geometry
geometric transformation
rotation
reflection
analytic geometry
linear algebra
matrix

The point $ (\minus{}3, 2)$ is rotated $ 90^\circ$ clockwise around the origin to point $ B$. Point $ B$ is then reflected over the line $ y \equal{} x$ to point $ C$. What are the coordinates of $ C$?

$ \textbf{(A)}\ ( \minus{} 3, \minus{} 2)\qquad \textbf{(B)}\ ( \minus{} 2, \minus{} 3)\qquad \textbf{(C)}\ (2, \minus{} 3)\qquad \textbf{(D)}\ (2,3)\qquad \textbf{(E)}\ (3,2)$

I am no good at this, but I will try:
I think B is (3,-2) and then C is (3,2)
E?

For me, I got $ B\equal{}(2,\minus{}3) ; C\equal{}(7,\minus{}8)$ There's no choice for my answer. After I look up in the internet, I see that
[quote]The point  $ (3,2)$ is rotated $ 90^\circ$  clockwise[/quote]
need to change to The point  $ (\minus{}3,2)$ is rotated $ 90^\circ$  clockwise. And the answer is $ B\equal{}(2,3) ; C\equal{}(3,2)$ which means $ \boxed{E}$

The original problem has the original point as $ (\minus{}3,2)$, as doabli said.

[hide="Rotation Matrix"]
The [url=http://mathworld.wolfram.com/RotationMatrix.html]rotation matrix[/url] for a rotation by $ \minus{}90^\circ$ counterclockwise around the origin is $ \begin{pmatrix}\cos{\minus{}90^\circ}&\minus{}\sin{\minus{}90^\circ}\\\sin{\minus{}90^\circ}&\cos{\minus{}90^\circ}\end{pmatrix}\equal{}\begin{pmatrix}0&1\\\minus{}1&0\end{pmatrix}$.

Thus, $ \vec B\equal{}\begin{pmatrix}0&1\\\minus{}1&0\end{pmatrix}\begin{pmatrix}\minus{}3\\2\end{pmatrix}\equal{}\begin{pmatrix}2\\3\end{pmatrix}$, so $ B\equal{}(2,3)$.

A reflection of this over the line $ y\equal{}x$ gives $ C\equal{}(3,2)\implies\boxed E$.
[/hide]
[hide="Polar Coordinates"]
In [url=http://mathworld.wolfram.com/PolarCoordinates.html]polar coordinates[/url], $ (\minus{}3,2)\equiv(r,\theta)$ for some $ r$ and $ \theta$, and since $ \minus{}3\equal{}r\cos\theta$, $ \cos\theta\equal{}\minus{}\frac3r$ and similarly, $ \sin\theta\equal{}\frac2r$.

After the rotation, $ B\equal{}(r,\theta\minus{}90^\circ)$, and since $ x'\equal{}r\cos(\theta\minus{}90^\circ)\equal{}r\sin\theta$ and $ y'\equal{}r\sin(\theta\minus{}90^\circ)\equal{}\minus{}r\cos\theta$, simply plugging in the values of $ \sin\theta$ and $ \cos\theta$ gives $ B\equal{}(2,3)$.

Then, the reflection is similar; we get $ C\equal{}(3,2)\implies\boxed E$.
[/hide]

Good catch, guys. In the future, if you see a problem with a typo like this, PM me, as I will not always go back to each individual thread.

[quote="doabli"]I can't find a simple way to find point C, I need to use too many calculate and I think for a AMC 12, problem B isn't good. Does someone have a nice way to find the reflection of a point through a line?[/quote]

It's good to know that the reflection of a point $ (x,y)$ through $ y \equal{} x$ is $ (y,x)$, and the reflection of a point $ (x,y)$ through $ y \equal{} \minus{} x$ is $ ( \minus{} y, \minus{} x)$.

You can see both of these relationships by drawing the feet from both points to both axes to get a pair of congruent rectangles. From this, it's more or less trivial.

[hide="To reflect a point over a line in general..."]
Let $ \vec v \equal{} \begin{pmatrix}x & y\end{pmatrix}$, the line be $ Ax \plus{} By \plus{} C \equal{} 0$, so we want the reflection, or $ \vec v' \equal{} \begin{pmatrix}x' & y'\end{pmatrix}$. The distance from the point to the line is $ \frac {|Ax \plus{} By \plus{} C|}{\sqrt {A^2 \plus{} B^2}}$.

It is easy to see that $ \|\vec v' \minus{} \vec v\| \equal{} \frac {2|Ax \plus{} By \plus{} C|}{\sqrt {A^2 \plus{} B^2}}$.

Now we need a direction for $ \vec v' \minus{} \vec v$: the vector that is normal to the line $ Ax \plus{} By \plus{} C \equal{} 0$ is $ \begin{pmatrix}A & B\end{pmatrix}$, so the unit normal in this direction is $ \frac {\begin{pmatrix}A & B\end{pmatrix}}{\|\begin{pmatrix}A & B\end{pmatrix}\|} \equal{} \frac {\begin{pmatrix}A & B\end{pmatrix}}{\sqrt {A^2 \plus{} B^2}}$**.

Finally, we have $ \vec v' \minus{} \vec v \equal{} \begin{pmatrix}x' \minus{} x & y' \minus{} y\end{pmatrix} \equal{} \frac {2|Ax \plus{} By \plus{} C|}{\sqrt {A^2 \plus{} B^2}}\cdot\frac {\begin{pmatrix}A & B\end{pmatrix}}{\sqrt {A^2 \plus{} B^2}}$

$ \equal{} \frac {2|Ax \plus{} By \plus{} C|\begin{pmatrix}A & B\end{pmatrix}}{A^2 \plus{} B^2}$

$ \implies(x',y') \equal{} \boxed{\left(x \plus{} \frac {2A|Ax \plus{} By \plus{} C|}{A^2 \plus{} B^2},y \plus{} \frac {2B|Ax \plus{} By \plus{} C|}{A^2 \plus{} B^2}\right)}$.

**The unit normal is only $ \frac {\begin{pmatrix}A & B\end{pmatrix}}{\sqrt {A^2 \plus{} B^2}}$ if the point that we are reflecting is below the line of reflection. If it's below it, then the unit normal is instead $ \frac {\begin{pmatrix} \minus{} A & \minus{} B\end{pmatrix}}{\sqrt {A^2 \plus{} B^2}}$, yielding $ \implies(x',y') \equal{} \boxed{\left(x \minus{} \frac {2A|Ax \plus{} By \plus{} C|}{A^2 \plus{} B^2},y \minus{} \frac {2B|Ax \plus{} By \plus{} C|}{A^2 \plus{} B^2}\right)}$.
[/hide]


Find the last two digits of N=11^10-9

someone delete this double post please

Please use $ \text{\LaTeX}$, with 400+ posts, it's time that you start.

[hide]$ 11^{10}-9=(10+1)^{10}-9=10^{10}+{10\choose1}10^{9}+\cdots+{10\choose9}10^{1}+1-9\equiv1-9\equiv-8\pmod{100}$, so the last two digits are $ \boxed{92}$.[/hide]


If  x,y,z are positive  numbers with $ xy \plus{} yz \plus{} zx \equal{} 1$ then
\[ \frac {x^4}{y^2} \plus{} \frac {y^4}{z^2} \plus{} \frac {z^4}{x^2}\ge \frac{(x^2 \plus{} y^2 \plus{} z^2)(x \plus{} y \plus{} z)^2}{3}
\]

[quote="nguoivn"]Using Am-Gm, we are easily to prove the stronger:
$ \frac {x^4}{y^2} \plus{} \frac {y^4}{z^2} \plus{} \frac {z^4}{x^2}\ge \frac {(x^2 \plus{} y^2 \plus{} z^2)^2(x \plus{} y \plus{} z)^2}{3}$
 :ninja:[/quote]

  nquoivn,can you detail , please?

[quote="nguoivn"]I using an old problem:
$ \frac {a^2}{b} \plus{} \frac {b^2}{c} \plus{} \frac {c^2}{a} \geq\ \frac {(a \plus{} b \plus{} c)(a^2 \plus{} b^2 \plus{} c^2)}{ab \plus{} bc \plus{} ca}$
(<=> $ x^2 \plus{} y^2 \plus{} z^2 \geq\ xy \plus{} yz \plus{} zx$
with $ x \equal{} ab^2, y \equal{} bc^2; z \equal{} ca^2$ )
 :)[/quote]

 Thanks nquoivn :)


geometry
inradius
circumcircle
incenter
similar triangles

Show that if $ r_a,r_b,r_c$ are the radii of the excircles of triangle $ ABC$, and $ r$ is the inradius then,
\[ \dfrac{1}{r_a} \plus{} \dfrac{1}{r_b} \plus{} \dfrac{1}{r_c} \equal{} \dfrac{1}{r}\]
Edit: Darn, I did forgot to divide, but yea that was what I meant.

If you already know that then here's something else similar:

\[ r_a\plus{}r_b\plus{}r_c\equal{}\dfrac{abc}{4R}\left(\dfrac{1}{s\minus{}a}\plus{}\dfrac{1}{s\minus{}b}\plus{}\dfrac{1}{s\minus{}b}\right)\]

Where $ R$ is the circumradius and $ s$ is the semi-perimeter.

[quote="tonypr"]Show that if $ r_a,r_b,r_c$ are the radii of the excircles of triangle $ ABC$, and $ r$ is the inradius then,
\[ \dfrac{1}{r_a} \plus{} \dfrac{1}{r_b} \plus{} \dfrac{1}{r_c} \equal{} \dfrac{a \plus{} b \plus{} c}{2r}\]
[/quote]
I think you mean  $ \dfrac{1}{r_a} \plus{} \dfrac{1}{r_b} \plus{} \dfrac{1}{r_c} \equal{} \dfrac{1}{r}$

That can't be right, since if it holds for $ \triangle ABC$, then for a similar triangle with proportions doubled (ie, side lengths $ 2a$, $ 2b$, $ 2c$), the LHS is halved, but the RHS doesn't change.

I believe the correct formula is $ \frac{1}{r_a}\plus{}\frac{1}{r_b}\plus{}\frac{1}{r_c} \equal{}\frac{1}{r}$, which is well-known.

(I'm a slow typist.)

^ Lol 3 minutes after i commented :)
Anyways
Let $ {A}$ represent the area of the triangle and $ s \equal{} \frac{a\plus{}b\plus{}c}{2}$  then
$ {A} \equal{} {r_{a}}({s} \minus{} {a}) \equal{} {r_{b}}({s \minus{} b}) \equal{} {r_{c}}({s} \minus{} {c})$
Or $ {r_{a}} \equal{} \frac{A}{s\minus{}a} , {r_{b}} \equal{} \frac{A}{s\minus{}b}$ and $ {r_{c}} \equal{} \frac{A}{s\minus{}c}$

Then  $ \frac{1}{r_{a}}\plus{}\frac{1}{r_{b}}\plus{}\frac{1}{r_{c}}\equal{} \frac{s\minus{}a}{A} \plus{} \frac{s\minus{}b}{A} \plus{} \frac{s\minus{}c}{A}$
After subbing in $ s \equal{} \frac{a\plus{}b\plus{}c}{2}$ and using $ A \equal{} rs$ where $ r$ is the inradius we get the RHS

[quote="tonypr"]And the other problem?[/quote]
Wasn't up there when i started to type :) lol
Here Let $ A$ represent area of the triangle
first $ \frac{abc}{4R} \equal{} A$
And $ {A}\equal{}{r_{a}}({s}\minus{}{a}) \equal{}{r_{b}}({s\minus{}b}) \equal{}{r_{c}}({s}\minus{}{c})$ or $ {s}\minus{}{a}\equal{} \frac{A}{r_{a}}, {s}\minus{}{b}\equal{} \frac{A}{r_{b}}, {s}\minus{}{c}\equal{} \frac{A}{r_{c}}$
Subbing this into the RHS we get
$ \frac{abc}{4R}(\frac{1}{s\minus{}a}\plus{} \frac{1}{s\minus{}b}\plus{} \frac{1}{s\minus{}b}) \equal{} \frac{abc}{4R}(\frac{r_{a} \plus{} r_{b} \plus{}  r_{c}}{A})$ and since $ \frac{abc}{4R} \equal{} A$, subbing that in gives the LHS

I've never worked with excircles before so I'm probably missing something simply, but how do you know that $ A\equal{}r_a(s\minus{}a)$?

The way I know how to prove it,

Let $ E_a$ and $ I$ be the ex-center(with respect to the $ A$ vertex) and the Incenter respectively, in triangle $ ABC$.

Extend the Line $ AC$, to the point where it is tangent to the excircle at $ P_2$ and to the incircle at $ P_1$.

We have that $ \triangle AIP_1 \sim \triangle AE_aP_2$.

By similar triangles we have, $ \dfrac{IP_1}{E_aP_2} \equal{} \dfrac{AP_1}{AP_2}$

Since $ IP_1$ and $ E_aP_2$ are tangent to the line $ AC$ and are drawn from the centers of the incircle and excircle, they are the radius and exradius(from vertex $ A$) respectively.

Also if we let $ a \equal{} y \plus{} z,b \equal{} x \plus{} z,c \equal{} x \plus{} y$, where $ AP_1 \equal{} x$, $ P_1C \equal{} z$,$ CP_2 \equal{} y$.

We have $ \dfrac{r}{r_a} \equal{} \dfrac{x}{x \plus{} y \plus{} z}\implies \dfrac{r}{r_a} \equal{} \dfrac{s \minus{} a}{s}\implies r_a \equal{} \dfrac{r\cdot s}{s \minus{} a}$.

And since $ rs$ is equivalent to the area, it is proven.

Edit: Sorry for the typo.

one little issue there: you assume that $ s\equal{}x\plus{}y\plus{}z$, but this would mean that $ z\equal{}CP_2$ equals the distance from $ B$ to one of its points of tangency with $ \circ I$.  Can you prove that?

If $ a \equal{} y \plus{} z,b \equal{} x \plus{} z,c \equal{} x \plus{} y$, then $ P_1C \equal{} z,CP_2 \equal{} s \minus{} b \equal{} y$
he probably made a typo

I see the typo now, but I don't get why $ AP_2 \equal{} s$... really the same issue as I stated before

ok I got it now... It just took a few extra steps to show $ CP_2\equal{}y$. (assuming $ P_1C\equal{}z$)

Let $ P_3$ be the point of intersection of the excircle and ray $ AB$.  Now $ AP_3 \equal{} AP_2$ and $ BC\equal{}BP_3\plus{}CP_2$.  So we have $ x\plus{}y\plus{}BP_3 \equal{} x\plus{}z\plus{}CP_2$ and $ y\plus{}z\equal{}BP_3\plus{}CP_2$.  Solving this system for $ CP_2$ yields $ CP_2\equal{}y$, so the final result follows as before.

Thanks for the help everyone!

yea, sorry for not showing that originally.


Wow! It's cool that you're posting on AoPS!  Maybe now we can start reviving this forum . . .

By the way, nice job at State - fourth place team is really good (a lot better than my team did back in '05).

wow this is an old topic....
but i did ^.^

So did I. Good luck next year to any 7th of 6th graders!


This was actually from the IMO, but it isn't that tricky. Proving is not necessary.

Evaluate $\cos \pi/7 - \cos 2\pi/7 + \cos 3\pi/7.$

http://www.kalva.demon.co.uk/imo/isoln/isoln635.html

Consider the roots of $x^7 + 1 = 0$. They are $e^{i\pi/7}$, $e^{i3\pi/7}$, ... , $e^{i13\pi/7}$ and their sum must have zero since there is no $x^6$ term. Hence, in particular, the sum of their real part is zero. But $\text{cos }7\pi/7 = - 1$ and the others terms are equal in pairs, because $\text{cos}(2\pi - x)=\text{cos }x$. So we get $\text{cos }\pi/7 + \text{cos }3\pi/7 +\text{cos }5\pi/7 = 1/2$. Finally, since $\text{cos}(\pi - x) = -\text{cos }x$, $\text{cos }5\pi/7 = -\text{cos }2\pi/7$.

[quote="puuhikki"]Consider the roots of $x^7 + 1 = 0$. They are $e^{i\pi/7}$, $e^{i3\pi/7}$, ... , $e^{i13\pi/7}$ and their sum must have zero since there is no $x^6$ term. Hence, in particular, the sum of their real part is zero. But $\text{cos }7\pi/7 = - 1$ and the others terms are equal in pairs, because $\text{cos}(2\pi - x)=\text{cos }x$. So we get $\text{cos }\pi/7 + \text{cos }3\pi/7 +\text{cos }5\pi/7 = 1/2$. Finally, since $\text{cos}(\pi - x) = -\text{cos }x$, $\text{cos }5\pi/7 = -\text{cos }2\pi/7$.[/quote]
This came directly from Kalva, right?

The above solution does seem to be from kalva.

However, the use of complex numbers really isn't necessary. Consider multiplying and dividing the expression by $\sin \pi/7$.

We then have \[ \frac{\sin \pi/7 \cos \pi/7 - \sin \pi/7 \cos 2\pi/7 + \sin \pi/7 \cos \ 3\pi/7}{\sin \pi/7} \]
\[ = \frac{\sin 2\pi/7 - \sin 0 - \sin 3\pi/7 + \sin \pi/7 + \sin 4\pi/7 - \sin 2\pi/7}{2\sin \pi/7} \]
\[ = \frac{- \sin 3\pi/7 + \sin \pi/7 + \sin 3\pi/7}{2\sin \pi/7} \]
\[ = \frac{\sin \pi/7}{2\sin \pi/7} \]
\[ = 1/2  \]

What is going on...  IMO problems in Intermediate???  :o  :o

[quote="amcavoy"]This came directly from Kalva, right?[/quote]
I found the proof myself. It is possible that Kalva knows the same proof. But honestly, I do not plagiarize any of my solutions anywhere. If I copy something nontrivial from somewhere, I will tell where I found the proof. Believe me. Sometimes it just happens that two mathematicians will found identical proofs for the same problem.

It's not a big deal, I was just wondering since your solution is word-for-word the solution Kalva provides.


Hi,

With this many people in the forum, and with that diversity in ages, I think this question may have interesting answers.

In my case, I am already married. My wedding was small, just my wife, the more closer family and friends and the judge (it was a civil ceremony, not religious!), there were around 20 people in that small party. If I would have to do it again, probably I would do it the same way.

Now, the question is, how, in respect to size (number of guests) is or was your dream wedding.

The option "Just the people strictly necesary", in the case of Costa Rica would be 5, the groom, the bride, two withnesses and the judge or the priest, but I suppose it is different in every place.

Best Regards,

Ha, an interesting question.  I voted for "small," which is definitely my preference.  I have an aunt and uncle who were married with "as few as necessary" wearing jeans and t-shirts, but I think it would be nice to have a little more ceremony than that.  My girlfriend is Indian, so we joke that if we get married, we will have to have two ceremonies: one small civil ceremony for me, one massive Indian-style ceremony with all of her relatives from India and a big party and fancy clothing and everything for her (or perhaps more, for her parents ;-) ).

[quote], one massive Indian-style ceremony..[/quote]
Ooohh..one of those! So is it  going to be  [url=http://www.msnbc.msn.com/id/5297284/]wedding festivities at Versailles?[/url] ...Or are we going to see the  [url=http://mid-day.com/news/world/2002/september/30404.htm]Replica of  Famed Hawamahal   of Jaipur[/url] built in Harvard Square ...or may be you are just going to  [url=http://dailypioneer.com/columnist1.asp?main_variable=Columnist&file_name=mitra29.txt&writer=MITRA&validit=yes]Fly in London Philharmonic Orchestra[/url] to play in the background? :) :)

[quote="shinwoo"]guys come on, do we have to talk about the whole future already? :D

I"m only in 7th grade and I find it really embarrasing to talk about it :D  :blush:  :P[/quote]

It is always good to talk about what you spect about the future :P . Besides don't feel afraid to talk about something, specially something as natural as this (most of the people eventually gets married, don't they?). What is totally acceptable is that you maybe don't really have an idea of what you want, but that is totally natural. If you are a seventh grader, you are around 12 or 13, that means, probably you'll be twice as old (or even older) when you'll think about getting married, therefore, probably it is not something that must be consuming your dream hours now (Math is!!!  :D ).

Best regards,

If I'd get married today, I'd have to make it big, if for no other reason but that some friends would be very angry to find out that I got married and I didn't call them to the party :blush:. And, of course, another reason would be to gather as many wedding presents as possible :D :D :D

[quote="Valentin Vornicu"] And, of course, another reason would be to gather as many wedding presents as possible :D :D :D[/quote]

Oh...I'd never thought of it that way. Hmm...perhaps "Big" isn't big enough...

i chose "small" in the poll but i have a feeling that my wedding would be big. my cousin had a wedding in Taiwan and they invited hundreds of people. the party was pretty sweet, though.

[quote="miraculouspostmaster"]I want to get married while skydiving.
'twould be quite awesome.[/quote]

AND DANGEROUS TOO!!  :o  :?

[quote="Valentin Vornicu"]And, of course, another reason would be to gather as many wedding presents as possible [/quote]

Ahhh!, unfortunatelly, in Costa Rica usually you don't receive wedding presents, or at least not from everyone, usually just your parents, brothers and sisters, and may be some close friends would give you presents!!

Best regards,

[quote]AND DANGEROUS TOO!![/quote]
Surely that's an exaggeration...NO wedding can be that dangerous. :)

[quote="djimenez"][quote="miraculouspostmaster"]I want to get married while skydiving.
'twould be quite awesome.[/quote]

AND DANGEROUS TOO!!  :o  :? [/quote]

Yeah, it'd be pretty dangerous, like what if you dropped the ring in mid-air? Not only would you definitely lose it, it might hit some dude on the ground on the head. Haha, I bet that person would be pissed. But then again, they just got a free wedding ring!

Must have been pretty hard to find a ring boy and flower girl for the skydiving wedding..
I would like to get married underwater.

[quote="aznness"]This is uberl33tmage's new acct.

I would like a huge underwater wedding, but I'd like the cake, dinner, and wedding presents kept above the water...

Soggy cake is not worth $1500.[/quote]

o_____o


trigonometry
inequalities
inequalities proposed

\[ \triangle ABC\Longrightarrow \overline {\underline {\left| \ \dfrac{3r}{R}\le \sin \frac A2+\sin \frac B2+\sin \frac C2\ \right| }}\le \frac 32  \]

Levi, the inequality $\dfrac{3r}{R}\leq sin\dfrac{A}{2} +sin\dfrac{B}{2} +sin\dfrac{C}{2}$ is in my book.
It is problem G.U.81., page 113 ;)

Can you present now, here, the your solution ?

Of course! Actually, what i said earlier was wrong. Actually it's not problem G.U.81. In fact it is problem G.U.27.

 Well, we suppose true the following identity: $\prod \sin\frac{A}{2}=\frac{r}{4R}$, so the inequality can be also
 written as $\sum\frac{1}{\sin\frac{A}{2} \sin\frac{B}{2}}\geq 12$ ;) .From here it is clear how to proceed.
 You can also check my book. :D


Find the integers like x,y that :
\[ x^2\plus{}(x\plus{}1)^2\equal{}y^4\]

Double posting: http://www.mathlinks.ro/viewtopic.php?t=241526

(By the way, I have to ask - why did you choose the subject "easy" for that problem, if you don't know the solution? Anyway, whether this is easy or not, that's not a very useful description of the problem. Please use more descriptive names in the future. "Integers" isn't much better).

Totally agreed@randomgraph
-> Locked
(And don't do this again!)


Suppose $ a_n > b_n > 0$ and $ a_n > a_{n \plus{} 1}$ and $ \lim_{n\rightarrow\infty}a_n \equal{} 0$.
Can we conclude that $ \sum_{n \equal{} 1}^{\infty}( \minus{} 1)^nb_n$ converges?

Nope: take $ b_n$ to be $ \frac{1}{n^2}$ if $ n$ odd and $ \frac{1}{n}$ if $ n$ even, and take $ a_n\equal{}\frac{2}{n}$. Then $ a_n$ decreases to 0 and is everywhere greater than $ b_n$, but the alternating series of the $ b_n$ diverges (the positive terms form a harmonic series, the negative terms converge).


calculus
integration
trigonometry
calculus computations

converge intergal :???  true or false
$\int_{0}^{+\infty}\frac{xdx}{1+x^{2}cos^{2}x}$

The integral diverges.

$\int_{0}^{\infty}\frac{xdx}{1+x^{2}(\cos x)^{2}}\geq \frac{xdx}{1+x^{2}}=\infty$.


[quote="Omid Hatami"]Suppose $P,Q\in \mathbb R[x]$ that $deg\ P=deg\ Q$ and $PQ'-QP'$ has no real root. Prove that for each $\lambda \in \mathbb R$ number of real roots of $P$ and $\lambda P+(1-\lambda)Q$ are equal.[/quote]

deg(P)>1.

First suppose P has at least one real root.
Let r_1,...,r_n be the real roots of P (they are all distinct  -of multiplicity 1- since if P(x)=0 then P'(x)<>0 because PQ'(x)-P'Q(x) <> 0).
We can obviously assume deg(P)>=1, and since PQ'-P'Q has no real root, then (Q/P)'= (PQ'-P'Q)/P^2 keeps the same sign + or - in (r_i,r_i+1) for all i in [1,n-1]. Now we know that for any real root r_i of P: Q'(r_i)<>0 by hypothesis, and P changes sign at r_i ( because P'(r_i) <> 0 ); as a result we can pick a neighborhood V_i of r_i such that Q(V_i) does not contain {0} and P(r_i+) and P(r_i-) have opposite signs. From here you can find the possible variations of Q/P over R-{r_1,...,r_n}, and considering the fact that deg(P)=deg(Q), you have lim(P/Q, +oo) = lim(P/Q, -oo) = constant.
There are only two possibilities for the variations of Q/P, and both of them will show you that the equation a*P+(1-a)*Q = 0 <=> Q/P = a/(a-1) ( the case a = 1 being trivial, and the case where P(x)=0 is ruled out since then Q(x)<>0 and the equation does not hold ) has exactly n solutions on R.

If P has no real root then Q/P will be strictly monotous as above but lim(P/Q, -oo) = lim(P/Q, +oo), which is absurd (and hence P always has at least one real root).

--
Julien Santini

Let's modify the conclusion to the statement that for all $t,\ Q+tP$ has the same number of real roots as $P$.

Assume WLOG that $PQ'-QP'>0$. This says that the function $\frac QP$ is strictly increasing on every interval on which it's defined. Notice that $P(Q+tP)'-(Q+tP)P'>0$ as well in this case, so it suffices to prove the result for $t=0$
, i.e. it suffices to prove that $P$ and $Q$ have the same number of real roots.

Let $a_1<\ldots<a_k$ be the distinct real roots of $P$. For $i\in\overline{1,k-1}$ the function $\frac QP$ goes to $-\infty$ as $x$ decreases to $a_i$ and to $\infty$ as $x$ increases to $a_{i+1}$, so $\frac QP$ and thus $Q$ has precisely one real root in each interval of the form $(a_i,a_{i+1})$. $\frac QP$ (and thus $Q$) also has precisely one root in exactly one of the intervals $(-\infty,a_1),(a_k,\infty)$, namely in the first one if the ratio of the dominant coefficients of $Q$ and $P$ is negative, and the second one if it's positive. This means that $\frac QP$ (and thus $Q$) has precisely $k$ real roots, as desired.

Edit:

Julien posted his message while I was typing mine.


calculus
integration
algebra
polynomial
calculus computations

Find the constant numbers $ u,\ v,\ s,\ t\ (s<t)$ such that $ \int_{\minus{}1}^1 f(x)\ dx\equal{}uf(s)\plus{}vf(t)$ holds for any polynomials $ f(x)\equal{}ax^3\plus{}bx^2\plus{}cx\plus{}d$ with the degree of 3 or less than.

I know this off the top of my head because it's a particular case of Gaussian Quadrature. (Try looking that up.) The answer will be that $ u\equal{}v\equal{}1,\ s\equal{}\minus{}\frac1{\sqrt{3}}$ and $ t\equal{}\frac1{\sqrt{3}}.$

Step 1: The polynomial $ 3x^2\minus{}1$ is orthogonal to all polynomials of the form $ ax\plus{}b.$ That is,

$ \int_{\minus{}1}^{1}(3x^2\minus{}1)(ax\plus{}b)\,dx\equal{}
\left.(x^3\minus{}x)(ax\plus{}b)\right|_{\minus{}1}^1\minus{}\int_{\minus{}1}^1(x^3\minus{}x)a\,dx\equal{}0.$

Step2: Any $ f(x)$ as in the problem can be written as $ f(x)\equal{}p(x)(3x^2\minus{}1)\plus{}q(x)$ where $ p(x)$ and $ q(x)$ are linear or constant polynomials. This is simply the division algorithm.

Step 3: If $ q(x)\equal{}ax\plus{}b,$ then

$ \int_{\minus{}1}^1q(x)\,dx\equal{}q\left(\minus{}\frac1{\sqrt{3}}\right)\plus{} q\left(\frac1{\sqrt{3}}\right)$

(This is a straightforward computation.)

Step 4:

$ \int_{\minus{}1}^{1}f(x)\,dx\equal{}\int_{\minus{}1}^1p(x)(3x^2\minus{}1)\plus{}q(x)\,dx\equal{} \int_{\minus{}1}^1q(x)\,dx$

$ \equal{}q\left(\minus{}\frac1{\sqrt{3}}\right)\plus{} q\left(\frac1{\sqrt{3}}\right)\equal{} f\left(\minus{}\frac1{\sqrt{3}}\right)\plus{} f\left(\frac1{\sqrt{3}}\right).$

[Basically, I've just particularized to this case all the steps for proving that Gaussian Quadrature works in general.]


I don't think this is a question of vital importance...especially since it has been discussed just recently on this forum.

[quote="kentliu"]now that i have your attension...

when they calculate your index score for AMC+AIME, how do they calculate if you do USAMTS instead of AMC?[/quote]

You get a 100 as your replacement AMC score.  Definitely take the AMC - don't just count on USAMTS.


function
geometry
parallelogram
geometric transformation
logarithms
complex analysis
complex numbers

1. Find all entire function such that $ |f(z)| \leq {|f(2z)|}^{\frac {1}{2}}$ for all $ z$ in $ C$.
2. Supposed that $ f$ is an entire function such that $ f(z\plus{}p)\equal{}f(z)\equal{}f(z\plus{}q)$ for all $ z$ in $ C$, where $ \frac {p}{q}$ isn't in $ R$, prove that $ f$ is constant for all $ z$ in $ C$.
3. Supposed that $ f$ is an real valued harmonic function such that $ f^2$ is also harmonic for all $ z$ in $ C$. Then prove that $ f$ is constant.
Thanks in advance

2) $ f(z)$ is bounded on the parallelogram with vertices $ 0,\ p,\ q,$ and $ p \plus{} q$, $ f$ is continuous and this is a closed set. The complex plane can be covered by translations of this parallelogram by $ ap \plus{} bq$ with $ a$ and $ b$ integers. Observe that the values of $ f$ on each of these translated parallelograms are the same as its values on the original parallelogram, so $ f$ is bounded on the complex plane. Liouville's theorem then implies that $ f$ is constant.

3) Since $ f(z)^2$ is harmonic everywhere, it is the real part of some entire function, $ g(z)$. Since $ f(z)$ is real, $ f(z)^2\ge 0$, so $ g(z)$ can never take any value with negative real part. This implies that $ g(z)\plus{}1$ has no zeros, so $ \frac{g(z)}{g(z)\plus{}1}$ is a bounded entire function. Hence, by Liouville's theorem, it is constant. The result follows.

For 1) we have : $ |f^2(z)|\leq |f(2z)|$

So the meromorphic map : $ \frac{f^2 (z)}{f(2z)}$ is bounded, thus all its singularities are removable and it is entire. And by Liouville, constant.Thus exists a constant $ C$ such that : $ f^2(z)\equal{}C f(2z)$
So multiplying by $ \frac{1}{C}$ we have to solve the following functional equation for f entire : $ g(2z)\equal{}g^{2}(z)$.
If $ g$ vanishes somewhere, let's say in $ z_0$ then $ \{z_0,\frac{z_0}{2},...,\frac{z_0}{2^n},...\}$ are also zeroes of g and then $ g\equal{}0$.("0" is not an isolated zero)

If not we can write $ g\equal{}e^{h}$ with $ h(2z)\equal{}2h(z)$.
let's derivate :$ h'(2z)\equal{}h'(z)$ so $ h(z)\equal{}\lambda z$

[quote="popolux"]If not we can write $ g \equal{} e^{h}$ with $ h(2z) \equal{} 2h(z)$.
let's derivate :$ h'(2z) \equal{} h'(z)$ so $ h(z) \equal{} \lambda z$[/quote]

This is correct, but some readers may misinterpret it, because $ h$ may be a multivalued function in this argument.  Here is the rigorous, explicit version:

We have to pick a small neighborhood $ U$ and define some branch $ l(z)$ of the logarithm on a neighborhood containing $ f(U)$. ($ U$ must be small to ensure that this neighborhood is simply connected)  Then $ h \equal{} l(g(z))$ on $ U$, so indeed $ h(z)\equal{}\lambda z$ and $ g(z)\equal{}e^{\lambda z}$ on $ U$.  Two entire functions equal on $ U$ are equal everywhere, so we can extend this equation to the whole plane.


Does anyone know of a proof for:
tan(A)+tan(B)+tan(C)=tan(A)*tan(B)*tan(C)
if A+B+C=180?

And none of the angles are right angles  ;)


induction
Galois Theory
superior algebra
superior algebra unsolved

Are there $a, b, c, d\in\mathbb{Q}$ such that \[ 1+\sqrt [3] {2}\in\mathbb{Q}(\sqrt{a},\sqrt{b},\sqrt{c},\sqrt{d})? \]

Nice titles ;)

$1+\sqrt[3]2$ has degree $3$, so it can't be contained in anything of type $\mathbb{Q}[\sqrt{a_1} , \sqrt{a_2} , ...]$ since their degrees would be a power of two by induction on the number of roots adjoined: Everytime when we adjoin a root, the degree grows by a factor $1$ (thus remains the same) or by a factor $2$ by multiplicativity of the degrees.

A more general approach for adjoining e.g. also something of degree $1337$ to something already having degree $481$:
$[LM: K]$ is always a divisor of $[L: K][M: K]$ since $[LM: K]=[LM: L][L: K]$, so that it suffices to show that $[LM: L]$ divides $[M: K]$; it seems to me like there is an easy and direct way, but the only thing I came up is te see that restriction gives an embedding of $Gal(LM|L)$ into $(M: K)$.

[quote="ZetaX"]... $[LM: L]$ divides $[M: K]$ ...[/quote]

This is wrong in general, as you surely know (play with the third roots of 2).
The only proof I know about in the case of galois extensions L/K and M/K uses the restriction of the galois group which you mentioned.

LiMa


I'm year 9 (15yo) and finding these abit tricky...

If s(x) = e^x - e^-x,    show that

a) s(-x) = -s(x)
b) {s(x)}^3 = s(3x) - 3s(x)}

and

Find {x: (x-2)^2 > 9}

and

Find {(x,y): x = 7 - 4y and y = 7 - 4x}

thanks!

For the first one, you just need to plug it in and show they're equal:
$ s(\minus{}x)\equal{}e^{\minus{}x}\minus{}e^{\minus{}(\minus{}x)}\equal{}e^{\minus{}x}\minus{}e^{x}$
$ \minus{}s(x)\equal{}\minus{}e^{x}\plus{}e^{\minus{}x}\equal{}e^{\minus{}x}\minus{}e^{x}\equal{}s(\minus{}x)$

That's pretty much it.

Oh right, it just looked complicated for some reason :S

Thanks!


induction
inequalities unsolved
inequalities

Going to assume it's $n \in \mathbb{Z}^{+}$ since $n = 0$ and $n =-1$ are counterexamples.

EDIT: Arg, too many $n$s. This solution is wrong.
[hide="Solution"]By induction:

If $n = 1$, then we have $3 \geq 2+1$, which is true.

Suppose $(2n+1)^{n}\geq (2n)^{n}+(2n-1)^{n}$. Then note that $(2n+1)^{n+1}> (2n+1)^{n}\cdot 2n \geq ((2n)^{n}+(2n-1)^{n}) \cdot 2n > (2n)^{n+1}+(2n-1)^{n+1}$, which proves the $n+1$ case, and we're done.[/hide]

:D
we can chek it for $n=1,2,3,4$
OK.
let denote $n+1: =m$
after few calculation easy seen that we only nust proof $e^{\frac{1}{2}-\frac{1}{8m}}-\frac{1}{\sqrt{e}}>1$ really it is true if $m\geq 5$
because we take that
$0,001..>0$
Done
I note that this is true but it's bad way.


[color=green]In the pattern, the $n$th letter of the alphabet is written $n$ times. What letter wll be in the 160th position?[/color]

   [color=blue]ABBCCCDDDDEEEE...[/color]

[hide] You can look as this as an arithmetically increasing sum - 1+2+3+4+5..., the $n$th term will be the $n$th letter of the alphabet, and the sum is the total # of letters in the pattern. Thus, using the formula for the sum of an arithmetic series, we get: $160 = \frac{n(n+1)}{2}$. Solving for $n$, we find that $n$ is somewhere between 17 and 18, so the letter will be the 18th letter, which is $\fbox{R}$. [/hide]

[quote="humblemastermind"][hide] You can look as this as an arithmetically increasing sum - 1+2+3+4+5..., the $n$th term will be the $n$th letter of the alphabet, and the sum is the total # of letters in the pattern. Thus, using the formula for the sum of an arithmetic series, we get: $160 = \frac{n(n+1)}{2}$. Solving for $n$, we find that $n$ is somewhere between 17 and 18, so the letter will be the 17th letter, which is $\fbox{R}$. [/hide][/hide][/quote]

[hide]I'm pretty sure the answer is $\fbox{R}$.[/hide]

[quote="indianamath"][hide]if it is between the 17 and 18, the answer would be the 18th :wink:

the pattern would be

and the 160th would be around here$\downarrow$
QQQQQQQQQQQQQQQQQ|RRRRRRRRRRRRRRRRRR
---------------------17th-------------------------------18th[/hide][/quote]

Oops  :blush:

Corrected


function
algebra
polynomial
number theory unsolved
number theory

Find all function $ f: N\rightarrow N$ and multiplicative polynomial $ P(x) \equal{} a_{n}x^{n} \plus{} ... \plus{} a_{0}$ ($ a_{0},...,a_{n}\in\mathbb{Z}$) which satisfy $ \sum_{d|n}f\left( d\right) \equal{} p\left( n\right)$ for all $ n\in\mathbb{N}$

the Mobius inversion formula tells us that

\[ f(n)\equal{}\sum_{d|n}\mu(d)P(\frac{n}{d}) \text{ for all }n\in \mathbb{N}.
\]

so $ f$ is determined by the choice of $ P$. so really the problem becomes to find out which polynomials are multiplicative.

for $ P$ to be multiplicative we should have $ P(x^2) \equal{} P(x)^2$ for all $ x$:
\[ a_nx^{2n} \plus{} a_{n \minus{} 1}x^{2n \minus{} 2} \plus{} \cdots \plus{} a_1x^2 \plus{} a_0 \equal{} (a_nx^n \plus{} \cdots \plus{} a_0)^2
\]
now, equating coefficients, we get

coeff. of $ x^{2n}$: $ a_n\equal{}a_n^2$, so $ a_n\equal{}1$. (we may assume the leading coefficient is non-zero.)

coeff. of $ x^{2n \minus{} 1}$: $ 2a_na_{n \minus{} 1} \equal{} 0$. so $ a_{n \minus{} 1} \equal{} 0$.

coeff. of $ x^{2n \minus{} 2}$: $ 2a_na_{n \minus{} 2} \plus{} a_{n \minus{} 1}^2 \equal{} a_{n \minus{} 1} \equal{} 0$. so $ a_{n \minus{} 2} \equal{} 0$.

coeff. of $ x^{2n \minus{} 3}$: $ 2(a_na_{n \minus{} 3} \plus{} a_{n \minus{} 1}a_{n \minus{} 2}) \equal{} 0$. so $ a_{n \minus{} 3} \equal{} 0$.

coeff. of $ x^{2n \minus{} 4}$: $ 2(a_na_{n \minus{} 4} \plus{} a_{n \minus{} 1}a_{n \minus{} 3}) \plus{} a_{n \minus{} 2}a_{n \minus{} 2} \equal{} a_{n \minus{} 2} \equal{} 0$. so $ a_{n \minus{} 4} \equal{} 0$.

and so on. you get the picture.

so our only option is $ P(x) \equal{} x^n$, which we can easily see is multiplicative.


There's a problem in my math book that I don't quite understand.

It's 5 > 4b + 9 >17

Now what I got as the answer was -1 > b > 2.

Is that a mistake by me or the book? Or is it supposed to be that answer? Cause b can't both be less than -1 and greater than 2.

Subtract 9 from both all 3 inequalities, and you get -4>4x>8.
Now divide everything by 4, and you get -1>x>2.

Mmmmm. That is weird, and 4b+9 can't be greater than 17, but less than 5?....
Are you sure it isn't 5<4x+9<17??? That would make more sense.

It's possible the inequality has no solutions.

You don't even have to solve for $ b$; the problem claims that $ 5>17$, which is clearly wrong :wink:

[quote="Utter Loser"]So either its a no solution, or the book typed the problem wrong???
Could it be that it meant x<-1 or x>2???[/quote]

or....
he typed the < > signs backwards...which is a possibility

[quote="leroyjenkens"]Thanks for the responses. Yeah, I guess the book just got it wrong. The > signs were actually greater than or equal to signs, but I didn't know how to type those on here, so I just made them > signs, since I didn't think it would matter. It doesn't, right?
Well here's a picture out of the book.
[img]http://img219.yfrog.com/img219/7669/inequality.jpg[/img]

I know it's probably petty, but it was just bugging me since I assumed it was right and I was missing something.
All the other questions up to #17 were right and I got them right.[/quote]

Isn't it still impossible?

[quote]does the book have an answer key?[/quote]
Unfortunately not. That's the first thing I looked for.
[quote]Isn't it still impossible?[/quote]
Yeah. I just posted it just in case I was missing something. I posted on another forum about a problem I had with a chemistry book too. But in that case, it wasn't something wrong with the book, it was something wrong with me.


parameterization
analytic geometry
geometry unsolved
geometry

Hello everybody

im from portugal and i work in GIS software.
i want analyze two polygons that intersects in 2 points. I know every vertex of the polygon A and B (see picture) but i dont know the two points of intersection?
I need a formula to make the algorithm. Is there any formula for this?

thanks

VitorJOL

If points $A(x_{A},y_{A}), B(x_{B},y_{B}), C(x_{C},y_{C}), D(x_{D},y_{D})$ are given, and we're looking for the intersection of SEGMENTS $AB$ and $CD$, then real parameters $\lambda$ and $\mu$ must satisfy

$x_{A}+\lambda(x_{B}-x_{A})=x_{C}+\mu(x_{D}-x_{C})$
$y_{A}+\lambda(y_{B}-y_{A})=y_{C}+\mu(y_{D}-y_{C})$
$0\leqslant\lambda\leqslant 1$
$0\leqslant\mu\leqslant 1$

So solve the above system of linear equations and check if the solution satisfies the last two conditions. Then the coordinates of the intersection $M$ are $x_{M}=x_{A}+\lambda(x_{B}-x_{A}), y_{M}=y_{A}+\lambda(y_{B}-y_{A})$.


In triangle ABC , <C=30 degrees.
From AC and BC , AD and BE are cut off respectively equal to AB ,
Prove that DE is perpendicular to IO as well as equal to IO, where I and O are respectively the incenter and circumcenter of triangle ABC .
[ i have a synthetic geometric solution, i am posting this as a challenge to solve the problem by complex-number  or vector method or rigid transformations]

Obviously, $\angle DIE=360^\circ-3\cdot\angle AIB=45^\circ$; $\angle BEO=90^\circ-\frac{\angle \ OBE}{2}=120^\circ-\frac{\angle \ B}{2}$; $\angle IEO=\angle BEO-\angle IEB=45^\circ$. Similarly $\angle IDO=45^\circ$, consequently $O$ is the orthocenter of $\triangle DIE$.
In any triangle with an angle of $45^\circ$ the distance from that vertex to the orthocenter is equal to the opposite side of that vertex.

Best regards,
sunken rock


AMC AIME trigonometry probability geometry inequalities USA(J)MO

i must score a 10 on AIME. now i usually get 3~4 on the practice tests.
what classes for AIME are on  now in AoPS online class.
i need to get enrolled fast so i can prepare. Which ones are the best. i'll do 1 least or two if possible.

i need to know which two i should take which could possibly increase my AIME score by 6~7 (for 10 on it) the most...

thanks. tell me fast pls because i need to enroll fast IF POSSIBLE.

intermediate Geo
intermediate Alg
   ''                 Trig/ complex
   ''                 counting/ probabaility
   "            number theory
olympiad geometry
olympiad problem solving
olympiad inequality

my guess (im not that good to be even considering usamo qualification) is that you should buy AoPS volumes 1 and 2 and take the AIME problem solving seminar (weekend before the test).
For the subjust classes, I think intermediate (any) is the best.

thanks. there's aime problem solving seminar before the test??

yeah.
It's new.
I just heard about it today.

Im not saying its impossible but 6 -7 point increase in less than a month is pretty hard to do.

I recommend aops books, intermediate aops classes, and past aime tests.

Do any past AIME problems you can find on AoPS. If you don't understand them, read other people's solutions. Also, work through all of AoPS Vol. II.

[quote="towersfreak2006"]The only class in which you can enroll before the 2006 AIME is the special AIME Problem Seminar. Most people who have taken it in the past say that it's more of a review session than introduction to new material.

Take some old exams, and choose a subject, not necessarily your weakest, to study. That seems to be the general idea.[/quote]

I took seminar last year even though I didn't get much out of it. In past, seminar had one mock exam but I think it's just review of materials.

Anyway, I agree with Calculus about raising score about 6-7. It is possible. I've heard about people making dramatic increase in short period of time. But it is very hard because not only you have short amount of time, you will still have to do school work and other activities. Unless you can fully devote all your time doing AIMEs, you probably not going to achieve it (maybe or maybe not).

Since you have limited time, I would recommend you just doing problems for now and when there is a week left, take a practice test everyday to get you ready. I can't do this because I don't have enough time each day but maybe, you do.

thanks everyone.
i believe i could devote most of my time on the preparation of AIME.
i don't do much of schoolwork these days, so it wouldn't be a problem.

3 months ago i got about 0 on practice exams. one month ago after some going over AoPS i got 3~4 on practice exams. 3 months ago i got 90~100 on AMC 12, but now i get 120~130 on it.

i'll devote about 2~3 hours each day for AIME, if i set my mind to it.
which way do you recommend me studying for it.

should i go over the AoPS 2 very carefully and solve the AIME practice tests. or should i do the practice tests only.

do you think there's any kind of professional help out there i could get help from? or is this forum the best source.

[quote="fred"]do you think there's any kind of professional help out there i could get help from? or is this forum the best source.[/quote]

Right now, no. There isn't any AoPS class-like things outside here. During summer, I plan to do similar things but you have to ask me for that by PM.

[quote="fred"]should i go over the AoPS 2 very carefully and solve the AIME practice tests. or should i do the practice tests only. [/quote]

This is pretty tricky question. If you know all the important theorems and ideas, I recommend that you should just jump into doing problems. But if you're not familiar with some concepts (say you don't know much about DeMoivre's --> complex numbers do not appear "that" often but I heard they're not "that" difficult when they do appear neither), I suggest you learning them first and then do problems. Either way, your score will depend on how many problems you did.

I recommend just going through past tests and when you come upon a concept/theorem/formula you don't know, read about it and learn how to apply it.

thanks for the advices.
i'm so happy now because i'm now getting 5~6 on the tests.
4 more to go.


AMC
USA(J)MO
USAMO
quadratics
inequalities
geometry
number theory

Well, USAMO is now over, and the solutions are up, so how well does everyone think they did?

Did anybody writing officially manage to do really well this year?  Perhaps some Canadian winners this time?  I got 5 out of the 6 problems (missed #2 on day 1; I hate number theory), but unfortunately, I wasn't writing officially, so I'll never even get to know my score.  :(

Did any other Canadians do well?

I've gotten first 5 for sure, I knew #6 had something to do with 9.. But not sure how many points I'm going to get on it.

Speaking of which, Elyot, I really appreciate you not telling me about you writing the test at UW.
Seriously, I felt so bad about screwing up my teacher's schedule, AND on the second day, Ian had to send me the problems because I could not download it from my school.. Thanks a lot.

Awww... I'm so sorry I didn't tell you.  I thought you had all your details worked out.  The reason I had to go to Waterloo was because none of my teachers were available for supervising.

That's great that you got 5 problems though...

My #6 went something like this:
When digits are carried, you lose 9 from your total (ie you lose 10 ones, and gain 1 ten when regrouping).  So only 9s can be subtracted off, so the digit totals for each number must be a multiple of 9.  This means that all numbers in S are divisible by 9, and f(n) is always divisible by 9.

So for f(n) where n<=10, you can just fill S with all 9s and k=9.  Since you need to regroup up to 99 times (>9 times) when 10<n<=100, you need k=18 so there are enough digits that regrouping can occur whenever a new number is added when up to 100 are added together.  A way to do this is to fill with 99s, for example, and k will be = 18.  etc etc

I used that argument (with a ton more explaining) to prove that f(n) cannot be less than 9*ceiling(log10(n)), and the above cases (filling with all 9s, all 99s, 999s, etc) to prove sufficiency.

So I ended up proving f(n) = 9*ceiling(log10(n)), and proving constraints after that is easy.

I'm not sure my explaination will be good enough to get the perfect 7 though... maybe 6/7.

5 problems on USAMO is still really awesome... you're probably more likely to make it to the IMO team than I am, given you're a grade 11.  Good luck!

Wow, good job Jenn and Elyot. You people are so ownage. With five problems, you should be guaranteed a free trip to Washinton. CONGRATULATIONS!
I did okay, with badly explained solutions for 1,3,4,5. My #2 has the square and 9th power thing, and mod 19, but i calculated 147^157 (mod 19) wrong so i probably will get 2 points for that one. I have nothing for #6. Once again, congratulations.

[quote="pengshi"]Wow, good job Jenn and Elyot. You people are so ownage. With five problems, you should be guaranteed a free trip to Washinton. [/quote]

But we were unofficial... :(

[quote="lotus87"]I thought problem 2 was the most fun...[/quote]

Yes, it was very interesting.  Are there any other solutions than the two posted (which are mod 13 and mod 19)?

Dr. Rice suggested something could be got from treating the first equation as a quadratic in x^3, but I'm not sure what he was talking about.

#6 actually was one of the most interesting problems I've seen in a long time.  You don't even need to know any difficult math to solve it.  Just logic.

Anyways, I'm in class now, so I should go do my work!

I'm pretty confident I solved problem 2 totally... and I did it by treating the first equation as a quadratic in x^3 by factoring it that way... then observing that x^3 + 1 must be a product of powers of 3 and 7.  But x^3 + 1 = (x+1)((x+1)^2 - 3x) so either x + 1 is 0 mod 3 or the factors are relatively prime.  But the latter yields a very simple contradiction.  After some more manipulation, I reduced the second equation modulo 19 since fermat's little theorem yields z^9 is -1 or 1 mod 19... and the result follows.

I never saw this square thing everyone's talking about lol... and I wonder how it can be solved mod 13.  I never thought of 13, though I remember reading that it's one of the most useful mods.

I did okay.

I solved #1, 3, 4, 5. Hopefully I didn't make any mistakes (especially on #1, which is easily prone to proofing errors).

I thought I solved #2, but I discovered shortly after the exam that it was flawed.

If I were to rank the difficulty of the problems, I would say $4<1<5<3<2<6$.

I honestly do not think #6 was so difficult... do you think perhaps there's a psychological element to it, given that it's the last problem on the paper, which most people attempt last?  I bet if it was #1, more people would have done well on it.  I don't think it really required any difficult math to solve (although maybe some effort to actually figure out what it was asking).

[quote="billzhao"][quote="Elyot"]I don't think it really required any difficult math to solve.[/quote]
But that's also true for every other problem (except perhaps #2).

I think what's difficult about #6 is that it's unorthodox (compare with last year's #3) and requires much ingenuity. A difficult problem often does not have a difficult solution, but one that is hard to find (e.g. APMO 2005 #2).[/quote]

Hence why I failed miserably at #2, which a problem that requires some experience to see what's going on.  I never would have thought to try mod 13 or 19.

#6 is just a problem that you can't practice for.  A lot of IMO-type people practice tons and tons of algebra problems, or inequality problems, or euclidean geometry problems, but nothing can really help you prepare for something like #6 on USAMO.  In a sense, it levels the playing field.  I know it's frustrating for some people, who perhaps studied tons and tons of inequalities in anticipation of a traditional inequality-type question, but IMO it adds to the challenge and interest of the contest.

I actually started #6 with about an hour to go in the contest, and confident that I wasn't going to get it, I just started writing it up as I went, making easy discoveries, some of which were irrelevant, but many of which actually led to the solution.  I was quite astonished when I realized that I had figured out what f(n) was, and I was pretty amazed to realize that I had actually solved it with about 10 minutes to go.

I love problems like that.  I think USAMO was my favourite contest this year because of having so many problems like that.  Especially something like #5, which a grade 6 or 7 kid could solve with enough ingenuity.

CONGRATULATIONS TO YUFEI AND PENG!!!!  :D  :D


[quote="amcavoy"][quote="Abhishek"]Let  X denote the set of all triples (a,b,c) of integers.

 Define a function f:X to X by
                f(a,b,c) = (a+b+c, ab+bc+ca, abc)

 Find all triples a,b,c in X such that
                f ( f (a,b,c) ) = (a,b,c)[/quote]
Just regarding the function:  Shouldn't that be $f: \mathbb{R}^{3}\rightarrow\mathbb{R}^{3}$?[/quote]

Well, you could just use $\mathbb{Z}^3$ instead of $X$.  :D

[quote="Abhishek"]Let  X denote the set of all triples (a,b,c) of integers.

 Define a function f:X to X by
                f(a,b,c) = (a+b+c, ab+bc+ca, abc)

 Find all triples a,b,c in X such that
                f ( f (a,b,c) ) = (a,b,c)[/quote]

[hide]If, $a+b+c=k$, $ab+bc+ca=k_1$, $abc=k_2$, $\{k,k_1,k_2 \in \mathbb{Z}\}$, we got:

$f(f(a,b,c))=f(a+b+c,\ ab+bc+ca,\ abc)=(k+k_1+k_2,\ kk_1+kk_2+k_1k_2,\ kk_1k_2)=(a,b,c)$

$\Longleftrightarrow k+k_1+k_2=a$ but: $k+k_1+k_2=(a+1)(b+1)(c+1)-1$ so:

$(a+1)(b+1)(c+1)=a+1 \implies (b+1)(c+1)=1$, since $b$ and $c$ are integers the equality holds when $b=c=0$ or $b=c=-2$.

$\Longleftrightarrow kk_1+kk_2+k_1k_2=b$ and $kk_1k_2=c$:

if $b=c=-2 \implies$  $k=a-4$, $k_1=-4(a-1)$, $k_2=4a$ then:

$(a-4)(-4(a-1))+(a-4)(4a)+(-4(a-1))(4a)=-2$ and the results follows...[/hide]

Posted and solved in [url]http://www.mathlinks.ro/Forum/viewtopic.php?t=73400[/url]


geometry
inequalities
inequalities unsolved

Prove that in all triangles  $ \sqrt {AI} \plus{} \sqrt {BI} \plus{} \sqrt {CI}\le3\sqrt {R}$ where I is the center of incircle and R the circumradius.

Sorry everybody now I edited.

[quote="ridgers"]Prove that in all triangles  $ \sqrt {AI} \plus{} \sqrt {BI} \plus{} \sqrt {CI}\le\sqrt {R}$ where I is the center of incircle and R the circumradius.[/quote]
It's obviously wrong! Try $ AI\equal{}BI\equal{}CI\equal{}R.$ :wink:

[quote="ridgers"]Prove that in all triangles  $ \sqrt {AI} + \sqrt {BI} + \sqrt {CI}\le3\sqrt {R}$ where I is the center of incircle and R the circumradius.

Sorry everybody now I edited.[/quote]

1. $ \boxed{AI + BI + CI \le \frac{1}{\sqrt3}( AB+BC+CA )}$

$ AB=x+y$ , $ BC=y+z$ , $ CA=z+x$

$ AI = \sqrt {r^2+x^2}$ ,$ BI = \sqrt {r^2+y^2}$ , $ CI = \sqrt {r^2+z^2}$

$ r^2= \frac{xyz}{s}$ , ( $ s=x+y+z$- semiperimeter $ \triangle ABC$ )

$ \Leftrightarrow \sqrt{\frac{xyz}{s}+x^2}+\sqrt{\frac{xyz}{s}+y^2}+\sqrt{\frac{xyz}{s}+z^2}\le \frac{2}{\sqrt{3}} s$

$ LHS \le \sqrt{xy+yz+zx+s^2} \le \frac{2}{\sqrt {3}}s$

2. $ \sqrt {AI} + \sqrt {BI} + \sqrt {CI} \le 3 \sqrt{\frac{AI+BI+CI}{3}} \le \sqrt[4]{3} \sqrt{2s} \le 3\sqrt{R}$ , $ ( 4s^2 \le 27R^2 )$

If $ x\equal{}sin\frac{A}{2} , y\equal{}sin\frac{B}{2} , z\equal{}sin\frac{C}{2}$ inequality equivalents to $ \sqrt{xy}\plus{}\sqrt{yz}\plus{}\sqrt{xz} \leq \frac{3}{2}$

$ \sqrt{xy}\plus{}\sqrt{yz}\plus{}\sqrt{xz} \leq x\plus{}y\plus{}z \leq \frac{3}{2}$ :wink:


geometry
parallelogram
projective geometry
geometry proposed

Let $ E$ and $ F$ be points on the sides $ AD$ and $ CD$ of parallelogram $ ABCD$ such that $ \hat{AEB}\equal{}\hat{AFB}\equal{}90^{\circ}$, and $ G$ be the point of $ BF$ for which $ EG \parallel{} AB$.If $ H \equal{} AF \cap BE$ and $ I \equal{} DH \cap BC$,prove that $ FI \perp GH$.

Strange problem :D

Let $ K$ denotes the intersection of $ FH$ and $ BI$.

We have $ \triangle BFK \sim \triangle HEA \Rightarrow \frac{KF}{AE}\equal{}\frac{KB}{AH}$

On the other hand, since $ DE//BI$, Thales's theorem yields:
$ \frac{KI}{AD}\equal{}\frac{KB}{AE}$

Hence, divide these two expression above side by side, we obtain:
$ \frac{KF}{KI}.\frac{AD}{AE}\equal{}\frac{AE}{AH}\Rightarrow \frac{KF}{KI}\equal{}\frac{AE^2}{AD.AH}$

We also have:
$ \frac{BG}{BH}\equal{}\frac{BG}{BF}.\frac{BF}{BH}\equal{}\frac{AE}{AD}.\frac{AE}{AH}\equal{}\frac{AE^2}{AD.AH}$

Thus: $ \frac{KF}{KI}\equal{}\frac{BG}{BH}$

But $ \widehat{FKI}\equal{}\widehat{FAE}\equal{}\widehat{HBG}$ so $ \triangle HBG \sim \triangle IKF$. This leads to $ \widehat{BHG}\equal{}\widehat{FIK}$.

Finally, let $ J$ denotes the intersection of $ IF$ and $ GH$, since $ \widehat{BHG}\equal{}\widehat{FIK}$, $ H,J,K,I$ are concyclic so $ \widehat{HJI}\equal{}\widehat{HBI}\equal{}90^o$, which ends our proof.

Dear Mathlinkers,
thanks to Mashimaru for the nice metric solution.
Sometimes it is possible to find a synthetic one...
1. Let U, V be the meet points of AD and BF, of DH and AB.
2. According to Pappus's theorem applied to the hexagon UDFIBVU, 	UV // FI.
3. Let T be the meet point of UV and AF.
4. According to Pappus's theorem applied to the hexagon UDFIBVU,	ET // VB ;
    in consequence,	ET = EG   and   E, T et G are collinear.
5. Let 1	the circle with diameter AB, 2 the circle with diameter UH,
Tu, Th the tangents to 2 at U, H,
S the second meet point of UV with 2
and G' the meet point of HS and UF.
6. H is the orthocenter of UAB and Tu, Th, AB are parallels.
7. According to Pascal's theorem applied to the degenerated hexagon Th SUTu FH, 	TG' // AB and G' = G.
8. We have	HSG   US  and USV // FI.
9. We are done.

Sincerely
Jean-Louis


How many 4-letter words with at least one vowel can be constructed from the letters A, B, C, D, and E?  (Note that A and E are vowels, any word is valid, not just English language words, and letters may be used more than once.)

Total words: $ 5^4$

Words without vowels: $ 3^4$

Words with vowels is thus: $ 5^4\minus{}3^4\equal{}625\minus{}81\equal{}\boxed{544}$.

That's what I did and then I tried doing it a different way.

I did 4 * 2 * 5^3 for 4 places to put one of the two digits and 5^3 for 5 choices of letters in the 3 remaining positions.  This gives 1000.  o_O  I know this method is wrong but I can't exactly word why.  Help please?

You can't do that--you have to use complementary counting.
In your solution, you're overcounting.

If you don't want to use complementary counting you can do casework. Truffles' method places a vowel in only one position, while there can be 1 to 5 vowels and multiple positions.


[quote="tkhoa2711"]please tell me more about this book !!!
thank you!!![/quote]
Maybe this will help:
[url]http://www.mathlinks.ro/Forum/viewtopic.php?t=83302[/url]
[url]http://www.mathlinks.ro/Forum/viewtopic.php?t=63226[/url]
[url]http://www.mathlinks.ro/Forum/viewtopic.php?t=10924[/url].  :D


$ \int {\frac{{dx}}{{\sqrt {2 \minus{} \cos x} }}}$

$ \int {\frac {{dx}}{{\sqrt {2 \minus{} \cos x} }}}\equal{}\int \frac{\sqrt{2\plus{}\cos x}}{\sqrt{4\minus{}\cos ^ 2 x}}dx$

[quote="kunny"]$ \int {\frac {{dx}}{{\sqrt {2 \minus{} \cos x} }}} \equal{} \int \frac {\sqrt {2 \plus{} \cos x}}{\sqrt {4 \minus{} \cos ^ 2 x}}dx$[/quote]
[color=blue][b]Is it a nice solution? But it is not still ended up perfectly. How can? May u finish ur solution?
Thanks and best regards,  [/b][/color]

[quote="ith_power"][quote="quocbao153"]$ \int {\frac {{dx}}{{\sqrt {2 \minus{} \cos x} }}}$[/quote]

I dont think the integral has solution in known elementary functions. :maybe:[/quote]

Yes, the integral is elliptic integral of the 1st kind.   :!:


geometry
trigonometry
calculus
search
Columbia
abstract algebra
MATHCOUNTS

Let's say you're serious about learning math. You don't have a lot of time to take courses, such as geometry, trigonometry, or calculus. Of course, you still have to get through all of those. So, you decide to read, study, and work hardcore at those topics and others.

Have you ever lived such a scenario?
Have you ever decided to get through math topics with self-study?

How long would you say it took you to get through a topic such as calculus or other with self-study?

Did you go through 30 or more pages a day?

I'd like to know your story if you've ever done self-studying and found a way to quickly excel at asorbing math and moving onto more math levels.

[quote="Genecks"]How long would you say it took you to get through a topic such as calculus or other with self-study?

Did you go through 30 or more pages a day?[/quote]
Saying something about pages is meaningless since different books have different amounts of content on each page. For example, some of the books on geometric measure theory that I've been reading lately take me 5 min or more per page, so I only go through a few pages each day.

Don't be so keen to blast through all of these topics. You need good foundations in elementary math in order to excel in advanced math. If your goal is to learn the BC Calculus curriculum in 1 month, you might be able to do it -- but would you retain the information well? Would you be able to do the problems quickly? Would you be able to figure out more difficult problems that use the same tools but go beyond anything you saw while studying? Be aware that there are serious costs associated with rushing through your studies.

i do a lot of self-study. if i didn't, i wouldn't know a fraction of what i do. sometimes, i'll search the internet or just read ahead in my math book.

i'm not really sure how long it takes me to study... I'd say when I was reading about calculus, i kind of stopped reading it for a while and then got back into it after like a month. so its hard to count the time.

most of what i learned from self-study was from the AoPS books; i'd say i read volume 2 in about a week or so.

edit- agreeing with what Xevarion just said, while the AoPS book had such great tools, I went through it rather quickly and didn't pay much attention to the problems, so i often find it difficult to retain some of the information i learned. i can think of a recent occurence where i missed a question of a competition because i could not remember a formula from it.

Well... I studied AP Stats in two weeks, took the AP test, and got a 5... of course now I remember nothing, so I wouldn't count it as actually learning.

I self-study lots of higher math, but I don't do it at a strict pace.  I take books out from the Columbia math library (not textbooks) and read them leisurely.  I do it for enjoyment and to develop a basic understanding of the subject, not to develop a textbook knowledge which would enable me to pass a test.  I figure the rigor will come when I study the subject in college.

I used to teach my myself A-level modules (of which you would take 6 in 2 years, if you were taking single maths) from the textbook, and take a practise exam paper when I thought I'd understood it all (sometimes less than a day if I worked hard enough).

That was actually suprisingly effective, and great when I needed to learn for an upcoming exam; but in the long run, of course I forgot it all.

Learning too much too quickly never worked for me. What I found helpful was looking through a topic (slightly above your head) you don't understand, trying to get familiar with it. Later on when I saw something related to that topic, I'd have some idea what was going on, until eventually I got it.

I find that I can learn a concept very quickly, but it takes time for it to solidify in my head as common knowledge (if that makes any sense?) So usually I will teach myself something, use it but not be 100% confident with it, and after some period of time has elapsed (it could be a week, a month, or even half a year as ended up happening with some aspects of calculus) I will go back and learn it rigorously and permanently and then even if I forget it, I am able to derive it. So as everybody else has said, take as much time as is needed to fully grasp a concept so that it truly becomes useful to you. I find (as is the case with my AoPS work right now) if I rush through too much math too fast, it all floats at the surface of my brain and it's all I can think about when solving contest problems even if the solution is more elementary than the stuff I just learned.

Yeah I got bored one day and decided what the big deal was with $\sin$ and $\cos$ was...

i read the chapters in my Alg 2 book and went on for 2 weeks until i reached half-angle forumlas...which i promtly took a break due to the Mathcounts competition  :D

[quote="Treething"]Well, when will you really need to memorize the Taylor series for cos(x) or the formula for bell curves? I see nothing wrong with cramming for some test and promptly forgetting, as once you've learned something, once you relearn it, it'll come back to you very very quickly.[/quote]

I totally hold with this.  In Calc B/C last year, I never bothered to memorize the formula Taylor series chicken toothbrush Congress whatever.  So 5 minutes before the AP test I memorized them really fast, and I was fine.  But 15 minutes after the end of the test, they were all gone, and right now I can barely tell you what a derivative is.  (Can I?  Actually, nope.)

Easier to derive the Taylor series for cos(x), if you need it.

[quote="Piggypi314"]I totally hold with this.  In Calc B/C last year, I never bothered to memorize the formula Taylor series chicken toothbrush Congress whatever.  So 5 minutes before the AP test I memorized them really fast, and I was fine.  But 15 minutes after the end of the test, they were all gone, and right now I can barely tell you what a derivative is.[/quote]
Wow piggy, I completely agree with you, cause I did the same thing basically.  So Zenecks, I guess you should just read, and practice a LOT of problems if you actually plan on remembering everything you read.  How long it takes depends on your current mathematical level.

[quote="Xevarion"]
Saying something about pages is meaningless since different books have different amounts of content on each page. For example, some of the books on geometric measure theory that I've been reading lately take me 5 min or more per page, so I only go through a few pages each day.
[/quote]
5 minutes to read a page of advanced math? That sounds like extreme speed to me. I think almost all literature I've read has been quite compact (usually a weeks worth of reading for a class has been at the very most 8 pages), but 5 minutes still sounds like amazing speed unless you're reading some crappy america-authored verbose exposition. Of course, most time "reading" is spent thinking.

Well, I wasn't counting if I stop and work out an exercise or try to prove some result the author mentions. If you add that in, some pages could take hours!


function
algebra
polynomial
Rational Root Theorem

Which of the following describes the 4 roots of the function?

$f(x)=x^4-2x^3+7x-4$

A. 4 Irrational
B. 2 Irrational and 2 Rational
C. 2 Irrational and 2 Real Roots
D. 2 Rational and 2 Non-real

How would do this? I could use the Rational Root Theorem and eliminate any possible rational roots (so D is out). But using Descrates' rule of sign, there could be 4 possible real roots. What do I do next?

thanks. practicing for upcoming regional competition.

[hide]According to the answer sheet, the answer is C.[/hide]

[hide="Hint"]The function factors to $f(x)=(x^2 -3x+4)(x^2 +x-1)$.[/hide]

oh woops. didn't even try to factor the equation. thanks.


Ring Theory
quadratics
parameterization
number theory unsolved
number theory

Can we find all integer solutions to $ a^2 \plus{} b^2 \equal{} 2c^2$
[hide="some solutions"] $ a\equal{}b\equal{}c\equal{}k$ is obviously a solution, but I couldn't generalize solutions like $ (a,b,c) \equal{} (7,1,5)$[/hide]

Write $ a^2 \minus{} 2c^2 \equal{} (1 \minus{} 2) b^2$.  $ \mathbb{Z}[\sqrt {2}]$ is a UFD, so we can write

$ (a \plus{} \sqrt {2} c)(a \minus{} \sqrt {2}c) \equal{} (1 \plus{} \sqrt {2})(1 \minus{} \sqrt {2})b^2$.

Let $ u \equal{} \frac {a \plus{} \sqrt {2} c}{1 \plus{} \sqrt {2}}$.  Then $ u \bar{u} \equal{} b^2$, and because of unique factorization we know that $ b$ must be of the form $ v \bar{v}$ for some $ v \in \mathbb{Z}[\sqrt {2}]$.  Writing $ v \equal{} p \plus{} q \sqrt {2}$ we have

$ u \equal{} (p \plus{} \sqrt {2}q)^2 \equal{} (p^2 \plus{} 2q^2) \plus{} 2pq \sqrt {2} \implies$
$ a \plus{} \sqrt {2} c \equal{} (p^2 \plus{} 4pq \plus{} 2q^2) \plus{} (p^2 \plus{} 2pq \plus{} 2q^2) \sqrt {2}$

so the set of solutions is $ (p^2 \plus{} 4pq \plus{} 2q^2, p^2 \minus{} 2q^2, p^2 \plus{} 2pq \plus{} 2q^2)$.  Alternately, there is a more general technique for any homogeneous quadratic using [url=http://www.artofproblemsolving.com/Forum/weblog_entry.php?t=197236]rational parameterization[/url].

[quote="gunes"]Can we find all integer solutions to $ a^2 \plus{} b^2 \equal{} 2c^2$
[hide="some solutions"] $ a \equal{} b \equal{} c \equal{} k$ is obviously a solution, but I couldn't generalize solutions like $ (a,b,c) \equal{} (7,1,5)$[/hide][/quote]
We can skip the solutions $ a\equal{}b\equal{}c$.. We are assuming $ c \in \mathbb{N}$
Assume $ a \ge b$. If $ a\equal{}b$ then $ a\equal{}b\equal{}c$, but we don't want these solutions so $ a>b$. The equation is equivalent to:
$ a^2\minus{}c^2 \equal{} c^2\minus{}b^2$. Obviously $ a>c>b$. So
$ (a\minus{}c)(a\plus{}c) \equal{} (c\minus{}b)(c\plus{}b)$. Let $ a \equal{} c\plus{}m$ and $ b \equal{} c\minus{}n$. $ n,m \in \mathbb{N}$ Then we get:
$ m(2c\plus{}m) \equal{} n(2c\minus{}n)$. So $ n > m$
It's equivalent to:
$ 2c(n\minus{}m) \equal{} m^2 \plus{} n^2$ So by $ \bmod 4$: $ n$ and $ m$ is even. Let $ n \equal{} 2x$ and $ m \equal{} 2y$:
$ c \equal{} \frac{x^2 \plus{} y^2}{x\minus{}y}$
So we want to find $ x$ and $ y$ so $ x\minus{}y \mid x^2 \plus{} y^2$ $ \iff$ $ x\minus{}y \mid 2x^2$
I'm not sure about whether there is an easy way to find all such $ x,y$. But once we have, we get the solutions:
$ a \equal{} c \plus{} m \equal{} c \plus{} 2y \equal{} \frac{x^2 \minus{} y^2 \plus{} 2xy}{x\minus{}y}$, $ b \equal{} c \minus{} n \equal{} c \minus{} 2x \equal{} \frac{\minus{}x^2 \plus{} y^2 \plus{} 2xy}{x\minus{}y}$
So all solutions is given by:
$ \left ( \pm \frac{x^2\minus{}y^2\plus{}2xy}{x\minus{}y}, \pm \frac{\minus{}x^2\plus{}y^2\plus{}2xy}{x\minus{}y}, \pm \frac{x^2\plus{}y^2}{x\minus{}y} \right )$. (and by changing $ a$ and $ b$)

t0rajir0u was a bit faster than me, but you can read it if you like to :P
(Btw. Can we use your solution to find all $ x,y$ such that $ x\minus{}y \mid x^2 \plus{} y^2$ ?)

[quote="nayel"]Here's a hint for a one-line solution: $ a^2 \plus{} b^2 \equal{} 2c^2$ can be written as $ \left(\frac {a \plus{} b}{2}\right)^2 \plus{} \left(\frac {a \minus{} b}{2}\right)^2 \equal{} c^2$. :wink:[/quote]
Lol... I feel so embarrased  :blush:  :blush:  :blush:


can you work out the greatest of the possible many (may be $ 1$ only) of less than one hundred digit numbers with all the digits identical (ie $ 2222222\cdots$ etc ) which is perfectly divisible by a three digit number having consecutive digits, most likely in ascending order (eg $ 123$ or $ 234$ or $ 345$ etc).

[quote="SimonM"]Claim, the largest possible number with identical digits $ < 10^{100}$ is $ 10^{100} \minus{} 1$ is divisible by 123

First we investigate $ 10^n \mod 123$ for small values of $ n$

We obtain $ 10, 100, 16, 37, 1$

Since $ 5|100$ then $ 10^{100} \equal{} 1$ (mod $ 123$).

Hence $ 10^{100} \minus{} 1 \equal{} 0$ (mod 123) $ \Rightarrow 123|(10^{100} \minus{} 1)$[/quote]

i wanted less than $ 100$ digits. $ 10^{100} \minus{} 1$ has $ 100$ digits.


This is my new ebook at Maths.vn forum. In this book, I present my collection about inequality of my friend: Tran Quoc Luat at his topic: http://www.maths.vn/forums/showthread.php?t=16721 (of course, I am the chef editor  :lol: ) with the  solutions contributed by

1) Vo Quoc Ba Can (nothing, can_hang2007) - me

2) Ngo Duc Loc (Honey_suck)

3) Tran Quoc Anh (nhocnhoc, nguoivn)

4) Seasky

5) Materazzi

You can enjoy it here:  http://canhang2007.wordpress.com/2008/12/30/new-ebook-of-me-at-mathsvn-forum/

Regards

Vo Quoc Ba Can

P/s: Please tell me your idea after reading it (about the content and designed). Thanks :)

Thanks!  :)
You always share your knowledge with us. This is just great.
Inequalities are very nice by the way. My favorite inequality in this ebook is number 10.

[b]Show that if $ a,b,c$, are positive real numbers, then:[/b]
\[ \frac {a}{a \plus{} b} \plus{} \frac {b}{b \plus{} c} \plus{} \frac {c}{c \plus{} a}\geq 1 \plus{} \sqrt {\frac {2abc}{(a \plus{} b)(b \plus{} c)(c \plus{} a)}}
\]

Thank you, Can! It is very nice of you to share your material for free  :)


why can't you do an two digit exponent (i.e. 2^(35)) in $ LaTeX$?

Use {} to denote the exponent $ 2^{35}$; oh and it's $ \text{\LaTeX}$

Also, when you're on the forum you can click on images others have created (or use the quote button) in order to see how they've done the things they've done.  Thus, to answer your question you could have just found a post in which there was a complicated exponent and used one of the methods I suggested in order to find out how to do it yourself.


inequalities
LaTeX
trigonometry
inequalities proposed

Prove that in any triangle ABC

     :sqrt: (bc)/a cos(A/2) +  :sqrt: (ca)/b cos(B/2) +  :sqrt: (ab)/c cos(C/2) >=(3 :sqrt: 3)/2

hello, can you write down your inequality in $ \text{\LaTeX}$ please?
Sonnhard.

I think the problem is like this

$ \frac{\sqrt{bc}}{a \cos \frac{A}{2}}\plus{}\frac{\sqrt{ca}}{b \cos \frac{B}{2}}\plus{}\frac{\sqrt{ab}}{c \cos \frac{C}{2}}\ge\frac{3\sqrt3}{2}$

isn't it ?

hello we need only AM GM and jensen inequality to kill it.


all $x_{i}$should be equal from cauchy
so $a=2007^{2}$ :wink:

[quote="maokid7"]i do not see how cauchy helps in this case.[/quote]

Use $\sum = \sum_{i=0}^{2007}$. By Cauchy \[a^{4}= \left(\sum x_{i}\right)\left(\sum x_{i}^{5}\right) \geq \left(\sum x_{i}^{3}\right)^{2}= a^{4},\] so equality occurs, so $\frac{\sqrt{a_{i}}}{\sqrt{a_{i}^{5}}}= \frac{1}{a_{i}^{2}}$ is constant, i.e. all $a_{i}$ are equal.

ok thanks. i for got to square the term on the LHS of the inequality.  :oops:

I had a question about a similar problem:
I will post it tomorrow, it is the last problem in chapter 13 of aops vol. 2
i think i found a 2 second solution with cauchy but i'm not sure it actually holds


7 states have posted in the Missouri Forum! YAY!!!! Keep it up! :clap2:  :thumbup:  :thumbup:

er...that still doesn't beat BA if the contest is still going


Imagine that one has 10 bags with unlimited supply of coins (as many as you want) in each bag. In each one of the bags, all of the coins have the same weight which is an integer  :le: 10g. Using only one weighing on a scale one must find the weight of the coins in each bag.

So we can have two bags both with coins weighing 5g?  I'm only asking this because I have no clue how to do this :-)

[quote="probability1.01"]So we can have two bags both with coins weighing 5g?[/quote]

Yes, you can.

Aha!

I know now, so I'll just make an assumption/hint:

[hide]0 can not be a coin weight, both for obvious and non-obvious reasons.[/hide]

Unless I missed a better solution.

Oh yeah, I didn't realize that you could [hide]do base 11, not just base 10.[/hide]

[hide]What if we take one coin from the first bag, 11 from the second, 121 from the third, 1331 from the fourth, etc.?[/hide]

Good job, ComplexZeta. The reason I posted this problem in the [b]Advanced Problems[/b] is I saw a similar problem (It has been moved) in this forum. This problem probably belongs to [b]Intermediate Topics[/b] too.

ComplexZeta, you must know base-11 numbers very well. How many fingers do you have? :-)

This problem is inspired from the fact that the coefficients of a polynomial f can be recovered from a [b]single value[/b] f(n) for some sufficiently large positive integer n provided that all the coefficients of f are non-negative integers.

You may view this as any positive integer has a [b]unique[/b] base-n representation.

One of my cats has six claws on one paw for a total of eleven claws on the front paws. :)

I basically tried to do something like what you were saying: I figured that I could take so many of each bag compared to all the previous ones that we could figure out all of them. Then I realized that powers of eleven were good enough.


The numbers a1, a2,...,an have the values of 1, 1/2, 1/3,..., 1/n
(not necessarily in that order), and so do b1, b2,...,bn.
Moreover, a1 + b1>eq a2 + b2>eq...>eq an + bn. Prove that
am + bm<eq 4/m for all m's from 1 to n

Let $A$ (resp. $B$ ) be the set of the subscripts ${k \in \{1,...,m}$ such that $a_k \leq b_k$ (resp. $b_k \leq a_k$).
Since $A \cup B = \{1,...,m \}$, it follows that at least one the sets $A,B$ contains at least $ \frac {m} {2}$ elements.
Wlog, we may assume that $|A| \geq \frac {m} {2}$. (1)

Now, let $b_p$ be the least of the $b_k$'s for $k \in A$.
Then $p \leq m$ and $a_p \leq b_p$.

Moreover, since $\{ b_1,...,b_n \} = \{ 1, \frac {1} {2},..., \frac {1} {n}$ and from (1), we deduce that $b_p \leq \frac {1} {\frac {m} {2}} = \frac {2} {m}$.
Thus :
$a_m+b_m \leq a_p+b_p \leq 2b_p \leq \frac {4} {m}$.

Pierre.

Our friend Iris Aliaj seems to have real love for russian problems :D

Pierre.

Well in fact i didn't know that this is a russian problem(and such an old one).As i told you i've got this from a site but...Albania has been a comminist country like Romania and many others and our parents and teachers always say that the russian educational system was a very strong,hard one.In addition, e few days ago I saw the IMO 2002 results and all the members of the russian team had got a gold medal.So why not try to profit from their experience?
I don't have real love only for russian problems but for all math problems. :)


[quote="Mr.JN2P"]For a, b, c >0. Prove that
$a^{2}+b^{2}+c^{2}+2abc+1 \geq 2ab+2bc+2ca$[/quote]

See Darij's post on

http://www.mathlinks.ro/Forum/viewtopic.php?search_id=575919256&t=19666

conceive


[quote="Nea"]Prove if $n,d,k \in N$ and $\left. d\right|n$
$\varphi (nd^k ) = d^k \varphi (n)$[/quote]
Is that supposed to represent the Euler phi function?

Given that $d | k$, we have $\phi(nd^k) = (nd^k) \times \prod_{p | nd^k} \frac{p-1}{p}$, where $p$ is prime.  We have $d^k \phi(n) = d^k (n) \times \prod_{p | n} \frac{p-1}{p}$.  Since the set of all $p$ such that $p | nd^k$ is equal to the set of all $p$ such that $p | n$, the $\sum$ in both equations is equal, and the factor $nd^k$ is also equal, so we have equality and the desired result.

[quote="Nea"]Why the set of all $p$ such that $p | nd^k$ is equal to the set of all $p$ such that $p | n$ ?[/quote]

We are already given that $d | n$, so $p | d \implies p | n$, which means that adding extra $d^k$ factors will not affect divisibility (remember that $p$ is prime).


How do you find the direction of a particle or when it changes direction if you were give the velocity function?
the velocity function could be -(x+1)sin((x^2)/2).
thankx

If $v(t)>0$, then the particle is moving to the positive direction at time $t$.
If $v(t)<0$, then the particle is moving to the negative direction at time $t$.

If $v(t)=0$ and $v'(t)\ne 0$, then the particle is changing its direction.
(For instance, if $v(t)=0$ and $v'(t)>0$, then the particle is changing its direction from negative to positive. This is because $v'(t)>0$ means $v(t)$ is increasing at time $t$.)

No, $v(t)$ is a velocity function (in terms of time $t$).

what is a negative direction and what is a positive direction positive and negative in respect to the position function or the velocity


x,y,z are positive reals  such that  \[ xyz\equal{}1\]  prove that:

\[ \frac{x^{2}}{(x\minus{}1)^{2}}\plus{}\frac{y^{2}}{(y\minus{}1)^{2}}\plus{}\frac{z^{2}}{(z\minus{}1)^{2}}\geq 1\]

:)

[hide="Solution"]Squares in numerators suggest using Titu's Lemma.

We have $ \frac {x^2}{(x \minus{} 1)^2} \plus{} \frac {y^2}{(y \minus{} 1)^2} \plus{} \frac {z^2}{(z \minus{} 1)^2}\ge \frac {(x \plus{} y \plus{} z)^2}{x^2 \plus{} y^2 \plus{} z^2 \minus{} 2x \minus{} 2y \minus{} 2z \plus{} 3}$.

It suffices to prove that
\[ \frac {(x \plus{} y \plus{} z)^2}{x^2 \plus{} y^2 \plus{} z^2 \minus{} 2x \minus{} 2y \minus{} 2z \plus{} 3}\ge 1\]

\[ \implies x^2 \plus{} y^2 \plus{} z^2 \plus{} 2xy \plus{} 2yz \plus{} 2zx\ge x^2 \plus{} y^2 \plus{} z^2 \minus{} 2x \minus{} 2y \minus{} 2z \plus{} 3.\]
Since $ xyz \equal{} 1$, we need to prove that $ \frac {2}{x} \plus{} \frac {2}{y} \plus{} \frac {2}{z}\ge 3 \minus{} 2x \minus{} 2y \minus{} 2z$, which is clear after moving $ \minus{} 2x \minus{} 2y \minus{} 2z$ to the LHS and applying AM-GM. $ \blacksquare$

Furthermore, equality never occurs.[/hide]

[quote="b.s.o"]x,y,z are positive reals  such that
\[ xyz \equal{} 1\]
prove that:
\[ \frac {x^{2}}{(x \minus{} 1)^{2}} \plus{} \frac {y^{2}}{(y \minus{} 1)^{2}} \plus{} \frac {z^{2}}{(z \minus{} 1)^{2}}\geq 1\]
:)[/quote]

this is an easy (by easy, I mean [b]really easy[/b]) version of [url=http://www.mathlinks.ro/viewtopic.php?t=215222]IMO 2008, Problem 2[/url].

notice that $ \frac{x^2}{(x\minus{}1)^2}\ge 1$ for $ x \ge \frac{1}{2}$. It is obvious to show that at least one of the three terms in the $ LHS$ is greater than $ 1$, and we are done.

the problem is really easy for an IMO one:
put that , $ a\equal{}\frac{x}{x\minus{}1}$ , $ b\equal{}\frac{y}{y\minus{}1}$ , $ c\equal{}\frac{z}{z\minus{}1}$
then the inequality becomes to prove that $ a^2\plus{}b^2\plus{}c^2\geq 1$ with $ a\plus{}b\plus{}c\equal{}ab\plus{}bc\plus{}ca\plus{}1$
wish true because we have $ a^2\plus{}b^2\plus{}c^2\equal{}(ab\plus{}bc\plus{}ca)^2\plus{}1\geq 1$
equality holds when $ ab\plus{}bc\plus{}ca\equal{}0\Leftrightarrow xy\plus{}yz\plus{}zx\equal{}3xyz$


probability
limit
function
calculus
integration
real analysis
Support

suppose that a “die” has infinitely many faces marked  $ 1, 2, 3, \ldots$. When the die is thrown,  the probability is $ a_i > 0$ that face $ i$ appears. Let $ p_n$ be the probability that after some number of tosses the sum of the numbers that have appeared is exactly $ n$. Find $ \lim_{n \to \infty} p_n$ when this limit exists.
(I think I have a solution, but I don't know how to determine when this limit exists ?)

Seems like a typical generating function problem to me :). If $ F(z)\equal{}\sum_{k\equal{}1}^\infty a_kz^k$ and $ G(z)\equal{}\sum_{n\equal{}1}^{\infty}p_n z^n$, then, considering where we could be one move before we reach $ n$, we get $ G(z)\equal{}F(z)(1\plus{}G(z))$ and $ G(z)\equal{}\frac{F(z)}{1\minus{}F(z)}\equal{}\minus{}1\plus{}\frac{1}{1\minus{}F(z)}$. Thus we are down to the question when the limit of the Taylor coefficients of the last expression exists and what it is. I do not want to spoil the fun completely, so I'll just make the first couple of steps. Note that $ \Re \frac 1{1\minus{}F(z)}$ is a non-negative harmonic function in the unit disk, so it must be the Poisson integral of some non-negative Borel measure. The Taylor coefficients we are looking for are essentially the same as the Fourier coefficients of that measure (up to a factor of $ 2$ or something like that). Now, the measure can have an absolutely continuous part, which doesn't bother us because of the Riemann-Lebesgue theorem, and a singular part. So here comes the main question: what can be said about the singular part of this measure?

No, Kent. The function is not in $ H^1$. Only the real part is non-negative...

yes, I was stuck here. It is clear that if one allows the probabilities $ a_i$ to be $ 0$, then it is possible that $ \lim p_n$ does not exist. And if the limit $ \lim p_n \equal{} L > 0$ exists and $ \sum k a_k < \infty$, it is quite easy to see that $ G(x) \approx \frac{1}{(1\minus{}x) \sum k a_k}$ and $ \frac{F(x)}{1\minus{}F(x)} \approx \frac{L}{1\minus{}x}$ near $ 1^\minus{}$, so $ L \equal{} \frac{1}{\sum_k k a_k}$ (which is quite intuitive since the jump sizes are like $ \sum k a_k$). But I have the feeling that if $ a_i>0$ for all $ i  \geq 1$, then the limit always exists ?

Actually, the condition that the greatest common divisor of all $ i$ with $ a_i>0$ is $ 1$ is sufficient for the existence of the limit. As I said, it boils down to the investigation of the singular part of some measure. If you are still stuck, let me be more specific in my last question. Assume that we know the Poisson integral of a finite positive measure (as a function in the open unit disk). How to determine the support of its singular part?

As one friend told me, this problem is well known among probabilists. Here is what I have just learned:

Suppose that $ \{ X_n \}_{n}$ are iid random variables such that $ P(X_n \equal{} k)\equal{}a_k$ and $ S_n \equal{} \sum_1^n X_k$. One wants to study the probability $ p_n \equal{} P( \exists k \geq 0 : S_k \equal{} n )$, and we will suppose that $ 0 < E(X_1) \equal{} \mu < \infty$. One can think of $ X_1,X_2, \ldots$ as the lifetimes of replaceable components in a system, such as light bulbs in a light socket. Keeping that in mind, one can define $ T(n) \equal{} \inf\{ m \geq 1 : S_m \geq n\plus{}1 \}$ (next time of break) and $ R(n) \equal{} T(n) \minus{}n$ (residual lifetime). This will be useful for us because $ p_n \equal{} P(R(n\minus{}1) \equal{} 1)$. So now it remains to study the random variables $ \{ R_n \}_n$ and show that $ \lim P(R_n\equal{}1)\equal{} \frac{1}{\mu}$. In fact something stronger appears: $ R_n$ converges in law towards a random variable $ R_{\infty}$ as $ n$ increases.

If the process $ \{ R_n \}_n$ were stationary (i.e. if every $ R_n$ had the same law) it would be trivial  :lol: So the idea is to choose a random variable $ S^*_0$ such that if one defines $ S^*_n \equal{} S^*_0 \plus{} \sum_1^n X^*_n$, where $ \{ X^*_n \}_{n}$ are iid random variables such that $ P(X^*_n \equal{} k)\equal{}a_k$, then the associated residual lifetime process $ R^*_n$ would be stationary. So suppose for a moment that one knows how to find $ S^*_0$ and that one knows the law of $ R^*_n \stackrel{law}{\equal{}} R^*_{\infty}$. A nice coupling argument allows to conclude easily: one can suppose that $ \{X^*_k\}_k$ is independant of $ \{X_k\}_k$, define $ N \equal{} \min\{n \geq 1 : S_n \equal{} S^*_n\}$ the first time that $ S$ meets $ S^*$ ($ N<\infty$ almost surely), and the coupled process $ \hat{S}$ defined by:
$ \hat{S}_n \equal{} S^*_n$ if $ n \leq N$ and $ \hat{S}_n \equal{} S_n$ if $ n > N$.
Then it is easy to show that $ \hat{S}$ is also stationary with the same law as $ S^*$. Because the two sequences $ S_n$ and $ \hat{S}_n$ coincide for $ n \geq N$, the residual lifetimes $ R_m$ and $ \hat{R}_m$ also coincide for $ m \geq S_N$. Hence $ |P(R_n\equal{}x) \minus{} P(\hat{R}_n\equal{}x)| \leq P(n < S_N)$ and $ P(n < S_N) \to 0$ because $ N < \infty$ almost everywhere. The problem is solved because one assumed that one knew the stationary law $ R^*_n \stackrel{law}{\equal{}} R^*_{\infty}$ associated to the stationary process $ S^*_n$: $ \lim_{n \to \infty} P(R_n\equal{}x) \equal{} P(\hat{R}_{\infty}\equal{}x)$

details:
1: check that $ S^*_0$ defined by $ P(S^*_0 \equal{} n) \equal{} \frac{1}{\mu} \sum_{n}^{\infty} a_k$ works i.e. for any time $ m \geq 1$ and any value $ n \geq 1$,  $ P(S^*_0 \equal{} n) \equal{} P(R^*_m \equal{} n)$

2: $ N<\infty$ almost surely because $ S_n \minus{} S^*_n$ is a random walk on $ \mathbb{Z}$ whose increment have $ 0$ expectation, so hits $ 0$ almost surely

3: check that $ \hat{S}$ has the same law as $ S^*$ (kind of Markov property applied to the stationary process $ S^*$ and the stopping time $ N$)

4: $ P(R^*_{\infty}\equal{}1)\equal{}\frac{1}{\mu} \sum_{1}^{\infty} a_k \equal{} \frac{1}{\mu}$, so the conclusion of the problem

Hope I did not bother you too much with these things, but I liked it. The same kind of coupling arguments can show many other convergence results (eg: convergence of Markov chains, same result as above but in the continuous case, etc ..)

Believe it or not, I came up with that coupling argument too, but somehow I thought you were interested in the case $ \mu \equal{} \infty$.

Fedja, could you show us your measure theoretic approach ? I think that everybody (at least me) is curious about it  :)

Here goes:

[b]Step 1[/b]. Let $ p_n$ be the probability that our random process will hit $ n$. It will be convenient to assume that we are at $ 0$ in the beginning, so $ p_0\equal{}1$. For $ n>0$, we can say that
\[ p_n\equal{}\sum_{k\equal{}0}^{n\minus{}1} p_ka_{n\minus{}k}\qquad(*)\]
(this merely means that the probability to hit $ n$ equals the sum over $ k<n$ of probabilities to hit $ k$ and then go directly from $ k$ to $ n$). Since $ p_n\le 1$ for all $ n$, the generating function $ G(z)\equal{}\sum_{n\equal{}0}^{\infty}p_n z^n$ is well-defined for all $ |z|<1$. The equation (*) can be rewritten as $ G(z)\equal{}1\plus{}F(z)G(z)$ where $ F(z)\equal{}\sum_{k\equal{}1}^\infty a_k z^k$, whence
\[ G(z)\equal{}\frac 1{1\minus{}F(z)}\,.\]

[b]Step 2[/b]. Note that if all $ a_k>0$, then $ |F(z)|<1$ for all $ |z|\le 1$ except $ z\equal{}1$. It follows from here that $ U(z)\equal{}\Re G(z)$ is a non-negative harmonic function in the open unit disk $ \mathbb D$. Note now that $ U(rz)\equal{}\int_{\mathbb T}\mathcal P(z,\zeta)U(r\zeta)\,dm(\zeta)$ where $ \mathbb T$ is the unit circumference, $ \mathcal P$ is the Poisson kernel, and $ m$ is the Haar measure on $ \mathbb T$. Also, the measures $ d\mu_r\equal{}U(r\zeta)dm$ are nonnegative measures of total mass $ U(0)\equal{}1$. Hence, we can choose a sequence $ r_k\to 1\minus{}$ such that $ \mu_{r_k}\to\mu$ weakly. Then $ U$ is the Poisson integral of $ \mu$. Also, for $ n>0$, the $ n$-th Fourier coefficient of $ \mu_r$ is $ \frac 12p_nr^n$, so, passing to the limit, we get $ p_n\equal{}2\widehat\mu(n)$.

[b]Step 3[/b]. Note that $ U(z)$ is continuous up to the boundary everywhere except the point $ 1$. Thus, for every neighborhood $ V$ of $ 1$ and every continuous function $ f: \mathbb T\to\mathbb R$ supported by $ \mathbb T\setminus V$, we can write $ \int_{\mathbb T} f\,d\mu\equal{}\lim_{k\to \infty}\int f\,d\mu_{r_k}\equal{}\int_{\mathbb T\setminus V}fU\,dm$, which immediately implies that $ \mu$ restricted to $ \mathbb T\setminus V$ is just $ U\,dm$. Since it is true for every $ V$, we conclude that $ \mu\equal{}\mu\{1\}\delta_1\plus{}U\,dm$ and $ \delta_1$ is the Dirac point mass at $ 1$.

[b]Step 4[/b]. We thus have for $ n>0$,
\[ p_n\equal{}2\widehat\mu(n)\equal{}2\widehat U(n)\plus{}2\mu\{1\}\]
and it remains to note that the first term (the Fourier coefficient of the integrable function $ U$) tends to $ 0$ by the Riemann-Lebesgue theorem as $ n\to\infty$, so the limit of $ p_n$ exists (and equals $ \mu\{1\}$).

[b]Step 5[/b]. Once we know that the limit exists, we can find it by considering the speed at which $ G(z)\equal{}\frac 1{1\minus{}F(z)}$ tends to $ \infty$ as $ z\to 1\minus{}$. This was clear from the beginning, so I'll not write down this part.

If you have any questions, feel free to ask :)

thank you! this was very interesting  :)

these days I was reading about Markov Chain stuffs, and as expected there is a very short solution based on it:
Suppoe that $ \{Y_k \}_{k \geq 0}$ is a sequence of iid random variables satifying $ P(Y_k \equal{} m) \equal{} a_n$ and such that the set $ \{n : a_n > 0\}$ is not included in $ d \mathbb{Z}$, $ d \geq 2$. Then one easily shows that
$ X_n \equal{} \inf\{ m \geq n : \exists k \geq 0, m\equal{}Y_1\plus{}Y_2\plus{}\ldots\plus{}Y_k  \} \minus{} n$
is an aperiodic markov chain on $ \mathbb{N}$ whose transition probabilities are:
$ P[X_{n\plus{}1}\equal{}i\minus{}1 | X_n\equal{}i]\equal{}1$ if $ i>0$ and $ P[X_{n\plus{}1}\equal{}k | X_n\equal{}0]\equal{}a_{k\plus{}1}$ for all $ k \geq 0$

Now suppose that $ E[Y]\equal{}\sum k P_k \equal{} \mu < \infty$. This shows that $ E_0[T_0]\equal{}\mu < \infty$ where $ T_0 \equal{} \inf \{ n \geq 0 :  X_n \equal{} 0 \}$ is the return time to the state $ 0$: the markov chain is positive reccurent. The convergence theorem also tells us that:
$ \lim_{n \to \infty} P[X_n \equal{} 0] \equal{} \frac{1}{E_0[T_0]}\equal{}\frac{1}{ \mu}$
this is exactly the conclusion one wanted to reach!

Ps: the usual proof of the Markov chain convergence theorem also uses a coupling argument - no big surprise here -


function
arithmetic sequence
algebra solved
algebra

Let f be a continuous function on R such that : f gets all possitive and negative values .
 Prove that : There exists an arithmetic progression a, b ,c satisfying :
  $ f(a) + f(b) + f(c) = 0 $

Just take $g:\mathbb R\times(0,\infty)\to\mathbb R$, $g(x,r)=f(x-r)+f(x)+f(x+r)$. $g$ takes both negative and positive values, and since $\mathbb R^2$ is connected, so is the image of $g$, which is thus an interval (the only connected sets in $\mathbb R$). This means that there must be $x\in\mathbb R,r>0$ s.t. $f(x-r)+f(x)+f(x+r)=g(x,r)=0$, Q.E.D.

I want to gine a more plain argumenation.
We can find $u$ and $v$ s.t. $f(u)<0$ and $f(v)>0$. Using continuity we will find $r$ s.t. $f(u\pm r)<0$ and $f(v\pm r)>0$. So $f(u-r)+f(u)+f(u+r)<0<f(v-r)+f(v)+f(v+r)$. It follows that continuous function $g(x)=f(x-r)+f(x)+f(x+r)$ takes both positive and negative values, so it takes 0 too.


advanced fields
advanced fields unsolved

For $ x$ positive, $ G(x) = \sum_{n\leq x}F\left(\frac{x}{n}\right)\quad\Longleftrightarrow\quad F(x) = \sum_{n\leq x}\mu(n)G\left(\frac{x}{n}\right)$, where $ \mu(n)$ is the Mobius function.

I can prove this (it follows from the Mobius inversion formula), but I'm having trouble coming up with an example(s) of $ F(x)$ and $ G(x)$. Does anybody have any ideas?


email
combinatorics unsolved
combinatorics

Hi everybody,

I hope you can help me with this. I found this problem quite interesting and certainly not too easy.

Here it is:

There is a circle, which is divided into 2n sectors. n of them are black and n are white. Now you have to number the white ones consecutively clockwise and the black ones counterclockwise.

You have to proof that there is always a way to divide the circle into to halves so you have the numbers from 1 to n in each of them, no matter in which way the black and white sectors are arranged and where you start to number them.

Have fun with it!

marcus

Email of "disciple" is A@gmx.de with A = marcuskreusch.
If it is not a fake, then I propose a nomination for "most brainless cheater of the year".

[size=200][b]@mods: CURRENT COMPETITION PROBLEM! PLEASE DELETE THE 2ND POST AND LOCK THE TOPIC.[/b][/size]

@ everybody else: the problem is old, but being posted by a German in this month it is obvious that it is taken from the [b]ongoing[/b] 2nd round of the Bundeswettbewerb Mathematik 2006.

  darij

Please, if it's possible, could you explain better to me what has happened? I can figure out part of it, but I'm a little bit confused. Sorry, I didn't intend to do anything wrong answering to this.

You did nothing wrong since you couldn't know (the rules forbid even to give the problems to non-participants, so it's impossible for others to know if it is from there).
But please delete the content of your first post here (click on Edit there, then just send an empty message or whatever).

What happened is:
disciple posted a problem from a german homework competition that is still running; so this is just a very poor attempt of cheating. Not more and not less.


Hazel walked 16 blocks east, turned right and walked 8 blocks, turned right and walked 4 blocks, turned right and walked 2 blocks, and turned right one last time and walked 1 block. Each block is $ \frac18$ mile. What is the number of miles in the straight line distance from Hazel's ending position to her original starting point? Express your answer as a decimal rounded to the nearest hundredth.

The total distance east traveled is $ 16 \minus{} 4 \plus{} 1 \equal{} 13$. The total distance south traveled is $ 8 \minus{} 2 \equal{} 6$. Since the directions of east and south are perpendicular to each other, drawing the segments represent net distance east, net distance south, and the vector of resultant distance traveled would produce a right triangle. Thus, we can use the Pythagorean Theorem to find the hypotenuse:

$ 6^2 \plus{} 13^2 \equal{} H^2$
$ H^2 \equal{} 205$
$ H \equal{} \sqrt {2005}$

Since each block is $ \frac {1}{8}$ of a mile, the distance in miles is $ \frac {1}{8}\sqrt {2005}$, which is approximately $ \fbox{5.6}0$.

EDIT: OH WHOOPS! OH WOW...

We get a right triangle with lengths 13 and 6.

Find the hypotenuse and divide by 8 to get approximately $ \boxed{1.79}$.


The sides of a triangle are in the ratio $ 6: 8: 9$. Then:

$ \textbf{(A)}\ \text{the triangle is obtuse} \qquad\textbf{(B)}\ \text{the angles are in the ratio } 6: 8: 9$
$ \textbf{(C)}\ \text{the triangle is acute}$
$ \textbf{(D)}\ \text{the angle opposite the largest side is double the angle opposite the smallest side}$
$ \textbf{(E)}\ \text{none of these}$

[hide="Click for solution"]
Since $ 6^2\plus{}8^2\equal{}100 > 9^2$, the triangle is acute, so $ \boxed{\textbf{(C)}}$ is correct.
[/hide]


inequalities
trigonometry
calculus
calculus computations

Solve the following inequality for $ 0\leq \theta <2\pi$.

$ |\sin \theta|>|\cos  3 \theta |$.

$ cos(3\theta) = cos(\theta)(4cos^2(\theta) - 3)$
the inequality can be written: $ \sqrt{1 - cos^2(\theta)} > \left|{cos(\theta)\right|\left|{4cos^2(\theta) - 3\right|}}$
let $ t = \left|cos(\theta)\right|$, $ u = t^2$, square both sides $ 1 - u > u(4u - 3)^2$, $ 16u^3 - 24u^2 + 10u - 1 < 0$, $ (2u - 1)(8u^2 - 8u + 1) < 0$
$ \left|cos(\theta)\right| \in (0, \frac{\sqrt{2 - \sqrt{2}}}{2}) U (\frac{\sqrt{2}}{2}, \frac{\sqrt{2 + \sqrt{2}}}{2}) = (0, a) U (b, c)$
if $ A = arccos(a), B = arccos(b), C = arccos(c)$
$ \theta \in (C, B) U (A, \pi - A) U (\pi - B, \pi - C) = I$, when $ \theta \in [0, \pi)$
$ \theta \in I U (I + \pi)$, when $ \theta \in [0, 2\pi)$

for a better feeling, you may draw the graphic of $ \left|cos(\theta)\right|$ and intersect this with horizontal lines y = a, y = b, y = c

You can solve, say, $ \cos \theta \equal{}\frac{\sqrt{2\plus{}\sqrt{2}}}{2}$.

$ cos(2\theta) \equal{} 2cos^2(\theta) \minus{} 1 \equal{} \frac{\sqrt{2}}{2} \equal{} cos(\frac{\pi}{4})$, $ \theta \in [0, \pi]$
$ C \equal{} \frac{\pi}{8}$
the same idea for $ B \equal{} \frac{2\pi}{8}, A \equal{} \frac{3\pi}{8}$


I have trouble solving this problem

[img]http://img393.imageshack.us/img393/6816/22127307bm4.jpg[/img]

Try conjugating.  (You could use l'Hopital's, but that's overkill here.)