What is the prime factorization of 1998?

Anirudh, I'm afraid you're wrong, it is : 2*3^3*37
I don't know why you've added those two "2"'s..  :roll:


limit
inequalities
calculus
calculus computations

[b]Find $ inf$ and $ sup$ of $ R\cap\left( \minus{} \infty,0\right]$ , where $ R$ is the set of solutions of the equation $ x \plus{} \frac {x \plus{} 3}{\sqrt {x \plus{} 1}} \geq 3$.[/b]

We can see, that $ R\subseteq( \minus{} 1,\infty) \Rightarrow R\cap\left( \minus{} \infty,0\right] \subseteq \left( \minus{} 1,0\right]$.

Putting $ x \equal{} 0$ into $ R$, gives $ 0 \plus{} \frac {3}{1} \equal{} 3 \geq 3 \Rightarrow sup( R\cap( \minus{} 1,\infty)) \equal{} 0$.

Now, here is my question. Since we know $ lim_{x\downarrow \minus{} 1}(x \plus{} 3) \equal{} 2$, $ lim_{x\downarrow \minus{} 1}(\sqrt {x \plus{} 1}) \equal{} 0$, $ lim_{x\downarrow \minus{} 1}(x) \equal{} \minus{} 1$, is it "legal" to say $ lim_{x\downarrow \minus{} 1}(x \plus{} \frac {x \plus{} 3}{\sqrt {x \plus{} 1}}) \equal{} \minus{} 1 \plus{} \infty > 3 \Rightarrow inf( R\cap\left( \minus{} \infty,0\right]) \equal{} \minus{} 1$?

I would accept this as a valid explanation. You may want to add some (rather trivial) details like "since $ \lim\equal{}\plus{}\infty$, there exist $ x$ arbitrarily close to $ \minus{}1$ for which the inequality holds, so...", etc. that will demonstrate that you know what you are talking about beyond any doubt, but there is nothing wrong with the solution even as is.


geometry
3D geometry
sphere
tetrahedron
similar triangles

[i] 3 congruent cones with height 115 and radius 10 are place beside one another (stuck together by their bases) in order to hold a sphere in place.  The top of the sphere does not exceed the tip of the cones.  What is the radius of the sphere?[/i]

Masoud Zargar

consider the lines connecting the points of tangency to the verticies of the cones, these 3 lines and the plane that the sphere is tangent to, make a tetrahedron (not regular), base equilateral with s=20, and lateral edges of$\sqrt{115^2+10^2}$, and the sphere is tangent to the plane, and tangent to the lateral edges, i am pretty sure that the rest is just using similar triangles using the right angles formed by pt of tangency to radius, and altitude to the pt the base


analytic geometry
trigonometry
inequalities unsolved
inequalities

Let $a , b \in R$ such that :
 $a^2+b^2 \leq 1$ and $c^2+d^2 \leq 1$ .
 Find the maximum and minimum value of
  $S =\sqrt{1-a^2-b^2}+\sqrt{1-c^2-d^2}-2\sqrt{1-\frac{(a+c)^2}{4}-\frac{(b+d)^2}{4}}$

Put  $a=rsinx,b=rcosx , c=rsiny , d=rcosy$  with  $-1\leq r\leq1$

Can you write down your solution , Nikolas ? I think we can use coordinate to solve this problem

I don't think you can do that, Nickolas, as it is not given that $a^2 + b^2 = c^2 + d^2$.

This problem is equivalently with the following geometrical problem:

"Let the fixed disc $w=D(O,R)$ and a mobile triangle $XOY$ for which $[XOY]\subset w$. Ascertain

$\max_{X,Y\in w} \left\{ \sqrt{R^2-OX^2}+\sqrt{R^2-OY^2}-2\sqrt{R^2-m^2_o}\right \}\ \ i.e.\ \ \max_{X,Y\in w} \left\{p(X)+p(Y)-2p(M)\right \}$

where $m_o$ is the length of the median $OM$ from the vertex $O$ in the triangle $XOY$ and $-p(U)$ is the power of the point $U$ w.r.t. the circle $C(O,R).$"

[u]Remark.[/u] This problem is equivalently with the following problem:

$\max_{0\le r_1,r_2\le 1,\ 0\le x,y<2\pi} \left\{ \sqrt{1-r^2_1}+\sqrt{1-r^2_2}-\sqrt{4-r^2_1-r^2_2-2r_1r_2\sin (x+y)\right\}\le}$

$\max_{0\le r_1,r_2\le 1}\left\{ \sqrt{1-r^2_1}+\sqrt{1-r^2_2}-\sqrt{4-r^2_1-r^2_2}\right \}\ \ (x+y=\pi ).$

This problem is still unsolved . Anybody , help me !


[hide]Explicitly writing out the sum gives $y_1 - y_0 + y_2 - y_1 +-... + y_{99} - y_{98} + y_{100} - y_{99}$.  All terms except the second and last cancel, giving $y_{100} - y_0 = \cos \pi - \cos 0 = -2$.[/hide]


[quote="KMK00009"]given 5a^4+4b^4+6c^4=90
find the maximum value if 5a^3+2b^3+3c^3[/quote]

Apply Holder directly,

$ (5a^4\plus{}4b^4\plus{}6c^4)^{\frac{3}{4}}(5\plus{}\frac{1}{4}\plus{}\frac{3}{8})^{\frac{1}{4}} \ge 5a^3\plus{}2b^3\plus{}3c^3$
$ \Rightarrow 45 \ge 5a^3\plus{}2b^3\plus{}3c^3$

Equality holds iff $ \begin{cases}5a^4\equal{}5 \times 16\\ 4b^4\equal{}\frac{1}{4} \times 16\\ 6c^4\equal{}\frac{3}{8} \times 16 \end{cases} \Leftrightarrow a\equal{}2,b\equal{}1, c\equal{}1$

So max. value of $ 5a^3\plus{}2b^3\plus{}3c^3$ is $ 45$


Let $a,b,c,d$ no-zero real numbers, which satisfy the inequality

$abc(a+b+c)(ab+bc+ca)>0$.

If $A=a^3+b^3+c^3+d^3$,  $B=a+b+c+d$ and  $C=\frac{1}{a}+\frac{1}{b}+\frac{1}{c}+\frac{1}{d}$

then prove that if two of the numbers $A,B,C$ are equal to zero then the third of them is also zero.

Alexandros

$\sum a^3-(\sum a)(\sum a^2-\sum ab)-3\sum abc=0$.
(guess how I found this ;) )


conics
parabola
symmetry
hyperbola
function
geometry
geometric transformation

The original problem reads:

Find the equation of the parabola (with axis parallel to the x-axis) whose vertex is (2, -4) and which passes through the origin.

Since this obviously makes no sense, I'm stemming it into two problems that are pretty similar.

a) Find the equation of the parabola (with axis of symmetry parallel to the y-axis) whose vertex is (2, -4) and passes through the origin.

b) Find the equation of the hyperbola (with axis parallel to the x-axis) whose vertex is (2, -4) and which passes through the origin.

is this inverse of $y = x^{2}$ not a parabola? my answer assumes that the question meant if the axis of [i]symmetry[/i] is parallel to the x axis..

[hide]
if the axis of symmetry is paralell to the x axis, then we have an inverse equation.

we know the vertex is at point $(2,-4)$ thus we can start basing our equation around that

$x = \pm\sqrt{y-2}-4$
now check to see if y = 0 when x = 0
$\pm\sqrt{y-2}= 4$
$y-2 = 16, y = 14$
it does not, so we must add a stretch/compression to the function, and also, in order for this functions vertex to be at point $x = 2$, the function must also be flipped in order for it to also pass through the origin

going from $y-2 = 16$, and making y = 0 what value $x$, multiplied by $(0-2) = 16$?

$x(-2) = 16$
$x =-8$

so now our function looks like:

$x = \pm\sqrt{-8(y-2)}-4$

and it checks out

$\pm\sqrt{-8(y-2)}-4 = 0$
$\pm\sqrt{-8(y-2)}= 4$
$-8(y-2) = 16$
$y-2 =-2$
$y = 0$

works for me  :| [/hide]

The original problem is doable.

Using the reflection symmetry on the line $ y\equal{}x$ and the equation of the parabola in vertex form, the equation is $ x\equal{}a(y\plus{}4)^{2}\plus{}2$ (with the axes parallel to the x-axis). Since the parabola passes through the origin, $ 0\equal{}16a\plus{}2$ or $ a\equal{}\minus{}\frac{1}8$; so, $ x\equal{}\minus{}\frac{1}8(y\plus{}4)^{2}\plus{}2$.

Its related to analytic geomtry. (original problem)

You do this: the equation has form (y+4)^2=4p(x-2)

we know that (0,0) passes through the graph, so 16=-8p and so p=-2.

therefore the equation of the parabola is (y+4)^2=-8(x-2)

[quote="Ignite168"]Find the equation of the parabola (with axis parallel to the x-axis) whose vertex is (2, -4) and which passes through the origin.

Since this obviously makes no sense[/quote]

Parabolas don't only take the form $ y \equal{} ax^{2}\plus{}bx\plus{}c$.  These are only the parabolas with axis of symmetry parallel to the $ y$-axis.  The parabolas with axis of symmetry parallel to the $ x$-axis take the form $ x \equal{} ay^{2}\plus{}by\plus{}c$.

In fact, more general parabolas can be written whose axes of symmetry are arbitrary.  :)


[quote="AnhHungThamLang"][i]Let[/i] $\large a,b,c\ge0$ [i]such that [/i]$\large a+b+c=3$[i].prove that[/i]
...............
$\large\frac{a^{2}}{\sqrt{bc+(a-1)^{2}}}+\frac{b^{2}}{\sqrt{ac+(b-1)^{2}}}+\frac{c^{2}}{\sqrt{ab+(c-1)^{2}}}\ge 3$
[b]@hungkhtn: anh xem bai nay co duoc khong a./neu duoc em se post dap an len./
@hungkhtn/toanhocmuonmau: tet nay hai co co ranh khong ve nha em choi./:rotfl:   [/b][/quote]

Let  $x,y,z\ge0$ such that $x+y+z=3$, then

$\sum\frac{x^{2}}{\sqrt{yz+(x-1)^{2}}}\ge 3\sum{2x^{3}+xyz \over (2x+y)(2x+z)}\geq 3$.

http://www.mathlinks.ro/Forum/viewtopic.php?t=132530


superior algebra
superior algebra theorems

I can't answer for the folowing question:

[b]If any valuations on the field K are trivial then K is a finite field [/b]

I think it's not true but I can't found a converse example,please help me !!!!

[quote="owk"]Take the algebraic closure of a finite field.
[hide="hint"]Every nonzero element is a root of unity.[/hide]
owk[/quote]

Thank you very much !


source: Book: Inegalitati; Authors:L.Panaitopol,V. Bandila,M.Lascu

Let $a_{1},a_{2}, \ldots,a_{n}$ and $b_{1},b_{2},\ldots,b_{n}$ be two real sequences with the properties: $a_{1}>a_{2}> \ldots >a_{n}$ and $b_{1}>b_{2}> \ldots >b_{n}$. Prove that: $a_{1}b_{1}+a_{2}b_{2}+ \ldots +a_{n}b_{n} > a_{1}b_{n}+a_{2}b_{n-1}+ \ldots +a_{n}b_{1}$.

Am I being stupid or is this just a case of the Rearrangement Inequality?


[quote="silouan"]Να υπολογιστεί το

$ A \equal{} \sum_{n_{60} \equal{} 0}^{2} \sum_{n_{59} \equal{} 0}^{n_{60}}\cdot \cdot \cdot \sum_{n_1 \equal{} 0}^{n_2}\sum_{n_0 \equal{} 0}^{n_1} 1$

 :D[/quote]

Πολύ μαγκιόρικη άσκηση. Ορίστε η λύση μου:

Η παράσταση μας δίδει τους δυνατούς τρόπους τοποθέτησης 3 αντικειμένων (0, 1 ή 2) με επανάληψη σε 61 θέσεις, ή αλλιώς οι επαναληπτικοί συνδυασμοί n αντικειμένων ανά k.
Είναι δηλαδή $ A \equal{} \frac {(3 \plus{} 61 \minus{} 1)!}{61! (3 \plus{} 61 \minus{} 1 \minus{} 61)!}$
(Είναι 3 + 61 - 1 συνδυασμοί ανά 61, αλλά δεν ξέρω πώς να το εισάγω σε Latex)
$ \equal{} \frac {63!}{61! (2!)}\equal{} \frac {62\cdot63}{2} \equal{} 31\cdot63 \equal{} 1953$


complex analysis
function
inequalities
absolute value
complex analysis unsolved

let $ C_1$ denote the unit circle. suppose f is analytic in a region containing the closed unit disk and $ f(C_1)\subset C_1\minus{}({1})$, then f is constant. i have a solution in mind but it looks too simple.

[quote="kouboy"]let $ C_1$ denote the unit circle. suppose f is analytic in a region containing the closed unit disk and $ f(C_1)\subset C_1 \minus{} ({1})$, then f is constant. i have a solution in mind but it looks too simple.[/quote]
Is $ C_1 \equal{}\{z: |z| \equal{} 1\}$ or $ C_1 \equal{} \{z: |z| < 1\}$ and what does $ (1)$ mean?

it is the unit circle and by that i mean just the boundary. and by $ C_1\minus{}({1})$ i meant the unit circle minus the point 1. i am sorry for not being clear earlier.

yes, that was exactly what i did too. :)

[quote="mlok"]It would be better to write $ C_1\setminus\{1\}$. By the [url=http://en.wikipedia.org/wiki/Argument_principle]argument principle[/url] $ f$ has no zeros in the unit disk. Apply the maximum principle to $ f$ and $ 1/f$.[/quote]
Can you provide more detail please?

what does the argument principle say? if you want to find the no of zeros of a holomorphic function "inside" a closed curve ( f is never zero on that curve) then look at the image of the curve under f, look at how many times it winds around the origin, thats the no of zeros. so with this mind, the winding no for our particular problem is 0. so there are no zeros inside .

now $ |f|$=1 on the unit circle and hence by the maximum modulus principle $ |f(z)|\leq 1$ for z inside unit disk. consider $ \frac{1}{f}$ . it is holomorphic on the closed unit disk ( since f has no zeros). therefore the inequalities in the previous line apply to 1/f. conclude that $ |f|$=1 identically. but a holomorphic function with this property has to be a constant ( absolute value 1)


Let  $OX_1 , OX_2$  be  rays in  the interior of a  convex  angle $XOY$ such that $\angle XOX_1=\angle YOY_1< \frac{1}{3}\angle XOY$. Points  $K$  on $OX_1$  and  $L$  on  $OY_1$  are fixed so that  $OK=OL$, and  points $A$, $B$  are  vary on rays $(OX , (OY$  respectively  such that  the area  of the pentagon  $OAKLB$  remains  constant. Prove that  the  circumcircle of the  triangle $OAB$  passes  from a  fixed  point,  other than  $O$.

AS the pentagon has a fixed area, OA + OB is constant.
If F is the second common point of the circle OAB with the bisector of <XOY, we have
OA + OB = 2.OF.cos<XOY; hence F is a fixed point.


calculus
derivative
function
analytic geometry
calculus computations

Given that x^2+y^2-2x+4y=-9

Find y' using implicit differentiation.

[hide="This is what I have. Pls check if it's right."]
dy/dx=(-x+1)/(y+2)
[/hide]

But then, I have to show that this derivative is nonsense because it does not meet the conditions of the Implicit Function Theorem.

I can't seem to find anything wrong with it.

Never mind  :blush:

I figured it out

For those of you who were wondering,
[hide="this is what happened"]
THE FUNCTION DOESN'T EVEN EXIST IN THE REAL COORDINATE GRID!!!
After you use complete the square, the equation becomes
 (x-1)^2+(y+2)^2=[b][u]-4[/u][/b]

[/hide]


I don't remember... but is there a formula for $1^k+2^k+3^k+...+n^k$ ???... maybe factorising that will help...

[hide="Solution"]Consider any even $n$ that doesn't divide $4$ (such as $n=2$ or $n=6$).

$1^{1987}+2^{1987}+3^{1987}+...+n^{1987}$ has an odd number of odd terms, and is therefore odd. $n+2$ on the other hand is even.

Since even numbers obviously can't divide odd numbers, $n+2$ cannot generically divide $1^{1987}+2^{1987}+3^{1987}+...+n^{1987}$.[/hide]

EDIT: Oops, I thought you meant for all n

[quote="The Zuton Force"][hide="hint"]"Proceed by contradiction." - PenguinIntegral's sig[/hide]

[hide="Solution"]Consider any even $n$ that doesn't divide $4$ (such as $n=2$ or $n=6$).

$1^{1987}+2^{1987}+3^{1987}+...+n^{1987}$ has an odd number of odd terms, and is therefore odd. $n+2$ on the other hand is even.

Since even numbers obviously can't divide odd numbers, $n+2$ cannot generically divide $1^{1987}+2^{1987}+3^{1987}+...+n^{1987}$.[/hide][/quote]
How about considering odd n?


is computerscience related physics  or computer science related mathmatics research similar to the stuff you do in AP computer science or is  it very different?  for example: biology research is very different from the stuff you do in a biology class.

thanks

Research in any science is by no means limited. If it was limited then it would not be interesting. Biology in AP is introductory. Biology research is surely higher since it concerns areas that are not fully covered. If you read the CLR(S) then I think you'll get a picture of the stuff that CS research deals with. CS in physics and maths is different, since computer acts as a means to help the research rather than the research itself.  But as I have said, research is not limited and CS can play a vital role in physics or math research. For example, a doctor can operate from US in Europe using a computer. So, research is about everything.


Harold tosses a nickel four times.  What is the probability that he gets
at least as many heads as tails?

The proabbility that he gets the same heads and tails is $ \frac{6}{16}$, and the probability that he gets more is $ \frac{5}{16}$, so we have $ \frac{11}{16}$/

How come the binomial theorem doesn't work here? ($ \binom{4}{2}\left(\frac{1}{2}\right)^4\equal{}\frac{3}{8}$)?

This is because it says "What is the probability that he gets at [i]least [/i]as many heads as tails?".  You only found the probability that he gets the same amount of heads as tails.  Instead, we should do $ \binom{4}{2}\left(\frac{1}{2}\right)^{4}\plus{}\binom{4}{3}\left(\frac{1}{2}\right)^{4}\plus{}\binom{4}{4}\left(\frac{1}{4}\right)^{4}\equal{}\frac{6}{16}\plus{}\frac{4}{16}\plus{}\frac{1}{16}\equal{}\frac{11}{16}$, as AIME15 showed.

Oh...wow...until now, I haven't noticed the phrase "at least"...epic fail...thanks for clarifying!

I think an easier way to do this is using complementary counting.  The problem is the same as "what is the probability that he does not get at least as many heads as tails".  The two cases here are three tails and one head, and four tails, for which there are 4 and 1 cases respectively.  This adds up to 5, and gives the probability as $\frac{5}{16}$ for not meeting the requirement.  From here we just subtract from one: $1-\frac{5}{16}=\frac{16}{16}-\frac{5}{16}=\frac{11}{16}$ which is your answer.

[hide=lol the alcumus sol was listing]We do casework on the number of heads.

Case 1: 2 heads
4c2/2^4

Case 2: 3 heads
4c3/2^4

Case 3: 4 heads
4c4/2^4

Add: (6+4+1)/16=11/16[/hide]

nah why is this lvl 21

[hide=sol]Based on the sentence provided, harold has to get at least two heads. Using binomial probability, we can quickly find the answer is $\dfrac{6+4+1}{16} = \boxed{\dfrac{11}{16}}$[/hide]


It says AMC office, but which address is AMC office?

ALso is s-iii the only form you need to send?

I'm wondering the same thing.  I'm afraid if I just send the form to the generic UNL address listed on the bottom of the letter that it will never reach the AMC office.

I would just fax it to the number that it specifies... if you don't have a fax machine, ask your teacher or go to kinko's... better to drive to kinko's than to have your form get lost :O

Read the USAMO Teachers Manual:
In Section III, on page 5, Item 1:  "We suggest faxing to 303/374-6339"

USAMO Teachers' Manual, page 15, on the top lineof the S-III form itself: "....returned by fax 303/374-6339"

This is all available at our website in addition to be in the hands of your contest manager.

Steve Dunbar
AMC Director


algebra
polynomial
function
vector
linear algebra
matrix
superior algebra

If the linear function f has the property that for all x in E (vectorial space) the family $ x,f(x),f^2(x),...,f^n(x)$ is not free (that is, there exists a linear combination with coefficients which are not all 0 such that the result is 0) and n is fixed then there is a polynomial of degree n that vanishes on f.

here what I think:
Let $x \in E$. $I=\{P \in K[X], P unitary, P(f)(x)=0\}$ is an ideal
generated by $P_x$. (and $deg(P_x) \leq n$)

1/ prove that if $F$ is a subspace of finite dimension then there exists
$x\in F$ such that $P_x$ is the minimal polynomial of the restriction of $f$
to $F$

2/If $dim(E) < +\infty$, it is finished.
  Suppose that $dim(E) = +\infty$ and that there is no minimal polynomial for $f$.
  Hence, for every $P \in K[X]$, there exists $x \in E$ such that $P(f)(x) \neq 0$.
  Take $E_1$ an arbitrary subspace of finite dimension, $P_1=P_{v_1}$ the minimal polynomial
  of the restriction of $u$ to $E_1$ (so $deg(P_1) \leq n$). Take $v_2$ such that $P_1(v_2) \neq 0$,
  $E_2=vect(E_1, v_2)$. Then $P_2$, the minimal polynomial of the restriction of $u$
  to $E_2$ is a multiple of $P_1$. Moreover, $P_2=P_{x_2}$ so $deg(P_1) < deg(P_2) \leq n$.
  Continue like this (n+1) times.This is a contradiction.

here what I think:
Let $x \in E$. $I_x=\{P \in K[X], P unitary, P(f)(x)=0\}$ is an ideal
generated by $P_x$. (and $deg(P_x) \leq n$)

1/ prove that if $F$ is a subspace of finite dimension then there exists
$x\in F$ such that $P_x$ is the minimal polynomial of the restriction of $f$
to $F$

2/If $dim(E) < +\infty$, it is finished.
  Suppose that $dim(E) = +\infty$ and that there is no minimal polynomial for $f$.
  Hence, for every $P \in K[X]$, there exists $x \in E$ such that $P(f)(x) \neq 0$.
  Take $E_1$ an arbitrary subspace of finite dimension, $P_1=P_{v_1}$ the minimal polynomial
  of the restriction of $f$ to $E_1$ (so $deg(P_1) \leq n$). Take $v_2$ such that $P_1(v_2) \neq 0$,
  $E_2=vect(E_1, v_2)$. Then $P_2$, the minimal polynomial of the restriction of $f$
  to $E_2$ is a multiple of $P_1$. Moreover, $P_2=P_{x_2}$ so $deg(P_1) < deg(P_2) \leq n$.
  Continue like this (n+1) times.This is a contradiction.

Thanks,alekk, I will study your solution. I think it is a very very hard problem.

let $E$ be a vector space of finite dimension, $f$ an endomorphism. One can prove that there exists $v \in E$ such that $P=P_v$ is the minimal polynomial of $f$. Suppose the contrary:
then $P_x | P$ and $P_x \neq P$ for every vector $x \in F$. But there is only a finite number of unitary polynomials that divise $P$, and are different of $P$: call $D$ this finite set of  polynomials . So $E = \cup_{Q \in D} Ker(Q(f))) $. But this is impossible because a vector space is not the union of a finite number of proper subspaces, if $K$ is infinite.

So if my solution worked(it's not sure), it only worked if $K$ is infinite. The problem is still open.

But the assertion that there exists x such that the minimal polynomial is $P-x$ I think it's always true, not matter if the filed is finite or not. I remember to have seen a proof a week ago and ithe condition of infinity of cardinal was not impossed. I will check.

and is the solution correct ? Moubi, what do you think ? Perhaps you know the official one ?

Theorem (Rational Canonical Form): Let $V$ be a finite-dimensional vector space over a field $F$, and let $A$ be a linear transformation. It is possible to find subspaces $V_1,V_2,\dots,V_r$ such that $V$ is the direct sum of these subspaces and $A(V_k)\subset V_k.$ In addition, there are polynomials $q_1,q_2,\dots,q_r$ such that $q_k$ is the minimal polynomial of $A$ over $V_k$, the degree of $q_k$ is $\dim V_k$, and $q_{k+1}|q_k.$
We can furthermore choose a basis for each $V_k$ so that the matrix of $A$ over that basis is of the form $\begin{bmatrix}0&0&0&\cdots&0&-c_o\\1&0&0&\cdots&0&-c_1 \\0&1&0&\cdots&0&-c_2\\\vdots&\vdots&\vdots&\ddots&\vdots&\vdots\\ 0&0&0&\cdots&0&-c_{m-2}\\0&0&0&\cdots&1&-c_{m-1}\end{bmatrix}$, with the $c_i$ the coefficients of the polynomial. The polynomial on $V_1$ is of course the minimal polynomial of $A.$
The polynomials are unique, although the subspaces and bases need not be.

This is standard stuff, if a little obscure. Most treatments of linear algebra prefer working in algebraically complete fields. It's in the graduate algebra textbook I have (Lang) under the heading of "Decomposition over one endomorphism". I did rephrase it a little, and I got the name elsewhere.


linear algebra
matrix
function
superior algebra
superior algebra solved

Let D={A belongs to M_n(R) | det(A)<>0} and the function f : D-->D, f(A)=A*
a) determine n natural, n>=2 so that f is bijective and then compute f^(-1).
b) does there exist A1,A2 in M_3(Z) invertible with A1<>A2 and A1*=A2*?

the function is defined on smilies??? :)
f:|)->D

for a)
since detA=/=0=>A* = det(A)A-1=>f(A) = det(A).A-1
let g = f-1=>
f(g(A))=I_n=>det(g(A))g(A)-1=I_n=>det(g(A))I_n = g(A)
=>det(g(A)) can be 1 or -1 if n is odd.
case I.det(g(A))=1=>g(A) = I_n. So f-1(A)=I_n(for every n natural n>1).
case II.det(g(A))=-1=>g(A)=-I_n(n odd)=>g(f(A)) = -I_n=/=I_n
The only solution is f-1(A) = I_n.
for b) since f is bijective results that those two matrices don't exist.

hope i'm not wrong :D

I guess that the function can be defined on smiles! :D

just joking, I guess that I wrote : too close to D and it became :D

I have reedited the post! :D


topology calculus derivative integration Princeton college geometry

I am near the end of the semester now and I have realized* that I understand very little of my real+complex Analysis course. *(I have been realizing this from the start actually)
I am slow or unable to read the proofs in my book, integrate information and understand the motivation. So i have been wondering why i find this math particularly difficult, aside from the answers: (more complex, more extensive theory, initially less intuitive). the main reason i came up with is that analysis is less adaptable to visual thinking, and i perhaps overuse visual imagery to understand math. So i was hoping to start a discussion about how we store and manipulate mathematical information in our brains.
For example i would guess that most people would invoke a visual image of a graph as points connected by lines in the process of solving a question of graph theory. It is often a natural thing for out minds to convert information to a visual format, and in this format to be able to extract information more readily. We are all familiar with this.

So this is one technique that is an aid in understanding and manipulating mathematical information, what are some others, discuss.

Ill also add a request for help with making concepts in analysis more intuitive. How can you help a fool like me? I don't know perhaps you can find a way? I have textbooks to read the proofs so I don't need that, but any 'side notes' (connections, simplifications, motivations) would be very helpful I think. The concepts I am talking about are the ones in a 'standard' 2nd year math course- R/C analysis. Concepts like convergence and its various criteria, sub-sequences: Bolzano-Weierstrass and completeness of $ C^n$, Double series... or other related topic, would enormously appreciate any help, thxs.

I also had the same problem when I took Real Analysis. One thing that helped me was reading the same material from different books - especially books that were good at explaining the motivation behind the theorems and the usefulness of the theorems. I'll look to see if I can remember what books those were for me.

As for techiques for understanding, having a body of examples and counterexamples in invaluable in Analysis. There are books on counterexamples in Real Analysis, which can be very useful.

I've heard that this book is particularly good in these respects, i.e. giving more examples and visual intuition: http://www.amazon.com/Real-Mathematical-Analysis-Charles-Chapman/dp/0387952977/ref=sr_1_5?ie=UTF8&s=books&qid=1243025973&sr=8-5.

In my case, I actually never really learned real analysis per se. I learned topology first, picked up topology of the real line, did a few proofs for differentiation and integration, and that was it. As a result, I never felt such difficulties. I can't be sure whether or not this leaves a gap in my knowledge, although I seem to understand almost everything in a typical real analysis syllabus. (I am thinking of a real analysis course like Math 215 at Princeton, which is probably not quite at the level of a course on R/C analysis.)

Ummm.... I actually find analysis to be pretty visual? I guess I don't really understand the problem here, but it might help to phrase questions in unrigorous intuitive terms first, give a hand-wavy explanation for it, and then try to formalize it later. Also, when dealing with R^n, it might help to think about things in R^3, which often captures most of the idea anyway.

gauss202: can you please name the books you found helpful? I also prefer texts that take the approach of motivating concepts. As for counter examples, you say there are whole books dedicated to this? Strange... I'll look into it.

calc rulz: I looked through the content page of the book you suggested, it looks interesting but its content is quite different to the syllabus in my course. Also, I generally dislike book that give lots of example, beacuse I find that the information just becomes more cluttered in my head. I prefer texts that unify concepts and motivate them, like Sheldon-Axler's Linear Algebra Done Right. About learning topology first, did you skip your real analysis course and just learn it yourself? What topology and how much of it is useful is understanding analysis? We did a little point-set topology in our course, you know open/closed sets and how we can describe limits and continuity with them.

probability1.01: yeah my post was a bit of a mix of two topics so I did not pose any clear questions in the end. I think that the first problem with lack of understanding is not knowing what exactly it is that needs to be understood. Not understanding there is a 'problem', but having difficulties with things related to the subject. That is my concern, analysis has been a very 'fuzzy' topic in my mind. Now I think I know what I need to do. I need to understand the individual concepts better and then to put it all in perspective. Perspective being that I see how the concepts link together better and understanding some of the motivation behind the theory.

I will post more specific questions in the analysis threads soon, come help me if u can. thanks.

I find analysis pretty visual (at least I myself think more in "pictures" than in "sentences" when solving analysis problems and, when presenting a long analysis proof in class, I keep in mind only a couple of key ideas and some "overall image of the argument" that is completely non-verbal and just translate those into the standard $ \varepsilon \minus{} \delta$ language). The main difficulty seems to lie in this translation from the standard symbolic language we use for writing proofs down to the internal "picture language" of the mind we use for creating/memorizing them and back. It is quite hard for an average student to learn this skill, indeed, but once it is mastered, everything becomes very clear and you get able to see the underlying structure behind the forest of symbols. As a matter of fact, you notice those epsilons and deltas no more than the shapes of separate letters in this text: your eye just glides over them and your brain translates all those curvy lines into something that has no relation whatsoever to their shape, something you call the "meaning". Once you got the meaning, the letters, words, and the entire sentences do not matter any more. The elementary story units became something like "main characters", "conflicts", "resolutions", etc. When asked to retell the story, you preserve those but not the exact words; as a matter of fact, you do not even remember the words in the original story; you just choose the appropriate ones from your vocabulary as the story line unfolds...

The skill of understanding/producung the proofs is very akin to this reading/writing skill, and, ideally, should be taught as such: you start with the "alphabet", proceed to simple "words", then try to make sense of "sentences", and so on, until you end up with the ablity to read/write/retell/summarize a good "essay". The simplest way to acquire this skill is to start with reading "children books" and working your way up. Unfortunately, very few "children books" in analysis have ever been written and even fewer are available. Frankly speaking, I do not know a single one in English. :(  So, you have to start with quite high level texts in the beginning and many people fail to overcome the "entrance barrier". To illustrate what I'm talking about, compare two questions.

1) A lady is working with two dogs on leashes. She starts to pull both leashes towards herself. What happens to the distance between the dogs? How long should the free parts of leashes be to ensure that the dogs are not more than 12 feet apart?

2) Prove that every convergent sequence is Cauchy.

In my mind there is absolutely no difference whatsoever between these two questions. Nevertheless, every student in my class answered the first one in no time and quite many had a hard struggle with the second one.

This post is getting rather long, so I'll stop here. I hope you see the point.


geometry
parallelogram
geometry unsolved

Let $G$ be a convex quadrilateral. Show that there is a point $X$ in the plane of $G$ with the property that every straight line through $X$ divides $G$ into two regions of equal area if and only if $G$ is a parallelogram.

All the Greek people should see this problem. It is the same with Greek competition ''Thales'' , pr 4 at A lyceum.

 BTW, when British 2005-2006 round 1 took place ?Thanks

简单题！

画图麻烦，直接写好了。

引理：给定直线l1,l2,l3.  l1//l2.  l3bu//l1.且l2的一部分夹在l1与l3之间.则在l2夹在l1与l3之间的部分上不存在点X，使过X做任意不重和的两直线与分别交得的两三角形面积相等。

证明：反证：设使过X做的任意不重和的两直线为L1,L2.L1与l1和l3分别交于A，D。L2与l1和l3分别交于B，C。设l3与l1,l2分别交于F，E。

由S（ABX）=S（CDX）得AX*XB=CX*XD，即AX/XD=CX/XB。而AX/XD=EF/ED，CX/XB=CE/EF。故EF/ED=CE/EF，即EF2=CE*ED。由于直线l1,l2,l3是给定的，故EF2为定值，而CE*ED随L1,L2变化，矛盾！故引理得证。

回到原题：

若ABbu//CD，过X点做EX//AB交CD于E。由上引理知过X做任意不重和的两直线与分别交得的两三角形面积不可能相等。这与题意不符！故AB//CD。同理可证AD//BC。故四边形ABCD是平行四边形，原命题得证！

Well it's rather an easy problem. Of course our point $X$ must be in the interior of $G$. In other case as $G$ is convex, we can take a line passing through $X$ and not intersecting $G$. So let $X$ be our point in the interior of quadrilateral $G$, which we will denote with $ABCD$. Let $l$ be a line passing through $X$ and intersecting $AB$ and $CD$ in points $P$ and $Q$ respectively. Let $K$ and $M$ be such points on $CD$ that $KQ=QM$ and $K\neq M$. Let lines  $KX$ and $MX$ intersect side $AB$ in $L$,$N$ respectively. By $S(XYZ)$ we will denote area of triangle $XYZ$ and the same thing for quadrilaterals... As lines $QP$,$KL$ and $MN$ are lines dividing $G$ into two parts of equal area, we obtain equations \[ S(KQX)+S(XPL)=S(QMX)+S(NPX) \] but as $S(KQX)=S(QMX)$ we have $S(XPL)=S(XNP)$ which means that $NP=PL$ .Now let $k$ be a line passing through $N$ and parallel to $CD$ this line intersects with lines $PQ$ and $KL$ in points $T$ and $U$ respectively. As we have $KQ=QM$ we have also $NT=TU$ ,but there is also equality $NP=PL$, and so on, if $k \neq AB$ it would mean that $TP||UL$ which is impossible. We obtained that $k=AB$, so $AB||CD$.
Analogically we may obtain that $AD||BC$ which means that $G$ is a parallelogram! :)  Now , if $G$ is a parallelogram for point $X$ we can take intersection point of it's diagonals  :)


inequalities
vector
trigonometry
calculus
function
triangle inequality
linear algebra

[b]Prove the Cauchy-Schwarz inequality:
|a•b| ≤ |a| |b|
for all pairs of vectors a and b.[/b]

[u][i]AND!![/i][/u]

[i](this is a seperate one)[/i]

[b]if a and b are arbitrary vectors, then

|a - b| ≥ |a| - |b|[/b]

a hint i received was to write a = (a - b) + b, then to apply the triangle inequality:  |a + b| ≤ |a| + |b| dunno where to go with this one :S

THANKS a bunch ya'll!

eagle

The first one is obvious (definition of the dot product).

The second one is not true (just consider negative b).

wht do u mean by "the second one is not true" :S

[quote="Phelpedo"]
The second one is not true (just consider negative b).[/quote]

Phelpedo, there's no such thing as "negative vector".

ive seen some proofs, but i don't want a complex explanation. Obviously, if one of the vectors is 0, then its true. What about non-zero vectors?

Help would be appreciated! :(

[quote="eagle"][quote="Rzeszut"]$|a-b|+|b|\geqslant |a-b+b|=|a|$, so $|a-b|\geqslant |a|-|b|$.[/quote]

ok? and??

sorry but i think thts an incomplete proof.[/quote]
From the triangle inequality: \[|a-b|+|b|\geqslant |(a-b)+b|= |a-b+b|= |a|.\] Thus \[|a-b|+|b|\geqslant |a|.\] Subtract $|b|$ from both sides: \[|a-b|\geqslant |a|-|b|.\] It is the same as
[quote="eagle"][b]if a and b are arbitrary vectors, then

|a - b| ≥ |a| - |b|[/b][/quote]
I hope it's clear and complete now.

[quote="eagle"]Thanks alot Rzeszut.

Anyone hav a simple proof to the Cauchy-shawrz?

Thanks![/quote]
It's well-known (or even used as definition of dot product) that $a\cdot b= |a|\cdot |b|\cdot \cos\left(\angle (a,b)\right)$. Using $\mathop\forall_{\alpha}\cos(\alpha)\leqslant 1$ we have $a\cdot b= |a|\cdot |b|\cdot \cos\left(\angle (a,b)\right)\leqslant |a|\cdot |b|$, so $a\cdot b\leqslant |a|\cdot |b|$ which is Cauchy-Schwarz.

Are you taking Multivariable Calculus and Linear Algebra? What book are you using?

I just had a test on this stuff :wink:

[quote="Rzeszut"]It's well-known (or even used as definition of dot product) that $a\cdot b= |a|\cdot |b|\cdot \cos\left(\angle (a,b)\right)$. Using $\mathop\forall_{\alpha}\cos(\alpha)\leqslant 1$ we have $a\cdot b= |a|\cdot |b|\cdot \cos\left(\angle (a,b)\right)\leqslant |a|\cdot |b|$, so $a\cdot b\leqslant |a|\cdot |b|$ which is Cauchy-Schwarz.[/quote]
That's more or less true. In the context of Linear Algebra, one first finds a proof of the Cauchy-Schwarz inequality, then defines the angle between two vectors by $\cos \measuredangle \left( \overrightarrow a, \overrightarrow b \right) = \frac{\overrightarrow a \cdot \overrightarrow b}{\left| \overrightarrow a \right| \cdot \left| \overrightarrow b \right|}$.

For a proof of the general case, see [url=http://www.cs.ut.ee/~toomas_l/linalg/lin1/node8.html]here[/url] (scroll down to Proposition 1.4.2). It's basically the same proof as in the real case: we consider a second degree function that is non-negative on $\mathbb R$, and conclude its determinant must be $\leq 0$.

:lol: yeaa i have the same book!! 6th edition.

hve u done the cauchy-shwarz thm!??

eagle


projective geometry
geometry proposed
geometry

Dear Darij
   I'm sorry.  I have read "Sawayama and Thebault's theorem", I cannot understand "Conclusion: using the converse of Pappus’s theorem ([17, Proposition 139] and [3, p.67]), applied to the hexagon PEIGQDP, the points P, I and Q are collinear." Do you help me? thank you!
Sincerely,
Fang-jh

[quote="Fang-jh"]I have read "Sawayama and Thebault's theorem", I cannot understand "Conclusion: using the converse of Pappus’s theorem ([17, Proposition 139] and [3, p.67]), applied to the hexagon PEIGQDP, the points P, I and Q are collinear."[/quote]

In fact, this question could have been posed to the whole board. Just in order to make it clear to others, we are talking about this paper: [url=http://forumgeom.fau.edu/FG2003volume3/FG200325index.html]Jean-Louis Ayme, [i]Sawayama and Thébault's Theorem[/i], Forum Geometricorum 3 (2003) pp. 225-229[/url].

The Pappus (or Pappos) theorem states:

[color=blue][b]Pappus theorem.[/b] If A, C, E are three collinear points, and B, D, F are three collinear points, then the points $AB\cap DE$, $BC\cap EF$ and $CD\cap FA$ are collinear.[/color]

Here, the abbreviation $g\cap h$ stands for the point of intersection of two lines g and h.

You can find some proofs of the Pappus theorem online.

The converse of the Pappus theorem used by the author is the following fact:

[color=blue][b]Converse Pappus theorem.[/b] If A, C, E are three points, and B, D, F are three collinear points such that the points $AB\cap DE$, $BC\cap EF$ and $CD\cap FA$ are collinear, then the points A, C, E are collinear.[/color]

The author applies this converse Pappus theorem to the three points P, I, Q and the three  collinear points E, G, D. Since the points $PE\cap GQ$, $EI\cap QD$ and $IG\cap DP$ are collinear (in fact, these three points all lie on the line at infinity, since PE || GQ, EI || QD and IG || DP), he concludes that the points P, I, Q are collinear.

Actually, the converse Pappus theorem is a trivial consequence of the Pappus theorem itself, applied to the collinear points B, D, F and the collinear points $CD\cap FA$, $BC\cap EF$, $AB\cap DE$.

  Darij


[quote="Zamfirmihai"]$ a,b,c>0$ cu $ abc=a+b+c$. Demonstrati ca:

$ (\frac{a}{b})^{2}+(\frac{b}{c})^{2}+(\frac{c}{a})^{2}\geq \sqrt{a^{2}+b^{2}+c^{2}}$[/quote]

Inegalitatea este echivalenta cu:

$ \left( \sum \left(\frac{a}{b}\right)^{2}\right)^{2}\geq \sum a^{2}$

Folosind inegalitatea lui Vasc $ (x^{2}+y^{2}+z^{2})^{2}\geq 3(x^{3}y+y^{3}z+z^{3}x)$ este suficient sa aratam ca:
$ 3\sum \frac{a^{3}}{b^{2}c}=3\sum \frac{a^{4}}{abc\cdot b}=3\sum \frac{a^{4}}{(a+b+c)b}\geq \sum a^{2}$ $ \Leftrightarrow$
$ 3\sum \frac{a^{4}}{b}\geq (a^{2}+b^{2}+c^{2})(a+b+c)$
Dar din CBS $ \Rightarrow$ $ 3\sum \frac{a^{4}}{b}\geq 3\frac{(a^{2}+b^{2}+c^{2})^{2}}{a+b+c}\geq(a^{2}+b^{2}+c^{2})(a+b+c)$
$ \Leftrightarrow$ $ 3(a^{2}+b^{2}+c^{2})\geq (a+b+c)^{2}$.Qed


Prove that there are no integers x and y for which
x^2 + 3xy − 2y^2 = 122.
 :!:

[hide="Solution"] $ 4x^2 \plus{} 12xy \minus{} 8y^2 \equal{} 488 \equal{} (2x \plus{} 3y)^2 \minus{} 17y^2$.  A prime $ p$ divides a number of the form $ a^2 \minus{} 17b^2$ if and only if either $ p | a, p | b$ or $ 17$ is a quadratic residue $ \bmod p$.  The only prime factor of $ 488$ other than $ 2$ is $ 61$, but

$ \left( \frac{17}{61} \right) \equal{} \left( \frac{61}{17} \right) \equal{} \left( \frac{10}{17} \right) \equal{} \left( \frac{2}{17} \right) \left( \frac{5}{17} \right) \equal{} (\minus{}1)^{ \frac{17^2 \minus{} 1}{8} } \left( \frac{17}{5} \right) \equal{} \minus{}1$

so this cannot occur.[/hide]

[quote="t0rajir0u"][hide="Solution"] $ 4x^2 + 12xy - 8y^2 = 488 = (2x + 3y)^2 - 17y^2$.  A prime $ p$ divides a number of the form $ a^2 - 17b^2$ if and only if either $ p | a, p | b$ or $ 17$ is a quadratic residue $ \bmod p$.  The only prime factor of $ 488$ other than $ 2$ is $ 61$, but

$ \left( \frac {17}{61} \right) = \left( \frac {61}{17} \right) = \left( \frac {10}{17} \right) = \left( \frac {2}{17} \right) \left( \frac {5}{17} \right) = ( - 1)^{\frac {17^2 - 1}{8} } \left( \frac {17}{5} \right) = - 1$

so this cannot occur.[/hide][/quote]

May be stupid, I can't understand

$ \boxed{\left( \frac {17}{61} \right) = \left( \frac {61}{17} \right) = \left( \frac {10}{17} \right) = \left( \frac {2}{17} \right) \left( \frac {5}{17} \right) = ( - 1)^{\frac {17^2 - 1}{8} } \left( \frac {17}{5} \right) = - 1}$

What is it?

The [url=http://en.wikipedia.org/wiki/Legendre_symbol#Properties_of_the_Legendre_symbol]Legendre symbol[/url] has several useful properties which make it very easy to compute whether a given residue is a quadratic residue modulo a given prime.  That computation demonstrates that $ 17$ is not a quadratic residue $ \bmod 61$.


Let $ D,E,F$ points on the sides $ BC,CA,AB$ in triangle $ ABC$ such that incircle of triangle $ ABD$ and $ CAD$, $ BCE$ and $ ABE$, $ CAF$ and $ BCF$ tangent externally
Prove that lines $ AD,BE,CF$ concurrent

[quote="Virgil Nicula"][color=darkred][b][u]An easy extension.[/u][/b]

Let $ ABCD$ be a convex quadrilateral. Prove that the following statements are equivalently :
$ 1\blacktriangleright$ the incircles of the triangles $ ABC$ , $ ADC$ are tangent .
$ 2\blacktriangleright$ the incircles of the triangles $ ABD$ , $ BCD$ are tangent .
$ 3\blacktriangleright$ the quadrilateral $ ABCD$ is circumscriptible, i.e. $ AB \plus{} CD \equal{} AD \plus{} BC$ .[/color][/quote]

[color=darkblue][b]Remark.[/b] Let $ ABDC$ be a convex quadrilateral. In the particular case
(to limit) when $ D\in (BC)$ the upper property becomes :[/color]

[quote="Vorelin Tancu"] [color=darkred]Let $ ABC$ be a triangle. Then for a point $ D\in (BC)$ , the incircles of the triangles $ ABD$ and $ ACD$ are tangent if and only if $ AB \plus{} DC \equal{} AC \plus{} DB$ , i.e. the ray $ [AD$ is the bisector of the angle $ \widehat {BAC}\ .$[/color] [/quote]


Let r be the number that results when both the base and the exponent of a^b are tripled, where a,b>0. If r equals the product of a^b and x^b where x>0 then x equals what?

When a and b are tripled, we have $(3a)^{3b} = (27a^3)^b = (ax)^b$.  So $x = 27a^2$


geometry
circumcircle
perpendicular bisector
geometry unsolved

For a point $P$ inside the equilateral triangle $ABC$ we have that $\measuredangle APC= 120^\circ$. Let $M$ be the intersection of $CP$ and $AB$, and $N$ is the intersection of $AP$ with $BC$. Find the locus of the circuncenter of triangle $MBN$.

Let $O$ be the circumcentre of triangle $ABC$ and $O'$ be the circumcentre of triangle $BMN$. Since $\angle{APC}=120^{o}=\angle{AOC}$, so $A$, $O$, $P$, $C$ lie on the same circle. Therefore $\angle{OPA}=\angle{OCA}=30^{o}$. Since $\angle{MPA}=60^{o}$, so $\angle{MPO}=30^{o}$. Because $\angle{MBO}=30^{o}$, thus $P$, $O$, $M$ and $B$ are concyclic. Similarly, we can show that $O$, $P$, $N$ and $B$ are concyclic. Hence $P$ and $O$ lie on the circumcircle of triangle $BMN$.

If we let the circumradius of triangle $BMN$ be $R$, then $O'O=R$. Since we also have $O'B=R$, and $B$ and $O$ are two fixed points, so we conclude that the locus of $O'$ is the perpendicular bisector of segment $BO$.

As $P \to C$, then $\angle{BO'C}\to 120^{o}$, thus the locus is within the circumcircle of $\triangle{ABC}$.


ratio
algebra
polynomial
induction
function
geometric sequence

(1+t)(1+t^2)(1+t^4)(1+t^8)(1+t^16)... right?

(1+t) = 1+ t
(1+t)(1+t^2) = 1+t+t^2+t^3
...(1+t^4) = 1+t^2+t^3+t^4+t^5+t^6+t^7
...(1+t^8) = 1+t^2+t^3...+t^13+t^14+t^15

each time you multiply it by (1+t^n) where n is one more than the last term of the result, so you have the original (1) and (t^n).
Therefore (1+t^(2^(k-1)))=1+t+t :^2: +t  :^3: +t^(2^k-1)

Lets test it with t=3 and k=4

(1+3)(1+9)(1+81)(1+6561)
4*10*82*6562
26320000

1+3+9+81 = 94??

Wait a minute. On the second part, you mean t^(k-1), not t^(2^k-1)?

I could see where you were going for a second, but now I'm confused.

EDIT:
Oh wait! I think you meant t^(2*k-1).
Okay same as before. (1+3)(1+9)... = 26320000.
then i need twice the amount for the second part, which is:
1+3+9+81+243+729+2187+6561 = 9814, but that still isnt right.

We can use the formula for a finite geometric sequnce:
g1(1-r^n)/(1-r), where r is the ratio, g1 is the first term, n is the number of terms, k.

1(1-3^4)/(1-3)
-80/-2
=40, but that still doesnt get me anywhere.

I still dont know. I can kindof see what you were getting at in the problem, but i guess i need more experience in proofing.

Hex -- you're off to a good start.  Your problem is that proving isn't the same thing as finding a pattern -- you have to show that your pattern will always continue to work.  There's a really useful mathematical tool for this called "induction."  Have you heard of it?  (Also, note that if you've shown something for t in general, you don't really have to check specific values, although sometimes it can help you in making sure you're right.  Also, note that t^(2^k - 1) when k = 4 is equal to t^15 -- I don't know why you stopped at t^4 when you were adding it up the other way.  xirtamax should have put in an elipsis (". . ."), because the sum isn't just those 5 terms, it's all powers of t from 0 to 2^k - 1.

Also, your idea with the geometric sequence is similar to another method for proving this problem, seperate from the induction.

[hide]alternative way to prove this :

These are monic polynomials, with the same degree, and the same roots.[/hide]

I could do it by induction, but what is the solution that belenos is
talking about?

What belenos was saying is this:

[hide]

We know two polynomials will be equal (up to multiplication by a constant) if they are of the same degree and any root of either is a root of the other.  So, if you can show that the roots of the LHS are distinct and each is a root of the right hand side, since they are both polynomials of the same degree, you will have shown that the polynomials are equal.[/hide]

Hint: [hide]multiply both sides by (1 - t)[/hide]

Or: [hide]binary + generating functions [/hide]. Actually, ComplexZeta posted this a little while ago, I think.


induction
number theory
relatively prime
absolute value
Diophantine Equations

Find all pairs $(x, y)$ of positive rational numbers such that $x^{y}=y^{x}$.

[quote="Peter"]Nice construction, but how do you know they are the only ones?[/quote]
You can find this problem in "American Mathematical Monthly".If someone wants,i could post a solution.

We surely do!

Also, can you please post the full reference? Then I can include it.

I'm sorry for my late reply. :blush:
Problems author:
[u]Oystein Ore[/u]
Problem:
Find all pairs of positive rational numbers $ a,b$,such that
$ a^b\equal{}b^a$
Solution:
Obviously $ a\equal{}b$,is the solution.
Now WLOG assume that $ b>a$.
Then let $ b\equal{}ra$,where $ r>1$,$ r\in\mathbb{Q_\plus{}}$,and $ a^{r\minus{}1}\equal{}r$,or $ a\equal{}r^{\frac{1}{r\minus{}1}}$.
Let $ \frac{1}{r\minus{}1}\equal{}\frac{u}{v}$,where $ u,v\in\mathbb{N}$ are relatively prime numbers.
Since $ a$ and $ r$ are rational numbers,then $ r\equal{}\frac{u\plus{}v}{u}$ should be $ v$-th power of a rational number.
But then numbers $ u$ and $ u\plus{}v$,as relatively primes,must be $ v$-th power of an integer number.
It is well-known and easy to prove,that an absolute value of the diffrence between the two $ v$-th powers of a natural numbers,is greater than $ v$,if $ v>1$.
Hence $ v\equal{}1$,and $ \frac{1}{r\minus{}1}\equal{}u$ must be an integer number.
So $ a\equal{}(\frac{u\plus{}1}{u})^u$ and $ b\equal{}(\frac{u\plus{}1}{u})^{u\plus{}1}$

1) There is an other interesting method by Sierpinski:

We are going to find all the solutions of
\[ (1)\hfill \qquad \qquad \qquad
x^{y} = y^{x},
\]
 in positive rational numbers, x,y with  $ x \ne y$.

Suppose that x,y is such a solution and x>y.Then $ r =  ( 1 + \frac{1}{r}  )x$ is a positive rational number and $ y = (1+\frac{1}{r})x$.
Therefore ${ x^{y} = x ^ (1+\frac{1}{r}) x}$ and, since
$ x^{y}=y^{x}$
we have also $ x^{(1+\frac{1}{r})}=y^{x}$
which proves that
$ x^{1+\frac{1}{r}}=y= (1+\frac{1}{r}) x$.
Hence  $ x^{\frac{1}{r}}= 1+\frac{1}{r}$   and
$ x=(1+\frac{1}{r} )^{r}$,
$ y=(1+\frac{1}{r} )^{r+1}$.

Let
$ r=\frac{n}{m}$,  with  $ (m,n)=1$ and
$ x=\frac{t}{s}$,  with $ (t,s)=1$. Since $ x=  ( 1+ \frac{1}{r})^{r}$ and
 we have $ \frac {m+n}{n})^{n\m}=\frac{t}{s}$
whence $ ( \frac{m+n}{n})^{n}=(\frac{t}{s})^{m}$.
Each side of this equality is an irreductible fractions.Since $ (m,n)=1$ then $ (m+n,n)=1$
and since $ (t,s)=1$ then ( $ t^{m}$ , $ s^{m}$)=1 .

Now, it follows that
$ (m+n)^{n}=t^{m}$ and $ n^{n}=s^{m}$.
By
$ (m,n)=1$ we infer that there exist natural numbers $ k$ and $ \ell$ such that
$ m+n =k^{m}$, $ n= \ell^{m}$ and  $ = \ell^{n}$.
Therefore  $ m+\ell^{m}=k^{m}$.
From this we deduce that
$ k\geq  \ell+1$.
If $ m>1$  we would have
 $ k^{m} >   ( \ell+1 ) ^{m} \geq \ell^{m}+ m \ell^{m-1}+1> \ell^{m}+m=k^{m}$
 which is impossible. So $ m=1$, whence $ r=\frac{n}{m}=n$. The conclusion is  that
\[ (2)\hfill \qquad \qquad \qquad
x=  ( 1+ \frac{1}{n})^{n}, \quad
y=  ( 1+ \frac {1}{n})^{n+1}
\]

with n natural number.Conversely, x and y verify the equation $ x^{y}=y^{x}$. Therefore, all the solutions of the equation in the rational numbers are given by (2)
with n natural number. It follows that n=1 is only value for which the equation has a solution in natural numbers $ (x,y)= (2,4)$.

2) There is an other possible solution but only for positive integers, using  :

Lemma: If $ a^{n}\mid b^{n}$ , then $ a\mid b$ for all  $ a, b, n$, positive integers.
Now let us suppose that in the equation
$ x^{y}=y^{x}$
one has $ y>x$ .Let $ y =x+t$ (t>0).Then (1) can be written as:$ x^{x+t}=y^{x}$
or $ x^ {x} \cdot x^{t} =y^{x}$ which implies:
$ x\mid y$.
So, by Lemma $ (a=x, b=y, n=x)$  we must  have $ x\mid y$.
Let $ y=kx$,  $ (k>1)$. Then the equation becomes $ x^{kx}=y{x}$  , yielding $ kx=x^{k}$ or $ x^{k-1}= k$. For $ k=2$ we get 2. For $ k\geq 3$ by induction we show that $ x^{k-1}$ >k ( for  $ x\geq 2$), so  the equation is impossible.Therefore $ k=2,x=2,y=4$.


hallo,
let $p(x) \in \mathbb{Z}[x]$  of the form $p(x)=x^{2}+ax+b$ and roots in $\mathbb{Z}$. Prove that if there $\exists c \in \mathbb{Z}$ s.t. $p(c)$ is prime, then $2p(c)+2=\max{\{p(c+1), p(c-1)\}}$.

It is obviosly, because $p(x+1)+p(x-1)=2p(x)+2 \ \forall x$ and from your condition following $p(c+1)=0$ or $p(c-1)=0$.
It is false if p(c)=-p - negative prime.


geometry
circumcircle
symmetry
angle bisector

ABC is an equilateral triangle. C lies inside a circle center O through A and B. X and Y are points on the circle such that AB = BX and C lies on the chord XY. Show that CY = AO.

Can you clarify "C lies inside a circle center O through A and B"?

i think it means: $\Gamma$ is a circle passing through $A$ and $B$. $O$ is the center of $\Gamma$, and $C$ is inside $\Gamma$.

[hide]BA = BC = BX, so the circumcircle of ACX has centre B.

So <YBA = <YXA     (concyclic points)

              = <CXA

              = (1/2)<CBA    (since B is centre of circumcircle of ACX)

Y lies on the angle bisector of <CBA.  Since ABC is equilateral, Y lies on the axis of symmetry through B, so CY = AY.

Also, <YOA = 2<YBA  (since O is centre of circumcircle of YAB)

                  = <CBA = 60

and we know OY = OA.

So YAO is an equilateral triangle.

Thus CY = AY = OY.

[/hide]

Nice problem -- where's it from?

We have $\angle BAY=\pi-\angle BXY=\pi-\angle BCX=\angle BCY$, so $BY$ is the bisector of $\angle ABC$ and $YA=CY$. It follows from here that $\angle AOY=2\angle ABY=2\cdot 30^{\circ}=60^{\circ}$, and since $OA=OY$, it means that $AOY$ is equilateral, so $OA=YA=CY$, Q.E.D.


conics
parabola
calculus
derivative
analytic geometry
graphing lines
slope

Find the equations of all lines through the origin tangent to the parabola $ y\equal{}x^2\minus{}2x\plus{}4$

Put $ y\equal{}x\equal{}0$ in $ y\minus{}y_{0}\equal{} k(x\minus{}x_{0})$ and express the gradient $ k$ from the equation. Since the line with this gradient which goes thorugh $ (0,0)$ should be the tangent to the curve, you can assume $ (x_{0},y_{0})$ lie in the curve and hence calculate the derivative of $ y(x)$ in this point, which should be the same to $ k$.

[hide]
So $ k\equal{}\frac{y_{0}}{x_{0}}\equal{}2x_{0}\minus{}2 \Leftrightarrow x^{2}_{0}\minus{}2x_{0}\plus{}4\equal{}2x^{2}_{0}\minus{}2x_{0} \Rightarrow x_{0}\equal{}\pm 2.$[/hide]

Little confused here. Never had any teachers use $ x^{2}_{0}$ before. How do I formulate this into the form $ y\equal{}expression1$, $ y\equal{}expression2$, etc?

[quote="ben_b"]Never had any teachers use $ x_{0}^{2}$ before.[/quote]
It's just notation to refer to some arbitrary point that lies on the line.  Recall that you need two points to determine a line, so $ (x_{0},y_{0})$ is one of them.  Rajiv's work is pretty simple to follow so try to figure out what he's doing.

[hide]Slope of tangent lines given by $ y' \equal{} 2x \minus{} 2$. Line has equation $ y \equal{} mx$ so $ x^2 \minus{} 2x \plus{} 4 \equal{} mx$ where $ m \equal{} y' \equal{} 2x \minus{} 2$.

$ x^2 \minus{} 2x \plus{} 4 \equal{} (2x \minus{} 2)x$
$ x^2 \minus{} 2x \plus{} 4 \equal{} 2x^2 \minus{} 2x$
$ \minus{} x^2 \plus{} 4 \equal{} 0$
$ \minus{} (x \plus{} 2)(x \minus{} 2) \equal{} 0$
$ x \equal{} 2, \minus{} 2$

Points are $ (2,4)$ and $ ( \minus{} 2,12)$ thus lines are $ y \equal{} 2x$ and $ y \equal{} \minus{} 6x$.[/hide]

[quote="Rajiv"]$ \text{Line passes through origin, hence,: }
\\
\\
(\minus{}y_0) \equal{} (2x_0 \minus{} 2)(x \minus{} x_0) $

[/quote]

Here's where I went wrong:

$ y_0\equal{}\sqrt{x_0^2\minus{}2x_0\plus{}4}$

But the method is correct, I'm sure.

Good work [b]ben_b[/b]!


[hide][quote="apratimdefermat"]The incircle of $ ABC$ touches $ BC, CA, AB$ at $ A', B', C'$respectively. The line $ A'C'$ meets
the angle bisector of $ A$ at $ D$. Find $ < ADC$[/quote]
I believe that the question is to find the measurement of angle ADC

You should immediately see that DA is the angle bisector of angle B'DC'
and note that if we denote K as the intersection of C'A' and AC , we have (ACB'K) = -1 , therefore DC is the angle bisector of angle B'DK which means the measurement of angle ADC is equal to 90 degrees[/hide]


superior algebra
superior algebra solved

Let $R$ be a ring where $\forall a\in R: a^2=0$. Prove that $abc+abc=0$ for all $a,b,c\in R$.

From $(x+y)^2=0,x^2=0,y^2=0$ it follows that $xy=-yx$, thus we have (associativity) that $abc=-bca=cab=-abc$, which proves the statement.

[quote=steppewolf]From the condition of the problem we have:

$$0 = (ab+c)^2 = (ab)^2 + abc +cab +c^2 = abc +cab,$$
$$0 = (bc+a)^2 = (bc)^2 + bca +abc +a^2 = bca +abc,$$
$$0 = (ca+b)^2 = (ca)^2 + cab +bca +b^2 = cab +bca.$$

Adding the first two equalities and using the third we get:

$$ 0 = abc+cab+bca+abc= (abc+abc) + (cab+bca) = abc+abc.$$[/quote]

Wow

We have:
\begin{align*}
(ab+c)^2=0&\Rightarrow abc+cab=0\\
(ca+b)^2=0&\Rightarrow cab+bca=0\\
(bc+a)^2=0&\Rightarrow bca+abc=0\end{align*}
Adding the first and the last implies $abc+abc+cab+bca=0$, giving the result. $\square$


algebra
polynomial
absolute value
algebra proposed

Find all polynomials $ P(x)$ with real coefficients such that
a) $ P(x) P(x \plus{} 1) \equal{} P(x^2)$
b) $ P(x)P(2x^2 \minus{} 1) \equal{} P(x^2)P(2x \minus{} 1)$.

[hide="a"]The polynomials of the form $ Lx^n$ ($ n\geq 0$) that work are 0 and 1 (if $ n\geq 1$, they're not equal at $ x\equal{}\minus{}1$). Otherwise, there are $ \alpha$ and $ \beta$, of minimal and maximal nonzero absolute value, respectively (since there are a finite number of nonzero roots).

If $ z$ is a root, then $ z^2$ is a root by plugging in $ x\equal{}z$. So $ |\alpha^2|\geq |\alpha|\Rightarrow |\alpha|\geq 1$. Similarly, $ |\beta|\leq 1$. This means all nonzero roots have absolute value exactly 1.

Also, if $ z$ is a root, then plugging in $ x\equal{}z\minus{}1$ gives $ (z\minus{}1)^2$ as a root. Either $ z\equal{}1$ or $ |z\minus{}1|\equal{}1$.

So each root can either be 0,1, or satisfy $ |z|\equal{}|z\minus{}1|\equal{}1$. In this last case,
\[ |z|\equal{}|z\minus{}1|\Rightarrow |z|^2\equal{}|z|^2\minus{}2\text{Re}(z)\plus{}1\Rightarrow \text{Re}(z)\equal{}\frac{1}{2}\]
This gives $ z\equal{}\frac{1\pm i\sqrt{3}}{2}$. But $ z^2$ (also a root if $ z$ is) is not of the required form. So 0,1 the only roots, and so $ P(x)\equal{}Lx^a(x\minus{}1)^b$ for nonnegative integers $ a,b$ and a real number $ L$. The original equation requires
\[ L^2(x\plus{}1)^ax^{a\plus{}b}(x\minus{}1)^b\equal{}L(x\plus{}1)^bx^{2a}(x\minus{}1)^b\]
Equating multiplicities of $ x\plus{}1$ gives $ a\equal{}b$ and equating leading coefficients gives $ L\equal{}0$ or $ L\equal{}1$. So our polynomials are $ \boxed{P(x)\equal{}0,P(x)\equal{}[x(x\minus{}1)]^a}$ for any nonnegative integer $ a$, and they all work.[/hide]


In how many ways can 81 be written as the sum of three positive perfect squares if the
order of the three perfect squares does not matter?

wait....it says that order does not matter.

Then, From the triplet(1,4,8) there would be 3*2*1 ways to add up to 81.

And the other two, there would be three ways, that is  3!/2!, times two, so another 6.

Then,in total, there would be 12 ways to add up to 81 if order does not matter

Let's do casework:

Largest square=81 or above-No way

64:

Other 2 sum to 81-64=17. This works with 16+1, so (1,16,64)

49

Other 2 sum to 81-49=32 so 16+16 so (16,16,49)

36

Other 2 sum to 81-36=45 so 9+36 so (9,36,36)

Clearly the largest must be a square above 27, otherwise we cannot have 81. Thus we have 3 possibilities:

(1,16,64)-6 arrangements
(16,16,49)-3
(9,36,36)-3 for a sum of 6+3+3=12

umm it's actually 3... order doesn't matter.

[quote=vectorfunction]Does 0 count as a perfect square? If so, there should be $\boxed{6}$ ways.[/quote]

[quote=Waterblastor][quote=ok71]But what about 9 squared plus zero squared plus zero squared? Wouldn't there be 4 ways?
9^2+0^2+0^2[/quote]

It says the sum of three POSITIVE perfect squares. 0 isn't positive.[/quote]

 :dry:

[quote=YIYI-JP]0 is neither posiive nor negative[/quote]

exactly. @vectorfunction asked if 0 was a perfect square, but it isn't even included in the set.

oh sorry I didn't see that part about the question

At least I second guessed myself at the last second :D


complex analysis
complex analysis unsolved

Can anyone point me in the right direction for solving this? Was passed around the office and no one has a clue. :D  Just trying to figure out what the amount was. Thanks in advance for your help.

[img]http://www.geocities.com/charlesemeredith/Mathquestion.bmp[/img]


Can $\frac{7}{17}$ be expressed in form $\frac{1}{a}+\frac{1}{b}$ where $a,b$ are natural numbers?

[hide]
$\frac1{a}+\frac1{b}=\frac7{17}$
$\frac{a+b}{ab}=\frac7{17}$
$17a+17b=7ab$
$7ab-17a-17b=0$
$49ab-119a-119b=0$
$49ab-119a-119b+289=289$
$(7a-17)(7b-17)=289$
Factors of 289: $1,17,289$
If $(7a-17)=289$, then $a=\frac{306}{7}$.
If $(7a-17)=17$, then $a=\frac{34}{7}$.
Since none of the possibilities work (we don't have to test one, since there would be no natural number b value), then $\frac7{17}$ cannot be expressed in the form $\frac1{a}+\frac1{b}$ where $a,b$ are natural numbers.[/hide]

I think we can solve the system:
$ab=17, a+b=7$. Since it has no real roots, so we cannot express

Yes, I know. This is an easy problem  :)


If $A,\,B,\,C$ are the angles of a triangle, show that:
$1.\quad\prod\sin\frac{A}{2}\le\frac{1}{8}$
$2.\quad\prod\sin\frac{A}{2}\le\frac{1}{8}\cos^{2}\frac{B-C}{2}$
When does equality occur?
$3.\quad\prod\sin\frac{A}{2}\le\frac{1}{8}\prod\cos\frac{B-C}{2}$

[quote="April"]If $A,\,B,\,C$ are the angles of a triangle, show that:
$3.\quad\prod\sin\frac{A}{2}\le\frac{1}{8}\prod\cos\frac{B-C}{2}$[/quote]
It's true for all reals $A,$ $B$ and $C$ such that $A+B+C=\pi.$ :wink:

We have:
$cos\frac{B-C}{2}.cos\frac{C-A}{2}.cos\frac{A-B}{2}\geq 8.sin\frac{A}{2}.sin\frac{B}{2}.sin\frac{C}{2}$ (1)
Definitely:
$(1)\Leftrightarrow 8.cos\frac{A}{2}.cos\frac{B}{2}.cos\frac{C}{2}cos\frac{B-C}{2}.cos\frac{C-A}{2}.cos\frac{A-B}{2}\geq 8.sinA.sinB.sinC$
$\Leftrightarrow (sinA+sinB).(sinB+sinC).(sinC+sinA)\geq 8.sinA.sinB.sinC$ (3)
And (3) easy to deduced by applying Cauchy' inequality

[quote="April"]$1.\quad\prod\sin\frac{A}{2}\le\frac{1}{8}$
$\Longleftrightarrow 4\left(\cos\frac{A-B}{2}-\sin\frac{C}{2}\right).\sin\frac{C}{2}\le 1$
$\Longleftrightarrow\left(2\sin\frac{C}{2}-\cos\frac{A-B}{2}\right)^{2}+\sin^{2}\frac{A-B}{2}\ge 0$[/quote]

$\prod{\sin{\frac{A}{2}}}=\frac{r}{4R}\leq  \frac{1}{8}$

The well known inequality $R\geq 2r$ occur for all triangle. The equality happen when triagle is $A=B=C=60^{0}$

[quote="hien"]The well known inequality $R\leq 2r$ occur for all triangle. The equality happen when triagle is $A=B=C=60^{0}$[/quote]

i think you meant $R\geq 2r$ :wink:


If the eigen values and eigen vectors are then it is possible to find its corresponding matrix?

suppose 1 is the eigen value and (1,2) is a eigen vector of 2x2 matrix..then how find this correponding 2x2 matrix..

   i need ur help..

Is 1 the only eigenvalue?  A $ 2\times 2$ matrix will have 2 eigenvalues, counting multiplicity.


conics
geometry
ratio
analytic geometry
geometry unsolved

Let $ ABC$ be an arbitrary triangle. The point $P$ is inside of $\triangle ABC.$

$ \frac{[ADF]}{[CDF]} \equal{} M \ , \  \frac{[ADE]}{[BDE]} \equal{} N \ , \  \sqrt{M} \plus{} \sqrt {N} \equal{} 1.$

what is the trajectory of point P ?

Sorry, but I don't understand what you mean by the "trajectory" of point P. Could you explain to me?

it means “locus” and this locus is a conic, an elliptical arc. :lol:

Parallel project arbitrary $ P$ inside $ \triangle ABC$ to triangle orthocenter. Label all projected points with primes. Area ratios are invariant.

$ \frac {S(\triangle A'D'F')}{S(\triangle A'P'B')} \equal{} \frac {A'D' \cdot A'F'}{A'P' \cdot A'B'} \equal{} \frac {C'D' \cdot C'F'}{C'P' \cdot C'B'} \equal{} \frac {S(\triangle C'D'F')}{S(\triangle C'P'B')}\ \Longrightarrow$

$ \frac {S(\triangle ADF)}{S(\triangle CDF)} \equal{} \frac {S(\triangle APB)}{S(\triangle CPB)}$ and similarly, $ \frac {S(\triangle ADE)}{S(\triangle BDE)} \equal{} \frac {S(\triangle APC)}{S(\triangle BPC)}.$

Let $ K, L, M$ be midpoints of $ BC, CA, AB.$ Parallel project $ \triangle ABC$ to equilateral triangle. Label all projected points with asterisks. Area ratios are again invariant. Let $ X, Y, Z$ be orthogonal projections of $ P^*$ to $ B^*C^*, C^*A^*, A^*B^*.$ Then

$ \frac {S(\triangle A^*D^*F^*)}{S(\triangle C^*D^*F^*)} \equal{} \frac {P^*Z}{P^*X}$ and $ \frac {S(\triangle A^*D^*E^*)}{S(\triangle B^*D^*E^*)} \equal{} \frac {P^*Y}{P^*X}.$

$ \sqrt {P^*Z} \plus{} \sqrt {P^*Y} \equal{} \sqrt {P^*X}\ \Longleftrightarrow$ $ P^*$ is on incircle arc $ L^*M^*$ of equilateral $ \triangle A^*B^*C^*$ opposite to $ K^*$. See
[url]http://www.mathlinks.ro/viewtopic.php?t=106446[/url]. $ P$ is therefore on Steiner inellipse arc $ LM$ of $ \triangle ABC$ opposite to $ K.$

Let $( x: y: z)$ be the barycentric coordinates of  $P$ with respect to $\triangle ABC.$ Thus we consider oriented areas throughout the calculations.

$ D \ (0: y: z) \ \ , \ \ E \ (x: 0: z) \ \ , \ \ F \ (x: y: 0)$

$ \frac {|\triangle ADF|}{| \triangle ABC|} \equal{} \frac {yz}{(x \plus{} y)(y \plus{} z)} \ \ , \ \ \frac {| \triangle CDF|}{| \triangle ABC|} \equal{} \frac {xy}{(x \plus{} y)(y \plus{} z)}$

$ M \equal{} \frac {| \triangle ADF|}{| \triangle CDF|} \equal{} \frac {z}{x} \ \ , \ \ N \equal{} \frac {|\triangle ADE|}{| \triangle BDE|} \equal{} \frac {y}{x}\Longrightarrow$

$\sqrt {\frac {y}{x}} \plus{} \sqrt {\frac {z}{x}} \equal{} 1 \Longrightarrow \  x^2 \plus{} y^2 \plus{} z^2 \minus{} 2(xy \plus{} yz \plus{} zx) \equal{} 0$

Locus of $P$ is the Steiner inellipse of  $\triangle ABC$ centered at its centroid $G \  (1:1:1).$

Let me tell you how I fabricate this problem，See that picture，I give a important conclusion：
[url]http://www.aoshoo.com/bbs1/dispbbs.asp?boardid=43&id=13143&page=0&star=1[/url]（second page，13th floor）

The circle ⊙$ AP,$⊙$ EF,$⊙$ BC$ have the same axis，so the $ Newton \minus{} theorem$ become easy .
The $ Orthocenter \minus{} Group, four \minus{} orthopole, four \minus{} symmetrical$ point of $ Mique$ totally 12 points are the axis of the Complete quadrilateral $ AEPF$ and ：

$ S$Δ$ BEF/S$Δ$ ABP \equal{} MN/LN \equal{} S$Δ$ CEF/S$Δ$ ACP \equal{} r$，
$ S$Δ$ AEF/S$Δ$ ABC \equal{} LM/LN \equal{} S$Δ$ PEF/S$Δ$ PBC \equal{} s$，
$ S$Δ$ APE/S$Δ$ BCE \equal{} ML/MN \equal{} S$Δ$ APF/S$Δ$ BCF \equal{} t$，$ r \plus{} s \equal{} 1,s/r \equal{} t$.

this is only basic conclusion，there are some Special circumstances，such as Orthogonal ，use this can solve [b]Bulgaria 46th Olympiad [/b]，[b]Bulgaria 46th Olympiad [/b] is the Special circumstances of three circle Orthogonal.


If integer $ a,b$ are s.t $ \gcd(a,b)\equal{}1$ then for any positive integer $ c$, there exists $ x$ s.t $ \gcd(a^2\plus{}b^2x,c)\equal{}1$.
[color=red]Latexed by moderator[/color]

$ gcd(a^2,b^2)\equal{}1$ gives $ |\{p\in\mathbb{N}: \ p\ is\ prime\}\cap\{a^2\plus{}xb^2\}|\equal{}\plus{}\infty$
so we take x such that $ a^2\plus{}xb^2$ is a prime$ >c$


If $ X$ and $ Y$ are independent standard normal random variables, determine the joint density function of:
$ U \equal{} X$ and $ V \equal{} \frac{X}{Y}$

Also show that $ V$ has a Cauchy Distribution.

$ E[f(X,\frac{X}{Y})]\equal{}\frac{1}{2\pi} \iint_{x,y} f(x,\frac{x}{y}) \exp(\minus{}\frac{x^2\plus{}y^2}{2}) \, dx \, dy$ and put $ (u,v)\equal{}(x,\frac{x}{y})$ - you will get the answer to both questions.


Well, is it because there is something happened inside the brain?? Or is it because of the influence of the social surroundings??
Please try to give some scientific answers.
Thanks!

mostly unknown.
however, the second (or later) male that a mother has has a higher chance of being homosexual.

yeah furious is right, yet, interestingly, the correlation he described is NOT found in women and daughters. kinda interesting.....

also, homosexuality is not something you can contribute solely to environment or to genetics. it is now known that homosexual men lack a nucleus of nerve cells in the hypothalamus that are presumed to contribute to a heterosexual orientation, scientists have theorized that this could be due to a genetic factor, certain concentrations of substances in the environment of the womb. the intricate complexity of life is made more difficult to understand by the effects of the world around us. essentially, we are born into the world with a certain genetic makeup, predisposed to many things.....temperament, some intelligence, yes, many things are seemingly hard-wired. but our genetic makeup that we have had since our day of conception is still very much able to be acted upon by the environment. here is a good analogy:

the young vine may born to grow toward the sky.....but if it is planted near a fence.....or in certain other conditions that dont kill it, it will adapt, much as we are adaptable. we bring ourselves into the environment with our set "prerequisites" and then our environment works on us.

The vine thing is a good analogy. I'm going to use that in the future when referring to genetics vs environment...credit to you.

yes, pendulum, you're correct. ive added a little note to remind everyone.

have any of you ever taken a look at some psychology texts in detail? from my experience, they have taught me a lot more about this subject than my biology texts.

[quote="pendulum"]Shobber, you're probably not aware of this but the term "homo" is considered derogatory.  You or the mod should change the word to "homosexual".[/quote]
I am sorry...
I never know the difference between these two, they seems to be one word. But obviously, i am wrong. So I apologize for that mistake and for those whom were possibally annoyed by that title.

Sorry.

its cool shobber, dont worry about it, its a common error. id probably venture to say that there havent been any homosexual people looking at it anyway, but just to be safe, you know? so i guess the ap test was today for biology.....i took it last year and got a 5. i wonder how this year's exam was

the exact reason for homosexuality could be the like and dislikes of the person which he has developed due to the different experiences he came across. one of the reasons could be some genes which are responsible for the control of attraction towards a sex e.g. females like men or men are attracted towards females in normal cases. But the exact answer for homosexuality even i will like to know

http://www.cbmw.org/news/ram150404.php

... ...
The case for a biological cause first gained credibility in 1991 through research by Michael Bailey, who studied patterns of male homosexuality among identical twins. The case was strengthened in 1993 when Hamer and colleagues claimed to have identified a specific genetic link to male homosexuality, and to have isolated the link to the X chromosome. Both studies received international media attention and coverage.

Rice and Ebers undertook their study to see if these claims could be confirmed. To the contrary, they found no link in the Xq28 region which could function with any significant influence. As Ebers stated, "there is no hint or direction of the initial observation."
... ...

No definitive direct link discovered yet.

Here's a great article on this topic:
[url]http://www.trueorigins.org/gaygene01.asp[/url]

I don't know a thing about biology, but my gut reaction is not to trust a website that thinks evolution is a myth. Not that that's an argument.

[quote="fleeting_guest"]It's a very good argument for mistrusting the web site.  The syllogism goes thus:

Anti-evolutionism = proxy for "out, loud and proud" US Christian fundamentalism;
this type of Christian fundamentalists tend to be anti-homosexual;
it is inconvenient for them if "God made" homosexuals so (ie. if it is genetically predetermined);
--->>  establish web site discrediting studies purporting a genetic link.

With that said, the evidence is scant in either direction.[/quote]

You should not reject this simply because they do not believe the same things as you do.  In my opinion, you should reject this paper on Homosexuality and the Gay Gene only if you have read it and it does not line up with the facts.  Of course there's no problem with being suspicious of the paper but, you should not dismiss without reading the facts.  Someone with your same line of reasoning could think

Pro-evolutionism = proxy for "out, loud and proud" US atheistic dogma;
this type of atheistic dogma tend to be Pro-homosexual;
it is inconvenient for them if "God made" them to be with a woman (ie. if it is genetically predetermined);
--->>  atheist discrediting studies purporting any scientific link to God.

and therefor ignore a possible scientific brake through.  You are supporting this blocking of any thing that does not support your opinion.  Thank God http://www.trueorigins.org has not reasoned like this and, researched things out instead.

[quote="Phydeaux"][quote]Anti-evolutionism = proxy for "out, loud and proud" US Christian fundamentalism [...][/quote]
Pro-evolutionism = proxy for "out, loud and proud" US atheistic dogma; [/quote]

The attempted analogy doesn't work.
Pro-evolutionism, as you call it, is the position of people of all sorts of religious beliefs and geographic origins; it is not limited to atheists, it is found worldwide, and is very common.
Anti-evolutionism, particularly "Intelligent Design" and political demands to remove evolution or include ID in school science lessons, is a 100 per cent US Christian operation, and is found mostly in a small and peculiar slice of the population.

If a person accepts or advocates evolution, there is very little basis to infer any additional information about their religious or political beliefs.  If a person is anti-evolutionist, that is extremely well correlated with being a US Christian fundamentalist, and the likely ideological implications.

[quote="fedja"]any uncontrolled dangerous deviation from Mother Nature's intent is an illness[/quote]
How in the world is homosexuality dangerous? Whats dangerous is the acts that some people do like killing them just because of their sexual orientation. And I would hope not killing is in your definition of "mother nature's intent."  And if you did not mean to put dangerous in there, which homosexuals obviously are not, then i guess we are all ill, right?
[quote="fedja"]...then homosexuality is an  illness by definition. Nobody should discriminate against ill people or ridicule them on the sheer ground of their illness[/quote]
I think by calling them "ill" and trying to find a medical cure for their "illness" is pretty discriminatory.You are almost as disgusting as spx2

Well, a dangerous biological trend is the one that, being followed by everybody, will result in population's dying out, even if it is not immediately life-threatening for the bearer of the trend.

Of course, killing people on the ground of their sexual orientation is as dangerous for society as any other murderous behavior. On the other hand, however strange it may seem, the instinct to get rid of a member of a group that somehow is different from yourself, be it homophobia in some form or the "disgust" towards people who for some reason are "against" homosexuality, (BTW, how about killing them instead, or putting them in jail, will that suit you more? :)) is biologically very natural. Fortunately or unfortunately, the norms of society we live in are much more sophisticated than biological norms.

By the way, in my humble opinion, in an ideal society everyone should be able to voice any opinion of his without immediately being killed, labeled as a "disgusting person", "being banned",  or whatever. If you are sure he mistakes and you are right, the correct reaction should be to try to correct and educate him in an open dialog. Declaring somebody "disgusting" or "banning him from the forum" will never convince him of anything. Remember also that person's opinions and emotions may be highly influenced by the culture he grew up in and that "blatant homophobia" was a society norm not so long ago and still is in many places on the globe, so I am not even sure on whose side the majority is now: your's, or spx2's, or mine. What I am sure about is that such conflicts of opinions should be resolved in calm discussions, not in flames and swearing.

As to recognizing something as an illness being a discrimination, I don't know what to say: the next thing you may say is that instead of trying to cure people with anything from obesity
to cancer, we should praise them as introducing diversity to the society. Probably, the best advice will be to look up the word "discrimination" in the dictionary.

[quote="fedja"]By the way, in my humble opinion, in an ideal society everyone should be able to voice any opinion of his without immediately being killed, labeled as a "disgusting person", "being banned",  or whatever.[/quote]
It is my humble opinion that you are very mean towards those that are different (I think) than you in sexual orientation. You act like they are a lesser race than you.
[quote="fedja"]What I am sure about is that such conflicts of opinions should be resolved in calm discussions, not in flames and swearing.[/quote]
I am sure that I am not flaming or swearing, if you were directing that towards me.
[quote="fedja"]As to recognizing something as an illness being a discrimination, I don't know what to say: the next thing you may say is that instead of trying to cure people with anything from obesity to cancer, we should praise them as introducing diversity to the society. Probably, the best advice will be to look up the word "discrimination" in the dictionary.[/quote]
discrimination: Treatment or consideration based on class or category rather than individual merit
calling them ill is a consideration based on their class/category of homosexuality. You seem like you think homosexuality is very similar to cancer or obesity. And who's praising who and whos telling you to? You're completely wrong. People die from cancer and obesity, not from homosexuality. And you are saying you want a cure for homosexuality and you haven't said anything aboout a cure for cancer when in reality cancer kills more people than homosexuality does (homosexuality has killed 0 people to this day, can't say the same about cancer). A cure for cancer would be infinitely times more helpful than a cure for homosexuality,

Slavery came about because whites thought that they were superior to Africans. This homosexuality argument fedja is posing is similar. As soon as you classify people as "diseased" or "inferior," it's that much easier to claim that discrimination towards that group is acceptable. This line of thinking is incredibly dangerous, and has caused countless problems throughout history. I suggest you rethink your beliefs; bigotry is not acceptable.

This is completely true, LordoftheMorons.

Also fedja, you say that homosexuality is a dangerous illnesss because if everyone was homosexual, humans would die out. Well thats not true because with the help of science people are able to reproduce (sperm bank, artificial insemination, etc.) without unwanted and/or potentially dangerous intercourse. This means that if everyone was homosexual, they would survive because they could reproduce.l

[quote="LordoftheMorons"]Slavery came about because whites thought that they were superior to Africans.
[/quote] The institute of slavery is much, much older than the clash between the white and the black races. Political doctrines that were brought forth to justify slavery were many and quite diverse, "white man superiority" being only one of them. The slavery come to the USA primarily because the agricultural South needed plenty of cheap labor force they couldn't get at home. Note that there was no comparable amount of slavery in England and in other parts of Europe at the same time, though the doctrine of "white man superiority" was at least as strong there as in the USA.
[quote]
As soon as you classify people as "diseased" or "inferior," it's that much easier to claim that discrimination towards that group is acceptable.[/quote]
Two really dangerous logical leaps here: firstly, "diseased" does not imply "inferior". Do you really think we should stop call people with flu or cancer "diseased" on the basis that it sounds like we make them "inferior" to healthy people? I agree that is the standard flaw of human thinking to think that if something is better in one respect, it means it is better (or, if you prefer, superior) in all other respects as well, but I assumed that the society learned to tolerate illnesses and treat ill people with respect long ago and the idea of their innate inferiority is something new to me.

Secondly, "inferiority" of some group (as a group) doesn't imply acceptability of discrimination. Once you raised the "African" versus "European/American" issue, let me just ask you a question. Are you seriously claiming that Central African culture and civilization are at the same level or even close to the European and American ones even today? The globalization and interaction between countries removed the most striking differences, but, in my humble opinion, Africans are still dozens of years behind in technology and centuries behind in such things as education and development of political systems. So what? Why should it immediately raise a desire to discriminate against them? The whole point of being civilized is not in closing eyes on the gaps in development and cultural differences, but in treating both the "inferior" and "superior" cultures with respect and trying to facilitate progress and development wherever and whenever possible. So, no, I do not agree that "inferiority" of some group in some or even many respects should "make it easier to claim that discrimination towards this group is acceptable".

the problem is that "diseased" and "inferior" have negative connotations in everyday usage. Also, people do discriminate against disabled and diseased people. The negative connotations of the specific words give the idea that something is wrong when nothing is(at least not according to american psychological association). For example, let's say I take an IQ test and I say everyone who has a lower IQ than me is "stupid". Maybe they are scientifically stupider than me, but using that word has more negative connotations beyond the scientific(or not scientific, I am not sure how useful IQ tests are) meaning.

In fact, your assessment of African cultures show that very problem. Its true that 3rd world countries may have inferior technology, however, that does not translate to cultural inferiority or intelligence inferiority or "inferiority of some group" as you seem to imply. You are the one making a leap of logic there. There's no reason why technology or liberal democracy should be the standard for how advanced a society is(at least, you don't provide one). What if everyone in that society was really happy without the technologies and stuff, suppose they have more utility per capita; are they still more inferior?

Essentially, just because one comes from a less technologically developed culture does not mean one is dumber.

[quote="agolsme"]
Essentially, just because one comes from a less technologically developed culture does not mean one is dumber.[/quote]
That's exactly the point I'm trying to make. Funny that nobody notices. The word "inferior" by itself is meaningless without specifying in what respect the comparison is made. What I'm saying is more: even if you make a lot of comparisons and some group loses in all of them (and the life in Central Africa is neither easy, nor happy, nor peaceful; if you want an example of a relatively happy non-technological society, you should look somewhere in Polynesia), it is still not a reason for discrimination.

As to  what should be the standard on how advanced the society is, it should involve many components, of course, but I'm inclined to believe that the level of ambition, technological advances, sophistication of arts, certain liberal democracy values (like equal rights for women, religious freedoms, etc.) and general happiness of its members are essential factors to look at (though this list is by no means complete). Truly, everything has its downside and at certain moments of history some of these values came into direct contradiction with other ones (especially it concerned technological progress versus general happiness), but we are slowly becoming able to add more and more to the list of advantages and make the list of disadvantages shorter and shorter. So, I'm inclined to believe that the modern Western civilization can serve as a current standard of an advanced society in many respects. The problem is that there is no easy way to reach this stage and there may be alternative ways leading to some different models of society (though we do not currently know of any: the great communistic experiment failed and the 3rd world countries can offer peaceful coexistence with nature at their best (no ambitions, no technological progress, just sit and wait until the Sun cools down or explodes or something else happens to quit the life on Earth) and bloody dictators, civil wars, famine and epidemias at their worst). China might finally build something completely different but it seems to lean more and more to the Western model as well, though here I may mistake.

As to words with negative connotation, I agree to you in respect that still too many people react to them in an improper way but I disagree that they should be eliminated from the dictionary. They just should be used in their very specific meaning without an attempt of making them all synonyms of "bad" and "wrong".

By the way, this thread should be, probably, split into at least 3 and some parts of it should be moved to the Round Table (my humble opinion, of course)

Oh well, I don't think they should be removed from the dictionary either. I think the problem is that even if you expect to use a word in it's scientific meaning(scientific meaning for inferior??), other people won't interpret it as such and since language is about communication, you should go use a different word in order to maximize understanding.

Also, I agree with natmc and lordofmorons on the use of those words to describe homosexuality.

I never said "inferior". All I said was "ill". Do you know a good synonym for "ill" without any negative connotation? If you do, please, consider me using it every time. Unfortunately, my English is not good enough to find such a synonym and I do not consider "health challenged" a valid English word. "Having (mild?) medical problems" is the best one I can offer. Will it suit you? If so, this will solve the issue of communication (By the way, I have always assumed that the responsibility for the effectiveness of communication lies on both sides and the desire to understand is equally important with the desire to be understood). This will, at least, bring us back from the question whether discrimination against homosexuals is admissible to the question whether it is a medical problem (physical or mental) or something else. Or, maybe, even to the original question of this thread: what causes it?

[quote="ROLEX4life"]About the notion of how later born males have a higher chance of being homosexual, i have heard an interesting theory. If a couple had 5 children that were all or mostly male, the later born males would have a higher chance of being homosexual because it's "nature's way of balancing out the gender ratio". The same phenomenon is not found in females because females are needed to give birth. This is just a theory that i found very interesting.[/quote]

But is there any biological reason why the (human) gender should be fine-tuned to a 1-1 balance?   Also, the mechanism you describe would not affect the balance of genders, it would reduce the number of eligible males as compared to females.

I agree with spx2 that the world has bigger problems than homosexuals. But of course gay and lesbain marriages should be accepted. There was no need to ban them in the first place. A man will sleep with whoever who wants to? If the other guy is ready, what to you and me?

In fact the institution of marriage is bogus.

CRAZY THEORY:

Everything supposedly has a purpose.  So I asked myself (not derogatorily) "what is the purpose of homosexuals?"  Let's assume God, or nature, or Buddha, or whoever, created homosexuals for a reason.  Then what's that reason?  To create a slim minority of academics who are not drawn to social lifestyles?? (This was based on the sterotype that I know 2 homosexual people, and both are the nerdy type)  I know, crazy and half stereotypical, but what is the purpose of homosexuals?  Is it an anomaly, because if it is, it's an awful big one.  Crazy theories welcome!  (I'm thinking cooky today!! :heli: )

It is probably to be an example of what not to be...(not trying to be rude or anything just replying with the first thing that came out of my mind.)


function
inequalities
integration
real analysis
real analysis unsolved

Let $ f: [0,1] \rightarrow \mathbb{R}$ be continuous and increasing on $ [0,1]$. Prove that for any $ n \in \mathbb{N}^*$, the following inequality holds:

$ \displaystyle\int^1_0 (t^n\plus{}t^{n\minus{}1})f(t)\, dt \ge \frac {2n\plus{}1}{n(n\plus{}1)} \displaystyle\int^1_0 f(t)\, dt$

It has already been posted [url=http://www.mathlinks.ro/viewtopic.php?t=266345]here[/url]


How'd you do?
...I failed with 123, since I misbubbled 6 and 7.

Yeah I misbubbled too. (assuming I misbubbled on the answer sheet as well as I did on test). Total of 133.5 :wallbash:

[quote="#H34N1"]I think 132 if I didn't mess up somewhere else :mad:

1 misbubble
2 misreadings :o[/quote]

really??/ cause there was no 8th grader that got even above 130...accroding the the offical AMC Ohio statistics

are we going to have the "I got 9 on AIME discussion again"? (http://www.artofproblemsolving.com/Forum/viewtopic.php?t=138972)  Cause  thats  what you said last year (scroll down).  The problem was on the AMC website, no Ohio middle scholar last year even made AIME.  :rotfl:...you might want to consider that before you say something like that.

Incase you guys go the there was no 129 (that was my official score, I apparently didn't misbubble one), I took it in Michigan, so I'm listed on there.

[quote="#H34N1"]It seems to be a mistake on UNL's side. I'm having my teacher write to them.[/quote]

ok ok i believe you.

Just like when you said you got a 9 on the AIME last year.
http://www.artofproblemsolving.com/Forum/viewtopic.php?t=138972
here

I bet that was a mistake on the UNL's side too...
Considering, looking at the AMC stats from 2007, no middle scholar qualified; looking at the AIME stats, no middle scholar qualified, and your name isnt on the USAMO list.
They must be good at making 3 crucial mistakes, especially the last one which would be obvious.

I like  how you said you got 7 on the AIME this year.
I cant wait until the stats come out


algebra
polynomial
linear algebra
matrix

A and B are n x n matrics that commute, (i.e. AB = BA)

i) Show that A^2 and B^2 commute

ii) Give a generalization of this result (without proof)

I suppose the generalization would be that $P(A)Q(B) = Q(B)P(A)$ for any polynomials $P$ and $Q$. The reasoning is the same.

Don't quite get it...

thought A and B were matrices...

$A,B$ are two matrices which commute (it's the hypothesis), so the normal commutative law does apply. What do you find wrong in this?

You can have a polynomial of a matrix just as well as of a number. Let $P(X) = X.X + X + I$, for example. That's a matrix-ish way of writing a polynomial, but it's really no different from $P(x) = x^2 + x + 1$. If the matrices $A$ and $B$ commute, then polynomials of $A$ and $B$ will commute as well. Is the other direction true? :)

oh yeah the commutative law applies if AB=BA

Sorry for the old thread revival, but could someone please explain how the normal commutative law applies. Thnx

[quote="Ramanujan"][quote="chaos_is_here"]A and B are, however, matrices, and not polynomials, so the normal commutative law does not apply...[/quote]

If $AB=BA$ it does .

If $Q(x)= \sum_{i=0}^{n}a_{i}x^{i}$ a polynomial then  $Q(A) =\sum_{i=0}^{n}a_{i}A^{i}$ .[/quote]

First, welcome to this forum. Second, what are you exactly asking? The original poster were referring to the fact that matrix multiplication is not, in general, commutative, contrary to multiplication in other sets. Or were you asking something else?

Hi Carcul,

I'm interested in the generalization of the result obtained after showing that the squares of the commutative matrices are equal, in particular how does this relate to the commutativity of polynominals?


Whenever I use the \fcolorbox command, the text flows into the margins if it is too long. Is there anyway to make the color box span multiple lines? I saw that you could use \parbox for multiline boxes, but I don't know how you color that.

Just combine the two. For example:
$ \fcolorbox{red}{green}{\parbox{5cm}{This is the first line \\ and this is the second line but the lines will also wrap automatically}}$

Thanks for the reply.

Now I also want to include a picture in the top left corner of the box (which I sucessfully did). However, the text isn't wrapping around the picture. Right now, I have the picture at the top left with blank space to the right of it, and then underneath is the text. Is there anyway to get text to the right of the picture and underneath it?

You would do it in the same way as you wrap text in a document eg using the wrapfig package.

I have this one multiline colored box that extends to more than a page. However, it keeps extending onto the margins at the bottom of the page. Is there anyway to make the box finish on the next page?

Make two boxes, one for each page but you will have to decide where to split the boxes which may change as the document is altered. Or you could try [url=http://tug.ctan.org/cgi-bin/ctanPackageInformation.py?id=longtable]longtable[/url] with just one column.

I have a better idea. The framed package allows frames to break across pages. For example:
[code]usepackage{framed}
...
\definecolor{shadecolor}{rgb}{0,1,0} % this is green
\begin{shaded}
Put your text here
\end{shaded}[/code]

Is there anyway to also get a borderline? Right now I'm using

\noindent\fcolorbox{black}{green}{\parbox{4.75in}{

text here

}}

Preferably, I'd like to have the black borderline too.

The shaded environment has a red border but it is easy to redefine that to use black instead. It's all in the documentation for the framed package which is inside framed.sty.
For the width you can't use \parbox as that won't break across pages, so use the MakeFramed environment. I have also added some code that redefines \fbox so that the top and left borders show better.
Do study the help in framed.sty for any other changes you want to make.
[code]\documentclass{article}
\usepackage{color}
\usepackage{framed}
\begin{document}
% redefine \fbox so that top and bottom border shows more clearly
\makeatletter
\def\@frameb@x#1{%
  \@tempdima\fboxrule \advance\@tempdima\fboxsep
  \advance\@tempdima\dp\@tempboxa
  \hbox{%
    \lower\@tempdima\hbox{%
      \setbox\@tempboxa\vbox{%
        \vskip\fboxrule % \hrule\@height\fboxrule
        \hbox{%
          \hskip\fboxrule % \vrule\@width\fboxrule
          #1%
          \vbox{%
            \vskip\fboxsep
            \box\@tempboxa
            \vskip\fboxsep}%
          #1%
          \vrule\@width\fboxrule}%
        \hrule\@height\fboxrule}%
      \@tempdima\ht\@tempboxa
      \@tempdimb\wd\@tempboxa
      \vbox{%
        \hbox{%
          \box\@tempboxa
          \hskip-\@tempdimb
          \vrule\@width\fboxrule
          \kern-\fboxrule
          \hskip\@tempdimb
        }%
        \vskip-\@tempdima
        \hrule\@height\fboxrule
        \vskip-\fboxrule
        \vskip\@tempdima
      }%
    }%
  }%
}
\makeatother
% end of redefinition

\renewcommand\FrameCommand{\fcolorbox{black}{green}}%
\begin{MakeFramed}{\hsize 4.75cm}
\noindent This is the first line\\
and this is the second line \vspace{10in}\\
and this is a new page
\end{MakeFramed}%

\end{document}[/code]

The compiler throws an error when I include code to add pictures. Is there anyway to modify that?

\begin{figure}[h] \begin{asy}
  include graph;
  size(1inch);
  filldraw(circle((0,0),1),yellow,black);
  fill(circle((-.3,.4),.1),black);
  fill(circle((.3,.4),.1),black);
  draw(arc((0,0),.5,-140,-40));
\end{asy}
\end{figure}

What error? Is it from asymptote or from LaTeX? Is it inside or outside the shaded environment? Which box are you using and does it go onto a new page? Floats (such as the figure environment) aren't designed to go over two pages.

Actually, I think asymptote is also the problem. I moved the code outside the box, and It isn't drawing quite correctly. Here are the errors messages I get

Input index file C:\Documents and Settings\ (then the rest of the file path name) not found.

Usage: C:\Program Files\MikTeX 2.7\miktex\bin\makeindex.exe [-ilqrcgLT] [-s sty] [-o ind] [-t log] [-p num] [idx0 idx1 . . .]

You give so little information away and then give contradictory statements. First you say Asymptote is causing the problem then you say the error messages come from makeindex.exe which has nothing to do with Asymptote.
I wasn't aware that there was an index in the document. What code are you using for the index? Where is the index file generated? What are its contents? A [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=157945]complete minimal document[/url] is essential for an analysis. It's not possible to know what is going on unless you tell us everything.

\documentclass{article}
\usepackage[pdftex]{graphicx}
\usepackage{asymptote}
\begin{document}
\begin{asy}
  include graph;
  size(1inch);
  filldraw(circle((0,0),1),yellow,black);
  fill(circle((-.3,.4),.1),black);
  fill(circle((.3,.4),.1),black);
  draw(arc((0,0),.5,-140,-40));
\end{asy}
\end{document}

I don't think I am even using an index so I thought it was some asymptote thing. Just compiling the above code with nothing else added throws the aforementioned errors.

That document is fine and produces no errors, warnings or bad boxes at all. I presume that you have checked the LaTeX log file?

I don't suppose you are running TexnicCenter in Windows by any chance (I did say you must tell us everything)? That runs makeindex automatically unless you turn it off in Define Output Profiles but of course you can ignore the output from makeindex if there's no index to make. If you leave it there then obviously it will tell you there's no .idx file. But then that's correct and not an error.

If that's not the case why are you running makeindex when there's no index to make? How are you compiling?

I am using a Windows, so I guess that solves that problem. But reason why i thought something was wrong was because when I replace the aforementioned code with

\documentclass{article}
\usepackage[pdftex]{graphicx}
\usepackage{asymptote}
\begin{document}
\begin{asy}
draw((0,0)--(3,7),red);
dot((0,0));
dot((3,7));
label("Produced with Asymptote "+version.VERSION,point(S),2S);
\end{asy}
\end{document}

The pdf isn't "refreshed" and the smilie face is still there. The way I compiled it was (1) compile/build (2) run asymptote then (3) compile again. Using the AOPS wiki guide, I also set up the process so that it does all three steps in just one compile/build, but that isn't working either.

Rename the file before compiling and check the asymptote log for errors. Check also that just compiling the asymptote file from the command line produces the correct image.

I renamed the file and compiled it using the smilie face code and also the code for the line. It works fine with the smilie face, but when I put in code for the line, I get these errors

Cannot execute 'C:\Program Files\MiKteX 2.6\miktex\bin/pdflatex' \scroll mode

Please put in config.asy:
import settings;
texpath="PATH";
where Path denotes the correct path to the directory containing your pdflatex engine (pdflatex).
Alternatively, set environment variable ASYMPTOTE_TEXPATH
or use the command line option -texpath="PATH"

When entered the code into command line, it produced the image just fine.

Is there any other information you need?

Thanks

I thoght you'd sorted the problem as the Asymptote error messages told you what to do
[quote]Please put in config.asy:
import settings;
texpath="PATH";
where Path denotes the correct path to the directory containing your pdflatex engine (pdflatex).
Alternatively, set environment variable ASYMPTOTE_TEXPATH
or use the command line option -texpath="PATH" [/quote]
See also Problem 3 at [[Asymptote: Help]].

Is there anyway to write asymptote code inside of the makeFramed box above? I get an error when I try to, and the figure isn't wrapping over multiple pages.

[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=157945]Minimal example[/url]? What error? Consider drawing the frame in the Asymptote code (the [url=http://www.artofproblemsolving.com/Forum/index.php?f=519]Asymptote forum[/url] will help you do this).