u]Case I:[/u] One of the integers is equivalent to $0 \pmod{n}$, so we're done.

[u]Case II:[/u] None of the integers are equivalent to $0 \pmod{n}$. We now only have $n-1$ integers to consider. By the pigeonhole principle, two of these integers must have the same remainder when divided by $n$, [color=red]so their difference, a sum of several of the original numbers, will be a multiple of $n$, and we are done. [/color]

The red part is what I don't understand. Could someone explain how knowing their difference is divisible by $n$ proves that a sum of several of them is divisible by $n$?

"The integers" doesn't refer to the $n$ original integers.  As far as I can tell, the argument goes something like this:

Let $a_{1}, a_{2}, ... a_{n}$ be the original integers.  Define

$s_{k}= \sum_{i=1}^{k}a_{i}$

Then either $n | s_{k}$ for some $k$ (in which case we are done), or the $n$ numbers $s_{1}, s_{2}, ... s_{k}$ take on $n-1$ residues $\bmod n$, so at least two of them (say, $s_{i}, s_{j}$) have the same remainder on division by $n$.  But then their difference, which [b]is[/b] a sum of several $a_{i}$, is divisible by $n$.

How do you get that their difference a sum of several $a_{i}$?

If $s_{j}\equiv s_{i}\pmod n$ for $i>j$, then clearly $s_{i}-s_{j}=a_{j+1}+\hdots+a_{i}\equiv 0\pmod n$.  This is natural approach to such problems.  This problem is shown as an example in Problem-Solving Strategies.

If n=5, consider a, a+b, a+b+c, a+b+c+d, a+b+c+d+e.

oh ok i get it now thanks everyone. cool im getting problem solving strategies soon :)


Hi!

It's Christmas, and I got a pretty hefty sum of money from my family.  Guess what I plan to do with it?=> Buy physics books of course! :lol:  I'm so excited, I never really had enough money to buy the classic physics books I wanted, and now, I am left with a very difficult choice.  I am a prospective physics major, and am enrolled in both Physics C and Calculus BC, but I also am studying differential equations and Calc III at home. Keep in mind that I'll buy a book for physics self study, so I won't have a teacher to guide me or solve problems for me.

My goal is to have a very strong understanding of physics, both quantitatively and qualitatively so that I ace my physics course once I go to college.  I know physics is not an easy subject, and would like to master it as well as people like MysticTerminator or JRav or Phuc on this forum (pardon me if I omitted some geniuses on here... there's so many of them!).  I want to be able to apply physics principles in unintuitive problems in physics.  Seeing how artfully you guys use concepts like torque etc... makes me salivate.

I already am using an intro book in physics C, which is Physics by Scientists and engineers by Paul A. Tipler.  I have narrowed my choices to the following:

- [b]Feynman's physics lectures[/b] (vol.I-III): I'm slightly concerned that this book will give me only superficial understanding b/c of the lack of practice exercises and mostly theory coverage.
- [b]Introduction to Classical Mechanics [/b]by David Morin:I'm concerned this book is too advanced for me and will discourage me.
- [b]Irodov problems in general physics
- Introduction to Mechanics by Kleppner and Kolenkow[/b] Both Irodov and K.&K. don't have solution manuals, which kind of sucks because I most probably can't solve all the problems. How would you describe the difference between Irodov and K.&K. problems? Are they about the same difficulty level?

Unfortunately, I don't have enough money to buy all of them  :( , I can only buy one of them.  Which one would you recommend me, knowing my goals and aspirations?

Thanks so much, you guys are all awesome! (AoPS is the best website ever after MIT ocw) :D

PS: if i were to buy irodov or K&K you guys will help me right?  :D

[quote="integration"]Actually, the choice now is between David Morin and Feynman!

I was able to borrow Irodov's from a local library!! So, so excited to start those problems.[/quote]
Try to get yourself a copy of S. S. Krotov's APTITUDE TEST: PROBLEM'S IN PHYSICS. These problems are tougher than those in Irodov but there are detailed and lucid solutions provided. Also the book is inexpensive (I bought it for $ Rs. 55$ $ \approx$ $ \$$ $ 1$!
You could also try D.C Pandey's Series (if you are in India).


Are high school students allowed to participate in Putnam? Is there a specific college they have to join (depending on their region)?

Allowed, but discouraged. There is a rule that limits any individual from taking the Putnam more than four times. All of the case in which I know of a high school student taking the Putnam (including Gabriel Carroll being a Putnam Fellow in 2000), the student was already taking courses at the college in question.

You can take it in high school if a university lets you, but you sacrifice a year of future eligibility.  So it's almost certainly not a good idea--unless, for example, your name is Gabriel Carroll.

The better thing to do might be to see if a nearby university will let you sit in for the exam unofficially, without turning in your answer sheets.  Putnam rules may or may not allow that, though.

If you don't turn in your answer sheets, Putnam rules don't care.  On the other hand, it seems rather silly to take the exam under strict testing conditions when you could just wait for the questions to be available online and do them at your leisure.


geometry
parallelogram
circumcircle
geometry proposed

[quote="Petry"]Hello!

Let $ ABCD$ be a quadrilateral circumscribed around a circle $ \left(I\right)$, $ \left\{V\right\} \equal{} AC\cap BD$, $ \left\{M\right\} \equal{} AB\cap\left(I\right)$,
$ \left\{N\right\} \equal{} BC\cap\left(I\right)$, $ \left\{P\right\} \equal{} CD\cap\left(I\right)$, $ \left\{Q\right\} \equal{} DA\cap\left(I\right)$ and $ \left(I_{a}\right)$, $ \left(I_{b}\right)$, $ \left(I_{c}\right)$, $ \left(I_{d}\right)$ are the circumcircles of
the triangles $ VQM$, $ VMN$, $ VNP$, $ VPQ$ respectivelly.
1. If $ \left\{V,S\right\} \equal{} \left(I_{a}\right)\cap\left(I_{c}\right)$ and $ \left\{V,T\right\} \equal{} \left(I_{b}\right)\cap\left(I_{d}\right)$ then prove that the points $ V,S,I,T$ are concyclic.
2. Prove that the quadrilateral $ I_{a}I_{b}I_{c}I_{d}$ is a parallelogram.

Best regards, Petrisor Neagoe :)[/quote]
Notice M,V,P  and  Q,V,N are collinear =>  the quadrilateral $ I_{a}I_{b}I_{c}I_{d}$ is a parallelogram