I got this from solving some geometry problems today. They're about #18 or lower (I'm not sure so correct me) on AMC-10.

1. AB is the hypotenuse of the triangle ABC. The length of median AD is 7 and the length of the median BE is 4. What is the length of AB?

A) 5 :sqrt: 3
B) 5 :sqrt: 2
C) 3
D) 2 :sqrt: 5
E) 2 :sqrt: 13

2. Find the length of the radius of a circle in which a chord of length 6 is twice as far from the center as a chord of length 12.

A) 3 :sqrt: 5
B) 5 :sqrt: 3
C) 2 :sqrt: 6
D) 6 :sqrt: 2
E) :sqrt: 30

Neither of them were very hard to me because I recognized the similar question on the other nation's country for number one and I remember doing the similar question, but harder one in AHSME for number two.

They're okay, I guess. :P

1.
[hide]
Let AC = x
sqrt( 16 - (2sqrt[sqrt(49-x^2)])^2 ) = x/2
16- 196 - 4x^2 = x^2/4

x^2 = 48
CD = 1, CB = 2
AB = x^2 + CB^2
AB = sqrt(52) = 2sqrt13

(E)
[/hide]

2.
[hide]
let x be the distance from radius to the chord further away.
radius r
sqrt(r^2 - x^2)  = 3
sqrt(r^2 - x^2/4) = 6

r^2 - x^2 = 9
4r^2 - x^2 = 144

3r^2 = 135
r^2 = 45
r = 3sqrt5

(A)
[/hide]

[hide="Solution"]
Let the distance in between the two chords be $x$. Draw the radius from the center to the endpoints of each of the chords forming two right triangles and giving the following equations:

\[r^2=4x^2+9\]
\[r^2=x^2+36\]

Subtracting the bottom from the top we get

\[3x^2-27=0\]
\[x=3\]

Thus

\[r=\sqrt{9+35}=3 \sqrt5\]

[/hide]

[hide="Solution"]
Let $BD=CD=x$ and $AE=EC=y$. From $\triangle ADC$ and $\triangle BEC$ we obtain

\[4y^2+x^2=49\]
\[4x^2+y^2=16\]
\[5x^2+5y^2=65\]

Since hypotenuse $AB=\sqrt{4y^2+4x^2}$ we find $\frac{4}{5} \cdot 5x^2+5y^2=4x^2+4y^2=52$. Hence, $AB=\sqrt{52}=2\sqrt{13}$

[/hide]


geometry
combinatorics solved
combinatorics

Suppose [i]x[/i] congruent circles C1, C2, ..., Cx are contained in circle A. Let C be the area of one circle Ci and S the area of circle A.
Given that x * C > n * S, where x > n, prove that there exist n+1 circles among the Ci's that have a common point.

This is a lemma used to prove a previous covering problem about rings and 650 points. I cannot get a rigorous solution.  :?

Use contradiction :
Suppose that none of the points in A is covered by more than n of the C_i's. Then, the total area of the C_i's is not more than n times the area of A.
Thus, x*C  \leq n*S.

Pierre.


linear algebra
matrix
function
induction
linear algebra unsolved

The Redheffer matrices $R_n$ are $n\times n$ matrices such that
\[ R_n(i,j)=\begin{cases} 1 &\text{ if }j=1\text{ or }i|j \\ 0 &\text{ otherwise} \\ \end{cases}  \]
Prove that
\[ \det(R_n)=\sum^n_{k=1}\mu(k)  \]
where $\mu$ is the Mobius function from multiplicative number theory.

Pretty neat result :).

We're interested in permutations $\sigma\in S_n$ which satisfy $k|\sigma(k)$ or $\sigma(k)=1$ for all $k\in\overline{1,n}$. Notice first that every such permutation for which $\sigma(1)\ne 1$ has exactly one cycle of length $>1$: if $\sigma(k)=1,\ k>1$, then $\sigma(k_1)=k$ for some divisor $k_1<k$ of $k$, then $\sigma(k_2)=k_1$ for some divisor $k_2<k_1$ of $k_1$, until we reach $1$, completing the cycle. For those numbers $u$ which do not belong to this cycle, we have $u|\sigma(u)\Rightarrow u\le\sigma(u)\ (*)$, but summing after $u$ we must get equality, so we have equality in each $(*)$.

This means that for each $k$, there are exactly as many permutations (that produce summands for the determinant) with $\sigma(k)=1$ as there are strictly increasing chains $k_0=1<k_1<\ldots<k_t=k\ (\#)$ with $k_{i-1}|k_i,\ \forall i\in\overline{1,t}$, and the signature of each such permutation is $(-1)^t$ (since it's a cycle of length $t+1$). The idea would now be to find the sum of those summands in the determinant, of the form $\varepsilon(\sigma)a_{1\sigma(1)}\ldots a_{n\sigma(n)}$, where $\sigma\in S_n$ and $\varepsilon$ is the signature, for which $\sigma(k)=1$. In fact, it would be great if we could prove that it actually equals $\mu(k)$.

Let's look at the case when there is a prime $p$ such that $p^2|k$. For each chain of the type $(\#)$, if $u$ is the smallest index for which $p^2|k_u$, but $k_{u-1}<\frac{k_u}p$, associate the chain we get by inserting $\frac{k_u}p$ between $k_{u-1}$ and $k_u$. If, on the other hand, $k_{u-1}=\frac{k_u}p$, associate to this chain the chain we get by deleting $k_{u-1}$. This should, unless I'm missing something, produce a bijection between the chains with odd length and those with even length, showing that the sum over all the permutations $\sigma$ with $\sigma(k)=1$ is indeed $0=\mu(k)$, as we expected.

Now we look at the case when $\mu(k)\ne 1$. Suppose $k=p_1\cdot p_2\cdot\ldots\cdot p_m$, for primes $p_1<\ldots<p_m$. We can prove that the corresponding sum we talked about is $\mu(k)=(-1)^m$ by induction on $m$. To each chain of the type $(\#)$ where $u$ is the smallest index for which, say, $p_1|k_u$, but $k_1\ne p_1,\ k_{u-1}\ne\frac{k_u}{p_1}$, associate the chain obtained by inserting $\frac{k_u}{p_1}$ between $k_{u-1}$ and $k_u$, and to each chain with $k_1\ne p_1$ for which $k_{u-1}=\frac{k_u}{p_1}$ associate the chain we get by deleting $k_{u-1}$. This produces a sum equal to $0$ for those chains in which $k_1\ne p_1$, so we're only left with those chains for which $k_1=p_1$. We erase $k_0=1$, divide everything by $p_1$, and we're left with the corresponding sum for $k=p_2\cdot\ldots\cdot p_m$. This sum is $(-1)^{m-1}$ by the idnuction hypothesis, so our initial sum was $(-1)^m$, since erasing those $1$'s had the effect of changing the sign.

Sorry if it's unclear (had I not written it myself, I probably would have found it a nightmare to read through it :)).


I wrote in the top 57 at nats. Will the problem series help me for next year a considerable amount?

The Problem Series class teaches you a lot of tricks for MATHCOUNTS. I would say yes

If you got in the top 57, I don't know...eg, I don't know what you covered and what you didn't. Try posting this in [url=http://www.artofproblemsolving.com/Forum/index.php?f=140]this forum[/url]

I took the class.  You would DEFINITELY benefit from the class.

However, if you have lots of material, such as practice tests that you haven't gone through, and a very limited time frame, then perhaps taking the class would not be as useful as doing those.

hmm.

I'd recommend cramming at the last moment, preferably for fifty hours within a three-day span of the test. Worked for me pretty good, and I'd recommend using any dirty tactics you know for Countdown. Making audience laugh is a nice plus too.


hi  :)
solve in  $ \mathbb{C}$
$ z^{3} \plus{} \sqrt {z} \plus{} 1\equal{}0$
thanks :)

hello, should the equation
$ z^3\plus{}\sqrt{z}\plus{}1\equal{}0$
be solved for $ z \in \mathbb{C}$?
Sonnhard.

[quote="Dr Sonnhard Graubner"]hello, should the equation
$ z^3 \plus{} \sqrt {z} \plus{} 1 \equal{} 0$
be solved for $ z \in \mathbb{C}$?
Sonnhard.[/quote]

is your question necessary to answer the problem :)  ?
thanks.

[quote="Tyl"]Your original post did not contain a problem. Dr Sonnhard Graubner proposes the most obvious fix to you. What is your answer?[/quote]

the only fix i have is to put $ \sqrt{z}\equal{}u$.

[quote="jmerry"]You still haven't answered the question. What you posted is not an equation, and therefore can't be solved.[/quote]

I fixed it now,sorry about that.


Can someone send me last years sprint and target? I want to see where I stand.

yea if somebody has it please pm it to me too.

thanks

my coach has one, but he didn't give it to us yet.

i would greatly appreciate getting a copy... please?

i'd like one too, please.
wow, everyone wants one, but nobody has one to give. :(

[quote="LuCy4EvA"]I got a copy from my state coach, but I don't know how to turn it into pdf... maybe y'all could ask your state coach?[/quote]
Our coach doesn't have it, you could scan it, a .JPG would be fine too, same to you pianoforte ;)

I have the sprint, target, team, countdown, answers but no scanner

[quote="jclarkemathL314159"]Let's make this fun. I will enter the 2006 National Competition on Thursday, May 11. If you are not on my team, you have the chance to approach me and ask me for a copy of last year's national. I'll bring around 10 copies. Its up to you to find out which room i'll be in, which gate i'll be arriving at, and most impornatly exactly who i am and where i'm from. All of these are going to be or already are somewhere in the forum. Oh and plus, I charge a buck
THATS A GOOD DEAL it costs about 25 bucks to buy it online, and you have to buy it with the state and chapter too.
Have fun
If you're on my team our coach is gonna give us one anyway[/quote]

You can get it from me after I buy him out also, only 50 cents from me, I see competition.

You might try to find it on tz's site.

Klebian... When you are at Nationals you need to have your mind on what you are there for!!!  You will not have time for this elaborate hide and seek game!!  If you want to meet them briefly great but you have to stay focused on the TEAM not the individual.

Less talk and more quiet action would be preferred.  Chance favors the prepared mind not the one that talks about it all the time.

Coach Boyd.....   :(  :(

Yes Klebian, luck favors the prepared mind, not the one who talks about it all the time. (Actually, there is a grammatical error, it should be luck favors the [i]one with [/i] prepared mind, otherwise it is the mind that talks) But like Bill Rustle or however you spell his last name said last year, it is better to be good than to be lucky.

[quote="Ignite168"][quote="pgpatel"]The 2005 written round has arrived. With it comes a question. SHould I scan and post? :D.[/quote]

Yes!!!!!!!!!!!!!!!!!!!!!!!!!! Is that even a question? :P[/quote]
It's more of a "should I be a *$&# and not scan it?" type question. I will eventually...

[quote="pgpatel"][quote="Ignite168"][quote="pgpatel"]The 2005 written round has arrived. With it comes a question. SHould I scan and post? :D.[/quote]

Yes!!!!!!!!!!!!!!!!!!!!!!!!!! Is that even a question? :P[/quote]
It's more of a "should I be a *&# and not scan it?" type question. I will eventually...[/quote]

"Eventually?"How long is that? I want it very badly.
(Don't worry, I won't beat you at the nationals. I'm not even going to be there. I just wants to prepare for next year.)

I sent them to biff. He should be putting them on the website. All I ask for in return is donations *cough cough state pins cough cough*. Sorry I have a cold :D. What score did you get for 196th place last year biffanddoc?

i got 186th :oops: and it was a 12.
Amazingly, after not seeing the questions for a year and truly not remembering them, i tied that score in target. Was that really the top target score last year in sergei? must have been a fluke.

It's much easier to do nationals tests at home, I guess. I got a 39 on the 2003 nationals :roll: . And last year, I only got a 20 or whatever 81st got, with 2 targets right! Maybe I've just improved a lot, all due to AOPS. :D

Is the team round included? Our coach has been looking for '96 Team and '05 Sprint, Target, and Team.

[quote="pgpatel"]When are you going to post the pdf biffanddoc?
Isn't it odd though? I posted this page asking for 05 and I end up providing it :D.[/quote]

Which website is he going to post it in?
(Sorry if I should've asked biffanddoc.) :P

[quote="Ignite168"]Rapidshare.[/quote]

Did he posted it already?
Can you please tell me the whole website address?
Thanks alot!!!! :D


linear algebra
matrix
function
inequalities
combinatorics proposed
combinatorics

We are given $n \times n$ table and positive integer in each of $n^2$ squares such that any two of them wich are in the same row or column are different. For $n$ numbers (or unit squares) we say that they are in [i]general position[/i] if any two of them are in different coloumn and different row. We are given that sum of any $n$ numbers in general positions is same.
Proove that if numbers, whose product is the smallest of all products of $n$ numbers in general position, are sorted (by their values) at one diagonal, then the numbers whose product is the greatest of all $n$ numbers in general position are on the other diagonal of the square.

Note that if we start with $n$ points in general position and $(i,j)$ and $(k,l)$ are two of them, replacing them by $(i,l)$ and $(k,j)$ gives another set of points in general condition.

This implies that (letting $a_{ij}$ be the entry in the $(i,j)$ square) that $a_{ij}+a_{kl}=a_{il}+a_{kj}$.  In particular, $a_{ij}=a_{1j}+a_{i1}-a_{11}$, i.e. the rest of the matrix is forced by the entries in the first row and column.  Furthermore, it implies we have a relationship $a_{ij}=f(i)+g(j)$, where $f$ and $g$ are some functions defined on $\{1, \dots , n\}$ (for example, we could take $f(i)=a_{i1}$, $g(j)=a_{1j}-a_{11}$).

Our minimized product condition implies that the product of $(f(i)+g(i))$ would be made smaller if we were to replace the two elements $(i,i)$ and $(j,j)$ by $(i,j)$ and $(j,i)$, i.e. that

$(f(i)+g(i))(f(j)+g(j)) \leq (f(i)+g(j))(f(j)+g(i))$.

Rearranging this inequality, we get that $(f(i)-f(j))(g(i)-g(j)) \geq 0$, i.e. that the values of $f$ and the values of $g$ are in the same ordering.  Since we're sorting them by increasing sum, this implies that $f$ and $g$ are both in increasing order.

Similarly, the product is maximized exactly when $f$ and $g$ are sorted opposite from each other (the maximum has to occur somewhere, and the argument above gives that it can't occur anywhere [b]except[/b]  where the two are in opposite orderings from each other).  But this is exactly what is happening on the other diagonal.


prove that ;

$ (\forall a.b.c \in \mathbb{R}^\plus{})$  ;$ \frac{a^3}{b^2}\plus{}\frac{b^3}{c^2}\plus{}\frac{c^3}{a^2} \ge \frac{a^2}{b}\plus{}\frac{b^2}{c}\plus{}\frac{c^2}{a}$

Clear denominators and then use Muirhead.

[quote="mathmen"]prove that ;

$ (\forall a.b.c \in \mathbb{R}^ \plus{} )$  ;$ \frac {a^3}{b^2} \plus{} \frac {b^3}{c^2} \plus{} \frac {c^3}{a^2} \ge \frac {a^2}{b} \plus{} \frac {b^2}{c} \plus{} \frac {c^2}{a}$[/quote]
we have:
$ \left( \frac {a^3}{b^2} \plus{} \frac {b^3}{c^2} \plus{} \frac {c^3}{a^2} \right)\plus{}\left( a\plus{}b\plus{}c \right)\equal{}\sum \left( \frac {a^3}{b^2}\plus{}b \right)\equal{}\sum \frac{{a}^{3}\plus{}{b}^{3}}{{b}^{2}}\geq \sum \frac{ab\left(a\plus{}b \right)}{{b}^{2}}\equal{}\sum \left(\frac{{a}^{2}}{b}\plus{}a \right)$
$ \Rightarrow  \frac {a^3}{b^2} \plus{} \frac {b^3}{c^2} \plus{} \frac {c^3}{a^2} \ge \frac {a^2}{b} \plus{} \frac {b^2}{c} \plus{} \frac {c^2}{a}$
@VIMF:
[hide]yes,It's me :D[/hide]

[quote="mathmen"]prove that ;

$ (\forall a.b.c \in \mathbb{R}^ \plus{} )$  ;$ \frac {a^3}{b^2} \plus{} \frac {b^3}{c^2} \plus{} \frac {c^3}{a^2} \ge \frac {a^2}{b} \plus{} \frac {b^2}{c} \plus{} \frac {c^2}{a}$[/quote]

$ \frac {a^3}{b^2} \plus{} \frac {b^3}{c^2} \plus{} \frac {c^3}{a^2} \minus{} \left( \frac {a^2}{b} \plus{} \frac {b^2}{c} \plus{} \frac {c^2}{a}\right)$

$ \equal{}\frac{(a\plus{}b)(a\minus{}b)^2}{b^2}\plus{}\frac{(b\plus{}c)(b\minus{}c)^2}{c^2}\plus{}\frac{(c\plus{}a)(c\minus{}a)^2}{a^2}$

$ \ge 0.$

[quote="mathmen"]prove that ;

$ (\forall a.b.c \in \mathbb{R}^ \plus{} )$  ;$ \frac {a^3}{b^2} \plus{} \frac {b^3}{c^2} \plus{} \frac {c^3}{a^2} \ge \frac {a^2}{b} \plus{} \frac {b^2}{c} \plus{} \frac {c^2}{a}$[/quote]

By applying Cauchy-Schwartz we get:

$ (\frac {a^3}{b^2} \plus{} \frac {b^3}{c^2} \plus{} \frac {c^3}{a^2})(b\plus{}c\plus{}a) \geq (\frac {a^2}{b} \plus{} \frac {b^2}{c} \plus{} \frac {c^2}{a})^2$
$ \Rightarrow$  $ \frac {a^3}{b^2} \plus{} \frac {b^3}{c^2} \plus{} \frac {c^3}{a^2} \geq \frac{(\frac {a^2}{b} \plus{} \frac {b^2}{c} \plus{} \frac {c^2}{a})^2}{a\plus{}b\plus{}c}$

Now we have to prove:

$ \frac{(\frac {a^2}{b} \plus{} \frac {b^2}{c} \plus{} \frac {c^2}{a})^2}{a\plus{}b\plus{}c} \geq \frac {a^2}{b} \plus{} \frac {b^2}{c} \plus{} \frac {c^2}{a}$, or after dividing all with RHS:

$ \frac {a^2}{b} \plus{} \frac {b^2}{c} \plus{} \frac {c^2}{a} \geq a\plus{}b\plus{}c$
By multiplying both sides with $ a\plus{}b\plus{}c$ we get:

$ (\frac {a^2}{b} \plus{} \frac {b^2}{c} \plus{} \frac {c^2}{a})(b\plus{}c\plus{}a) \geq (a\plus{}b\plus{}c)^2$

What is true by Cauchy-Schwartz.

[quote="MathUniverse"]
By applying Cauchy-Schwartz we get:

$ (\frac {a^3}{b^2} \plus{} \frac {b^3}{c^2} \plus{} \frac {c^3}{a^2})(b \plus{} c \plus{} a) \geq (\frac {a^2}{b} \plus{} \frac {b^2}{c} \plus{} \frac {c^2}{a})^2$  [/quote]
Nice, but better write this step more suggestively as

$ (\frac {a^3}{b^2} \plus{} \frac {b^3}{c^2} \plus{} \frac {c^3}{a^2})(a \plus{} b \plus{} c ) \geq (\frac {a^2}{b} \plus{} \frac {b^2}{c} \plus{} \frac {c^2}{a})^2$ :!: :)

[quote="khanhsy"][quote="mathmen"]prove that ;
$ (\forall a.b.c \in \mathbb{R}^ + )$  ;$ \frac {a^3}{b^2} + \frac {b^3}{c^2} + \frac {c^3}{a^2} \ge \frac {a^2}{b} + \frac {b^2}{c} + \frac {c^2}{a}$[/quote]

I think : We have:
$ (\forall a.b.c \in \mathbb{R}^ + n\in Z^ + )$  ;${{ \sum_{i = 1}^n} \frac {a_i^{n + 3}}{b_i^{n + 2}}\ge \frac {\sum_{i = 1}^na_i}{\sum_{i = 1}^nb_i}.\sum_{i = 1}^n} \frac {a_i^{n + 2}}{b_i^{n + 1}}$ :D[/quote]

yes.

and I have a generaly one... :maybe:

[quote="khanhsy"]
I think we have:
$ (\forall a.b.c \in \mathbb{R}^ + n\in Z^ + )$  ;${{ \sum_{i = 1}^n} \frac {a_i^{n + 3}}{b_i^{n + 2}}\ge \frac {\sum_{i = 1}^na_i}{\sum_{i = 1}^nb_i}.\sum_{i = 1}^n} \frac {a_i^{n + 2}}{b_i^{n + 1}}$ :D[/quote]
This a special case of Marin Banco's lemma...see e.g. [url=http://www.mathlinks.ro/viewtopic.php?t=310721]here[/url]  :wink:


vector
induction
algebra
polynomial
linear algebra
linear algebra unsolved

Hi,

This seems to be an easy question on rank, but somehow I can't get it.

Let U be a linear operator on a finite-dimensional vector space V. Prove:

If rank(U^m)=rank(U^m+1) for some posiive integer m, then rank(U^m)=rank(U^k) for any positive integer k>=m.

It's in the section introducing Jordan canonical forms, so I assume the proof involves that. I tried induction and got to 'dim(U^m+p+1)<=rank(U^m) where p>1,' but I'm not sure how useful that is.

Also, I'm having trouble with this problem

Let T be a linear operator on a finite-dimensional vector space V whose characteristic polynomical splits.

Suppose B is Jordan basis for T, and let lambda be an eigenvalue of T. Let B'=B union K(lambda). Prove that B' is a basis for K(lambda). (K(lambda) is the generalized eigenspace corresponding to lambda)

I'd definitely appreciate some help with this!

That first question isn't about Jordan form at all. The rank of $U^{k}$ is the dimension of $U^{k}V$. For each $k$, we have $U^{k+1}V=U^{K}(Uv)\subset U^{k}V$, so $U^{m+1}V=U^{m}V$ by the equality of ranks. This means that $U(U^{m}V)=U^{m}V$, and applying $U$ arbitrarily many times won't change the space.


[hide]

Radius of smaller circle is 2, slant height is 6. by pythagorean theorem, height is 4:rt2:

[/hide]

[quote="anirudh"][quote="Syntax Error"]A circular sheet of paper of radius 6 inches is cut into 3 equal sectors, and each sector is formed into a cone with no overlap.  what is the height in inches of each cone?[/quote]

4 inches[/quote]

no, I think theSouthPaw is right.  :|


[quote="GameBot"]The circumference of a circle is $ 14\pi$ centimeters. What is the number of square centimeters in the area of the smallest square that contains all the interior points of the circle?[/quote]

The circumscribed square has side length 14, so $ 14^2\equal{}196$.

Can you explain it a little more detail?

We want the square to contain the circle, so it to make it the smallest, we have the circle inscribed inside the square. The diameter of the circle is the same length as the square's diagonal (you can see this by drawing it out.) Since the circle's circumference is $ 14\pi$, its diameter is $ 14$, so the square's side length is $ \frac{14}{\sqrt{2}}$. Therefore, the square's area is 98.

(I got a different answer than BOGTRO.)

Isabella2296: I think you misread the question.  An inscribed circle doesn't contain all the points in the circle-the circumscribed circle does.  As jxl28 points out, the side of the square equals the diameter which is 14, so the area is 196.

:lol: okay, i sorta get it, but one more question... how do you know the diamiter from the circumference. you just do the pi times radius cubed backwards? well, ill let u guys explain that. : )

The formula for circumference is $ diameter$ times $ \pi$. The formula for area is $ radius^2$ times $ \pi$. To find out what you need, you just plug in numbers and solve.


geometry
incenter
circumcircle
analytic geometry
graphing lines
slope
trigonometry

Given: a triangle with the verticies $(x_1,y_1)$, $(x_2,y_2)$ and $(x_3,y_3)$
Find: the incenter of the triangle

i know how to approach the problem, but it gets very messy, so anybody have a clever way to do it

How would you approach this problem Altheman? Would you simply use the distance formulas and use substitution, and then use $rs=A$ and then coordinatize those points to find the incenter (which is VERY messy) or would you do it another way like finding the circumcenter and find the distance between incenter and circumcenter using $d^2=R(R-2r)$ and then coordinatize those points? Either way, I haven't found out a clever and easy way to do this :(

i am not saying that i have a great way to do it, but you could find the angle biscectors of the angles and find the intersection

Well, this is quite easy.  Just use mudpoints, slopes, linear equations, and find the intersection.

Find the midpoint of the line connecting the points $(x_1,y_1)$ and $(x_2,y_2)$, which is
\[ M=\left(\frac{x_1+x_2}{2},\frac{y_1+y_2}{2}\right)  \].
The slope of teh line connecting $(x_1,y_1)$ and $(x_2,y_2)$ is
\[ m=\frac{\delta y}{\delta x}=\frac{y_2-y_1}{x_2-x_1}  \]
The equation of the line is
\[ y=\frac{y_2-y_1}{x_2-x_1}x+b,  \]
where b is determined as follows
$\\y=\frac{y_2-y_1}{x_2-x_1}x+b \\y_1=\frac{y_2-y_1}{x_2-x_1}x_1+b \\b=\frac{y_2-x_1(1+\delta y)}{\delta x} \\\implies y=\frac{y_2-y_1}{x_2-x_1}x+\frac{y_2-x_1(1+y_2-y_1)}{x_2-x_1}$
The equation of the line perpendicular to $y=\frac{y_2-y_1}{x_2-x_1}x+\frac{y_2-x_1(1+y_2-y_1)}{x_2-x_1}$ and passing though $(x_3,y_3)$ would be $y=\frac{x_1-x_2}{y_2-y_1}x+b$, where $b$ is determined as follows:
$\\y=\frac{x_1-x_2}{y_2-y_1}x+b\\y_3=\frac{x_1-x_2}{y_2-y_1}x_3+b\\b=y_3-\frac{x_1-x_2}{y_2-y_1}x_3\\b=\frac{y_3(y_2-y_1)-x_3(x_1-x_2)}{y_2-y_1}\\\implies y=\frac{x_1-x_2}{y_2-y_1}x+\frac{y_3(y_2-y_1)-x_3(x_1-x_2)}{y_2-y_1}$

The centre of the triangle is then the intersection of $y=\frac{y_2-y_1}{x_2-x_1}x+\frac{y_2-x_1(1+y_2-y_1)}{x_2-x_1}$ and $y=\frac{x_1-x_2}{y_2-y_1}x+\frac{y_3(y_2-y_1)-x_3(x_1-x_2)}{y_2-y_1}$.
Where they are both different equations with different properties.

And the result follows.
$\mathbb{QED}$

Hopefully, I do not have any typos or small mistakes. :)

EDIT: I'm still pondering about a clever way, however, I believe if we consider my above results as a formula, it could be used a fast method.

Masoud Zargar

Masoud, if you use \Delta instead of \delta you get $\Delta$ as opposed to $\delta$.

boxedexe, the incenter is the intersection of the angle biscectors, not the altitudes

[hide="an approach"]

but to find the slopes of the angle biscectors, it might be a bit more complicated...

we can find the slopes easily, and consider the angle that the line makes with the x-axis,  (also consider starting the angle with the x-axis, and counterclockwise is positive, so clockwise would be a negative angle)

this is pretty standard, but i am trying to be clear

so we have $\tan{\alpha}=m_1$ and $\tan{\beta}=m_2$

so the angle that the slope that would be inbetween them would be $\tan{\frac{\alpha+\beta}{2}}$, so we want to find that in terms of $m_1$ and $m_2$

if you expand that using the half angle formulas and substituting $m_1$ and $m_2$, you get your new slope to be

$\frac{ \sqrt{(m_1^2+1)(m_2^2+1)} +1-m_1m_2 }{m_1+m_2}$

then putting that into point slope form, and finding the intersections would give an answer...[/hide]

as for having a really simple solution, i do not know if that exists or not, but we will see, doing this with numbers is not that bad...

the only other approach that i can see would be to find the inradius with $A=rs$, then doing the line to line distance formula to find the line parallel to a of the triangle side, the inradius away from the side, and do that with 2 sides, and find the intersection

EDIT- Altheman has already said the same thing (so I deleted) :P

By the way, I think we can use the following fact:

Given $\triangle ABC$, and $P$ is on $BC$ such that $AP$ bisects $\angle A$.

Then we have $\frac{AB}{AC}=\frac{BP}{BC}$

(Actually I believe kunny's solution uses this property.)

[quote="Altheman"]boxedexe, the incenter is the intersection of the angle biscectors, not the altitudes[/quote]

Yes, sorry.  I'll fix that later.  Well, my approach would be to find the slope of the midpoint and point $(x_n,y_n)$.  Then since we know teh slope and midpoint of other lines, we'll be able to evaluate the centre.  This will lead us to a fomula, which could be used in these cases. :)

P.S.  I like theses ideas.

Masoud Zagar

[quote="kunny"]Let $A(x_1,y_1)$,$B(x_2,y_2)$ and $C(x_3,y_3)$, and $I$ be the incenter of $\triangle{ABC},$
we have
\[{ \overrightarrow{OI}=\frac{a\overrightarrow{OA}+b\overrightarrow{OB}+c\overrightarrow{OC}}{a+b+c}}.  \][/quote]

i understand do not understand how you got this, could you please explain?

Altheman, Sorry for that I can't explain now,because I'm busy now. Please wait for a while.

kunny

Altheman, I'm sorry that I have kept you waiting for.

Here is the desired proof.

Let $BC=a,CA=b$ and $AB=c.$ use the following [b]angle bisector theorem[/b],

[quote="frt"]Given $\triangle ABC$, and $P$ is on $BC$ such that $AP$ bisects $\angle A$.

Then we have $\frac{AB}{AC}=\frac{BP}{BC}$[/quote]
From $BD: DC=c: b,$ we have $\overrightarrow{AD}=\frac{b\overrightarrow{AB}+c\overrightarrow{AC}}{c+b}.$

Since $BI$ bisects  $\angle B$, use the theorem we have $AI: ID=BA: BD=c: \frac{ca}{c+b}=b+c: a,$ thus we have

$\overrightarrow{AI}=\frac{b+c}{b+c+a}\overrightarrow{AD}=\frac{b\overrightarrow{AB}+c\overrightarrow{AC}}{a+b+c}.$ Then

$\overrightarrow{OI}=\overrightarrow{OA}+\overrightarrow{AI}=\overrightarrow{OA}+\frac{b(\overrightarrow{OB}-\overrightarrow{OA})+c(\overrightarrow{OC}-\overrightarrow{OA})}{a+b+c}=\frac{a\overrightarrow{OA}+b\overrightarrow{OB}+c\overrightarrow{OC}}{a+b+c}.$

Just a suggestion, we could try using the two area formulas:
$A=rs$, then $r=\frac{A}{s}$ where we could find the side lengths using the distance formula,
and the formula that is 1/2*determinant of a 3x3 matrix with columns $(1, x_1, y_1), (1, x_2, y_2), and (1, x_3, y_3)$
I would try posting a solution but I am unsure of the latex for matrices.

Well, since I had a mistake in my original proof, here is the new (correct) version.

Find the midpoint of the line connecting the points $(x_1,y_1)$ and $(x_2,y_2)$, which is
\[ M=\left(\frac{x_1+x_2}{2},\frac{y_1+y_2}{2}\right)  \].
The midpoint of the line connecting $(x_3,y_3)$ and $(x_2,y_2)$ is
\[ M=\left(\frac{x_2+x_3}{2},\frac{y_2+y_3}{2}\right)  \].
The equation of the passing through $(x_3,y_3)$ and $\left(\frac{x_1+x_2}{2},\frac{y_1+y_2}{2}\right)$ is determined by first knowing the slope of the line connecting $(x_3,y_3)$ and $\left(\frac{x_1+x_2}{2},\frac{y_1+y_2}{2}\right)$; the slope is determined as follows:
$\\m=\frac{y_3-\frac{y_1+y_2}{2}}{x_3-\frac{x_1+x_2}{2}}=\frac{2y_3-(y_1+y_2)}{2x_3-(x_1+x_2)}$
This line has an equation, which is determined as follows:
$y=\frac{2y_3-(y_1+y_2)}{2x_3-(x_1+x_2)}x+b$, where $b$ is determined as follows:
$\\y_3=\frac{2y_3-(y_1+y_2)}{2x_3-(x_1+x_2)}x_3+b\\b=y_3-\frac{2y_3-(y_1+y_2)}{2x_3-(x_1+x_2)}x_3\\\implies \boxed{y=\frac{2y_3-(y_1+y_2)}{2x_3-(x_1+x_2)}x+y_3-\frac{2y_3-(y_1+y_2)}{2x_3-(x_1+x_2)}x_3}$.
The slope of the line connecting $(x_1,y_1)$ and $\left(\frac{x_2+x_3}{2},\frac{y_2+y_3}{2}\right)$ is
\[ \\m=\frac{\frac{x_2+x_3}{2}-x_1}{\frac{y_2+y_3}{2}-y_1}\\=\frac{x_2+x_3-2x_1}{y_2+y_3-2y_1} \].
The equation of the line connecting $(x_1,y_1)$ and $\left(\frac{x_2+x_3}{2},\frac{y_2+y_3}{2}\right)$ is
\[ y=\frac{x_2+x_3-2x_1}{y_2+y_3-2y_1}x+b, \]
where $b$ is determined as follows:
$\\y=\frac{x_2+x_3-2x_1}{y_2+y_3-2y_1}x+b\\ y_1=\frac{x_2+x_3-2x_1}{y_2+y_3-2y_1}x_1+b\\ \implies b=y_1-\frac{x_2+x_3-2x_1}{y_2+y_3-2y_1}x_1\\ \implies \boxed{y=\frac{x_2+x_3-2x_1}{y_2+y_3-2y_1}x+y_1-\frac{x_2+x_3-2x_1}{y_2+y_3-2y_1}x_1}$

The centre of the triangle is then the intersection of $\\\boxed{y=\frac{x_2+x_3-2x_1}{y_2+y_3-2y_1}x+y_1-\frac{x_2+x_3-2x_1}{y_2+y_3-2y_1}x_1\text{ and }\\ y=\frac{2y_3-(y_1+y_2)}{2x_3-(x_1+x_2)}x+y_3-\frac{2y_3-(y_1+y_2)}{2x_3-(x_1+x_2)}x_3}$.

$\mathbb{QED}$

Hopefully, I do not have any typos or small mistakes. :)

Masoud Zargar[/quote]


inequalities
function
vector
calculus
derivative
induction
inequalities theorems

Hi there,

Here comes another one that is making my mind colapse.

Show that , if Xi≥0, i = 1,2...n , then:

1/nΣXi ≥ (ΠXi)1/n), Π represents the product of Xi's.

In other words , show that the arithmetical mean for a set of nonnegative numbers is always greater than their geometrical mean.

Clue: Consider the problem:

Minimize ΣXi subject to ΠXi =c where c is a constant and Xi ≥0 with i = 1,2...n

This is a very useful inequality, and its proof can be found for example in [url=http://planetmath.org/encyclopedia/ProofOfArithmeticGeometricHarmonicMeansInequality.html]Planetmath[/url].

[quote="henrik"]You have left out something important it seems.

1/2(2+3) = 2.5
1/2(2*3) = 3

You mean if each 0 <= xi <= 1 ?
.[/quote]
Thankyou for the answer you lead me to the solution. In the problem I mistyped the power signal in the geometric mean, "  1/nΣXi ≥ (ΠXi) ^ 1/n), Π represents the product of Xi's."

So for instance 1/2(2+3) = 2.5 > (2*3)^1/2 = 2.44 and the inequality holds for all instances.

Regards
Claudio

Yes, I realize now what you meant. I should have guessed.

The proof goes through largely as before. Consider the hyperplane a(x) = c'. Decreasing c' means moving the plane parallell. Do so until you are about to leave the surface S. At that point the plane must be tangent to S.

The calculation of that point remains as above. You need to be a little careful, though. It's not always true that a linear function has a minimum on a manifold. Consider the curve y = 1/x and the linear function y. y = c either intersects the curve or is disjoint from it; it is never tangent to it.

You must reason as follows: a(x) is positive on the surface so has an infimum c0. Let c0 < c1. Let C be the set bounded by a(x) = c1 and xi = 0 i =1,2,...,n. It is compact. a(x) attains its infimum on S if and only if it attains it on S ^ C (intersection). But the latter is compact so it does indeed attain it. A minimal point must be critical and you compute as above. I'm sure you've found the solution on your own. I only wanted to clean
this up.

[quote="henrik"]  .[/quote]

Thanks again Henrik.

I have another doubt in a problem similar to this one. I've already  posted it on this forum, but unfortunattely, so far, I had no answer, probably you can help me. Here it comes:

Consider x* as the solution of :

minimize Σfi(xi) subject to Σxi=1 with xi ≥ 0 , i =1,2...,n

consider also that ζi= δfi(x*)/δxi, where δ represents the partial derivative operator.

Show that , there is a scalar k such that ζi≥k and (ζi-k)x*=0 , i= 1,2....n.

I am clueless

Hmmm. Do you mean to say that you've got functions f1,...,fn : Rn -> R? And you want to minimize f1(x1,...,xn)+...+fn(x1,...,xn)?

I'm mystified by the conclusion: (zeta(i) - k)x* = 0 ??

x* is a vector in Rn, right? You can't make it 0 by multiplying with a scalar other than 0.  I must have misunderstood something. Probably some inner product is meant.

[quote="henrik"]Hmmm. Do you mean to say that you've got functions f1,...,fn : Rn -> R? And you want to minimize f1(x1,...,xn)+...+fn(x1,...,xn)? .[/quote]
No, I want to minimize f1(x1)+f2(x2)+f3(x3)...fn(xn) subject to x1+x2+x3+...+xn=1. considering that all xs are greater or equal than zero.

[quote="henrik"]

I'm mystified by the conclusion: (zeta(i) - k)x* = 0 ??

x* is a vector in Rn, right? You can't make it 0 by multiplying with a scalar other than 0.  I must have misunderstood something. Probably some inner product is meant.[/quote]

You are right, I forgot the index "i" in the vector, the correct is  (zeta(i) - k)x(i)* = 0.

Where x* is the point of mínimum and x(i)* is its ith element, zeta(i) seems to be the ith element of the gradient of f(.) in the point x(i)*.

Is it possible to solve it now, with these definitions?

Yes, now I think I can solve it.

Put g(x1,...,xn) = f1(x1)+...+fn(xn)

Consider the surface g(x) = c. Decreasing c we will sooner or later be about to leave the plane x1+...+xn = 1, i.e. no intersection any more for xi >= 0. We have called the point where that happens x*. If x* is an interior point of the plane bounded by xi >= 0 then the gradient of g at x* must be parallell to the plane normal:

(dg/dx1(x*),...,dg/dxn(x*)) = k(1,...,1)

But dg/dxi(x*) = dfi/dxi(x*) = zeta(i), so in this case we're through: zeta(i) = k => (zeta(i)-k)x*(i) = 0   i=1,..,n.

The problem is that x* may lie on the boundary. That happens if and only if some x*(i) = 0. Let I be the subset of {1,...,n} for which this happens. In I the conclusion is trivially satisfied.

Now, fix xi at 0 for i in I and solve the problem for the remaining n-#I variables. Then you get an interior point just like above and the result is proved.

[quote="henrik"]You have left out something important it seems.

1/2(2+3) = 2.5
1/2(2*3) = 3

You mean if each 0 <= xi <= 1 ?

The equation g(x) = x1x2...xn = c determines a hypersurface in Rn. We consider a(x) = x1+...+xn a function on this surface. At a minimal point its gradient vector in Rn must project to 0 in the tangent plane on the surface, i.e. it must be parallell to the surface normal, which is given by the gradient vector of g at the point.

The gradient of a is just the uniform (constant) vector field (1,1,...,1). That of g has i'th component x1x2...^xi....xn which means that xi has been omitted.

For these to be parallell we must have x1x2...^xi....xn = k for some k != 0. But then xi/xj = 1 for all i,j i.e. xi = xj = x .

At such a point a(x) = nx while g(x) = x^n. But for 0 <= x <= 1 nx >= x >= x^n .

Therefore we only need to give one example of a non-critical point where the inequality holds. Take x1 = 1 and x2=...=xn=0. a(x) = 1 while g(x) = 0.

This technique is called Lagrange multipliers.[/quote]
How about
$x+y=x-z+y+z$ but $xy<(x-z)(y+z)$ if $y<x$. Now use induction on the number of variables.


algebra
polynomial
function
limit
complex analysis
advanced fields
advanced fields unsolved

Let $f$, $g$ be two entire functions such that \[ \lim_{z\to \infty} f(g(z))=\infty. \]
Prove that both $f$, $g$ are polynomials.

Hi,...
well it is a consecuence of Casorotti Weierstrass theorem that says if an entire function $f$ is not a polinomial then $f(\mathbb{C}-K)$ is dense in $\mathbb{C}$ for any compact set $K$. (in fact is because $f$ has a esential singularity at infinity) it is easy to prose using the maximun principle.
About the problem if $\lim_{z\rightarrow\infty}f(g(z))=\infty$ exists an $M>0$ such that
\[|z|>M\Longrightarrow |f(g(z))|>1.\]

If $g\notin\mathbb{C}[z]$ then $g(\{z>M\})$ must be dense, which is a contradiction, then $g$ must be in $\mathbb{C}[z]$. for $f$ is analogous.


I formulated this question on my own and lets see if you can get it, its more mathematical than anything else.

Consider a particle of mass m and charge q at the origin of co-ordinates. An electric field $ \vec E \equal{} E_0\vec i$ is present alongwith a magnetic field $ \vec B \equal{} B_0 \vec j$. the particle is projected with a velocity $ \vec v \equal{} v_0(\vec j \plus{} \vec k)$ at time t=0. Also, the particle is subjected to a force, which varies with the y co-ordinate as $ \vec F \equal{} \minus{} F_0y \vec j$. Trace out the trajectory of the particle. Assume $ E_0$ is very small.

The most fun part of this question is that you need not even write down equations, but just try and draw out the path the particle follows as time passes. It is really very interesting.
 :)

hey hell ever. i am only a 11ther. I have no idea abt magnetic and electric fields. :(
Anyways ur question is rockin particularly as no one was able to solve it. :rotfl:


analytic geometry
topology
real analysis
real analysis unsolved

Let $ E\equal{}\{x_n : x_n \in l_2  \mbox{and every coordinate for} x_n \mbox{is rational}\}$. $ \dim E\equal{}?$

Depends on what $ \dim$ means. There are dozens of ways to define "dimension" of something. I tend to use the Hausdorff dimension most often: it is equal to $ 0$ for $ E$ since $ E$ is countable. On the other hand, as a linear space over $ \mathbb Q$, $ E$ is infinite dimensional.

Yes.
a) [b]Hausdorff[/b] dimension
b) [b]Minkowski[/b] dimension.
c) [b]A topological space[/b]  has at most  $ k$ dimension when if it has a basis ,where boundary of every element in this basis has  at most $ k\minus{}1$ dimension.

Minkowski dimension is infinite. Indeed, the set of rational points in $ \mathbb R^n$ has Minkowski dimension $ n$. You can't cover rational points with balls of equal size without covering pretty much everything.

Topological dimension is zero, since the set is totally disconnected.

I think topological dimension is not zero because not exits a basis with open-closed elements.


calculus
integration
function
inequalities
real analysis
real analysis unsolved

[color=green] [b] If $f: [a,b] \to [0,\infty)$ is convex on $[a,b] ,$ then for all $x,y$ such that  $a\le x<y\le b$ the inequality
\[ \begin{array}{|c|} \hline \\ \displaystyle \left(\frac{1}{y-x}\int\limits_{x}^{y}\left|f(t)\right|^2 \; dt \right)^{\frac{1}{2}} \le \displaystyle \left(\frac{ f^{\frac{4}{3}}(x)+ f^{\frac{4}{3}}(y)}{2}\right)^{\frac{3}{4}} \\ \\ \hline \end{array}  \]
is verified. [/b] [/color]

Remark $|f(x)|=f(x)$. Take $g(x)=f^2(x)$. We have that $g$ is also convex. The ineq can be written:

\[ \frac{1}{y-x}\int_x^y g(t)dt \le \left(\frac{g(x)^{\frac 2 3}+g(y)^{\frac 2 3}}{2}\right)^{\frac{3}{2}} \]
1)From power means ineq we can prove that:
\[ \frac{g(x)+g(y)}{2}\le \left(\frac{g(x)^{\frac 2 3}+g(y)^{\frac 2 3}}{2}\right)^{\frac{3}{2}} \]

2)On the other hand we have that
\[ I=\int_x^yg(t)dt=\int_x^yg(x+y-t)dt \Rightarrow I=\int_x^y\frac{g(t)+g(x+y-t)}{2}dt\le \int_x^y\frac{g(x)+g(y)}{2}dt=(y-x)\frac{g(x)+g(y)}{2} \]

The last ineq hold since $g$ is convex and ,so , $g(x)+g(y)\ge g(x+t)+g(y-t) \Leftrightarrow g(x+t)-g(t) \le g(y-t+t)-g(y)$. Which comes from the monotonicity of  $h(x)=g(x+t)-g(x)$.

From thesetwo inequalities the requested one follows.

but the inequality that you have use with power mean isn't correcte , because 1>2/3 ( so it's inversed )

I see:D Here's a start for good solution:

Take the function $g(t)=at+b$ such that $g(x)=f(x)$ and $g(y)=f(y)$. f being convex we have that $f\le g\Rightarrow \int_x^yf^2(t)dt \le \int_x^yg^2(t)dt$. So it is sufficient to prove that
\[ \left(\int_x^yg^2(t)dt\right)^{\frac{1}{2}} \le \left(\frac{g(x)^{\frac{2}{3}}+g(y)^{\frac{2}{3}}}{2}\right)^{\frac{3}{2}} \]
Finding a and b, this one can be reduced to an algebric inequality only in $g(x),g(y)$. Which I'm sure that will hold :D  because f can get very close to g. I will try to finish it tomorow.

[quote="flip2004"][color=green] [b] If $f: [a,b] \to [0,\infty)$ is convex on $[a,b] ,$ then for all $x,y$ such that  $a\le x<y\le b$ the inequality
\[ \begin{array}{|c|} \hline \displaystyle (\frac{1}{y-x}\int\limits_{x}^{y}|f(t)|^2 \; dt )^{\frac{1}{2}} \le \displaystyle (\frac{ f^{\frac{4}{3}}(x)+ f^{\frac{4}{3}}(y)}{2})^{\frac{3}{4}} \\ \hline \end{array}  \]
is verified. [/b] [/color][/quote]
[b]SOLUTION.[/b]  For  $a\le x<y\le b$ let  $L(t): =\frac{t-x}{y-x}f(y)+\frac{y-t}{y-x}f(x) \; ,\; t\in {\mathbb R}$ so that the set $\{(t,z)\in {\mathbb R}^2\; ; t\in [a,b] \; , z=L(t) \}$  is the chord,   of the graph of  $f$  , determined by the points   $(x,f(x))$ and       $(y,f(y)) .$ Because $f\ge 0$ observe that
\[ (1)\; \; \; \; \;f^2(t)\le L^2(t) \; ,\; \forall t\in[a,b] .  \]
 Also
\[ (2)\; \; \; \; \; \frac{1}{y-x}\int\limits_{x}^{y} L^2(t)\; dt=\frac{f^2(x)+f(x)f(y)+f^2(y)}{3}\; ,\; (a\le x<y \le b ).  \]
Note that  $(\frac{1+z^3 +z^6}{3})^2\le (\frac{1+z^4}{2})^3\; ,\; \forall z\in [0,\infty).$ In order to prove this inequality, remark that the difference
  RHS - LHS = $: D(z)$ is $D(z)=(z-1)^4 (z^8+4z^7+10z^6+3z^5 +z^3(z-1)^2 + 3z^3+10z^2+4z+1).$
  Select   $z=\displaystyle (\frac{f(y)}{f(x)})^{\frac{1}{3}} \; ,\; f(x) \ne 0 .$ One finds
\[ (*)\; \; \; \; \; (\frac{f^2(x)+f(x)f(y)+f^2(y)}{3})^{\frac{1}{2}}\le \frac{f^{\frac{4}{3}}(x)+ f^{\frac{4}{3}}(y) }{2})^{\frac{3}{4}}.  \]
From (1)--(2) and (*)   we conclude with
\[ \frac{1}{y-x}\int\limits_{x}^{y}|f(t)|^2\; dt\le \frac{1}{y-x}\int\limits_{x}^{y}|L(t)|^2\; dt\le \displaystyle (\frac{ f^{\frac{4}{3}}(x)+ f^{\frac{4}{3}}(y)}{2})^{\frac{3}{4}}  \]
which completes the solution.


geometry
3D geometry
sphere
function
vector
complex numbers
absolute value

Here is a very nice problem: take $x_{n}$ a sequence on the unit sphere in $R^{2}$ and let $X$ be the union of the closed segments connecting the origin and $z_{n}$. Prove that any continuous function from $X$ to $X$ has a fixed point. And if the family of points $z_{n}$ is no longer countable?

As a slight generalization, we can show that this set has the fixed point property whenever $\{x_{i}\ |\ i\in I\}$ has empty interior (as a subset of the unit circle):

Let $0$ be the origin of the plane (all of this works just as well in any dimension), and let $f: X\to X$ be a continuous map. If $f(0)=0$, we're done. Otherwise, $f(0)$ belongs to a unique half-open segment $(0,x_{i_{0}}]$. Because of our assumption on the set $\{x_{i}\}$, the path connected component of $X\setminus\{0\}$ to which $f(0)$ belongs is $(0,x_{i_{0}}]$. There are now two possibilities:

(a) There is a smallest $t\in(0,1)$ such that $f(tx_{i_{0}})=0$.

In this case, for $0\le s\le t,\ f(sx_{i_{0}})\in[0,x_{i_{0}}]$. The map $s\mapsto f(sx_{i_{0}})-sx_{i_{0}}$ defined on $[0,t]$, whose values are vectors along the line $0x_{i_{0}}$, takes values of opposite orientations in $0$ and $t$, so it must have the value $0$ for some $s\in[0,t]$. For this $s$ we'll have $f(sx_{i_{0}})=sx_{i_{0}}$.

(b) There are no $t$ such as in (a).

In this case the restriction of $f$ to $I=[0,x_{i_{0}}]$ is a self-map of $I$, and since segments have the fixed point property, we are, again, done.

In order to get a counterexample to the second question, take, for example, our set $\{x_{i}\}$ to consist of the complex numbers of absolute value $1$ with arguments in the interval $[0,\pi)$. Topologically, $X$ will be the same as a closed square (together with the enclosed area) with a deleted edge, and it's easy to see that this space does not have the fixed point property.


Find the possible least value of $ x^4y^2\plus{}x^2y^2\plus{}8x^2y\plus{}x^2\plus{}2008$

$ x,y\in \mathbb {R}$

:P

[quote="vinskman"]Find the possible least value of $ x^4y^2 \plus{} x^2y^2 \plus{} 8x^2y \plus{} x^2 \plus{} 2008$

$ x,y\in \mathbb {R}$

:P[/quote]
$ x^4y^2 \plus{} x^2y^2 \plus{} 8x^2y \plus{} x^2 \plus{} 2008 \equal{} x^2y^2 \plus{} 2x^2y \plus{} x^2 \plus{} x^4y^2 \plus{} 6x^2y \plus{} 9 \plus{} 1999 \equal{} x^2(y \plus{} 1)^2 \plus{} (x^2y \plus{} 3)^2 \plus{} 1999\ge 1999$
Equality when $ y \equal{} \minus{} 1,x \equal{} \sqrt {3}$


geometry
incenter
angle bisector
geometry unsolved

P is a point that in the triangle ABC, ∠APB-∠ACB=∠APC-∠ABC , and D is the incenter of APB.E is the incenter of APC.
 Proof that AP,BD,CE all intersect at one point

I'm puzzled at how to use the condition

Let $X,Y,Z$ be the orthogonal projections of $P$ onto $BC,CA,AB.$ Then

$\angle APB=\angle ACB+\angle XZY \ , \ \angle APC=\angle ABC+\angle XYZ$

$\angle APB-\angle ACB=\angle APC-\angle ABC \Longrightarrow \ \angle XZY=\angle XYZ$

Therefore, pedal triangle $\triangle XYZ$ of $P$ with respect to $\triangle ABC$ is isosceles with apex $X$ $\Longrightarrow$ $P$ lies on the A-Apollonius circle of $\triangle ABC$ $\Longrightarrow$ $\frac{_{AC}}{^{PC}}=\frac{_{AB}}{^{PB}}.$ By angle bisector theorem, it follows that bisectors $BD,CE$ of $\angle ABP$ and $\angle ACP$ cut $AP$ at the same point.


HCSSiM
Columbia
geometry
geometric transformation
percent
MATHCOUNTS

hehe, first time in the Chinese thingy thing...

 Kristy, I think u know the answer to this but... Wuxi, China, and the second one Vestavia Hills, Alabama, US....

heheheh have fun

我生在美国，也?在美国。但是我父?生在中国。

因为Valentin Vornicu说我们因该用中文谈?, 所以我把我以?的回答用中文?讲一??,

我在中国, 无锡(江??) 出生, 现在?在 Vestavia, Birmingham, Alabama (我?知?怎么用中文说 Vestavia, Birmingham, Alabama)

Hehe... I am not really Chinese but you know how Vietnamese and Chinese got to stick together right? I just want to meet some asian mathelete friends since there are not so many in my school.

Oh yeah.. and I know Nanjing. We just learned about that place in history  :P

NJ to both of them.

About your avatar, I can only say one thing.

omfg...

[quote="KBabe"]yea, i  did, lucy.

anyways, i was born in Nanjing, China
and i''m in Vestavia Hills, AL[/quote]

我生在南京(江苏), 现在使在vestavia hills, AL.

[quote="RC-7th"]我生在美国，也?在美国。但是我父?生在中国。[/quote]

你的中文怎么学的??

Beijing...

Houston (I mean the polluted place)...

sry on a labtop rite now can't type chinese characters...

:D 我是江??镇江市丹阳县人，出生在常州市，现在在?州读高中。
我是?自农?的 :D

:D 楼上的?志都是海外?人?？
我想?人都一定?学会且学好汉语，这一点犹太人是我们汉?人的榜样

我想我?算是海外?人?..  :D  香港在1997年早已回歸了..
qxy，請?你幾年級？

我生在长春(?林?), 现在?在Acton, MA, USA.

我的数学很差 :(...

我出生在纽约，现在住在弗罗里达。我父母是中国人。

I was born in america, but my parents came from Hong Kong I live in New york now and always

I was born in New York City, but I live in Hong Kong now. :D

I was born in Taiwan and I still live here :)

i was born in the town next door to the town i live in right now...
i'm half chinese though :)

America. That doesn't mean I'm not Chinese though...

I live on the other side of the world from where I am born

[quote=AlastorMoody][quote=KBabe]where were u born? where are u now??[/quote]

I was born at hospital and right now I am on Earth[/quote]

XD

[quote=steven2015]IM GONNA REPORT THIS THREAD FOR ASKING PRIVATE INFO[/quote]

This only asks for info the users are willing to share. If this thread makes you uncomfortable, you are welcome to leave.

[quote=steven2015]IM GONNA REPORT THIS THREAD FOR ASKING PRIVATE INFO[/quote]

This is probably older than you...

我在广州出生，现在在马里兰上学


The sum of the squares of two positive integers is 193. The product of the two integers is 84. What is the sum of the two integers?

You can approach this problem two ways.

[b]Method I: Solve a system of equations. [/b]

$ xy\equal{}84$
$ x^2\plus{}y^2\equal{}193$

Substituting the first equation into the second, we get $ x^2\plus{}(\frac{84}{x})^2\equal{}193$.
$ x^2\plus{}\frac{7056}{x^2}\equal{}193$
$ x^4\plus{}7056\equal{}193x^2$
$ x^4\minus{}193x^2\plus{}7056\equal{}0$

The two positive integer solutions for $ x$ are $ 12$ and $ 7$, and $ y$ must be the other, and $ 12\plus{}7\equal{}\fbox{19}$.

[b]Method II: Find factors of $ 84$ with both squares less than 193. [/b]

The factor pairs for $ 84$ are:
$ (1,84)$
$ (2,42)$
$ (3,28)$
$ (4,21)$
$ (6,14)$
$ (7,12)$

The last case, $ (7,12)$, is the only case for which both squares are less than $ 193$. In $ (6,14)$, $ 14^2\equal{}196>193$, and in all other cases, the larger square is even greater. Thus, the pair that in being sought must be $ (7,12)$, and $ 7\plus{}12\equal{}\fbox{19}$.

Better yet:

We have the equations $ a^2 \plus{} b^2 \equal{} 193$ and $ ab \equal{} 84$. The first equation factors as $ (a \plus{} b)^2 \minus{} 2ab \equal{} 193$, so we can substitute $ 84$ for $ ab$. This gives us $ (a \plus{} b)^2 \equal{} 361$, so $ a \plus{} b \equal{} \boxed{19}$.


[color=darkblue]Let $ABC$ be a triangle for which $a>b$ and $a>c.$ Prove that :

$\boxed{\ A=90^{\circ}\Longleftrightarrow (\sqrt{a+b}+\sqrt{a-b})\cdot (\sqrt{a+c}+\sqrt{a-c})=2p\sqrt 2\ }\ ,$

where ${2p=a+b+c\ }\ .$[/color]

[hide="Not sure what you mean by $p$, buy here is what I got:"] \[\left(\sqrt{a+b}+\sqrt{a-b}\right)^{2}=a+b+a-b+2\sqrt{a+b}\sqrt{a-b}=2a+2\sqrt{a^{2}-b^{2}}\\ =2a+2c\] Similarly for other term, we arrive at: \[(\sqrt{a+b}+\sqrt{a-b})\cdot (\sqrt{a+c}+\sqrt{a-c})=2\sqrt{a+b}\sqrt{a+c}=2\sqrt{a^{2}+\sum_{cyc}ab}\] Now we use a trick: \[a^{2}+\sum_{cyc}ab=a^{2}+\frac{(a+b+c)^{2}-a^{2}-b^{2}-c^{2}}{2}, b^{2}+c^{2}=a^{2}\implies \\ \implies a^{2}+\frac{(a+b+c)^{2}-a^{2}-b^{2}-c^{2}}{2}=a^{2}+\frac{(a+b+c)^{2}-2a^{2}}{2}=\frac{(a+b+c)^{2}}{2}\implies \\ \implies (\sqrt{a+b}+\sqrt{a-b})\cdot (\sqrt{a+c}+\sqrt{a-c})=2p\sqrt 2\ = 2\sqrt{a+b}\sqrt{a+c}= 2\sqrt{a^{2}+\sum_{cyc}ab}= 2\sqrt{\frac{(a+b+c)^{2}}{2}}= \boxed{(a+b+c)\sqrt{2}}\] Nice but really easy problem, how did you dream it up?[/hide]

[color=darkblue]And reciprocally, i.e. $\Longleftarrow\ ?$
[hide="Only for Me@home."]You are from Portland, I live in Yakima, Wa., at my daughther and two grandsan ;
therefore, I gladden, we are neighbours ! After three weeks, I"ll return in my country, Romania.[/hide]This problem is mine. And my solution has one trick, but which works with the chain of equivalencies !

[b]Proof.[/b] Denote $2p=a+b+c\ ,\ 2\Delta=b^{2}+c^{2}-a^{2}$ and $E=(\sqrt{a+b}+\sqrt{a-b})(\sqrt{a+c}+\sqrt{a-c})\ .$
Prove easily that the following interesting chain of identities (here is the [b][u]trick[/u][/b]):
$\boxed{\ 2p^{2}-(a+b)(a+c)=2(p-a)^{2}-(a-b)(a-c)=2(p-b)^{2}-(a-b)(a+c)=2(p-c)^{2}-(a+b)(a-c)=\Delta\ }\ \ (*)\ .$
The  relations $(*)$ solve immediately the our proposed problem !
$1.\blacktriangleright \Delta<0\Longrightarrow \sqrt{(a+b)(a+c)}> p\sqrt 2\ ,\ \sqrt{(a-b)(a-c)}>(p-a)\sqrt 2$ a.s.o $\Longrightarrow E> 2p\sqrt 2\ .$
$2.\blacktriangleright \Delta>0\Longrightarrow \sqrt{(a+b)(a+c)}<p\sqrt 2\ ,\ \sqrt{(a-b)(a-c)}<(p-a)\sqrt 2$ a.s.o. $\Longrightarrow E<2p\sqrt 2\ .$
$3.\blacktriangleright \Delta=0\Longrightarrow \sqrt{(a+b)(a+c)}=p\sqrt 2\ ,\ \sqrt{(a-b)(a-c)}=(p-a)\sqrt 2$ a.s.o. $\Longrightarrow E=2p\sqrt 2\ .$[/color]


algebra
polynomial
search
linear algebra
linear algebra unsolved

$ A\in M_{2}(Z)$  and for one $ n$ , $ A^n\equal{}I_{2}$.Prove that exist $ 1 \le k \le 6$ s.t. $ A^k\equal{}I_{2}$

thx Moubinool , but can you be more explicit pls ?

$ A$ satisfies the polynomial $ A^n \minus{} I_2 \equal{} 0$, hence its minimal polynomial must be a cyclotomic polynomial $ \Phi_d(x)$ where $ d | n$; this polynomial divides its characteristic polynomial, which is of degree at most $ 2$, so $ d \equal{} 1, 2, 3, 4, 6$.

In a book I found this solution: the eigenvalues
$ x_{1}$ and $ x_{2}$ of $ A$ are solution of ecuation $ x^n \equal{} 1$ so $ x_{1},x_{2} \in ( {1, \minus{} 1} )$ and continues with : n odd and even etc.What i dont' understand is why $ x_{1},x_{2} \in ( {1, \minus{} 1} )$....why  $ x_{1}$ and  $ x_{2}$ can not be any other unit roots?

Edit :thx t0rajir0u and Moubinool...now I understand the solution with cyclotomic polynomial !

You can also solve it if you don't know what a cyclotomic plynomial is. Here is the sketch of an elementary proof.
$ A$ is diagonalizable in $ \mathcal{M}_2(\mathbb{C})$
denote $ a,b$ its eigenvalues, $ a^n\equal{}b^n\equal{}1$, and $ \mbox{tr}(A)\in \mathbb{Z}$, so $ \mbox{tr}(A)\in \{\minus{}2,\minus{}1,0,1,2 \}$.
Then you treat each of these 5 cases, and with a bit of calculation you conclude.
Here is a post related to that, with a nice article posted by harazi http://www.artofproblemsolving.com/Forum/viewtopic.php?search_id=18757968&t=100736

As ZetaX might ask, what makes you categorize cyclotomic polynomials as non-elementary?  :P

You are right.

Less elementary than the above I posted if more accurate.


Cunoaste cineva un exemplu de functie derivabila f:I->R astfel incat

             f(xy+yz+zx)=f(x)+f(y)+f(z)  pentru orice x,y,z din I ?

   Aici I este un interval , de preferinta cat mai mare, eventual chiar R.

Ce zici de $f(x)=0$ pentru orice $x\in I$? :)

Am uitat sa precizez ca functia trebuie sa fie neconstanta.
 O asemenea functie ar fi pentru mine un fel de analog al logaritmului(acesta transforma
 produsele in sume). Mai general, m-ar fi interesat o functie f:I-R ,neconstanta desigur, care sa transforme
 S_k(x_1,x_2,....x_n) in S_p(f(x_1),f(x_2),....f(x_n)) unde S_k este suma simetrica de ordinul k.
In particular, logaritmul transforma S_n in S_1.

Daca o asemenea functie ar avea o forma suficient de simpla ar putea avea destule aplicatii practice.

Imi cer scuze ca nu am folosit latexul,sper sa invat cat de curand sa-l utilizez.

Pentru $x=y=z=0$ obtinem $f(0)=0$. Pentru $z=0$ rezulta $f(xy)=f(x)+f(y)$, si, de aici, $f$ e logaritmica. Dar atunci nu e verificata relatia initiala.


(a)Suppose there exist an infinite set of positive integers such that,for any finte nonempty subset $ T$ of $ S$,there exists an integer $ >1$ which divides every element of $ T$.Show that there exists an integer $ >1$ which divides every element of $ S$.

(b)Let $ k$ be a fixed positive integer.Show that there exists an finite set $ S$ satisfying the following condition:
i)every subset $ T$ of $ S$ having at most $ k$ elements has a common divisor $ >1$
ii)no integer $ >1$ divides every element of $ S$

I really need your help.Thanks.

(a) Let $ a>1$ be fixed in $ S$. Then $ a$ has a finite number of prime divisors, say $ p_1, \cdots, p_n$.
Assume that none of the $ p_i$'s divides each element of $ S$ :
Then, for each $ i$, we may find $ x_i \in S$ such that $ p_i$ does not divide $ x_i$.
It is easy to verify that $ T \equal{} \{a,x_1, \cdots, x_n\}$ is a finite subset of $ S$ such that there is no common divisor of all the elements of $ T$. A contradiction.

(b) Let $ p_1, \cdots, p_{k\plus{}1}$ be pairwise distinct primes. Let $ P \equal{} p_1p_2 \cdots p_{k\plus{}1}$ and $ P_i \equal{} \frac P {p_i}$.
Let $ S$ be the set of positive integers which are divisible by one of the $ P_i$'s but not by $ P$, so that each element of $ S$ is divisible by exactly one of the $ P_i$'s.
Note that $ S$ contains each of the $ P_i$'s so that there  is no common divisor to all elements in $ S$, which proves the second desired condition.
Now, let $ T$ be a nonempty subset of $ S$ with at most $ k$ elements. Since each of them is divisible by exactly one of the $ P_i$'s and because there are $ k\plus{}1$ numbers $ P_i$, it follows that at least one of the $ P_i$'s, say $ P_1$, has no multiple in $ T$. Therefore each element of $ T$ is divisible by $ p_1$ and we are done.

Pierre.


The noble race of 1337ians have one flaw: They feel the need to write every rational number in the form $ \frac{1337}{x}$, where $ x$ is a common fraction or an integer. Let the irritating-to-writeness of a representation of a number be the number of digits it takes to write it down. A random rational number such that the numerator and denominator are both positive integers less than or equal to 10. Let $ p$ be the irritating-to-writeness as a common fraction and $ s$ be the irritating-to-writeness in the 1337ian way. What's the probability that $ p\minus{}s\ge 7$?

Er, sorry, I meant $ s\minus{}p$, not $ p\minus{}s$.


Prove that gcd(4012a+1,4012b+1)=1 for natural numbers a,b such that $ a>b>1$

[quote="thegod277"]Prove that gcd(4012a+1,4012b+1)=1 for natural numbers a,b such that $ a > b > 1$[/quote]
it is not true...
for example if $ a\equal{}7,b\equal{}2$ then both of them are divisible by $ 5$...


vector
LaTeX
topology
integration
Functional Analysis
real analysis
real analysis unsolved

1. Let $ f$ be a compact vector field on a closed ball $ B'(0,r)$ in a Hilbert space $ H$ such that $ \langle f(x),x \rangle > 0,\forall x, \|x\| \equal{} r$
Then there is $ u$ in $ B(0,r)$ such that $ f(u) \equal{} 0$.

2. Let $ D$ be an open bounded subset with smooth boundary in $ R^3$ and $ g\in L^3(D)$. Then there is a weak solution in $ W^{1,2}_0(D)$ to the following equation $ \minus{} \triangle u \plus{} \frac {1}{4}u^4 \equal{} g$.

Unsolicited latex suggestion: "\langle a, b \rangle" gives $ \langle a, b \rangle$, a much prettier inner product than "< a , b >" ($ < a , b >$).  You can get the same effect (at least in the forum) by using "\left< a, b \right>".

I don't understand your idea, please tell me more specific, thanks

My idea has nothing to do with your problem, it has to do with LaTeX (your formatting): I suggested two ways you could avoid producing ugly-looking expressions like $ <f(x),x>>0$.  (Also, in that same formula, you wrote "\forallx" -- you need a space ("\forall x") if you want LaTeX to interpret it correctly.)

I just edited JBL's suggestions, along with one other (\|x\| for the norm of x) into kidmath's original post.

No one has said anything substantive yet.

[quote="kidmath"]1. Let $ f$ be a compact vector field on a closed ball $ B'(0,r)$ in a Hilbert space $ H$ such that $ \langle f(x),x \rangle > 0,\forall x, \|x\| \equal{} r$
Then there is $ u$ in $ B(0,r)$ such that $ f(u) \equal{} 0$.[/quote]

We now prove the following auxiliary lemma.

[b]Lemma.[/b] Let $ H$ be a Hilbert space and $ F$ a compact operator ona bounded open set $ \Omega \subset H$  into $ H$. Assume that $ o \in \Omega$ and
\[ \left\langle {g\left( x \right),x} \right\rangle  \leqslant \left\| x \right\|^2
\]
for each $ x \in \partial \Omega$. Then $ g$ has a fixed point in $ \overline \Omega$.

[b]Proof.[/b] Suppose not and there is no $ t_0 \in [0,1]$ and $ x_0 \in \partial \Omega$ such that $ x_0 \equal{} t_0 g(x_0)$ then just consider
\[ G\left( {t,x} \right): \equal{} x \minus{} tg\left( x \right).
\]
By the homotopy invariance property
\[ \deg \left( {I \minus{} G,B\left( {o,r} \right),o} \right) \equal{} \deg \left( {I,B\left( {o,r} \right),o} \right).
\]
Since
\[ \deg \left( {I,B\left( {o,r} \right),o} \right) \equal{} 1
\]
then
\[ \deg \left( {I \minus{} G,B\left( {o,r} \right),o} \right) \equal{} 1
\]
which proves that
\[ x \minus{} g\left( x \right) \equal{} o
\]
has a solution $ x \in \overline \Omega$, a contradiction.

Now there is some $ t_0 \in [0,1]$ and $ x_0 \in \partial \Omega$ such that $ x_0 \equal{} t_0 g(x_0)$. By assumption
\[ \left\| {x_0 } \right\|^2  \equal{} \left\langle {x_0 ,x_0 } \right\rangle  \equal{} \left\langle {t_0 g\left( {x_0 } \right),x_0 } \right\rangle  \equal{} t_0 \left\langle {g\left( {x_0 } \right),x_0 } \right\rangle  \leqslant t_0 \left\| {x_0 } \right\|^2
\]
Thus $ t_0\equal{}1$. That is to say $ x_0\equal{}g(x_0)$.

In order to prove your problem, consider
\[ g\left( x \right) \equal{} x \minus{} f\left( x \right)
\]
then
\[ 0 \leqslant \left\langle {f\left( x \right),x} \right\rangle  \equal{} \left\langle {x \minus{} g\left( x \right),x} \right\rangle  \equal{} \left\| x \right\|^2  \minus{} \left\langle {g\left( x \right),x} \right\rangle .
\]
OK?

[quote="kidmath"]2. Let $ D$ be an open bounded subset with smooth boundary in $ R^3$ and $ g\in L^3(D)$. Then there is a weak solution in $ W^{1,2}_0(D)$ to the following equation $ \minus{} \triangle u \plus{} \frac {1}{4}u^4 \equal{} g$.[/quote]

Define
\[ \begin{gathered}
  \left\langle {F\left( u \right),v} \right\rangle  \equal{}  \minus{} \frac{1}
{4}\int\limits_D {u^4 vdx}  \hfill \\
  \left\langle {S\left( u \right),v} \right\rangle  \equal{} \int\limits_D {uvdx}  \hfill \\
  \left\langle {G,v} \right\rangle  \equal{} \int\limits_D {g\left( x \right)vdx} . \hfill \\
\end{gathered}
\]
Next we define a homotopy
\[ T_\tau  \left( u \right) \equal{}  \minus{} \Delta u \minus{} \left( {1 \minus{} \tau } \right)F\left( u \right) \minus{} \tau \lambda S\left( u \right) \plus{} \left( {\tau  \minus{} 1} \right)G
\]
for $ \tau \in [0,1]$ and $ u \in H_0^1 \equal{} W_0^{1,2}$.

By indirect argument, we can show that there exists a $ R > 0$ large enough such that this homotopy is admissble with respect to the ball $ B(o, R)$. This yields
\[ \deg \left( { \minus{} \Delta \minus{} G \minus{} F,B\left( {o,R} \right),o} \right) \equal{} \deg \left( { \minus{} \Delta \minus{} \lambda S,B\left( {o,R} \right),o} \right).
\]
Since $ \deg \left( { \minus{} \Delta \minus{} \lambda S,B\left( {o,R} \right),o} \right)$ is an odd number then
\[ \deg \left( { \minus{} \Delta \minus{} G \minus{} F,B\left( {o,R} \right),o} \right) \ne 0
\]
which completes our proof.


induction
number theory proposed
number theory

Let $ M$ be the set of those positive integers which are not divisible by $ 3$.  The sum of $ 2n$ consecutive elements of $ M$ is $ 300$. Determine $ n$.

It's very easy to prove by induction that the sum of $ 2n$ consecutive integers satisfying the mentioned condition is $ 3n^{2}$ thus $ 3n^{2}\equal{}300\Rightarrow n\equal{}10$ :wink:
Proof:
Check $ n\equal{}1$
Let us have for some n $ S(2n)\equal{}3n^{2}$
Now $ S(2n\plus{}2)\equal{}S(2n)\plus{}(3n\plus{}1)\plus{}(3n\plus{}2)\equal{}3n^{2}\plus{}6n\plus{}3\equal{}3(n\plus{}1)^{2}$ thus the induction ends and the conclusion follows :wink:


calculus
integration
number theory proposed
number theory

[b]Prove:[/b]

\[{(x+y)}^p\equiv x^p + y ^p \mod p\]

Such easy problem. :D
$(x+y)^p=\sum_{i=0}^{p}x^iy^{p-i}\binom{p}{i}$.Note that for $i=1,2...p-1$,we have $p \mid \binom{p}{i}$.

Expand $\binom{p}{i}$ as a product/quotient of factorials.  It is integral by interpretation, and clearly the numerator is divisible by $p$ and the denominator is not.  The result follows.

[quote="yunorman"]But can you prove $\displaystyle p \mid \binom{p}{i}$?[/quote]
This is really trivial... [i]blahblahblah[/i] has explained why, but to me, a much simpler way to show that $p \mid \binom{p}{i} \forall i = 1, 2, \ldots, p-1$ is to notice that $\binom{p}{i} = \frac{p}{i} \binom{p-1}{i-1}$.  :)


counting
distinguishability
combinatorics unsolved
combinatorics

Give a combinatorial proof that: $\binom{n}{m}S(m,m)+\binom{n}{m+1}S(m+1,m)+...+\binom{n}{n}S(n,m)=S(n+1,m+1)$
Note: $S(n,m)=$ The number of ways to distribute n distinguishable items into m unlabeled bins, at least 1 item per bin

this is again an application of the local-gobal counting principle:
the RHS of course is the Stirling number of the (second?) kind.
for the LHS, consider the partition of $\{1,2,\ldots,n+1\}$ into $m+1 $ sets by first taking apart the set that contains $n+1$. so firstly choose  $k$ elements from $\{1,2,\ldots,n\} (k\geq m)$ and form a partition of those $k$ elements into $m$ sets and the elements that were not chosen together with $n+1$ will form the last partition. this counts the LHS.


The way I graph parabolas was taught (big surprise ;)) by AoPS. See if you can find the length of the segment with both endpoints on the parabola parallel to the directrix that goes through the focus. Then use that to help you graphing the parabola.

For the second question, I suppose you just plug in.

actually,
if u already know (h,k), the focus, the directrix, p, yadda yadda
y don't u just pick numbes?
let x=-1,0,1

seems the easiest to me


Obviously one color will not work. If we have two colors, then tiling the figure like a checkerboard, alternating colors, works. Thus the answer is $ \boxed{2}$


geometry
3D geometry
Pythagorean Theorem

A bug sits on one corner of a unit cube, and wants to crawl to the diagonally opposite corner. The bug has to stay on the surface of the cube (or else the problem is too easy :D ). What is the length of the shortest path?

I came up with an answer, and I am not sure how to prove it. Please prove your answers!
[hide="My answer"]$\sqrt{10}$[/hide]

[hide] Open the cube into a net. Then the distance is equal to the distance from one corner to the opposite corner of 2 unit cubes. Thus, $\sqrt{1^{2}+2^{2}}=\boxed{\sqrt5}$.  :maybe: [/hide]

That's a really nice way, but can you prove it formally?

[hide="REALLY BIG HINT"]Imagine a net of a cube.[/hide]

[hide="solution"]We take two adjacent unit squares and connect two opposite corners. we then compute the length. $\boxed{\sqrt5}$[/hide]

[hide]First, imagine a net of a cube. Then, we see the shortest path is from 2 units' opposite corners. Using Pythagorean Theorem, we get the distance to be $\sqrt{5}$.[/hide]

[hide]Imagine a net of a cube.  It can look like a t, or like a zigzag.

Imagine two vertices.

If you unfold either one of nets, you will get that the distance between the two points is $\sqrt{1^{2}+2^{2}}=\sqrt{5}$[/hide]


I searched and searched and searched some more but I can't find the answer to my questions. Firstly, some background. I was making around 120's on my practice AMC 10s. I got a really really low score (for my expectations), a 69.5. I want to improve. I have all the intros, CPIG (a gift from my uncle, came just a few days ago!) and Vol 2. Should I do Volume 2 and AIME problems while doing AMC 10s? Or just focus on the intros and just the AMC 10? Should do the AMC 10 timed or untimed? What other competition problems should I do? Any marathons I should follow or do?

thx

I think that you are in the same scenario as me. I had as well, taken many practice AMC10's and scored 120+ on most of them, but in the real contest, I only scored a less-than-desirable 100.5.

It sounds to me that you have all the talent and smarts to score 120+ (as you've already been doing it), but you just didn't too well on the test in real life. Focus mainly on doing problems slowly to start out (to avoid any stupid mistakes) and then try and increase your speed to what you want.

And yes, definitely take the AMC10 under timed conditions, and try to make the situation as real as possible! Its EXTREMELY important to do that.
As for problems to do...
it really matters on how far you want/think can go. If you want to just make AIME, then solely work on AMC10, but if you want to make AIME, and make USAMO too, then perhaps you want to study some harder mathematics (such as those covered in Volume 2).

since you r not yet completely confident making aime from the 10, i would master the intro series. if you do that, then you may or may not want to get aops 1. after the intro series, you have the potential to get a perfect on the amc 10 and ~5 on the aime. the key to actually get them is to do problems.

I did the whole of the intro series twice already!


calculus
integration
limit
trigonometry
symmetry
Putnam
calculus computations

$\hspace* \quad \lim_{\alpha \rightarrow \frac{\pi}{2}}\int_{0}^{\alpha}\;\frac{dx}{1+tan^{2007}x}$

Answer:
[hide]$\frac{\pi}{4}$[/hide]

Observe that

$I = \int_{0}^{\pi/2}\frac{\cos^{2007}x }{\sin^{2007}x+\cos^{2007}x}\, dx = \int_{0}^{\pi/2}\frac{\sin^{2007}x }{\sin^{2007}x+\cos^{2007}x}\, dx$

by the symmetry of sine and cosine function. Then we have $2I = \frac{\pi}{2}$.

You mean the symmetry on that interval.

nice  :)

[b]misan [/b] what did your solution look like?

this is similar to a recent putnam problem (B1 on 2005 if i am correct)
but the $\tan^{2007}(\theta)$ doesnt matter. the answer is the same for any power.


Hehe, my school looks great, there's soooo  much green around.

[quote="djimenez"]Hello,

Specially for all those who are in the US or Ireland, HAPPY SAINT PATRICK'S DAY. Did you know that over 37 million people in the US claim to have Irish heritage. That is more than ten times the population of Ireland  :what?: .

Best,[/quote]

Hehe, thx djimenez, that is one big tree heritage! ;)
Well, Happy St.Patrick's Day!
Hey you, that Leprechaun gimme your pot of gold! :D  :rotfl:

[color=green]Happy St.Patrick Day!
And for my green: :D  :D  :D  :D [/color]

[quote="mcalderbank"][quote="plokoon51"]Happy St. Patricks day everybody

at my hs, nobody wore green, lol :D what spirit, huh???[/quote]

Our school had super spirit, especially girls who were wearing shirts saying, "Kiss me, I'm Irish." :D  :D

Whoever started that slogan is a genius :P[/quote]

one of my friends did the exact same thing, except he got a sheet of large green construction paper and wrote "Kiss me, I'm Irish" and he taped it onto his back. lol

tonight was opening night for our school musical, and the cast has been doing crazy spirit day things all week, and today we decked it out in green. i had 7 different shades of green on, and one guy had a green shiny cape that he made. it was amazing!  :D  :D  :D

I have green shoes(converses) so I'm wearing green all the time(or until I take them off) :P

[quote="ankur87"]I told people I have a skin disease that turns my skin green if I get pinched...

Didn't work.[/quote]
Ha! Well it was worth a try...

[quote="djimenez"][quote="filletwho"]I wore all orange and red. got pinched a couple of times. But then I started telling ppl I had red-green colorblindness.[/quote]

And is that true?[/quote]

No... not to my knowledge. How do ppl know if they have red-green colorblindness? I mean, all red would be green and all green would be red. right?

[quote="filletwho"]No... not to my knowledge. How do ppl know if they have red-green colorblindness? I mean, all red would be green and all green would be red. right?[/quote]
Good point. How would you know? Unless you look at stop lights and someone explained it to you.

Well, I don't have an idea how they notice, but they do. My father in law is collor blind. It is somehow funny. The first appartment my wife and I had after we got married had a bathroom with a nice green in the walls, and the first time he went he said "How nice is that gray in the bath walls!"  :D  :D

Best,


AQUI hay un nuevo foro: http://olimpiadadelasmatematicas.mundoforo.com/

aorita solo ay un problema, ojala y algunos de aqui pongan soluciones,
este es, determinar todos los enteros n tales que $n^{4}-4=p$
donde $p$ es primo

Disculpa Tony, no kerras decir
$n^{4}+4=p$ ???

 :lol:

bueno,  :D  creo, k solo es factorisar jeje.
$(n^{4}+4n^{2}+4)-(2n)^{2}=(n^{2}+2n+2)(n^{2}-2n+2)=P$, como $p$ es primo, entonces su factores solo son 1 y p, entonces igualando.
$a)$
        $n^{2}-2n+2=1$--->$n=1\Rightarrow p=5..cumple$
$b)$
        $n^{2}+2n+2=1$--->$n=-1\Rightarrow p=5..cumple$
por lo tanto los pares que cumplen son: $+1 y-1$


analytic geometry
Euler
geometry
3D geometry
tetrahedron
algebra open
algebra

Because he could not paint my avatar (I hope it's there) on a square grid so that ALL corners and intersections are on gridpoints
The arcs are circles with center in the left down corner.The small circle intersects the intersection of the horizontal and vertical lines.
Kweenie

PS: I don't think that Mondrian knew this is a famous unsolved problem.

[quote="kweenie"]Because Mondrian could not paint my avatar (it's there) on a square grid so that ALL corners and intersections are on gridpoints.
[/quote]
1) This thread was posted to show a simple flat drawing of this famous unsolved 3 dimensional problem, in the hope that it helps to find a solution. Either prove or disprove that the construction can be done on gridpoints. I should rather go for the "disprove" option, as it is proved (by exhaustion) not to be possible even for VERY huge grid-dimensions.
It's possible if ONE of the coordinates of the intersections is not a gridpoint.
2) What "open" problem are we talking about, in your opinion?
kweenie
 :spider:

This was about the perfect Euler brick or perfect cuboid.

Perhaps the perfect Euler brick problem is related to the following:

Does there exist a rectangular tetrahedron (a corner-cut from a cube) where besides the distances also all 4 area's are integers?

I guess hat it is related to it but really don't know?

kweenie


Ok, basically the situation is like this: I am one of the millions of students that hated mathematics to death while in high school but want to study computer science. As a result of this passionate hate I now find myself in university and although I can deal with the math of the course I really want to get a better understanding of most of the concepts and ideas in discrete mathematics, linear agebra, number theory, etc. However, it sounds ineffective that I just pick a subject, go to wikipedia, read about that, pick another subject, do the same, and so on. So, what I am asking is suggestions for books covering the concepts that a person who wants to understand maths better needs to be ok with.

Not trying to be a plant, but the Intro AoPS books are a great resource. This site as well as the series of books places a great deal of emphasis on discrete math. Grab the intro counting and intro number theory books. They are excellent. And most importantly, do the problems. Math is a participation sport.

Graph theory is a great deal of interest to computer science. The intermediate counting book deals with subject.

AoPS Vol 1 & 2 cover the same material but more at a review level.


trigonometry
vector
function
calculus
calculus computations

Good day everyone. I have some problems here regarding motion that I have no idea how to solve. Here they are:

1) The center of a dartboard is $6$ ft. from the floor. If a woman $8$ ft. away throws a dart with her hand $5$ ft. above the ground at an angle of $\cos^{-}{1}4/5$, what initial velocity is required to hit the bull's eye?

2) Given the position vector $r(t) = <t^{2}, \cos t+t\sin t, \sin t-t\cos t>$ with $t \geq 0$ of a particle. Give the position and speed of a particle if it has moved $\frac{\sqrt{5}}{8}\pi^{2}$ units from its starting pt. $(0,0,1)$.

Any hint on how to solve this problems would be a nice start. Thanks.

1)  Break the initial velocity into a horizontal and vertical component.  The horizontal component is a constant function of the initial velocity, whereas the vertical component varies by time because of gravity.  Use the horizontal component to calculate $t$, the amount of time needed for the dart to move $8$ feet, and solve for $v_{0}$ by setting the final $y$-position where it needs to be.

Thanks sir for your reply.

By the way, in no. 1 the angle should be $\cos^{-1}\frac{4}{5}$. I have solved for the initial velocity to be $40 feet per second$.

Is this the right answer? Hope someone could check. Thanks again.


[quote="khashi70"]Farz Konid se khate $AD , BE , CF$ dar dakhele mosalase $ABC$ hamres dar noghteye $P$ bashand be tori ke $BE$ nimsaze $\angle ABC$ bashad . agar $K$ mahale talaghie $EF$ ba $AD$ bashad va  $L$ mahale talaghie $DE$ va $CF$ bashad . sabet konid $BE$ nimsaze $\angle LBK$ ast .[/quote]

farz kon $KL$ va $FD$ dar $X$ hamdigar ro ghat konand.va hamchenin $FD$ va $AC$ dar $Y$ hamdigar ro ghat konand.darim 4 shoae $BC,BE,BA,BY$ hamsazand, va hamchenin 4 shoae $BD,BE,BF,BX$ hamsazand pas $X\equiv Y$,farz konid $KL\cap BE=Z$ darim $(XZ,KL)=-1$ va choon $XB\perp BE$ pas $BE$ nimsaze $\angle KBL$ ast.


parameterization
number theory unsolved
number theory

Determine a parametric form for the integer solutions of the equation " x^2 + y^2 = z^2 + y^2 "

$ x\equal{}z$ or $ x\equal{}\minus{}z$??
Can you edit the problem?

I think the OP meant $ x^2 \plus{} y^2 \equal{} z^2 \plus{} w^2$.  Anyway, write $ x^2 \minus{} w^2 \equal{} z^2 \minus{} y^2$; the rest should be more or less clear.

[b]Edit:[/b]  To be more precise, write $ x \plus{} w \equal{} pq, x \minus{} w \equal{} rs, z \plus{} y \equal{} pr, z \minus{} y \equal{} qs$; then

$ x \equal{} \frac{pq \plus{} rs}{2}$
$ w \equal{} \frac{pq \minus{} rs}{2}$
$ z \equal{} \frac{pr \plus{} qs}{2}$
$ y \equal{} \frac{pr \minus{} qs}{2}$.

To get uniqueness we need to specify certain relative primality conditions and to make sure $ x, y, z, w$ are integers we need to specify certain parity conditions; these considerations should be familiar from the Pythagorean triple parameterization, which can be solved by a similar argument, as well as the Pythagorean quadruple parameterization.

I have always found this article really interesting, hope you like it
http://reflections.awesomemath.org/2007_6/factoring_lemma.pdf


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

I want to make USAMO this year. what would be the safest bet, to do AMC 10 or AMC 12? I am just starting Vol 2 and am done with all the intro's. Also, advice to prepare would be very much appreciated.

[quote="pythag011"]AMC 10 is harder to pass, but it is much easier to get to USAMO with AMC 10 than it is with AMC 12.[/quote]

hang on

what do you mean 'harder to pass'?

EDIT: ohh i get it

thanks

It's harder to get to AIME on the 10 but easier to get to USAMO.

[quote="pythag011"]AMC 10 is harder to pass, but it is much easier to get to USAMO with AMC 10 than it is with AMC 12.[/quote]

I disagree. The AMC10 is not harder to pass. On the AMC12, you need to get 14 questions correct (assuming that the rest are skipped). On the AMC10, you need only 19 questions correct. So, you need 5 more questions right on the AMC10 than you do the AMC12.

But this is quite negligible in comparison because the AMC12 is so much harder than the AMC10.  Plus, certain topics are excluded from the AMC10 like higher level geometry and complex number and their graphs. (There's more to this list, I can't remember them all).

Well anyway,
USAMO invitees are largely determined by an individual's index.
You calculate index by [AMC score] + 10[AIME score], and you generally want to aim for around a 200. Now, the AMC 10, being an easier test that is scaled to the same grading, will generally get you a higher index. So it's your best bet to go with an AMC10.

Yes, but some peopel haven't learned those advanced topics.

For qualifying for AIME, amc 12 is best.  I was 1 problem off for qualifying with AMC 10 A, but I made it through AMC 12 B.

[quote="pythag011"][quote="soltixnj"]

But this is quite negligible in comparison because the AMC12 is so much harder than the AMC10.  Plus, certain topics are excluded from the AMC10 like higher level geometry and complex number and their graphs. (There's more to this list, I can't remember them all).
[/quote]

what? I find AMC 12 to be barely harder than AMC 10... just covers more advanced topics...[/quote]
i agree. if given two different tests (one AMC 10 the other AMC 12) and i'm not told which is which i'll probably won't know.

If you hypothetically take both the 10 and the 12 and get and qualify for AIME on both of them and manage to get a high score on the AIME, which AMC will be used to qualify for USAMO?

Your "index" will be the higher of the two, but really... it doesn't matter anymore once you make it to the USAMO!


geometry
geometric transformation
integration
calculus
derivative
function
trigonometry

Hello,

Who can help me getting started on this equation? $y''-4y=tanh(2x)$

Greets Thanks a lot.

Edit DV must DE

just a dutch-english translation remark: "DV" = "DE" in english. :)

('differentiaal vergelijking'->'differential equation')

Let $K(x)$ be the solution to the homogenous differential equation $y'' + ay' + by = 0$ that fulfills the $y(0) = 0, y'(0) = 1$. If $g$ is continuous then a particular solution to $y'' + ay' + by = g$ is
$y(x) = \int_0^x K(x-\xi)g(\xi)d\xi$.

You show this by simply differentiating $y(x)$ and substituting in. For the differentiation you need Leibniz' formula for differentiating $I(t) = \int_{\phi(t)}^{\psi(t)} f(x,t) dx$:
$I'(t) = \int_{\phi(t)}^{\psi(t)} f'_t(x,t)dx + f(\psi(t),t)\cdot\psi'(t) - f(\phi(t),t)\cdot\phi'(t)$

This can be established by the use of the the standard formula for differentiating under the integral sign and applying the general chain rule on
$F(t,b,a) = \int_a^b f(x,t) dx$

note: In this special case it is possible to evaluate the integral quite easily if you use $\tanh(2x) = \frac{e^{4x}-1}{e^{4x}+1}$

[quote]note: In this special case it is possible to evaluate the integral quite easily if you use $tanh(2x)=\frac{e^{4x}-1}{e^{4x}+1}$ [/quote]

Where can I find those indenties?

it is the definition of tanh? at least the one we used.

$\tanh(x)=\frac{e^{2x}-1}{e^{2x}+1}$.

[quote]it is the definition of tanh? at least the one we used.
[/quote]

of course but can you also defind al the othter goniometric function? like [tex] sin(x) \ \ tg(x) ... [tex] ans so on ?

If yes where can you find a list of they? Greets.

convert(sin(x),exp) in maple

or look them up at wikipedia.

But I find it easiest to derive them on the spot. The idea is simple: $e^{i\theta}=\cos\theta+i\sin\theta$, and everything follows from that.

[quote]Did you try to find a particular solution using the method I gave?
[/quote]

I will try it


geometry
circumcircle
ratio
geometry proposed

Let $M$ be a point on the side $|BC|$ of the acute-angled triangle $ABC$. What are those points $M$ for which the common area of the circumcircles of $ABM,ACM$ is minimal?

[yes, I know it's easy, but I thought it was sort of cute :)]

I think the position of M which minimize the common area is the foot of the altitude from A.

In fact if we take a different point M' on BC we have that AM' > AM.

We can construct a similar figure (including circles) rotating through A of the angle <M'AM and dilating of the ratio AM'/AM.

For this reason the ratio betwenn the latter common area and the former is (AM'/AM)^2 > 1.

I know could be not easy to understand my idea, but this is the better I can do whith my english.

[quote="sprmnt21"]
...
I know could be not easy to understand my idea, but this is the better I can do whith my english.

[/quote]

Perhaps is better to clarify some think   :) .

If M is a point on BC different from H (where H is the ortogonal projection of A on BC) let's trace the ortogonal line p through M wrt AM. Let B' be the point where c(AMB ) cuts p and let C'=c(AMC)^p. As <ABM = <AB'M and <ACM=<AC'M, the traingles ABC and AB'C' are similar, so the ratio of the common area of c(AMB') and c(AMC') wrt the common area of c(AHB) and c(AHC) is equal to (ABC)/(AB'C')=(AM/AH)^2.

Don't worry, man, the idea is pretty clear: because of angle equalities, the shape common to the two circles is is always the same (i.e. all such shapes are similar), so its minimal area is reached when $AM$ is minimal.


Fie $A$ un inel in care $x^2=0$ pentru orice $x$ din $A$.
 Sa se demonstreze ca $abc+abc=0$ oricare ar fi $a,b,c$ din $A$


counting
distinguishability
group theory
abstract algebra
combinatorics unsolved
combinatorics

In how many ways can one build k groups with N people?

Each group has to contain at least one person. The groups are distinguished only via their members.

the answer is the sterling number (of the second kind) $S(n,k)$, which has no closed form expression. not quite an "intermediate topic"

Let us find a method to calculate $S(n,k)$.  We want to partition a set of $n$ elements into $k$ groups.  Let the set be $\{x_{1},...,x_{k}\}$.  Now, find all possible ways to write $n$ as a sum of k positive integers.  Each number represents the number of elements in one of the $k$ partitions.

Example:  We have 5 elements.  We want to find the number of ways to partition these 5 distinguishable elements in a set into 3 groups, non of which are empty.  Since $5=1+1+3=1+2+2$, we have that the answer is $\frac{1}{2}\binom{5}{1}\binom{4}{1}\binom{3}{3}+\frac{1}{2}\binom{5}{1}\binom{4}{2}\binom{2}{2}=25$.

I hope this is correct, and moreover useful.

Masoud Zargar

Thank you for your response. I think that the final response is given by summing the multinomial coefficients over all number i_1,...,i_k such that their sum is N and not counting the permutation of such indexes and finaly dividing by the product of the respective number of equal subgroup sizes.

for example, with N=10, k=4 and for the configuration
2,2,3,3 we have to divide the multinomial coefficient (10; 2,2,3,3) by 2! 2!


trigonometry
trig identities
cosine double angle

Just trying to get my feet wet with oly problems. http://web.archive.org/web/20040821051342/www.kalva.demon.co.uk/irish/soln/sol882.html
I tried an this problem for about 30-45 mins and then I tried to look at the solution. I understood it until the part where it says "2 lhs".
1) I have no idea what "2 lhs" means, but I know it has something to do with the left hand side
2) I don't understand how he got the following.

Could someone please help?

To show that both sides are equal, he multiplied both sides by 2, calculated each of them separately, and showed that they were equal. For the 2 lhs, he used the cosine double angle formula, and for 2 rhs, he used a product to sum formula.

[hide="hint for a possibly easier solution"]
Use Ptolemy's on PABC and PBAD[/hide]


modular arithmetic
algebra
polynomial
analytic geometry
number theory
relatively prime
number theory proposed

Let $ n \geq 2$ be a prime, and let $ b_1, \ldots, b_k$ be pairwise relatively prime positive integers, none of which is the $ n \minus{}$th power of an integer. Prove that there is a prime $ p$ such that none of the $ k$ equations
\[ \begin{array}{c} X^n \equiv b_1 \pmod p \\
\vdots \\
X^n \equiv b_k \pmod p\end{array}
\]
has a solution.

[i]Change:[/i] $ n$ is a prime.

I don't think you can assume $ k\equal{}1$, as $ b_1,b_2,\dots,b_k$ are given positive integers. :maybe:

You may want to tell me what you mean by "given"  :wink:

Easy problem, when n=2. We can consider only $ b_i$ -squarefree $ 4|p\minus{}1$. Then problem equavalent to $ (\frac{p}{b_i})\equal{}\minus{}1$.
Let $ (\frac{x_i}{b_i})\equal{}\minus{}1$, then exist $ x$, suth that $ x\equal{}x_i\mod b_i$. If $ p\equal{}x\mod 4b_1b_2...b_n$ we get need prime p.

I changed the problem so that $ n$ is required to be prime, although all we really need is that $ b_i$ is not the $ q\minus{}$th power of an integer, for any prime $ q$ dividing $ n.$

All that polynomials $ x^n\minus{}b_i$ are irreducible by http://www.mathlinks.ro/viewtopic.php?t=228729 (note that in the exception $ x^4\plus{}4a^4$, we would need negative $ b_i$). So the smallest field extension of $ \mathbb Q$ containing a root of one of these polynomials must have a degree $ \geq n$. Especially, none of them has a root in $ K \equal{} \mathbb Q[\zeta_n]$ with $ \zeta_n$ a primitive $ n$-th root of unity (as that extension has degree $ \varphi (n) <n$), so $ b_i$ isn't a $ n$-th power in $ K$. Thus if we take the biggest $ n_i$ such that $ b_i^{n_i}$ is an $ n$-th power, we get $ n_i>1$.

Now consider the the splitting field $ L$ over $ K$ of all these polynomials $ x^n\minus{}b_i$: By standard Kummer theory, we have that $ \mathrm{Gal}(L|K) \cong \prod_i (\mathbb Z / n_i \mathbb Z)$ where the $ i$-th coordinate $ a \mod n_i$ acts by sending $ \sqrt[n] {b_i} \to \zeta_{n_i}^a\sqrt[n] {b_i}$. Especially, we can take the element $ \sigma \equal{} (1,1,...,1)$ fixing none of the $ \sqrt[n_i] {b_i}$ as $ n_i > 0$. Now by Chebotarev's theorem, we find infinitely many unramified prime ideals $ \mathfrak p \subset \mathcal O_L$ such that $ \sigma$ is the Frobenius automorphism of $ \mathcal O_L  /\mathfrak p$.
Thus $ x^{n_i}\minus{}b_i$ has no roots $ \mod \mathfrak p$, and the same holds $ \mod p$ for the prime $ p$ with $ (p) \equal{} \mathfrak p \cap \mathbb Z$. This is what we wanted.


linear algebra
matrix
function
parameterization
vector
probability and stats

Hello everybody,

I have a problem and I am really in need of your comments. My problem can be described as follow.

[b]INPUT:[/b] Many sets of observed temporal data with $ T$ elements ($ T$ is constant). Each set corresponds to a pair of number of entities and size, denoted by $ n$ and $ s$. For example, if $ n \equal{} 50, s \equal{} 1000$, and $ T \equal{} 1000$, I  have one set.

[b]OUTPUT:[/b] A statistical model illustrating the observed data. In addition, that model should be able to predict value of output data with different pairs of $ (n, s)$. It means that if we test $ n \equal{} 60$ and $ s \equal{} 1500$, we should have  output with same shape as observed data and appropriate values.

I use a nonlinear regression model with normal distributed error term and the model has some, e.g. 2 $ (a, b)$, coefficients. These coefficients can be expressed as functions of $ (n, s)$ with new coefficients, however. Therefore, these "mother" coefficients have "sibling" coefficients. For example, consider  coefficient $ a \equal{} f_1(n, s) \equal{} a_1n \plus{} a_2s$; that means $ a_1$ and $ a_2$ are  sibling coefficients of $ a$.

I want to use Markov chain Monte Carlo method to generate values for these coefficients. In MCMC method, suppose that $ (a, b)$ are multivariate normal distributed, we need a covariance matrix. There are some studies on this stuff. Nevertheless, the posed situation in my problem is a little different as it has siblings. Because of this reason, I do not know how to compute covariance matrix, which needs to take care of all sibling coefficients, not just 2 mother coefficients.

Could you please kindly recommend me something to do with this problem?

I would highly appreciate every comment that you can give me. Thank you very much.

Regards,
Iaoh.

[quote="alekk"]it is not very clear to me (sorry). Could you give a concrete description of what you are trying to model ?[/quote]

Well, I have some sets of simulation results and when I draw them, all figures follow one shape. Simulations are conducted with the same length of time, e.g. $ T\equal{}1000$, under various scenarios, i.e. different simulation size $ s$ and population size $ n$. I want to have a statistical model which describes these data. In addition, I wish that model to be capable of predicting the output data if I pass other values of the pair $ (n, s)$. For example, suppose that the pair $ (n\equal{}60, s\equal{}2000)$ which has not been used for simulation, it is desired that if I pass these "new" values into the model I will have a new prediction for output.

I propose a model of nonlinear regression for the drawn figures and use Bayesian inference to guess parameters (consisting of coefficients of regression function and error term). I do believe that you know this step better than me  :) Using Bayesian inference, it is possible to obtain unknown coefficients for the regression model. Suppose that these coefficients are multivariate normal, we need a covariance matrix in order for MCMC method to work.

Since I want my model to be able to predict output for new $ (n, s)$, it is obviously that $ n, s$ have to be expressed somewhere in my model. I find that each coefficient of regression function is dependent on $ n$ and $ s$. In other words, each coefficient is a function of $ n$ and $ s$. These functions have coefficients, too. We call them "sibling". (You can refer to my previous post for example about this.)

Consider one "mother" coefficient (coefficient of regression function). Arrange all siblings of this coefficient as elements of a vector. We have some, e.g. 2, vectors at all. The problem arises here, I do not know how to build the covariance matrix for the the posed situation, each vector has more than one element.

Does this explanation give you clearer information? If you feel something unclear, please tell me, I will try my best to clarify it.

Best regards,
Iaoh.

P.S: I am sorry, English is not my native language so my post is obscure.


I'm looking for a book from which I can learn some basic C++ data structures. My local university uses "Data Structures Using C++" by Malik. Is this any good? If not can anyone recommend some others? Thanks.

Big part of CLRS is devoted to Data Structures. It's really worth reading. You can easily translate their pseudocode to C++.