how do I join a class?

should I pay first?

Click "Online Classes" on the sidebar.  THers info there.

http://www.artofproblemsolving.com/Classes/howclasseswork.php

http://www.artofproblemsolving.com/Classes/AoPS_C_FAQ.php

Shad0w, can you move this to Classes Information? Thanks.

EDIT: Shad0w, edit the second portion of AIME15's post.


superior algebra
superior algebra solved

Let $G$ be a finite group and $f$ an isomorphism s.t. $f(f(x))=x$ for all $x\in G$ and if $f(x)=x$ then $x=e$.
Prove that $G$ is commutative.

Let's prove that for every x in G $f(x)=x^{-1}$. The fisrt step we need is showing that the product f(a)a^{-1} takes all the
values in G --- supposing contrary $f(a)a^{-1}=f(b)b^{-1}$ it easy to verify that it follows $f(b)^{-1}f(a)=b^{-1}a$, and since $f$ is isomorphism $f(b^{-1}a)=b^{-1}a$ and obtain that $b^{-1}a=e$ thath's means $b=a$.
So we have already proved that the product $f(a)a^{-1}$ takes all the values in G together with a, that means that every $g\in G$ culd be represent in form $g=f(a)a^{-1}$ Thus $f(g)=f(f(a))f(a^{-1})=af(a)^{-1}=(f(a)a^{-1})^{-1}=g^{-1}$, so we see that $f(g)=g^{-1}$ for all $g\in G$. So $ab=f(a^{-1})f(b^{-1})=f(a^{-1}b^{-1})=(a^{-1}b^{-1})^{-1}=ba$ for every pair $(a,b)$ from G.

uau... nice solution eugene .. thank you
I tried to prove that $xf(x)=f(x)x$ (from which follows imediately that $f(x)=x^{-1}$) but didn't manage yet :(

Nice solution! But it seems a little out of the blue! Is this a common technique I don't know of?

uhm.. I also found this problem in "Bazele algebrei by C. Nastasescu, C. Nita, C. Vraciu"...


Let $A = (a_1, a_2, \cdots ,a_{2000})$ be a sequence of integers each lying in the interval $[-1000,1000]$. Suppose that the entries in A sum to $1$. Show that some nonempty subsequence of $A$ sums to zero.

if $a_i=0$ the problem is easy
=>$a_i \not=0$
$s_0=0$
$b_1>0,s_1=b_1$
$b_2<0,s_2=b_1+b_2$
             .
             .
             .
$s{}_2{}_0{}_0{}_0=b_1+...+b{}_2{}_0{}_0$
now if $s_i>0$=>$b_i_+_1<0,s_i_+_1<s_i$
now if $s_i\leq 0$=>$b_i_+_1>0,s_i_+_1>s_i$
now we have $2001$ numbers from $200$ different places
so there exists $i,j$ such trhat $i<j$ and $s_i-s_j=0$ so $\sum_{k=i+1}^j a_k=0$
sorry if I didnt explain the soloution as well as possible  :oops:  ;)

See newly posted topic at [url]http://www.mathlinks.ro/viewtopic.php?p=1636442#1636442[/url]

f $a_i=0$ the problem is easy $ \rightarrow a_i \ne 0$|
Now construct sequence $\{s_n\}_{n=0}^{2000}$ of integers, such that:
$s_0=0$
$b_1>0,s_1=b_1$
$b_2<0,s_2=b_1+b_2$
...
$s_{2000}=b_1+...+b_{2000}$
Moreover, if $s_i>0$ then we choose $b_{i+1}<0,s_{i+1}<s_i$. If $s_i < 0$, then we will choose $b_{i+1}>0,s_{i+1}>s_i$
now we have $2001$ numbers $\left(s_0, ... , s_{2000}\right)$ from $2000$ $\left(\text{they all are from } -1000 \text{ to } 1000, \text{ due to construcrion, but it's impossible to have both } -1000 \text{ and } 1000 \text{ as partial sums, becuase in this case we do have both } -1000 \text{ and } 1000 \text{ as original numbers, and thier sum is equal to } 0\right)$ different places. So there exists $i,j$ such trhat $i<j$ and $s_i-s_j=0$ so $\sum_{k=i+1}^j a_k=0$

[hide=My solution]Assume none of term of $A$ equal $0$ (if no then pick it). We will pick $B=(b_1,b_2,...,b_{2000})$ by picking from $A$ with the following rule: $b_1>0,b_2<0$ and $b_i$ is reverse sign with $s_{i-1}=b_1+b_2+...+b_{i-1}$ and you got $s_1,s_2,...,s_{2000}$ in $[-999,1000]$. If one of $s_i=0$ then done. If not, then by Dirichlet there is $i<j$ such that $s_i=s_j$. And then $b_{i+1}+...b_j=0$, as desired

Let celebrate the 540th post.[/hide]


given $u,v \in \mathbb{K}^{n \times 1}$

Prove that:

$||u v^H||_F=||u||_2 ||v||_2$ (F meaning frobenius norm)
$||u v^H||_2=||u||_2 ||v||_2$ (2 meaning 2-norm or standard euclidian norm)

Is there really nobody who can help me with this problem? I've seriously tried to solve it but cannot find it and I would love to know how I can solve this.

Thx in advance :)


geometry
analytic geometry
graphing lines
slope
geometry unsolved

Hi everyone I'm a new member in this forum. I'm dealing with a geometry problem so I need your help. Sorry if I use wrong geometrical terms. The problem is: We have a semicircle AB and K is it's centre. C is a random spot. DC is vertical to AB. AC and CB are also semicircles. ε1 and ε2 are tangents. We have to prove that the ε1 and ε2 are parallels (ε1//ε2). There is an image in order to help you. I appreciate your help.

[url=http://img255.imageshack.us/my.php?image=geometryproblemlj9.png][img]http://img255.imageshack.us/img255/9908/geometryproblemlj9.th.png[/img][/url]p[/img]

Construct a xy plane with origin K, then with the given things we can get the slopes of the two lines.

Let $M,N$ be the centers of the two semicircles$AC,BC$,$X,Y$ be the two points at which $\epsilon 1$ tangents the two semicircles$AC,BC$ respectively.

Now,we draw $MZ\perp NY$ at $Z$ and have

1)$\angle DCK=\angle MZN=90^{o}$,
2)$MN=MC+CN=r+\rho=R=DK$,
3)$NZ=NY-ZY=NY-MX=\rho-r=R-2r=KA-CA=KC$,

$\Rightarrow \triangle DCK \~=\triangle MZN\Rightarrow \angle DKC=\angle MNZ\Rightarrow KD//NZ$.

Finally,$\epsilon 1//\epsilon 2$!


circles
geometry
modular arithmetic
Intersection
combinatorics
IMO Shortlist
double counting

Let ${n}$ and $k$ be positive integers. There are given ${n}$ circles in the plane. Every two of them intersect at two distinct points, and all points of intersection they determine are pairwise distinct (i. e. no three circles have a common point). No three circles have a point in common. Each intersection point must be colored with one of $n$ distinct colors so that each color is used at least once and exactly $k$ distinct colors occur on each circle. Find all values of $n\geq 2$ and $k$ for which such a coloring is possible.

[i]Proposed by Horst Sewerin, Germany[/i]

That's amazing!!!!
I was thinking about creating a problem to submit for IMO, and I 'invented' exactly this problem!

Pierre.

I think the answer is all $n,k$ for $n\geq k\geq 3$. Is it the true answer?

no, it isn't. it is clear that $k=1$ is not possible. maybe there's a typo?

Yes there was a typo that I have corrected now. I've spent impossibly much time on this problem, but the solution is so simple (if it's correct)

[quote="iura"]I think the asnwer is all $n,k$ for $n\geq k\geq 3$. Is it the true answer?[/quote]
possibly no-for instance when $n=4$,(4 circles, each circle has 6 pts on it),$k=6$ is possible,though for 3 circles,$k=4$ is not.so your answer may not be 'complete'.( of course,$k \leq 2(n-1)$ since there are so many pts on each circle when there are $n$ circles in all.)

Well, since there are used $n$ colors, we get $k\leq n$ isnt't it?

oh - i didn't see that stipulation in the problem.

Well, I guess our German friends (Darij, Peter) should know the solution. I'm waiting for a confirmation or an infirmation of my answer. If it's correct - I show my solution, if no - sorry for my $4$ useless posts

I haven't read the proposed solution, since this problem is the one I'm least interested in, but the answer is: either $2\leq k\leq n\leq 3$ or $3\leq k\leq n$. So your answer is almost completely correct.

  Darij

Well I guess $2\leq k \leq n$ holds only for small $n$.
Here's the proof that $k\neq 2$ for $n\geq 4$
Firstly we draw a (non-symmetric) matrix $(A_{i,j})$ with entries $1,\cdots,n$ s.t.  for $i\neq j$ $a_{ij},a_{ji}$ are the colors of the intersections of circle $\#i$ and $\#j$ and $a_{ii}=0$. Denote $R_i$ the row number $i$ and by $C_i$ the column number $i$. The we must have for every $i$ in $R_i \bigcap C_i$ exactly $k$ distinct non-zero numbers.
Let's prove $k\neq 2$.
Firstly draw $R_1,C_1$ containing WLOG colors $1$ and $2$. Now at step $i$ we add the cells from $R_i \bigcap C_i$ which are not already added. We see that for $2\leq i\leq n-1$ these cells can add at most one color that never met before (if it would add two colors, this color and the color of $a_{1i}$ would be three colors in $R_i\bigcap C_i$
Moreover, we don't perform step $n$ since all cells of $R_n\bigcap C_n$  are already drawn.
Hence we get at most $2+(n-2)=n$ colors. Since there are exactly $n$ colors, equalitites hold everywhere so every step adds a new color. But if step $i$ adds a new color, this means that all already drawn elements of $R_i \bigcap C_i$ (i.e. $a_{1i},\cdots, a_{i-1,i},a_{i1},\cdots,a_{i,i-1})$ have same color.  For a given $n>i>1$, denote $U_i$ these cells. Then $U_i$ is monochromatic.
Moreover $U_i$ partition the cells of the matrix (except the diagonal which we completely ignore, and except $R_n \bigcap C_n$).
But every $U_i$ intersects $R_1\bigcap C_1$ hence the colors of $U_i$ are at most two. Also, $R_n \bigcap C_N$ can have at most one more color hence $n \leq 3$. The case $n=2,3$ is easily handled.

Now let's prove that if $3\leq k \leq n$ we can draw such a matrix.
If $k=n$ it's easy so let's look at $k<n$.
Consider $R_1=0,1,\cdots,1,2,3,\cdots,k$ ($1$ meets $n-k$ times).
Now let $R_i$ be $R_1$ shifted cyclically by $i-1$ positions to the right.
We easily see that the matrix satisfies our conditions but we have only $k$ colors. To make their number $n$ we proceed like this: Choose four cells $a_{i,j},a_{j,i},a_{n+1-i,n+1-j},a_{n+1-j,n+1-i}$ which have the same color.
If this color is not $1$ and meets elsewhere in the table, the we can change the color of this four cell into $k+1$ and the conditions will hold. Then make the same procedure for color $k+2$ etc. It's kid's play to prove that we have enough colors to  perform all these operations
QED

BTW Darij, why do you say this problem was least interesting to you? If you are weaker in some section of mathematics, you should pay more attention especially to it if you want to increase your IMO performance.
 How many persons solved this problem at the contest? Me, I wouldn't solve it because I've spent so much time on it..

[quote="iura"]
BTW Darij, why do you say this problem was least interesting to you? If you are weaker in some section of mathematics, you should pay more attention especially to it if you want to increase your IMO performance.
 How many persons solved this problem at the contest? Me, I wouldn't solve it because I've spent so much time on it..[/quote]

Darij is aware of the fact that he also should devote some times to other sections of math olympiads than geometry and inequality but you always do that what you like best. Especially it will not help him if he gets yet still more mature in his favourite topics. You just cannot get more than 7 points per problem.  :D But I guess it is the same with you about combinatorics.  ;)

BTW what happened to Sasha. I haven't seen him on AoPS/ML for a while ?

The official solution can be found on the second page of [url=http://www.bundeswettbewerb-mathematik.de/imo/aufgaben/pdf/loes_awkl2_05.pdf]this .pdf-file.[/url] But Darij gave already a link but not yet so specific.  :)

Hey people, if I'll ever start with combinatorics, then surely not with such case-distinction problems...

  darij

Thank you Orl, but it seems to me that I know German only from some songs and they do not sing mathematics.. I could just check the answer.
About Sasha, nothing happened to him. he just doesn't enter the site so often.
I understand him, in autumn I also stopped entering the site for a while because it can be really a drug if you can't stop..
I remember me, for instance: I opened about $20$ pages, then closed Internet and started to solve them. Since I'm home at $5$ pm, until I finished with all these pages there was time for bed and I didn't have time to study anything else about mathematics. But everyone needs some books, so I decided to leave the site for a while. It's a good site if you want to practice your solving skills, but there is some theory you have to do..

About combinatorics, I keep my opinion: You, Orl, said that Darij will get his gold medal in Mexico. But he can eassily miss it if he can't solve a combinatorics problem because it's not his style, and another problem simply because it's the hardest..

[quote="iura"]
About combinatorics, I keep my opinion: You, Orl, said that Darij will get his gold medal in Mexico. But he can eassily miss it if he can't solve a combinatorics problem because it's not his style, and another problem simply because it's the hardest..[/quote]

Yeah. You are right. Even [url=http://www.mathlinks.ro/Forum/viewtopic.php?p=22677#p22677]Myth[/url] thinks like that. Maybe we should motivate Darij to practise some combinatorics.  ;)

[quote="iura"]Now let's prove that if $3\leq k \leq n$ we can draw such a matrix.
If $k=n$ it's easy so let's look at $k<n$.
Consider $R_1=0,1,\cdots,1,2,3,\cdots,k$ ($1$ meets $n-k$ times).
Now let $R_i$ be $R_1$ shifted cyclically by $i-1$ positions to the right.
We easily see that the matrix satisfies our conditions but we have only $k$ colors. To make their number $n$ we proceed like this: Choose four cells $a_{i,j},a_{j,i},a_{n+1-i,n+1-j},a_{n+1-j,n+1-i}$ which have the same color.
If this color is not $1$ and meets elsewhere in the table, the we can change the color of this four cell into $k+1$ and the conditions will hold. Then make the same procedure for color $k+2$ etc. It's kid's play to prove that we have enough colors to  perform all these operations
QED[/quote]

iura, I'm sorry, maybe I'm just quite sleepy right now, but I don't see how you can guarantee that there exist enough of such cells $a_{i,j},a_{j,i},a_{n+1-i,n+1-j},a_{n+1-j,n+1-i}$ which have the same color that is not $1$, such that you can add an additional $n-k$ colours.  In fact, for $n=8$ and $k = 4$, I can't even find two elements $a_{i, j}, a_{j, i}, i \neq j$, which have the same color that is not $1$.  Perhaps this is because I've misread what you've written.  Would you mind giving an example to illustrate it more clearly?  Thanks!

It's me sleepy now, but I'm able too realize you are right. I don't know what I thought with then, but now I think maybe we should correct it like this: Place the '0' in $R_i$ at place $i$ (to be on the diagonal) and the rest cells be the row $R_1$ shifted $i-1$ positions to the [b]left[/b]. Now the matrix must be simmetric. Now is it OK? (Remember, I am sleepy, I can't check at the moment)

I don't see how to get a symmetric matrix by doing a cyclic shift of the row $R_1$ (excluding $0$ - is that what you meant?) to the left... :?

anyway, the final result and the method of the proof are correct - but i just wrote a paper and therefore am too lazy to post the correct proof(i had to ask christian sattler sitting next to me)

Let $f(n,k)=1$ if it's possible and $0$ otherwise; similarly define $g(n,k)=1$ if we can find a coloring such that $C_i$ has color $i$ (this allows us to modify the number of distinct colors per circle more easily), so that $f(n,k)\ge g(n,k)$. Obviously $g(2,2)=g(3,2)=g(3,3)=1$.

For $k=3$ and $n\ge4$, we can label $C_1C_2,C_2C_3,\ldots, C_nC_1$ with
\[\{1,2\},\{2,3\},\{3,4\},\{4,4\},\ldots,\{n,n\}\]and then set $C_iC_j=2$ if $j-i\not\equiv\pm1\pmod{n}$, so $g(n,3)=1$ for $n\ge4$ as well.

Moreover, $g(n+1,k+1)\ge g(n,k)$. Indeed, if $g(n,k)=1$, then taking $C_{n+1}C_i=\{i,n+1\}$ for $1\le i\le k$ and $C_{n+1}C_i=\{n+1,n+1\}$ for $k+1\le i\le n$, we have $g(n+1,k+1)=1$.

Thus $f(n,k)=g(n,k)=1$ for $2\le k\le n\le 3$ and $3\le k\le n$.

It remains to show that for $n\ge4$, $f(n,2)=0$, so assume for the sake of contradiction that $f(n,2)=1$. WLOG, say that $C_1$ has colors $1$ and $2$, $C_2$ has colors $x\in\{1,2\}$ and $3$, and $C_3$ has colors $y\in\{1,2\}$ and $4$ ($k=2$, each color is used at least once, and all circles intersect $C_1$). Considering $C_2C_3$, we need $x=y$, so WLOG say $x=y=1$. Then for $4\le i\le n$, $C_i$ must have color $1$ by analyzing $C_1C_i,C_2C_i,C_3C_i$. But each color must be contained in at least two circles, so we need at least $2(n-1)>n$ circles, a contradiction.

[quote=iura]OK, all I've said yesterday is real non-sense so let me construct the example in another way(the notations are from the first solution):

Call a $n$-'free' set of cells a set of $n$ cells s.t. in every line and column lies exactly one of the set.
We want to color the matrix in $k$ colors, s.t. the sets of cells colored in each of the color $2,\cdots,k$ is free.
Then, observe that if $a_{i,j}, a_{j,i}$ have the same color $c>1$ then we can change their colors to any new color, so we can eventually add new colors until we reach $n$ colors. Basing upon this main, idea, let's try to construct the example.
Lemma: if from a $m \times m$ subtable we deleted some $m$-free sets, the we can pick up a new $m$ free-set unless we already deleted all cells. Proof is easy from Marriage Lemma.
Now using this lemma, let's construct the example:
Choose the inverse diagonal $x+y=t$ where $t$ is odd (so that it doesn't intersect the main diagonal).
Then we can color all elements of this diagonal in color $i$. Using the lemma, we color more cells in color $i$ to yield a $n$-free set. Now taking $t$ every time as close to $b$ as possible (to reach a more cells on the diagonal), we can ensure that we have enough monochromatic pairs $a_{i,j}, a_{j,i}$ to be changed. The other cells not on the main diagonal shall be colored $1$. This ends the proof

I'm sure I didn't explain clearly the proof, so let me give an example:
$n=8, k=4$.
Firstly the empty matrix:
$0.......$
$.0......$
$..0.....$
$...0....$
$....0...$
$.....0..$
$......0.$
$.......0$
Now add color $2$ on the diagonal $x+y=9$ (closest to $8$):
$0......2$
$.0....2.$
$..0..2..$
$...02...$
$...20...$
$..2..0..$
$.2....0.$
$2......0$

Then we add color $3$ on the diagonal $x+y=7$ and complete it with $2$ cells to produce a $n$-free set.
$0....3.2$
$.0..3.2.$
$..03.2..$
$..302...$
$.3.20...$
$3.2..0..$
$.2....03$
$2.....30$

Finally, we add color $4$ on the diagonal $x+y=11$ (actually, it doen't matter anymore, since we already have enough pairs)
$04...3.2$
$40..3.2.$
$..03.2.4$
$..302.4.$
$.3.204..$
$3.2.40..$
$.2.4..03$
$2.4...30$
The rest we complete with $1$.
$04111312$
$40113121$
$11031214$
$11302141$
$13120411$
$31214011$
$12141103$
$21411130$
Finally, to add colors $5$,$6$,$7$,$8$ we change monochromatic pairs symmetric with respect to main diagonal.
$04111312$
$40113171$
$11081614$
$11805141$
$13150411$
$31614011$
$17141103$
$21411130$

We are done.[/quote]

Does this proof work when n and k are very close? For example if (n, k) = (8,6) you would just run out of diagonals.

Solution from [i]Twitch Solves ISL[/i]:

Rephrasing: [quote]Given the graph $G = K_n$, 	we wish to color its edges with two of $n$ colors 	(possibly the same color) such that every color is used, 	and every vertex $v$ has exactly $k$ colors used at it. [/quote]
The answer is that: [list] 	[*]when $k=2$, only $n \in \{2,3\}$ are possible. 	[*]when $k=3$, any $n \ge k$ is permissible. [/list] Let's prove this. In what follows, $S(v)$ will denote the set of colors at a vertex $v$.
For $k=2$, the construction of $n=2$ and $n=3$ are routine: [list] 	[*]For $n = k = 2$, let $V(G) = \{v_1,v_2\}$. 	Then set $S(v_1) = S(v_2) = \{A,B\}$ and paint the edge with $AB$. 	[*]For $n = 3$ and $k = 2$, let $V(G) = \{v_1,v_2,v_3\}$. 	Then set $S(v_1) = \{A,B\}$, $S(v_2) = \{B,C\}$, $S(v_3) = \{C,A\}$ 	and color $v_1v_2$ with $BB$, $v_2v_3$ with $CC$, $v_3v_1$ with $AA$. [/list]
[b][color=red]Claim:[/color][/b]  	For $k=2$, we have $n \le 3$.
[i]Proof.[/i]  	Assume $n > 3$. 	Each $S(v)$ is a two-element set. 	So we get a natural graph $\Gamma$ on the set of colors 	where the edges join two colors if they appear as some $S(v)$. 	This graph $\Gamma$ has the property any two edges intersect. 	This means it is either a $3$-cycle (hence $n=3$) or $K_{1,n-1}$. 	For $n > 3$, the latter case cannot work, 	since every color must appear in at least two $S(v)$'s. $\blacksquare$

[b][color=red]Claim:[/color][/b]  	For $k \ge 3$ there is a construction for all $n \ge 3$.
[i]Proof.[/i]  	If $n=k=3$ let $V(G) = \{v_1,v_2,v_3\}$. 	Then set $S(v_1) = S(v_2) = S(v_3) = \{A,B,C\}$ 	and color $v_1v_2$ with $AB$, $v_2v_3$ with $BC$, $v_3v_1$ with $CA$.
Now suppose $n \ge 4$. 	Let $V(G) = \{v_1, v_2, \dots, v_n\}$ 	and let the $n$ colors be $C_1$, $C_2$, \dots, $C_{n-1}$, $C_{\text{special}}$ 	(breaking symmetry). 	We will create a construction with 	\begin{align*} 		S(v_1) &= \left\{ C_{\text{special}}, C_1, C_2 \right\} \\ 		S(v_2) &= \left\{ C_{\text{special}}, C_2, C_3 \right\} \\ 		&\vdotswithin{=} \\ 		S(v_{n-2}) &= \left\{ C_{\text{special}}, C_{n-2}, C_{n-1} \right\} \\ 		S(v_{n-1}) &= \left\{ C_{\text{special}}, C_{n-1}, C_1 \right\} \\ 		S(v_n) &= S(v_1) = \left\{ C_{\text{special}}, C_1, C_2 \right\}. 	\end{align*} 	Assign colors to the following edges: 	[list] 		[*]Assign $v_1v_2$ colors $C_{\text{special}} C_2$. 		[*]Assign $v_2v_3$ colors $C_{\text{special}} C_3$. 		[*]\dots 		[*]Assign $v_{n-2}v_{n-1}$ colors $C_{\text{special}} C_{n-1}$. 		[*]Assign $v_{n-1}v_{n}$ colors $C_{\text{special}} C_{1}$. 		[*]Assign $v_{n}v_{1}$ colors $C_1 C_2$. 	[/list] 	Then assign $C_{\text{special}} C_{\text{special}}$ to all other edges. $\blacksquare$

[b][color=red]Claim:[/color][/b]  	Given a construction for $(n,k)$ 	we can get a construction for $(n+1, k+1)$.
[i]Proof.[/i]  	We do the following procedure. 	[list] 	[*]Create a new color $C_{\text{new}}$ and add it to every $S(v)$. 	[*]Add a new vertex $v_{\text{new}}$. 	[*]Let $C_1$, \dots, $C_k$ be any old colors 	and let $v_1$, \dots, $v_k$ be vertices with $C_i \in S(v_i)$. 	[*]Set $S(v_{\text{new}}) = \{C_{\text{new}}, C_1, C_2, \dots, C_k\}$ 	[*]The edges $v_{\text{new}}v_i$ are colored $C_iC_{\text{new}}$. 	[*]Any other edges are colored $C_{\text{new}} C_{\text{new}}$. 	[/list] 	This completes the construction. $\blacksquare$

[b][color=red]Remark:[/color][/b]  	I have to say I found most of the conditions in the problem 	rather unnatural to work with. 	For example, there is no real reason why the number of colors 	should be equal to the number of circles, 	and in fact the construction has to ``work around this'' 	by breaking the symmetry somewhat in the first main claim.

We claim that the answers are $k=2, n=2, 3$, and $3\le k\le n$ for any $n\ge 3$. The cases $k=2, n=2,3$ are trivial to check. It is obvious that $k\le n$.

First, we prove that when $n>3$, $k\le 2$ is impossible. Let the [i]exposability[/i] of a color denote how many circles contain at least one of that color. Since each color appears at least once, the exposability of every color is at least $2$, and therefore the minimum possible overall exposibility is $2n$. For a particular $k$ to be possible, the overall exposability must be $nk$. Thus $k=1$ is impossible. $k=2$ is only possible if the exposability of every color is $2$. This would mean that each color could only appear at most twice (the two intersections of any two circles). This implies $$2n\ge 2\binom{n}{2}\Rightarrow n\ge \binom{n}{2}=\frac{n(n-1)}{2}\Rightarrow n\le 3,$$ and thus $k=1,2$ is impossible for $n>3$.

Next, we will prove that when $n\ge 3$, any $k$ satisfying $3\le k\le n$ works. Number the circles from $1$ to $n$ in the colors from $1$ to $n$.

[b][color=#f00]Claim:[/color][/b] When $n\ge 3$, $k=3$ always works.
[i]Proof.[/i] Take color and circle numbers modulo $n$. This is trivial when $n=3$. When $n\ge 3$, the following construction works:

1. For $1\le i\le n$, color one intersection from circle $i$ to circle $i+1$ with color $i$.
2. Color one intersection between circles $2$ and $n$ with color $n$.
3. Color one intersection from circle $1$ to circle $3$ with color $3$.
4. Color all other intersections with color $1$.

This ensures every circle has $3$ colors as

1. If $3\le i\le n$, the circle $i$ has colors $i-1, i, 1$.
2. Circle $1$ has colors $1,3,n$.
3. Circle $2$ has colors $1,2, n$. $\blacksquare$

[b][color=#f00]Claim:[/color][/b] If a particular pair $(k,n)$ (with $n\ge 3$) work with the configuration having the property that circle $i$ contains color $i$ for all $1\le i\le n$ ,then the pair $(k+1, n+1)$ works.
[i]Proof.[/i] Use the configuration given for $(k,n)$, and add another circle $n+1$. Then use this procedure to color in the intersections with this new circle:

1. For circle $n+1$ and circle $i$, for $1\le i\le k$, make one intersection color $n+1$ and one intersection $i$.
2. For circle $n+1$ and a circle numbered greater than $k$, make both intersections color $n+1$.

All circles $1$ through $n$ now intersect exactly one new color $n+1$, and circle $n+1$ has color $n+1$ in addition to colors $1$ through $k$, so this works. $\blacksquare$

Now given an $(n,k)$ with $n\ge 3$ and $3\le k\le n$, by using the above claim repeatedly we conclude that it is sufficient to showing the pair $(n-k+3, 3)$ works, which is true by the first claim.

[b]Answer[/b]: For $n=2$ the only such $k$ is 2; for $n=3$, $k=2, 3$ works; for $n\geq 4$ there is a construction for any $3\leq k\leq n$.

The small cases $n=2, 3$ can be verified by hand and I won't do that here. Assume henceforth $n\geq 4$.

[b]Proof of impossibility of $k=1, 2$.[/b] If $k=1$ all intersections must be of the same colour, which by assumption is not true. Now, arguing indirectly, assume we have a construction for $k=2$. Construct a graph $G$ with vertex set the $n$ colours, and edges representing the circles, which we can do since any circle contains exactly two colours. It is given that i) there are $n$ edges and any two edges share a vertex and ii) every vertex is the endpoint of some edge. This is impossible for $n\geq 4$ because 1) By i) the graph is not a tree so it contains a cycle, 2) If the cycle has four or more edges we contradict ii), so the graph contains a triangle, 3) Every edge is adjacent to at least one vertex of this triangle. If we have an edge that is only adjacent to one of them, then it shares no endpoint with the opposite edge in the triangle, contradicting ii) again. Hence $n\leq 3$. $\square$

[b]Construction for $n\geq k\geq 3$[/b]. Let the colours be $x_1, \dots, x_n$, and the circles $c_1, \dots, c_n$. Set $c_n$ aside and consider the other $n-1$ circles with indexes modulo $n-1$. Now, colour the two element set $c_i\cap c_j$ as follows:

1. For $1\leq i\leq k-1$ let $c_n\cap c_i\{x_i, x_n\}$, and for $i\geq k$ let $c_n\cap c_i=\{x_n, x_n\}$

2. If $n\not\in \{i, j\}$, then $c_i\cap c_j=\{x_j, x_n\}$ if $i<j\leq i+k-2$ (if we cross $c_n$ between $c_i, c_j$ we add $n-1$ to the indexes when studying the inequality), and $c_i\cap c_j=\{x_n, x_n\}$ if $j$ does not lie in this interval.

Circle $c_n$ contains the colours $c_1, \dots, c_{k-1}, c_n$, and for $i\neq n$ circle $c_i$ contains the colours $c_i, c_{i+1}, \dots, c_{i+k-2}$ (indexes modulo $n-1$) and $c_n$; in both cases, the circles contain exactly $k$ colours, and each colour is used at least once, so this works. $\square$.


function
calculus
integration
complex analysis
complex analysis unsolved

Let $f$ be an entire function that is bounded on every line that goes through the origin. Is $f$ constant ?

I may be missing something, I don't see how this is any different from saying f is bounded, in general?  Pick some point in the complex plane--then it has to be on a line through the origin, so it has to be less than the supremum of the bounds of all the lines. So, f is bounded and entire.

In other words, consider a contour integral.  Liouville is just integrating over a counter of expanding radius.  If it's bounded at every point over the contour since every point is on a line, then Liouville still applies, since as the contour expands the value of f' goes to 0.

[quote="gka"]I may be missing something, I don't see how this is any different from saying f is bounded, in general?  Pick some point in the complex plane--then it has to be on a line through the origin, so it has to be less than the supremum of the bounds of all the lines. So, f is bounded and entire.
.[/quote]
I think you may miss something.
Bounded on every line that goes through the origin doesn't means bounded in the whole complex.
For example,f(z)=1/(z-1),z not in the line y=x.f(z)=0,y=x is bounded on every line that goes through the orign.But is not bounded in the whole complex plane.

[quote="ffrogg"]But your example is not an entire function.[/quote]
I just say the proof of $gka$ is wrong,but I didn't say that this problem is wrong...

Although I believe an entire function with this property is bounded throughout the complex plane, with respect to my initial claim you're completely, right.  The general argument I stated still applies, though, I think:  When you integrate over the circle, you still get |f'(z)| < M/r^n, since every point on the edge of the circle lies on a line through the origin, and thus every point must be less or equal to the supremum of the upper bounds through all lines.  Since the function is entire there's no residue, so the contour integral approaches zero as r approaches infinity.

there is no finite upper bound, a priori !

I think I see the problem.  As I said I'm new to this, but there's no harm in trying.

My train of thought was to assume that the supremum over all lines is finite. Assuming you agree with that, then every point is less than the supremum over all lines of upper bounds.  Suppose f took a value greater than this supremum at some point.  Then f of this point is greater than the supremum of the upper bound over all lines.  But f is on a line through the origin since the function is entire.  Contradiction.

The issue, as you point out, is showing that the supremum over all lines is finite.  I can't give such a proof in a great degree of rigor, but here's a possible sketch.  I might guess that the best way to do this would be to use a method of bisection, or some complex analogue.  Suppose that the upper bound is not finite, then between two theta (theta_1,theta_2) on complex plane f does not have a finite upper bound.  Take the midpoint of these theta, then either between theta_1 and theta_m or theta_2 and theta_m there does not exist a finite upper bound.  These bisections of theta limit to some theta.

From here, I have a guess on how to proceed: If it's true that the maximums on the line are a continuous function (since f is holomorphic, this would seem to be true--arbitrarily close to the f(theta,r) that takes a maximum, f(theta_prime,r) has to take a value arbitrarily close), take the limit of the maximums, which is equal to f(theta), which is finite by supposition. Contradiction.

Probably it is time for a
[hide="Hint"]
There exist such non-constant functions. One can, for instance, construct a non-constant function that is bounded everywhere outside the horizontal strip $1<Im\,z<2$. Clearly, such a function would be bounded on every line passing through the origin.
[/hide]


calculus
integration
analytic geometry
trigonometry
logarithms
calculus computations

[quote="Kent Merryfield"][quote="paladin8"]This looks like it's screaming for a polar substitution to me...[/quote]
The most important part of that would be identifying the set $D$ in polar coordinates so as to have the limits of integration. The condition $x^{2}+y^{2}\le1$ puts you inside the unit circle, so $r\le 1.$ Then the condition $x\ge 0$ restricts you to the right hand half plane. The boundaries of that half plane - the $y$-axis - are rays and rays can be described by equations of the form $\theta=c.$ I would wind up writing it in the form

$\int_{0}^{1}\int_{-\frac{\pi}2}^{\frac{\pi}2}(\text{whatever })\,d\theta\,dr$

Can you finish it from there?[/quote]
My solution would be this:

First polar substitution: $(x,y)\mapsto(r\cos{\varphi},r\sin{\varphi})$ with Jacobian $r$. The half disk (which you described above) $D$ would map to a strip $E$ where $0<r<1$ and $-\pi/2<\varphi<\pi/2$. Neglecting the null sets ($D$ could be made into a open, finite set with null-set-boundary)
we have
\[\iint_{D}\ldots dxdy=\iint_{E}\frac{r\cos{\varphi}}{1+r^{2}}=\int_{0}^{1}\int_{-\pi/2}^{\pi/2}\frac{r}{r^{2}+1}\cos{\varphi}d\varphi dr=2\int_{0}^{1}\frac{r}{r^{2}+1}=[\ln{(r^{2}+1)}]_{0}^{1}=\ln{2}\]

Is this correct? I appreciate some comment on the calculation.

[quote="Kent Merryfield"]That is correct. That is what I would have done (apart from minor notational conventions).[/quote]
Thanks! :lol:


Henry's Hamburger Heaven orders its hamburgers with the following condiments: ketchup, mustard, mayonnaise, tomato, lettuce, pickles, cheese, and onions. A customer can choose one, two, or three meat patties, and any collection of condiments. How many different kinds of hamburgers can be ordered?

$ \textbf{(A)}\ 24 \qquad \textbf{(B)}\ 256\qquad \textbf{(C)}\ 768\qquad \textbf{(D)}\ 40,\!320\qquad \textbf{(E)}\ 120,\!960$

[hide] There are 8 condiments, so you can have 0 out of the eigh, one out of the eight, two out of the eight... up to eight out of the eight.  There are 8 choose 0+8 choose 2+8 choose 3...+8 choose 8=256 ways to have this.  Now, there are 3 possibilties for the patty, so there are 256*3=768 total possibilies, which is $C$.  [/hide]

instead of adding all the choosing thins up, realize that by pascal's traingle:
$\binom{n}{0}+\binom{n}{1}.......+\binom{n}{n-1}+\binom{n}{n}=2^{n}$

[quote="#H34N1"]Henry's Hamburger Heaven orders its hamburgers with the following condiments: ketchup, mustard, mayonnaise, tomato, lettuce, pickles, cheese, and onions. A customer can choose one, two, or three meat patties, and any collection of condiments. How many different kinds of hamburgers can be ordered?

(A) 24 (B) 256 (C) 768 (D) 40; 320 (E) 120; 960[/quote]

[hide]There are 8 condiments and each one can either be included or not included in the hamburger.  This gives us $2^{8}=256$ possibilities for the condiments.  $3\cdot 256=\boxed{768}$.[/hide]


function
trigonometry
algebra unsolved
algebra

[quote="bjh7790"]find x when

$ sin61 \plus{} sin47 \minus{} sin 25 \minus{} sin11 \equal{} cosx$

Thanks.[/quote]

Do you mean degree? If so, then

$ (\sin61 \plus{} \sin47 \minus{} \sin 25 \minus{} \sin11)(2 \cos18)\equal{}2 \cos x \cos18$
$ \Leftrightarrow (\sin 79\plus{}\sin 43)\plus{}(\sin 65\plus{}\sin 29)\minus{}(\sin 43\plus{}\sin 7)\minus{}(\sin 29\minus{}\sin 7)\equal{} 2 \cos x \cos 18$
$ \Leftrightarrow \sin 79\plus{}\sin 65\equal{} 2 \cos x \cos 18$
$ \Leftrightarrow 2 \sin 72 \cos 18\equal{}2 \cos x \cos 18$
$ \Leftrightarrow \sin 72\equal{}\cos x$
$ \Leftrightarrow x\equal{}(360n \pm 18)^{\circ}$, where $ n$ is any integer.

umm... I think that it is $ 2sin72cos7$ instead of $ 2sin72cos18$...


Two blocks $A$ and $B$ of equal masses are released from an inclined plane of inclination $45degrees$ at $t=0$.Both blocks are initially at rest.The coefficients of Kinetic friction between the block$A$ and the inclined plane is $0.2$ while it is $0.3$ for block $B$.Initially the block $A$ is $\sqrt{2}m$ behind block $B$.When and where their front faces will come in a line.($g=10m/s^2$)

P.S--don't just post answers,it is the method that matters ;)

is it

$t=\sqrt{\frac{\sqrt{2}}{gcosA(tb-ta)}}$
where tb and ta is coefficient friction for B and A
i do jee last year online  (2004) in 1h30 . in INDIA , do people feel it hard

Setting the the plane of incline as the $x$-axis,
\[ s_a = \frac{1}{2}g \{\sin \theta - \mu_{k;a} cos \theta\}t^2 \]
\[ s_b = \frac{1}{2}g \{\sin \theta - \mu_{k;b} cos \theta\}t^2 + \sqrt{2} \]
Since $s_a = s_b$,
\[ \frac{1}{2}g\{\left(\mu_{k;b} - \mu_{k;a}\right) cos \theta\}t^2 = \sqrt{2} \]
\[ t = \sqrt{\frac{2\sqrt{2}}{g(\mu_{k;b} - \mu_{k;a}) cos \theta}} \]
\[ t = 2s \]
\[ s_a = s_b = 8 \sqrt{2} m. \]


linear algebra
matrix
linear algebra unsolved

can someone pls help? need to determine the signature of the form <A,B>=trace(AB) on the space R^(nxn) of nXn real matrices (n times n)

cheers!

Well, the trace of $ A^TB$ would be positive definite. I'd suggest looking at the basis consisting of elements like $ \begin{bmatrix}1&0&\cdots&0\\ 0&0&\cdots&0\\ \vdots&\vdots&\ddots&\vdots\\ 0&0&\cdots&0\end{bmatrix}$, $ \begin{bmatrix}0&1&\cdots&0\\ 1&0&\cdots&0\\ \vdots&\vdots&\ddots&\vdots\\ 0&0&\cdots&0\end{bmatrix}$, and $ \begin{bmatrix}0&1&\cdots&0\\ \minus{}1&0&\cdots&0\\ \vdots&\vdots&\ddots&\vdots\\ 0&0&\cdots&0\end{bmatrix}$, separating symmetric and antisymmetric matrices.

Let $ E_{i j}$ be the matrix that is zero everywhere, except for a 1 on the i,j position. These $ n^2$ matrices are a good basis.

One checks that $ trace ( E_{ij } E_{k l})$ is always zero,except for $ trace( E_{ij } E_{ j i}) \equal{} 1$.

First put all the $ E_{ i i}$ together in your basis, followed by all pairs $ E_{i j } , E_{j i}$ with $ i\neq j$

You should find that there are $ (n^2 \plus{} n)/2$ positive eigenvalues


[quote="AMC question"]Triangle ABC is a right triangle with $\angle ACB$ as its right angle, $m\angle ABC=60$, and $AB=10$.  Let $P$ be randomly chosen inside $\displaystyle \triangle ABC$, and extend $BP$ to meet $AC$ at D.  What is the probability that $BD > 5\sqrt{2}$?[/quote]

(See picture below.)
When $BC=5$ and $BD=5\sqrt{2}$ then $m\angle CBD=45$.  Since $m\angle ABC=60$, then $m\angle ABD=15$ when $BD=5\sqrt{2}$.  Then since we want $BD > 5\sqrt{2}$, why can't we just say that that is satisfied when $m\angle ABD < 15$, making the probability $\frac{15}{60}=\frac{1}{4}$?

Why not? Because the [b]point[/b], was chosen at random, not the [b]angle.[/b]

(Side issue: "at random" is a fairly nebulous concept, but "chosen according to a uniform distribution" would have to be taken as the default meaning.)

To choose a point "at random" in a two-dimensional figure is to choose it in such a way that probability is proportional to area. Your picture then explains it all. The required probability is

$\frac{\mathrm{area}(\triangle ABD)}{\mathrm{area}(\triangle ABC)} = \frac{5\sqrt{3}-5}{5\sqrt{3}}=1-\frac1{\sqrt{3}}.$

Your answer was based on the assumption that $\angle ABP$ was chosen according to a uniform distribution of angles within $\angle ABC$. That's simply not what the problem said.


Okay, eigentlich bin ich am wenigsten derjenige der das hier posten sollte, weil mein Beitrag zu dem ganzen wohl im Endeffekt eher destruktiv ausgefallen ist, aber da der Rest des Teams grad noch in Gunzenhausen ist stell ich einfach mal die Aufgaben rein.

Die Idee war es, einen Matboj (= Math Fight = Teamwettbewerb mit mündlicher Besprechung der Aufgaben) innerhalb des QED-Vereins durchzuführen, mit selbstgesammelten oder im Idealfall -erfundenen Aufgaben. Die Organisatoren waren v. a. Daniel Harrer und ich, wobei ich gelinde gesagt die eigentliche Organisation wenig positiv beeinflusst habe (ich musste wegen einer Erkältung kurzfristig absagen). Der Wettbewerb fand auf einem von Christoph-Simon Senjak organisierten QED-Seminar in Gunzenhausen statt.

Danke (und Respekt!) an Daniel und Christoph-Simon für die eigentliche Organisation. Die Aufgaben findet ihr in der beiliegenden PDF-Datei. Außerdem habe ich die Geometrie- und Ungleichungsaufgaben (für den Teil war ich mehr oder weniger zuständig) in den internationalen Teil von ML gepostet:

5 (G1): http://www.mathlinks.ro/Forum/viewtopic.php?t=60123
2 (G2): http://www.mathlinks.ro/Forum/viewtopic.php?t=60124
9 (G3): http://www.mathlinks.ro/Forum/viewtopic.php?t=60125
14 (G4): http://www.mathlinks.ro/Forum/viewtopic.php?t=60126
6 (U1): http://www.mathlinks.ro/Forum/viewtopic.php?t=60127
12 (U2): http://www.mathlinks.ro/Forum/viewtopic.php?t=60128

[b]NEU: Die Musterlösungen[/b] der Geometrie- und Ungleichungsaufgaben sind auf http://www.cip.ifi.lmu.de/~grinberg/Dreigeom/Inhalt.html#qedmo zu finden.

  Darij

Nachtrag zu Darij's post: nun auch Zahlentee und Kombi:

4 (Z1): http://www.mathlinks.ro/Forum/viewtopic.php?t=60226
8 (Z2): http://www.mathlinks.ro/Forum/viewtopic.php?t=60227
11 (Z3): http://www.mathlinks.ro/Forum/viewtopic.php?t=60225
1 (Z4): http://www.mathlinks.ro/Forum/viewtopic.php?t=60224

7 (C1): http://www.mathlinks.ro/Forum/viewtopic.php?t=60235
3 (C2): http://www.mathlinks.ro/Forum/viewtopic.php?t=60233
10 (C3) http://www.mathlinks.ro/Forum/viewtopic.php?t=60230
13 (C4): http://www.mathlinks.ro/Forum/viewtopic.php?t=60243


inequalities
geometry
perimeter
area of a triangle
Heron\u0027s formula
inequalities solved

If a,b,c are the lengths of the sides of a triangle and A=area,P=perimeter
prove that

 (ab) 2 +(bc) 2 +(ca) 2  \geq 12A 2 +P^4/108

[i]unknown source[/i]

I have only a partial solution.

From Heron's formula for the area of a triangle

 \sum a^2b^2 =8A^2 +1/2 \sum a^4

we have the equivalent inequality

  8A^2 +1/2 \sum a^4 \geq 12A^2 +(P^4)/108

or 4A^2 \leq 1/2 \sum a^4 -(a+b+c)^4/108

If you can prove that (a^4+b^4+c^4)/4 \geq 4A^2  (*)

it remains to prove that

(a^4+b^4+c^4)/4 \leq 1/2 \sum a^4 -(a+b+c)^4/108

which reduces to  a^4+b^4+c^4 \geq (a+b+c)^4/27

which is true by the power mean inequality.

To complete the proof it remains to prove (*)

But (*) holds trivially from the well-known inequality a 2 + b 2  + c 2   \geq  4A \sqrt 3, and using the inequality between means of order 2 and order 4 : a 4  + b 4  + c 4   \geq (a 2 + b 2  + c 2)2 /3.

Pierre.


First of all, what exactly is the meaning of uniform motion?

Secondly, what is the best way to learn how to solve word problems regardless of difficulty level?

QUESTION BElOW.

Two cars enter the Florida Turnpike at Commercial BLVD at 8am, each heading for Wildwood.  One car's average speed is 10 miles per hour more than the other's.  The faster car arrives at Wildwood at 11am, 1/2 hour before the other car.  What is the average speed of each car?  How far did each car travel?

Also, is there a formula for average rate or speed?

[hide]$d=rt$ :D
There are two vehicles so we have
$d_{1}=r_{1}t_{1}$
$d_{2}=r_{2}t_{2}$
$d_{1}=d_{2}$
$r_{1}=r_{2}+10$
It looks like you also know the times. You should be able to solve.[/hide]


Solve the equations $ x^3\minus{}7x^2\minus{}204x\plus{}1260\equal{}0$ and $ x^3\minus{}15x^2\minus{}394x\plus{}840\equal{}0$ knowing that the first equation has a root in which its value equals three times the value of a root from the second.

hello, for the first equation we get $ x\equal{}6$ as a solution and the second one has $ x\equal{}2$ as one solution.
Sonnhard.

Factor the equations as :

$ (x\minus{}6)(x^2\minus{}x\minus{}210)\equal{}(x\minus{}6)(x\minus{}15)(x\plus{}14)\equal{}0$ and $ (x\minus{}2)(x^2\minus{}13x\minus{}420)\equal{}(x\minus{}2)(x\plus{}15)(x\minus{}28)\equal{}0$

Now it's easy to get the answers which Dr. Graubner got.

Hint on how to factor: look at the divisors of the constant term


p is a prime number greater than 2. Are there two natural numbers x and y such that sqrtx + sqrty = sqrt2p ?

[hide]No.

[hide]x = (sqrt(2p) - sqrt(y))^2  will mean    2py has to be a perfect square,  means y =2p (and factor of 4p^2) > 2p
similary x > 2p

So no dice.[/hide][/hide]


Let $G$ be the additive group of $\mathbb{Z}[x]$ and let $H$ be the multiplicative group of all positive rational numbers.Prove that $G$ isomorphism with  $H$

[quote="Christian Hirsch"]$x^n\to p_n$ where $p_n$ is the $n$-th prime ?  :?[/quote]

I can't understand your solution. :(

[quote="ZetaX"]Let $p_0,p_1,p_2,...$ be the primes in any order. Then mapping $a_0+a_1x+a_2x^2+...+a_nx^n$ to $p_0^{a_0}p_1^{a_1}p_2^{a_2}...p_n^{a_n}$ works.[/quote]

Me too  :) (I find it,yesterday)


We know that   a quadratic expression is a perfect square if its discriminant is zero .but n^2 -1 is perfect square at n=1 even its discriminant is not zero.   why. Pl explain this concept I am little bit confused.

 Hence prove that there is no +ve  integer  for which  (n+1)^.5+ (n-1)^.5  is a rational number

I'm not sure but I guess it's something like that:

$ n^2 \minus{} 2n \plus{} 1$ is a perfect square and its discriminant is $ 0$.
$ n^2 \minus{} 2n \plus{} 1$ => $ n^2 \minus{} 1 \equal{} 2n \minus{} 2$
For $ n \equal{} 1$, $ 2n \minus{} 2 \equal{} 0$. So $ n^2 \minus{} 1$ is indeed a perfect square because $ 2n \minus{} 2$ equals $ zero$

[quote="basted"]I'm not sure but I guess it's something like that:

$ n^2 \minus{} 2n \plus{} 1$ is a perfect square and its determinant is $ 0$.
$ n^2 \minus{} 2n \plus{} 1$ => $ n^2 \minus{} 1 \equal{} 2n \minus{} 2$
For $ n \equal{} 1$, $ 2n \minus{} 2 \equal{} 0$. So $ n^2 \minus{} 1$ is indeed a perfect square because $ 2n \minus{} 2$ equals $ zero$[/quote]

I don't really get this, but to answer ayush_2008's question, when the discriminant of a quadratic is equal to zero, then it can be [b]factored[/b] as a perfect square in the form $ (ax\plus{}b)^2$.  Just because a quadratic can be a perfect square doesn't mean it can be factored as such.


Let $x,\ y,\ z$ be positive integers that are coprime each other such that $\frac{1}{x}+\frac{1}{y}=\frac{2}{z}.$ If $z$ is a odd number, prove that $xyz$ is a square number.

Are you sure that the numbers are "coprime each other" instead of "coprime"? The problem becomes much easier.
[hide="the former case"]$\frac{1}{x}+\frac{1}{y}=\frac{2}{z}$
$\Rightarrow z(x+y)=2xy$
$\Rightarrow z|2xy$
Since $x,y,z$ are pairwise relatively prime, it follows that $z|2$. So $z=1$.
Now $x+y=2xy$
$\Rightarrow (2x-1)(2y-1))=1$
So $x=y=1$.
Therefore $xyz=1$.[/hide]
[hide="the latter case"]$z(x+y)=2xy$
Let $x=dx_{0}, y=dy_{0}$ where $gcd(x_{0},y_{0})=1$. We also have $gcd(d,z)=1$. Substituting above,
$z(x_{0}+y_{0})=2dx_{0}y_{0}$
Since $gcd(x_{0},y_{0})=1$, $x_{0}$ and $y_{0}$ cannot divide $x_{0}+y_{0}$, and so
$x_{0}y_{0}|z$
Let $z=kx_{0}y_{0}$, where $k$ is odd. Substituting
$k(x_{0}+y_{0})=2d$
$\Rightarrow k|2d \Rightarrow k|d$
But $k|z$, so $k=1$.
Therefore $xyz=(dx_{0})(dy_{0})(x_{0}y_{0})=(dx_{0}y_{0})^{2}$.[/hide]

Yes,　latter part is right, thank you, leepakhin.

kunny


Prove by mathematical induction that the sum of two consecutive numbers is always an odd number.

I think this post should be moved to getting started or something.

All right. I will try

[hide]
If k=1, then  it works, because 1+2=3, which is odd.

Assume that k+(k+1) is odd. then

$(k+1)+(k+2) = k+k+1+(1+1) = k+k+1+2 = (k+k+1)+2 = odd+even = odd

[/hide]

that was poorly written. I was in a rush :blush:

[quote="xxreddevilzxx"](k+1)+(k+2) = k+k+1+(1+1) = k+k+1+2 = (k+k+1)+2 = odd+even = odd[/quote]It's simpler to write $(k+1)+(k+2)=2(k+1)+1$, which is odd. In fact, your proof assumes that the sum of an odd and an even number is odd, which would by itself lead immediately to the result.

[quote="qweretyq"]Even though this problem might be simple I dont think induction is a "Getting Started" topic so I think its ok here[/quote]

Exactly.

Additionally, don't bother asking for a problem to be moved from Intermediate to Getting Started.  Most of the time it won't be moved ([i]especially[/i] if it's a proof)


limit
integration
logarithms
real analysis
real analysis unsolved

show :D

$ \lim_{x\to 0^{\plus{}}}\;\;\frac{e^{\minus{}2\,\textbf f(x)}}{x^{2}}\;\;\equal{}\;\;\boxed{e^{2\,\gamma}}$

where $ \textbf f(x)\;\equal{}\;\int_{x}^{\infty}\;\frac{e^{\minus{}y}}{y}\;\textbf dy$......... and $ \gamma$ is euler's constant

I will omit the Limit notation to simplify

$ \frac{e^{\minus{}2 f(x)}}{x^{2}}\equal{} e^{2\gamma }$

$ \int_{x}^{\infty }\frac{e^{\minus{}y}}{y}\, dy \equal{}\Gamma (0,x)$

MathWorld For more information about those integrals.

$ \frac{e^{\minus{}2\Gamma (0,x)}}{x^{2}}\equal{} e^{2\gamma }$

$ \Gamma (0,x)\plus{}\log (x) \equal{}\minus{}\gamma$

Now a part that i'm not very sure about rigor. 'cause i had assumed that $ x$ is small so the error would be small.

$ \Gamma (a,x) \equal{}\Gamma (a)\minus{}\sum_{k \equal{} 0}^{\infty }\frac{(\minus{}x)^{k}x^{a}}{(a\plus{}k) k!}$

$ \Gamma (0,x) \equal{}\lim_{a\to 0}\,\left(\minus{}\frac{x^{a}}{a}\plus{}\Gamma (a)\plus{}O\left(x^{a\plus{}1}\right)\right)$

$ \lim_{x\to 0}\, (\Gamma (0,x)\plus{}\log (x)) \equal{}\lim_{x\to 0}\,\left(\lim_{a\to 0}\,\left(\minus{}\frac{x^{a}}{a}\plus{}\Gamma (a)\plus{}\log (x)\right)\right)$

Note: If i change the order of the limit the solution would be (ignoring the log term, cause the above limit of a is independent of x, it will cancel):

$ \lim_{a\to 0}\,\left(\Gamma (a)\minus{}\frac{1}{a}\right) \equal{}\minus{}\gamma$

That i dont know how to proof

Note: I found :) $ \Gamma (a) \equal{}\frac{1}{2}\left(\gamma^{2}\plus{}\frac{\pi^{2}}{6}\right) a\plus{}O\left(a^{2}\right)\minus{}\gamma\plus{}\frac{1}{a}$
And the rest i give as a exercise


The circle $ (O_1) $ with center $ O_1  $ , radius $ R_1 $ cuts the circle $ (O_2) $ with center $ O_2 $ , radius $ R_2 $ at the points A and B . The tangent to  $ (O_1) $ at A cuts $ (O_2) $ at C . The tangent to  $ (O_2) $ at A cuts $ (O_1) $ at D . Let M be the midpoint of CD . Prove that :  $ <CAM = <DAN $

I assume that we need to prove <CAM =<DAB.

We will prove that sin DAB/sin BAC = sin CAM/sin DAM.
Using < DAB + < BAC = < CAM + < DAM. = < CAD.
It will be sufficient condition to prove < CAM = < DAB.

DM/sin DAM = AD/sin DMA, CM/sin CAM=AC/sin CMA
As a result: sin CAM/sin DAM = DM*AC / CM*AD = AC/AD

From the other part < DAB = <BCA, <CAB = < ADB. < ABD = <ABC.
Hence AD/sin ABD = AB/sin ADB, AC/sin ABC = AB/sin BCA,
or  sin BCA/sin ADB = sin DAB/sin CAB = AC/AD.

Finally sin DAB/sin CAB = sin CAM/sin DAM = AC/AD.

Another solution: what we must show is that $B$ lies on the $a$-symmedian of $ACD$. In order to do this we need $\frac{B,CA}{B,AD}=\frac{CA}{AD}$, but this follows immediately from the similarity of triangles $BCA,BAD$, which can be proven by observing $\angle BAC=\angle BDA,\ \angle BAD=\angle BCA$.

You probably meant < CAM = < DAB.

An "insider" solution goes as follows: Speaking from the viewpoint of triangle DAC, the point B is the point of intersection (apart from A) of the circle through A and D touching the side AC with the circle through A and C touching the side AD. Hence, the point B is the A-vertex of the second Brocard triangle of the triangle DAC. Consequently, this point B lies on the A-symmedian of triangle DAC. On the other hand, the point M, being the midpoint of the side AC, lies on the A-median of triangle DAC. Since the A-median and the A-symmedian are symmetric to each other with respect to the angle bisector of the angle DAC, it follows that < CAM = < DAB.

In order to get a more elementary solution, use inversion with center A.

  Darij


[quote="Cezar Lupu"]Sa se rezolve in $\mathbb{R_{+}^{*}}$, ecuatia:

                                             \[ 2^{x+\frac{4}{y^{2}}}+2^{y+\frac{4}{x^{2}}}=16.  \][/quote]
   Cam pentru ce clasa e problema?Specifica si tu!

Nu e adevarat ca Inegalitatea lui Cauchy pentru medii se invata in a VII-a?Eu m-am gandit la o solutie mai lunga...totusi asa imi da si mie:x=2,y=2.N-o mai postez, mai ales ca nu stiu nici LaTex :blush: Oricum, este o solutie eleganta cea propusa de tine. :)

Functia exponentiala se invata conform programei numai in clasa a X-a.

Mah, vad ca tu esti persistent rau de tot. Chiar te rog, ia posteaza o solutie acestei probleme diferita de cea data de Metru care este identica cu a mea. Incep sa cred ca am mai vazut dintr'astia asa ca tine care se dau mari viteji, dar de fapt....mai bine nu zic. Mai bine n-ai spune nimik si ai posta o solutie cum face orice om de pe site-ul asta.  :spam:

Nu ma dau mare viteaz cum zici tu!!! :mad: Daca eu stiu, ai ceva impotriva?Cred totusi ca ai ceva impotriva mea.
Acum ar trebui sa zici ca tu esti cel mai bun de pe-aici, ca nimeni de-a saptea nu-ti face probelmele...am vazut ce-ai propus pentru O.N.M.2005 la a VII-a si sa stii ca nu era cine stie ce...Nu pot sa spun ca eu sunt foarte bun, dar in nici foarte prost.Sunt un elev mediu ca sa zic.Trebuie sa-ti imaginezi si tu ca totusi in Bucuresti posibilitatiile sunt mult mai mari decat in Slatina(unde locuiesc eu.s-ar putea totusi sa ma mut in Rm Valcea).Asa ca nu mai jigni copii din provincie..te rog frumos...si apoi iti arat si solutia mea...

Robertux comes back, this time with another hat...  ;)

What are you talking about?I don't understand.Who is Robertux?Please, explain me!Don't enter in this trouble!Only the administrator can do something about this! :spam:

acum intru si eu in discutie, desi poate ca nu ar fi cazul. sorin, asa cum spune toata lumea e bine ca dai dovada de entuziasm, asta lipseste multora azi. totusi ceea ce incearca Cezar sa spuna e ca postezi de multe ori doar ca sa-ti mai incrementezi nr de posturi.
oricum, Cezar e moderator, nu e chiar administrator, dar are dreptul sa protesteze atunci cand faci spam(si chiar sa-ti stearga posturile ;) ). totodata, daca tie ti se reproseaza asta nu ar trebui sa crezi ca toti ceilalti fac acelasi lucru.(vezi postul anterior acestuia).
parerea mea: daca tot esti plin de entuziasm: invata latex, si posteaza cat mai mult pe forum legat de probleme si de solutiile acestora.
in afara de asta, ma indoesc ca Cezar ar avea ceva cu oamenii din provincie. si eu sunt din provincie si asa sunt si multi alti omeni care frecventeaza site-ul si nu s-a simtit cineva lezat din cauza asta. data viitoare incearca sa ai si argumente solide atunci cand faci o afirmatie.
in concluzie: mathlinks e un forum care isi propune sa-i ajute pe toti sa invete. si e normal ca fiecare sa invete de la altii. asculta si tu de cei mai "batrani" si analizeaza situtia. poate chiar au dreptate...

Domane fereste, n-am nimik cu cei din provincie, mai ales ca si eu sunt tot din provincie si nimic in mod special cu Sorin. Apreciez efortul lui, insa ar fi mult mai instructiv daca s-ar limita strict doar la solutii si probleme. Nu cred ca este nevoie de comentarii inutile. Oricum deja nu mai are rost sa mai zic nimic. ;)

$1.\blacktriangleright$ Sorine, am impresia ca ai un usor complex al celui care traieste departe de capitala. Iti recomand sa treci in revista (si sa-i numeri) pe toti profesorii universitari, doctorii sau docentii, academicienii (in Mathe bineinteles, dar anticip ca acelasi lucru se intampla si la celelalte discipline) si vei constata ca nu trebuie sa sti sa numeri decat cel mult pana la o suta ! Mie imi place Tuca (moderatorul de la T.V.). Si nu m-a interesat niciodata sa stiu de unde este, desi acum stiu pentru ca a spus-o el. In general, a intreba in State pe cineva de unde este, pare o indiscretie si deranjeaza (este asimilata cu o violare a intimitatii), mai ales ca aici sunt din toate colturile lumii. Complexul provincialitatii este in lumea de azi un atavism. La urma urmei, poti alege orice punct din plan ca origine a unui sistem (de vectori, de numere complexe, reprezentant al unei clase de resturi etc).
$2.\blacktriangleright$ Dragul meu, meseria se fura , zice un om simplu. Asculta, priveste, miroase etc , antreneaza-ti toate simturile on line intr-o singura directie : fantana ta fara fund, cu sediul in creier, sa capteze cat mai multe informatii ! Insa pentru aceasta, trebuie "sa-ti pacalesti" maestrul prin amabilitate, sa-i pui intrebari (cu premeditare) chiar naive pentru a exprima ca esti un incepator, acorda-i respect etc , doar-doar iti va spune si tie ceva din tainele meseriei pe care vrei ... s-o furi. Esti cam mititel pentru a fi asa de orgolios. Pentru fiecare individ exista cel putin unul mai bun si cel putin unul mai slab decat el. Asa ca realizeaza un coctail intre modestie si orgoliu (fifty-fifty). Nu este bine nici sa fi prea modest, dar nici sa fii prea orgolios !

Te voi urmari cu atentie in continuare. Imi place in mod deosebit la tine "foamea" de informatii. Incearca sa te organizezi putin, fiseaza aceste informatii pe sectiuni etc, intr-un cuvant, putin management informational ! In felul acesta, accesul la informatia dobantita devine mai accesibil si rapid. Gata, stop. Am luat-o razna, devin plicticos si pe undeva pedant.

[u]Unde gasim lista aia scurta de la Nationala ?[/u] Asta este mult mai interesant decat aceasta dezbatere cu sens unic $\rightarrow\mathrm{\ Sorin\ .}$

P.S. Cred ca-i ajunge atata ... sapuneala. L-am lustruit destul. Ii amintesc in final ca in dictionarul limbii romane exista cuvinte care intre timp s-au cam uitat (sau au intrat in derizoriu). De exemplu : multumesc, scuze, iarta-ma, dupa parerea mea, esti amabil, solutia ta este mai buna decat a mea, te rog, ce parere ai despre ... desi eu cred ca... , sper sa nu fie o greseala in ceea ce spun, dar ...etc.


trigonometry
integration
induction
function
calculus
real analysis
real analysis solved

Show that for x>0 \int{0->x}sin(et)dt>0.

Let f(x)= \int {0->x}sin(e t )dt.It's easy to see that
f(x)=\int {0->e x }(sin(t)/t)dt.We will prove first that f>0 for every x in the interval (0..ln(2 \pi)).Then by "induction" is obvious that that f>0 for every x>0.
When x is in the interval (0.. ln(\pi) ) the function (sin(t)/t) >0 (t is in (0.. \pi ))so the integral f(x)>0.For x in [ ln(\pi) ..ln(2 \pi) ) we have:
x=lny and
f(x)=\int {0-> \pi  }(sin(t)/t)dt+\int { \pi -> y }(sin(t)/t)dt=
\int {0-> y- \pi  }(sin(t)/t)dt + \int {y- \pi -> \pi  }(sin(t)/t)dt+ \int {0-> y- \pi }(sin(t+ \pi )/(t+ \pi ))dt=\int {y- \pi -> \pi  }(sin(t)/t)dt {greater then 0}  + \int {0 -> y-\pi  }((sin(t)/t-sin(t)/(t+ \pi ))dt {also greater then 0} =>  f(x)>0 for every x in [ ln(\pi) ..ln(2 \pi) ).
For X >0 we take an interval of the form [ln((2k+1) \pi ...ln((2k+2) \pi  ))  which contains X.And we divide f(x) in intergrals of the form  \int {(2k+1) \pi->(2k+2)\pi} which are >0.(1).So f(x) is >0 for every x>0.
I still have to make the same proof for (1).But it's same with the begining part.....

You're almost right, but f(x)=\int{1->ex}(sin(t)/t)dt, so it starts from 1, not 0. The induction is Ok, but we have to show that f(ln(2\pi))>0. Try it, it's not that hard.

For the integral  \int {0->2 \pi } we use the same idea:
We divide the integral into 2 parts :one from 0 to  \pi  and the other one from  \pi  to 2 \pi .And we make a variable change into the second one(t->t+ \pi ) and the rest is the same ....with the begining of the solution.

Ok, I got that, but e0=1 and eln(2\pi)=2\pi, so we need to show that \int{1->2\pi}(sin(t)/t)dt>0, which is a bit harder than \int{0->2\pi}(sin(t)/t)dt>0.

See what I mean?


hey guys
thank you for viewing this thread, i really need some HELPPPPP plz!!!!!  :(  :(  :(

i got the acceptance email of summer camp lik 3 weeks ago, then i've been waiting for the OFFICIAL LETTER to come

the thing is, it's still not here!

is there such a thing for this math camp?
i mean, i'm a canadian resident holding a chinese passport,and i need to get a USA visa in order to go; to apply for a visa, i'll need to show the proper documentation and explain why i need to go to USA. the thing is, i don't think the email acceptance counts as a proper document and i've been waiting for 3 month for the hardcopy to come!

now i just noticed that the deadline for paying is june 10.....and its may 27 now!

i'll need a few days to get my visa, too

now what's troubling me is that, how do i get an official letter saying i can go? before paying or AFTER PAYING?
if i pay it before i get my visa, and my request for visa gets rejected, can i have a full refund?

anyone has this kind of experience before?

HELP NEEDED!!! T_T_T_T

MANY THANKS[/b]

I think you need to e-mail the coordinators of the camp about this.

I don't think you get an official acceptance letter, just an email. I was accepted a while ago, but I never got the official letter. Then again, I haven't sent the check yet either, so maybe you get it afterwards.


ratio
geometry
parallelogram
area of a triangle

What is the ratio of the height of the parallelogram to the height of the
triangle if each has the same area? Express your answer as a common
fraction.

[asy]draw((10,0)--(30,0)--(20,10)--(0,10)--cycle);
draw((0,10)--(-20,30)--(20,10));[/asy]

Umm mewto could you stop rushing to answer these questions?

As in, why does the same base length make one-half?

This would be because the area of a triangle is $ \frac{bh_t}{2}$ and the area of a parallelogram is $ \frac{bh_p}{2}$.  Setting these two equal would show that $ \frac{h_p}{h_t} \equal{} \frac{1}{2}$, which is our answer...

Common sense, combined with the bh/2 and bh formulas, shows that one height is half the other (since the bh cancels) so we see that the triangle's height is obviously the larger one. Better?

Obviously that’s exactly what I wrote, but you know people are asking these questions for a reason…you should actually try to explain how to do it.

Yes but ih people are like me, they would want a prod hint, not a full-fledged soltuion. WHich reminds me, should we post these in hide-tags?


inequalities
geometry
inequalities proposed

Prove that in any tringle $ABC$ the following inequality is valid:

               \[ \frac{1}{(b+c-a)^{2}}+\frac{1}{(c+a-b)^{2}}+\frac{1}{(a+b-c)^{2}}\geq\frac{1}{4r^{2}}.  \]

Am - Gm after multiplying by $E^4=(area\, of\, the\, triangle)^4$. It is pretty straightforward .  :)

It is a nice and easy inequality. [u]See[/u] the my reply from http://www.mathlinks.ro/Forum/viewtopic.php?t=87054

[b][u]Proof.[/u][/b] $\{x,y,z\}\subset (0,\infty )\ ,\ xy\cdot yz+yz\cdot xz+xz\cdot xy\le (xy)^2+(yz)^2+(xz)^2$ $\Longrightarrow$ $xyz(x+y+z)\le x^2y^2+y^2z^2+z^2x^2$ $\Longrightarrow$ $\frac{x+y+z}{xyz}\le \frac{1}{x^2}+\frac{1}{y^2}+\frac{1}{z^2}\ \ (*)\ .$ For $x: =p-a$ a.s.o. we obtain $x+y+z=p$, $S=\sqrt{xyz(x+y+z)}$ and $r=\sqrt{\frac{xyz}{x+y+z}}$. Thus, the inequality $(*)$ becomes the proposed problem.

$\boxed {\ \frac{1}{R^2}\ ,\ \sum \frac{1}{a\sqrt{bc}}\le \frac{1}{2Rr}\le \sum \frac{1}{bc}\le \sum \frac{1}{a^2}\le \frac{1}{4r^2}\le \sum \frac{1}{(b+c-a)^2}\ }\ .$

[b][u]Question :[/u][/b] Is truly $\sum \frac{1}{a\sqrt{bc}}\le \frac{1}{R^2}$ or $\frac{1}{R^2}\le \sum \frac{1}{a\sqrt{bc}}$ or the both are falsely ?

[b]Remark.[/b] $\sum \frac{1}{a\sqrt{bc}}=\frac{1}{abc}\sum \sqrt{bc}\le \frac{a+b+c}{abc}=\frac{1}{2Rr}\Longrightarrow \sum \frac{1}{a\sqrt{bc}}\le \frac{1}{2Rr}\ .$


calculus integration trigonometry logarithms

Evaluate
[tex]
\displaystyle \int_0^{\frac{\pi}{2}} \frac{x \ dx}{1+\sin x}
[/tex]

Everything on this forum is a challenge :)

I see.
* But integration is quite cheatable. *

[quote="ComplexZeta"]So don't cheat.[/quote]

Right, who would cheat on a challenge?  It would certainly be more fun to solve it.

I solved it... my lengthy tex solution will be typed out now...

Let [tex]\displaystyle{f(a)=\int_0^{\frac{\pi}{2}}\frac{x}{1+\sin{(ax)}}\; dx}[/tex]
[tex]\displaystyle{F(a)=\int_0^{\frac{\pi}{2}}\int \frac{x}{1+\sin{(ax)}}\; da\;dx}[/tex]
[tex]\displaystyle{F(a)=\int_0^{\frac{\pi}{2}}x\int\left(\sec^2{(ax)}-\tan{(ax)}\sec{(ax)}}\right)\; da\;dx}[/tex]
[tex]\displaystyle{F(a)=\int_0^{\frac{\pi}{2}}x\left(\frac{\tan(ax)}{x}-\frac{\sec{(ax)}}{x}\right)\;dx}[/tex]
[tex]\displaystyle{F(a)=\int_0^{\frac{\pi}{2}}\left(\tan(ax)-\sec{(ax)}\right)\;dx}[/tex]
[tex]\displaystyle{F(a)=\frac{-\ln|\cos(\frac{\pi}{2}a)|}{a}-\frac{\ln|\sec(\frac{\pi}{2}a)+\tan(\frac{\pi}{2}a)|}{a}}[/tex]
[tex]\displaystyle{F(a)=\frac{-\ln|\sin(\frac{\pi}{2}a)+1|}{a}}[/tex]

Now we just turn to the quotient rule for diff and we get our desired result.

[tex]\displaystyle{f(a)=\frac{-\frac{\pi}{2}\left(\frac{a\cos(\frac{\pi}{2}a)}{\sin(\frac{\pi}{2}a)+1}\right)+\ln\left(\sin(\frac{\pi}{2}a)+1\right)}{a^2}}[/tex]
[tex]\displaystyle{f(1)=\frac{0+\ln(2)}{1}}[/tex]
[tex]\displaystyle{f(1)=\ln2}[/tex]

that took forever!

what is really interesting about this problem is that if we were solving for f(2)... the result would take on a whole different form!

Great problem.


yeah, West Lafayette's on spring break right now, but we go back to school on Monday. i just wanted to know when ur guys' schools get on spring break...

:o we get on spring break the first week of april

really? i just realized today that our school didn't take the AIME I because we were on spring break... we're doing it on March 28. It's a half-day, and we're doing it from 8-11, so i basically get to go home after i finish it :P ! Actually, i might have to go to math class, since i have it 8th period, but who cares, i sleep through it anyways :D . lol.

p.s. does ur math teacher favor u over other ppl cuz u did well in mathcounts? just wondering :whistling: ...

We get spring break in the first week of April too, but we won't get to do much because I have a science competition the last Saturday of Spring Break.


group theory
abstract algebra
superior algebra
superior algebra unsolved

[quote="Albanian Eagle"]You are taking the union over $ \frac {|G|}{|H|}$ subgroups with $ |H|$ elements each. So our fact would be true only if all those subgroups where disjoint. This is not the case as they all contain $ e$ the identity[/quote]
Actually it's $ |G: N_G(H)|\le \frac {|G|}{|H|}$ subgroups
and they are disjoint if there is only one conjugate.
Don't forget that case.

If H is normal, then doesn't $ x H x^{ \minus{} 1} \equal{} H$? Does he mean for any proper subgroup, not just a normal one?

[quote="isomorphism"][quote="soltanpoor_aut"]if G is a group
|G|<∞ ∧ H<G⟹G≠ ∪_(x∈G) xHx^(-1)[/quote]

I think he means

"Let H be a normal subgroup of a finite group G, then prove that $ \displaystyle G \neq \cup_{x \in G} xHx^{ \minus{} 1}$"[/quote]

$ N_G(H) \equal{} G \Leftrightarrow H \triangleleft G$ and $ N_G(H) \equal{} \{ x \in G\ : \ xHx^{\minus{}1} \equal{} H\}$

If $ H \triangleleft G$ we have that $ \bigcup_{x \in G} xHx^{ \minus{} 1} \equal{} \bigcup_{x \in N_G(H)} xHx^{ \minus{} 1} \equal{} H \neq G$

But it's trivial...

soltanpoor means: "Let $ H$ be a subgroup of a finite group $ G$, then prove that $ G \neq \bigcup_{x \in G} xHx^{ \minus{} 1}$"

Anyway Albanian Eagle and Olorin doesn't consider $ H$ normal.


When cycloheptatriene is treated with chloroform in the presence of t-BuOK, and the resulting mixture is heated to 140ºC, two products A ($ C_8H_7Cl$) and B ($ C_8H_8Cl_2$) are obtained with 46% and 10% yields, respectively. Propose structures for A and B and explain their formation.

I got it:

First CCl2 is formed which undergoes addition with alkene followed by ring exapansion through elimination of HCl.

Both structures are still wrong. Try again.

Maybe is it like first there's a carbene addition to form a cyclopropane ring and after that a ring expansion to relieve the strain? :maybe:

Yes, first there is carbene addition to a double bond, that's natural. But what happens next is a mistery...

Carcul that's what I have drawn for B. Or is it selectivity that u want.

Does the next step involve the formation of a eight membered ring?  :maybe:

Yes, compound B is initially formed. But in the presence of base and with heating it gives two products, and you are to find these two.

But why does ring expansion not take place? :huh:

to relieve itself of the strain? i remember my sir doing a similar problem :)

The same point I was trying to make carcul sir. I think ring expansion should take place and so the product must be 1-chloro cyclooctatetraene.

correct me if im wrong, but the 3 double bonds of cycloheptatriene arent equivalent, so how do you know which double bond is attacked. and the anion of cycloheptatriene is antiaromatic so ring expansion cannot take place.

That's a good point. But perhaps the stereochemical structure of cycloheptatriene (which I have already posted on this forum) may help. Organic chemistry is a 3D science, never forget that.

P.S. [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=185688]Look here[/url] (last post)

maybe ring contraction takes place and u get a benzene derivative? :huh:


AMC
AIME
USA(J)MO
USAMO
ARML
AMC 12
AMC 10

this is something that has been bothering for a little while...im thinking of ordering a bunch of previous amc 10 and amc 12 tests and they give you the option to buy either/both the A and B tests....is there a difference between the two or are they both the same test?

Two AMC 10s and 12s are given every year: A and B. Both are intended to be roughly the same difficulty but are made to give contestants two times/chances to take them. It doesn't matter which of the two you order. They are the same difficulty; at least they were made to be that way.

are they the same test though...like same questions?

No; the questions are completely different.

Check out the [url=http://www.artofproblemsolving.com/Forum/resources.php]contests section[/url] of this website.  I hope the CAMC isn't strapped for funds :maybe: .

Wait...are you saying that you can take for an example AMC 10 A & B in the same year?

Yes, You can take up to one A test and one B test a year

But you cannot take, say both 10A and 12A oficially, because they share questions.

Would it be recommended then to take one AMC 10 and one AMC 12 or to take both versions of the same grade level? (for a rising freshman)

it's easier to qualify for AIME through the 12 but easier to qualify for USAMO through the 10, so it all depends on what your goals are. also, I am going to take both because the 10 will help me get into USAMO, but the 12 score is useful for many other contests (iTest, ARML selection, etc.)

For freshmen, I would recommend taking the 10A, and, if you feel ready, taking the 12B.

For sophomores, generally I would start with the 12A and take the 10B if you are satisfied with your 12A score. If not, you can take the 12B and try to do better.

And don't take the 10 thinking that it will help you qualify for the USAMO. First, that's more or less a tiebreaker, and second, lessons you learn from the 12 will carry over more to the AIME than those from the 10 will; i.e., AMC 12 problems are more similar to AIME problems. The AIME is by and far the most important component of USAMO qualification.

Actually, it'd probably be better to take the AMC10. There are already people who missed qualifying this year because they took the 12 instead of the 10. Whatever experience you get from solving AMC12 problems as opposed to AMC10 can be made up by doing the problems afterwards or the many years of AIME problems posted on this forum.

I have never taken an AMC 10 officially and never will, and I don't regret it.  The AMC 10 is intended for people not ready for the AMC 12, so if you're not one of those people, why take it?

To stoke your ego with a perfect score on the 10 :P. But if you're ready for the 12, don't take the 10 more often than the 12.


The equation $ |z\minus{}1\minus{}3i| \plus{} |z\minus{}9\minus{}3i|\equal{}10$ corresponds to an ellipse in the argand diagram.

i) Find the center (in x+iy form)

ii) state the lengths of major and minor arcs

-----------

let z = x+iy

$ \sqrt{(x\minus{}1)^2 \plus{} (y\minus{}3)^2}\plus{} \sqrt{(x\minus{}9)^2 \plus{} (y\minus{}3)^2}\equal{}10$

how can I proceed from here.

The algebraic solution (which I recently demonstrated on the board) is tedious at best.

[hide="A geometric hint"] Given two foci $ f_1, f_2 \in \mathbb{C}$, the ellipse with major axis $ 2b$ is given by

$ |z \minus{} f_1| \plus{} |z \minus{} f_2| \equal{} 2b$ [/hide]


Let $ V$ be a finite-dimensional vector space and $ A$ and $ B$ two linear transformations of $ V$ into itself such that :
$ A^2\equal{}B^2\equal{}0$ and $ AB\plus{}BA\equal{}I_n$
1)Prove that if $ N_A$ and $ N_B$ are the respective null spaces of $ A$ and $ B$ then $ N_A\equal{}AN_B$ and $ N_B\equal{}BN_A$ and
$ V\equal{}N_A \bigotimes N_B$
2) Prove that the dimension of $ V$ is even .

1) $ (A\plus{}B)^2\equal{}I$ then $ A\plus{}B\equal{}S$, a symmetry. $ A|_{N_B}\equal{}S|_{N_B}: N_B\rightarrow{N_A}$ is one to one, then $ dim(N_B)\leq{dim(N_A)}$. By the same way $ dim(N_A)\leq{dim(N_B)}$; thus $ dim(N_B)\equal{}{dim(N_A)}$, $ A(N_B)\equal{}N_A,B(N_A)\equal{}N_B$.
2) $ Ax\equal{}Bx\equal{}0$ implies $ Sx\equal{}0$ and $ x\equal{}0$. Moreover if $ x\in{V}$ then there exists $ y\in{V}$ s.t. $ x\equal{}Sy\equal{}Ay\plus{}By$. Finally $ N_A,N_B$ are supplementary subspaces and $ dim(V)$ is even.
3) If we choose a $ V$-basis as the union of a $ N_A$ basis and a $ N_B$ basis then $ A,B$ are in the form: $ A\equal{}\begin{pmatrix}0&U\\0&0\end{pmatrix}$ and $ B\equal{}\begin{pmatrix}0&0\\U^{\minus{}1}&0\end{pmatrix}$.


[quote="shadysaysurspammed"]$1$)Let r1,r2....,rn be n real numbers each > 0.Prove that for  any real x>0,
$(x+r1)(x+r2)....(x+rn)\geq (x+\sqrt[n]{r1r2...rn})^n$

$2)$Find smallest n such that
$(x^2+y^2+z^2)^2\leq n(x^4+y^4+z^4)$
for all real numbers x,y,z.[/quote]
1.Easy.Use Holder
2.$n=3$ because $(x^2+y^2+z^2)^2\leq 3(x^4+y^4+z^4)$ and if $n<3$ let $x=y=z=1$

What's holder?

and for the second one,how did you directly say n=3?Isn't there any proper way to proceed?

[quote="shobber"]Cauchy for the second one.[/quote]

Yup,
$(x^2+y^2+z^2)^2\leq (1^2+1^2+1^2)(x^4+y^4+z^4)$
so n=3


A barn with a flat roof is rectangular in shape, $ 10$ yd. wide, $ 13$ yd. long and 5 yd. high. It is to be painted inside and outside, and on the ceiling, but not on the roof or floor. The total number of sq. yd. to be painted is:

$ \textbf{(A)}\ 360 \qquad\textbf{(B)}\ 460 \qquad\textbf{(C)}\ 490 \qquad\textbf{(D)}\ 590 \qquad\textbf{(E)}\ 720$

[hide]
The walls, which are $ 10 \times 5$, $ 10 \times 5$, $ 13 \times 5$, and $ 13 \times 5$, need to be painted twice each (once inside and once outside), so there is $ (10 \times 5 \plus{} 13 \times 5)\times4\equal{}460$ square yd.
Adding the ceiling, which is $ 13 \times 10\equal{}130$ square yards, we have $ 590\equal{}\boxed{D}$.
[/hide]


algebra
polynomial
email
inequalities
IMO
IMO 2007

Also in the already well known tradition, we open the thread about our users who are going to be present at IMO 2007 (most of you of course contestants and many obviously medalists :D). Some countries have already selected their teams, so let's hear it guys!

I will be there again :blush:  ,but I promise that this is the last time I will participate in IMO as a student :D .

I will back in VIetnam and join to the IMO as an home old contest :d. If anyone needs a guide, I am always available (in French prefered and in English). I am sure that you will pass the most fantastic IMO in Vietnam.
For more informations about Vietnam and Hanoi city, send me a mail, at my adress truongnguyen1986(at)gmail.com.
Welcome to Vietnam!!!

India has its TST's in May - there are 5 tests on the 15th , 18th 21st , 24th and 28th of May. I'm writing the TST's so I'll wait and see - I might make it though :)
  Ashwath

[quote="Tiks"][quote="RDeepMath91"]
@Tiks : Do you know a little boy from Armenia who also got silver medal in Zhautykov Olympiad 2007 ? I forgot his name but will he participate in IMO 2007 ? Thx..[/quote]

Hello RDeepMath91,

I think you are talking about the little boy from this picture
[url]http://www.mathlinks.ro/Forum/album.php?mode=pic&id=5744[/url]

aren't you? :D

Yes,he is coming to Vietnam too. He was 4th with his results  in our TST :wink: .[/quote]

Yup, that's right.. ^^

[quote="Peter Scholze"]I will come, too (well, ok, there is a theoretical chance i dont, but im quite significantly the 1st in the TST's till now). Heading for a perfect score with special prize ^^ ;).[/quote]We don't expect anything less from you Peter! :D

I am going as member of Polish team, which is a big surprise for me.

[quote="hungkhtn"][quote="Peter Scholze"]I will come, too (well, ok, there is a theoretical chance i dont, but im quite significantly the 1st in the TST's till now). Heading for a perfect score with special prize ^^ ;).[/quote]

Just for my curiousness. Are you in Germany as Darij? I think you must be a teacher , not a member of IMO team (so young!)  :lol:[/quote]
What do you mean with this? peter comes from germany. right. darij "only" lives here ;) and he has finished school last year. but peter is actually in year 13 so he cannot be a teacher or something similar but only a little member of a IMO team ;)

Greetings from east-Germany, Naphthalin

I am also going to imo 07 :-d.I have been selected for Bangladesh imo team.I am the most junior member of the team.Bangladesh is a very new member of imo.So the result wasnt outstanding last year.Our two members got honorable mention and one got 11.
But this year i want to win a medal,the 1st medal for bangladesh.

Moon

how young are you

BTW the nz teams are anounced, im not going

It's funny how the credits of the students from the UK team consist in "Balkan" titles :P

Here is the list of the team members for this IMO:

[b]Whan Ghang[/b] (Hansung Science High School 2) IMO '05(silver), '06(gold)
[b]Yoon Joo Kim[/b] (Seoul Science High School 3)
[b]Jeong Soo Sim[/b] (Seoul Science High School 2)
[b]Jeha Yang[/b] (Seoul Science High School 2) IMO '06(silver)
[b]Seok Hyeong Lee[/b] (Seoul Science High School 2) IMO '06(gold)
[b]Soo Hong Lee[/b] (Choong Ang High School 1).

I'm one of the observers of Korean team.  See you in Hanoi!

[quote]I noticed that the Indian team hasn't been put up in this forum. So,...
Team:
1. P.N. Bhargav
2. Manas Joglekar
3. Atul Shekhar
4. Prashant Sohani
5. Utkarsh Tripathi
6. Abhishek Dang
Team Leader: Dr. C.R Pranesachar
Deputy Leader: Prof.Udayan Prajapati
Observer B: Dr. Sharad Sane[/quote]

Have you noticed that USA team also has a guy with the family name Tripathi? Any relative relations?

'tripathi' is an indian surname..

and arnav tripathi's parents are indians (from near my place).
but he is in the USA from birth...
got gold medal in IMO 2006

going also to IPhO 2007(am i right?) and IMO 2007.

[quote="SOUMYASHANT NAYAK"]
going also to IPhO 2007(am i right?) [/quote]

No he isn't . :noo: [url=http://www.mathlinks.ro/viewtopic.php?t=142961&start=40]Here's[/url] the list

[b]It is really wonderful to go to IMO five times[/b]

it should be wonderful being 5-6 years older than other teammates  :rotfl:

[b]Whan Ghang (Hansung Science High School 2) IMO '05(silver), '06(gold) [/b]
he is a great guy and plays soccer very well and I know how to spell his name  :lol:

I will be there for the second time , and lets see,

gl all mathlinkers ;)

[quote="Moonmathpi496"][quote="Little Gauss"]
But I'm 13 years old![/quote]

So... whats wrong? :P

[color=indigo]See u in Vietnam my friend :) [/color][/quote]

he's saying that though he's in 1st year of high school he's only 13 years old...

[quote="ihavethreeeyes"][quote="Little Gauss"][quote]Whan Ghang (Hansung Science High School 2) IMO '05(silver), '06(gold)
Yoon Joo Kim (Seoul Science High School 3)
Jeong Soo Sim (Seoul Science High School 2)
Jeha Yang (Seoul Science High School 2) IMO '06(silver)
Seok Hyeong Lee (Seoul Science High School 2) IMO '06(gold)
Soo Hong Lee (Choong Ang High School 1).

[/quote]
But I'm 13 years old![/quote]

He is the youngest boy in our team. In fact, I'm 18 years old!! :lol:[/quote]

Ya i know it and i also know him,see u also. :)
[i am also the youngest member of my team but not so young may be]

[quote="Armenia"][quote="RDeepMath91"]Well, as I said in another topic, I'll be there.. After last year my friend (username : BoesFX) and I couldn't go since the visa can't be finished on time..  :( Anyway, he has graduated to Singapore and superbly breaks through the top 10 in IMC's selection of Singaporean Team...  :)

@Tiks : Do you know a little boy from Armenia who also got silver medal in Zhautykov Olympiad 2007 ? I forgot his name but will he participate in IMO 2007 ? Thx..[/quote]
I am the little boy from Armenia!!!!!!!!!!!!!!! :roll: My name is Tigran Hakobyan!!!!!!!!!
And what is your name???????? :huh:[/quote]

My name is Raymond Christopher (Indonesia), and I've seen and met you in Kazakstan (2006,2007) and in Hanoi, too. But maybe you didn't see me here, did you ?  :D

[quote="RDeepMath91"]Well, as I said in another topic, I'll be there.. After last year my friend (username : BoesFX) and I couldn't go since the visa can't be finished on time..  :( Anyway, he has graduated to Singapore and superbly breaks through the top 10 in IMC's selection of Singaporean Team...  :)

@Tiks : Do you know a little boy from Armenia who also got silver medal in Zhautykov Olympiad 2007 ? I forgot his name but will he participate in IMO 2007 ? Thx..[/quote]

Don't exaggerate too much, I am top ten in IMC selection of NTU Team, but only ten that are interested in joining the selection (And I really sucks. :( )


The midpoint of triangle's side and the base of the altitude to this side are symmetric w.r.t. the touching point of this side with the incircle. Prove that this side equals one third of triangle's perimeter.

[i]Author: A.Blinkov, Y.Blinkov[/i]

Let  $ H,M,X$ be the foot of the A-altitude, midpoint of $ BC$ and  tangency point of the incircle $ (I)$ with $ BC.$ If $ H,M$ are symmetric WRT $ X,$ then the tangency point of $ (I)$ with the 9-point circle is the antipode of $ X$ on $ (I).$ Thus, Feuerbach point $ F_e$  satisfies the barycentric equation of the A-Nagel ray $ AY \equiv (p \minus{} c)y \minus{} (p \minus{} b)z \equal{} 0$

$ F_e \ ((p \minus{} a)(b \minus{} c)^2: (p \minus{} b)(a \minus{} c)^2: (p \minus{} c)(a \minus{} b)^2)$

$\Longrightarrow (p \minus{} c)(a \minus{} c)^2(p \minus{} b) \minus{} (p \minus{} b)(a \minus{} b)^2(p \minus{} c) \equal{} 0$

$ \Longrightarrow 2a \equal{} \frac {b^2 \minus{} c^2}{b \minus{} c} \Longrightarrow \  3a \equal{} a \plus{} b \plus{} c$


Annie estimated that her household recycles five cubic feet of materials per collection day.  There are 26 collection days per year in her town of 16,000 households.  If every household is similar to Annie's, how much recycled material is collected in one year in Annie's town?

$ 5\cdot26\cdot16,000\equal{}2,080,000$
That doesn't seem right to me.


how can I type a formula using latex in this forum? thanks

Hi,
Use tags. Eg: [code][tex]\psi[/tex][/code] gives $\psi$.
SSN.


A fair die is rolled four times. Find the probability that each number is no smaller than the preceding number.

I get something different from Kalva's solution, which confuses me. Can someone explain to me how to find the answer? :  [color=white]Given the pattern of values (one 6, two 5s, one 3, for example), there is unique admissible sequence of throws giving those values (eg 3, 5, 5, 6). The no. of patterns is the same as the number of ways of placing 4 identical balls in 6 urns. That is the same as the number of ways of placing 4 balls and 5 dividers in a row, or 9C4 = 126. Hence prob = 126/6^4 = 7/72.[/color]

Is that the Kalva solution you posted?  Which part don't you understand?

This is how I understood and did it.

[hide]If you use nCr format then there is no order involved so you can automatically place it in greatest to least order. When he says, 4 balls in 6 urns he's using pigeonhole principle where a ball is a roll and the urns are the vvalues that you can receive. So,  2 balls in the 6 urn and one ball in each the fourth and third urn would be the same as rolling two sixes, one four, and one three.  So this would be the same thinking as 1+1++1+1+ where the 1's are the repective balls and the +'s divide up theurn categories. Thus thhe amt. of ways to arrange this is 9C4 and the denominator is 6^4. So 9C4/6^4 should simplfy to 7/72[/hide]

You can also do it this way...[hide]The problem is just the same as 1 :le: a_1  :le:  a_2  :le: a_3 :le: a_4 :le: 6, find number of possible (a_1,a_2,a_3,a_4). We can imagine a possible sequence of 6 A's and 4 B's and the sequence must start with A. The 4 B's divides the sequence that the value of a_n correspond to the number of A's before the nth B (for example, AABABAABBA, a_1=2, a_2=3, a_3=5, and a_4=5. ABBAAABAAB, a_1=1, a_2=1, a_3=4, a_4=6) Since the sequence must start with A, so the answer is (10-1)C4=9C4. 9C4/6^4=7/72. so 7+72=79.  [/hide]

How come you are not allowed to calculate all the possible sequences of rolls which satisfy the conditions and then divide by 6^4? Like:
1,1,1,2
1,1,2,2
1,1,2,3
etc...
When I do this I get 11+9+7+5+3+1+(9+7+5+3+1)+(7+5+3+1)+(5+3+1)+(3+1)+(1)=91 possible sequences.

[quote]How come you are not allowed to calculate all the possible sequences of rolls which satisfy the conditions and then divide by 6^4?[/quote]

You definitely can. I started counting and found a pattern fairly quickly; the total number of sequences is (21+15+10+6+3+1)+(15+10+6+3+1)+(10+6+3+1)+(6+3+1)+(3+1)+(1)=126, so 126/6^4 = 7/72. You probably just missed a few possible sequences.

haha, I see my mistake now. Thanks a lot for you help!


induction
number theory unsolved
number theory

Is it true that $ \forall m\in\mathbb{N}$ ,

$ 3^{m\plus{}1}\parallel{}(4^{3^m}\minus{}1)$ ?

( the symbol || means $ 3^{m\plus{}1}|(4^{3^m}\minus{}1)$ but $ 3^{m\plus{}2}\not|(4^{3^m}\minus{}1)$ )

Yes it is true.Prove by induction.
$ m\equal{}1$ then it is true.
Suppose it is true for $ n$ we prove for $ n\plus{}1$
$ x\equal{}4^{3^m}$ then $ 3^{m\plus{}1}\parallel{}x\minus{}1$ (From induction )
Now we have $ x^3\minus{}1\equal{}(x\minus{}1)(x^2\plus{}x\plus{}1)$
because $ 3\parallel{}x^2\plus{}x\plus{}1$ so $ 3^{m\plus{}2}\parallel{}x^3\minus{}1$
This problem claim.


Most people have heard the story of Hardy calling 1729 a dull number.  Ramanujan said "No, it is a very interesting number; it is the smallest number
expressible as a sum of two cubes in two different ways."

I was wondering what is smallest number that can be expressed as a sum of two numbers raised to the 4th in two different ways?

Any Ideas?

Known since Euler's days (By Euler :)) . Of course the smallest  such number is:

$635318657 = 133^4 + 134^4 = 158^4 + 59^4$

BTW - such numbers are called "Taxi-cab Numbers" (legend has it the number 1729 was that of a Taxi which Hady took when he visited Ramanujan in the Hospital)

Ok, so i might as well be the one to say this, what about the smallest number that can be expsessed as two different n'th powers?

I heard the taxi-cab story, but my friend says 1729 was the number from a hospital room.

Is there a method to find such pairs? I remember an ML question from a few years ago that gave quite a few numbers that can be expressed as the sum of two different cubes in two different ways.

You  may like to look at: [url=http://mathworld.wolfram.com/TaxicabNumber.html]Mathworld's  Taxicab numbers.[/url]

Also, there is a cube that is the sum of a square and a power of 5.

100 + 243 = 343

After a quick search (that may not be that acurate), I find that the smallest solution for the second power is $50=5^2+5^2=7^2+1^2$, the firstone on which the two numbers on each pair are different is $65=8^2+1^2=7^2+4^2$, and the first on which non of the numbers in the pairs are different from 1 is $85=7^2+6^2=9^2+2^2$. Can someone find smaller numbers of some of those cathegories.

Best,

[quote]Can someone find smaller numbers of some of those cathegories.
[/quote]
The number of ways you can  represent a number  as a sum of two squares is to count the difference between its  (4k+1) type factors and (4k-1) type factors. (Actually mutliply this number by 4  to get all (-ive numbers are also included)

( Factors of 2 don't count - if  n is such number so is 2n)

Thus  all you  have to do is to multiply (4k+1) type primes... (5,13,17,29 ...etc)
and you will get $25 = 3^2+4^2 = 5^2+0^2$  other numbers would be 5*13, 5*17  etc .. and 5*13*17 would have  8 different ways ..

Hope this is fun ...


I assume the web page of Gabriel Carrol called "Gastropods realm"
has been taken down long ago.

But does anyone have the material that was on the page?
Any cached version of it?

Please reply.

Been about a year, but here you go: http://www.mit.edu/~jhscott/iql/


Let $ C_1, C_2, C_3$, and $ C_4$ be four circles of general position through a point $ P$. Let $ P_{ij}$ be the second intersection of the circles $ C_i$ and $ C_j$. Let $ C_{ijk}$ be the circle $ P_{ij}P_{ik}P_{jk}$. Then the four circles $ C_{234}, C_{134}, C_{124}$, and $ C_{123}$ all pass through the point $ P_{1234}$.

Well known: Invert the figure in a circle (P) centered at P, with arbitrary radius,  to get four lines $ c_1,\ c_2,\ c_3,\ c_4$ in general position, the inversion images of the four circles $ \mathcal C_1,\ \mathcal C_2,\ \mathcal C_3,\ \mathcal C_4$ passing through the inversion center P and intersecting again at $ P_{ij} \equiv \mathcal C_i \cap \mathcal C_j.$ The four lines $ c_1,\ c_2,\ c_3,\ c_4$ intersect at $ Q_{ij} = c_i \cap c_j,$ the inversion images of $ P_{ij}.$ Circumcircles $ \mathcal C_{ijk}$ of the $ \triangle P_{ij}P_{jk}P_{ki}$ go into circumcircles $ \mathcal D_{ijk}$ of the $ \triangle Q_{ij}Q_{jk}Q_{ki}.$ The circumcirles $ \mathcal D_{123},\ \mathcal D_{124}$ meet at $ Q_{12}$ and again at $ Q_{1234},$ the inversion image of $ P_{1234}.$ Since $ Q_{1234} \in \mathcal D_{123},$ the feet $ F_1,\ F_2,\ F_3$ of normals from the point $ Q_{1234}$ to the lines $ c_1,\ c_2,\ c_3$ are collinear, on a Simson line $ F_1F_2F_3$ of the $ \triangle Q_{12}{Q_{23}Q_{31}.}$ Since $ Q_{1234} \in \mathcal D_{124},$ the feet $ F_1,\ F_2,\ F_4$ of normals from the point $ Q_{1234}$ to the lines $ c_1,\ c_2,\ c_4$ are also collinear, on a Simson line $ F_1F_2F_4$ of the $ \triangle Q_{12}{Q_{24}Q_{41}.}$ Thus $ F_1 F_2F_3 F_4$ is the common Simson line of the first two $ \triangle Q_{12}{Q_{23}Q_{31},\ \triangle Q_{12}{Q_{24}Q_{41}.}}$ This means that $ F_1 F_2F_3 F_4$ is also the common Simson line of the remaining two $ \triangle Q_{23}{Q_{34}Q_{42},\ \triangle Q_{13}{Q_{34}Q_{41}}}$ and its pole $ Q_{1234}$ lies on their respective circumcircles $ \mathcal D_{234},\ \mathcal D_{134}.$


algebra
polynomial
number theory unsolved
number theory

Find all polynomials $f(x)$ such that:

$f(k)+3^{k}=n\cdot 2^{k}$ for $n\in\mathbb{Z}$

For a fixed $n$?  Or do you just mean $2^{k}| (f(k)+3^{k})$?


inequalities
induction
function
logarithms
inequalities proposed

Let $x_{1},x_{2},...,x_{n}$ be positive real numbers with product $1$, prove that
\[\sqrt{(n-1)(x_{1}^{2}+x_{2}^{2}+...+x_{n}^{2})}+n-\sqrt{n(n-1)}\ge x_{1}+x_{2}+...+x_{n}.\]

Moreover, prove that $n-\sqrt{n(n-1)}$ is the best estimation possible,

Dear VASC, Arqady and Mathlinkers,

Could you prove this one? I think is is so close to the famous Sleifer's inequality in Crux and another inequality of VASC here.

Thank you for your gift. I'll try it.
Tell me, hungkhtn, Shleifer's inequality  is already proved? If yes, give please the link. Thank you.

By $n-1$ EV Theorem, it suffices to consider the case $x_{2}=...=x_{n}\ge 1$.
Next we have to do some ugly calculations. For $x_{1}\rightarrow 0$ it follows that the constant is the best.

[quote="silouan"]I know two inequalities that are close to this (I didn't check if they are strongers, could someone check? )
1)(Vasc) If $a_{1}a_{2}...a_{n}=1$ then
$(n-1)(a_{1}^{2}+a_{2}^{2}+...+a_{n}^{2})+n(n+3)\geq (2n+2)(a_{1}+a_{2}+...+a_{n})$

(this can be solved by RCLF)

2)(harazi)
if $a_{1}a_{2}...a_{n}=1$ then
$a_{1}^{2}+a_{2}^{2}+...+a_{n}^{2}-n\geq\frac{2n}{n-1}\sqrt[n]{n-1}(a_{1}+...+a_{n}-n)$[/quote]

The famous shleifer is
\[(n-1)\sum a_{i}^{2}+n\sqrt[n]{\prod a_{i}^{2}}\ge (\sum a_{i})^{2}. \]
In other words, if $a_{1}a_{2}...a_{n}=1$ then
\[(n-1)\sum a_{i}^{2}+n \ge (\sum a_{i})^{2}. \]
I solve my problem just be elementary mathematics method. I use induction.

[quote="silouan"]This inequality Hunghtn is a consequence of Shleifer's inequality  :wink:
Indeed , if $a_{1}+...+a_{n}=s$ we have to prove that
$\sqrt{s^{2}-n}\geq s-n+\sqrt{n(n-1)}$  or equivalently

$2ns-2s\sqrt{n(n-1)}+2n\sqrt{n(n-1)}-2n^{2}\geq 0$ or equivalently

$(s-n)(n-\sqrt{n(n-1)})\geq 0$ which is true .[/quote]

Thank you very much, Silouan. I never think before that this problem is just a consequence. Now return to the Shleifer's inequality. I have two elementary solutions to it, one is induction, one is Karamata inequality. The rest is using SMV, UMV theorem or EM theorem but it is not so elementary.

Does any one have a nice solution to this one?

[quote="silouan"]hungkhtn could you post your solution with the Karamata's inequality to the Shleifer's inequality ? Thanks.[/quote]

Ok, silouan. I will post it next week. It use the Symmetric Majorization criterion of Darij (It is one of the most beautiful and hard solution of the Vietnamese book). But I still wait an other solution to Shleifer's inequality. Induction is really good, isn't it?

[quote="silouan"]Also the Zaimi -Marku inequality gives the following (for k=2-n
$(n-1)(a_{1}^{2}+...+a_{n}^{2})+a_{1}...a_{n}(a_{1}^{2-n}+...+a_{n}^{2-n})\geq (a_{1}+a_{2}+...+a_{n})^{2}$[/quote]

But it seems to be not as strong as the initial one?

You are right but I posted for shake of completenes of the inequalities of the
form $\sum a_{i}^{2}-(\sum a_{i})^{2}\geq smt$   :P

(smt=something  :D )

Ok. Silouan. I say I will post the solution next week but your diligent attitude stimulate me. So, the following solution is dedicated for you

Note. The theory of karamata in my book is constructed entirely according to the symmetric majorization criterion (to prove some arrays majorize the other, I only use the symmetric criterion). Darij post it here, maybe

http://www.mathlinks.ro/Forum/viewtopic.php?t=14975

File for solution

[quote="silouan"]Thank you very much Hunghtn nice solution.
Vasc in his book has also a nice solution .Here it is :
By Generalised Popoviviu's inequality we have
$f(a_{1})+f(a_{2})+...+f(a_{n})+\frac{n}{n-2}f(a)\geq\sum_{1\leq i\leq j\leq n}f(\frac{a_{i}+a_{j}}{2})$

where $a=\frac{a_{1}+...+a_{n}}{n}$

Now use this for the convex function $f(x)=e^{x}$ and replacing
$a_{1},a_{2},...,a_{n}$ by $2\ln x_{1},...,2\ln x_{n}$

 :wink:[/quote]

Yes, I see. In my book, I inclined to use the symmetric majorization criterion in applying Karamata inequality for $n$ numbers (this is different from VASC's book), since I think the problem of all absolute signs (as I show) in really lovely. As you see, my proof attached here is directly referring the Generalized Popoviciu's inequality in VASc's book.


What is the value of (1-w^2+w)(1+w^2-w)
where 1, w, w^2 are the cube roots of unity?

Since $\omega$ is the cube root of unity, we have $\omega^2+\omega+1=0$ or $\omega^2+1=-\omega$ and $\omega+1=-\omega^2.$ Substituting that, we get

\begin{eqnarray*}(1+\omega^2-\omega)(1+\omega-\omega^2) &=& (-\omega-\omega)(-\omega^2-\omega^2)\\ &=& -2\omega\cdot (-2\omega^2)\\ &=& 4\omega^3\\ &=& 4\end{eqnarray*}

I'm trying to figure out how to attach a latex file. This is my first time. I have the same answer, 4

You don't need to attach a $\text{\LaTeX}$ file. Just enclose your $\text{\TeX}$ in $\$$ signs. Example: $\$5\text {x}^2\$$ produces $5x^2$.

[hide]Expand to get $1-w^4+w(w^2-1+w^2+1)-w^2$

This equals $1-w^4-w^2+2w^3=3-w^2-w^4 =3-w(1+w)=3-w-w^2$

Now notice that $-(w+w^2)=-(-1)=1$ either by computing the two roots and using algebra or by utilyzing the symmetry in the unit circle in the complex plane.

Thus $3-w-w^2=4$[/hide]

thanks for the help with latex


Twenty-five percent of what number is 12?

Set up an equation: $ 25\%x\equal{}\frac x4\equal{}12\implies x\equal{}\boxed{48}$.

Or, you know that $ 25 \%$ is just $ \frac{1}{4}$ of the number, so we multiply $ 12$ by $ 4$ to get $ \boxed{48}$.


How to estimate square roots:

1) Pick a number to the nearest tenth that would be a reasonable estimate for the square root.

2) Divide the number in the square root by the number you estimated and round to the nearest hundreth.

3) Average the quotient and the origional number you picked, and this should be the square root to the nearest hundreth :lol:

i think that most contests would allow a calculator for square roots to the hundredths place but it's good knowing that

You'd be amazed.

I've seen some countdown problems and state level problems that require you to know your sqaure roots.

you can actually follow that estimation process and countinue the square root infinately.

Id think its easier just to know three basic ones that come up a lot.

 :sqrt: 2=1.414
 :sqrt: 3=1.732
 :sqrt: 5=2.236

Unless they ask you for :sqrt: 51 to the nearest hundreths.

 :D


Let  $ q\equal{}\frac{1\plus{}\sqrt{5}}{2}$, $ f: N \to N, \forall n \in N^*$ and
$ |f(n)\minus{}qn|<\frac{1}{q}$
Prove that:$ f(f(n))\equal{}f(n)\plus{}n$
 :lol:  :D

I guess that the original problem was to determine whether there exist a function $ f: \mathbb{N}\rightarrow\mathbb{N}$,such that
1.$ f(1)\equal{}2$
2.$ f(f(n))\equal{}f(n)\plus{}n$.
3.$ f(n\plus{}1)>f(n)$.
And it was at IMO 1993,the fifth problem.
Am i right?
And the one solution that i know was based on the following idea:
Let $ f(n)$ be the natural number such that $ |f(n)\minus{}\alpha n|<\frac{1}{2}$,where $ \alpha\equal{}\frac{1\plus{}\sqrt{5}}{2}$.Then
$ f(f(n))\equal{}f(n)\plus{}n$.
Proof:
Let $ S_n\equal{}f(n)\minus{}\alpha n$,from the definition of $ f$,we conclude that $ S_n<\frac{1}{2}$.
At first we will prove that $ f$ is monotonically increased function,indeed:
$ f(n\plus{}1)\minus{}f(n)\equal{}\alpha\plus{}S_{n\plus{}1}\minus{}S_n\geq a\minus{}(|S_{n\plus{}1}|\minus{}|S_n|)>a\minus{}1>0$.
Now consider the diffrence:
$ A\equal{}f(f(n))\minus{}f(n)\minus{}n\equal{}\alpha(\alpha n\plus{}S_n)\plus{}S_{f(n)}\minus{}\alpha n\minus{}S_n\minus{}n$.It is well-known that
$ \boxed{\alpha^{2}\minus{}\alpha\minus{}1\equal{}0}$,hence

$ |A|\equal{}|(\alpha\minus{}1)S_n\plus{}S_{f(n)}|<\frac{\alpha\minus{}1}{2}\plus{}\frac{1}{2}<1$.
But $ A\in\mathbb{Z}$,from where we conclude $ \boxed{A\equal{}0}$,and we are done.

[quote="Erken"]
And it was at IMO 1993,the fifth problem.
[/quote]
visit the following link for more solutions:

[url]http://www.mathlinks.ro/viewtopic.php?p=372306#372306[/url]