LaTeX
inequalities
quadratics
graph theory
combinatorics solved
combinatorics

Square board $n\times n$ is filled with '0' and '1'. It is known that for each '0' there are at least $n$ '1' in the same row and column. Prove that there are at least $n^2/2$ '1' on the board.

consider a bipartite graph G described as follows:
let the vertex classes be the rows and columns of the array;then row i and column j are adjacent iff the (i,j) entry of the array is 1.the given condition means,that if vertices i and j' are non adjacent,then d(i)+d(j')>=n.(i'll use primed indices for the vertices denoting the columns). what we want is that such a graph has e(G)>=n^2/2.
  now consider the sum of (d(i)+d(j')) over all pairs (i,j') such that vertices i and j' are nonadjacent. by the above this sum is >= n(n^2-e(G)). on the other hand, d(i) is counted (n-d(i)) times and similarly for d(j'). so this sum is also
       = 2ne(G) - sum(d(i)^2) - sum(d(j')^2)
 (sorry-i still haven't figured to use LaTex!) :blush:
   now just use the inequality
                  sum(a(i)^2)/n>= (sum(a(i))/n)^2. and write the inequality out. rewrite the quadratic in e(G) and solve the inequality. that finally gives what we want.


From your camp ground you walk 1 mile east. Sqrt(2) miles northeast and 2 miles south. What is the distance from you and your campground?

[hide="hint"]
Use the Pythagorean Theorem.
[/hide]

[hide="solution"]let ur starting point be A. u move one mile east. then moving one mile northeast. u can draw the legs of a 45-45-90 triangle. thus two miles south is really like 1 mile south of the vertical line where u started. let ur resulting point be C. C forms a right triangle with A with legs length 2 and 1. thus ur distance from A is $ \sqrt{5}$[/hide]

i probably made some stupid mistake.

[hide="Vectors?"]
$ \vec{r}\equal{}\left( 1\plus{}\sqrt{2}\cos\frac{\pi}{4}\right)\hat{i}\plus{}\left(\sqrt{2}\sin\frac{\pi}{4}\minus{}2\right)\hat{j}\equal{} 2\hat{i}\minus{}\hat{j}$.
$ \|\vec{r}\| \equal{}\sqrt{5}$.
$ \theta \equal{}\arctan\frac{\minus{}1}{2}\approx\minus{}22.5^{\circ}$.
[/hide]

Same Answer.

can u do a more indepth explanation of ur vector method astro?

thanks!

[hide="Explaination"]
Used unit vectors to note the changes in the $ x$ and $ y$.  With unit vectors $ \hat{i}$ is the $ x$ part and $ \hat{j}$ is the $ y$.
So it is just trying to find the $ x$ and $ y$ components for each movement and adding them, north and east are positive while south and west are negative, etc.
For the $ x$ part first movement $ \plus{}1$, second movement is just the $ \cos$ of the angle which is $ \frac{\pi}{4}$ for northeast times the length of the movement, and third movement nothing on the $ x$.
For $ y$, first movement nothing, second movement just the $ \sin$ of the same angle and same length, and third is $ \minus{}1$ for the south direction.
The magnitude of $ \vec{r}$ is just the Pythagorean Theorem.
The direction is just the definition of tangent.
[/hide]


sa se arate ca exista $A\subset \mathbb{R}^{2}$ finita, astfel incat pentru orice $X\in A$ exista $Y_{1},Y_{2},\ldots,Y_{2007}\in A$ astfel incat distanta dintre $X$ si $Y_{k}$ este egala cu $1$ pentru orice $k$, si aceste 2007 puncte sunt singurele cu aceasta proprietate.


function
real analysis
real analysis unsolved

hi

plz correct me if i'm posting this topic on the wrong place

my question concerns dirichlet series
a dirichlet series is a real( or complex) function of form
F(s)=  sum( a(n) /n^s,n=1..infinity)

in the book hardy and wright they proof that if two dirichlet series converge for s>s0 and are equal for all such values of s, all the coefficients have to be equal

they mysteriously do it without ever mentionning absolute converge  :? , and their proof doesn't really seem that convergent

so how about this uniqueness theorem, can it be proven without absolute convergence?

thx,

frederic

I have the same problem, well kinda,.... in Apostoll's book of number theory, it skip the details of such proof....


Consider this pattern where
the positive, odd integers
are arranged in a triangular
formation. The 1st through
4th rows are shown; each row
has one more entry than the
previous row. What is the sum
of the integers in the 15th row?

[asy]label("1",(0,0),S);
label("3",(-5,-5),S);
label("5",(5,-5),S);
label("7",(-10,-10),S);
label("9",(0,-10),S);
label("11",(10,-10),S);
label("13",(-15,-15),S);
label("15",(-5,-15),S);
label("17",(5,-15),S);
label("19",(15,-15),S);
dot((0,-22));
dot((0,-20));
dot((0,-24));[/asy]

Find the sum for the first row, then the second row, then the third. They are 1, 8, and 27, respectively. From here, assume that the $ n$th row has a sum of $ n^3$, giving an answer of $ 15^3\equal{}\boxed{3375}$.

We must make a formula to solve for the sum of the numbers in row $n$.

Take for example $n = 4$ as shown in the picture.

We can find the sum of the numbers in in row $n$ as the sum of the numbers in the red triangle minus the sum of the numbers in the blue triangle.

The number of numbers in the red triangle is triangular number 4 and the number of the numbers in the blue triangle is triangular number 3 in row 4 so the number of the numbers in the red triangle is triangular number $n$ and the number of the numbers in the blue triangular number $n - 1$ in row $n$.

We know that the sum of the first $n$ odd numbers is $n^2$ so:

SUM OF NUMBERS IN RED TRIANGLE: $tr (n)^2$

SUM OF NUMBERS IN BLUE TRIANGLE: $tr (n-1)^2$

THE SUM OF NUMBERS IN ROW $N$: $tr (n)^2 - tr (n-1)^2 = (\frac{n(n+1)}{2})^2 - (\frac{(n-1)n}{2})^2$

To simplify $(\frac{n(n+1)}{2})^2 - (\frac{(n-1)n}{2})^2$, we can see that $a^2 - b^2 = (a-b)(a+b)$ so:

$(\frac{n(n+1)}{2})^2 - (\frac{(n-1)n}{2})^2 = (\frac{n(n+1)}{2}-\frac{(n-1)n}{2})(\frac{n(n+1)}{2}+\frac{(n-1)n}{2}) = (\frac{n^2+n}{2}-\frac{n^2-n}{2})(\frac{n^2+n}{2}+\frac{n^2-n}{2})$
$=(\frac{(n^2+n)-(n^2-n)}{2})(\frac{(n^2+n)+(n^2-n)}{2}) = (\frac{n^2+n-n^2+n}{2})(\frac{n^2+n+n^2-n}{2}) = (\frac{2n}{2})(\frac{2n^2}{2}) = n*n^2 = n^3$

The sum of the numbers in row $n$ is $n^3$ so the sum of the numbers in row 15 is $15^3 = \boxed{3375}$


geometry
rectangle
analytic geometry
parameterization
geometry unsolved

Let ABC be an acute angled triangle, and inscribe a rectangle DEFG in it so that D is on AB, E is on AC, and F and G are on BC. Describe the locus (the curve occupied by) the intersections of all possible diagonals of DEFG.

Could someone show me how to do an analytic solution to this problem?

Thanks.

[b]Lemma:[/b]  The line segment from $ A$ to $ B$ is parameterized $ tA \plus{} (1 \minus{} t)B$, where $ t$ varies from $ 0$ to $ 1$.

Give the points coordinates $ B(0, 0), A(a, b), C(c, 0)$.  We can then describe the points $ D(ta, tb), E(ta \plus{} (1 \minus{} t)c, tb), F(ta \plus{} (1 \minus{} t)c, 0), G(ta, 0)$ analytically for some parameter $ 0 \le t \le 1$.  The intersection of the diagonals is given by the average of any pair of opposite points, which is

$ \left( \frac {2ta \plus{} (1 \minus{} t)c}{2}, \frac {tb}{2} \right)$.

This is a line from $ \left( \frac {c}{2}, 0 \right)$ to $ \left( a, \frac {b}{2} \right)$ - in other words, from the midpoint of $ BC$ to the midpoint of $ A$ and the foot of the perpendicular from $ A$ to $ BC$.


inequalities
geometry
circumcircle
inradius
Euler
vector
geometry solved

Let $ABC$ be a triangle with centroid $G$ and circumcentre $O$, prove that
with any right triangle $KOG$ (right at $O$) then
$KA+KB+KC \ge 3R \ge GA+GB+GC$  :)  :)

Hei My stars isn't as much as U have ,May be I have to add to them :D

Hi saeid I wait to see your star  :D  :D ,but can anybody help me!

Nice inequality, easier than I used to think...

[color=blue][b]Problem.[/b] Let ABC be a triangle with circumcenter O and centroid G, and let P be a point in the plane such that $OP\perp OG$. Then, prove that $6r\leq GA+GB+GC\leq 3R\leq PA+PB+PC$, where R is the circumradius and r is the inradius of triangle ABC.[/color]

[i]Solution.[/i] Let H be the orthocenter of triangle ABC. Then, according to the Euler line theorem, the centroid G of triangle ABC is a point on the segment OH. Also, the centroid G of triangle ABC obviously lies inside the triangle ABC. Thus, according to http://www.mathlinks.ro/Forum/viewtopic.php?t=19316 , we have $6r\leq GA+GB+GC\leq 3R$. Hence, it only remains to prove the inequality $3R\leq PA+PB+PC$.

Upon multiplication by R, this inequality becomes $3R^{2}\leq R\cdot PA+R\cdot PB+R\cdot PC$. Now, using vectors and their scalar products, we have $\overrightarrow{OA}\cdot\overrightarrow{PA}\leq\left|\overrightarrow{OA}\right|\cdot\left|\overrightarrow{PA}\right|=OA\cdot PA$. But

$\overrightarrow{OA}\cdot\overrightarrow{PA}=\overrightarrow{OA}\cdot\left(\overrightarrow{PO}+\overrightarrow{OA}\right)=\overrightarrow{OA}^{2}+\overrightarrow{OA}\cdot\overrightarrow{PO}$
$=\left|\overrightarrow{OA}\right|^{2}+\overrightarrow{OA}\cdot\overrightarrow{PO}=OA^{2}+\overrightarrow{OA}\cdot\overrightarrow{PO}$.

Thus, $\overrightarrow{OA}\cdot\overrightarrow{PA}\leq OA\cdot PA$ becomes $OA^{2}+\overrightarrow{OA}\cdot\overrightarrow{PO}\leq OA\cdot PA$. Now, OA = R, because O is the center and R is the radius of the circumcircle of triangle ABC. Thus, this rewrites as $R^{2}+\overrightarrow{OA}\cdot\overrightarrow{PO}\leq R\cdot PA$. Similarly, $R^{2}+\overrightarrow{OB}\cdot\overrightarrow{PO}\leq R\cdot PB$ and $R^{2}+\overrightarrow{OC}\cdot\overrightarrow{PO}\leq R\cdot PC$. Hence, in order to prove the inequality $3R^{2}\leq R\cdot PA+R\cdot PB+R\cdot PC$, it will be enough to show the identity

$3R^{2}=\left(R^{2}+\overrightarrow{OA}\cdot\overrightarrow{PO}\right)+\left(R^{2}+\overrightarrow{OB}\cdot\overrightarrow{PO}\right)+\left(R^{2}+\overrightarrow{OC}\cdot\overrightarrow{PO}\right)$.

This simplifies to

$0=\overrightarrow{OA}\cdot\overrightarrow{PO}+\overrightarrow{OB}\cdot\overrightarrow{PO}+\overrightarrow{OC}\cdot\overrightarrow{PO}$,

and thus to

$0=\left(\overrightarrow{OA}+\overrightarrow{OB}+\overrightarrow{OC}\right)\cdot\overrightarrow{PO}$.

Since G is the centroid of triangle ABC, we have $\overrightarrow{OA}+\overrightarrow{OB}+\overrightarrow{OC}=3\cdot\overrightarrow{OG}$, so this is equivalent to

$0=3\cdot\overrightarrow{OG}\cdot\overrightarrow{PO}$.

But this is true, since $OP\perp OG$ yields $\overrightarrow{OG}\cdot\overrightarrow{PO}=0$. This completes the solution of the problem.

  Darij

yes that's very nice solution Darij! WE can extent this problem for any line! :)


I don't know what exactly this problem means...
this is my homework,, please help me...

there is a triangle shown below...

$ a$ and $ b$ are constant
$ x$ changes as time change of time $ x(t)$

write $ \frac{d\theta}{dt}$ as a function of $ x$,$ \frac{dx}{dt}$,$ a$,$ b$

thanks

[quote="mhar_teens"]I don't know what exactly this problem means...
this is my homework,, please help me...

there is a triangle shown below...

$ a$ and $ b$ are constant
$ x$ changes as time change of time $ x(t)$

write $ \frac {d\theta}{dt}$ as a function of $ x$,$ \frac {dx}{dt}$,$ a$,$ b$

thanks[/quote]
$ \theta(t)\equal{}Arccos\frac{x(t)}{b} \minus{} Arccos\frac{x(t)}{a}$
Then you can find $ \frac {d\theta}{dt}$ as a function of $ x$,$ \frac {dx}{dt}$,$ a$,$ b$


Is there you can make LaTeX choose its page height "as necessary"? The intention is that everything should be on one page, no matter how high that page becomes then. Like a book would be 1 huge page of heigt 20-30 meters high, but just a formula doesn't need more than a few cm high.

The reason I need it is to use with ImageMagicka. I'd want to convert some tex files to an images, mostly very small, but some quite large tex files.
And just putting page height on 100 meters by default would make it increadibly heavy to handle with IM, even when you trim, right?

I don't see how this is possible. LaTeX is a document publishing system which fits the text and images on the page size and margins you give it. I don't see how it can lay out the document without knowing the dimensions of the page first.

Ok.

Is there something like a trim-option for documents? So that a long 10 meter heigh page can be reduced to the space actually containing text? I know you can convert everything to a pixel-image and then apply trim to it, but that takes vaste amounts of memory for large pages. Is there something like a vector-based trim that can trim the document before converting it to a pixel-image?

I don't understand how there can be such an option in LaTeX. You have to tell LaTeX the page size so it knows what to do. Even the best the [url=http://tug.ctan.org/cgi-bin/ctanPackageInformation.py?id=savetrees]savetrees package[/url] can do is alter margins etc but it can't decide the page size.

You need to look outside LaTeX for a way to manipulate pdf documents or
if you are into batch file writing you could write one that:
1. compiles a tex file
2. looks at the log file for overfull boxes - if none then end
3. alters the tex file so it has a different page size
4. back to 1
This automates what would be simpler to do manually.

I am indeed looking for batch file processing, and indeed, it doesn't matter that it is outside LaTeX (as long as it can be done without difficult-to-install software).

If there would be a (batch) program that trims the document to the right size, outside LaTeX, that would be perfect. Maybe I wasn't clear enough in that, but that would do exactly what I want.

If that fails, I could start from a low page height $ h$, and then if document has $ n>1$ pages, recompile it with height $ nh$. Then I'd have to filter out all ways to do \newpage etc., a bit dangerous for loopholes and abuse, but an option nonetheless.


Let $ \{M,A,I,B,N\}$ be five different points so that $ \left\|\begin{array}{c} IA \equal{} IB \equal{} r \\
 \\
AM\perp AI \\
 \\
BN\perp BI\end{array}\right\|$ .

Prove that $ \boxed {\ IM\perp AN\ \Longleftrightarrow\ AM^2 \plus{} BN^2 \equal{} MN^2\ \Longleftrightarrow\ IN\perp BM\ }$ .

[b]Particular cases.[/b]

$ 1\blacktriangleright$ Let $ ABCD$ be a trapezoid so that $ AD\parallel BC$ and $ AB\perp AD$ . Denote the

midpoint $ I$ of $ [AB]$ . Then $ \boxed {\ CI\perp BD\Longleftrightarrow DI\perp CA\Longleftrightarrow BC^2 \plus{} AD^2 \equal{} CD^2\ }$ .

[u]Proof[/u]. Choose $ \{D,A,I,B,C\}$ in the above general problem.

$ 2\blacktriangleright$ The incircle $ w \equal{} C(I,r)$ of $ \triangle ABC$ touches the sides $ [BC]$ , $ [CA]$ , $ [AB]$ at $ D$ , $ E$ , $ F$

respectively. Denote $ P\in BC\cap EF$ . Then $ PI\perp AD$ and $ PD^2 \plus{} AE^2 \equal{} PA^2$ (see [url=http://www.mathlinks.ro/viewtopic.php?t=284023][color=red][b]here[/b][/color][/url]).

[u]Proof[/u]. Choose $ \{A,E,I,D,P\}$ in the above general problem.

$ 3\blacktriangleright$ The $ A$-exincircle $ w \equal{} C(I_a,r_a)$ of $ \triangle ABC$ touches $ BC$ , $ CA$ , $ AB$ at $ D_a$ , $ E_a$ , $ F_a$

respectively. Denote $ P_a\in BC\cap E_aF_a$ . Then $ P_aI_a\perp AD_a$ and $ P_aD_a^2 \plus{} AE_a^2 \equal{} P_aA^2$ .

[u]Proof[/u]. Choose $ \{A,E_a,I_a,D_a,P_a\}$ in the above general problem.

[quote="Virgil Nicula"]Let $ \{I,A,B,M,N\}$ be five different points so that $ \left\|\begin{array}{c} IA \equal{} IB \equal{} r \\
 \\
AM\perp AI \\
 \\
BN\perp BI\end{array}\right\|$ .

Prove that $ \boxed {\ IM\perp AN\ \Longleftrightarrow\ AM^2 \plus{} BN^2 \equal{} MN^2\ \Longleftrightarrow\ IN\perp BM\ }$ .
[/quote]
Very interesting lemma !
Let me solve it  :)
$ MN^2 \equal{} AM^2 \plus{} BN^2\Leftrightarrow MN^2 \equal{} AM^2 \plus{} IN^2 \minus{} IB^2$
$ \Leftrightarrow MN^2 \minus{} IN^2 \equal{} AM^2 \minus{} IA^2\Leftrightarrow NA\perp MI$
Similarly, $ MB\perp IN$

See [url=http://www.mathlinks.ro/viewtopic.php?p=1537850#1537850][color=red][b]here[/b][/color][/url] another application of this problem !


find $k$ is a positive number so that $k \leq 2000$ and \[ N= \frac {1!2! \cdots 2000!}{k!}=p ^2\]

if you do not impose extra conditions on p it is obvious:
1!2! ... 2000!=x^2*1000!, where x=1!*3!...*1999!*2^500. so k=1000 works ...

[quote="S_Alex"][quote="vinoth_90_2004"]if k<= 1997, no such k exists:
suppose we can find such a k. note 1!2!..1997!=x^2*998!*1997! where x=1!*3! ... *1995!*2^449. so 998!*1997!/k! is the square of a integer. if k < 1949, then there is a factor of the prime 1949 in 998!*1997!/k!, a contradiction. so k >= 1949. i checked values of 1949<k<=1997 with maple and there are  no solutions (there probably is a good way of doing it, but ... ).[/quote]
I think there is an easier way  :lol: . You see $1997$ is a prime. So suppose $k<1997$ that means that $1997$ does not divide $k!$ but it divides $Q=\prod_{i\leq1997} {i!}$ and furthermore $1997^2$ does not divide Q so. $1997|p^2$ but $1997^2$ does not divide $p^2$. Thus we have obtained a contradiction (if we take into account that $1997$ is a prime. It remains to tackle with the case k=1997  :) .[/quote]

I'm such an IDIOT. For k=1997. Then $p^2={(1!3!..1995!)*2^{499})}^2 *(998!)$ so  we should have $998!$ is a perfect square but exp $2$ in $998!$ is odd (hope I compute it right $[\frac {998}{2}] + [\frac {998}{4}] +\dots$). So k=1997 doesn't work. And because k<1997 doesn't work either, there are no solutions.

your solution is good  we can't have k so that [tex] p^2*k!=1!*2!...1997!


geometry
rectangle
real analysis
real analysis unsolved

Is the mapping $ f: \mathbb{R}\times\mathbb{R}\to\{0,1\}$ defined by $ f(x,y)\equal{}\left\{\begin{array}{rl}1 & x\equal{}y \\ 0 & \mbox{otherwise}\end{array}\right.$ measureable?

[quote="jmerry"]Yes. It's zero almost everywhere.[/quote]
Excellent but why does the set $ \{(a,a): a\in\mathbb{R}\}$ have measure 0? I understand that it is a line in $ \mathbb{R}^2$ but I don't see why it must have measure 0.

First, slice it up- we can look at the piece between $ (0,0)$ and $ (1,1)$, then take a union of translates of it.

That depends on exactly how you define measure in $ \mathbb{R}^2$. If you base it on aligned rectangles, consider a covering by squares with opposite corners $ (\frac kn,\frac kn)$ and $ (\frac{k\plus{}1}n,\frac{k\plus{}1}n)$.

More abstractly, you can fit infinitely many disjoint translates (of the segment) into a $ 2\times 1$ rectangle. If $ \infty\cdot \mu(S)\le 2,$ $ \mu(S)\equal{}0$.

See also [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?&t=132440]here[/url]


search
superior algebra
superior algebra unsolved

let H be a subgrp of finite index in G
show that there exists a subgrp N in H which is normal in G and of finite index in G
And estimate its upper bound(index)

Surely posted before.

Take the intersection of all $xHx^{-1}$ with $x \in G$.

why that intersection is of finite index?what about its size>?

A search for "finite index normal" in Superior Algebra yields 25 results. At least two of them contain proofs of the result you're asking about: [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?&t=109423]these[/url] [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?&t=19724]threads[/url].

Do a search before you ask a question like this; it's not hard.

Let said intersection be known as $K$.

Well, since $eHe^{-1}= H$ is one of the sets in the intersection, then $\forall y \in K$, $y \in H$.

Therefore, $K \subseteq H$.

And therefore $K$ has finite index, since $H$ has finite index.


I think he means $\gcd (p,q)=1$, p and q are relatively prime.

$(p,q)=1$ is shorthand for $\gcd(p,q)=1$ or "$p$ and $q$ are coprime".

[hide="Proof"]If the prime factorizations of $p$ and $q$ are $p=u_1^{\alpha_1}u_2^{\alpha_2}\dots u_r^{\alpha_r}$ and $q=v_1^{\beta_1}v_2^{\beta_2}\dots v_s^{\beta_s}$, then we have $u_i\neq v_j$ for all $i,j$, since $p$ and $q$ are coprime.

Hence in the product $pq$ there will be no simplification of exponents, i.e. $a^2=u_1^{\alpha_1}u_2^{\alpha_2}\dots u_r^{\alpha_r}v_1^{\beta_1}v_2^{\beta_2}\dots v_s^{\beta_s}$, which means that all $\alpha_i$ and $\beta_i$ must be even, since otherwise $a^2$ can't be a perfect square.

But if all $\alpha_i$ and $\beta_i$ are even, then $p$ and $q$ are perfect squares.[/hide]

Ah, ok. That makes sense then. Well, factorization of $a^2$ is firstly $a \times a$, and because p and q are coprime, a must be composite, with some two factors say b and c, where $b^2=p$ and $c^2=q$ since $pq=a^2$. Therefore they are perfect squares. I don't really know how to express it in a better way, but all it really has to do with is (as aforementioned) the fundamental theorem of arithmetic, that a can be broken down into a product of primes.
Edit: Is there shorthand for lcm(p,q)?

[quote="silouan"]Let  positive integers  $p,q,a$   such that  $ ( p,q\ )=1$
and $pq=a^2$

Are the $p,q$ both perfect squares ?Show your proof[/quote]

[hide="Rather simple"]
$a^2 = p_1^{2a_1} p_2^{2a_2} ... p_k^{2a_k}$
Because $(p, q) = 1$ they cannot split any of these prime factors and hence they must each take all of the prime factors separately, which will all have even powers as a result.  Hence $p, q$ must be square.
[/hide]

[hide="Contradiction"]

Suppose $p$ is not a square. Then there exists a prime $k$ such that $k^{\alpha}|p$ and $\alpha$ is odd. But since $(p,q) = 1$, $k \not| q$. Therefore $a^2$ contains a factor of $k^{\alpha}$ for $\alpha$ odd and cannot be a square. $\rightarrow \leftarrow$.[/hide]

[quote="Hokkage"]Edit: Is there shorthand for lcm(p,q)?[/quote]
$[p,q]$


Hello!

$ \triangle ABC$ is a triangle; $ AD, BE$ and $ CF$ are the altitudes. Assume that the altitudes $ AD, BE,CF$ meet the circumcircle in $ G,H,I$ respectively. Prove that

\[ {{AG}\over{AD}}\plus{}{{BH}\over{BE}}\plus{}{{CI}\over{CF}}\equal{}4.\]

Let $ X$ be orthocenter.Use $ DX\equal{}DG,EX\equal{}EH,FX\equal{}FI$ and $ \frac{XD}{AD}\plus{}\frac{XE}{BE}\plus{}\frac{XF}{CF}\equal{}1$.


Let's each say a phrase whenever we want, and define it.

I start off with one:

de nada - You're welcome!

Hola amigos, como esta? - Hello friends, how are you?

I don't have accents and stuff on my keyboard...

No hablo español. - I do not speak Spanish.

Me llamo Prime Factorization. Me gusta uno fajita muy caliente.

Hopefully, I just said "my name is Prime Factorization. I am eating a very hot fajita."

Estoy sintiendo presioné por ninguna razón evidente. - I'm feeling depressed for no apparent reason.

No sé español tampoco, yo estoy utilizando un traductor en línea. - I don't know Spanish either, I'm using an online translator.

¡“Último del final de los individuos de Niza” - Leo Durocher
no tengo siempre razón, sino que nunca soy incorrecto!

my sig$ \Uparrow$

bonjour is french. In that case:

Bonjour monsieur. Comment ca va?

Getting off topic, don't make me delete your posts...

back to spanish!

tengo hambre! <- I'm hungry.

Tipo o goma aquí para traducir el texto…=Type or paste here to translate text...

"Oye hombre, usted es bastante una fruta. Que tal usted anda calle abajo y estira sus armamentos mientras yo me escapo de usted. "

hey man, you are quite a fruit. How about you walk down the street and stretch your arms while I run away from you.

i Ay, señor ! ¿i¿i Qué pasa !?!? Debemos chill juntos este fin de semana.

Hey you! What's up? We need to chill together this weekend.

[quote="Nexmus"]This point has been  uninteresting[/quote]

Tiene de hecho. - Indeed it has.

Don't be racist azhang. Maybe the mexican's and the american/british/french/german....   people all know how to speak English

我觉得，这个讨论题目太没有意思拉。

:)  ----------------------〉:| ---------〉:(
Start--------------------------------------〉End

Notice: The end of the topic comes faster than the start to middle. I wonder why....

uhh..

I thought that, this discussion topic too does not have the meaning to
pull.

?????

Spanish = Tengo un perro caliente.
English = I have a hot dog.


$(1)$ Condisder $\triangle ABC$, let $h$ be its orthocenter. Let $H_{1}, H-{2}, H_{3}$ be the refelctions of H across $AB,AC, BC$ prove that $H_{1}, H-{2}, H_{3}$ lie on the circumcircle of $\triangle ABC$. :D

$(2 Hungarian 1995)$ Let $A,B,C,D$ be points on a plane so that no three of them are collinear. $AB$ intersects $CD$ at $E$, $BC$ and $DA$ at $F$. Prove that either the circles with diammeter $AC,BD, EF$ passes through a common point or not two of them have any common point

[hide="hint for $(2)$"]Use your solution to $(1)$ to solve it :P [/hide]

No one's interested :maybe:  :huh:  :ninja:

The first problem is very well-known and can be done by angle chasing.
See the attached picture. I will show that $H_{1}$ lies on the circumcircle; the proof for $H_{2}$ and $H_{3}$ is analogous.
Since the angle $BMH$ is right, it sufficies to prove that $|\angle HBM|= |\angle H_{1}BM|$. But obviously $|\angle H_{1}BM|= |\angle H_{1}BA|= |\angle H_{1}CA|$, so we have to prove that $|\angle HBM|= |\angle HCA|$. And this is almost trivial, since $|\angle HBM|= 90^{\circ}-|\angle BAC|= |\angle HCA|$.

I will post hte solution to $(2)$ tomorrow if anyone is interested :maybe:

[quote="D.P.L"]

$(2 Hungarian 1995)$ Let $A,B,C,D$ be points on a plane so that no three of them are collinear. $AB$ intersects $CD$ at $E$, $BC$ and $DA$ at $F$. Prove that either the circles with diammeter $AC,BD, EF$ passes through a common point or not two of them have any common point

 [/quote]
in general you can prove they are coaxal.

prove the power of orhocenter of $AED$ wrt these circle are equal.
like this prove it for  orthocenter of $FCD$ ,now we are done.

also with menelaous theorem you can prove that midpoints of $AC,BD,EF$  are collinear.


let M={x1,x2..} be a set containing 30 distinct positive real numbers ,and let An denote the sum if products of elements of M taken n at a time ,1<=n<=30,prove that if A15>A10 ,then A1>1


quadratics
geometry
3D geometry
complex numbers
superior algebra
superior algebra unsolved

Suppose that x and y are real numbers such that [Q(x,y):Q] = 2.
Show that (x,y) is constructible.

[quote="ZetaX"]It's not hard: the point is $z=x+iy$ and you have that there is an equation $z^2+az+b=0$.
It's clear how to add complex numbers in a constructive way (just draw it when not clear ;) ), so it's enough to be able to multiply (and divide, but not needed here) and to take square roots (since with all that, we could solve quadratics as usual).
Both can be done by Thales theorem, try it :)[/quote]

ok, that's what I thought Zetax. The problem is that I thought that this argument could also be used with [Q(x,y):Q] = 3. Am I wrong??

Thanks

[quote="ZetaX"]No, exactly those points with algebraic degree a power of $2$ can be constructed (when initially only given some rationals).
The problem for degree $3$ is that there is no way to construct cube roots (which you would need), whereas this is possible for square roots (needed above).
One way to see the impossibility:
when you intersect line/circle with line/circle, then the degree of the equation of the intersection points will never be higher than $2$, with coefficients the already constructed points; thus adding the new point to our field will either remain it's dimension/degree (over $\mathbb{Q}$) unchanged or will double it; but by that, we can only reach dimensions/degrees that are powers of $2$.[/quote]

Are you saying that a point (x,y) is constructible if and only if [Q(x,y):Q] = power of 2 ??

Yes, the only if part is sketched above, the other direction uses a bit more theory.

Edit: wrong direction ^^ (if <-> only if)

[quote="ZetaX"]Yes, the if part is sketched above, the other direction uses a bit more theory.[/quote]

Thanks ZetaX :)


pigeonhole principle
number theory
relatively prime

Given any $n$ elements in the set $\{1, 2, \ldots, 2n-1\}$, prove that one can always find two that are relatively prime.

[quote="julien_santini"]You pick $n$ elements from a set with $2n-1$ elements. This guarantees you to find two consecutive positive integers among those $n$ elements. But two such integers are always relatively prime.[/quote]

[hide="not quite"]
If n = 3, and I select 1, 3, 5, there are still two that are relatively prime, but they aren't necissarily consecutive, but it's pretty easy to tie off that loop hole:

[hide="solution"]

Given the set of $n$ numbers, each one will be placed in one of the following n-1 categories:

{1,2,3}
{4,5}
{6,7}
.
.
.
{2n-2,2n-1}

By pigeonhole, two numbers must land in the same category.  Two cases:

Case 1,2,or3: These are all relatively prime.

Case k,k+1: $\gcd(k,k+1) = \gcd(k,1) = 1$.

QED
[/hide]

[/hide]

Or we can just assume 1 is not chosen, because $\gcd (1,n)=1$ for all positive integers $n$.


Hello everybody. :D

I wanted to make a movie. Like i have movies clips, audio, and background music. I have about 100 dollars to spend and I was wondering if you guys knew any good video editing software that would be in that price range. ( I mean like really good ones). I know of WMM and iMovie but....You can also just tell me some video editing software even if I can't buy it.....if they are really good. :D

Thanks everybody

quicktime pro and iMovie together for editing

adobe premier...you could always download it off of special places  ;)

[quote="bubka"]You can make a good movie with a mediocre m. e. software if you have the thoughts. Equipment is secondary.

By the way how old are you? You don't sound mature enough to make a movie![/quote]

Whats ME software?

I am in 8th grade. I just wanted to make a movie for projects such as the ones in Science, Math, History, English,etc. But I had wanted it to look as professional as possible and Windows Movie Maker....well....What is ME software?

[quote="noneoftheabove"]adobe premier...you could always download it off of special places [/quote]

I hear that adobe premier is pretty good...where do I go to download it?

You could just make a movie in Flash, seems like easiest option

Yea...its simple but not really professional. AND it takes to much time

No, definitely use Flash for what you want. Although, if you aren't artistic, you might as well just give up. Flash can be "bought" off of places you surely know, and tutorials can also be found. I'm actually working with ActionScript (Flash programming language) right now.

Adobe Premier Pro is good...

so is:
Sony Vegas
and if u wanan spend several thousand dollars,
Avid Liquid

For not actual movies:
Flash,
ImageReady

Wait....doesn't Avid Liquid only cost 1 thousand dollar and not thousands of dollars. I think your think about the AVID adrenaline things that they used 9 of to make KING KONG....yea.....Avid adrenaline HD is not cheap....

either way its still way out of price range.
i'd suggest flash. i've worked with it and its pretty easy.
it may take time, but i'd rather spend time than money.


calculus
integration
function
calculus computations

something related to Gamma function, I guess.

The answer can be expressed as $\displaystyle \frac{\Gamma(5/3)\Gamma(1/2)}{2\Gamma(13/6)}$

Of course, by the usual Calculus II-Calculus III understanding of the idea of "closed form", only $\Gamma(1/2) = \frac{\sqrt{\pi}}2$ would count as "closed form", not $\Gamma$(5/3) or $\Gamma$(13/6).

In the original integral, make the substitution $u=x^2$. With that substitution, the original integral becomes

$\displaystyle \frac12 \int_0^1 (1-u)^{\frac23}u^{-\frac12} du$.

Now go look up "Beta function" or "Beta integral."

:D Oh. Thanks for my lord.
I don't know what is gamma function or beta function. However I will try to undetstand it. Thanks. !!! ;)

[quote="phongthong"]Hi. Does any one can solve this problem by hand.? I try many ways but can't solve. :D

integrate (1 - x2)2/3 dx
integrate from x = 0 to x = 1 :maybe:[/quote]

$ \int^1_0 \sqrt[3]{(1 \minus{} x^2)^2}dx$

Hmmmmmm.... A tough one!! Maybe some method I'm not aware of..........

I believe Professor Merryfield has already showed that this integral can be solved (in a confortable way) through the use of the concept of Beta function.


algebra
polynomial
number theory
group theory
superior algebra
superior algebra unsolved

If a monic polynomial $ f(x)$ has a root in $ \mathbb F_p$ for all primes $ p$ , then it is reducible over $ \mathbb Z[x].$

OK, OK, degree $ \ge 2$.

Note that being reducible mod every prime is insufficient; $ x^4\plus{}1$ factors mod any prime, but not in $ \mathbb{Z}$.

An office-mate with whom I shared a variation of this question e-mails:
[quote="S. Sam"]Here is an article which constructs integer polynomials which are
reducible mod p for all p, but are irreducible over the integers:
http://www.jstor.org/stable/2323681

This paper (Example 1):
http://www.ams.org/proc/1996-124-06/S0002-9939-96-03210-8/S0002-9939-96-03210-8.pdf

gives a polynomial with no integral solutions, but which has solutions
modulo all integers: $ P(x) \equal{} (x^2 \minus{} 13)(x^2 \minus{} 17)(x^2 \minus{} 221)$.[/quote]

... which still doesn't answer the question posed here, with the stronger hypothesis and weaker conclusion. I would have used $ (x^2\plus{}1)(x^2\plus{}2)(x^2\minus{}2)$ as the example, to parallel my example for the other variant.

Yes, but I thought the examples were interesting enough to post even once I realized that I'd asked a different question than the one in this thread.

[quote="t0rajir0u"]I believe this is true by [url=http://en.wikipedia.org/wiki/Chebotarev%27s_density_theorem]Chebotarev's density theorem[/url], but I don't know enough algebraic number theory to be sure I know what I'm talking about.  (The argument I have in mind is as follows: suppose that $ f$ is irreducible.  Then primes that don't split over $ \mathbb{Z}[x]/(f(x))$ can't divide an integer value of $ f(n)$, and by Chebotarev these primes have positive density.)[/quote]

Yes, in fact it follows from the simpler Frobenius density theorem. A nice introduction is the
exposition [1] by Stevenhagen and Lenstra. There you'll find the following remarks on p.  32:

 "With a little group theory one can deduce several charming consequences from
Frobenius's  theorem. For example, if $ f$ modulo $ p$ has a zero in $ \mathbb{F}_p$ for every prime number $ p$,
then $ f$ is either linear or reducible. Also, the number of irreducible factors of $ f$ over $ \mathbb{Z}$ is equal
to the average number of zeroes of $ f$ modulo $ p$ in $ \mathbb{F}_p$, averaged over all $ p$ (in an obvious way)."

--Bill Dubuque

[1] Stevenhagen, P.; Lenstra, H.W. Jr.
Chebotarev and his density theorem.
Math. Intelligencer 18 (1996), 26-37.
http://websites.math.leidenuniv.nl/algebra/chebotarev.pdf
http://www.math.leidenuniv.nl/~hwl/papers/cheb.pdf

[2] Wyman, B.F. What is a Reciprocity Law?
Amer. Math. Monthly, Vol. 79, No. 6, 1972, pp. 571-586.
http://home.comcast.net/~billwgd/What_is_a_reciprocity_law_--_Wyman.pdf
http://dx.doi.org/10.2307/2317083
Correction, ibid, Vol. 80, No. 3, 1973, p. 281
http://dx.doi.org/doi:10.2307/2318450

[3] Franz Lemmermeyer. Class Field Theory: notes & links
http://www.fen.bilkent.edu.tr/~franz/cft.html

[4] Pesiri: The Chebotarev Density Theorem Applications
http://www.mat.uniroma3.it/scuola_orientamento/alumni/laureati/pesiri/sintesi.pdf


function
algebra
polynomial
summer program
Mathcamp
real analysis
rational function

Let f be a smooth function (infinitely differentiable) on the real line (to the reals).  If f(x) is rational for all rational values of x, must f be a rational function? (actually polynomial if the range is the entire real line instead of an interval)

Compare with:

Quantum M247:
Does a nonlinear function exist that is defined and differentiable for all real numbers and takes rational values at rational points and irrational values at irrational points?

Old MathCamp selection test / old Japan Olympiad?
If P(x) is a polynomial that takes rational values at rational points and irrational values at rational points, determine all such polynomials P(x).

Let $(q_{n})_{n\ge 1}$ be an enumeration of the rationals. Then, if the sequence $(a_{n})_{n}$ of rationals is chosen properly (that is, if it decreases fast enough), then the function $f(x)=\sum_{n\ge 1}a_{n}(x-q_{1})\cdot\ldots\cdot(x-q_{n})$ can even be made entire, and it clearly takes rational values in rational points.

The third one has been posted before, and it should be somewhere in the "Number Theory" section, and I don't really know what the answer is for the second problem :).

Neat idea!  However there are some details I am having trouble following.  To show that a sequence of continuous functions is continuous you can use
(1) uniform convergence
(2) Arzela-Ascoli Thm

similarly in complex analysis you use Montel's Thm for sequences of analytic functions.  If you restrict to the unit interval, your function can be uniformly bounded (by choosing the $a_{n}$ to decrease rapidly as you state) but its not clear (to me at least ) that you can do this for the entire real line.

4th variation:

Does a nonlinear function exist that is defined and infinitely differentiable (smooth) for all real numbers and takes rational values at rational points and irrational values at irrational points?

This would be a strong generalization of the classic problem where the function is assumed to be a polynomial.  I think smooth should be enough, but maybe the correct condition is real or complex analytic.

You can construct the sequence $(a_{n})_{n}$ of rationals inductively.

The coefficients of the polynomial $\sum_{k=1}^{n}a_{k}(x-q_{1})\cdot\ldots\cdot(x-q_{k})$ are sums of numbers of the form $a_{j}q_{i_{1}}\ldots q_{i_{t}}\ (*)$. Suppose you've chosen $a_{1},\ldots,a_{n}$ such that for $k=\overline{1,n}$, the absolute values of the numbers $(*)$ which make up the coefficient of $x^{k}$ in this polynomial sum up to less than $\frac 1{k!}$. Clearly, you can choose $a_{n+1}$ small enough such that this condition still holds for the polynomial $\sum_{k=1}^{n+1}a_{k}(x-q_{1})\cdot\ldots\cdot(x-q_{k})$, this time for $k$ up to $n+1$. This should be good enough to ensure that you end up with an entire function.


There are equal numbers of pennies, nickels, dimes, and quarters in a bag. Four coins are pulled out, one at a timek and each coin is replaced before the next is drawn. What is the probability that the total value of the four coins will be less than 20 cents?

it matters how many are in the bag

because if there's only 1 of each, then u get a different answer, hence a different amount of cases

are we assuming there's 4 or more?
-jorian

But they're replaced each time, so it doesn't matter. The probability that you draw any given coin will always be the same.

ok, sorry i read over that.

in that case[hide]

well...we have 4 factorial ways to pick the coins (24)

lets list all of the ways to go under 20 cents

pppp
pppd
pppn
ppnn
ppnd
pnnn

so 1/4

g2g, sry couldn't finish[/hide]
-jorian


Here's a problem I wrote about a month ago.  Let me know if it's too easy in which case I will move it.

Circles K and A1 have equal radii and are externally tangent to each other and to line m.  Circles A2, A3,  have the property that circle Ai is externally tangent to circles K, Ai-1, and line [i]m[/i].  If all such circles are constructed, what is the ratio of the circumference of K to the sum of the circumferences of A1, A2, A3, ?

Very nice -- I've seen the same basic diagram set-up quite a lot, but I don't think I've ever seen this particular problem.  And, of course, the answer is nice :)

I like this problem!  Here I have outlined a solution in spoiler:[hide]

1. Suppose the radius of K and A1 is 1.

2. Perform an induction that guarantees that the radius of An is 1/n; this is doable in a number of ways, I chose the Descartes Circle Theorem (Yes, a line is, for the purposes of this theorem, a BIG circle!)  Since this useful tool is remarkably little known, I summarize its generalized form here:  Suppose [tex][/tex] hyperspheres are mutually externally tangent in [tex][/tex]-space. We let their radii be [tex]r_{i} \forall i \in \{ 1, \dots , n\}[/tex], and let [tex]\kappa_{i} = \frac{1}{r_{i}}[/tex] denote their curvatures. Then:

[tex]

\displaystyle{n \cdot \left( \sum_{i=1}^{n} \kappa_{i}^{2} \right) = \left( \sum_{i=1}^{n} \kappa_{i} \right)^{2}

[/tex]

where [tex]\kappa_{j} = \frac{-1}{r_{j}}[/tex] if the [tex][/tex]-th hypershere contains all the others.

3. Use the fact that [tex]\sum_{n=1}^{\infty} \frac{1}{n^{2}} = \frac{\pi^{2}}{6}[/tex] to help compute the value of [tex]\sum_{i=1}^{\infty} C_{i}[/tex].[/hide]

Ya, Descartes' circle theorem is the most useful but least known theorem that I know! I guess it's a well kept secret or something.


geometry
3D geometry
tetrahedron
sphere
trigonometry
analytic geometry
trig identities

21. What is the length of one edge of a regular tetrahedron inscribed in a sphere of unit radius?
(A) sqrt(3)
(B) sqrt(6)/2
(C) sqrt(3)/2
(D) 2*sqrt(2)/3
(E) 2*sqrt(6)/3

Source: Univ. Maryland 1998 Round 1

I was looking over those problems today, and have no clue how to start this one, so i cheated a bit.

CH4 method:
Carbon is the center of the tetrahedron, and the Hydrogens are 109 degrees apart. The legs to hydrogen is 1. By law of cosines, the hydrogens are 1.63 units apart.

E is the closest.
pick E

[edit]
oops. just noticed that they have a discussion on this in the solution page

Here's a very strange looking tip: In dealing with a problem with a regular tetrahedron, consider the following points in $\mathbb{R}^4:$ $(1,0,0,0),(0,1,0,0),(0,0,1,0),(0,0,0,1).$ They form a regular tetrahedron, with the tetrahedron confined to the three-dimensional hyperplane $x+y+z+w=1.$ The advantage of this simple and highly symmetrical set of coordinates is that it is very easy to find coordinates for the center of the tetrahedron, the center of one face, and so on.

Each edge of this tetrahedron has legth $\sqrt{2}.$

The center of the tetrahedron is $\left(\frac14,\frac14,\frac14,\frac14\right).$

The radius of the circumscribed sphere is the distance from the center to one of the vertices, say $(1,0,0,0).$ That is, $r^2=\left(\frac34\right)^2+\left(\frac14\right)^2 +\left(\frac14\right)^2 +\left(\frac14\right)^2$ or $r=\sqrt{\frac{12}{16}}=\frac{\sqrt{3}}2.$

For the problem at hand, multiply everything by $\frac2{\sqrt{3}}$ so that the radius will be 1. Then the edge length is $\frac{2\sqrt{2}}{\sqrt{3}}=\frac{2\sqrt{6}}3.$

If you're not comfortable with $\mathbb{R}^4$, you can use the following set of vertices in $\mathbb{R}^3:$ $(1,0,0),(0,1,0),(0,0,1),(1,1,1).$

Incidentally, the method just outlined says that the methane bond angle, which your chemistry book says is approximately 109.5, can be written exactly as $\cos^{-1}\left(-\frac13\right).$

Or you can inscribe the tetrahedron in a cube and that's another method.


1. Master Hamtaro-sama is taken as a god among humans. Even in his childhood, he'd already showed great strength and intelligence, virtues of a great warrior. In his early days, he traveled the world overthrowing kingdoms, defeating emperors and destroying armies, while freeing millions of people from opression. But now, tired of his bloody career, he dedicates his life to training his pupils in his fighting style.
An ancient legend tells us that, one day, during a heavy training session, one of his disciples protested against the rigorosity of his master and challenged him to solve the following system:
$ x \plus{} y \plus{} z \equal{} \displaystyle \frac {121}{12}$
$ xy \plus{} yz \plus{} xz \equal{} \displaystyle \frac {265}{8}$
$ xyz \equal{} \displaystyle \frac {425}{12}$
Feeling insulted by such an easy question, the master, at the blink of an eye, gave the solution $ \left( \frac 52, \frac {10}{3}, \frac {17}{4} \right)$
After verifying the correctness of the answer, the apprentice dared once more and asked his master to find the roots of the polynomial
\[ P(x) \equal{} 24x^6 \minus{} 242x^5 \plus{} 867x^4 \minus{} 1334x^3 \plus{} 867x^2 \minus{} 242x \plus{} 24\]
With an ironic smile, the master immediately told his student the six roots. However, the scribe who was in charge of documenting this fact unfortunately forgot to write the answer. Are you capable of completing the manuscript by finding the roots?

hello, it must be $ P(x)\equal{}(x\minus{}2)(x\minus{}3)(x\minus{}4)(2x\minus{}1)(3x\minus{}1)(4x\minus{}1)$
Sonnhard.

Given the first half of the problem, maybe a solution using Vieta's Formulas is the one intended.


group theory
abstract algebra
calculus
integration
invariant
geometry
geometric transformation

Prove that any finite subgroup of the linear group $GL_2(R)$ is either cyclic or dihedral.

Come on, guys, this is just beautiful!!! Hint: first prove it for unitary matrices subgroups.  ;)

I've just read a proof (using the Haar measure and integration on groups) of the fact that every linear representation of degree $n$ of a compact group is equivalent to a representation in the group of unitary matrices of order $n$, and this is just an adaptation of what I read there to the finite case :).

We first prove that every finite subgroup $G$ of $GL_2(\mathbb R)$ is a conjugate of a subgroup of $O_2(\mathbb R)$, and then prove the result for finite subgroups of $O_2(\mathbb R)$ (this works for any $n$, not just $2$, of course).

Let $\langle\cdot,\cdot\rangle$ be the usual scalar product in $\mathbb R^2$, and define another bilinear form $(\cdot,\cdot)$ in terms of the old one as follows: $(x,y)=\frac 1{|G|}\sum_{g\in G}\langle gx,gy\rangle$. It's easy to see that $(\cdot,\cdot)$ is a scalar product invariant under every element of $G$, so, since every two Euclidean structures on $\mathbb R^2$ are equivalent by a similarity, $G$ can be regarded as a subgroup of $O_2(\mathbb R)$, which is what we wanted.

From now on we can assume $G$ consists of orthogonal matrices. Let $H\lhd G$ be the subgroup of $G$ formed by the orientation-preserving transformations, which in our case are simply rotations around the origin. The fact that $H$ is cyclic is obvious, so I won't insist :). Now, if $H=G$, we're done: $G$ is cyclic. Otherwise, $[G: H]=2$, so there must be an element $g\in G\setminus H$. This $g$ is an orientation-reversing orthogonal transformation, so it must be a symmetry wrt some axis $\ell$ through the origin. This means that our group $G$ is generated by a rotation of angle $\frac{2\pi}n$ for some $n$ and a reflection in an axis $\ell$. This is the symmetry group of the regular polygon $A_1A_2\ldots A_n$, where $A_1$ is one of the intersections between $\ell$ and the unit circle, so $G\cong D_n$.


The sum of three different positive unit fractions is $ \frac67$. What is the least number that could be the sum of the denominators of these fractions?

Um....is a unit fraction one with numerator 1 and denominator integer?

I had asked about this. I believe the answer was like 78 or something.

if we write \[ \frac 1a \plus{} \frac 1b \plus{} \frac 1c \equal{} \frac 67 \Longleftrightarrow ab \plus{} bc \plus{} ac \equal{} \frac{6abc}{7}\] where $ a,b,c$ are integers, we note that by closure, the LHS is an integer, so the RHS is as well, so at least one of $ a,b,c$ must be a multiple of 7 in order for the RHS to be an integer. I shall try to work on this tomorrow, since I must go to bed now.

hm....the LEAST number that could be sum of denominators...I think they got the answer wrong.

2,5,70 works, giving 78, which is the answer they gave

but

2,6,21 also works. (meaning the answer might be $ \boxed {29}$?

too lazy to try more right now.

but the above two possibilities were found on pure luck: I was try to find bounds on $ \max \{a,b,c \}$, so I noted that $ \frac 57 \equal{} .\overline{714285}$ and I made was like WLOG $ a < b < c$ and I made a = 2 and b = 5, maximizing b when a=2, and it somehow WORKED. Then a=2, b=6 also worked.

for future reference the answer is 47.

I checked

blah oops, I did the majority of the work above in my head while trying (and failing) to go to sleep that night, and I thought the sum of 5/7........


its all in the title...a poll to see what file format do ppl prefer  :D

-actually, if mathlinks had PostScript or LaTeX capabilites, i personally think that would be a good thing... :?

hmm...actually i was searching the net for a while for LaTeX, but i couldn't figure out what i needed to download and install. Most sites reccommended i install many different files or go to CTAN, but does anyone know what files i need, and where can i go to get them?

-In my opinion downloading and installing the necessary files to view PostScript files was a lot easier (just 2 files- GhostScript and GSview)

if you run windows, search for and download miktex for your LaTex viewing.

yeah, i just finished downloading the thing---> its a huge download...took 2 and a half hours on cable for 500 MB (if your gonna download, might as well get the entire thing in my opinion  :D )

-hmm...now all thats left is to install the thing

Hi splintercell (btw great game though  :D )
what plain-text editor do you use for your miktex? just wondering.
bye

hehe..yes it is 3x.lich, but anyway I am kinda new to miktex, so im not sure what you mean by plain-text editor  :blush: because I just installed it today :D

-but if you mean plain-text editor as in something like Word, I use that :). btw what plain-text editor do you use?

Sorry to say this splintercell but I think you got multiple topics talking about on the same thing. Please delete the topic 'Installing Latex---Help!'. Less spam = good stuff.  :D  thx.

no problem 3x, i just realized that myself...oh well anyway its all gone now :)


modular arithmetic
inequalities
function
logarithms
number theory proposed
number theory

Solve the equation $2p^{q}=7+q^{p}$ in prime numbers.

By little Fermats we have $7+q\equiv 0 \pmod{p}$ and $2p-7\equiv 0 \pmod{q}$. From the 2nd congruence we have $q=3$ if $p=2$ or $p=5$ and $p=3$ is impossible. By checking we have solutions $p=2, q=3$ and $p=5, q=3$

Now let $p\ge 7$. From the above congruencies we have $7+q \ge 2p$ and $2p-7 \ge q$  because the cases $7+q = p$ and $2p-7 = 0$ are evidently impossible. From this inequalities we have $q = 2p-7$ and in particular $q \ge p$
Since the function $\frac{\ln x}{x}$ is decreasing in $(e;+\infty)$ we have $p^{q}\ge q^{p}$. Because of evident  $p^{q}> 7$ we conclude that $2p^{q}> 7+q^{p}$

Answer: $p=2, q=3$ and $p=5, q=3$

Now I think that playing with the calculater to obtain the equation wasn't such a bad idea :blush:


inequalities
function
LaTeX
Functional Analysis
real analysis
real analysis unsolved

Suppose that $1<p<2$, $\frac{1}{p}+\frac{1}{q}=1$ and $f,g\in L^{p}$. Then \[\parallel \frac{f+g}{2}\parallel ^{q}+\parallel \frac{f-g}{2}\parallel ^{q}\leq (\frac{\parallel f\parallel ^{p}+\parallel g\parallel ^{p}}{2})^{\frac{1}{p-1}}\] .

Sorry, but just a remark about latex: use \| for the norm instead of \parallel:
$\left \| \frac{f+g}{2}\right\|^{q}$

[b][color=red]Cordialement.[/color][/b]

The proof is not hard but long. I get reference alone: Sobolev S. L. Some applications of functional analysis in mathematical physics. Clarcson inequalities are situated at the begining of this book.


geometry
rectangle
geometry proposed
U sviti matematyky

In triangle $ ABC$, construct externally squares $ BCC_1B_2,CAA_1C_2,ABB_1A_2$. Let $ O_A,O_B,O_C$ be respectively the centers of these squares. Let $ A_0\equal{}C_1A_2\cap B_2A_1, B_0\equal{}C_2B_1\cap A_1B_2,C_0\equal{}C_1A_2\cap C_2B_1$.

Prove that the lines $ O_AA_0,O_BB_0,O_CC_0$ are concurrent.

It is easy to know OaOc∥C1A2  OaOb∥B2A1  ObOc∥B1C2
(sin∠A0OaQ/ sin∠A0OaM )* (sin∠C0OcF/ sin∠C0OcP)*( sin∠B0ObN/ sin∠B0ObE)=
A0Q/A0M*C0F/C0P*B0N/B0E=1
=> OaA0 ObB0 OcO0 are concurrent

We can also say that the triangles $ \bigtriangleup O_{a}O_{b}O_{c},\ \bigtriangleup A_{o}B_{o}C_{o},$ are perspective and the problem is true in general case of arbitrary erected rectangles, instead of squares.

The triangles $ \bigtriangleup ABC,\ \bigtriangleup A_{o}B_{o}C_{o}$ are also perspective, but the triangles $ \bigtriangleup ABC,\ \bigtriangleup O_{a}O_{b}O_{c},$ are perspective, only in the case of erected squares or similar rectangles.

Kostas Vittas.

[quote="discredit"][quote="vittasko"]

The triangles $ \bigtriangleup ABC,\ \bigtriangleup A_{o}B_{o}C_{o}$ are also perspective,
[/quote]
Really? Can you please show it?  :maybe:[/quote]
A solution of [b][size=100]Virgil Nicula[/size][/b], by vectorial method, has all ready been posted in the topic [url=http://www.mathlinks.ro/Forum/viewtopic.php?t=167669]A nice and difficult concurrency[/url].

I will post here next time, the two synthetic proofs I have in mind ( after the problem's birthday ).

Kostas Vittas.


limit
inequalities
calculus
calculus computations

Hope i didnt make a silly mistake here :blush:
It is obvious that $a_{k+1}>a_{k}$. Assume that $a=\lim a_{k}<\infty$. Thus we can find $k_{0}: \forall k\geq k_{0}\ |a_{k}-a|<\delta$. Then
\[\frac{2}{b_{k}+c_{k}}<2\delta, \ b_{k}+c_{k}>\frac{1}{\delta}\]
so one or both sequences $b_{k}$, $c_{k}$ are unbounded. Supposing only $b_{k}$ tends to $\infty$, we see $b_{k}\approx \frac{k}{a+c}$, thus $\lim a_{k}$ is bounded below by $C\sum_{k}\frac{1}{k}$ and cant be finite. Now assume both $b_{k}$, $c_{k}$ tend to $\infty$. Then
\[\sum_{k=1}^{\infty}\frac{1}{c_{k}}=\infty \]

\[\sum_{k=1}^{\infty}\frac{1}{b_{k}}=\infty \]

\[\sum_{k=1}^{\infty}\frac{1}{c_{k}+b_{k}}<\infty \]
But $b_{k}-c_{k}\approx b_{k-1}+\frac{2}{c_{k-1}}-c_{k-1}-\frac{2}{b_{k-1}}=(b_{k-1}-c_{k-1})(1+\frac{2}{b_{k-1}c_{k-1}})$, so $b_{k}>c_{k}$ or $c_{k}>b_{k}\ \forall k$. But then $\sum_{k=1}^{\infty}\frac{1}{c_{k}+b_{k}}$ is bounded below by one of
\[\sum_{k=1}^{\infty}\frac{1}{2c_{k}}\]

\[\sum_{k=1}^{\infty}\frac{1}{2b_{k}}\]
And now we conclude that $\lim a_{k}$ must be also $\infty$.

[quote="Yustas"] we see $b_{k}\approx \frac{k}{a+c}$, .[/quote]
What are $a,c$?

The (supposed) limits of the sequences $a_{i}, c_{i}$.

Ok, I think so. But
[quote="Yustas"]
\[\sum_{k=1}^{\infty}\frac{1}{c_{k}}=\infty \]

\[\sum_{k=1}^{\infty}\frac{1}{b_{k}}=\infty \]

\[
[/quote]
Why?  :maybe:

[quote="N.T.TUAN"]
Why?  :maybe:[/quote]
Because, as $\lim a_{n}<\infty$, and when $k$ is big enough, we can write $b_{k+1}>b_{k}+\frac{2}{a+c_{k}}>b_{k-1}+\frac{2}{a+c_{k-1}}+\frac{2}{a+c_{k}}+\dots$, so as we supposed $\lim b_{k}=\infty$, there must be
\[\sum_{k=1}^{\infty}\frac{1}{c_{k}}=\infty \]

May be you pay to much attention to details  :) I just had to show that $b_{k}-c_{k}$ has always the same sign for sufficiently big $k$. You can make it accurate by calculating \[b_{k}-c_{k}=b_{k-1}-c_{k-1}+\frac{2}{c_{k-1}+a_{k-1}}-\frac{2}{b_{k-1}+a_{k-1}}=b_{k-1}-c_{k-1}+\frac{2}{c_{k-1}+a-\varepsilon}-\frac{2}{b_{k-1}+a-\varepsilon}=(b_{k-1}-c_{k-1})(1+\frac{2}{(c_{k-1}+a-\varepsilon)(b_{k-1}+a-\varepsilon)})\] without writing $' \approx '$

[quote="Yustas"] I just had to show that $b_{k}-c_{k}$ has always the same sign for sufficiently big $k$. [/quote]
This is what I want know  :oops: I surprised when you wrote $\forall k$. Thank you very much. Now, I will post a solution i know, but not of mine  :oops: .

Denote $M_{k}=\max{(a_{k},b_{k},c_{k})},m_{k}=\min{(a_{k},b_{k},c_{k})}$ forall $k=0,1,2,...$. We will prove $\lim_{k\to+\infty}\frac{M_{k}}{m_{k}}\in\mathbb{R}$(1) and $\lim_{k\to+\infty}M_{k}=+\infty$(2). From (1) and (2) we have done.

[i]Proof of [/i] (2). We have $\sum a_{k+1}^{2}>\sum a_{k}^{2}+4\sum \frac{a_{k}}{b_{k}+c_{k}}\geq 6+\sum a_{k}^{2}$, then $3M_{k}^{2}\geq \sum a_{k}^{2}>6k$, and have (2).

[i]Proof of [/i] (1). From hypothesis we have $m_{k+1}\geq m_{k}+\frac{1}{M_{k}}$ and $M_{k+1}\leq M_{k}+\frac{1}{m_{k}}$ therefore $M_{k+1}m_{k}\leq (M_{k}+\frac{1}{m_{k}})m_{k}=M_{k}m_{k}+1=M_{k}(m_{k}+\frac{1}{M_{k}})\leq M_{k}m_{k+1}$, then $1\leq\frac{M_{k+1}}{m_{k+1}}\leq \frac{M_{k}}{m_{k}}$, and have (2).

[quote="N.T.TUAN"]
[i]Proof of [/i] (2). We have $\sum a_{k+1}^{2}>\sum a_{k}^{2}+4\sum \frac{a_{k}}{b_{k}+c_{k}}\geq 6+\sum a_{k}^{2}$, then $3M_{k}^{2}\geq \sum a_{k}^{2}>6k$, and have (2).
[/quote]
Not very clear, especially everything going after the second inequality :maybe:

[quote="Yustas"]I dont understand it anyway  :wink:[/quote]
i.e $\sum_{k=1}^{n}(u_{k+1}-u_{k})=u_{n+1}-u_{1}$

Ok, that's what i see:
$\sum a_{k+1}^{2}>\sum a_{k}^{2}+4\sum \frac{a_{k}}{b_{k}+c_{k}}=\sum a_{k}^{2}+2\sum a_{k}(a_{k+1}-a_{k})$, but how you get $6$ and everything else? If you write a solution, you shouldn't miss nontrivial estimations(may be, they are nontrivial only for me  :oops: )

Oh, Sorry! I used $\sum\frac{a}{b+c}\geq\frac{3}{2}$ for $a,b,c$ are positive reals. :)

[quote="N.T.TUAN"]$\sum\frac{a}{b+c}\geq\frac{3}{2}$ for $a,b,c$ are positive reals[/quote]
What do you mean? If you mean what you wrote then it is wrong.

Not all :) The statement $\sum\frac{a_{k}}{b_{k}+c_{k}}\geq \frac{3}{2}$ is not obvious.

Ok, This is prove of statement that. By AM-GM $(\sum a)(\sum\frac{1}{a})\geq 9$ therefore $(\sum (b+c))(\sum\frac{1}{b+c})\geq 9$, also $(a+b+c)(\frac{1}{b+c}+\frac{1}{c+a}+\frac{1}{a+b})\geq\frac{9}{2}$ equivalent to $3+\frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{a+b}\geq\frac{9}{2}$, and done! :)


Has anyone participated in this or is going to this year?

Here's a link to the info:

http://students.imsa.edu/org/mao/jhmc.html

Probably not going, nor have I heard about it.

it's not terribly difficult, although the test is written by students so you might have to protest something. they might also lose your answer sheet...

lol, IMSA students can be trusted as responsible...that's why they organized and wrote the contest?

Oh yeah I remembered that I can't do it because that's during spring break for my school..

HEY I helped make it and ALL of our answers were right (except for a typo  :P) and we didn't lose ANYTHING.  :ninja:


Let G be a connected graph such that each vertex has even degree. Show that the number of components of (G-v) is at most (d(v))/2 for each vertex v in G, where d(v) denotes the degree of v.

Within any component $C$, the total number of edges is $\frac{1}{2}\sum_{v \in C}d(v)-\frac{1}{2}e(C)$, where $e(C)$ is the number of edges joining our deleted vertex to any vertex in $C$.  Since the whose expression is an integer and $\frac{1}{2}\sum_{v \in C}d(v)$ is an integer, we must have that $\frac{1}{2}e(C)$ is an integer.  Consequently, $e(C)$ is at least 2, so each component is joined to our removed vertex by at least 2 edges, so there are at most $d(v)/2$ components.


geometry parallelogram trigonometry MATHCOUNTS ratio

[quote="#H34N1"]A parallelogram  [i]ABCD  [/i]is such  [i]AB[/i] = 14 in, [i]AD [/i]= 30 in  and the measure of the angle [i]ABC [/i]is 150 degrees  . What is the number of square inches in the area of the parallelogram  [i]ABCD [/i]?[/quote]

[b]Hint[/b]

 [hide]Use the formula :

   [size=75][color=red]<No trigonometry in the MATHCOUNTS Forum, please.  -- moderator RCV>[/color][/size]

 which gives the area of triangle $ABC$.

 There are many other ways ![/hide]

No trigonojmetry in the mathcounts forums.

If the title doesn't give this away:

[hide="Hint"]
The 30-60-90 triangles go in $1: \sqrt{3}: 2$ ratio.[/hide]

[hide]No no I meant 210 root 3 sorry[/hide]

ijmmom
Really? I said 210 first then I changed it which one's right?

[quote="#H34N1"]A parallelogram  [i]ABCD  [/i]is such  [i]AB[/i] = 14 in, [i]AD [/i]= 30 in  and the measure of the angle [i]ABC [/i]is 150 degrees  . What is the number of square inches in the area of the parallelogram  [i]ABCD [/i]?[/quote]

[hide]$\angle ABC=150^\circ$

Then $\angle BAD$ is a supplement of $\angle ABC$, thus $\angle BAD=30^\circ$.

Applying the ratio in $30-60-90$ triangle,
we find out that the altitude to $AD$ from $B$ has length $7$.

∴$\text{Area}=30 \cdot 7=\boxed{210}$(square inches).[/hide]


Addition:Find the real roots of the equation. Real roots only.

$ x(2008 \minus{} x^{2007}) \equal{} 2008x \minus{} x^{2008}$. As $ x^{2008} > 2008x$ for $ x\ge 2$, then $ x < 2$. One solution is $ x \equal{} 1$. Let's look for other solutions:

$ x$ can't be negative, because $ A \equal{} 2008x \minus{} x^{2008}$ is then negative, so $ 0 < x < 2$. So $ x \equal{} 1$ is the only one.

EDIT: I did integer solution

I don't think  nobody can do this problem.  :maybe:  :maybe:  :maybe:

[hide="Solution"]It's easy to see that $ x\equal{}1$ is one root and $ 0<x<2$.
Assume $ x\neq 1$. The equation is equivalent to
$ (x^{2007}\minus{}1)x\equal{}2007(x\minus{}1)$
$ \Rightarrow (x^{2007}\plus{}x^{2006}\plus{}...\plus{}x)(x\minus{}1)\equal{}2007(x\minus{}1)$
$ \Rightarrow x^{2007}\plus{}x^{2006}\plus{}...\plus{}x\equal{}2007$
If $ x>1 \Rightarrow LHS>RHS$
If $ x<1 \Rightarrow LHS<RHS$
Thus the equation has only one root $ x\equal{}1$[/hide]
[hide="to Nguyen Ngoc Linh"]Hello Nguyen Ngoc Linh, do you know Phan Chi Thanh :D[/hide]


Can someone show me how to use the sigma notation in terms of math?

Sigma Notation is a easier way of writing a sum.

I guess the best way to learn is by examples:

$%Error. "displaytextstyle" is a bad command.
\sum_{i=1}^{n}i=1+2+3+...+n$

$%Error. "displaytextstyle" is a bad command.
\sum_{i=1}^{n}i^2=1^2+2^2+...+n^2$

I'm sure someone can give a better definition than me  :P


hello(salaaam)
in which countries people speak savahili?
thanks

hi (salaam) ;)

savalihi?
i've never heard this word! :D

i guess he meant "swahili", it's spoken in many african countries...you can find about it on http://www.infoplease.com/ce6/society/A0847370.html


geometry rectangle rotation function LaTeX

Here's a problem I found in Fermat 2006 (no.10). By the way, Fermat is a math competition in Canada for grade 11 students.

What is the maximum number of [i]finite[/i] points that any two rectangles can intersect at?

[hide]It would help to make diagrams[/hide]

Isn't it just 4?

Take one rectangle and put it on top of another.

Rotate the top rectangle 45 degrees and we have 4 intersection points?

it's not 4. Try different sized rectangles.

I think you are thinking along the lines of tangency. Think max intersections.

Just in case you are stumped:
[hide]8[/hide]

[quote="Daniel Hartmann"]I apologize to enter in the Pakistan Forum like this, but I tried to solve the last problem and I found the same result showed by option [b]C[/b]. I am right?[/quote]
C? Yes that would be correct  :D
You don't have to apologize for entering the Pakistani forum, feel free to post whenever you like. :lol:
And here's another problem, thought this is harder still. enjoy! (Fermat 2003)

Hello again. Thank you for give me a "green card" to this forum! Well, the angle TSB must be 27º (it's the nearest to 30º)...

I have something to ask you: do you know how to put those math symbols (like belongs, functions, integers, etc) ?

See ya!

You mean Latex? just click the link in the top-left corner of the page. Remember to put the dollar signs around what you want to be in Latex.
example: $\Sigma, \Pi, \sqrt{abc}, \equiv, \left( \begin{array}{cc} 2\tau & 7\phi-frac5{12} \\ 3\psi & \frac{\pi}8 \end{array} \right) \left( \begin{array}{c} x \\ y \end{array} \right) \mbox{~and~} \left[ \begin{array}{cc|r} 3 & 4 & 5 \\ 1 & 3 & 729 \end{array} \right]$, etc.  :)

These problems are much to easy. Here's one that might take a little while to do. (Fermat 2001)  :idea:

I just realized this is easy too :o . All you need is one strategy and the whole question unfolds (most geometry questions are like that). Hint...
[hide]put the radii in the right postion and use similar triangles.[/hide]

[quote="Daniel Hartmann"]Well, I'm in the third grade of Brazillian High School. Latelly I'm without time to try to solve these problems (I'm taking the school final exams...). But tomorrow I'll take a big look in this last! Just one last thing: mustn't I have a program (I downloaded MiKtEx) to right in "Latex"?[/quote]
I'm not sure, but you just write in the message area. Remember the dollar signs :). You're having your exams? I had mine a few weeks ago, and I did really good. :lol:


Does anyone have any idea of what AMC scores should get you what AIME scores?  Like let's say a person got a 120 on the AMC 12, what should he expect for the AIME?

All depends on the kind of math present on the AIME.

Two years ago, I got a 116 AMC, and an 8 on the AIME.
Last year, A comparable 113.5 on the AMC, but a dreadful 3 on the AIME. It varies wildly.

I don't think there's a very good correlation between AMC scores and AIME scores.  The pacing of the AIME is quite different from that of the AMC (especially now that the AMC is down to 75 minutes to AIME's 3 hours).  Also, with the AMC, there are always these strategic concerns about whether you try to answer a hard problem or not; on the AIME, you just try to solve as many of the (usually quite challenging) problems as you can and see where it goes.

Each format favors different kinds of students.  When I was in high school, I never could get my AMC score above a mid-130 (if that), and it didn't vary that much.  But I was able to improve my performance on the AIME quite a bit from soph. to junior year.

They are very different tests.  There's really no way to say.  The fact that it's 3 hours, not multiple choice, and non-calculator changes a lot about the test.

[quote="Ultimatemathgeek"]If I remember correctly,

USAMO index = [10*AIME + AMC] ?[/quote]
Yup

[quote="Kent Merryfield"][quote]USAMO index = [10*AIME + AMC][/quote]
Note that this in practice gives the AIME substantially more weight than the AMC. The difference between the best possible score (150) and the minimal AIME qualifying score (100) is 50 points; for the same people, the difference between best and worst possible AIME score is 150 points.

More to the point, we could take a large group of you with AMC scores in the 115-145 range (just a 30 point spread). Some of you will make 2's on the AIME, quite a few of you will make 4's and 5's and with just that difference, 115-5 and 145-2, both quite possible, are the same index. But 145-5 (probably) won't get you into the USAMO. Then a few of you will make 10, 12, even 14 on the AIME, which makes a huge difference to the index. And 115-11 will (probably) get you into the USAMO.[/quote]

I believe US citizenship or green card is required to take the test, in addition to good test scores.

My greencard in progress, and I got 107 on AMC.  No hope.......

[quote="Treething"]What USAMO index do you need to qualify?[/quote]

This has been asked so many, many times...noone knows yet.  But you don't need to care abou USAMO index but rather just AIME since you are 10th or less in grade level.  So what ever the floor value is.

[quote="Treething"]What USAMO index do you need to qualify?[/quote]

Usually the USAMO cutoff index is around 200-210.

[quote="probability1.01"]I also agree that AMC-AIME correlation is not strong.  Among other things stated above, there is also a psychological factor sometimes.  For example, last year, I didn't study too much for AMC, knowing that I had a 99% chance of making it to AIME.  Then, I studied like 40+ hours for the AIME, because it was my first time.  Meanwhile, I know a lot of classmates who study a bunch just to make the AIME and then relax because they believe they have no hope for USAMO.  Furthermore, some people suffer extreme volatility on test days.[/quote]

I agree whole-heartly, half-heartly, quarter-heartly


function
limit
algebra
domain
real analysis
real analysis unsolved

Let $ f,g: [0,\infty]\to{R}$, $ f$ continuous and $ g$ is increasing and unbounded. For every sequence $ (x_n)_{n\in{N}}$ of rational numbers with $ \lim_{n\to\infty}{x_n}=\infty$ we have $ \lim_{n\to\infty}{f(x_n)g(x_n)}=1$. Prove that $ \lim_{x\to\infty}{f(x)g(x)}=1$

Given a real sequence $ \{ y_n \}$for which $ \mathop {\lim }\limits_{n \to \infty } y_n \equal{} \infty$, define two [i]rational [/i]sequences $ \{ a_n \}$and $ \{ b_n \}$by choosing their nth terms from the following intervals (which are to be intersected with the domain): $ a_n \in (y_n \minus{} {\textstyle{1 \over n}},y_n )$, $ b_n \in (y_n ,y_n \plus{} {\textstyle{1 \over n}})$.

Assuming the limit of f at infinity exists,
$ \forall \varepsilon > 0\exists N(n > N \Rightarrow |f(a_n ) \minus{} f(b_n )| < \varepsilon )$

Fixing such $ \varepsilon$, we can then write the following thanks to the fact that g is increasing:
$ f(y_n )g(a_n ) \le fg(y_n ) \le f(y_n )g(b_n )$, for each n > N.

The estimate provided show that the extreme terms tend to 1 and the squeeze theorem give the result.


In how many ways can the letters of the word CLASSROOM be arranged such that the new word does not begin or end with a vowel?

[hide="Hint"]For this, Im assuming that you know how to count the number of ways to arrange the letters in, say, the word MISSISSIPPI.

Ok, so:
There are 6 consonants and 3 vowels.

fix the first and last letters with consonants, since it cannot have vowels (but ignore the letter S for the moment...pretend that S cant be at the end or beginning). Then find the number of ways to arrange the middle, (dont forget to correct for overcounting in the middle)

Finally consider all the cases where S is at the beginning or end (or both) then find the number of ways to arrange the middle and correct for overcounting in the middle.

The reason we do these separately is because S is used twice, so it creates problems.

[/hide]


symmetry
calculus
derivative
function
induction
linear algebra
matrix

Hello,

I'm having some trouble with multivariable differential calculus. First, let's clear some notation:
Let $ E$ and $ F$ be Banach spaces and $ X\subset E$ open and $ f: X\to F$. Let $ \partial f(x)$ denote the total derivative of $ f$ at $ x$ (and $ \partial f: X\to\mathcal{L}(E,F)$ denotes the derivative function, where $ \mathcal{L}(E,F)$ is the space of all continuous linear operators.) The higher derivatives are defined in the same fashion:
$ \partial^0: \equal{}f$ and $ \partial^mf(x_0): \equal{}\partial(\partial^{m\minus{}1}f)(x_0)\in\mathcal{L}(E,\mathcal{L}^{m\minus{}1}(E,F))\cong\mathcal{L}^m(E,F)$ where $ \mathcal{L}^m(E,F)$ is the space of all multi-linear maps $ E^m\to F$ which are continuous. The last isomorphism is an isometric isomorphism which is the reason why we can identify these spaces.

Now the theorem in question is:
\[ \partial^mf(x)\in\mathcal{L}_{sym}^m(E,F)\] i.e. the $ m$-th derivative is (if it exists) is a [b]symmetric[/b] $ m$-linear operator.
Proof: There was a lemma in advance where the case $ m\equal{}2$ has been proven. So this is an invitation to induction over $ m$. It is easy to prove:
\[ \partial^{m\plus{}1}f(x)[h_1,h_2,...,h_{m\plus{}1}]\equal{}\partial^{m\plus{}1}f(x)[h_1,h_{\sigma(2)},...,h_{\sigma(m\plus{}1)}]\] (arguments of multilinear operators are in square brackets) for every permutation $ \sigma$ of $ {2,...,m\plus{}1}$. So this leaves to prove
\[ \partial^{m\plus{}1}f(x)[h_1,h_2,h_3,...,h_{m\plus{}1}]\equal{}\partial^{m\plus{}1}f(x)[h_2,h_1,h_3,...,h_{m\plus{}1}]   (*)\]
Now the only words the authors looses about this equation is "Because $ \partial^{m\plus{}1}f(x)\equal{}\partial^2(\partial^{m\minus{}1}f)(x)$ this equation follows from the $ m\equal{}2$-case".
It is clear, that (*) would follow from the $ m\equal{}2$-case and the following equation
\[ \partial^{m\plus{}1}f(x)[h_1,h_2,h_3,...,h_{m\plus{}1}]\equal{}(\partial^2(\partial^{m\minus{}1}f)(x)[h_1,h_2])[h_3,...,h_{m\plus{}1}]\],
but I'm not able to prove this equation. I would appreciate any help.

Thanks in advance,
laxton

The object on the RHS of
[quote="laxton"]\[ \partial^{m \plus{} 1}f(x)[h_1,h_2,h_3,...,h_{m \plus{} 1}] \equal{} (\partial^2(\partial^{m \minus{} 1}f)(x)[h_1,h_2])[h_3,...,h_{m \plus{} 1}]\][/quote]is not well-defined, but this object \[ \partial^{m \minus{} 1}\left(x\mapsto (\partial^2f)(x)[h_1,h_2]\right)[h_3,...,h_{m \plus{} 1}]\] is. Have a closer look at the brackets.  :maybe:  For your induction step you need the identification you mentioned above which is tacitly used most of the time.

Does this hint give you the right kick? ;)

Why should the object on the RHS be not well-defined? $ f$ is a function $ X\to F$, so $ \partial^{m\minus{}1}f$ is $ X\to\mathcal{L}^{m\minus{}1}(E,F)$, so $ \partial^2(\partial^{m\minus{}1}f)(x)[h_1,h_2]\in\mathcal{L}^m(E,F)$ ergo $ RHS\equal{}(\partial^2(\partial^{m\minus{}1}f)(x)[h_1,h_2])[h_3,...,h_{m\plus{}1}]\in F$ as well as $ LHS\in F$.

I agree that your identity ($ \partial^{m\plus{}1}f(x)[h_1,h_2,h_3,...,h_{m\plus{}1}]\equal{}\partial^{m\minus{}1}\left(x\mapsto (\partial^{2}f)(x)[h_{1},h_{2}]\right)[h_{3},...,h_{m\plus{}1}]$) would prove (*) but nevertheless I have no clue how to prove your identity.

laxton

OK, I had another identification in mind. The identification should be tracked to have a proper proof of your theorem.

So here is what I can think of:

1 - The identification process of multilinear forms always works through the last variable. To be more precise $ \mathcal L(E,\mathcal L^{m-1}(E,F))\to \mathcal L^m(E,F)$ is defined by \[ \beta\mapsto ((h_1,\ldots,h_m)\mapsto \beta[h_m][h_1\ldots,h_{m-1}])\]

2 - A linear continuous map $ \Phi: E'\to F'$ with Banach spaces $ E', F'$ is smooth with constant total differential: \[ \partial\Phi(x')=\Phi\]for all $ x'\in E'$.

3 - Let $ f: X\to E'$ be smooth and $ \Phi: E'\to F'$ a continuous linear map, then the chain rule gives, for every $ x\in X\subset E$
\[ \begin{matrix}
\partial(\Phi\circ f)(x)&=&\partial\Phi(f(x))\circ\circ partial F(x)\\
&=&\Phi\circ\partial f(x)
\end{matrix}\]
This can be reworded that the differential is homogeneous with respect to continuous linear maps.

4 - In the situation of (3) let $ E'=\mathcal L^m(E,F)$, $ F'=F$ and, given $ h=(h_1,\ldots,h_m)\in E^m$, the evaluation map $ \Phi: E'\to F'$ defined by \[ \Phi(\beta)=\beta(h)\] is linear and continuous. Then, by the the result of (3), we see that evaluation and differentiation commute: Because of $ (\Phi\circ f)(x)=f(x)[h]$ this implies \[ \partial(x\mapsto f(x)[h])(x_0)=\Phi\circ \partial f(x_0)=(\partial f(x_0))[h]\]

5 - Now let $ g: X\to F$ be smooth and define $ f=\partial^2g: X\to\mathcal L^2(E,F)$. Given $ (h_1,h_2)\in E^2$, we have
\[ \begin{matrix}
\partial(x\mapsto f(x)[h_1,h_2])(x_0)[h_3]&=&(\partial f(x_0)[h_3])[h_1,h_2]\\
&=&(\partial (\partial^2g)(x_0)[h_3])[h_1,h_2]\\
&=&\partial^3g(x_0)[h_1,h_2,h_3]
\end{matrix}\]
This is the identity I mentioned above for a starting point of a proof by induction. Using the described technique you can deduce by induction flipping forth and back the fixed point in some $ E^k$ where the maps are evaluated.

Just try it. If you need some help, give me a shout.

Hello,

I'm struggling with the concept of introducing another identification especially because I don't think we need it. In any case we need to be very careful.
Is this the induction step you thought of (using [b]your[/b] identification)?
\[ \partial^{m\plus{}1}g(x_0)[h_1,...,h_{m\plus{}1}]\equal{}(\partial(\partial^mg)(x_0)[h_{m\plus{}1}])[h_1,...,h_m]\equal{}(\partial f(x_0)[h_{m\plus{}1}])[h_1,...,h_m]\] with $ f: \equal{}\partial^mg$
\[ \equal{}\partial(x\mapsto f(x)[h_1,...,h_m])(x_0)[h_{m\plus{}1}]\]
thus we can permute the first $ m$ arguments of $ \partial^{m\plus{}1}g(x_0)$. But we could already permute $ m$ of the $ m\plus{}1$ variables (see my first post).

laxton

The choice of the identification is crucial in the proof. Two identifications are linked by a $ \sigma\equal{}(\sigma_n)\in \prod_{n\in\mathbb N}S_n$. The theorem simply states that the higher derivatives do not depend on the choice of $ \sigma$. This means that if you have proven your theorem for one single $ \sigma$, then it is proven for all identifications. Most of the time this is denoted by canonical identifications which I think is always a very vague notion.

Assuming [b]my[/b] choice of identification throughout of the following, you can prove
\[ \partial^{m\plus{}n}f(x)[h_{1},h_{2},h_{3},...,h_{m\plus{}n}]\equal{}\partial^{n}\left(x\mapsto (\partial^{m}f)(x)[h_{1},\ldots,h_{m}]\right)[h_{m\plus{}1},...,h_{n\plus{}m}]\] by induction on $ n$ (use the evaluation trick). Taking $ n\equal{}2$ we have
\[ \partial^{m\plus{}2}f(x)[h_{1},h_{2},h_{3},...,h_{m\plus{}n}]\equal{}\partial^{2}\left(x\mapsto (\partial^{m}f)(x)[h_{1},\ldots,h_{m}]\right)[h_{m\plus{}1},h_{m\plus{}2}]\]
Since the RHS is symmetric in $ (h_{m\plus{}1},h_{m\plus{}2})$, by the so-called $ m\equal{}2$-case, $ \partial^{m\plus{}2}f(x)$ is symmetric in the last two variables. By the induction hypothesis  $ \partial^{m\plus{}2}f(x)$ is symmetric in the first $ m\plus{}1$ variables (using [b]my[/b] identification), it is proven that $ \partial^{m\plus{}2}f(x)\in \mathcal L^{m\plus{}2}_{sym}(E,F)$.

Thank you very much, Third Edition,

now I understand every single step in your proof and it's not good that this fairly long and non-trivial calculation isn't done in a beginner's textbook. But still I don't see a reason why "the choice of identification is crucial in the proof". Why is the proof wrong if using the identification of [b]my first post[/b].

laxton

Hmm, actually I could not see any explicit identification in your first proof.  :maybe:  But just try to run through the proof with any - or your special - identification to see that it should work.


What is the largest integer which must evenly divide all integers of the form $ n^5\minus{}n$?

I understand that 30 will evenly divide all integers of that form, since taking the expression modulo 2, 3, and 5 will all turn out to be zero... but how do you know there are no larger numbers? The book's only explantion is plugging in 2 for n yields 30...

It is impossible for an integer $ >30$ to divide $ 30$.

you could use that argument for all integers...

Taken $ \bmod2,3,5$ this is $ 0$, so the least possible largest integer is $ \text{lcd}(2,3,5) \equal{} 30$. For $ n \equal{} 2$, the expression is $ 30$, and for all other $ n$, it can be expressed in the form $ 30k$ for some integer $ k$ (if negative $ n$ are allowed; it doesn't matter though), and thus $ 30$ works. Clearly though, anything larger than $ 30$ will not divide $ 30$, so $ 30$ is indeed the maximum, as it is achievable and satisfied by all integer $ n$.

Yes. You've shown that $ 30$ is the minimal such integer and the fact that $ 2^5\minus{}2\equal{}30$ shows that it is the maximal such integer (all integers divide zero).


How can I enumarete theorems like this:
Definition 1.1
Definition 1.2
Lemma 1.3
Lemma 1.4
Theorem 1.5
I tried this:
[code]
\newcounter{thm}
\newcommand{\numb}{\ \thesection.\thethm}

\def\sec#1{\stepcounter{section}\vskip 3mm\leftline{\bf \thesection #1}\vskip 1mm \setcounter{thm}{1}}
\def\th#1{\vskip 1mm\noindent{\bf #1\numb}\quad \addtocounter{thm}{1}}
\def\thn#1{\noindent{\bf #1 \numb}\quad \addtocounter{thm}{1}}

\newcommand{\F}{\mathbb{F}}
\newcommand{\N}{\mathbb{N}}
\newcommand{\Q}{\mathbb{Q}}
\newcommand{\Z}{\mathbb{Z}}
\newcommand{\R}{\mathbb{R}}
\newcommand{\C}{\mathbb{C}}
\newcommand{\card}{\operatorname{card}}
\newcommand{\Hom}{\operatorname{Hom}}
\newtheorem{defn}{Definition}[section]
\newtheorem{nota}{Notation}[section]
\newtheorem{lem}{Lemma}[section]
\newtheorem{cor}{Corollary}[section]
\newtheorem{prop}{Proposition}[section]
\newtheorem{theo}{Theorem}[section]
\newtheorem{note}{Note}[section]
[/code]

If you load the amsthm package then you can tell LaTeX which counter to use when you define the lemma, definition etc and say that they should use the section counter.
[code]\newtheorem{theorem}{Theorem}[section]
\newtheorem{definition}[theorem]{Definition}[section][/code][section] says use the section counter to get 1.1, 1.2 ... instead of 1, 2 ...
and in definition, [theorem] says use the theorem counter.
You will then get
Theorem 1.1.
Definition 1.2.


can anybody send me some books of complex numbers or some difficult exercises ?

Thank you!

can you post it here,to be useful for evry mathlinkers


Can you help me solving this problem?

Find all natural number solutions for the following equality:

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

I started like that:    We want that $ b^3$ be divisible by 5 when substracted 3. Using modular algebra

If  $ b \equiv 0 (mod 5)$  -------    $ b^3 \equiv 0 (mod 5)$
     $ b \equiv 1 (mod 5)$ ------     $ b^3 \equiv 1 (mod 5)$
      $ b \equiv 2 (mod 5)$ ------     $ b^3 \equiv 3 (mod 5)$    *
        $ b \equiv 3 (mod 5)$ ------     $ b^3 \equiv 2 (mod 5)$
        $ b \equiv 4 (mod 5)$   -----   $ b^3 \equiv 4 (mod 5)$

Thus $ b$ is a number from the serie $ 5k\plus{}2$

Then $ 5a^2\plus{}3\equal{}(5k\plus{}2)^3$   but now i don´t know how to continue  :(

Any help?

The general rule about cubic Diophantines is that if there is at least one solution, it is very difficult to classify all solutions, and here we have the solution $ (\pm 1, 2)$.

This problem is rather easy!!
first we write the equation
b^3 - 3 = 5a^2
meaning that
b^3=3 (mod 5)
solving this cubic congruence you get 2.
So the solution for b is 2.
Plugging it in for b and solving the equation for a, yields us the answer.
(1,2) and (-1,2)

I'm pretty sure that would be write

[quote="AlphaBetaTheta"]So the solution for b is 2.[/quote]
This only tells you $ b \equiv 2 \bmod 5$.

It does seem to be the only solution, though -- at least, there are no others with $ a \leq 125000$.

Well.....i know "2"is a solution for b, but i would like to know a method for finding all the natural-number solutions.

AlphaBetaTheta i know that b should be congruent 2 in modulo 5 in order to the equation have a solution, but you still need to find for which values it is a sqare when substracted 3 and divided by ""5".

?

Any idea? (Excuse me for double posting; is only for having more chances of having an answer)

The best-case scenario for cubic Diophantines with at least one solution is to use factorization in some ring of integers to reach a contradiction.  That doesn't look likely to me for this problem; another general technique is to use a computer algebra program to compute the [url=http://en.wikipedia.org/wiki/Elliptic_curve#The_structure_of_rational_points]rank[/url] of the curve.  If it has rank zero, one can use [url=http://en.wikipedia.org/wiki/Nagell%E2%80%93Lutz_theorem]Nagell-Lutz[/url] to find the integer points.

how do we prove that there are no other solutions other than 1 and 2


Whoa... that is pretty ridiculous! Although it's not nearly as exciting as "Perfect numbers proven to be even only."

But Today is 4th of April and this number is small therefore I thinks ... :P

Uh... $22021 = 19^{2}\cdot 61$ so the actual prime factorization is

$3^{2}\cdot 7^{2}\cdot 11^{2}\cdot 13^{2}\cdot 19^{2}\cdot 61$

So it doesn't really work....

[i](or does this have something to do with April fools day. It's still March 31 where I live)[/i]

[quote="JAM"]Uh... $22021 = 19^{2}\cdot 61$ so the actual prime factorization is

$3^{2}\cdot 7^{2}\cdot 11^{2}\cdot 13^{2}\cdot 19^{2}\cdot 61$

So it doesn't really work....[/quote]
I am true  :rotfl:

I totally forgot about that! Considering it was Gyan, I should have known..  :blush: . That's what happens when you're too busy doing math and not paying attention to the real world.

Sorry to point out the obvious, but atleast for  today, $22021$ is prime. ...
To show it's false factors (and that too by  not keeping it under the "hide" tag)  is not good ..I am :o

[quote="Gyan"]Story  Just in!

Mathematician have long looked for perfect numbers. Perfect numbers , as every one knows, are the one that if you add all its  divisors, except the number itself, the result it the number itself.

For example $6=1+2+3$ is a perfect number.
Next perfect number is:
 $28 = 1+2+4+7+14$

There are others
$496 = 1+2+4+8+16+31+62+124+248$

But all known perfect numbers have been even.

Until now... A high school student  just discovered an odd perfect number.
Mathematicians say it is the first know odd perfect number.

The number is  $3^{2}\cdot 7^{2}\cdot 11^{2}\cdot 13^{2}\cdot 22021$

It is easy to check that the product is: $198585576189$
And sum of all the    factors is $=397171152378$ but since we do not count  $198585576189$ itself  it leaves us   $198585576189$

--- By our special correspondent.[/quote]

A math April Fools' joke... this is better than DHMO.

Did you come up with this yourself?

[quote]Did you come up with this yourself?[/quote]
The "story"  is published in AoPS only..(kind of spur of the moment thingie..)

(Actually while I was typing to "show"  using $\sigma$ I realized that $\sigma (7^{2}) = 57$ is divisible by 19 which will give the game away.. so I edited the line back)

The  semi-perfect numbers like $198585576189$  are , of course, fairly well known  (and not too difficult to makeup)

(60, if you consider 4 as prime, 90 if you consider 9 as prime etc :) etc.. are such "semi perfect" or pseudo perfect numbers )


superior algebra
superior algebra solved

Suppose that for a ring $ R$ we define $ N(R)\equal{}\{a\in R, a^n\equal{}0 \  \text{for \ a \ natural \ number } n\}$ Prove that

$ N(Z_m) \equal{} \{0\}$ iff $ m$ is not divided by the square of a number.

($ Z_m$ is the set of integers $ mod m$).

Alexandros

$ k^n \equiv 0 (mod m) \Leftrightarrow m | k^n$

So if m is squarefree $ m | k^n$ implies $ m | k$ and this implies $ k \equiv 0 (mod m)$.
On the other hand if m is p is a prime with $ p^2 | m$ the class of $ \frac {m}{p}$ is nilpotent in $ \mathbb{Z}_m$.


$\lfloor x \rfloor$ . $\lfloor y \rfloor$ = $x+y$

[quote="olorin"][quote="eLeM"]
All solutions are: $(x,y)=(0,0), (2,2), (2+a,4-a), (3+a,3-a), (-1+a,-1-a), (-2+a,-a)$, where $0<a<1$[/quote]
Almost. The last $2$ are wrong.
The solutions are: $(x,y)=(0,0), (2,2), (2+a,4-a), (3+a,3-a), (a,-a), (a-1,1-a)$ with $0<a<1$.[/quote]

You are right. I have to learn how to avoid stupid mistakes :)

[quote="olorin"][quote="eLeM"]
All solutions are: $(x,y)=(0,0), (2,2), (2+a,4-a), (3+a,3-a), (-1+a,-1-a), (-2+a,-a)$, where $0<a<1$[/quote]
Almost. The last $2$ are wrong.
The solutions are: $(x,y)=(0,0), (2,2), (2+a,4-a), (3+a,3-a), (a,-a), (a-1,1-a)$ with $0<a<1$.[/quote]
even all the above $(y,x)$ for the above $sol_{n}$


What is your average score in mathcounts for chapter when you either practice or when you competed. Please only have people who are going to participate this year vote in the poll. Please tell the truth.

My average is around 44.

44-46, usually. Mostly 46s, a few 45s, 1 44.

[quote="lotsofmath"]I score between 36 and 40 most of the time. In the minicomp in Houston, I got a 37. Maybe I'll make it to State, but this is only my first year (6th) and I have 2 more years.[/quote]

Kind of funny, the bad thing with me is that I am in the same situation as you (first year) but I am in eighth grade. I just learned about it in October or so. So I only have one shot.   :|

[quote="karatemagic7"]you guys or girls scores seem a bit on the high side ... I mean, if everyone were getting perfect scores on their chapt. tests, there would be a lot of ties to get to states and a lot of ties to get to nats, and I really don't think hardly Anyone scored that high on at least last year's chapters, and if they did, they were far and few between ... are you guys who are claiming such high scores all new at mathcounts? Where were you all last year?[/quote]

well I've heard that chapter gets harder and harder as the years go on. We are talking about the averages over all including practice.

[quote="karatemagic7"]you guys or girls scores seem a bit on the high side ... I mean, if everyone were getting perfect scores on their chapt. tests, there would be a lot of ties to get to states and a lot of ties to get to nats, and I really don't think hardly Anyone scored that high on at least last year's chapters, and if they did, they were far and few between ... are you guys who are claiming such high scores all new at mathcounts? Where were you all last year?[/quote]

Most people that go to AoPS are definitely of the better problem solvers (on average) compared to your normal wannabe Math Geek.

Take Ubemaya for example.

I should start a Meggy-Meg appreciation club.

For some reason, I do much better on sprints than on targets... Maybe I always make a mistake on the calc. I got like a 25 on sprint, but only 6 on target in the minicomp. (And my 37 was good enough for 23rd place :rotfl:, I wasn't even the best 6th grader!) (I was 2nd best in 6th grade, the best 6th grader had a 39.)

[quote="alkjash"][quote="karatemagic7"]I was basing my assessment on my own experience with MATHCOUNTS last yr - I mean, averages aren't exactly posted or easy to find, but knowing the numbers of the top 4 individuals in a few states, I just find it hard to believe that 3 out of 4 people responding here got such high scores -- even on practice tests, but if so, good for all of you! (then again, did you all TIME yourselves when you took, or did you give yourself limitless time to complete - this makes a BIG diff. you know ... I'm curious to know - how many of you time yourselves??[/quote]

I time myself and average 28 on sprints.
I don't usually do targets, but I doubt I'd get more than one wrong.
So about 44 or 42 average.[/quote]

Just a suggestion, if you're like me, maybe doing targets is good; I averaged 8 (no, not 7.9, 8) on Nats targets in practice and guess what happened at nats?  3/8 :).

"you guys or girls scores seem a bit on the high side" - two things: 1.  the real thing is VERY different (i.e. scores drop dramatically for most people), 2. We are AoPS (i.e.  a lot of the top mathcounts kids go to this website; you can check out the "AoPS Community Awards" thing on wiki and see)

I took my second chapter test last night..and I got a 46 (1st time=44)! And it is true that contests are always different when you're there.  Sometimes, you do better if you're prepared and if you are well-rested and aren't too nervous.  But usually, your nerves take over-and when you are struggling you get frustrated and are scared that you won't win the contest-leading to even more mistakes.

[quote="karatemagic7"]I was basing my assessment on my own experience with MATHCOUNTS last yr - I mean, averages aren't exactly posted or easy to find, but knowing the numbers of the top 4 individuals in a few states, I just find it hard to believe that 3 out of 4 people responding here got such high scores -- even on practice tests, but if so, good for all of you! (then again, did you all TIME yourselves when you took, or did you give yourself limitless time to complete - this makes a BIG diff. you know ... I'm curious to know - how many of you time yourselves??[/quote]

I always time myself

I've only done 3 chapter sprints+target and I got a 35, a 40, and a 46 with my first, 2nd, and 3rd try respectively. I know the average from that isn't 44, but I think that I can get a perfect score average from now on.

[quote="karatemagic7"](then again, did you all TIME yourselves when you took, or did you give yourself limitless time to complete - this makes a BIG diff. you know ... I'm curious to know - how many of you time yourselves??[/quote]

I actually use less than 40 minutes...

This is just my chapter average, states, well, I haven't done any recently, probably will soon.

Actually, I would expect most of us in the MC forum are 12-14.

[quote="wanflove"]i didn't take mine yet.
im probably the youngest person here at 12.

... :blush:

question:is mathcounts hard?   Y/N
if so, then how do i prepare fer it?   Y/N[/quote]

wrong.

First question: for some, yes. For others, no. It's moderately challenging.
Second question: Everyone needs to prepare.

Who really cares about chapter at this time of year?

oooo....
lots o nerds!!!
(kiddin...)
anyway will anyone answwer my questions????!!!!

I usually get 44-46 due to 1 careless mistake on sprint or target. chapters are always easy though. My state average plummets to like a 40

Yes that seems to suck but i get it over with

[quote="illusiat"]2007 chapter was killer[/quote]

It was. I got a 41 in my chapter and got first  :o

(I was averaging 44s last year on my practices)

Now I'm averaging 46s, but i still make stupid mistakes occasionally (like on this year's target). I only got a 44 this year....
 :wallbash_red:  :wallbash_red:  :wallbash_red:

Allen, what score did you get on NATS last year?

Also, I don't do practice chapters, so I've only done 2.
6th grade, I think I got around 40, but they don't give the scores back in my chapter. I got 46 this year.

[quote]i didn't take mine yet.
im probably the youngest person here at 12.[/quote]
... :blush:

Two things. One, I'm 11 years old.

Two, I was so stupid back in my post in January. Now I avg about 44 in chapter in practice and 36 in state. I've never taken a natl test before.

[quote="ZhangPeijin"][quote="gauss1181"][quote="ZhangPeijin"]Me.
Since it makes me feel better about failing at states.
:wink:[/quote]
Why would chapter make you feel better about failing at states?[/quote]
Because
$ \mbox{Chapter: Success}$
$ \mbox{State: Failure}$
And as we all know,
$ \mbox{Success} > \mbox{Failure}$[/quote]

Well, considering that chapter only determines if you go on to states, and we pretty much secured it for you by getting 2nd, it's not really that big of a success.

at states, you set your expectations to the point where you'd probably end up not coming close.

What do you mean?
I wasn't on the team during Chapter.


Let f and g two polynomials in a variable and j a polinomial in 2 variables such that:
f(x)-f(y)=a(x,y)(g(x)-g(y))
Prove that we can find a polynomial h such that f(x)=h(g(x)).

I hope you're not working on finite fields too, because in this case my solution does not work. Anyway, here is one way to do it: take some large $N$ (larger than $deg(f)$) and some $x_{0},...,x_{N}$ such that $g(x_{i})$ are all distinct. Take some polynomial $h$ of degree at most $N$ such that $h(g(x_{i}))=f(x_{i})$ for all $i\leq N$. Then $f(x)-f(x_{i})$ is divisible by $g(x)-g(x_{i})$. But so does $h(g(x))-f(x_{i})$, thus $h(g(x))-f(x)$ is a multiple of all $g(x)-g(x_{i})$, which are pairwise relatively prime. Thus $f(x)-h(g(x))$ is a multiple of the product of all $g(x)-g(x_{i})$. Compare the degrees and you are done.


calculus
integration
analytic geometry
trigonometry
calculus computations

Hello,

Here is an integral which I'm trying to solve :

$ \int\int_{D}\sqrt{4\minus{}x^2\minus{}y^2}dxdy,\ \ \ D \equal{}\{(x,\ y)\in\mathbb{R}^{2}| (x\minus{}1)^2\plus{}y^2 \leq 1 \}$

Any suggestions?

If you replace the region $ D$ with $ \{(x,\ y)\in\mathbb{R}^{2}| x^2\plus{}y^2 \leq 1 \}$ then the
problem is classical.

[quote="JRav"]You're integrating over a circle that has been translated.  Try polar coordinates, but be careful about your angles.[/quote]

make a transformation of polar coordinate.Let $ x=r\cos \theta,y=r\sin  \theta$,note that $ \theta$ goes from $ -\frac{\pi}{2}$ to  $ \frac{\pi}{2}$,and $ r$ goes from $ 0$ to $ 2\cos \theta$.

     The given integral can be rewritten as
     $ \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}}\int_{0}^{2\cos \theta} \sqrt{4-r^{2}} r dr d\theta=-\frac{1}{2}\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}}\int_{0}^{2\cos \theta} \sqrt{4-r^{2}} d({4-r^{2}) d\theta=-\frac{8}{3}\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}}(|\sin\theta|^{3}-1) d\theta=-\frac{16}{3}\int_{{0}}^{\frac{\pi}{2}}(\sin^{3}\theta-1) d\theta}$=${ -\frac{64}{9}}+\frac{16\pi}{3}$

Thanks!

But,
$ \int_{{0}}^{\frac {\pi}{2}}\sin^{3}(\theta) d\theta = \int_{{0}}^{\frac {\pi}{2}} \sin(x)\frac {1 - \cos(2x)}{2} = \frac {1}{2}\int_{{0}}^{\frac {\pi}{2}}\sin(x) - \frac {1}{2}\int_{{0}}^{\frac {\pi}{2}}\sin(x)\cos(2x) = \frac {1}{2} + \frac {1}{3} = \frac {2}{3}$

so the right term in the last equality(in the chain of equalities) has to be replaced by,

${ \frac {1}{2}( - \frac {64}{9}} + \frac {16\pi}{3})$.

So now one can solve another problem, using the value of the integral $ \int_{ - \frac {\pi}{2}}^{\frac {\pi}{2}}\int_{0}^{2\cos\theta}\sqrt {4 - r^{2}}r dr d\theta$, stated as follows :

Determine the volume of the body given by the following relations

$ x^2 + y^2 + z^2 \leq 4, \ \ 1 \leq (x - 1)^2 + y^2$


algebra
polynomial
induction
function
algebra proposed

Assume that $p(x)$ is a polynomial of degree $n$ with real coefficients. If $a\geq3$ is a real number, prove that \[\max_{0\leq j\leq n+1}|a^{j}-p(j)|\geq1\]

It is not correct. I think you forget about $p(x)=x^{n}+a_{n-1}x^{n_{1}+..}$, t.e. $a_{n}=1$.

If it's not true then give a counter-example. :)

It is not true, because $p(x)=a^{n+1}$ give $max(a^{j}-p(j))=0$.
 I found for best polinom approximation \[\forall j \ |a^{j}-p(j)|=(\frac{|a-1|}{2})^{n+1}\]

That was a little typo I corrected it. :)  I have the bound \[\max_{0\leq j\leq n+1}|a^{j}-p(j)|\geq\left(\frac{a-1}{2}\right)^{n+1}\]

It is not best bound. For example when n=0 must be $|\frac{a-1}{2}|$, your result give (not best) bound 1.

I had a typo. This is my best bound. Also another question : What is the best bound?

P.S.
Instead of posting useless posts, solve the problem.

Finally this is my solution.
We prove with induction on $n$ that $\max_{0\leq i\leq n+1}\geq\left(\frac{a-1}2\right)^{n+1}$ Problem is obvious when $n=0$. Suppose $Q(x)=\frac{p(x+1)-p(x)}{a-1}$. Obviously $deg(Q)=n-1$, Thus by induction hypothesis $\exists0\leq i\leq n\left|\frac{p(i+1)-p(i)}{a-1}-a^{i}\right|\geq\left(\frac{a-1}2\right)^{n}$. So \[|(p(i+1)-a^{i+1})-(p(i)-a^{i})|\geq2\left(\frac{a-1}2\right)^{n+1}\] Thus \[\max\{|p(i+1)-a^{i+1}|,|p(i)-a^{i}|\}\geq\left(\frac{a-1}2\right)^{n+1}\] and we are done.

Yes, your prove is shorter. But I found polinom $p_{0}(x)$, suth that
$\forall j=0,1,...,n+1 \ |a^{j}-p_{0}(j)|=(\frac{|a-1|}{2})^{n+1}.$

I didn't notice that. Fantastic !!! :lol:
It was question for me wether such function does exist or not.


This seems rather "silly", but can you give an answer for it?  ;)

At the Olympic Games a number of medals is given like this:
In the 1st day, one medal plus 1/13 from the rest of the medals are awarded.
In the 2nd day, two medals plus 1/13 from the new rest of the medals are given, and so on in the 3rd day... and in each of the days of the Olympic Games until the last day of the competition.
The question is how many days last the Olympics and how many medals are awarded in each day of the games?
I like this a lot and I hope that you will enjoy it!

I found it as an ancient IMO problem with $\frac{1}{7}$ instead of $\frac{1}{13}$ :
http://www.kalva.demon.co.uk/imo/imo67.html


What is the median of the first nine prime numbers?

The median of the first $ 9$ prime numbers is the $ 5$th prime number.
Counting up, we have $ 2, 3, 5, 7, 11$, so $ \boxed{11}$ is the answer.

how do you know it's the median is the fifth?

[quote="isabella2296"]Of any 9 numbers, the middle number is the 5th one. Try it and see.[/quote]

Note: Median is the 5th number when sorted in [b]increasing or decreasing order, not for any[/b].

The number that is the median in a group of n numbers when arranged in numerical order is (n+1)/2.
If n is even i.e. 8, and you get 4.5, it is the average of the 4th and 5th numbers.


The big, tall left side is 8, and the tall right side is 6. The little extension is 1. (These are all lengths in the given diagram.) Now we only need to find the base. Notice that this is simply 1 + 1 + 2 = 4, so we have 8 + 6 + 1 + 4 = 19.