What would the 100th digit be of  :pi:  3.141592654....?

If you don't include the 3, then 9. If you do, then 7.

I just know it. You can look it up pretty easily.

[quote="Osiris"]There isn't exactly an algorithm (in the sense that given the first n digits you can find the n+1th digit) for calculating digits of pi.[/quote]

But see   [url]http://mathworld.wolfram.com/BBPFormula.html[/url].

The subject of this thread reminds me of a completely unrelated question.  I think this question appeared on a contest I took centuries ago.  (Perhaps AHSME.)

Without using a calculator, which is larger e^pi or pi^e.

Is there any way to do this problem without just manually looking at the digits of pi? I'm pretty sure there isn't without using any obscure theorems.


linear algebra
matrix
integration
logarithms
calculus
calculus computations

Let $a,\ b,\ c$ be real numbers such that $1\leq a\leq 2,\ 1\leq b\leq 2,\ 1\leq c\leq 2$ which satisfy

$\left(\begin{array}{ccc}1 & a & 0 \\0 & 1 & 0 \\0 & 0 & 1\end{array}\right) \left(\begin{array}{ccc}1 & 0 & 0 \\ 0 & 1 & b \\ 0 & 0 & 1\end{array}\right) \left(\begin{array}{ccc}1 & c & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1\end{array}\right) =\left(\begin{array}{ccc}1 & 0 & 0 \\0 & 1 & x \\0 & 0 & 1\end{array}\right) \left(\begin{array}{ccc}1 & y & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1\end{array}\right) \left(\begin{array}{ccc}1 & 0 & 0 \\ 0 & 1 & z \\ 0 & 0 & 1\end{array}\right).$

Let $K$ be the solid formed by the point $(x,\ y, z).$ Find the volume of $K.$

An uninspired solution: from
\[\begin{pmatrix}1& a+c & ab \\ 0 & 1 & b \\ 0 & 0 & 1\end{pmatrix}=\begin{pmatrix}1& y & yz \\ 0 & 1 & x+z \\ 0 & 0 & 1\end{pmatrix}\]
find that $y=a+c$, $x+z=b$, and $z=ab/(a+c)$. To simplify the computations $u=x+z$; the Jacobian $\frac{\partial (x,y,z)}{\partial (u,y,z)}$ is equal to one. Therefore,
\[\frac{\partial (x,y,z)}{\partial (a,b,c)}=\frac{\partial (u,y,z)}{\partial (a,b,c)}=\begin{vmatrix}0& 1 & 0 \\ 1 & 0 & 1 \\ z_{a}& z_{b}& z_{c}\end{vmatrix}=z_{a}-z_{c}=\frac{b}{a+c}\]
Thus, the volume of $K$ is
\[\int_{1}^{2}\int_{1}^{2}\int_{1}^{2}\frac{b}{a+c}da\,db\,dc\]
which is equal to $\frac32(3\log 5-4\log 3+\log 2)$ in the unlikely event of my computations being correct.

Thank you for your reply, mlock.
Regrettably you seemed to have made a mistake in calculation.

P.S. Japanese high school students can't understand your solution, but your solution is interesting to me. :)


I remember watching a show called the Condor heroes. it was awesome!


algebra
polynomial
modular arithmetic
number theory proposed
number theory

let $p$ be a prime number and $f_1(x),\dots,f_n(x)$ be polynomials with integer coffecient and for every $n$ polynomials $g_1(x),\dots,g_n(x)$ with integer coffecient $f_1(x)g_1(x)+\cdots+f_n(x)g_n(x)\equiv 0\ (\mod p\ )$
has answer.
Prove that exist integer$x_0$ such that $f_1(x_0)\equiv f_2(x_0)\equiv\dots \equiv f_n(x_0)$

Assume that there is no such $x_0$, then
for every $s\in [0,p-1]$ there exists such index $i_s$ that $f_{i_s}(s)\not\equiv 0\pmod p$.

Consider a polynomial \[ F(x) = \sum_{s=0}^{p-1} Q_s(x) f_{i_s}(x)  \] where $Q_s(x)$ is a polynomial with integer coefficients defined by $Q_s(x) \equiv \frac{x^p - x}{x - s}\pmod p$. Important property of $Q_s(x)$ is that

$Q_s(t)\not\equiv 0\pmod p$ if and only if $t\equiv s\pmod p$.

The polynomial $F(x)$ obviously has the form $f_1(x)g_1(x)+\dots+f_n(x)g_n(x)$ for some $g_1(x), \dots, g_n(x)$ with integers coefficients.

Note that for every $m\in [0,p-1]$ we have
\[ F(m) \equiv Q_m(m) f_{i_m}(m) \not\equiv 0\pmod p,  \]
that is a contradiction to the fact that $F(x)$ must have a root.

The contradition proves that the required $x_0$ exists.


I messed up the A pretty badly.  I hope to do better on the B!

Good luck guys!

And do you guys know when the results for the 10A will come out.  I dont think I passed (114) but I still might have a chance if they lower it.

i think they will be out next week. but idk.


ABCDEF  is  ahexagon. Each interior angle of the hexagon  is 120. Prove that

AB + FA = CD + DE.

oops  :rotfl:

i 4got about the angles, they could be any measure of sides.

[quote="now a ranger"]Since each interior angle of ar egular hexagon is 120 degrees, than this hexagon also must be a regular hexagon.[/quote]

That's plain wrong, there are lots of different hexagons with all angles 120°.

The problem is a slight restatement of http://www.mathlinks.ro/Forum/viewtopic.php?t=16758 part [b]a)[/b] (the equation AB + FA = CD + DE rewrites as AB - DE = CD - FA).

  Darij


geometry
circumcircle
trigonometry
geometric transformation

Let a triangle $ABC$ and $O$ its circumcenter .Let $D$ the midpoint of $AC$ , $M$ the midpoint of $BC$ and $K$ the midpoint of $OM$ . The perpendicular from $D$ to $AB$ meets $AB$ at $E$ . Prove that $KE=KC$

Dear silouan,
especially for you, a solution using spiral similarities  :wink:

Point $F$ is on $AB$ so that $CF \bot AB$

[b][u]Proof:[/u][/b]

We can easiley prove that $\Delta AFC$ and $\Delta OMC$ are similar.
Therefore, there exists a spiral similarity $S$ with center $C$, factor $k$ and angle $\alpha$ that maps $\Delta OMC$ to $\Delta AFC$.

$S(O)=A$ and $S(K)=E$ so clearly $\Delta EKC$ and $\Delta AOC$ are similar. This implies that $\left|EK\right|=\left|CK\right|$.
\[\mathbb{QED}\]
[img]http://img441.imageshack.us/img441/5346/silouan060207mlab4.jpg[/img]
[list]
[b][u]Notes:[/u][/b]
[*]It is easiley checked that $k$ equals $\frac{R}{\left|AC\right|}$, but we don't have to specify $k$ to solve the problem.
[*]It is easiley checked that $\alpha$ equals $\frac{\pi}{2}-\angle ABC$, but yet again, we don't have to specify this angle.
[*]$S(K)=E$ because they are midpoints of segments that are mapped to each other[/list]

Thank you it's very nice.  :)
Could you post a solution without spiral similarities ?

Of course, we can find loads of proofs for this easy problem.

Here's a trigonometric solution:

$P$ is the orthogonal projection of $K$ on $AB$, and $Q$ is the orthogonal projection of $O$ on $AB$.
All we have to prove is that $\left|AE\right|=2\left|QP\right|$, because that is equivalent to saying that $PK$ is the perpendicular bissector of $BE$, so then $\left|KC\right|=\left|KB\right|=\left|KE\right|$

[b][u]Proof:[/u][/b]

First, we note that $\left|OM\right|=\frac{\left|BC\right|}{2\tan{BAC}}$
Now, obviously, $\angle OKP = \angle ABC$, so $\left|QP\right|=\sin{ABC}.\frac{\left|BC\right|}{4\tan{BAC}}$

Because of the sine law in $\Delta ABC$, we can reduce this expression to
\[\left|QP\right|=\cos{BAC}.\frac{\left|AC\right|}{4}\]
But $\left|AE\right|=\frac{\left|AC\right|.\cos{BAC}}{2}$, so we are done here.
\[\mathbb{QED}\]


what are the domain and range of the function y=xsquared +3 :blush:

Think about the graph of $y=x^2+3$. What do we know about it?

Well, it is a parabola, and we know what that looks like. What else? It is just the base $y=x^2$ parabola, shifted up three units along the y-axis. So now, what's the lowest point on this parabola? $y=3$, when $x=0$. What's the highest? There is no highest, the parabola stretches without bounds! And how far to the left/right can it go? Same idea, it can go infinitely far in either direction.

So the domain and range is that all points are real numbers, where $y\geq 3$. You can get all this just by looking at a picture :D


Suzanne's birthday is tomorrow (March 19).  Suzanne's father, Rick, has his birthday on April 1 (no fooling!).  On March 18, 1999,  Rick was three times as old as Suzanne.  Today (March 18, 2000), Rick is 32 years older than Suzanne.   (In this problem, all "ages" are whole numbers.)

What will be the sum of their ages on April 2, 2000?

A.64 years  B.65 years	C.66 years     D.67 years   E.68 years

[hide] Let $ S$ be Suzanne's age on March 18, 1999, and $ R$ be Rick's age on March 18, 1999. Then we have a system of equations.

$ 3S\equal{}R$
$ S\plus{}32\plus{}1\equal{}R\plus{}1 \implies R\equal{}32\plus{}S$.

Substituting,

$ 3S\equal{}S\plus{}32 \implies S\equal{}16 \implies R\equal{}48$.

They"ll both have had two more birthdays by April 2, 2000, so $ 18\plus{}50\equal{}68$. Thus, the answer is E.[/hide]


algebra
polynomial
parameterization
algebra proposed

I think this will help
g(x)=x is a solution and f(x)=x^2007+2 x +1

and

f(x) and g(x) are both monic polynomial with degree m and n such as m n = 2007
thus m,n= 1, 3, 9, 223, 669 or 2007.

[quote="Thaakisfox"]Find all those $f(x)$ and $g(x)$ polynomials, with integer coefficients, for which:

$f(g(x)) = x^{2007}+2x+1$[/quote]

$\frac{d}{dx}f(g(x))= \frac{d(fg(x))}{d g(x)}* \frac{dg(x)}{dx}$
Or $h(x)=f(g(x)) ,h'(x)=f'(g(x))* g'(x)$
Thus ,$2007 x^{2006}+2=u(x)v(x)$ where $u,v$ are both polynomial.
but By Eisenstein Thereom(I don't know if I spell correctly :blush: ,who can tell me?Thanks)
$2007 x^{2006}+2$ is irreducible.
So $g'(x)=Const$
EDIT: Thanks for the help of ZetaX. :)

[quote="ZetaX"]Eisenstein
For those speaking german: http://fachschaft.mathematik.uni-freiburg.de/dinge/eisenstein.html[/quote]

I've failed to send this hint yesterday because of connection problems.

[hide="for those who do not speak german"]The polinomial $2007x^{2006}+2$ is irredussible under Q.
The answer includes 4 infinite series depending on 1 integer parameter c. [/hide]
 :oops_sign: After edided Hawk Tiger's post it only remains to write

[hide="answer"]1) $g(x)=x+c, \ \ f(x)=(x-c)^{2007}+2(x-c)+1$
2) $g(x)=-x+c, \ \ f(x)=(c-x)^{2007}+2(c-x)+1$
3) $f(x)=x+c, \ \ g(x)=x^{2007}+2x-c+1$
4) $f(x)=-x+c, \ \ g(x)=-x^{2007}-2x+c-1$[/hide]


Al, Bert, and Carl are the winners of a school drawing for a pile of Halloween candy, which they are to divide in a ratio of $ 3 : 2 : 1$, respectively. Due to some confusion they come at different times to claim their prizes, and each assumes he is the ﬁrst to arrive. If each takes what he believes to be his correct share of candy, what fraction of the candy goes unclaimed?

$ \textbf{(A)}\ \frac {1}{18} \qquad \textbf{(B)}\ \frac {1}{6} \qquad \textbf{(C)}\ \frac {2}{9} \qquad \textbf{(D)}\ \frac {5}{18} \qquad \textbf{(E)}\ \frac {5}{12}$

[hide]Al will take $ \frac {1}{2}$ of the candy, Bert will take $ \frac {1}{3}$ of the remaining candy, and Carl will take $ \frac {1}{6}$ of the candy remaining after Bert.

$ 1 \minus{} \left(\frac {1}{2} \plus{} \frac {1}{2}\cdot\frac {1}{3} \plus{} \frac {1}{2}\cdot\frac {2}{3}\cdot\frac {1}{6}\right) \equal{} 1 \minus{} \frac {13}{18} \equal{} \frac {5}{18}\quad\Rightarrow\quad\mathrm{D}$[/hide]


geometry
circumcircle
ratio
geometry proposed

[color=darkred]Let $ABC$ be a right triangle so that $AB\perp AC$. Let $M,N$ be two mobile points so that $\{\begin{array}{c}MB\perp AB\ ,\ (CM)\cap AB=\emptyset\\\ NC\perp AC\ ,\ (BN)\cap AC\ne\emptyset\end{array}$
and $\widehat{MAB}\equiv\widehat{NAC}$. Prove that the lines $BN$, $CM$ and the tangent in the point $A$ to the circumcircle of $\triangle ABC$ are concurrently.[/color]

Thank toy dear [b][size=100]Virgil[/size][/b] for a nice problem.

Let $A'$ be, the antidiametric ( = diametric opposite ) point of $A,$ with respect to the circumcircle $(O),$ of the given triangle $\bigtriangleup ABC$ and we denote as $S,$ the intersection point of the segment line $A'S,$ from the line through $A$ and perpendicular to $AA'.$ Also, we denote as $M',$ $N',$ the midpoints of the segments $BA',$ $CS,$ respectively.

Because of $AB\perp AC$ and $AA'\perp AS,$ we have the similarity of the right triangles $\bigtriangleup BAA',$ $\bigtriangleup CAS$ $($ $\angle BAA' = \angle CAS$ $)$ and so, we conclude that $\angle BAM' = \angle CAN'$ $\Longrightarrow$ $AM'\perp AN'.$

We will prove that the point $P\equiv BN'\cap CM',$ lies on the segment line $AS.$

We denote the points $P\equiv AS\cap BN',$ $P'\equiv AS\cap CM'$ and it is enough to prove that $P\equiv P'.$

From $AB\parallel N'S$ $\Longrightarrow$ $\frac{PA}{PS}= \frac{AB}{N'S}$ $,(1)$

Apply the [b][size=100]Menelaus theorem[/size][/b], in the right triangle $AA'S,$ taken as its transversal the segment line $CM'P',$

we have that $\frac{P'A}{P'S}\cdot \frac{CS}{CA'}\cdot \frac{EA'}{EA}$ $\Longrightarrow$ $\frac{P'A}{P'C}= \frac{CA'}{CS}\cdot \frac{EA}{EA'}$ $,(2)$ where $E\equiv AA'\cap CM'P'.$

From $(1),$ $(2)$ we conclude that it is enough to prove that $\frac{AB}{N'S}= \frac{CA'}{CS}\cdot \frac{EA}{EA'}$ $\Longrightarrow$ $\frac{EA}{EA'}= \frac{CS}{N'S}$ $,(3)$ $($ because of $AB = CA'$ $).$

But the equality $(3)$ is true because of $\frac{EA}{EA'}= 2$ $($ easy to prove $)$ and $\frac{CS}{N'S}= 2$ $($ from $N'$ as the midpoint of $CS$ $).$

So, we conclude that $P'\equiv P$ and hence, we have that the point $P\equiv BN'\cap CM',$ lies on the segment line $AS,$ as the tangent line of the circumcircle of $\bigtriangleup ABC,$ through its vertex $A.$

$\bullet$ We denote the points $M$ $($ between $B,$ $M'$ $)$ and $N$ $($ between $C,$ $N'$ $),$ as the problem states
$($ $\angle MAB = \angle NAC$ $)$ and it is easy to show that $AM\perp AN.$

We will prove that the intersection point $Q,$ of $BN,$ $CM,$ lies on the segment line $AS.$

The pencil $C.BMM'A',$ is intersected from the segment line $AS$ and so, we have that $(C.BMM'A') = (R,Q,P,S)$ $,(4)$ where $R\equiv AS\cap CB$ ab\nd $Q\equiv AS\cap CM.$

Because of $AB\perp AC,$ $AM\perp AN,$ $AM'\perp AN'$ and $AA'\perp AS,$ we conclude that the homologous rays of the pencils $A.BMM'A',$ $A.CNN'S,$ form equal angles and so, they have equal cross ratios.

That is we have $(A.BMM'A') = (A.CNN'S) = (B.CNN'S)$ $,(5)$

The pencil $B.CNN'S,$ is intersected from the segment line $AS$ and so we have $(B.CNN'S) = (R,Q',P,S)$ $,(6)$ where $Q'\equiv AS\cap BN.$

From $(4),$ $(5),$ $(6)$ and because of $(A.BMM'A') = (C.BMM'A'),$ we conclude that $(R,Q,P,S) = (R,Q',P,S)$ $,(7)$

From $(7)$ we conclude that $Q'\equiv Q$ and the proof is completed.

Kostas Vittas.

PS. We say as $(A.BMM'A'),$ the cross (=double) ratio of the pencil $A.BMM'A'$ and as $(R,Q,P,S) =\frac{PR}{PQ}\div \frac{SR}{SQ}$ the cross ratio of the collinear points $R,$ $Q,$ $P,$ $S,$ in the order as they are written.

A friend of mine Kostas Helatiotis, said me that we don't need so mach theory to solve this problem.

So, we denote the points $Q\equiv AS\cap BN,$ $Q'\equiv AS\cap CM$ and we can easy to prove that $Q'\equiv Q,$ as follows:

From $AB\parallel NS$ $\Longrightarrow$ $\frac{QA}{QS}= \frac{AB}{NS}$ $,(1)$

From the triangle $\bigtriangleup AA'S,$ with respect to $CM$ as its transversal, by [b][size=100]Menelaus theorem[/size][/b] we have that

$\frac{Q'A}{Q'S}\cdot \frac{CS}{CA'}\cdot \frac{EA'}{EA}= 1$ $\Longrightarrow$ $\frac{Q'A}{Q'S}= \frac{CA'}{CS}\cdot \frac{EA}{EA'}$ $,(2)$ where $E\equiv AA'\cap CM.$

From $(1),$ $(2)$ we conclude that it is enough to prove that $\frac{AB}{NS}= \frac{CA'}{CS}\cdot \frac{EA}{EA'}$ $\Longrightarrow$ $\frac{EA}{EA}= \frac{CS}{NS}$ $,(3)$ $($ $AB = CA'$ $).$

Because of $AB\perp AC$ and $AA'\perp AS,$ we conclude the similarity of the right triangles $\bigtriangleup BAA',$ $\bigtriangleup CA'S$

$($ $\angle BAA' = \angle CAS$ $)$ and so we have that $\frac{CS}{NS}= \frac{BA'}{MA'}$ $,(4)$ $($ because of $\angle MAB = \angle NAC$ $).$

We denote the point $F\equiv AB\cap CM$ and from $BF\parallel CA'$ $\Longrightarrow$ $\frac{MB}{MA'}= \frac{BF}{CA'}$

$\Longrightarrow$ $\frac{MB+MA'}{MA'}= \frac{BF+CA'}{CA'}= \frac{AB+BF}{CA'}$ $\Longrightarrow$ $\frac{BA'}{MA'}= \frac{AF}{CA'}= \frac{EA}{EA'}$ $,(5)$

From $(4),$ $(5)$ $\Longrightarrow$ $\frac{EA}{EA'}= \frac{CS}{NS}$ $\Longrightarrow$ $Q'\equiv Q$ and the proof is completed.

Kostas vittas.

PS. Dear all my friends, please ignore all I wrote in my first message, as unnecessary prattling.


function
limit
real analysis
real analysis solved

what properties should the function $f$ have so $\lim_{x\to 0} xf(x)=0$?

Very strange question: I have always thought that $\lim_{x\to 0}xf(x)=0$ is a very simple property that doesn't require any further attempts to reformulate it. It is almost like asking "What properties should $f$ have to ensure that $f(0)=0$?". Perhaps, I am just misunderstanding what is being asked.   :? Can you clarify your question, mathematica?

No properties !It can be bounded but and  not bounded as well.( $f(x) = \frac{1}{\sqrt x }$. You must ask something else , as <<post 1 >> asks you.

 Babis

It can easily be proved that for every function that is bounded in zero's neighbourhood this limit equals zero. But how about unbounded functions? For which group of unbounded functions is this limit correct?

I don't think these functions can be "described" in any way. I mean, there are some obvious solutions (such as $f(x) = x^a$ and all bounded functions) but there are tons of weird, non-trivial solutions.

Well, my opinion , as I have allready  said  , is that we can not decide about this family of functions. The question must be reformulated ! The condition << Bounded around 0 >> is sufficient , but not necessary.I hope that nobody missunderstood my first post !

Babis

$f$ should have the property $f(x)=o\left(\frac1x\right)$ as $x\to0.$

... which is, of course, just repeating what fedja said in post #2.


Find all $ f: \mathbb{R}\to\mathbb{R}$ be a function such that
$ f(x\plus{}y)\plus{}f(x\minus{}y)\equal{}2f(x)f(y)$ for all $ x,y\in\mathbb{R}$ and $ \lim_{x\to\infty}f(x)\equal{}0$

This is one of the famous d'Alambert's functional equations, whose one of its continuous solutions is the called "Prostapheresis" formula (I don't know how is the name in english, so I guessed).
Since you did not impose any continuity or differentiability condiction, that becomes non-trivial, I think, and even stranger, the limit you posted do not exist to trigonometric functions! Weird, isn't it?

I think prostaférese is simply called "sum-to-product" trigonometric formula in English.

$ P(x\plus{}a,x): f(a)\equal{}2f(x\plus{}a)f(x)\minus{}f(2x\plus{}a)$.
Now fix $ a$ and let $ x\rightarrow \infty$.
$ f(a)\equal{}0,\forall a$


I don't know how to post problems in the contest archive. I have all of the AMC 10 from 2000-2006, but I don't know how to put them up. I also don't know how to make a picture in latex. Should I send the problems to a moderator, or what?

You post problems in coreponding box, mods will post in Resources later, i think  :wink:

what corresponding box?

P.S. Do you think that if I post AMC 10 problems in High School Basics, do you think that a moderator will post them in the archive???

[quote="nutz_for2.718281828"]

P.S. Do you think that if I post AMC 10 problems in High School Basics, do you think that a moderator will post them in the archive???[/quote]

Yes, i think  :wink:

... or something similar so they will be easy find when you search, someone will probably do it.


function
parameterization
geometry
geometric transformation
algebra
polynomial
vector

Fermat's last theorem:If $ (x, y)\equal{}1, (x, z)\equal{}1, (y, z)\equal{}1,
n$ is odd prime, $ x, y, z\in\mathbb{Z}$, then there is no such
solution of Fermat equation $ x^n\plus{}y^n\equal{}z^n$.
Proof: Suppose $ (x, y, z)$ is such solution. Since $ n$ is odd,
we can let $ (z, 2^{k\plus{}1})\equal{}2^k, k>0$, then let $ P(z)\equal{}k$. Since $ x,
y, n$ are odd, we can verify that $ P(x\plus{}y)\equal{}nk$. Let
$ P(\frac{rz\minus{}x\minus{}1}{2})\equal{}2nk$, then $ P(\frac{rz\plus{}y\minus{}1}{2})\equal{}nk\minus{}1$. Let
$ t\equal{}(\frac{rz\minus{}x\minus{}1}{2}, \frac{rz\plus{}y\minus{}1}{2})$, then $ P(t)\equal{}nk\minus{}1,
0\equal{}(\frac{z}{2})^n\minus{}(\frac{x}{2})^n\minus{}(\frac{y}{2})^n\equiv
(\frac{z}{2})^n, \bmod t$. As a result, we verify
$ P((\frac{z}{2})^n)\equal{}n(k\minus{}1)<nk\minus{}1$, which yields a contradiction.                                        $ \square$

$ r\equal{}2^ab, \minus{}k<a, b$ is odd.

Fermat's last theorem:If $ (x, y) \equal{} 1, (x, z) \equal{} 1, (y, z) \equal{} 1, n$ is odd prime, $ x, y, z\in\mathbb{Z}$, then there is no such
solution of Fermat equation $ x^n \plus{} y^n \equal{} z^n$.
Proof: Suppose $ (x, y, z)$ is such solution.
$ Step$ $ 1$: We can let $ z$ is even.
Suppose $ (a, b, c)$ is such solution. If $ c$ is even, then let $ z \equal{} c$.
If $ a($ or $ b)$ is even, since $ n$ is odd, we can get
$ c^n \plus{} ( \minus{} b)^n \equal{} a^n$, then let $ z \equal{} a$.
$ Step$ $ 2$: We denife a function P(t).
If $ (t, 2^{k \plus{} 1}) \equal{} 2^k, k > 0$, then let $ P(t) \equal{} k$.
$ Step$ $ 3$: We verify $ P(z), P(x \plus{} y)$.
Since $ z$ is even, we can let $ P(z) \equal{} k, k > 0$. Since $ x, y, n$ are odd,
we can verify that $ \frac {x^n \plus{} y^n}{x \plus{} y}$ is odd. Hence
$ P(x \plus{} y) \equal{} nk$.
$ Step$ $ 4$: We denife a parameter $ r$.
Let $ l \equal{} y \minus{} 2^{nk}$. $ P(x \plus{} l) \equal{} nk$.Soppose $ x \plus{} l \equal{} 2^{nk}a, z \equal{} 2^kb$. We know that there is some
$ r$ to be the solution of $ rb\equiv 2^{nk \minus{} k}a \bmod 2^{2nk \minus{} k}$.
Hence $ P(\frac {rz \minus{} x \minus{} l}{2}) \equal{} 2nk$.
$ Step$ $ 5$. We yield a contradiction.
We can verify $ P(\frac {rz \plus{} y \minus{} l}{2}) \equal{} nk \minus{} 1$.
Let $ t \equal{} (\frac {rz \minus{} x \minus{} l}{2}, \frac {rz \plus{} y \minus{} l}{2})$, then we get
$ P(t) \equal{} nk \minus{} 1, \frac {x}{2}\equiv \frac {rx \minus{} l}{2}, \frac {y}{2}\equiv \minus{} \frac {rx \minus{} l}{2} \bmod t$. Since $ n$ is odd, hence
$ 0 \equal{} (\frac {z}{2})^n \minus{} (\frac {x}{2})^n \minus{} (\frac {y}{2})^n\equiv (\frac {z}{2})^n, \bmod t$. As a result, we verify
$ P((\frac {z}{2})^n) \equal{} n(k \minus{} 1) < nk \minus{} 1$, which yields a contradiction.
$ \square$

You have misunderstood completely the properties of your $ P$ function. $ P(x\plus{}1) \equal{} nk$ should be $ P(x\plus{}1) > nk$, for instance. Similar errors plague the rest of your proof from there.

It's $ x \plus{} l$ not $ x \plus{} 1$ in the 2nd version. Thank you!
$ Step$ $ 4$: We denife a parameter $ r$.
Let $ l \equal{} y \minus{} 2^{nk}$. $ P(x \plus{} l) \equal{} nk \plus{} m, m > 0$.Soppose $ x \plus{} l \equal{} 2^{nk \plus{} m}a, z \equal{} 2^kb$. We know that there is some
$ r$ to be the solution of $ rb\equiv 2^{nk \minus{} k \plus{} m}a \bmod 2^{nk \minus{} k \plus{} m\minus{}1}$.
Hence $ P(\frac {rz \minus{} x \minus{} l}{2}) \equal{} nk \plus{} m\minus{}1$.

Fermat's last theorem: If $ (x, y)\equal{}1, (x, z)\equal{}1, (y, z)\equal{}1,
n$ is odd prime, $ x, y, z\in\mathbb{Z}$, then there is no such
solution of Fermat equation $ x^n\plus{}y^n\equal{}z^n$.
Proof: Suppose $ (x, y, z)$ is such solution. Suppose $ (a, b, c)$ is such solution. If $ c$ is even, then let
$ z\equal{}c$. If $ a($ or $ b)$ is even, since $ n$ is odd, we can get
$ c^n\plus{}(\minus{}b)^n\equal{}a^n$, then let $ z\equal{}a$. If $ (t, 2^{k\plus{}1})\equal{}2^k, k>0$, then
let $ P(t)\equal{}k$. Since $ z$ is even, we can let $ P(z)\equal{}k, k>0$. Since
$ x, y, n$ are odd, we can verify that $ \frac{x^n\plus{}y^n}{x\plus{}y}$ is
odd. Hence $ P(x\plus{}y)\equal{}nk$. Let $ l\equal{}y\minus{}2^{nk}$. $ P(x\plus{}l)\equal{}nk\plus{}m, m>0$.
Suppose $ x\plus{}l\equal{}2^{nk\plus{}m}a, z\equal{}2^kb, a, b$ are odds. We know that there
is some $ r$ to be the solution of $ rb\equiv 2^{nk\plus{}m\minus{}k}a \bmod
2^{nk\plus{}m\minus{}k}$. If $ P(r)\equal{}nk\plus{}m\minus{}k$, then $ P(\frac{rz\minus{}x\minus{}l}{2})>nk\plus{}m\minus{}1$.
If $ P(r)>nk\plus{}m\minus{}k$, then $ P(\frac{rz\minus{}x\minus{}l}{2})\equal{}nk\plus{}m\minus{}1$. Hence
$ P(\frac{rz\minus{}x\minus{}l}{2})\geq nk\plus{}m\minus{}1$. We can verify
$ P(\frac{rz\plus{}y\minus{}l}{2})\equal{}nk\minus{}1$. Let $ t\equal{}(\frac{rz\minus{}x\minus{}l}{2},
\frac{rz\plus{}y\minus{}l}{2})$, then we get $ P(t)\equal{}nk\minus{}1, \frac{x}{2}\equiv
\frac{rz\minus{}l}{2}, \frac{y}{2}\equiv \minus{}\frac{rz\minus{}l}{2} \bmod t$. Since
$ n$ is odd, hence
$ \frac{x}{2}\plus{}\frac{y}{2}|\minus{}(\frac{x}{2})^n\minus{}(\frac{y}{2})^n$. It
follows that
$ 0\equal{}(\frac{z}{2})^n\minus{}(\frac{x}{2})^n\minus{}(\frac{y}{2})^n\equiv
(\frac{z}{2})^n, \bmod t$. As a result, we verify
$ P((\frac{z}{2})^n)\equal{}n(k\minus{}1)<nk\minus{}1$, which yields a contradiction.
$ \square$
$ Acknowledgement$
When I was going to give up from trying to get a chance to
advanced study, it's Simon Singh's $ FERMAT'S$ $ LAST$
$ THEOREM^{[1]}$, The story of a riddle that confounded the world's
great minds for $ 358$ years, that encouraged me to succeed once
more. I have made lots of mistakes and I give my thanks to those
who have encouraged and helped me in the past four years.
$ Reference$
$ [1]$: Simon Singh, $ FERMAT'S$ $ LAST$ $ THEOREM$, The story of a riddle that confounded the world's great minds for
$ 358$ years, chinese edtion translated by Mi Xue, Shanghai
Translation Publishing House, Shanghai, 1998

Could you check it now?

I have read http://research.microsoft.com/~cohn/Thoughts/advice.html
and will do my best to follow it.

Frey curves
　　Suppose there were a nontrivial solution of the Fermat equation for some number n, i. e. nonzero integers a, b, c, n such that
　　Then we recall that around 1982 Frey called attention to the elliptic curve
　　Call this curve E. Frey noted it had some very unusual properties, and guessed it might be so unusual it could not actually exist.
　　To begin with, various routine calculations enable us to make some useful simplifying assumptions, without loss of generality. For instance, n may be supposed to be prime and 5. b can be assumed to be even, a 3 (mod 4), and c 1 mod 4. a, b, and c can be assumed relatively prime.
　　The "minimal discriminant" of E, can be computed to be - a power of 2 times a perfect prime power. One unusual thing about E is how large the discriminant is.
　　The conductor is a product of primes at which E has bad reduction, which is the same as the set of primes that divide the minimal discriminant. However, the exact power of each prime occurring in the conductor depends on what type of singularity the curve possesses modulo the primes of bad reduction. The definition of the conductor provides that p divides the conductor only to the first power if x(x-a )(x+b ) has only a double root rather than a triple root mod p. Now, any prime can divide only a or b but not both, since otherwise it would also divide c, and we have assumed a, b, and c are relatively prime. Hence the the polynomial will have the form x (x+d) mod p, where (p,d) = 1. Hence there is only at most a double root modulo any prime, and therefore the conductor is square-free. In other words, E is semistable.
　　There are other odd things about E, which have to do with specific properties of its Galois representations. Because of these, Ribet's results allow us to conclude that E cannot be modular.
　　Proof of Fermat's Last Theorem from the Taniyama-Shimura conjecture
　　After Frey drew attention to the unusual elliptic curve which would result if there were actually a nontrivial solution to the Fermat equation, Jean-Pierre Serre (who has made many contributions to modern number theory and algebraic geometry) formulated various conjectures which, sometimes alone and sometimes together with the Taniyama-Shimura conjecture, could be used to prove Fermat's Last Theorem.
　　Kenneth Ribet quickly found a way to prove one of these conjectures. The conjecture itself doesn't really talk about either Frey curves or FLT. Instead, it simply states that if the Galois representation associated with an elliptic curve E has certain properties, then E cannot be modular. Specifically, it cannot be modular in the sense that there exists a modular form which gives rise to the same Galois representation.
　　We need to introduce a little additional notation and terminology to explain this more precisely. Let S(N) be the (vector) space of cusp forms for (N) of weight 2. "Classical" theory of modular forms shows that S(N) can be identified with the space of "holomorphic differentials" on the Riemann surface X (N). Furthermore, the dimension of S(N) is finite and equal to the "genus" of X (N). "Genus" is a standard topological property of surfaces, which is intuitively the number of holes in the surface. (E. g. a torus, such as an elliptic curve, has genus 1.)
　　But there are relatively simple explicit formulas for the genus of X (N). These formulas, developed long ago by Hurwitz in the theory of Riemann surfaces, involve the index of (N) in G. A fact of crucial importance is that for N <. 11, the genus of X (N), and hence the dimension of S(N), is zero. In other words, S(N) contains only the constant form 0 in that case. We shall use this fact about S(2) very soon.
　　There are certain operators called Hecke operators, after Erich Hecke, on spaces of modular forms, and for the subspace S(N) in particular, since they preserve the weight of a form. Hecke operators can be defined concretely in various ways. There is a Hecke operator T(n) for all n 1. There are formulas that relate T(n) for composite n to T(p) where p is a prime dividing n, so T(p) for prime p determine all T(n).
　　All T(n) are linear operators on S(N). If there is an f in S(N) that is a simultaneous eigenvector of all T(n), i. e. T(n)(f) = (n)f, where (n) C, f is called an eigenform. (Nontrivial eigenforms need not exist, e. g. if S(N) has dimension 0.) f is said to be normalized if its leading Fourier series coefficient is 1. In that case, the eigenvalues (n) turn out to be the Fourier series coefficients in the expansion
　　It can be shown that if f(z) is a cusp form which is a normalized eigenfunction for all T(p), then there is an Euler product decomposition for the L-function L(f,s). This is obviously of great technical usefulness in relating L-functions of forms and those of elliptic curves (which are Euler products by definition).
　　If f S(N) is a normalized eigenform of all Hecke operators, it can in fact be shown that the coefficients in the Fourier expansion are all algebraic numbers and that they generate a finite extension K of Q.
　　Prime ideals of the ring of integers of K are the analogues of prime numbers of Q. In the case that f is a normalized eigenform it is possible to carry out the construction of a Galois representation (f, ) of Gal( /Q) for any prime ideal of the ring of integers of K.
　　At last we can describe what Ribet proved. Suppose E is a semistable elliptic curve with conductor N and that its associated Galois representation (E,p) for some prime p has certain properties. Suppose 2 divides N (which is true for Frey curves). If E is modular, then there is a normalized eigenform f and a prime ideal lying over p (i. e. one of the prime factors of p in the extension field generated by the Fourier coefficients of f) such that the Galois representation (f, ) is (E,p). Ribet showed that it is possible to find an odd prime q p which divides N such that there is another f' S(N/q), and a corresponding prime ideal ' of the ring of integers in the field generated by the coefficients of f' such that (f', ') gives essentially the same Galois representation. This is known as the "level lowering" conjecture since it asserts that under the right conditions there is an eigenform of a lower level that gives essentially the same representation.
　　But this process can be repeated as long as N has any odd prime factors. It is important that the curve E is semistable so that N is square-free. This means that all odd prime factors of N can been eliminated, so there must be a nontrivial eigenform of level 2, i. e. in S(2), that gives essentially the same Galois representation. And that is a contradiction, since S(2) has dimension 0, hence contains no non-trivial forms. The contradiction means that E can't be modular.
　　Now we invoke the "unusual" properties of the Frey curve resulting from a solution of FLT. These properties allow it to be shown that the associated Galois representation has the properties required to apply Ribet's result. Hence the Frey curve can't be modular.
　　But the Frey curve is semistable, so the semistable case of the Taniyama-Shimura conjecture, which Wiles proved, implies the curve is modular. This contradiction means that the assumption of the existence of a nontrivial solution of the
　　Fermat equation must be wrong, and so FLT is proved.
　　Proof of the semistable case of the Taniyama-Shimura conjecture
　　Not very surprisingly (since it was such hard work), the proof is quite technical. However, the outline of it is relatively simple. In the following, we assume that E is a semistable elliptic curve with conductor N. We have to prove E is modular.
　　We know we can construct a Galois representation (E,p ): G->GL (Z ) for any prime p. To show that E is modular, we have to show this representation is modular in a suitable sense. The wonderful thing is, this needs to be done for only one prime p, and we can "shop around" for whatever prime is easiest to work with.
　　To show (E,p ) is modular involves finding a normalized eigenform f in S(N) with appropiate properties. The properties required are that the eigenvalues of f, which are its Fourier series coefficients, should be congruent mod q to trace( (E,p ) ( )) for all but a finite number of prime q. ( G is the "Frobenius element".) We know that the trace is, for q prime to pN, the coefficient a = q + 1 - #(E(F )) of the Dirichlet series of L(E,s).
　　The longest and hardest part of Wiles' work was to prove a general result which is roughly that if (E,p) is modular then so is (E,p ). In other words, to show that E is modular, it is actually sufficient just to show that (E,p): G->GL (Z/pZ) is modular.
　　This is called the "modular lifting problem".
　　The problem boils down to assuming that (E,p) is modular and attempting to "lift" the representation to (E,p ). This is done mainly by working with the theory of representations as much as possible, without specific reference to the curve E. The proof uses a concept called "deformation", which suggests intuitively what goes on in the process of lifting.
　　The outcome of this part of Wiles work is:
　　Theorem: Suppose that E is a semistable elliptic curve over Q. Let p be an odd prime. Assume that the representation (E,p) is both irreducible and modular. Then E is a modular elliptic curve.
　　At this point, all we have to do is find a single prime p such that (E,p) is irreducible and modular. But Langlands and
　　Tunnell had already proven in 1980-81 that (E,3) is modular.
　　Unfortunately, this isn't quite enough. If (E,3) is irreducible, we are done. But otherwise, one more step is required. So suppose (E,3) is reducible. Wiles then considered (E,5). That may be either reducible or irreducible as well. If it is reducible, Wiles proved directly that E is modular.
　　So the last case is if (E,5) is irreducible. Wiles showed that there is another semistable curve E' such that (E',3) is irreducible, and hence E' is modular by the above theorem. But Wiles could also arrange that the representations (E',5) and (E,5) are isomorphic. Hence (E,5) is irreducible and modular, so E is modular by the theorem.

http://cgd.best.vwh.net/home/flt/flt01.htm
The Mathematics of Fermat's Last Theorem

1、http://www.pbs.org/wgbh/nova/proof/
For over 350 years, some of the greatest minds of science struggled to prove what was known as Fermat's Last Theorem -- the idea that a certain simple equation had no solutions. Now hear from the man who spent seven years of his life cracking the problem, read the intriguing story of an 18th century woman mathematician who hid her identity in order to work on Fermat's Last Theorem, and demonstrate that a related equation, the Pythagorean Theorem, is true.
2、http://fermatslasttheorem.blogsp ... m-proof-for-n3.html
topic blog
3、http://rendezvous.com/tangledweb/conferences/fermat/index.html
A Conference on the proof of Andrew J. Wiles
4、http://www-groups.dcs.st-and.ac.uk/~history/HistTopics/Fermat's_last_theorem.html
Fermat's last theorem
5、http://math.stanford.edu/~lekheng/flt/index.html
Papers

Do you sure about the last part?
The power of 2 in right psrt of this division is smaller than left part.
\frac{x}{2}+\frac{y}{2}|-(\frac{x}{2})^n-(\frac{y}{2})^n

Thank you! It can be deleted. I am sure we can get  $ \minus{} (\frac {x}{2})^n \minus{} (\frac {y}{2})^n\equiv0 \bmod t$ from
$ \frac {x}{2}\equiv \frac { \minus{} l}{2}, \frac {y}{2}\equiv \frac {l}{2} \bmod t$.

full version:
Fermat's last theorem: If $ (x, y) \equal{} 1, (x, z) \equal{} 1, (y, z) \equal{} 1, n$ is odd
prime, $ x, y, z\in\mathbb{Z}$, then there is no such solution of
Fermat equation $ x^n \plus{} y^n \equal{} z^n$.
Proof: Suppose $ (a, b, c)$ is such solution. If $ c$ is even, then let
$ (x, y, z) \equal{} (a, b, c)$. If $ a($ or $ b)$ is even, since $ n$ is odd,
we can get $ c^n \plus{} ( \minus{} b)^n \equal{} a^n$, then let $ (x, y, z) \equal{} (c, \minus{} b, a)$. If
$ (t, 2^{k \plus{} 1}) \equal{} 2^k$, then let the $ 2$-adic order function
$ \nu_2(t) \equal{} k$. Since $ z$ is even, let $ \nu_2(z) \equal{} k, k > 0$. Since $ x, y, n$ are odd, we can verify that $ \frac {x^n \plus{} y^n}{x \plus{} y}$ is odd.
Hence $ \nu_2(x \plus{} y) \equal{} nk$. Let $ l \equal{} y \minus{} 2^{nk}$. $ \nu_2(x \plus{} l) \equal{} nk \plus{} m, m > 0$.
Hence $ \nu_2(\frac { \minus{} x \minus{} l}{2}) \equal{} nk \plus{} m \minus{} 1, \nu_2(\frac {y \minus{} l}{2}) \equal{} nk \minus{} 1$.
Let $ t \equal{} (\frac { \minus{} x \minus{} l}{2}, \frac {y \minus{} l}{2})$, then we get
$ \nu_2(t) \equal{} nk \minus{} 1, \frac {x}{2}\equiv \frac { \minus{} l}{2}, \frac {y}{2}\equiv \frac {l}{2} \bmod t$. It follows that
$ 0 \equal{} (\frac {z}{2})^n \minus{} (\frac {x}{2})^n \minus{} (\frac {y}{2})^n\equiv (\frac {z}{2})^n, \bmod t$. As a result, we verify
$ \nu_2((\frac {z}{2})^n) \equal{} n(k \minus{} 1) < nk \minus{} 1 \equal{} \nu_2(t)$, which yields a
contradiction. $ \square$

[quote="daplikos"]Is not necessery that [b]t[/b] is a integer?
Becuase this t isn't integer.[/quote]

Since $ l\equal{}y\minus{}2^{nk}, \frac { \minus{} x \minus{} l}{2}, \frac {y \minus{} l}{2}$ are both integers, Hence $ t \equal{} (\frac { \minus{} x \minus{} l}{2}, \frac {y \minus{} l}{2})$ is integer


If you know what a math marathon is, you only have to read the problem. For those of you who don't know,

 :arrow: You have a current math problem to solve. Hide your solution, but be sure to post it.
 :arrow: Please keep the questions at the Basics level. You don't have to tell the source of your question, but you can't post a question until the current one is solved correctly.
 :arrow: Please post the problem number. If there have been 6 problems posted and solved, you post a big "Problem 7" And your problem.

I'll start.

[color=indigo][size=200]Problem 1[/color][/size]

Four delagate types are sitting in a theatre. Each type fills a whole row. They must sit in the backmost rows. How many ways can they sit, if the position of delagates doesn't matter, but the types do?

Marathons, nowadays, are disliked.... please do not start one

It's best you read these things before posting in a forum....
[quote="MCrawford"]For now, we are locking marathon threads and asking that students discontinue creating them.  The reasons are that (1) many of the problems are growing too difficult for the level of the forum, (2) the depth of discussion of the solutions is low, and (3) it's hard to search/identify the problems in any way.  Please keep posts to one problem per thread (except where another problem relates to the original problem discussed).[/quote]
This topic will die soon.


[quote="Bottema"]Prove that sqrt((n+1)n(n^2+1)) is irrational for all n natural![/quote]

Did you mean n > 1? Sqrt((1+1)(1)(1^2+1)) = 2.

[hide]A hasty try :

sqrt((n+1)n(n^2+1)) is rational iff it is integer.

So the problem is showing that (n+1)n(n^2+1) is never a square (n >= 3).

n+1=n(n+1) - (n-1) so gcd (n+1,n+1)= 1 if n even, 2 otherwise.

gcd(n,n+1)=gcd(n,n+1)=1.

So the only case possible is n square and n+1 and n+1 have only 2 as prime factor => n=1.[/hide]


function
geometry
geometric transformation
real analysis
real analysis unsolved

let $ f: [a,b]\to\mathbb R$ be a continuous function,and let $ f(a) \equal{} A,f(b) \equal{} B$. prove that for every real number $ C$ between $ A,B$,there exists a real number $ c\in [a,b]$ such that $ f(c) \equal{} C$.

I know that this theorem is pretty well-known,and I also know a complicated "non-trivial" solution for it,I was just wondering and looking for a simpler solution.

It's called the "Intermediate Value Theorem."

It's Intermediate Value Theorem (or in my literature: Bolzano-Cauchy), and the proof available on Wiki isn't too complicated:

http://en.wikipedia.org/wiki/Intermediate_value_theorem

and the proof of it special case (Bolzano's Theorem) can be found here

http://www.cut-the-knot.org/fta/brodie.shtml

and can be used for proving the general result.

//EDIT: JBL was a sec faster.

thanks JBL and hsiljak,the solution on Wiki is much simpler than the solution I had in mind,the solution I knew was somehow beautiful but a bit complicated:
first it proves the theorem for the special case where "$ g: [0,1]\to\mathbb R$ and $ g(0) < 0,g(1) > 0$ and $ C \equal{} 0$",by writing every number in $ [0,1]$ in base 2,i.e. in the form:

$ r \equal{} 0.n_1n_2n_3\cdots$

where $ n_i$ is either 0 or 1.

and then out of no where he comes out with:

$ g(t) \equal{} f\left((1 \minus{} t)a \plus{} tb\right) \minus{} C$

and finishes the proof.

The uniqueness of binary expansion of reals is a somewhat complicated result and more powerful than you need to prove the IVT, which is a consequence of the way continuous functions translate (edit:) connected sets and is far more general.  The definition of $ g(t)$ is not "out of nowhere": it's a translation and stretching of $ f$.  The special case contains all of the features of the general case (in other words, this is a WLOG argument).

[quote="t0rajir0u"]continuous functions translate [b]compact[/b] sets[/quote]  I think you want "connected" here :)

there is also another problem which the solution uses the same argument (binary expansion of reals),is there a simpler solution for this one too?:
[b]every continuous function $ f: [a,b]\to\mathbb R$ has a "maximum" and "minimum" point on $ [a,b]$.[/b] :maybe:

Now that's Weierstrass' Extreme Value theorem: and again, the proof given in Wiki isn't too hard to understand-although it could have been written better, I suppose.

http://en.wikipedia.org/wiki/Extreme_value_theorem
http://www.cut-the-knot.org/fta/fta_note.shtml


I can't see the Mandelbrot set on the virtual classroom page.

Please see the image below.

[img]http://www.freewebs.com/a1b9/problem1.PNG[/img]

that is the mandelbrot set...(black circle-thingys, red background...)

lol, it also took me a while to realize what "Mandelbrot Set" that I"m suppose to see is  :D

If you want to test it, you can click on it and zoom in on different areas.

Click Start -> Default Programs -> Add or Remove Programs -> Java 5.0 Update 9.
Uninstall.
Oh, and if you're using Windows 2000, go to program files instead.


find all pirme numbers p for which number (2^(p-1) - 1)/p is a square of natural number.
 :huh:

That's some former QEDMO shortlisted problem  :wink:

Write it as $ 2^{p\minus{}1}\minus{}1 \equal{} px^2$. Now check $ p\equal{}2$ and factor the right hand side if $ p$ is odd. By unique factorisation, you end up with $ 2^m\plus{}1$ or $ 2^m\minus{}1$ being a perfect square ($ m\equal{} \frac{p\minus{}1}2$).
The second one is ruled out $ \mod 4$ for $ m>1$, the first one means $ 2^m \equal{} x^2\minus{}1$, again factor the left hand side to reach a contradiction for $ m>1$.
So $ m\equal{}1$ is left, corresponding to $ p\equal{}3$, being a solution.

[quote="ZetaX"]
Write it as $ 2^{p \minus{} 1} \minus{} 1 \equal{} px^2$. Now check $ p \equal{} 2$ and factor the right hand side if $ p$ is odd. By unique factorisation, you end up with $ 2^m \plus{} 1$ or $ 2^m \minus{} 1$ being a perfect square ($ m \equal{} \frac {p \minus{} 1}2$).[/quote]

Sorry, I don't understand this factorisation... Could you write it more approachable? :)

You have $ px^2 \equal{} (2^m\plus{}1)(2^m\minus{}1)$ ($ m\equal{}\frac{p\minus{}1}2$ as before). Now $ 2^m\plus{}1$ and $ 2^m\minus{}1$ are coprime, so they don't share a common prime factor. But except $ p$, every prime factor in $ px^2$ occurs an even number of times. Take the factor $ 2^m \pm 1$ not divisible by $ p$, it needs to have every prime divisor an even number of times. This means: it is a square.

Oh, I see  :lol:
Thank you very much  :!:

ZetaX u overlooked the fact that any 2 powers of 2 can also differ by 2, giving x=3 and hence p=7

Sorrym the contradiction in the first case only occures if $ m>3$ (or $ x>3$).


Find all positive integers $ a,b,n$ such that :
\[ {a^2} \plus{} {b^3} \equal{} {5^n}
\]

Observing this in $ \mod5$ gives us solutions for $ (a,b)$ as $ (5l,5k), (5l\plus{}1,5k\plus{}4), (5l\plus{}2,5k\plus{}1), (5l\plus{}3,5k\plus{}1), (5l\plus{}4,5k\plus{}4)$ and $ n$ just depends on each set of solutions of $ (a,b)$. I'm guessing there are infinitely many solutions?

[quote="modularmarc101"]Observing this in $ \mod5$ gives us solutions for $ (a,b)$ as $ (5l,5k), (5l \plus{} 1,5k \plus{} 4), (5l \plus{} 2,5k \plus{} 1), (5l \plus{} 3,5k \plus{} 1), (5l \plus{} 4,5k \plus{} 4)$ and $ n$ just depends on each set of solutions of $ (a,b)$. I'm guessing there are infinitely many solutions?[/quote]

Maybe you made a confusion between $ a^2 \plus{} b^3 \equal{} 5^n$ and $ a^2 \plus{} b^3 \equal{} 5n$. Else, what could be $ n$ for the set of $ (a,b)$ built with $ l \equal{} k \equal{} 1$ in your proposal ?

BTW, we surely have infinitely many solutions as soon as we have one : $ a^2 \plus{} b^3 \equal{} 5^n$ $ \implies$ $ (5^{3p}a)^2 \plus{} (5^{2p}b)^3 \equal{} 5^{n \plus{} 6p}$ and so for example $ (2\cdot 5^{3p})^2 \plus{} (5^{2p})^3 \equal{} 5^{6p \plus{} 1}$

The question is to get all of them.


Show that each number in the sequence
49, 4489, 444889, 44448889,...
is a perfect square.

[hide="Hint"] If $n^{2}$ is a perfect square, then $n^{2}-1 = (n+1)(n-1)$.

[hide="Then one notices that..."] $4488 = 44 \cdot 102 = 22 \cdot 2 \cdot 34 \cdot 3 = 66 \cdot 68$ [/hide] [/hide]

[hide="solution"]Let's subtract one from each to get

48
4488
444888
44448888

We divide each by 666...666, where there are n 6's(when there are n 6's, that means that there are n 4's and n 8's.

And we find out that each number is 666...666(n 6's)*666.....668(n-1 6's and an 8.), and, like t0rajir0u said, we add one to that to get perfect squares, a.k.a. the original numbers.[/hide]


Μιας και πλησιάζει ο Ευκλείδης βάζω μια εύκολη Συνδυαστική έτσι για εξάσκηση,μιας και από τότε που γράφτηκα στο φόρουμ έπεσε πολύ νέκρα.
Έστω ένας πίνακας  NxN.Διαλέγουμε έναν τυχαίο υποπίνακά του.Ποια η πιθανότητα να είναι τετραγωνικός?

Σημείωση:Για παραδειγμα σε έναν 3x3 πίνακα τα στοιχεία στις θέσεις (1,1)(1,2),(1,3),(2,1)(2,2),(2,3) ορίζουν έναν υποπίνακα του αρχικού,ενώ τα στοιχεία (1,1),(1,2)(2,1),(3,3) δεν ορίζουν(με το παράδειγμα θέλω να πω ότι δεν παίρνεις σκόρπια entries του πίνακα και φτιάχνεις έναν υποπίνακα,προς αποφυγή παρεξηγήσεων).

Τελικα ειναι πραγματι πολυ ευκολο :P

Οι i*j υποπινακες του δωσμενου πινακα ειναι $ (k \plus{} 1 \minus{} j)(k \plus{} 1 \minus{} i)$ στο πληθος (Αυτο επηδη υπαρχει bijection απο το συνολο ολων των πινακων με σταθερες διαστασεις με το συνολο ανω γωνιων των πινακων αυτων, οποτε τους λαμβανουμε ολους μετακινοντας την ανω γωνια σε ολες τις δυνατες θεσεις)

Οποτε ολοι οι υποπινακες ειναι:

$ A \equal{} \sum_{i \equal{} 1}^{N}{\sum_{j \equal{} 1}^{N}{(N \plus{} 1 \minus{} j)(N \plus{} 1 \minus{} i)}} \equal{} \sum_{i \equal{} 1}^{N}{\sum_{j \equal{} 1}^{N}{ij}} \equal{} (1 \plus{} 2 \plus{} ... \plus{} N)^2 \equal{} (\frac {(N \plus{} 1)N}{2})^2$

Ενω οι τετραγωνικοι ειναι:

$ B \equal{} \sum_{i \equal{} 1}^{N}{(N \plus{} 1 \minus{} i)^2} \equal{} \sum_{i \equal{} 1}^{N}{i^2} \equal{} \frac {N(N \plus{} 1)(2N \plus{} 1)}{6}$

Οποτε η πιθανοτητα ειναι:

$ \frac {B}{A} \equal{} \frac {2(2N \plus{} 1)}{3N(N \plus{} 1)}$.

Έτσι πολύ σωστά.Λίγο διαφορετικά για τον παρονομαστή:Κάθε ορθογώνιο καθορίζεται μοναδικά από 2 κάθετες παράλληλες ευθείες και δύο οριζόντιες.Κάθε ζεύγος καθέτων δύναται να εκλεγεί κατά Ν+1 ανά 2 τρόπους οπότε συνολικά (Ν+1 ανά 2)^2.


function
Asymptote
algebra
domain
quadratics
rational function

From II, we know that the domain is $(-\infty , 5) \cup (5 , \infty)$.
From III, we know that -4 and 3 are not in the domain either. That limits the domain to
$(-\infty ,-4) \cup (-4 , 3) \cup (3 , 5) \cup (5 , \infty)$

That's all I can figure out. See if that can help you at all.

[hide]from I we know $y=x+20+\frac{bleh}{bleh}$
from II,III,V
$y=x+20+\frac{k(x-6)(x+5)}{(x+4)(x-3)(x+5)}$
we can find $k$ with IV but im too lazy[/hide]

$2 = 0+20+\frac{k(-6)(5)}{(4)(-3)(5)}$
$-18 = \frac{k}{2}$
$k = \boxed{-36}$
Substituting that into above equation should give you your rational function.

[quote="paladin8"][quote="junggi"][hide]from I we know $y=x+5+\frac{bleh}{bleh}$
from II,III,V
$y=x+5+\frac{k(x-6)(x+5)}{(x+4)(x-3)(x+5)}$
we can find $k$ with IV but im too lazy[/hide][/quote]

Not sure why the oblique asymptote being $x+20$ gives you an $x+5$...[/quote]
woops sorry, my bad
fixed

Ok, guess I'll have to change my solution as well. I wasn't really paying attention  :wink:

[quote="brady00n"]Find the rational function such that:

I. The oblique asymptote is $y=x+20$
II. There is a hole at $x=-5$
III. Vertical asymptotes are located at $x=-4, x=3$
IV. The y-intercept is $(0,2)$
V. The only x-intercept is $(6,0)$.[/quote]

this is obviously not uniquely defined...

for vertical asymptotes, you can have $\frac{1}{(x-3)^{5}}$, for example

Also, after graphing the equation, the X intercept is not at (6,0).
I think we need to re think the approach.

[quote="junggi"]$y=x+20+\frac{k(x-6)(x+5)}{(x+4)(x-3)(x+5)}$[/quote] This gives the point $(6,26)$.  :P
[hide] $\frac{p(x)(x-6)(x+5)}{(x+4)(x-3)(x+5)}$ would work with a quadratic $p(x)$ such that $\frac{p(x)(x-6)}{(x+4)(x-3)}$ gives a remainder of $x+20$ and a constant remainder $k$ (which doesn't really matter).
EDIT: I believe that $p(x)=x^{2}+26x+c$ will work for any $c$. So [b]one[/b] answer is $\frac{x(x+26)(x-6)(x+5)}{(x+4)(x-3)(x+5)}$.
EDIT2: Oh, and $c=4$ for the y-intercept to be $(0,2)$. :D
$\boxed{f(x)=\frac{(x^{2}+26x+4)(x-6)(x+5)}{(x+4)(x-3)(x+5)}}$[/hide]

:wallbash:
I meant something like
from I we get something liek $y=x+20+baskdgjasdgasklgd$
then from other stuff we get
$y=\frac{k(asgdadsgadsg)}{dsglasdg}$


First, multiply both sides with $(a+b+c)$.

Now,

\[\sum_{cyc}{\frac{1}{b(a+b)}(a+b+c) \geq \left(\sum_{cyc}{\frac{1}{\sqrt{a+b}}}\right)^{2}}\]

Now use power mean in the form $\frac{x_{1}+x_{2}+...+x_{n}}{n}\geq \sqrt{\frac{n}{\frac{1}{x_{1}^{2}}+\frac{1}{x_{2}^{2}}+...+\frac{1}{x_{n}^{2}}}}$

To prove that

\[\left(\frac{\sum_{cyc}{\frac{1}{\sqrt{a+b}}}}{3}\right)^{2}\geq \frac{3}{2(a+b+c)}\]

this is wrong

[quote]$\frac{1}{b(a+b)}+\frac{1}{c(b+c)}+\frac{1}{a(c+a)}\ge\frac{9}{2\sum{ab}}$[/quote]

and even if you used cauchy, you would get $\sum{a^{2}}\le \sum{ab}$ anyways

[quote="pohoatza"][quote="parinaz.sa"]prove this:$a,b,c>0: \frac{1}{b(a+b)}+\frac{1}{c(b+c)}+\frac{1}{a(c+a)}\ge\frac{27}{2(a+b+c)^{2}}.$[/quote]

just cauchy

$\frac{1}{b(a+b)}+\frac{1}{c(b+c)}+\frac{1}{a(c+a)}\ge\frac{9}{2\sum{ab}}$

so it remais us:

$\frac{9}{2\sum{ab}}\geq \frac{27}{2\sum{a^{2}}+4\sum{ab}}$

$\Longleftrightarrow \sum{a^{2}}\geq \sum{ab}$[/quote]
This was my solution, the problem appears in Hojoo Lee inequalities problems
it is quite easy :wink:

Now try this one:
\[\text{When }\sum_{cyc}a^{2}=1\text{, prove that }\sum_{cyc}\frac{a}{b^{2}+1}\geq \frac{3}{4}\left(\sum_{cyc}a\sqrt{a}\right)^{2}\]
I warn you: it is harder than it looks at first sight :o
(this is the very next problem in the packet, it is from Greece I believe)

edit:
[hide="@splashD"]
$\frac{1}{b(a+b)}+\frac{1}{c(b+c)}+\frac{1}{a(c+a)}\ge\frac{9}{2\sum{ab}}$
is in fact correct by Cauchy, do you understand?

\[\left(\frac{1}{b(a+b)}+\frac{1}{c(b+c)}+\frac{1}{a(c+a)}\right)\left(\sum_{cyc}b(a+b)\right)\geq 9\text{ \ \ This is clearly from Cauchy}\]
now, $\sum_{cyc}b(a+b)=2\sum_{cyc}ab$. Do you get it now?

Your second comment is just absurd.
[/hide]

[quote="me@home"]
$\text{When }\sum_{cyc}a^{2}=1\text{, prove that }\sum_{cyc}\frac{a}{b^{2}+1}\geq \frac{3}{4}\left(\sum_{cyc}a\sqrt{a}\right)^{2}$
[/quote]
We have: $\left(\sum a^{2}\right)^{2}\ge 3\sum a^{2}b^{2}\implies\sum a^{2}b^{2}\le\frac{1}{3}$
Then: $\frac{4}{3}\left(\sum\frac{a}{b^{2}+1}\right)\ge\left[\sum a^{2}\left(b^{2}+1\right)\right]\left(\sum\frac{a}{b^{2}+1}\right)\ge\left(\sum a\sqrt a\right)^{2}$
$\Longrightarrow\left(\sum\frac{a}{b^{2}+1}\right)\ge\frac{3}{4}\left(\sum a\sqrt a\right)^{2}$

Just use Cebisev and Berstrom( angel, Andreescu) and the ineq is solve.

[quote="parinaz.sa"][quote="sen"]i just want to know, how do you prove;

$\frac{a_{1}^{2}}{b_{1}}+\frac{a_{2}^{2}}{b_{2}}+\frac{a_{3}^{2}}{b_{3}}+...+\frac{a_{n}^{2}}{b_{n}}\geq \frac{(a_{1}+a_{2}+a_{3}+...+a_{n})^{2}}{b_{1}+b_{2}+...+b_{n}}$[/quote]
by induction!![/quote]

Uhm...What? :lol:

By Cauchy on $\left(\frac{a_{i}}{\sqrt{b_{i}}}\right)$ and $\left(\sqrt{b_{i}}\right)$, we get
\[\left(\sum_{i=1}^{n}\frac{a_{i}^{2}}{b_{i}}\right)\left(\sum_{i=1}^{n}b_{i}\right) \geq \left(\sum_{i=1}^{n}a_{i}\right)^{2}\ \Rightarrow\ \sum_{i=1}^{n}\frac{a_{i}^{2}}{b_{i}}\geq \frac{\left(\sum_{i=1}^{n}a_{i}\right)^{2}}{\sum_{i=1}^{n}b_{i}}\]

By the way, [quote="me@home"]now, $\sum_{cyc}b(a+b)=2\sum_{cyc}ab$. Do you get it now?

Your second comment is just absurd.
[/quote]
I guess you'd better check that equality before talking about absurdity  :wink:

[hide="@mychrom"]: what is Berstrom inequality? (is it rearrangement?)[/hide]
[hide="@splashD, Godel"]I am really embarassed :blush: I got confused when I read his post, because it is so similar to the correct method; replace his $2\sum ab$'s with $\sum a(b+a)$'s and it will read correctly...
Kurt: I guess I just caught you peeking into hide tags not written for you :D but if you hadn't, my mistake wouldn't have been caught...

If you are not splashd or godel, I just caught you!!![/hide]

The Bergstrom ineq is the same with Angel or Andreescu ineq. It's a second form of Cauchy. U can find it whit this four names. It is what Sen wrote aftre me. :)

me@home, you should still check your claim, I think it's wrong!

I'll have to repeat SplashD's comment:

If you use Cauchy, you will get this:

\[LHS \geq \frac{9}{a^{2}+b^{2}+c^{2}+ab+bc+ac}\]

Now, if you say that

\[\frac{9}{a^{2}+b^{2}+c^{2}+ab+bc+ac}\geq \frac{27}{2(a+b+c)^{2}}\]

then you are wrong, because this is equivalent with $ab+ac+bc \geq a^{2}+b^{2}+c^{2}$. I think you flipped the sign there  :wink:

And, me@home, if you want to send a message to sbd without having other people reading it: I present you the wonderful invention of the Personal Message!  :lol:


να βρεθουν οι φυσικοι αριθμοι a,b για τους οποιους ισχυει
$ 8[a,b]\equal{}105(a,b)\plus{}30$

Hint

[hide]Let $ (a,b) \equal{} d$. There exist $ x,y\in\mathbf{N}$, such that:

$ a \equal{} dx,\quad b \equal{} dy\quad\wedge\quad (x,y) \equal{} 1$

Notice that:

$ (a,b)\cdot [a,b] \equal{} ab$

The rest is trivial.

Stelios[/hide]

kali skepsi stelio!
another hint gia mia diaforetiki direction
[hide]estw (a,b)=x, [a,b]=y tote to problima anagetai se lysi diofantikis me sinthiki x|y :wink:  again trivial?[/hide]
ILIAS


AMC AIME probability ARML trigonometry geometry USA(J)MO

I've seen a lot of discussion about the two AIMEs, the potential difference in difficulty, etc., so here are my two cents.

In my experience (and I have a reasonable amount of experience), the two AIMEs each year have almost always been of approximately equal difficulty (dating back to 2000 when they started producing two per year).

It seemed to me that if you had a significant discrepancy in your score, it was more due to your weakness to a particular type of problem that may have appeared on one contest and not the other.

Having said that, if there was ever a year where the two tests could actually have a significant discrepancy in difficulty, it would be this year. As MCrawford said, the first AIME was too easy due to a variety of gimmicky tricks (assumptions that are obvious to make but not easy to prove, such as for the semi circle in problem 11), and he also said that he wasn't sure if the committee actually intended for the contest to be as easy as it turned out to be.

As a result, it seems possible that the second AIME could be significantly more difficult than the first one. I really don't see how they could make #15 easier (the only easier one I can remember off the top of my head was the problem from 1995 when they asked for the probability of flipping 5 consecutive heads before 2 consecutive tails ... an ARML level problem). The same goes for problems 11 and 12, both of which required one fairly obvious assumption or some very common knowledge. With 13 having been pretty much completely brute-forceable, and nothing truly spectacular in problems 1 - 10 (unlike the star problem from the 2004 AIME1 which caused A LOT of people to make a silly mistake somewhere), I really don't see how the second AIME could possibly be easier, or even as easy, as the first.

It will be interesting to see what happens.

It will be more intersting, however, if AIME II is harder but people have better score, or  AIME II is easier(doubt) but people bomb it.

Actually as far as topics go, I believe the committee try to balance the 2 tests. But yes, I agree the AIME Alt may very easily be much more difficult (in this non-proof format) than the AIME I is.

I agree about what you said about the balancing of the topics.

However I am very curious to see what they'll pop as a #12 or #13 this time.

Maybe they'll have some nice trig ones this time (or any trig at all!  :P ).

[quote]Maybe they'll have some nice trig ones this time (or any trig at all! Razz ).[/quote]

Good point ... I was thinking something was weird with the problems. Even #15 didn't really have a trig solution.

Wait, are you allowed to take both AIMEs, or just 1?

Also, shouldn't they have already been finished writing the AIME 2. That wouldn't give them time to change its difficulty in response to AIME 1 scores.

i am really happy that i di take AIME I bebecause it had no areas of the problems where i lack at.

AIME II shoudl have some of those... complex #, trig, more counting(and harder...) :)

You can only take one of the two AIMEs. Also, the AIME II is set in stone (I think) despite the feedback from the AIME I scores.

Just 1 question: are the cutoffs for USAMO separate? Or are they combined that averaged to make one USAMO cutoff?

Anyways I wouldn't make it since I am such a cripple at geometry. I didn't even have the cofidence to do #1 on the AIME. (Even though I could've gotten it!!!!!)

AIME floor has in the past been the same for both AIMEs.

The AIME II has definitely been written and printed. In fact, with the test next week it should've been mailed out already.

Seeing as how we all agree on the lack of trig problems on the AIME I, I doubt there'll be a significant number of them on the AIME Alt. #14 does have a rather nice complex num. solution though, which opens the door for the possibility of one or two trig problems on the AIME Alt.

[quote="DPopov"]I think they mail the tests out the day before the competition. I'm pretty sure that's how it was with the AMC at least.[/quote]

Nope. I understand there was some trouble at A&M Consol this year, but most people received it a week if not more in advance. In particular, I know my proctor said everything was ready 3 days before the AIME when my parents inquired about it.

[quote="MithsApprentice"]AIME floor has in the past been the same for both AIMEs.
[/quote]

So are the floors  made seperately or combined from two tests?

To our knowledge, combined. Yet combined in a weird way because of that little clause about A and B proportionality. It's all quite confusing. The moral of the story is to expect the same floor, even though no one here except Mr. Dunbar can tell you for sure that'll be the case.


let a,b,c,d>0 such that $ a^2\plus{}b^2\plus{}c^2\plus{}d^2$=1.Prove that
S= $ \sqrt{1\minus{}a}  \plus{} \sqrt{1\minus{}b}  \plus{} \sqrt{1\minus{}c}  \plus{} \sqrt{1\minus{}d}  \ge  \sqrt{a}  \plus{} \sqrt{b}  \plus{} \sqrt{c}  \plus{} \sqrt{d}$

[quote="math man"]let a,b,c,d>0 such that $ a^2 \plus{} b^2 \plus{} c^2 \plus{} d^2\equal{}1.$Prove that
S= $ \sqrt {1 \minus{} a} \plus{} \sqrt {1 \minus{} b} \plus{} \sqrt {1 \minus{} c} \plus{} \sqrt {1 \minus{} d} \ge \sqrt {a} \plus{} \sqrt {b} \plus{} \sqrt {c} \plus{} \sqrt {d}$[/quote]
See here:
http://www.artofproblemsolving.com/Forum/viewtopic.php?t=179699


Hai folks, I have a query pertaining to Imagination.

I have this ability of multiplying two four digit numbers in my head using only my Imagination,I also have vivid, explicit imagination, is this a great ability? Can anyone multiply two four digit numbers in their head?

I would like to seek your suggestions, replies to this particular query, please don't misinterpret me  as a conceited, vainglorious person in asking such a query.

I Thank you all In advance....................

Of course it's great to have a vivid imagination and to be able to multiply four-digit numbers.

If you're asking if you're a [i]savant[/i] of some kind, this is not the place to get an accurate answer. If you seriously think that you are a savant, or that your natural way of thinking is different from that of most people, then pursue it with a professional.

In the meanwhile, move on to seven-digit numbers  :P

How long does it take you to multiply the 4 digit numbers? That's sort of important. IF you can do it really fast, I agree with TZF that you should pursue with a professional whether your a savant.

TZF, I am not a savant, I just wanted to know if everyone could multiply two 4 digit numbers in their head using only their imagination or was it a great gift, well, I took some 6 minutes to arrive at the answer......... :lol:

I can multiply two four-digit numbers in my head! Ready: 1000*1000=1,000,000. lol sorry just felt like posting that.

There are probably many more people on this board than me who are also able to do this. [url=http://www.math.hmc.edu/~benjamin/]This guy[/url] can do 5 digit multiplication in his head in a few minutes, using math-based shortcuts (the same way I do it).

We had him (Art Benjamin) as our guest speaker one year for our local contest (CSULB Math Day at the Beach). His act was well appreciated.

Dang, you guys are trying to be all tricky... I just use raw memorization of intermediate steps most of the time.

FOLKS, I don't use any shortcuts, I apply only my imagination, I use the same algorithm  that is applied for multiplication (in general).

Are you sure "imagination" is the right word in this situation? If you mean setting up the multiplication in your head... that's not bad, for 4 digits. That's beyond what most people can do.

If it's your imagination, then you're exceptional.

All depends how you look at it  :wink: .


[quote="MCrawford"]If you start with 3 and count by 7s, you get the terms of the sequence 3, 10, 17, ..., 528 where 3 is the 1st term, 10 is the 2nd term, 17 is the 3rd term, and so forth up to 528 which is the Nth term.  What is the value of N?[/quote]

[hide] 76[/hide]

(528 - 3) / 7 finds the number of terms of the arithmetic progression if you were to start at 0. Then since, 3 was the first term, add one to get 76.

It is the 76th term

This is interesting, I will do this in a few minutes

$3,10,17...528$

Since 528 is a multiple of 7 (after you subtract three) then just divide by 7.

$\frac{528-3}{7}=\frac{525}{7}=75$

Generally, the formula for the $n$th term of an arithmetic sequence is $a_n=a_1+(n-1)d$, where $a_n$ is the $n$th term, $d$ is the difference between consecutive terms, and $n$ is the number of the term you want to find. In this case, $528=3+7(n-1)$, or $n=76$.

[quote="236factorial"][hide]3 + 3(N-1) = 528
3N = 528,
N = [color=darkblue]176[/color].[/hide][/quote]
[hide]i think you made a careless error, it should be 76 not 176.[/hide]

[quote="GoBraves"][quote="math92"][hide="easy"]76th term.  528-3/7=525/7=76[/hide][/quote]

How did you end up with 76 as your answer when you did 525/7....which equals 75.[/quote]

Consider this arithmatic progression:

$0, 3, 6, 9, 12, 15$

When you divide 15 by 3, you get 5.  Because you need to count 0, you add 1


probability
inequalities
triangle inequality

A point is randomly selected inside a square. What is the probability that at least three of the lengths of the four perpendiculars dropped from the point to the sides of the squares determine a non-degenerate triangle? [wording changed by me]

What does non-degenerate mean for triangles?  Is it meaning it works for the Triangle Inequality Theorem?  Because if it is, then the answer is just 1...

How?
I got 1.... :roll:  :?
EDIT i meant 1

answer is 1. call the perpendiculars to the sides p, q, 1-q, 1-p, wlog in increasing order.

q+(1-q)>1-p, so we are done.

OMGNORE yet again I changed my answer from right to wrong! :wallbash: I think I did that on about 5 problems. no exaggeration


what kind of plates can you microwave and safely eat from? is there some list on the internet of what u can and cannot?

are paper plates fine?
styrofoam?
ceramic?

It's not a good idea to microwave paper or styrofoam, since they can catch fire and melt.  Ceramic plates are usually the safest.

what's the worst that happens if the styrofoam or paper melts slightly, possibly unnoticeable-ee,  and you eat off it?

y are ceramic plates the safest?

if the melting is unnoticable, it probably won't cause too many problems
the real danger is if it catches fire, in which case you might be injured

ceramic plates are safest because there is no risk of it catching fire or making you sick
if you eat too much melted styrofoam, you will probably have an unhappy stomach for a while

With plastic type plates, continuous microwave irradiation may cause carcinogenic products to leak from the plate (specially compounds that are used in polymerizations). Personaly I use glass plates.


!!این جا برای مساله های ترکیبیاته حالا دیگه لطف کنید سوالهاتون رو با همون سیستم بفرستید دیگه

salam be hameye azizan...
man mikham az ye donbale ke(esmesho taghyir dadam)sohbat konam:
donbaleye "fiboghaati" ro injoori taarif mikonim:
$g_n=g_{n-1}+g_{n-2}+g_{n-3}, n\geq3,g_0=g_1=0,g_2=1$
formoole sarihe in donbale ro benevisid.(ba hallesh).kheili manoon va motshakker.bedrood

man ino ghablan post kardeh boodam, bekhatereh hamin Ingilisiye:

just use polynomials.
let $F(x)=a_0+a_1x+a_2x^2+....$
here $a_j$ are the fibonacci numbers. note that we don't care what $x$ is, we just want to use $F(x)$ to find the $a_j$
$xF(x)=a_0x+a_1x^2+a_2x^3+a_3x^4....$
$x^2F(x)=a_0x^2+a_1x^3+a_2x^4+a_3x^5....$
now if you group these you get:
$F(x)-xF(x)-x^2F(x)=a_0+(a_1-a_0)x+ (a_2-a_1-a_0)x^2+(a_3-a_2-a_1)x^3+(a_4-a_3-a_2)x^4+.....$
the importance of the fibonacci sequence is that $(a_2-a_1-a_0), (a_3-a_2-a_1), (a_3-a_2-a_1),....=0$ so the above's equation is so simple.
$F(x)-xF(x)-x^2F(x)=a_0+(a_1-a_0)x$
we also know that $a_0=a_1=1$ so $(1-x^2-x)F(x)=1$ or
$F(x)=\frac{1}{1-x-x^2}$. now if you factor out the denominator you get:

$F(x)=\frac{1}{[1- \frac{-2x}{1-\sqrt{5}}][1-\frac{-2x}{1+\sqrt{5}}]}$
now with use of some algebra you get:
$F(x)=\frac{5+\sqrt5}{10}*[\frac{1}{[1+\frac{2x}{1-\sqrt{5}}]}]+\frac{5-\sqrt5}{10}*[\frac{1}{[1+\frac{2x}{1+\sqrt{5}}]}]$
the above parts that are in a bracket can be rewritten by the summation formula:
$F(x)=\frac{5+\sqrt{5}}{10}.\sum_{n=1}^{\infty} [\frac{-2x}{1-\sqrt{5}}]^{j}+\frac{5-\sqrt{5}}{10}.\sum_{n=1}^{\infty}[\frac{-2x}{1+\sqrt{5}}]^{j}$
now if you group the similar powers of $x$  you get:
$F(x)=\sum_{n=1}^{\infty}[\frac{5+\sqrt{5}}{10}.[\frac{-2x}{1-\sqrt{5}}]^{j}+\frac{5-\sqrt{5}}{10}.[\frac{-2x}{1+\sqrt{5}}]^{j}]$
now remember we started the problem by saying
$F(x)=a_0+a_1x+a_2x^2+....$
so now we have two differen formulas for $F(x)$ these two MUST equivelant to each other, especially the coefficents of the same powers of $x$. so we conclude that:
$a_j= \frac{5+\sqrt{5}}{10}.[\frac{-2x}{1-\sqrt{5}}]^{j}+\frac{5-\sqrt{5}}{10}.[\frac{-2x}{1+\sqrt{5}}]^{j}$
now we are almost there, all we need to do is to replace some of the stuff up there by equivalent of them.
$\frac{5+\sqrt{5}}{10}=\frac{1}{\sqrt{5}}.\frac{1+\sqrt{5}}{2}$ also $\frac{5-\sqrt{5}}{10}=\frac{-1}{\sqrt{5}}.\frac{1-\sqrt{5}}{2}$ and $\frac{-2}{1-\sqrt{5}}=\frac{1+\sqrt{5}}{2}$, and $\frac{-2}{1+\sqrt{5}}=\frac{1-\sqrt{5}}{2}$
if we replace these values for the $a_j$ formula we get:
$a_j=\frac{[\frac{1+\sqrt{5}}{2}]^j- [\frac{1-\sqrt{5}}{2}]^j}{\sqrt{5}}$
so, we are done, (don't ask me how long it took me type this in latex!! :wacko:)


Do you mean $-1 < a_i < 1$?

Or do you mean $a_i \in \{ -1, 1 \}$?

The first it's untrue and the second it's obvious.

well... $a_k \cdot a_{k+1}\in$  {$1,-1$}
so there must be equal number of $1's$
and equal number of $-1's$ .
so if we group the 1's and the -1's we have
$\sum_{i\in A} a_i a_{i+1}=k$
$\sum_{i\in B} a_i a_{i+1}=-k$
it is clear that $2|n$ because there are 2 such sets.
and because for everyone of these sets is formed by elements wich
have 2 elements namely $a_i,a_{i+1}\Rightarrow 4|n$


Hello,

I have written another mock AMC. Please post if you would like to take it, and what would be a good time for you. I am hoping to do it either Tuesday or Saturday of next week, but it depends on who can do it on those days. I am also planning to let people take it early, but that will probably start on Saturday.

Thanks,
JTG

[i]Moderator's note: removed sticky.  -mathfanatic[/i]

Cool!  yet another amc!  I should be able to do it as long as it is on a weekday and is not in the middle of the night.

I'll take it.  School starts up next week and I don't get home until about 6.  Then I have an algebra class at 7 on tuesday so saturday would be best for me.  Or I could take it early if you can't do it saturday.

good job J

i'll bet you 2 dollars my results will be better than yours  ;)

i'll take it
whoa nosepuppy do you like losing money?  ;)
since he kinda made the test....

[hide]shhhh... i plan on having him dq'd for having spent more than 75 minutes on the problems[/hide]

I think Nosepuppy was talking to joml88.

I'll try too.  But I'm kind of getting scared that we'll run out of questions, since there's like 1 mock AMC a week.

I'll take it! If somebody needs help preparing another AMC, I'm your man.

I'll take it but I'd prefer to take it early

I'll take it.  I don't know when I can do it but I'll let you know on Tuesday.

I don't know - that would have to be 26 tests, and that's 650 problems. It might take a while if we only do 2 every month. We would still run into next year summer before finishing.

[quote="Magnara"]How much longer until we run out of letters and just have to do AMC X_27, AMC X_28,...,AMC X_n?[/quote]

No, it would go AMC X, AMC Y, AMC Z, AMC AA, AMC BB ... AMC AAAAAAA, AMC BBBBBBB

tuesday would be fine as long as it starts at 4:00 or after

[quote="JGeneson"]I am going to post the problems for Mock AMC D in another thread. It is looking like the test will be Tuesday at 5:00 instead of Tuesday at 4:00. If this is less convenient for you, PM me and you can take it early.[/quote]

5:00 PM EST Tuesday would be fine with me

I'd like to participate. What number what you give this test?

For those of you curious about the difficulty...

I took it early...

I must say, the early questions are much harder than the early questions in the other Mock AMCs...

But there are some late questions that are considerably easier than the late questions in the other ones...

Overall, Beta's was definitely easier... I'll have to see my score before I can compare to the other two... I answered more questions than I usually do but probably made more mistakes.

EDIT: My Score = 87.5, for reference my other 3 scores were 90, 90, 93.5 so it's about an average result for me.

[quote="JGeneson"]Well, since the test is over, I would like to find out who voted my test was easy :). Or was it by accident?[/quote]

I voted your test was easy.  (By no mean, this is intended to be an insult)
I got about 68 on AMC A, about 86 AMC B, and on your test (AMC D) I got 90.

I think it was easiER than the first two.

[quote="beta"]My score went like 98, 110, 93.5 (A, B, D). So it's definitely harder for me too.[/quote]

I seriously thought this was definitely easy because I made especially more mistakes on AMC D then any other tests and still got 90 :)

[quote="Pork_Chop8"]That doesn't make sense. I usually get 140+ on AMC's and after looking over the test there is no way I would've got to 140. Maybe your expertise wasn't touched on in the other tests, but I think it was just a stroke of luck that the D was easier for you. The test was much harder than the average AMC and most scores show it to be so. The highest score is under 130, it was not easier than the rest, just easier for you.[/quote]

Sure, I am not going to argue with you :)
Hmm, I was very relaxed this time taking the test.

Who is the other guy voted Easy?

The test touched all my weaknesses(Hyperbola, Euler's Formula). This is the first time I skipped a #2 on an AMC.

Yeah I know, I didn't like the conic stuff.  The vertices, faces, etc. problem wasn't too bad... what I usually do is look at a cube.  I know that the formula involves vertices, faces, and edges.  I also know that there is a 2 in there  ;)

I don't think that you even need to know Euler's Formula for AMC/AIME
Here is an AIME question:
10.  A polyhedron has 32 faces, each of which has 3 or 5 sides. At each of it s V vertices it has T triangles and P pentagons. What is the value of 100P + 10T + V? [b]You may assume Euler's formula (V + F = E + 2, where F is the number of faces and E the number of edges). [/b]

I didn't understand number 7. Which regular polyhedron fulfills it?

The nice thing about that problem is that it does not matter which regular polyhedron fulfills it. You just have to plug into the formula. If you are curious, though, I think you can draw something with two parallel hexagons and a point above and below them and connect the dots. That should give you what you want.


Hi (Revives forum :D)

I once got into the COMC gold medallist, in the 2007 - 2008 year, and received a Medal for that achievement.
The Next year, I got into the same criteria, but didn't receive any medal..
Could anyone tell me what happened?
The first year I got this, I even got a letter from waterloo to my school regarding this.

Just in case confirmation is required..
[hide]My name is David Lee from Westdale Secondary School, Hamilton, as written in the official COMC Results pdf for 2007 and 2008
http://cemc.uwaterloo.ca/contests/past_contests/2007/2007-08COMCResults.pdf
http://cemc.uwaterloo.ca/contests/past_contests/2008/2008-09COMCResults.pdf
[/hide]

Yay for reviving forum!!
That is strange; perhaps it got lost in the mail or there was some glitch on the senders end. Or maybe your school lost it (my school lost an invitation for someone to National Camp several years ago).
Hey, I know you  :maybe:.

Oh yay, COMC time. maybe i can make it this year to CMO. I think i get an advantage of regional basis because i'm from nova scotia and no one from there ever makes it, but maybe im just being delusional.  Anyways, i'll get there (even if its not this year, though im sure it will be) and haha bacteria that sounds like something my prevcious school would do
and good luck to you guys


Alpha and beta particles, and gamma rays, I think.

[quote="Rushil"][size=150][b]What are cosmic rays made up of?!?!?!?!?![/b][/size][/quote]

X-ray and $\gamma$ - ray photons.

they're just energetic particles from outer spaces.


geometry
circumcircle
incenter
geometry unsolved

$ I$ is the incenter of $ \triangle ABC$ and $ A_1,B_1,C_1$ on $ AI,BI,CI$. The perpendicular bisectors of $ AA_1,BB_1,CC_1$ form a triangle $ \triangle A_2B_2C_2$. Prove that the circumcenters of $ \triangle ABC$ and $ \triangle A_2B_2C_2$ are the same iff $ I$ is the orthocenter of $ \triangle A_1B_1C_1$.

Just after I posted it I found the problem not very hard but pretty good,
a nice solution exists.

If it was posted, please tell me where it is  :lol:
I just didn't find this problem in the forum. :maybe:

[quote="shalex"]

If it was posted, please tell me where it is  :lol:
[/quote]

It was  :D
http://www.mathlinks.ro/viewtopic.php?p=19007#19007

Thanks a lot.

Several hours ago we found the second solution which is similar to the one you showed. Now we have two solutions, which are both beautiful. :lol:


[quote="chien than"]Find $x;y$ such that:
$\frac{x+1}{\sqrt{x}}=\frac{4(y-1).\sqrt[3]{y-1}+4}{\sqrt[3]{(y-1)^{2}}}=10$[/quote]
For $\frac{x+1}{\sqrt{x}}=10$ you need slove $(x+1)^{2}=100x$, easy?

For $\frac{4(y-1).\sqrt[3]{y-1}+4}{\sqrt[3]{(y-1)^{2}}}=10$, put $\sqrt[3]{y-1}=z$ then you need solve a polynomial equation of degree $4$ of $z$, easy so ? :D

[quote="chien than"]Find $x$ such that:
$x=\sqrt{x}+2(1-\sqrt{1-\sqrt{x}})^{2}$[/quote]
Put $y=\sqrt{1-\sqrt{x}}$, you will have a polynomial equation of degree $4$ of $y$.  :wink:

[quote="chien than"]Find $x$ such that:
$x+4\sqrt{x+3}+2\sqrt{3-2x}=11$[/quote]
Square both sides. Maybe this is last post of mine in this topic.  :lol:

Sorry first problem is:
Find $x;y$ such that:
$\frac{x+1}{\sqrt{x}}+\frac{4(y-1).\sqrt[3]{y-1}+4}{\sqrt[3]{(y-1)^{2}}}=10$
Help me,please!


If I have a matrix $ A$, and I use $ n$ different row operations of this form: $ a_kR_i + R_j \rightarrow R_i$ to construct a new matrix $ B$, what is the determinant of $ A$ in terms of $ B$?

${ |A| = \frac {|B|}{\sum_{k = 1}^n a_k}}$

Is this the right answer?

Edit: Confirmation Anyone?!

[quote="Carcul"]Adding a row (column) multiplied by a scalar to another row (column) does not change the determinant. Perhaps the simple technique for evaluating determinants that you are looking for is Gaussian Elimination.[/quote]
Are you sure? Then what's wrong with this proof that seems to prove you wrong?

$ |A|$ will be defined from the formula for the determinant of a 3 by 3 matrix.

$ A \equal{} \begin{bmatrix} a & b & c \\
d & e & f \\
g & h & i \end{bmatrix}$-----     $ |A| \equal{} a\left|\begin{array}{cc}e & f \\
h & i\end{array}\right| \minus{} d\left|\begin{array}{cc}b & c \\
h & i\end{array}\right| \plus{} g\left|\begin{array}{cc}b & c \\
e & f\end{array}\right|$

B will be defined by the following to row operations on A:

$ \minus{} \frac {a}{d}R_2 \plus{} R_1 \rightarrow R_2$

$ \minus{} \frac {a}{g}R_3 \plus{} R_1 \rightarrow R_3$

(let's denote the two coefficients on the row operations $ r_n$, so $ r_1\equal{}\minus{} \frac {a}{d}$ and $ r_2\equal{}\minus{} \frac {a}{g}$.)

$ B \equal{} \begin{bmatrix} a & b & c \\
0 & \frac {db \minus{} ae}{d} & \frac {dc \minus{} af}{d} \\
0 & \frac {gb \minus{} ah}{g} & \frac {gc \minus{} ai}{g} \end{bmatrix}$

We then find $ |B|$:

$ |B|\equal{}a[\frac{(db\minus{}ae)(gc\minus{}ai)\minus{}(dc\minus{}af)(gb\minus{}ah)}{dg}]$

Foiling we have:

$ |B|\equal{}a[\frac{dbgc\minus{}aegc\minus{}dbai\plus{}a^2ei\minus{}dbgc\plus{}afgb\plus{}dcah\minus{}a^2fh)}{dg}]$

Factoring an $ a$ out and canceling the term $ dbgc$ we have:

$ |B|\equal{}a^2[\frac{\minus{}egc\minus{}dbi\plus{}aei\plus{}fgb\plus{}dch\minus{}afh)}{dg}]$

Grouping in a particular way we have:

$ |B|\equal{}a^2[\frac{aei\minus{}afh\minus{}dbi\plus{}dch\plus{}fgb\minus{}egc)}{dg}]$

Factoring an $ a$ out of the first two terms, a $ \minus{}d$ out of the second two, and a $ g$ out of the third two we have:

$ |B|\equal{}a^2[\frac{a(ei\minus{}fh)\minus{}d(bi\minus{}ch)\plus{}g(fb\minus{}ec)}{dg}]$

Transforming the arguments inside the innermost three parenthesis into determinant functions we have:

$ |B|\equal{}a^2[\frac{a\left|\begin{array}{cc}e & f \\
h & i\end{array}\right|\minus{}d\left|\begin{array}{cc}b & c \\
h & i\end{array}\right|\plus{}g\left|\begin{array}{cc}b & c \\
e& f\end{array}\right|}{dg}]$

Noticing that the numerator is actually the definition of $ |A|$ mentioned at the beginning of the post, we have:

$ |B|\equal{}a^2[\frac{|A|}{dg}]$

Solving for $ |A|$ we have:

$ |A|\equal{}|B|\frac{dg}{a^2}$

So at this point we see that $ B$, after being manipulated from $ A$ using only row operations specified in the first post of this thread, does not have a determinant equal to that of $ |A|$. They differ by some coefficient $ \frac{dg}{a^2}$, which is actually just the reciprocal of the product of $ r_1$ and $ r_2$, which were both defined earlier.

Thus, in general, one can assume that if you use $ n$ row substitutions of the form $ a_{k}R_{i}\plus{}R_{j}\rightarrow R_{i}$ on a matrix $ A$ to obatain a matrix $ B$, the relation between $ |A|$ and $ |B|$ is:

$ |A|\equal{}|B|\prod^{n}_{k\equal{}1}\frac{1}{a_{k}}$

@Epkid08: I have not checked your proof, but the formula ${ |A| = \frac {|B|}{\sum_{k = 1}^n a_k}}$ in post #1 should be corrected to  ${ |A| = \frac {|B|}{\prod_{k = 1}^n a_k}}$. In fact, each of your operations $ a_kR_i + R_j \rightarrow R_i$ multiplies the determinant by $ a_k$ (because it is the composition of the row operation $ a_kR_i\rightarrow R_i$, which multiplies the determinant by $ a_k$, with the row operation $ R_i+R_j\rightarrow R_i$, which leaves the determinant invariant).

  darij

[quote="epkid08"]Are you sure?[/quote]

Yes, I am sure, but I see that you are referring to another operation, namely, substituting one row by the sum of that same row multiplied by a scalar and another row, while I was referring to substituting one row by the sum of that same row with another row multiplied by a scalar. Those are different operations: in your case, yes, each those operations multiplies the determinant by a scalar, while in my case the determinant is invariant. Anyway, Gaussian elimination is the simple method for evaluating determinants, except, of course, in special cases like triangular matrices.

So you're saying if I'd just use substitutions of the form $ R_i + a_kR_j \rightarrow R_i$, I wouldn't have to worry about using my formula; the determinant wouldn't change?

[quote="darij grinberg"]@Epkid08: I have not checked your proof, but the formula ${ |A| = \frac {|B|}{\sum_{k = 1}^n a_k}}$ in post #1 should be corrected to  ${ |A| = \frac {|B|}{\prod_{k = 1}^n a_k}}$. In fact, each of your operations $ a_kR_i + R_j \rightarrow R_i$ multiplies the determinant by $ a_k$ (because it is the composition of the row operation $ a_kR_i\rightarrow R_i$, which multiplies the determinant by $ a_k$, with the row operation $ R_i + R_j\rightarrow R_i$, which leaves the determinant invariant).

  darij[/quote]

Yeah I can't edit it anymore. Check the bottom of the proof for the real answer.

[quote="Carcul"]Yes, that's right. In fact, that is why Gaussian elimination is such a nice way for the (hand) computation of determinants.[/quote]

I see! That's why I got all the answers wrong on those tests, I was using the wrong type of substitution! :mad:


floor function
number theory
prime factorization

Let $ p$ be a prime,and $ a$ be an integer with $ 1\le a\le p^2\minus{}2$.

Show that  $ \binom{p^2}{a}\plus{}\binom{p^2\minus{}1}{a}\plus{}\binom{p^2\minus{}2}{a}\plus{}\cdots \plus{}\binom{a\plus{}1}{a}\plus{}\binom{a}{a}$ is dividible by $ p$.

Hockey Stick Identity says that
\[ \sum_{k\equal{}a}^{p^2}{i\choose a}\equal{}{{p^2\plus{}1}\choose{a\plus{}1}}
\]

And we have $ {{p^2\plus{}1}\choose{a\plus{}1}}\equal{}\frac{(p^2\plus{}1)!}{(p^2\minus{}a)!(a\plus{}1)!}$

Our fraction is a multiple of $ p^2$, so $ p$ must be a factor of the fraction as well.

[quote="grn_trtle"]Our fraction is a multiple of $ p^2$[/quote]  This is not true: for example, with $ p \equal{} 3$ and $ a \equal{} 2$, we have $ \binom{p^2 \plus{} 1}{a \plus{} 1} \equal{} \binom{10}{3} \equal{} 120$ which is not divisible by 9.  Just because the fraction has $ p^2$ in the numerator ....

[hide]\[ {{p^2\plus{}1}\choose{a\plus{}1}}\equal{}\frac{(p^2\plus{}1)!}{(p^2\minus{}a)!(a\plus{}1)!}
\]

the number of factors of $ p$ in the numerator is $ p\plus{}1$.

I don't know an easy way to count for the denominator but if you take $ a$ in intervals of $ k\cdot p \le a <(k\plus{}1)\cdot p$, it is easy to see that there are $ p$ factors of $ p$ in the denominator.[/hide]

[hide="Showing that binomial coefficient is divisible by p"]Note that $ \nu_p(n)$, the exponent of $ p$ in the prime factorization of $ n$, satisfies
$ \nu_p(n!) \equal{} \left\lfloor \frac{n}{p} \right\rfloor \plus{} \left\lfloor \frac{n}{p^2} \right\rfloor \plus{} \left\lfloor \frac{n}{p^3} \right\rfloor \plus{} \ldots$

Using this formula, for the numerator we get an exponent of
$ \left\lfloor \frac{p^2 \plus{} 1}{p} \right\rfloor \plus{} \left\lfloor \frac{p^2 \plus{} 1}{p^2} \right\rfloor \equal{} p\plus{}1$.
For the denominator, note that $ a\plus{}1$ and $ p^2 \minus{} a$ are both less than $ p^2$ (requires using both bounds on $ a$), so we get
$ \left\lfloor \frac{p^2 \minus{} a}{p} \right\rfloor \plus{} \left\lfloor \frac{a\plus{}1}{p} \right\rfloor \leq \left\lfloor \frac{p^2 \plus{} 1}{p} \right\rfloor \equal{} p$.[/hide]


a,b,c,d adaade haghighi hastand &
(a^2+1)(b^2+1)(c^2+1)(d^2+1)=16
saabet konid:
 -3 =< ab+ac+ad+bc+bd+cd-abcd =< 5

[hide="A hint"]Setting $ a=\tan x,b=\tan y,c=\tan z, d=\tan t$.[/hide]

we have to show that:
$ -3\leq ab+ac+ad+bc+bd+cd-abcd\leq 5$
$ -4\leq ab+ac+ad+bc+bd+cd-abcd-1\leq 4$
$ (ab+ac+ad+bc+bd+cd-abcd-1)^{2}\leq 16$
$ (ab+ac+ad+bc+bd+cd-abcd-1)^{2}\leq (a^{2}+1)(b^{2}+1)(c^{2}+1)(d^{2}+1)$
which is easy to see...

[quote="navid"]This problem purposed by Gabriel Dospinescu (Harazi) It has many solutions try Mixing variables.[/quote]
this forum is for [i]iranian MLs[/i] so its ok if you post in persian,which I think you should...  :)

[color=white][size=59]P.S:this is my 500 th post in ML  :lol: [/size][/color]


I'm not sure but my AoPS book said "Sam V_________(can't spell his last name) and authors worked to build the Mandelbrot, four round high school competition"  Does that mean they actually created books AND competition?

Correct me if i'm wrong

Sam, Sandor, and I created the Mandelbrot in 1990.  Sandor and I wrote the books in 1993-4.  Sandor and I stepped out of the 'Brot in 1997, I think it was.  Fortunately, Sam continued it on his own.

So, it's true that you were in part of making competition?

Amazing...

His homepage - http://www.math.yale.edu/mandelbrot/home.htm
A biography - http://www-groups.dcs.st-and.ac.uk/~history/Mathematicians/Mandelbrot.html

[quote="rrusczyk"]Yes, that's true.  The three of us built the contest right after my first year of college.  I called Mandelbrot himself from a payphone outside my dorm at Princeton for permission to use his name.[/quote]

Just as a curiosity, is there any particular reason you wanted to call it the Mandelbrot, as opposed to, say, the Jacobi competition?

(OK, I'll admit, Mandelbrot has more of a ring than Jacobi, but still...)


AMC
AIME
function
algorithm
probability
geometry
analytic geometry

So, what do you think are good numbers to guess on AIME? (for the impossible problems)

I'm taking AIME II so I'll probably know 15 less numbers that are probably not on the second one.  :D

006, 026, 206, 008, whatever. Those are the ones that I'll guess :P

Wow this is pretty fun! Let's see... for now I'll say (I'm taking the AIME II)

042 for number 14 and 298 for number 15. I'll guess the rest if I can't get them  ;)

And just for kicks:

000, 123, 456, 789.

You can guess 1000. Or you could actually solve the problem :P.

[quote="JesusFreak197"][quote="calc rulz"]42, 87, 91, 63, 679, 541, 987, 245, 677, 28, 4, 12, 687, 107, 98

My guesses for all 15 problems  :sleep2:[/quote]

Let's see... $\frac{1}{10^{45}}$ chance of getting a perfect score! :D[/quote]Errr...don't you mean $\frac{1}{1000^{15}}$?

Edit: stupid, stupid, me.  They're the same value.  :blush:

Won't it be really funny if we guess correctly?! :rotfl:

[quote="Iversonfan2005"]well, on the 2005 AIME, here were my guesses:

942, 12, 109, 294, 630, 45, 150, 113, 74, 47, 544, 25, 83, 936, 38...and by some miracle by God... :rotfl:  :rotfl:[/quote]

:rotfl:   :rotfl:  :rotfl:  :rotfl:

Seriously if anyone guessed those answers, they'd be best off meditating for the first 2 1/2 hours and then just "recieving" the answers.... now that would be cool...

I'm guessing 025 for everything no matter what. Last year on the day before AIME I planned to guess 025 for all, except I IMed someone the day before and he told me to guess 406 instead...and that's what I did!

Had I guessed 025 for all problems I didn't solve last year I would've guessed #12 right :( I am not making the same mistake :D

[quote="chess64"]I'm guessing $487$ because I just made a program to generate 999 random numbers and compute their average, it was $487.251$ :lol:[/quote]

How do you create a random number? Isn't there theoretically no way to generate a random number with a computer?

He definately has a lump of radioactive material that he is using to generate random numbers.

Uhh...it's called Javascript, folks :D

[code]function sortNumber(a, b)
{
return a - b
}

var rnd = new Array()

for (var t=0; t<1000; t++)
{
rnd[t] = Math.floor(Math.random()*1000);
}

document.write(rnd.sort(sortNumber));

var sum = 0

for (var t=0; t<1000; t++)
{
var sum = sum + rnd[t]
}

var average = sum/1000

document.write("<br> " + average);[/code]

It's random enough for me ;)

Oh speaking of guessing, I know someone from last summer who is really really good at guessing (he like guesses something right on almost every competition, like he spent an hour and guessed #15 right on the AIME last year and got a 13)

I have no idea how he does it but...

[quote="qweretyq"][quote="beta"]BTW I'm going to guess 025 for the next 3 AIME's I have left before graduating, look 025 is an extremely popular answer it showed up in

1998 AIME
1999 AIME
2000 AIME I
2002 AIME I
2005 AIME I

And it should show up some time in the next 3 years  :lol:[/quote]

You shouldn't have said this. Now AoPS is going to send this to the AMC people and they'll make sure not to put it.[/quote]

[b]OR[/b] beta is in cohorts with AMC, you just don't know it. He's trying to drop us some hints, jesus

Beta is definitely in cohorts.

No, no random number generator is random. The computer has an algorithm that it sends to the ALU along with a "seed" number that influences the outcomes. It is possible to <i>approach</i> randomness by using a couple of random numbers AS the seed number...

[quote="Iversonfan2005"][quote="probability1.01"]Well, I think I'm a better guesser than the person beta was talking about. Last year I didn't solve any problems, but I worked really hard to guess a bunch of problems and I think I ended up getting like 11 right or something like that.[/quote]

How does that work out?  :o , you succesfully guessed 11 problems right!! :o[/quote]

Haha that's because probability faked guessing. The person I was talking about guessed with real skills. I know it sounds like I'm just kidding but I'm not kidding (or maybe the person is lying to me).

It's actually very easy to guess the right answer.  The question itself is in code, and if you spend time figuring that out, the answer becomes apparent.

See, what most people don't know is that this is all a conspiracy.  It's not about the math...it's actually a government program developed to create future human codebreakers that will surpass the limitations of current technology.   :P

Because counting problems usually don't have prime numbered answers?

Its not easy to guess a question correctly.

For geometry, I'm drawing them and measuring with ruler or protractor.  :P

[quote="Adunakhor"][quote]For geometry, I'm drawing them and measuring with ruler or protractor.[/quote]

Lol, didn't Mr. Rusczyk say that he did that on one of the AIME problems once, and got it right?[/quote]Yep. THat's why I'm doing that instead of randomly guessing. :D

On one practice AIME, I drew the problem exactly, estimated the measurements a whole bunch and got the answer off by 1 :P

Hmm... maybe I should buy a new ruler then, the ones my school provides are all like worn out a bit, a couple precious millimeters off.

Are you allowed to use graph paper? That stuff can be a lifesaver sometimes, especially on coordinate problems.

So the most common answers on the AIME are 025, 012, and 083.  And the average of all the answers since 1997 is 333.

Why not try the answers that didn't appear yet?

I guessed 169 on number 15 for the AIME I last year. Figures that it would be the answer to number 15 on the AIME II.


Concurrency here can be proven using mass points, though is there a cleaner way? (mass points isn't so bad)
After that, it follows from the reatios established with, say, ME and MF, that the thing is parallel to EF, and thus to BC.

[color=red][b]Very nice problem, Virgil Nicula! But now, I only have an ugly proof for it :([/b] [/color]
[quote="Virgil Nicula"][color=darkred]Let $ G$ be the [b]centroid[/b] of the triangle $ ABC$ and let $ M$ be the [b]middle[/b] of the side $ [BC]$. Define the points $ \{\begin{array}{c}E\in (AC)\ ,\ GE\parallel AB\\\ F\in (AB)\ ,\ GF\parallel AC\end{array}$.
Prove that : the lines $ BG$, $ ME$, $ CF$ are concurrently (in $ K$) ; also and the lines $ CG$, $ MF$, $ BE$ are concurrently (in $ L$) ; $ KL\parallel BC$.[/color][/quote]
[color=darkblue][b]Proof.[/b] Easy to show that $ \frac{AF}{AB}=\frac{AE}{AC}=\frac{1}{3}$

In triangle $ \triangle ABC,$ we have $ \frac{MB}{MC}\cdot\frac{EC}{EA}\cdot\frac{FA}{FB}=1,$ so $ AM,\,AE,\,CF$ are concurrent at a point $ S$

By Menelaus' Theorem for triangle $ \triangle ABE,\,F,\,S,$ and $ C$ are collinear, we have $ \frac{SB}{SE}\cdot\frac{CE}{CA}\cdot\frac{PA}{PB}=1\implies\frac{SB}{SE}=3$

Denote $ K=EM\cap CF,\,N=BK\cap AC$
By Ceva's Theorem for triangle $ \triangle BCE$ and $ BN,\,CS,\,EM$ are concurrent, we have $ \frac{MB}{MC}\cdot\frac{NC}{NE}\cdot\frac{SE}{SB}=1\implies\frac{NC}{NE}=3\implies NA=NC$
i.e. $ BG,\,ME,\,CF$ are concurrent at $ K$

By Menelaus's Theorem for triangle $ \triangle BSC$ and $ E,\,K,\,M$ are collinear, we have $ \frac{MB}{MC}\cdot\frac{KC}{KS}\cdot\frac{ES}{EB}=1\implies\frac{KC}{KS}=4$

Similarly, we also have $ CG,\, MF,\, BE$ are concurrent at $ L$ and $ \frac{LB}{LS}=4$
Hence $ KL\parallel BC$[/color]

[quote="Virgil Nicula"][color=darkred]Let $ G$ be the [b]centroid[/b] of the triangle $ ABC$ and let $ M$ be the [b]middle[/b] of the side $ [BC]$. Define the points $ \{\begin{array}{c}E\in (AC)\ ,\ GE\parallel AB\\\ F\in (AB)\ ,\ GF\parallel AC\end{array}$.
Prove that : the lines $ BG$, $ ME$, $ CF$ are concurrently (in $ K$) ; also and the lines $ CG$, $ MF$, $ BE$ are concurrently (in $ L$) ; $ KL\parallel BC$.[/color][/quote]
[color=darkblue][b]Proof.[/b] Denote $ \{\begin{array}{c}N\in (AC)\ ,\ NA=NC\\\\ P\in (AB)\ ,\ PA=PB\\\\ K\in ME\cap CF\\\\ L\in MF\cap BE\end{array}$. Observe that $ \{\begin{array}{c}EF\parallel BC\ ,\ BC=3\cdot EF\\\\ \frac{KM}{KE}=\frac{MC}{EF}=\frac{3}{2}\\\\ MN\parallel AB\parallel GE\end{array}$. Apply the [b]Menelaus' theorem[/b] to $ \triangle AME$

and to the points $ \{\begin{array}{c}G\in AM\\\ K\in ME\\\ N\in EA\end{array}$. Therefore,  $ \frac{GA}{GM}\cdot\frac{KM}{KE}\cdot\frac{NE}{NA}=2\cdot\frac{3}{2}\cdot\frac{1}{3}=1$ $ \Longleftrightarrow$ $ K\in GN$ $ \Longleftrightarrow$ $ K\in BG$ $ \Longleftrightarrow$ $ BG\cap ME\cap CF\ne\emptyset$.

Similarly show that $ CG\cap MF\cap BE\ne\emptyset$. From the relations $ \frac{KM }{KE}=\frac{MC}{EF}=\frac{MB}{EF}=\frac{LM}{LF}$ obtain $ KL\parallel EF$, i.e. $ KL\parallel BC$.[/color]

[color=purple][b][u]Remark.[/u][/b] Try an extension of this problem to any point $ M\in BC$ and to any point $ G\in AM$ ! What become the points $ N$, $ P$ in the general case ?

I posted later an extension at http://www.mathlinks.ro/Forum/viewtopic.php?t=159276 [/color]

[hide="Solution"]
Let $ BG$ intersect $ AC$ at $ P$, and $ CG$ intersect $ AB$ at $ Q$.
Because $ GE \parallel AC$, we have $ \triangle BGF \sim \triangle BPA \implies \frac{BF}{FA}=\frac{BG}{GP}=2$. Similarly, $ \frac{CE}{AE}=2$. By the converse of Ceva on $ ABC$, we see that $ BE$, $ CF$, $ AM$ are concurrent (call the point of concurrency $ N$). (Concurrent because 1/2 * 2/1 * 1/1 = 1) Also, we see that $ FA = \frac{AB}{3}$ and $ FL = \frac{AB}{2}-\frac{AB}{3}= \frac{AB}{6}$.

Now, we use Menelaus and get $ \frac{PA}{AB}*\frac{BM}{MC}*\frac{CN}{NF}$. Plugging in, we see that $ \frac{1}{3}*\frac{1}{1}*\frac{CN}{NF}=1 \implies \frac{CN}{NF}=3$.

Finally, applying converse Ceva to $ \triangle CFB$, we see that $ \frac{FL}{LB}*\frac{BM}{MC}*\frac{CN}{NF}= \frac{AB/6}{AB/2}*\frac{1}{1}*3 = \frac{1}{3}*1*3 = 1$, so $ CL, BN, MF$ are concurrent.

[/hide]


group theory
abstract algebra
superior algebra
superior algebra solved

Prove that if H <= G and G <= K then H <= K.

Since $H \subset G$ is a subgroup of the group $G$ iff it satisfies the group axioms with the 'operator' (sorry, I don't know the correct word) defined on $G$.
In this case trivially $H \subset G \subset K$ and it fulfills the axioms since both $G$ and $K$ have the same 'operation'.