Let $ x,\ y,\ z$ be positive numbers and $ p,\ q,\ r$ be positive constants.

Prove that $ p\left(\frac{x}{y\plus{}z}\plus{}1\right)\plus{}q\left(\frac{y}{z\plus{}x}\plus{}1\right)\plus{}r\left(\frac{z}{x\plus{}y}\plus{}1\right)\geq \frac{1}{2}(\sqrt{p}\plus{}\sqrt{q}\plus{}\sqrt{r})^2$.

The inequality becomes

$ 2\left(x\plus{}y\plus{}z\right)\left(\frac{p}{y\plus{}z}\plus{}\frac{q}{a\plus{}c}\plus{}\frac{r}{x\plus{}y}\right) \ge \left(\sqrt{r}\plus{}\sqrt{p}\plus{}\sqrt{q}\right)^2$

It is just an application of Cauchy-Schwartz.

$ \left[(y \plus{} z) \plus{} (z \plus{} x) \plus{} (x \plus{} y)\right]\left(\frac {p}{y \plus{} z} \plus{} \frac {q}{z \plus{} x} \plus{} \frac {r}{x \plus{} y}\right)\ge\left(\sqrt {(y \plus{} z)\frac {p}{y \plus{} z}} \plus{} \sqrt {(z \plus{} x)\frac {p}{z \plus{} x}} \plus{} \sqrt {(x \plus{} y)\frac {r}{x \plus{} y}} \right)^{2}$

Does it satisfy you, kunny?  :)

Thank you, I obtained the result as well. :)

How did you get this result?

[quote="great math"]The inequality becomes

$ 2\left(x \plus{} y \plus{} z\right)\left(\frac {p}{y \plus{} z} \plus{} \frac {q}{a \plus{} c} \plus{} \frac {r}{x \plus{} y}\right) \ge \left(\sqrt {r} \plus{} \sqrt {p} \plus{} \sqrt {q}\right)^2$.[/quote]

$ LHS\equal{}\sum_{cyc}p\left(\frac{x}{y\plus{}z}\plus{}1\right)\equal{}\sum p\frac{x\plus{}y\plus{}z}{y\plus{}z}\equal{}(x\plus{}y\plus{}z)\sum \frac{p}{y\plus{}z}$
Hence we have to prove : $ 2(x\plus{}y\plus{}z)\sum\frac{p}{y\plus{}z}\geq (\sqrt{p}\plus{}\sqrt{q}\plus{}\sqrt{r})^2$


modular arithmetic
number theory unsolved
number theory

The question is:

Show that if p is an odd prime and ord(p^a)a=2t, then
a^t== -1 mod p^a

First, I used ord(p^a)a to mean "order of a, mod p^a" and the == sign means congruent.

So first, I tried a few examples. Let p=3, a=2
Since ord(9)2=6, then t=3 and:
2^3 == -1 mod 9 TRUE

I continued with different values of p and a. Here is a table(sorry it looks weird):

p--a--t--p^a
3--2--3--9
5--2--10--25
11--2--55--121
13--2--78--169
17--2--68--289

It seems that p|t in all of my examples but I'm stuck....THANKS!

Let p be an odd prime and $ord_{p^{\alpha}}a=2t$, where $\alpha$ is not necessarily equal to a. Let us denote $b=a^{t}$ then $b^{2}\equiv1\pmod{p^\alpha}$. It follows from here that $b\equiv\pm1\pmod{p^{\alpha}}$(it's because p is odd.) But $b\equiv1$ is impossible.

Let me add some words about p|t in your examples. Let a=2, $ord_{p^{2}}2=d$. Then d|p(p-1). If (d,p)=1, then d|p-1, but the last leads to $2^{p-1}\equiv1 \pmod{p^{2}}$. Such p are called Wieferich primes. Only 2 such primes are known: http://primes.utm.edu/glossary/page.php?sort=WieferichPrime . That's why it's not surprise that in your examples p|t


modular arithmetic
number theory proposed
number theory

Which number is bigger : the number of positive integers not exceeding 1000000 that can be represented by the form $2x^{2}-3y^{2}$ with integral $x$ and $y$ or that of positive integers not exceeding 1000000 that can be represented by the form $10xy-x^{2}-y^{2}$ with integral $x$ and $y?$

[i]Proposed by A. Golovanov[/i]

First note that $10xy-x^{2}-y^{2}=2(x+y)^{2}-3(x-y)^{2}$ and therefore if a positive integer can be represented by this form it can be represented by the other form. Also note that $6=2.3^{2}-3.2^{2}$ and $10xy-x^{2}-y^{2}=8xy-(x-y)^{2}\equiv 0,1\pmod{4}$ so 6 cannot be represented and thus the number of positive integers of the form $2x^{2}-3y^{2}$  is bigger.


AMC
AIME
search
USA(J)MO
USAMO
AIME I
AIME II

I have not seen it, but am curious about it. Does anyone know the estimated # of AIME I vs. AIME II takers last year and this year? Does AMC publish this sort of stats?

I'm sure there's more information elsewhere on the site, but quick search found this:
[url]http://www.unl.edu/amc/e-exams/e7-aime/e7-1-aimearchive/2008-aa/T13-A,stats,ScDist.pdf[/url]

totals for each test are on the top left of the page

I'm surprised that so many people take the Alternate AIME since they are about equal and you have to pay for the II

[quote="worthawholebean"]For many schools, the I happens during a break or other inopportune time.[/quote]
For instance, the sophomores at my school had to take the CAHSEE (California High School Exit Exam) on the day of the AIME I. They had to take the AIME II.

My school also had to take the CAHSEE on the same day...It prevented me from taking AIME I, and from getting a much higher score =(

Not from CA so we are not sure, is CAHSEE a required test for every student in CA? Is this is the case, then the largest AIME student body will have to take AIME II, which will have a bigger impact on the overall USAMO index.....

CAHSEE is a required test in California, we have to pass it to graduate, even though it is super easy.  However, some schools, not mine unfortunately, took the AIME in the afternoon since the CAHSEE had to be taken in the morning.

yes it is a required test. however, i know of some schools that took the AIME I and then are taking a makeup CAHSEE sometime in April.

This year, according to a thread in this forum, a majority of AoPSers believe that the AIME II was indeed more difficult than the AIME I, and interestingly enough, statistics from past years indicate that the AIME II has been harder in the past, on average; however, these differences can most likely be attributed to chance.

it's not on purpose. the maa makes 2 aime's and randomly picks one of them to be the I.

I'm pretty sure this is the case...but I found this year's AIME surprisingly suspicious...

Like it was on April Fool's Day...and one of the answers was 000, and like 5 answers contained the numbers 4 and 1 (114, 141, 401, 041, 014)...


Alright so here's the situation: I am a junior, I'm alright at math (MV Calc, Diff EQ, Math Structures, and Number Theory this year all at a university), I scored 105.5 on AMC 12B last year, and 2 on the AIME. I just took the practice AMC 12 from 2000, and got 107.5, a score I was disappointed with. I was hoping after one year of really intense math study, I would have improved. What is the best way to improve your AMC score in a time crunch like I am currently in?

Also, what are some little techniques to go faster? I mean like do you check your answers over? If so, how much time do you have left when you start checking your answers. Also, how do you know when to guess or not to guess? Thanks everyone!

I don't know for how EFFECTIVE that is but if you want the archives of the problems, do the AMC B day :D I post the three problems everyday and I think it's level is around AMC-10 to AMC-12, one time it even had AIME (although really easy).

I will be taking both days.

(Thats still allowed, right?)

And btw Silverfalcon, thank you for responding to all my questions so quick!

Vinny, what he means is his little thread called AMC B day, in which he posts various problems for practice.  Wow, how do you get into Number Theory as a junior!...I thought the Universities won't allow that...

Do AMC tests to get familiar with the types of problems that appear on each test. Also, taking upper level math classes like you are currently enrolled in won't necessarily help you on the AMC since it tests over material covered in Pre-Cal and below. Another test taking tip; Only do problems you are sure of getting right unless you have not answered enough questions since missing a question due to stupid guessing or careless mistakes can be very painful (I got a 99.5 because of two careless mistakes on AMC 12 A as a fish last year. However, there are always exceptions to the case such as my friend Brian who guess on 4 and got them all correct giving him a 118). Lastly, go to bed early and don't try to cram last minute the night before, it helps!

Thanks for the advice! Statistically, isn't it better to guess if you narrow it down to two since 6/2 > 2.5?

Yes, Number Theory is a 400 level class, but is not very difficult (or at least doesn't seem to be since I'm starting that and diff eq this semester). The professor is also extremely confident in me since I had him for Math Structures which was only a 300 level class last semester. I think the ease with which I got into these classes is that I live in Arizona and the local university is trying to entice me to going there when I grow up.


geometry
calculus
integration
trigonometry
calculus computations

I think you're right

if you weren't sapose to use double integration i think you should have gotten full credit

Isn't that you have to take the big one (3cos(theta) subtract with the small one (1+cos(theta)...
[/list]

No, $1+\cos\theta>3\cos\theta$ on this interval.

The problem is that this form doesn't work when either radius is negative. Since $3\cos\theta<0$ when $\frac\pi2<\theta<\frac{3\pi}2$, we need the following integral instead:

$\int_{\frac\pi3}^{\frac\pi2}\frac12((1+\cos\theta)^2-(3\cos\theta)^2)\,d\theta +\int_{\frac\pi2}^{\frac{3\pi}2}\frac12(1+\cos\theta)^2\,d\theta$
$+\int_{\frac{3\pi}2}^{\frac{5\pi}3}\frac12((1+\cos\theta)^2-(3\cos\theta)^2)\,d\theta$


superior algebra
superior algebra unsolved

Hi,
I need some help with this exercise:

Find for each prime p in R = Z[√−61] a local generator at p of the non-principal R-ideals
(2, 3 + √−61) and (5, 3 + √−61).

Thanks a lot  :)

[hide="Solution"]$ 1$ if $ 2\nmid p$, $ 2$ otherwise.[/hide]

Third Edition: No this isn't true.
In fact, the only exception happens for localisation at $ (2,3\plus{}\sqrt{\minus{}61})$ itself, and then ist is generated by e.g. by $ 3\plus{}\sqrt{\minus{}61}$ but not by $ 2$.

[quote="ZetaX"]Third Edition: No this isn't true.
In fact, the only exception happens for localisation at $ (2,3 \plus{} \sqrt { \minus{} 61})$ itself, and then ist is generated by e.g. by $ 3 \plus{} \sqrt { \minus{} 61}$ but not by $ 2$.[/quote]Apologies for the late reply. Certainly my fault. My piece of paper says $ 2\mathbb Z[\sqrt{\minus{}61}]\equal{}(2,3 \plus{} \sqrt { \minus{} 61})^2$ with trivial residue field extension. ... I cannot remember what I was thinking while posting my solution. Explicit calculations are not my favourites - always different results.  :whistling:

The solution for the second ideal is still open:
[quote="jdie"]Find for each prime p in R = Z[√−61] a local generator at p of the non-principal R-ideals ... (5, 3 + √−61).[/quote][hide="Solution"] 3 + √−61 for p=5, 1 otherwise. Note that there are exactly two primes sitting above 5. ... Hopefully I haven't made a fatal error in my calculations here again. ;)[/hide]


geometry
ratio
geometric transformation
reflection
trapezoid
similar triangles

[i]To this classic problem I would like to see your different and nice   solutions.[/i]

  Let  a circle whith centre  $O$   , an exterior point  $S$ , the tangents  $SB , SC$  to the circle ,  the diameter  $BOA$  and the perpendicular $CD$  on  $AB$.( Point  $D$  is on  $AB$ )
  If  $AS$  meets  $CD$  at $M$ , prove that  $MC = MD$.

[u]Babis[/u]

A Menelaus bash is easy enough, but let's try to get more style points:

Let X and Y be the midpoints of AC and BC. Let P be the intersection of XY and CD. It suffices to show S, P, and A collinear. But since XCY and OCS are similar triangles, we have AX/SY = OY/SY = XP/PY, so we have SYP ~ AXP and we're done.

$MC=MD \Longleftrightarrow \triangle{ACS}=\triangle{ADS}$ so area ratio also works for this problem.

[quote="dondigo"]Where is point $D$?[/quote]

 On the diametre $AB$

 [u]Babis[/u]

Let $P=l_{BS} \cdot l_{AC}$
$Q=l_{BC} \cdot l_{SO}$
$\angle{ACB}=\angle{BQO}=\frac{\pi}{2} \Rightarrow l_{AP} \parallel l_{SO}$, since $BO=OA$ we have $BS=SP$.
Triangles $\triangle DAC$ , $\triangle BAP$ are homothetic, so $DM=MC$

I like this nice and simple proposed problem.
[hide="Here is an elementary proof (in four steps)."]
[color=blue]Denote the reflection $R$ of the point $B$ w.r.t. the point $S$.
$1.\blacktriangleright SB=SC=SR\Longrightarrow CB\perp CR\ .$
$2.\blacktriangleright CB\perp CR\ \wedge \ CB\perp CA\Longrightarrow R\in AC\ .$
$3.\blacktriangleright CD\perp AB\ \wedge \ RB\perp AB\Longrightarrow CD\parallel BR\ .$
$4.\blacktriangleright$ In the trapezoid $CDBR$ the point $A\in BD\cap RC$ and the middlepoints of the parallel sides $DC$ and $BR$ are collinearly, i.e. the point $M$ is the middlepoint of the segment $CD\ .$

[u]Remark.[/u] If denote the intersection $I\in AS\cap BC$ then the points $A,M,I,S$ are collinearly (a harmonical division) ![/color][/hide]

Here is my solution . SA is the symmedian at the triangle ABC and CD is the altitude of the right-triangle ABC so CD is also symmedian .So M is the Lemoine point . But it is known that the Lemoine point of a right-angled triangle is the mid point of the altitude (the one from the angle of 90) and we are done .

Let $\{P\}\in AB\cap CS$; obviously, $CD$ is the polar of $P$ w.r.t. $\odot ABC$, hence $\frac{AP}{PB}=\frac{AD}{BD}\iff \frac{AP}{AD}=\frac{BP}{BD}$, but $\frac{BP}{BD}=\frac{PS}{CS}$, hence applying [b]Menelaos[/b] to the triangle $\Delta DPC$ with the transversal $\overline{AMS}$, with the previous equalities we get the desired $CM=MD$.

Best regards,
sunken rock


floor function
geometry
rectangle
3D geometry
analytic geometry
algebra
factorization

Suppose that you have a rectangular $m$x$n$ grid of points. How many squares can be formed with vertices as grid points? There are no orientation restrictions on the squares.

I came up with a moderately ugly double summation that appears to work in the general case, but I need confirmation and I was wondering if there is a way to simplify it. Here is the number of "slanted" squares:

${ \sum_{a+b< min(m,n)}\sum_{i=1}^{\lfloor \frac{min(m,n)}{a+b}\rfloor}(m-i(a+b))(n-i(a+b)) }$

where $a$ and $b$ are positive integers. This can be rewritten as

${ \sum_{k=2}^{min(m,n)-1}(k-1) \sum_{i=1}^{\lfloor \frac{min(m,n)}{k}\rfloor}(m-ik)(n-ik) }$

 And of couse the number of nonslanted squares is

$\sum_{i=1}^{min(m,n)-1}(m-i)(n-i)$

[quote="digger"]Let (a,b) be left-lower vertex and (c,d) be opposite right-upper vertex.
Then we have to find the number of (a,b,c,d) with property $0\le a < c \le m-1$ and $0\le b < d \le n-1$
This is the problem?
If yes then its simple. Choose pair (a, c) and independently (b, d).
So the answer is $\binom{m}{2}\cdot \binom{n}{2}$

P.S. I think my guess was right because the problem became trivial in case we consider rectangles (a,b,c,d) and (a+h,b+h,c+h,d+h) be the same. Then we simply move $(a,b,c,d) \rightarrow (0,0,c-a,d-b)$ and so have $(m-1)\cdot (n-1)$ distinct rectangles.[/quote]

We are looking for squares, not just rectangles. I'm not sure that the result works. For example, with m=4 and n=5 we have 10 slanted squares and 20 nonslanted squares. You're expression gives 60.

Can a recursion be useful for the slanted squares? [ex. when m=n] You can have a_{n+1}=4a_n+b_n where b_n denotes the number of squares that where x (or y) component has a distance of n+1

When $m = n$, see http://www.research.att.com/~njas/sequences/A002415

For rectangles when $m = n$, see http://www.research.att.com/~njas/sequences/A085582

Say that the m=n is sequence 1, then the link says that it is equal to seq2+seq3

they give you seq2, but then they say that seq3=seq1-seq2

That is such bs. haha

Here's an idea for the horizontal ones:

WLOG, say $m\le n.$ Then you can pick the size of the square (and location on the "m axis") in $\binom{m}{2}$ ways.

Then, you need only select a starting point along the "n axis." This can be done in $\frac{(n-1)!}{(n-m+1)!}$ ways, so for the total, just multiply the two together.

I think...

[hide="The Key"]The four vertexes of any square can be described as being points on a larger square which is parallel to the grid lines.[/hide]

[hide="Turning the Key"]
Consider square ABCD (labeled counterclockwise), with sides parallel to the gridlines, and k dots on each side, inclusive of vertexes. Suppose we start with vertex A, and, moving counterclockwise, mark every (k-1)th point, until we return to A. These points happen to be B, C, and D, and together form the whole square.

Instead if we move one point counterclockwise of A and repeat the procedure, we will find that these four vertexes form a different square. We can repeat this for the rest of the points on AB, and each will define a different square. Hence, there are k-1 squares formed by a k by k root square whose sides parallel to the gridlines (we're subtracting one because we don't want to count both vertexes as starting points).[/hide]

[hide="Opening the Door"]
Now, all that is left is to calculate the number of k by k point squares (|| to axes) there are in an m by n lattice.
This is simply (m-k+1)(n-k+1).

Each of these squares is actually the root square of (k-1) different squares.
So, for each k, there are (m-k+1)(n-k+1)(k-1) different squares.

Now, WLOG say $m \leq n$.

Then,

$S_{m,n}=\sum_{k-2}^{m}(m-k+1)(n-k+1)(k-1)$, which can easily be broken down with the sum of squares and sum of cubes formula.

It simplifies to $S_{m,n}= \frac{(2n-m)(m+1)(m)(m-1)}{12}$, where $m \leq n$

Or, if you prefer, the somewhat cleaner:
$\boxed{S_{m,n}=\binom{m+1}{3}\left( n-\frac{m}{2}\right)}$

[/hide]

[quote="Altheman"]err...how can that be correct if it is not symmetric in $m$ and $n$?[/quote]
[quote="The Zuton Force"]
Now, WLOG say $m \leq n$.[/quote]
:)

Okay I think I will finally read the posts and offer a solution. Zuton Force had a great idea!

In essence, for each square, we can construct another square by drawing lines parallel to the coordinate axes through the verticies.

Suppose our squares are circumscribed by a $j$ by $j$ square, then we have: $4(j)$ border points on the square, and a forth of these determine a square, for a total of $j$ squares.

Now lets consider a $m$ by $n$ square. [WLOG $m\ge n$] The largest square that we can have is $n$ by $n$. Suppose we have a $j$ by $j$ square. We have $(m+1-j)*(n+1-j)$ $j$ by $j$ squares that are parallel to our axes.

Hence, we want the sum: $\sum_{j=1}^{n}(m+1-j)(n+1-j)(j)$.

Which I just plugged into my calculator [this is trivial...sum of cubes, squares, etc]

Then I factored on my calculator.

$\frac{(2m-n+1)(n)(n+1)(n+2)}{12}$

[which is obviously positive since $m\ge n$.]

[Did I do anything wrong? Does this work for small values? I feel like we should not be able to solve something that was not solved on the OEIS.]

Well your result seems to agree with mine (you interchanged $ m$ and $ n$ and replaced $ n$ with $ n+1$, or something like that - most likely, you used side-lengths and I used lattice points - you might want to clarify this), and your approach was pretty much what I did.

So I'm assuming you got what I did, which works for every example I bothered to try.

I can't believe that OEIS wouldn't have it...

Anyway, I'm after a combinatorics oriented solution. I can see why there might be a $ \binom{m+1}{3}$ is coming from (in my version of the formula).

To choose the root squares setting the $ n$ coordinate of the lower-left corner to $ 1$ means we need to pick 2 points of the $ m$.
To choose a slanted square originating from such a root square, we need to choose 3 points from $ m$: the $ m$ bounds of the root square and the seed point in between from which we will count off and get the other vertexes.

And then, adding them, $ \binom{m}{3}+\binom{m}{2}=\binom{m+1}{3}$.

However, I can't figure out how to complete the proof with this approach - can anyone help? Or offer a different combinatorics solution?

[quote="Altheman"][Did I do anything wrong? Does this work for small values? I feel like we should not be able to solve something that was not solved on the OEIS.][/quote]

[quote="The Zuton Force"]I can't believe that OEIS wouldn't have it...[/quote]

I've tried all of the standard input forms that I would expect (as a triangle by rows, as a square array by diagonals) and wasn't able to find it, so it seems it's just not there.  It's actually not that shocking -- there are lots and lots of integer sequences out there :)
I've submitted it now, including a link back to this thread -- my past experience is that it takes a few days to a week for a submission to show up online.  I'll probably make another post on this thread when it does appear.  (I submitted it in Altheman's form because the only difference is that TZF's has a lot of zeroes.)

This now exists in the OEIS: http://www.research.att.com/~njas/sequences/A130684


Prove that for all $ n\in\mathbb{N}$, $ \displaystyle\frac{n^2\plus{}n\minus{}1}{n^2\plus{}2}$ is irreducible  :D .

[quote="pelao_malo"]Prove that for all $ n\in\mathbb{N}$, $ \displaystyle\frac {n^2 \plus{} n \minus{} 1}{n^2 \plus{} 2}$ is irreducible  :D .[/quote]
We see that the value is equal to $ 1\plus{}\frac{n\minus{}3}{n^{2}\plus{}2}$. For $ n \geq 4$ the top and bottom are positive integers, clearly with the top less than the bottom, not resulting in integer. We can check 1,2,3, from which we get n=3 is only value for which it isn't irreducible.

[quote="pelao_malo"]Prove that for all $ n\in\mathbb{N}$, $ \displaystyle\frac {n^2 \plus{} n \minus{} 1}{n^2 \plus{} 2}$ is irreducible  :D .[/quote]
$ \gcd(n^2 \plus{} n \minus{} 1,n^2 \plus{} 2) \equal{} \gcd(n \minus{} 3,n^2 \plus{} 2) \equal{} \gcd(n \minus{} 3,11)$
So this fraction is reduciple if and only if $ n\not\equiv 3 (\mod 11 )$


calculus
integration
trigonometry
complex numbers
complex analysis

Show $\int_{0}^{2\pi}e^{\cos\theta}\cos(n\theta-\sin\theta)d\theta=\frac{2\pi}{n!}$,
$n \in N$.

The integral is equal to the real part of the integral:

$\int_{0}^{2 \pi}e^{\cos \theta}e^{i n \theta-i \sin \theta}d\theta$

$=\int_{0}^{2 \pi}e^{i n \theta+e^{-i \theta}d\theta}$.

Now let $u= e^{-i \theta}$ and then $d\theta = \frac{i du}{u}$.  This transforms the integral to:

$-\oint_{|u|=1}u^{-n}e^{u}\frac{i}{u}du$.

(Minus because $e^{-i \theta}$ traverses the unit circle clockwise).

Now the only residue is at $0$.  Expanding the power series for $e^{u}$ and looking at the order $n$ term shows that the residue of $f(u)=\frac{i e^{u}}{u^{n+1}}$ at $0$ is $\frac{i}{n!}$.  Therefore your integral is equal to the real part of:

$-2\pi i \frac{i}{n!}$  which is just $\frac{2\pi}{n!}$.

My solution:
$\oint_{|z|=1}\frac{e^{z}}{z^{n+1}}dz=\frac{2\pi i}{n!}$ by Cauchy's integral formula.
But $z=e^{i\theta}$, $dz=ie^{i\theta}d\theta$, hence
$\int_{0}^{2\pi}\frac{e^{e^{i\theta}}}{e^{i(n+1)\theta}}ie^{i\theta}d\theta=\frac{2\pi i}{n!}$.
This implies
$Re\left(\int_{0}^{2\pi}\frac{e^{e^{i\theta}}}{e^{i(n+1)\theta}}e^{i\theta}d\theta\right)=\int_{0}^{2\pi}e^{\cos\theta}\cos(\sin\theta-n\theta)d\theta=$
$=\int_{0}^{2\pi}e^{\cos\theta}\cos(n\theta-\sin\theta)d\theta=\frac{2\pi }{n!}$.


Can I add *.jpg or *.gif files in a *.tex file using Texniccenter. I tried that but it seems to show some errors. If it's not possible to add the above 2 formats, which other graphic formats can be added and where can I get free software on net to write files in these formats. Also, which program can be used to convert .jpg or .gif files to those formats which are usable in Texniccenter???

Yes to jpg, no to gif.

See here: [url]http://www.artofproblemsolving.com/LaTeX/AoPS_L_PictHow.php[/url]


Find : $ \lim_{x \rightarrow 0}[ \lim_{n \rightarrow \infty} \frac {1}{x^{2}}(A^{n} - E)]$.Where $ A \in M_{2}\mathbb{(R)}$ [b] E is identity 2x2 matrix [/b] .

 $ A = \left [\begin {array} {11} 1 & \frac {x^{2}}{n} \\
- \frac {x^{2}}{n} & 1 \end {array} \right ],$
 :P  nice A - A , Do math to Love Math !!!

[quote="JRav"]What is $ E$?  To compute matrix limits, you need to have specific eigenvalues or else the limit blows up.[/quote]

I think hE means identity :P


Show that the complex exponential function takes the segment [a +ic, b+ic] over the segment [e^a * e^ic, e^b * e^ic]

[hide]$e^{a+ci}=e^{a}e^{ci}$

$e^{b+ci}=e^{b}e^{ci}$

so $\left[a+ci,b+ci\right]\rightarrow\left[e^{a}e^{ci},e^{b}e^{ci}\right]$ because $f\left(x\right)=e^{x}$ is strictly increasing.[/hide]


ashegh! i know yoy are a silly boy :rotfl:
and shoma ri.....!!! :blush:

hilbert 2005 (farzin kashany) i will kill u if u dont stop spamming. :lol:

Guys, behave yourselves, please. And ashegh, maybe you should ask yourself the question whether you are spamming?

Please follow Anre's advice. Here we write and do  only mathematics, free and respecting the others opinion.

Otherwise Vornicu will be very angry , many others  too , and he will cansel you from members ! But we need and love all members.

 Both ,     enjoy solving good math problems !!!

Babis

i didnt say any thing...

i just invite hilbert 2005,not to spamm.,,


The object of the game is to post the number closest to two-third the mean of all of the posted numbers. You may post as many times as you want, and you may edit your previous posts as much as you want (no double-posting)

$0$

Can we start a new set besides 0?  We cannot do very much with that.  If so, my number is:  $\fbox{35760}$  Are we always posting integers?

[quote="Chocoboom"]i dont understand this game, please explain :D[/quote]

So everyone picks a number.  Then randomdragon will take the mean, and mulypily it by $\frac{2}{3}$.  Whoever's number is closest to two-thirds the mean wins.

12214

8128 (there's no way i could pass up a chance to post a large-ish perfect number)

OK. I get this game. The numbers strictly decrease, but they are nonnegative. So I call the 100000000th post and post 0 in there. :winner_first:


calculus
search
linear algebra
number theory

Hi. How are you?
I entered now in college and i am studying physics with minor in math. I would like to play a lot with calculus, which books do you suggest? (I have Spivak) And  for Algebra and Linear Algebra? Number Theory? Which are your favorites? Thanks.

Hi  :)
Search in http://www.lookforbook.com  :wink:


In the circle shown, chords $ \overline{AB}$ and $ \overline{CD}$ intersect at $ E$. If two of the six segments shown are selected at random, what is the probability that the segments are contained in the same line? Express your answer as a common fraction.

[asy]pair A,B,C,D,E;
A=(-12,16); C=(16,12); B=(14.1,-14.1); D=(-20,0); E=(-1.5,11.5);
draw(Circle((0,0),20));
draw(A--B); draw(C--D);
label("$A$",A,NW);
label("$B$",B,SE);
label("$C$",C,NE);
label("$D$",D,SW);
label("$E$",E,S);[/asy]

Either on AB line or on CD line.

AB line: 3 ways
CD line: 3 ways
Total possible: 15 ways

probability: 2/5


Isn't this too easy?

\[\begin{aligned} & 9\left(x^2 + yz\right)\left(y^2 + zx\right)\left(z^2 + xy\right) \\ & \leq 9\left(\frac{x^2 + yz + y^2 + zx + z^2 + xy}{3}\right)^3 \\ & = \frac{1}{3} \cdot \left(\left(x^2 + y^2 + z^2\right) + \left(xy + yz + zx\right)\right)^3 \\ & \leq \frac{8}{3} \cdot \left(x^2 + y^2 + z^2\right)^3 \\ & \leq 8\left(x^3 + y^3 + z^3\right)^2 \end{aligned}\] where the last step follows from the Power mean inequality.

Yes, it is very easy. I gave the same solution. ;)  It had been proposed in Crux Mathematicorum. :D

[quote="cezar lupu"]Let $x, y,z\geq 0$.Prove that:

         $8(x^3 +y^3 +z^3 )^2\geq 9(x^2 +yz)(y^2 +zx)(z^2 +xy)$ ;)[/quote]

This is also pretty old:

http://www.mathlinks.ro/Forum/viewtopic.php?t=6507
http://www.mathlinks.ro/Forum/viewtopic.php?t=25455

  darij


$ f(x,y)$ satisfies $ f(0,y)\equal{}y\plus{}1, f(x\plus{}1,0)\equal{}f(x,1), f(x\plus{}1,y\plus{}1)\equal{}f(x,f(x\plus{}1,y))$ for all nonegative integers $ x$ ad $ y$.  Evaluate $ f(4,2008)$.

If I recall correctly, this is the well-known Ackermann function.
$ f(4, 2008) \equal{} \begin{matrix}\underbrace{{2^2}^{{\cdot}^{{\cdot}^{{\cdot}^2}}}} \minus{} 3 \\\mbox{ 2011 twos}\end{matrix}$


Let $ a,b,c,d\in\mathbb{N}$ such that $ \frac{a}{a\plus{}b}\plus{}\frac{b}{b\plus{}c}\plus{}\frac{c}{c\plus{}d}\plus{}\frac{d}{d\plus{}a}\in\mathbb{N}$. Prove that $ abcd$ is a perfect square.

Maybe the statement holds if the numbers a,b,c,d are distinct because if $ a \equal{} c \equal{} k\in N$ and $ b \equal{} 1$ , $ d \equal{} 3$ then
the $ \sum\frac{a}{a\plus{}b}\equal{} 2$ and $ abcd \equal{} 3k^{2}$ which is not a square ....

Yeah, there definitely needs to be an extra condition. $ a \equal{} c$ or $ b \equal{} d$ makes the expression always a natural number (2, in fact). But then you have freedom to make the other two variables whatever you want.

You are right! $ a,b,c,d$ are distinct. I'm sorry.

Now it is an easy job ...
We have that $ 1<\sum\frac{a}{a\plus{}b}<4$ so it is 2 or 3 .
The case that $ \sum\frac{a}{a\plus{}b}\equal{}3$ is clearly impossible after expanding while the case
$ \sum\frac{a}{a\plus{}b}\equal{}2$ gives $ \frac{(a\minus{}c)(b\minus{}d)(ac\minus{}bd)}{(a\plus{}b)(b\plus{}c)(c\plus{}d)(d\plus{}a)}\equal{}0$ and because the numbers are distinct
we have ac=bd so $ abcd\equal{}(ac)^{2}$  , qed  :)


A bowl contains only red marbles, blue marbles and green marbles. The probability of selecting a red marble from the bowl is $ \frac{3}{13}$. The probability of selecting a blue marble from the bowl is $ \frac{2}{5}$. There are fewer than 100 marbles in the bowl. What is the probability of selecting at random and without replacement, a green marble and then a red marble from the bowl on the first two selections? Express your answer as  common fraction.

probability of a red marble is 1-2/5-3/13=24/65.

this means there are 65 total marbles.  the rest is simple.

Yeah, you should arrive at the answer of
[hide]$ \frac {9}{104}$.[/hide]

[quote="pascal_1623"]A bowl contains only red marbles, blue marbles and green marbles. The probability of selecting a red marble from the bowl is $ \frac {3}{13}$. The probability of selecting a blue marble from the bowl is $ \frac {2}{5}$. There are fewer than 100 marbles in the bowl. What is the probability of selecting at random and without replacement, a green marble and then a red marble from the bowl on the first two selections? Express your answer as  common fraction.[/quote]

[hide]
haha this is a nice trick question (remember this one) :)

so 24/65*15/64
cross out some...
3/13*3/8
9/104[/hide]
mamathbug

[hide]Since there are only three colors of marbles, the probability of drawing a green marble is
\[1 - \frac{3}{13} - \frac{2}{5} = \frac{24}{65}.\]
This tells us that the total number of marbles in the bowl must be a multiple of 65. But we are told that the number of marbles in the bowl is less than 100, so it must be 65. Hence, there are $3/13 \cdot 65 = 15$ red marbles, 24 green marbles, and $2/5 \cdot 65 = 26$ blue marbles in the bowl.

Therefore, the probability of selecting a green marble and then a red marble is
\[\frac{24}{65} \cdot \frac{15}{64} = \boxed{\frac{9}{104}}.\]

Notice that the answer is not $\dfrac{24}{65} \cdot \dfrac3{13}$ because the stated probability for a red marble is only valid if the bowl has its initial setup. Since we do not replace the first marble drawn, this changes the composition of the bowl, and hence, the probability of drawing a red marble.[/hide]


function
calculus
integration
floor function
algebra unsolved
algebra

[quote="arkhammedos"]show that for all $m>3$ reel .we have $[\frac{m(m+1)}{2(2m-1)}]=[\frac{m+1}{4}]$
[\cdot]=integral part.[/quote]

That doesn't seem to be correct.  If the equation is

$\left\lfloor \frac{m(m+1)}{2(2m-1)}\right\rfloor = \left\lfloor \frac{m+1}{4}\right\rfloor$,

then it is not true for, say, m = 6.6.

\[\left\lfloor \frac{m(m+1)}{2(2m-1)}\right\rfloor =\left\lfloor \frac{m+1}{4}+\frac{m+1}{4(2m-1)}\right\rfloor = \left\lfloor \frac{m+1}{4}\right\rfloor\]
equavalent to $\{\frac{m+1}{4}\}+x<1, \ x=\frac{m+1}{4(2m-1)}$.
Obviosly it is true, if m is integer and m>2.
If m>3 $\frac{1}{8}<x<\frac{1}{5}$, but can be $\{\frac{m+1}{4}\}\ge \frac{7}{8}$, when it is false.


[quote="bluecarneal"]Nice! :D :lol:  :!:[/quote]

then..

[quote="The screen"]Too many smilies in your post: [b]4[/b].
You can have at most [b]3[/b] smilies in this message. Please remove some smilies before submitting this message. [/quote]

I only see three.

Does the one in your signature count? You can try deactivating your signature to test that limit.

I have two in my signature, so something would still be wrong.  (it only says I had four, including signature I have 5, without, 3)

Searching for "too many smilies" in the "Site & General Information" forums yields http://www.artofproblemsolving.com/Forum/viewtopic.php?search_id=663802694&t=245654


function
number theory unsolved
number theory

If $f(n)$ is the sum of divisors of $n$ and $g(n)$ is the number of divisors of $n$

Find all $n\in N$

such that

1) $n=f(n)^{2}$

2) $n=g(n)^{2}$

[hide]
1 is trivially $n=1$, since $f(n)>n$ for $n>1$.

For 2, we first see that $n$ is a square, so $g(n)$ odd, so $n$ odd. Then the number of divisors of $n$ that are at most $\sqrt{n}$ is at most $\frac{\sqrt{n}+1}{2}$, equality only if all odd numbers that are at most $\sqrt{n}$ divide $n$. Each divisor $d$ greater than $\sqrt{n}$ can be paired with $\frac{n}{d}$. Then we have
\[g(n)\leq \sqrt{n}\]
In order to have equality, all odd numbers that are at most $\sqrt{n}$ divide $n$. This is only true for $n=1$ and $n=9$ (otherwise, we arrive at a contradiction: Bertand's Postulate is strong enough for this, since we can consider greatest prime less than $\sqrt{n}$)[/hide]

Sorry..
For (1) it should be $n^{2}=f(n)$

[quote="Yosh..."]Sorry..
For (1) it should be $ n^{2}=f(n)$[/quote]

So there is again no solution except $ n=1$, since, if $ n=\prod p_{i}^{n_{i}}$, then $ f(n)=\prod\frac{p_{i}^{n_{i}+1}-1}{p_{i}-1}$ and the equation $ \prod p_{i}^{2n_{i}}=\prod\frac{p_{i}^{n_{i}+1}-1}{p_{i}-1}$ is impossible since it is easy to show that $ p^{2n}>\frac{p^{n+1}-1}{p-1}$ for any $ p>1$ and $ n>0$

Another method is to write, for $ n>1$ $ f(n)\leq 1+2+\cdots+n=\frac{n(n+1)}{2}<n^{2}$

Sorry but I still don't understand with the solution of number (2),probably It cause by my poor English :blush:

So can you give explanation again?
thx..


Let a,b,c be the positive numbers such that a+b+c=3
Prove that
$\frac{a\sqrt{b+ca}}{c+a}+\frac{b\sqrt{c+ab}}{a+b}+\frac{c\sqrt{a+bc}}{b+c}\leq \frac{3\sqrt{2}}{2}$[/hide]

:roll: This is not true, for counterexample, $\,a=\frac{5}{3}, \, b=\frac{1}{34}, \, c=\frac{133}{102}.$


inequalities
AMC
AIME
AMC 10
Cauchy Inequality

Hello all. I'm in grade 7 in Singapore and I recently bought Intermediate Algebra (let's call this IA). I also have AoPS volume 2 but it's super super tough! I don't understand (most topics; I can understand only a few)!

IA is definitely much more simpler for me. I can solve many problems at the back, while for AoPS I can't solve practically anything.

Is IA meant to be easier? It seems. Many topics like Cauchy-Scwharz (how u spell it? lol) are in both books but they're presented in a simpler manner.

Should I read both books simultaneously [i.e. after reading a topic in IA I read AoPS vol 2] or do I finish the whole of IA?

I'm also intending to buy Contest books for AMC if they're available. How do I use them? I'm not very familiar with these books because I'm not from USA.

I'm intending to get a much higher score on the AMC 10 next year and get invited to the AIME. I got a super low score this year.

Your advice? Most of my classmates haven't even heard of AoPS. I think AoPS is the best book! Except that IA is rather thick. I don't know why they don't consider separating it into two volumes; oh well.

PS: Any forums on the AoPS site good for my level?

Thanks (A LOT)!

Because AoPS vol 2 serves as an introduction, you can read whatever relevent topics from vol 2 that applies to IA. If you have questions on the Cauchy Inequality, you can consult vol 2 from IA or vice versa. Generally, IA questions can be posted in HSB (only the really basic ones), and most of them belong in Intermediate. AoPS vol 2 most nearly always go in Intermediate.

As for contests, you can go to the Contests forum, where they have AMC, AIME, etc.

Intermediate Alg covers things more in depth, while AoPS V2 moves through a lot of information pretty quickly.  It's good to use the inter and intro books to supplement the info from volumes 1 and 2

if u are preparing for the 10 u should read aops v1 not v2 because v2 isnt on the 10.

I have already read v1, but not very thoroughly though. What do I do now? I plan to prepare for the AIME too as I think I should be able to be invited.

mayb read v1 thoroughly then do v2 and practice AIMES


ratio
geometry
circumcircle
geometry unsolved

Let $ M,N$ be distinct points in the plane of the triangle $ ABC$ such that $ AM: BM: CM\equal{}AN: BN: CN$. Prove that the line $ MN$ passes through the circumcenter of triangle $ ABC$.

http://www.mathlinks.ro/Forum/viewtopic.php?t=46344

  darij


Analytic Geometry (Charles Theorem)

My question is about Charles Theorem which says that for each 3 points: A,B and C on the x'ox axis:

segment AB + segment BC = segment AC

Is this theorem true for n points (n> or = to 4)) on the axis or not? If so, how can I prove it for n> or = to 4?

uhh well there for $n\ge4$ points there will be $n-1$ segments made, but it should work

for example $n=4$, you have points $A,B,C,D$

you would just modify the theorem for $n=3$ like so...

$AB + BC + CD = AD$

or you can look at it as...

$AB+BC+CD = AD \implies AC+CD = AD \implies AB + BD = AD$

so either way it works

does that answer ne thing?


[size=200]. . . . .
. . . . .
. . . . .
. . . . .[/size]

(That should be a 4x5 grid of dots.)  Assuming each point is exactly one unit from its nearest neighbors, how many squares can be formed by connecting groups of four points?

Ok, then, there are
12 1x1 squares
6   2x2 squares
2   3x3 squares
_______________
20 squares

But then, there are squares that's not straight, they go diagonally.
I counted:
6 sqrt(2)xsqrt(2) squares
4 sqrt(5)xsqrt(5) squares
________________________
10 squares

20 + 10 = 30 squares

My answer: 30 squares (Please tell me if I'm wrong)

i get 30 squares also, if it means anything by now that is. :shock:

20 normal, 10 diagonal = 30.

im 2 years late to solve this problem, but whatever im just going through all of them  :lol:


email
ARML
MATHCOUNTS
AMC
AMC 8
AMC 10
AIME

Hello everyone!

Anyone can use this forum, but mostly for anyone from North Carolina. Any competition that takes place in North Carolina can be discussed, as well as preparation for national competitions such as ARML. I hope more North Carolinians get to know each other through this forum, so please post who you are and something about yourself. You do not have to say your name if you do not wish to.

I moderate this forum, and if you have any questions, you can contact me by pm or email at soccerboy1992@gmail.com.

I have been to ARML once (this year), I have participated in mathcounts for the last two years, and will participate again next year, and I am going into 8th grade and help other students at my school in North Carolina for our math club.

WOOOOOOOOOOOOOOOOOOT!!!!!!!!!!!!!!!!!!!!!!!!!!!!!

NORTH CAROLINA FORUM!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!

GO B-FLAT!!!!!!!!!!!!!!!!!!!!!!!!!

 :ninja:

About me: Mathcounts:2x state qualifier, 1x countdown  :mad: ,
20, 24 on AMC8, 143 on AMC10, took AIME  :ninja: , 1st Algebra II contest with 65%, lol, and ARML 2006 B team.

Yes, I suck.

Hello, b-flat.  I pmed you that I wanted to join.  I don't check my pms very much, so I responded rather late.

NO

MATHCOUNTS = BAD

but um I thought I should post something

[quote="MysticTerminator"]MATHCOUNTS = BAD[/quote]

Only for me  :mad:

Mathcounts nurtures stupidity in Daesun  :ninja:

but not in Neal Wu...

Wooah. I totally missed this until I looked in b-flat's sig. Anyway, I think everybody knows who everybody else is. If not, I live in Wilmington. That narrows me down to one person.

Arnav: get all those people from East Chapel Hill on AoPS!!! We need more high schoolers, and you are the NC high school math factory (that's singular you).

jeremy = n^4+4
noah "lurks" on the forum
I can't remember any other ECHHSers at the moment

Does anyone know who was 5th at states this year? Apparently beat Eric McCabe in Countdown.

Steven (Stephen?) Hunt. He's going to be at my high school this year.

I was 6th  :(

The last problem in Countdown was hilarious.

It was \[\frac{109^{2}-100^{2}}{9}\] and they both did it the stupid way.

I wasn't much better though...

I did $109^{2}= 100^{2}+100*18+81$...

 :oops:

My mom was, like, "Use difference of squares" and I just about killed myself...

It was pretty funny.

EDIT: BTW, Steven (?) only beat me because he raised his hand about .1 seconds before mine.  :mad:

Heh. Did you see Daniel beat me two years ago? That was funny. First question, I raised my hand before he rang in... Second and third, we rang in at almost exactly the same time. Anyway, this is getting a bit off topic now.

dude yeah seriously stop with the discussion of these infantile matters

I'm offended  :D

I beat Daniel at chapter because he beat me to the wrong answer

Sorry, Arnav. This is not THAT infantile...

NO WE MUST CEASE AND DESIST WITH THIS UNSPEAKABLE "COMPETITION"

there's nothing worse than a dead forum.

look the forum temporarily died again :o

wtf crosswalks???

yes, we have math competitions (Duke Math Meet, w00t)

of course we do, we won ARML, wtf man

.......

[quote="funcia"]yeah i feel bad for you people, well you do have nifty crosswalks in that one place. do you have math competitions there?[/quote]

If you check the stickied thread, you will find a list of several of the contests in North Carolina.

aw man spam tastes good

but cold vienna sausage is disgusting (i know this from experience)

TSINGHUA > IIT >> EAST $\approx$ ENLOE > UNC > DUKE >> NCSSM  :D

but yeah - maybe we should start predicting USAMO questions.  we'd get a LOT of traffic...

Man, Tsinghua sucks. Those guys who get in there are 1337 test-takers and that's about it. I guess they'd do well on math contests.

Seriously, guys, what is the reason for putting down NCSSM in so many posts? If it wasn't for NCSSM, I wouldn't be writing this right now, because I never would have had the opportunity to participate in any math contests at all.

Why all the hatred?

(I had a brief discussion with Vitek about this at SMC, but I want to see what the rest of you think.)

Wait, you don't want us to dis NCSSM, but now you are asking us to do the exactly that? If you really want to read it and let yourself down, go find the thread devoted to it. As we make very clear, it is an excellent school for some but not all.

I didn't say I didn't want you dissing NCSSM, I said that I wanted to know [i]why[/i] everyone was dissing NCSSM. If there are a lot of good reasons, go ahead and dis it all you want.

Yes, I've read the NCSSM thread before, but it was mostly incoherent Arnav "SUCKS!!" without many good reasons. The reasons that were valid (inability to take college courses, heavy workload, etc.) don't seem to merit the constant bashing that NCSSM is constantly getting on this forum. Sure, NCSSM might not be the best place for some of the top math students in the state, since most are living in vibrant cities with excellent high schools and top-notch universities nearby (Chapel Hill, Durham, etc.), but if a small-town guy like me finds this forum and reads through some of it, he might be turned away from what might be a huge opportunity for him, NCSSM.

Succinctly, what justification do you have to disrespect NCSSM to the degree that it has happened in this forum?

Ok. So there are two separate things going on here. Many might be surprised by our perceived lack of awe for NCSSM as it has traditionally been held to be higher than all other schools; as you have no doubt noted, we do not feel that to be the case as John lucidly explained above. This lack of complete admiration may come as a shock to some who could perceive it as a lack of respect and so on and so forth. The other factor is probably a general resentment towards NCSSM for still being treated as the greatest thing ever in spite of the earlier point: for example, the fact that they send their whole school to the state math contest whereas all the other sites have to in fact turn away MOPpers is patently ridiculous, and many feel strongly on such points.

See the unusually coherent post above. Also, there are definitely good points in the thread. That is why it exists. If you want to know why we are dissing it, I repeat: look there.

And we are mostly responding to questions by those who we feel NCSSM is not appropriate for certain people.

Please limit the spam.  I have deleted several post that are spam.

Thanks.

Since I am too lazy to create a new thread on this, I'd like to note that Mr. Batterson is now a site admin... wow! Good luck with that :D

Also I think (@ very old discussion above that is only tangentially related to this topic) NCSSM seems to be regaining some of its luster. Now if Calvin would go...

$\color{blue}\text{LaTeX}$ $\color{blue}\text{i was born in elizabeth town NC buhya}$


vector
geometry
geometric transformation
rotation
trigonometry
trapezoid
circumcircle

On a circle we take the points A,B,C,D,E,Z (cyclic in the order ), such that the bows AB,CD,EZ are 60. Let K,L,M the midpoints of the chords BC, DE, ZA. Prove that the triangle KLM has KL=LM=KM.

Take the center of the circle as the origin; let the affixes of A, C, E be $2a, 2c, 2e$, respectively; rename Z as F (what was that about anyway...?); if $\omega = e^{\frac{i \cdot \pi}{3}}$, the affixes of B, D, F will be $2a\omega, 2c\omega, 2e\omega$, respectively. Then the affixes of  K, L, M will be $k = a\omega + c, l = c\omega + e, m = e\omega + a$. Now, $k - m\omega + l\omega^2 = a\omega + c - e\omega^2 -a\omega - c + e\omega^2 = 0 \Rightarrow (k - l) = (m - l)\omega$, so the triangle is equilateral.

Excellent! A geometric solution?Or a trigonometric? I wait

[quote="silouan"]On a circle we take the points A,B,C,D,E,Z(cyclic in the order ), such that the bows AB,CD,EZ are 60.Let K,L,M the midpoints of the chords BC, DE, ZA. Prove that the triangle KLM has KL=LM=KM.[/quote]

That's an old problem. The condition that the arcs AB, CD, EZ contain the angle 60° is equivalent to the condition that the lengths of the chords AB, CD, EZ are equal to the radius of the circle (in fact, if a chord in a circle has the chordal angle 60°, then its length is the radius of the circle), and thus the problem is the same as the one posted at http://www.mathlinks.ro/Forum/viewtopic.php?t=16908 .

  darij

[quote="MysticTerminator"]rename Z as F (what was that about anyway...?)[/quote]

Haha well, in Greek (I believe Silouan is Greek) the alphabet starts \[\alpha,\;\beta,\;\gamma,\;\delta,\;\eta,\;\zeta\] so Silouan translated the sixth letter, which is pronounced as [i]zeta[/i] (or actually [i]dzeta[/i] I think) as a "Z".

Silouan.
 this is really a very good problem with many interesting solutions( vectors, complex numbers , tranformations , and pure syntetic geometry). In Greek books you can find easily a solution (syntetic). For example , see the journal  APOLLONIUS , issue 2 , 3( or call me !)

 [u]Babis[/u]

But not only !

 You will see a solution from N. Dergiades with vectors ( rotations)  and a new synthetic solution. See the other issues too.

 [u]Babis[/u]

[quote="silouan"]Yes I have this issue of Appolonios and a very nice proof comes from mr. Cristos Patilas(trigonmetric one).[/quote]
Can you please post one or two of these proofs? I am interested in an geometrical and a trig solution.

Thanks.

The plain geometric solution i found.

Because arc$AB$=arc$CD$=60  the  $ABCD$  is an isosceles trapezoid.

Let $K$ the midpoint of $BC$ and $P$ the midpoint of $DA$ and $T$ the midpoint of $AB$. Easily we can find that $KPT$ is equitaletar triangle.

Also $LP=\frac{AE}{2}$  and $LP//AE$  and $TM=\frac{BZ}{2}$ and $TM//BZ$.

Because $ABEZ$ is an isosceles trapezoid we have $BZ=AE$ so $LP=TM$

Now $\widehat{LPK}=90+\widehat{DAE}=90+\frac{arcDE}{2}$.

Also $\widehat{KTM}=\frac{arcAB+arcZEDL}{2}=\frac{60+60+arcDE+60}{2}=90+\frac{arcDE}{2}$
So $\widehat{LPK}=\widehat{KTM}$.
Now the triangles $KLP$  and $KMT$  are equal so  $KM=LK$  .
The conclusion follows

Let $X \equiv DE \cap FA,\ Y \equiv FA \cap BC,\ Z \equiv BC \cap DE.$ Inversion in the hexagon circumcircle (O) carries the lines $YZ \equiv BC,\ ZX \equiv DE,\ XY \equiv FA$ into the circumcircles $(O_{1}),\ (O_{2}),\ (O_{3})$ of $\triangle OBC,\ \triangle ODE,\ \triangle OFA$ and the vertices X, Y, Z of $\triangle XYZ$ into the pairwise intersections $X' \equiv (O_{2}) \cap (O_{3}),\ Y' \equiv (O_{3}) \cap (O_{1}),\ Z' \equiv (O_{1}) \cap (O_{2})$ other than O. According to the problem [url=http://www.mathlinks.ro/Forum/viewtopic.php?t=22976]MOP 1996[/url], the triangle $\triangle X'Y'Z'$ is equilateral and congruent to the equilateral triangles $\triangle OAB,\ \triangle OCD,\ \triangle OEF.$ But $\triangle KLM$ is a pedal triangle of the triangle $\triangle XYZ$ with respect to O, the inversion center. Therefore, the triangles $\triangle KLM \sim \triangle X'Y'Z'$ are similar, i.e., both equilateral.

Is the following a new proof of the asymmetric propeller theorem?

[b][i]Let the equilateral triangles $\Delta OAD,\Delta OBE,\Delta OCF$ and $M,N,P$ be the midpoints of $AF, BD, CE$ respectively. All triangles are counterclockwise noted.
Prove that $\Delta MNP$ is equilateral.[/i][/b]

Proof:
$\Delta AOB\equiv\Delta DOE$ and a $60^\circ$ rotation about $O$ will map the two triangles, similarly $\Delta BOC\equiv\Delta EOF$ and $\Delta COA\equiv\Delta FOD$ and $\widehat{BC,EF}=\widehat{AC,FD}=60^\circ$, consequently after a $60^\circ\ (\ *\ )$ rotation about $O$, the triangles $ABC$ and $DEF$ will map.
Construct the equilateral triangles $BCX, CAY, ABZ$ all extrenal to $\Delta ABC$. By [b]Fermat theorem[/b] $AX=BY=CZ\ (\ 1\ )$ and angles between any two of those segments is $60^\circ\ (\ 2\ )$.
With $(*) \wedge (1)$ we find $FEXC$ a parallelogram and $P$ the midpoint of $FX$ as well, consequently $MP=\frac{AX}{2}$; similarly $NP=\frac{CZ}{2}$. From $(1) \wedge (2)\implies MP=NP$ and $\angle MPN=60^\circ$, so $\Delta MNP$ is equilateral.

Best regards,
sunken rock


Hi all, I guess most AoPSers might be aware of the   [b] "International Congress of Mathematicians" (ICM) [/b] -- a congress in which the prestigious fields medal is conferred upon mathematicians whose contributions are recognized as being profound and which have significant or far-reaching implications for the field of Mathematics.

The ICM is recognized as being the largest congress in the Mathematics community.

The next ICM is going to be held at Hyderabad-- which is located in the state of Andhra Pradesh, India. Hyderabad is the capital of the state of Andhra Pradesh, India.

For more information pertaining to ICM 2010, check the link below:

http://www.icm2010.org.in/geninfo.php

I would like to know how many AoPSers (if any) are attending the next ICM-- which is going to be held at Hyderabad, India.  Is anyone attending it as a delegate, etc?

I think some AoPSers might be contemplating of attending it.

[b]P.S.:[/b] Though this may be extraneous, I would like to pay my deep condolences to the bereaved family of [b] "The Chief Minister of Andhra Pradesh" [/b] who died yesterday (or probably the day before yesterday) in a horrific helicopter crash. A Chief Minister is the elected head of government of a sub-national (e.g. constituent federal) state, notably a state (and sometimes a union territory) of India.

If you really insist on knowing the answer to this question, it is that there will be at least one AoPSer there. But what's the point? You do not expect a flock of middle and high-schoolers (which constitute an overwhelming majority of the AoPSers) to fly there, do you?

No, I don't expect "middle-schoolers/high-schoolers" to attend the ICM. :D  I'm pretty keen on meeting some AoPSer(s) in real life. I'm contemplating of attending the ICM. (as a spectator, etc)

I personally think it would be 'great' if you could attend the ICM. I just posed this question so as to know how many AoPSer(s) may be attending the ICM -- I'm pretty keen on meeting some fellow AoPSer(s). (Despite the fact that most of them are complete strangers to me). I think it would be pretty nice to meet people with similar interests.

[quote="Gen8"]Oh man! Isn't even a single AoPSer going to attend the International Congress of Mathematicians 2010 at Hyderabad?[/quote]

I'm not going, nor are any of the undergraduate or graduate students I know.  There simply isn't much in the way of accessible talks and travelling to India from North America is expensive.  I'm sure it would be an interesting experience but when I travel on my supervisor's dime I prefer it to be something that I feel I will gain a lot from, e.g. conferences or workshops or summer schools in my field of research.

[b] fedja [/b], You're right! I'm sure at least one AoPSer is going to attend ICM'10 which is going to be held at Hyderabad. The AoPSer whom I'm referring to is from MIT. I surmise many Indian Americans are going to attend the ICM next year.

[quote="blahblahblah"]
I'm not going, nor are any of the undergraduate or graduate students I know.  [/quote]
[b] Oh no! [/b] I thought you would be attending the ICM next year. Probably, I could have received you at the airport, and might have invited you for lunch or dinner the next day. I'm extremely keen on meeting an absolute 'genius' like you. I pray or rather hope that you would change your mind--There's still enough time for registration.

[quote="blahblahblah"] There simply isn't much in the way of accessible talks and travelling to India from North America is expensive.  I'm sure it would be an interesting experience but when I travel on my supervisor's dime I prefer it to be something that I feel I will gain a lot from, e.g. conferences or workshops or summer schools in my field of research. [/quote]
Well, IMO, I guess the commute is pretty expensive but it's definitely not exorbitant. I consider myself to be pretty lucky because I don't have to pay even a single dime for the commute.

I wonder who're going to be the recipients of next year's Field's medal. I wouldn't be surprised if Terence Tao manages to win another Field's medal. He's still pretty young and is doing profound research. I think he wouldn't be 40 years old even at ICM '14.

[quote="Gen8"]

I wonder who're going to be the recipients of next year's Field's medal. I wouldn't be surprised if Terence Tao manages to win another Field's medal. He's still pretty young and is doing profound research. I think he wouldn't be 40 years old even at ICM '14.[/quote]
I doubt if it is appropriate for anyone to get Fields medals,I don't see any exception of this case in history.
and I hear that if someone is invited to speak out the report in this congress he will has a high probability to award the fields medal,so it will be great if prof Ngo Bao Chau(who is one of the two such that)  wil be the first vietnamese to get the Fields medal  :idea:

[quote="nkht-tk14"]
I doubt if it is appropriate for anyone to get Fields medals,I don't see any exception of this case in history.[/quote]
You mean "to get Fields medal [b]twice[/b]"? So far there hasn't been any such case. I'm not sure if it is written somewhere in the official rules that you cannot do it but, given that the number of worthy candidates every year certainly exceeds the number of awards, I doubt the Fields medal committee will ever prefer a previous recipient to a new worthy candidate.

[quote]
and I hear that if someone is invited to speak out the report in this congress he will has a high probability to award the fields medal[/quote]
As a plenary speaker, that is. There are 20 this year. Dropping those over the age of 40, you get quite a small pool of potential awardees, so, yes Chau's chances are pretty good.

I hope Jacob Lurie wins one. He is my mathematical idol.  :oops:

Probably, Chau does have a high probability of winning the Field's medal next year. He had also been the recipient of many IMO medals. But, I guess the Abel prize is somewhat more prestigious than the Fields medal. The Abel prize is most often likened to be the 'Nobel prize of Math' -- which I guess is primarily due to its high endowment to the recipient.

But, I would be pretty happy (or rather ecstatic) if more AoPSers could attend the ICM next year. (especially [b] JBL [/b] :D)

[quote="Gen8"]But, I guess the Abel prize is somewhat more prestigious than the Fields medal. [/quote]  I don't think so -- more lucrative, and conciously modeled after the Nobels, but probably not so prestigious as the Fields (incidentally, there's no apostrophe in "Fields" -- the name has an "s" in it), which has its age going for it, if nothing else.  The two awards are quite different, though -- the Abel Prize is (in application, at least) much closer to a "lifetime achievement" prize, and so far has only been won by one person born since 1940.

I suspect that probably some other AoPSers will be there, but it seems unlikely right now that I will be among them.  (Sorry to disappoint, and thanks for the kind words :) )

[quote="JBL"] I suspect that probably some other AoPSers will be there, but it seems unlikely right now that I will be among them.  (Sorry to disappoint, and thanks for the kind words :) )[/quote]

Oh no! That's highly disappointing. You might have exacerbated the disappointment that arose from [b] blahblahblah's [/b] probable non-appearance at the ICM. Anyways, do attend if you could find some time. :D

Well, Now, I would like to know how many [b] AoPSers [/b] have "pre-registered or registered "themselves for the International Congress of Mathematicians (ICM) '10 which is going to take place at Hyderabad in August.

Set up a poll with just one option. Whoever wants to tell you that he'll be there will just click the button.


the incircle of a tri angle ABC touches BC, CA and AB at A', B', C' respectively. prove that
\[ AB'^2  \plus{} AC'^2  \plus{} BC'^2  \ge AB'.B'C' \plus{} BC'.C'A' \plus{} CA'.A'B' \ge B'C'^2  \plus{} A'C'^2  \plus{} B'A'^2 \]when does equality occur?

To mod: Please lock or hidden this topic! The deadline is 15/04/2008!


If $2^{25.964.951}-1$ is a prime number, find the remainder when we divide $2^{25.964.951}-1$ by $8$.

[hide="Solution"]
$2^{k+3} - 1$ can be written in the form $8 (2^{k}) - 1$ for any nonzero integer $k$, so the remainder on division by $8$ is $\boxed{7}$.
[/hide]

[quote="t0rajir0u"]
$2^{k+3} - 1$ can be written in the form $8 (2^{k}) - 1$ for any nonzero integer $k$.
[/quote]

Ok till there, could you please explain your conclusion? I still trying to figure out :wallbash: .

Well, since 2^25964051 is divisible by 8, and we know anything divisible by 8 minus 1 gives a remainder of 7 when divided by 8. So 2^25964051-1 gives a remainder of 7 when divided by 8.

Thanks Scrambled, i need some sleep.  :blush:

Which relies on the the fact our power of 2 can be expressed in terms of 2 cubed, and hence; essentially the same as torijarou's* solution.

Well but i was missing the last part, so now it's clear.  :)


Find all positive integers $n$ such that $n^{2006}+n+1$ is  a  prime number.

let $\omega$ be a root (distinct from 1) of the equation $x^3-1=0$.

then, $\omega^{2006}+\omega+1=\omega^{3\cdot668+2}+\omega+1=\omega^2+\omega+1=0$

so $n^2+n+1|n^{2006}+n+1$ and $n^2+n+1>1$ for every positive integer, so we must have $n^2+n+1=n^{2006}+n+1$, i.e. $n=1$  :D


geometry
power of a point
geometry proposed

Given two altitudes $ BE$ and $ CF$ in triangle $ ABC$ and point $ N$ on it's circumcircle. Let $ X\equal{}NB\cap CF$ and $ Y\equal{}NC\cap BE$. Prove that $ EF$ bisects $ XY$.

[hide="My solution"]
Assume $ N$ is on arc $ BC$ without point $ A$ (the proof in other cases is similar). Let $ S\equal{}CN\cap EG$ where $ G$ is the foot of A-altitude. Since $ ABGE$ is inscribed quadrilateral with diameter $ AB$ we have $ \angle (BN; SN)\equal{}\angle (AB; AC)\equal{}\angle (GC; GE)$ hence $ BSNG$ is inscribed quadrlateral. Now since $ FBGH$ ($ H$ is orthocenter of $ ABC$) is inscribed quadrilateral using power of a point with respect to the circle we obtain $ CN\cdot CS\equal{}CG\cdot CB\equal{}CH\cdot CF$ hence $ FSHN$ is inscribed quadrilateral (1). Now using fact that $ AFHE$ is inscribed we obtain $ \angle (NX; NY)\equal{}\angle (AB; AC)\equal{}\angle (HX; HY)$ hence $ HNXY$ is inscribed quadrilateral (2). Using (1) and (2) we obtain $ \angle (YN; YX)\equal{}\angle (HN; HX)\equal{}\angle (SN; SF)$ which means that $ XY\parallel{}FS$ (3). Now we see that pencil $ E.YZCS$ is harmonic (because $ E.BFAK$ is harmonic), hence using (3) we obtain $ (Y,X;M,\infty )\equal{}1\Rightarrow MX\equal{}MY$ QED[/hide]

[quote]Let $ ABC$ be a triangle and $ \mathcal{C}(O)$ its circumcenter.  Let $ BE$, $ CF$ the altitudes through vertices $ B$, $ C$, respectively, and let $ H$ be their intersection. $ M$ is an arbitrary point on $ \mathcal{C}(O)$. Denote by $ K$ the intersection of $ MB$ and $ CF$, $ L$ the intersection of $ MC$ and $ BE$. Prove that $ EF$ bisects the segment $ KL$.[/quote]

We have $ \angle KML\equal{}\angle BAC\equal{}\angle KHL$, so quadrilateral $ KLHM$ is cyclic. It implies $ \angle BKF\equal{}\angle CLE$, therefore $ \triangle BKF\sim\triangle CLE$. So $ \frac{FK}{EL}\equal{}\frac{BF}{EC}$, i.e. $ \frac{FK}{EL}\equal{}\frac{HF}{HE}$ $ (*)$

Now, denote by $ I$ the intersection of $ EF$ with $ KL$. According to the Menelaus theorem applied to $ \triangle BCG$ for the transversal $ FEI$, we find that $ \frac{FK}{FH}\cdot\frac{EH}{EL}\cdot\frac{IL}{IK}\equal{}1$. Combining with $ (*)$, we conclude that $ \frac{IL}{IK}\equal{}1$. This completes our solution. $ \hfill\square$

See: http://www.mathlinks.ro/viewtopic.php?p=220179#220179 and [url=http://reflections.awesomemath.org/2008_5/MR_4_2008_Solutions.pdf]Solutions for problem S92[/url]


The numbers from 1 to 150, inclusive, are placed in a bag and a number is randomly selected from the bag. What is the probability it is neither a perfect square nor a perfect cube? Express your answer as a common fraction.

there are 12 squares and 5 cubes, however, there are numbers that are squares and cubes (1,64)

so subtract 2 from 17

so the answer is 1/10

It's 9/10 because that you are subtracting what you don't want from the total. 1-1/10=9/10

The answer is 9/10

R@D!OACT!VÉ

╥ tastes good.


1) Prove that in equilateral triangle ABC with point O somewhere in the interior of ABC, AB+BC>AO+BO+CO

2) Prove that in 120-30-30 triangle ABC where angle C=120 given any point O, AO+BO+CO>AC+BC

[hide="hint"] None of these problems require any trigonometry[/hide]

the easiest way to do this would be just to place it on a coordinate plane and do the distance formulas

Actually, i dont think that will really work, sincethat our equilateral triangle coordinates cannot be rational numbers.

why would that matter? there is no restriction against $(0,0), (s,0), \left(\dfrac{s}{2},s\sqrt{3}\right)$

then, um, can i see something that gives me a little more insight on how you plan to do that?

given a point (x,y) in the region of the triangle, show
$2s>\sqrt{x^2+y^2}+\sqrt{(x-s)^2+y^2}+\sqrt{(x-\dfrac{s}{2})^2+(y-s\sqrt{3})^2}$
i guess that is difficult to show...i did not really think it through to well

ok, that is definitely not the easiest way to go about this


When Timmy was asked his age, he responded, “My age is 24 less than my father’s age.” When Timmy’s father was asked his age, he responded, “My age is three times Timmy’s age.” How old is Timmy?

We can solve this problem by setting up two equations.
From Timmy's statement, $ x\plus{}24\equal{}y$, where x is Timmy's age and y is his father's age.
From Timmy's dad's statement, we know that $ 3x\equal{}y$. Therefore, by substitution, we get $ x\plus{}24\equal{}3x$ and consequently $ x\equal{}12$. Therefore, Timmy is 12 years old.

Set up equation x-24 for timmy.  X is his dad.  Multiply x -24 by 3 because his dad is 3 times as old as him which is 3x-72.  Now since we have two equations for the dad 3x-72 and x, we can solve by using 3x-72=x  solve and you get x as 36 so the dad is 36. But you need to find what timmys age is so subtract 24 and you should get 12.


Hi guys....
Can anyone please tell me the mechanism of catalyst?...i.e How exactly a catalyst works?..what are the principles behind it?...
For example in preperation of alkanes from unsaturated hydrocarbons we use Pt, Pd or Ni as the catalyst... what role does these elements play..or more importantly WHY ARE THESE ELEMENTS USED..why not other...
Similarly in Wurtz reaction(preperation of alkane from alkyl halides using sodium) we use ETHER as the catalyst..WHAT PROPERTIES OF ETHER  make it suitable to be used as a catalyst?..
please help me guys!
(This may be trivial..but my chemistry is pathetic..so please dont mind if u find these doubts foolish/ridiculous)

Trivial perhaps. :D Maybe you should refer to NCERT Class XII textbook - Surface Chemistry :D :D

Id rather think that ether is a solvent there than a catalyst :) .

PS : One of the best resources for solvolysis is M&B Chapter 7 i think :D


Ας υποθέσουμε ότι $ ABC$ είναι τυχαίο τρίγωνο και $ O$ ένα τυχαίο εσωτερικό σημείο του τριγώνου. Aν $ x_1,x_2,x_3$ είνα οι αποστάσεις του $ O$ από τις κορυφές του τριγώνου και $ d_1,d_2,d_3$ από τις πλευρές του τριγώνου τότε να αποδείξετε ότι $ x_1\plus{}x_2\plus{}x_3>\equal{}2(d_1\plus{}d_2\plus{}d_3)$.

It is Erdos-Mordell theorem!
[url]http://mathworld.wolfram.com/Erdos-MordellTheorem.html[/url]


Primero consideremos una semicircunferencia de diámetro $AB$. Trazemos las perpendiculares por $A$ y por $B$ a $AB$, además trazemos una tangente por un punto cualquiera $D$ ($D$ diferente de $A$ y $B$) de la circunferencia, esta tangente intersecta a la perpendicular por $A$ en $X$ y a la perpendicular por $B$ en $Y$. Demostrar que $BX$, $AY$ y la perpendicular desde $D$ hacia $AB$ son concurrentes.

En resumen la solución sería:

Sea $Z$ el punto de intersección de $BX$ y $AY$. Tenemos que $\frac{BZ}{ZX}=\frac{BY}{AX}=\frac{YD}{XD}$ entonces $DZ$ es paralelo a $BY$, de donde $DZ$ es perpendicular a $AB$.

$Tipe$

Además $Z$ es el punto medio de la perpendicular desde $D$ hacia $AB$, este es uno de los lemas de Arquímides.

[quote="M4RI0"]Además $ Z$ es el punto medio de la perpendicular desde $ D$ hacia $ AB$, este es uno de los lemas de Arquímides.[/quote]Creo que este problema es bueno para quienes quieren practicar con el teorema de Thales y semejanzas de triángulos como técnicas para resolver problemas (en un nivel iniciante). Para establecer que Z es punto medio, ya no son necesarias la semicircunferencia ni los ángulos rectos, por causa del siguiente resultado:

Sea ABCD un trapecio, cuyas diagonales se intersecan en un punto O, y cuyos lados paralelos (bases) $ \overline{AB}$ y $ \overline{CD}$ tienen longitudes a y b, respectivamente. Considere una recta paralela a estas bases, que pasa por el punto O e interseca a los lados $ \overline{AD}$ y $ \overline{BC}$ en los puntos $ P$ y $ Q$, respectivamente. Entonces
\[ PO \equal{} QO \equal{} \frac {ab}{a \plus{} b}
\]
Demostración
[hide]En virtud de la semejanza de los triángulos ABD y POD (con los vértices en ese orden), el teorema de Thales y algo de aritmética:
\[ \frac a{PO} \equal{} \frac {AB}{PO} \equal{} \frac {DB}{DO} \equal{} \frac {DO \plus{} BO}{DO} \equal{} 1 \plus{} \frac {BO}{DO} \equal{} 1 \plus{} \frac {AB}{DC} \equal{} 1 \plus{} \frac ab \equal{} \frac {a \plus{} b}b
\]
entonces $ PO \equal{} \dfrac{ab}{a \plus{} b}$. Análogamente, $ QO \equal{} \dfrac{ab}{a \plus{} b}$[/hide]

Aparte $ BX$ y $ AY$ concurren en el punto medio $ M$  del segmento de perpendicular desde $ D$ a $ AB.$ Lo que también produce este interesante resultado:

Si $ D$ es un punto variable en una semicircunferencia de diámetro $ AB$ y la tangente por $ D$ corta a las perpendiculares a $ AB$ por $ A$ y $ B$ en $ X,Y,$  siendo $ M \equiv AY \cap BX,$ entonces el lugar geométrico de $ M$ es una elipse que tiene el círculo de diámetro $ AB$ como círculo pedal.


[quote="happyme"]The product of four different integers, exactly three of which are odd, is $7!$. The sum of the four integers is $63$. What is the largest of the four integers?[/quote]
[hide]7!=7*6*5*4*3*2*1, since three numbers are odd, my odd numbers are 3,5, and 7 and my even number is 2*4*6 or 48. Since 3+5+7+48=63, the largest of the four integers is 48.[/hide]

[quote="cat810930"]
[hide]7!=7*6*5*4*3*2*1, since three numbers are odd, my odd numbers are 3,5, and 7 and my even number is 2*4*6 or 48. Since 3+5+7+48=63, the largest of the four integers is 48.[/hide][/quote]Mind enlightening us on how you immediately jumped to decide that the numbers were 3,5,7,48 instead of 5,7,9,16 or 1,7,15,48 or something?[/hide]

Well he could've gotten that by:

a. Bruteforcing
b. Prime factorizing and then brute forcing

As this was a problem that was easily brute forced  :wink:


geometry
geometric transformation
reflection
parallelogram
circumcircle

Let $ABC$ be an acute scalene triangle whose orthocenter is $H$. Let $M$ be the midpoint of segment $BC$. $N$ is the point where segment $AM$ intersects the circle determined by $B,\, C$ and $H$. Show that $HN\perp AM$.

I liked your problem, it's quite nice.  :)

Reflect $A$ and $H$ in $M$, and call the reflections $A'$ and $H'$ respectively.
Call $\omega$ the circle containing $B,C,H$. Because $\angle CHB = \pi-\angle BAC$, we know that $A'$ is on $\omega$.

It is well known that $H''$, the reflection of $H'$ in $BC$, lies on $\omega$.
Because $H$ and $H''$ symmetric wrt $OM$, and knowing that they both lie on $\omega$, we conclude that $\angle H''A'B = \angle HA'C$.

So the isogonal conjugate of $H$ lies on $HA'$, but this isogonal conjugate is $O$, so $HA'$ is a diameter of $\omega$.
We conclude that $\angle HNM = \frac{\pi}{2}$.
\[\mathbb{QED}!\]

[img]http://img242.imageshack.us/img242/9529/perpendicularnobody1ee9.jpg[/img]

[hide="Check it!"]Construct parallelogram $CABK$. It's easy to show that $C,N,H,B,K$ all lie on a circle.
So $\widehat{ACB}=\widehat{CBK}=\widehat{CNM}$
On the other hand, we have $\widehat{HBC}+\widehat{HNC}=\pi$. Hence, $HN\perp AM$[/hide]

[asy]size(200); pair A,B,C,H,M,N,K; A=(10,30); B=origin; C=(50,0); K=(40,-30); H=orthocenter(A,B,C); draw(A--B--C--cycle); draw(B--K); draw(C--K); draw(B--H); M=(25,0); draw(A--M); draw(circumcircle(B,H,C)); N=intersectionpoint(A--M,circumcircle(B,H,C)); draw(H--N); draw(C--N); dot(M); dot(A); dot(B); dot(C); dot(H); dot(N); dot(K); label("K",K,SE); label("N",N,S); label("A",A,NW); label("B",B,SW); label("C",C,SE); label("M",M,S); label("H",H,S);[/asy]

beautiful solution. :lol:  but could you explain the following please? how does the conclusion follow? :blush:
[quote="hikaru123"]So $\widehat{ACB}=\widehat{CBK}=\widehat{CNM}$
On the other hand, we have $\widehat{HBC}+\widehat{HNC}=\pi$. Hence, $HN\perp AM$ [/quote]

[color=indigo][b]Nice but not hard problem![/b][/color]
[quote="nobody1"][color=darkred]Let $ABC$ be an acute scalene triangle whose orthocenter is $H.$ Let $M$ be the midpoint of segment $BC.$ $N$ is the point where segment $AM$ intersects the circle determined by $B,\, C$ and $H.$ Show that $HN\perp AM.$[/color] [/quote]
[color=indigo][u][b]Proof.[/b][/u] Denote: $O$ be the circumcenter of $\triangle BHC$ and $P=AM\cap HO$
Easy to show that: $OM=\frac{1}{2}AH$ (by let $Q=BO\cap (O)$ and prove that $AHQC$ is a parallelogram) $\Longrightarrow PO=\frac{1}{2}PH$
And so $HP$ is a diameter of $(O)\Longrightarrow\angle HNP=90^\circ\Longrightarrow HN\perp AM\quad\mathbb{QED}$[/color]

[quote="hikaru123"]Ok, it' easy to see. Notice that $AC$ is parallel to $BK$, $CNBK$ and $CNHB$ are inscribed in a circle.
Hence $\widehat{HBC}+\widehat{HNC}=\pi$. But $\widehat{HBC}+\widehat{MNC}=\widehat{HBC}+\widehat{ACB}=\frac{\pi}{2}\rightarrow\mathbb{QED}$
P/s: thanks for your illustrate. :)[/quote]

thank you very much! its all clear now. great solution! :D

[b]NOTE[/b]

[color=red]      After your nice solutions , I have to add this :[/color]

  Let $K$ the symmetric of $A$ wrt $M$.As mentioned, $K$ lies on the circle $(B,H,C)$.Since $BH \perp AC, AC \parallel BK$ , we conclude that $BH\perp BK$. This means that $HK$ is a diameter of the circle , so $\angle HNK = 90 \circ$ , that is $HN \perp NM$
[u]
 Babis   [/u]


function
algebra
domain
vector
Functional Analysis
real analysis
real analysis unsolved

If $E$ and $E'$ are compact metric spaces and $\gamma: E \to E'$ is a continuous function, map $C(E')$ into $C(E)$ by sending each $f\in C(E')$ into $f \circ \gamma \in C(E)$. Show that this map is a continuous linear transformation.

Can anyone help? I don't really quite get the problem.

Given a function $f\in C(E')$ - that is, $f$ is a continuous real-valued (at least, I'm presuming real-valued unless told otherwise) function with domain $E'.$

We define the operater $T$ by $g=T(f)$ means that $g(x)=f(\gamma(x)).$

We have the following things to check:

1. This is well-defined and maps $C(E')$ into $C(E).$ There's no ambiguity about the definition of $g.$ $g$ is a function whose domain is $E.$ It's real valued (assuming $f$ was real valued). And finally, $g$ is the composition of two continuous functions, hence continuous - which means that $g\in C(E).$

From now on, I'll say $T(f)$ instead of $g,$ and $T(f)(x)$ if I have to.

2. We need to check that this is linear. We note that $C(E)$ and $C(E')$ are vector spaces - more than that, they are Banach spaces. What is $T(\alpha_{1}f_{1}+\alpha_{2}f_{2})?$

$T(\alpha_{1}f_{1}+\alpha_{2}f_{2})(x)=\alpha_{1}f_{1}(\gamma(x))+\alpha_{2}f_{2}(\gamma(x))= \alpha_{1}T(f_{1})(x)+\alpha_{2}T(f_{2})(x).$

That's linearity.

3. Finally, we need to show that $T$ is continuous. The most direct way to show that is to show that it is a "bounded" operator from one Banach space to another. That is, we want to show that there exists $M$ such that $\|Tf\|\le M\|f\|$

Here the norms that we will use are the uniform norm on $C(E)$ and $C(E')$ - the usual norms that make these Banach spaces. That is, $\|f\|=\sup_{y\in E'}|f(y)|.$

But $\|T(f)\|=\sup_{x\in E}|f(\gamma(x))| \le \sup_{y\in E'}|f(y)|=\|f\|.$

Hence this is a bounded operator with constant no larger than one, hence uniformly continuous.


inequalities geometry circumcircle trigonometry algebra polynomial

I don't have a nice solution for this one. You do have, do you?
 Prove that if $ a,b,c$ are nonnegative and add up to 2 then $ (a^2+ab+b^2)(b^2+bc+c^2)(c^2+ca+a^2)\leq 3 $.

I know you don't like Muihead's inequality, but it seems to work here (if I made no mistake). But, that's probably why you said you didn't have a nice solution.... ;)

Pierre.

How about geometry? With three angles of 120 degrees?

Harazi and Pierre, how did you prove the inequality with Muirhead? I don't think it can be proven by Muirhead, or by Schur either, since in Muirhead and Schur, equality holds only for a = b = c, while in your inequality, we have equality for a = 0 and b = c = 1 (for instance), while a = b = c doesn't give equality. So I don't think Muirhead will help (a good inequality should not be Muirhead-bashable, BTW). If you still want to try, here is an equivalent form of the inequality:

[tex]13P_{2,2,2}+52P_{3,2,1}+13P_{4,1,1}-19P_{4,2,0}-2P_{3,3,0}+18P_{5,1,0}+\frac{3}{2}P_{6,0,0}\geq 0[/tex],

where $\displaystyle P_{i,j,k}=\sum\limits_{\text{all permutations}}a^{i}b^{j}c^{k}$ (symmetric sums, e. g. $P_{2,2,2}=6a^{2}b^{2}c^{2}$).

Quick, your idea is VERY interesting. Curiously enough I still don't see any simple reason why $PA \cdot PB \cdot PC$ is maximized in a side midpoint, but I think this is definitely the way to the solution.

  Darij

Hmm...I've made it wrong since I confused that (4,1,1) majorizes (3,3,0) but obviously not.

Pierre.

Ideas base on geometry is really nice.But I am really stupid at it,so I just have this ugly solution:
Take [tex] c=min\lbrace{a,b,c}\rbrace [/tex] and
[tex] f(a,b,c)=(a^2+ab+b^2)(b^2+bc+c^2)(c^2+ca+a^2) [/tex].
Then we can prove that:[tex] f(\frac{a+b}{2},\frac{a+b}{2},c) \geq f(a,b,c) [/tex].This is equal to:
[tex] \frac{(\frac{(a+b)}{2})^2+\frac{(a+b)c}{2}+c^2)^2}{(b^2+bc+c^2)(c^2+ca+a^2)} \geq \frac{(a^2+ab+b^2)}{\frac34(a+b)^2} [/tex]
If my computation is not wrong then it is equal to:
[tex] \frac{(a-b)^2}{3(a+b)^2} \leq \frac{(a-b)^2[(\frac{a+b}{2})^2+ab+c(a+b)-c^2]}{4(b^2+bc+c^2)(c^2+ca+a^2)} [/tex]
Luckily,we have
 [tex] 3[(\frac{a+b}{2})^2+ab+c(a+b)-c^2](a+b)^2 \geq 12(ab+ac+bc)ab\geq 4(b^2+bc+c^2)(c^2+ca+a^2) [/tex]
(Just expand it)
Now we must solve the problem with$ f(a,a,b)$,but this is easy.
However,I really hope that someone can prove this nice one by geometry. :)

[quote="Quick "]In that case, you have to prove that if $P$ is inside an equilateral triangle $ABC$, then the maximum of $PA.PB.PC$ is obtained when $P$ is the midpoint of one of the triangle's sides...  Is this obvious ?[/quote]

I have found a different problem.

Let us consider an equilateral triangle $ABC$ of side $ l=\frac{4\sqrt{3}}{3} $ and a point $ P $ inside it . If we denote with $a,b,c$ the distances $PX,PY,PZ$ of $P$ from $BC,CA,AB$ then it is easy to prove that $a+b+c=2$.

So our algebraic inequality reduces to prove that $(XY)(YZ)(ZX) \leq 3$ which is equivalent to prove that minimum of $(XY)(YZ)(ZX)$ occurs when the point $P$ is the midpoint of one of the triangle's sides $(a=1,b=1,c=0 ; a=1,b=0,c=1 ; a=0,b=1,c=1) $.

Klamkin has written several articles on inequalities between the elements of a  triangle inscribed in an equilateral triangle ( I have found only one of these articles ) so I am sure he created this VERY NICE inequality on the grounds of a geometric inequality.

Perhaps in the recent book "Recent advances on geometric inequalities " there is some useful  hint or reference  to solve this problem. :maybe:

However this proplem, proposed by Harazi, is another beautiful challenging for all people keen on inequalities. ;)

I wish you Good Work  :lol: in solving it !!!

[quote="manlio"][quote="Quick "]In that case, you have to prove that if $P$ is inside an equilateral triangle $ABC$, then the maximum of $PA.PB.PC$ is obtained when $P$ is the midpoint of one of the triangle's sides...  Is this obvious ?[/quote]

I have found a different problem.

Let us consider an equilateral triangle $ABC$ of side $ l=\frac{4\sqrt{3}}{3} $ and a point $ P $ inside it . If we denote with $a,b,c$ the distances $PX,PY,PZ$ of $P$ from $BC,CA,AB$ then it is easy to prove that $a+b+c=2$.

So our algebraic inequality reduces to prove that $(XY)(YZ)(ZX) \leq 3$ which is equivalent to prove that minimum [b][it should be "maximum" - Darij][/b] of $(XY)(YZ)(ZX)$ occurs when the point $P$ is the midpoint of one of the triangle's sides $(a=1,b=1,c=0 ; a=1,b=0,c=1 ; a=0,b=1,c=1) $.[/quote]

It's really the same problem as Quick suggested. Indeed, since < AYP = 90° and < AZP = 90°, the points Y and Z lie on the circle with diameter AP. In other words, the segment AP is a diameter of the circumcircle of triangle AYZ. Hence, after the Extended Law of Sines, applied to the triangle AYZ, we have $YZ = AP \cdot \sin \measuredangle YAZ = AP \cdot \sin \measuredangle CAB = AP \cdot \sin 60^{\circ}$. Similarly, $ZX = BP \cdot \sin 60^{\circ}$ and $XY = CP \cdot \sin 60^{\circ}$. Thus,

$YZ \cdot ZX \cdot XY = \left(AP \cdot \sin 60^{\circ}\right)\cdot\left(BP \cdot \sin 60^{\circ}\right)\cdot\left(CP \cdot \sin 60^{\circ}\right)$
$ = \left(PA\cdot PB\cdot PC\right)\cdot\sin^3 60^{\circ}$.

Consequently, maximizing $YZ \cdot ZX \cdot XY$ and maximizing $PA \cdot PB \cdot PC$ is the same thing.

  Darij

I'm not sure if it's easier the geometric way. I know that Klamkin used the algebraic inequality to solve the geometric one. But who manages to prove it geometrically is really great. Anyway, you don't want to know how Klamkin proved the algebraic one. Believe me...

Oh,yes.....I would be very happy to see the solution :lol:

Thanks a lot, Harazi.

Here is a simple but (very) ugly solution to the first algebraic inequality : first homogenize using [tex] (a+b+c)=2 [/tex], so that the inequality becomes equivalent to

[tex]3 (a+b+c)^6 - 64 (a^2+ab+b^2)(b^2+bc+c^2)(c^2+ca+a^2) \ge 0 [/tex]

Expanding the left hand side and grouping terms, you can come up with

[tex] 3 \Bigl[ \sum_{sym} a (a^2-b^2-c^2) \Bigr]^2 + 4 \sum_{sym}a b (a-b)^4 + 20 \sum_{sym} a b (a^2-b^2)^2 + 20 \sum_{sym} a^4 b c + 52 \sum_{sym} a^3 b c (b+c) + 60 a^2b^2c^2 \ge 0 [/tex]

which is obviously true and shows that equality occurs if and only if one variable is zero and the other two are equal to each other.

Fanfan

[quote="Fanfan"][tex] 3 \Bigl[ \sum_{sym} a (a^2-b^2-c^2) \Bigr]^2 + 4 \sum_{sym}a b (a-b)^4 + 20 \sum_{sym} a b (a^2-b^2)^2 + 20 \sum_{sym} a^4 b c + 52 \sum_{sym} a^3 b c (b+c) + 60 a^2b^2c^2 \ge 0 [/tex][/quote]

How do you people come up with these things?!

There's a regular contributor to Crux, Li Zhou, who always rewrites inequalities as a giant and asymetrical expression that is "obviously non-negative". I know that it kinda defeats the purpose of inequalities, but I'm still curious of it's done.

Let me first point out that I find the idea behind Anh Cuong's solution much more elegant than mine.

Now about my solution : an advantage is that it allows to strengthen the original inequality a little : rearranging some terms you can find

[tex] 3 \Bigl[ \sum_{cyc} a (a^2-b^2-c^2) \Bigr]^2 + 4 \sum_{cyc}a b (a-b)^4 + 20 \sum_{cyc} a b (a^2-b^2)^2 + 10 (a+b+c)^3 + 10(a+b+c)(a^2+b^2+c^2) + 12 \sum_{cyc} a^2(b+c) \ge 0 [/tex]

(by the way, I incorrectly used the [i]sym[/i] summing convention in my previous post, where I really meant [i]cyc[/i])

If you plug it back, you can find that

[tex] (a^2+ab+b^2)(b^2+bc+c^2)(c^2+ca+a^2) + \frac54 abc  \le 3 [/tex]

or the even stronger

[tex] \prod_{cyc} (a^2+ab+b^2) + abc \Bigl( \frac54 + \frac{5}{16} (a^2+b^2+c^2) + \frac{3}{16} \sum_{cyc} a^2(2-a) \Bigr] \le 3 [/tex]

Now how do you find such a decomposition ? In my case, a computer was involved in two ways: to perform the annoying algebraic expansions (but this could also have been done by hand) and to help me identify the correct terms to include in the sum. The second role is the most interesting, and is linked to some theory.

It is well-known that all nonnegative single-variable real polynomials can be expressed as a sum of (at most two) squares. Unfortunately, this result does not extend to the multi-variable case, since the are nonnegative polynomials that cannot be expressed as a sum of squares. An example is the so-called [i]Motzkin[/i] polynomial
 [tex] 1 + x^2 y^2 (x^2 + y^2 - 3) [/tex]
However, these examples are quite uncommon, and most polynomials do admit a sum of squares decomposition.

All of this is in fact related to Hilbert's 17th problem, which was solved by Artin in 1927 when he proved that any nonnegative (multi-variable) polynomial can be expressed as a sum of squares of [i]rational[/i] functions (i.e. ratios of two polynomials).

So all of this means that in principle any true polynomial inequality can be proved by expressing the difference between its sides as a sum of squares of polynomials or rational functions. The question is now : how do you find these sums ?

Well, it turns out that an efficient computational procedure was proposed a few years ago to compute the sum of squares decomposition of a nonnegative polynomial, if it exists. (see e.g. Nesterov, Parrilo, Lasserre, I can provide specific references if somebody is interested).

To apply this technique to our homogenized inequality, I had first to replace a, b, c by x^2, y^2, z^2 to account for the fact that the variables are nonnegative. Thus I had to express the difference
[tex] 3(x^2+y^2+z^2)^6-64(x^4+x^2y^2+y^4)(y^4+y^2z^2+z^4)(z^4+z^2x^2+x^4)[/tex]
as a sum of squares. The output of the algorithm was not that nice : first there were numerical inaccuracies, then the decomposition that was returned had 11 terms (in fact, most of these sos decompositions are not unqiue). However, by having a close look at these terms, I was able to guess a correct expression for the first three terms (admittedly, the first one is probably the most difficult to find) and notice that the remainder was obviously nonnegative.

To conlude with a few words about the general applicability of this technique, let me mention that you can also use it to find sums of squares of rational functions, if you know (or guess) an a priori upper bound on the degree of the denominators.

I also have to add that this algorithm is computationally nontrivial (both with respect to cpu time and numerical inaccuracy) and one cannot reasonably expect to solve problems involving more than a few dozen monomials. All of this is the subject of ongoing research in the applied mathematics community.

[quote="Michael Lipnowski"]Fanfan: Can you please provide specific references for algorithms on expressing non-negative polynomials as a sum of squares of rational functions?  I'm very interested.[/quote]

You can download one of the following research packages for MATLAB :
yalmip available at http://www.cds.caltech.edu/sostools/
gloptipoly available at http://www.laas.fr/~henrion/software/gloptipoly/gloptipoly.html
sostools available at http://control.ee.ethz.ch/~joloef/yalmip.msql

All of these require the resolution of "semidefinite optimization problems" as a subroutine, which can be done with one of the following
SeDuMi available at http://fewcal.kub.nl/sturm/software/sedumi.html
SDPT3 available at http://www.math.cmu.edu/~reha/sdpt3.html

Many articles are available on the subject : see for example the paper "Semidefinite programming relaxations for semialgebraic problems" by P. A. Parrilo (which you can download at the bottom of the sostools page linked above) ; however some knowledge of optimization theory is probably necessary to understand them thoroughly.

You can contact me if you need any help.

[quote="harazi"] But who manages to prove it geometrically is really great. [/quote]
Thank you for your compliment harazi.It makes me  :blush:
We will use complex numbers to show that P must be on the sides for maximum. We wil take one of the points on the real axis, and let a be its afix, then the side of the equilateral triangle is $a\sqrt 3$.
$\Rightarrow |p-a||p-a\varepsilon||p-a\varepsilon^2|\leq a^3\frac 98\Rightarrow (x=\frac pa)|x-1||x-\varepsilon||x-\varepsilon^2|\leq \frac 98\Rightarrow |x^3-1|\leq \frac 98$. Now draw the circle centered at $z=1(Z)$ and radius $\frac 98$. $|x^3-1|$ is the distance from $Z$ to a point on a certain segment, if we fix the argument of p and thus of x. Let X be the other extreme of this segment besides O. Than the maximum of the distance is achieved at O or at X, but at O $|x^3-1|$ is $1$ so we shall only study what happens in that X. Notify that raising the module of p we raise the module of x. Raising the module of p means bringing it on the proper side, if it is not in the center(this is a simple case to analyze). We shall raise the module of x until P gets on the side. So that X is actually the case when P gets on the side. On any side we now have that product equals $y(l-y)\sqrt{l^2+y^2-ly}$, where y is the distances from P to the closest vertice so it is $\leq l/2$. Just divide that by $l^3$ and get that the maximum of each of $\sqrt{l^2+y^2-ly}$ and $y(l-y)$ is achieved for the middle of that side. It's done. Thank you again harazi.
The product made me remember of the Romanian TST 5/2004 / first problem, and that's why I used complex numbers. Why on Earth didn't I think of using complex numbers then also? It would have bringed me to the imo...:((

[quote="darij grinberg"]Harazi and Pierre, how did you prove the inequality with Muirhead? I don't think it can be proven by Muirhead, or by Schur either, since in Muirhead and Schur, equality holds only for a = b = c, while in your inequality, we have equality for a = 0 and b = c = 1 (for instance), while a = b = c doesn't give equality. So I don't think Muirhead will help (a good inequality should not be Muirhead-bashable, BTW). If you still want to try, here is an equivalent form of the inequality:

[tex]13P_{2,2,2}+52P_{3,2,1}+13P_{4,1,1}-19P_{4,2,0}-2P_{3,3,0}+18P_{5,1,0}+\frac{3}{2}P_{6,0,0}\geq 0[/tex]
[/quote]

If your calculation is correct (I didn't verify), then it can be proved by Muirhead and Schur. Here is how it goes:

$\displaystyle\frac{1}{2}P_{6,0,0} + \frac{1}{2}P_{4,1,1} \geq P_{5,1,0}$ by Schur with r=4.

$\displaystyle \frac{3}{2}P_{6,0,0} + \frac{3}{2}P_{4,1,1} + 18 P_{5,1,0} \geq 21P_{5,1,0} \geq 19P_{4,2,0} + 2P_{3,3,0}$.

To make the equality to hold,  many leftover terms have to disappear which forces one of the variables to be 0.

sorry can anyone tell me where can i find anything about Muirhead?? i don't know his inequality

Fanfan's algorithmic method maybe powerful for fixed number of variables. Can it prove inequalities with variable number of variables? For a simple example, Cauchy-Schwarz inequality:

\[
\left(\sum_{i=1}^n a_i^2\right)\left(\sum_{i=1}^n b_i^2\right) \geq
 \left(\sum_{i=1}^n a_ib_i\right)^2.
\]

it seems like darij forgot that equality is for $a=0, b=c$ in schur.

one can prove cauchy-schwarz writing it as a sum of squares(though i think a computer wouldn't find a general decomposition, at least not the algorithms that are normally used):
$(\sum a_i^2)(\sum b_i^2)-(\sum a_ib_i)^2=\sum_{sym}(a_ib_j-a_jb_i)^2$

Peter

[quote="Peter Scholze"]one can prove cauchy-schwarz writing it as a sum of squares[/quote]

This is exactly what I mean, human can write it as a sum of squares easily, but computer may not be able to do it since the number of variables is not fixed.

[quote="fuzzylogic"]Fanfan's algorithmic method maybe powerful for fixed number of variables. Can it prove inequalities with variable number of variables? [/quote]

[quote="Peter Scholze"](though i think a computer wouldn't find a general decomposition, at least not the algorithms that are normally used)[/quote]

[quote="fuzzylogic"]This is exactly what I mean, human can write it as a sum of squares easily, but computer may not be able to do it since the number of variables is not fixed.[/quote]

Indeed, the fact that the number of variables is not fixed is an obstacle to the application of the method ... but you can always try it for a large arbitrary number of variables, and hope you (=the human) will be able to generalize the result easily (which might not be the case since the solution is often not unique, and the computer could return a specific solution that is not generalizable). You can also in principle add a constraint that stipulates that the sum of squares decomposition must be cyclic or symmetric in some of its variables, although I have never tried this myself.

[quote="ikap"]We will use complex numbers to show that P must be on the sides for maximum. We wil take one of the points on the real axis, and let a be its afix, then the side of the equilateral triangle is $a\sqrt 3$.
$\Rightarrow |p-a||p-a\varepsilon||p-a\varepsilon^2|\leq a^3\frac 98\Rightarrow (x=\frac pa)|x-1||x-\varepsilon||x-\varepsilon^2|\leq \frac 98\Rightarrow |x^3-1|\leq \frac 98$. Now draw the circle centered at $z=1(Z)$ and radius $\frac 98$. $|x^3-1|$ is the distance from $Z$ to a point on a certain segment, if we fix the argument of p and thus of x. Let X be the other extreme of this segment besides O. Than the maximum of the distance is achieved at O or at X, but at O $|x^3-1|$ is $1$ so we shall only study what happens in that X. Notify that raising the module of p we raise the module of x. Raising the module of p means bringing it on the proper side, if it is not in the center(this is a simple case to analyze). We shall raise the module of x until P gets on the side. So that X is actually the case when P gets on the side. On any side we now have that product equals $y(l-y)\sqrt{l^2+y^2-ly}$, where y is the distances from P to the closest vertice so it is $\leq l/2$. Just divide that by $l^3$ and get that the maximum of each of $\sqrt{l^2+y^2-ly}$ and $y(l-y)$ is achieved for the middle of that side. It's done. Thank you again harazi.[/quote]

Your proof is very smart, Ikap, but it has a little mistake:

[quote="ikap"]the maximum of each of $\sqrt{l^2+y^2-ly}$ and $y(l-y)$ is achieved for the middle of that side.[/quote]

That's wrong. The term $\sqrt{l^2+y^2-ly}$ achieves its [b]minimum[/b] at the middle of the side (i. e., for $y=\frac{l}{2}$).

Anyway, this mistake doesn't wreck your solution, since the product $y\left(l-y\right)\sqrt{l^2+y^2-ly}$ for $y\in\left[0;\;\frac{l}{2}\right]$ still achieves its maximum at $y=\frac{l}{2}$. This can be proved as follows:

Since the product that we are considering is homogeneous, we can WLOG assume that $l=1$. Then, what we have to prove is the following fact: The term $y\left(1-y\right)\sqrt{1+y^2-y}$ for $y\in\left[0;\;\frac12\right]$ achieves its maximum at $y=\frac12$. Proving this is equivalent to verifying the inequality $y\left(1-y\right)\sqrt{1+y^2-y}\leq\frac{\sqrt3}{8}$ for all $y\in\left[0;\;\frac12\right]$ (in fact, note that $\frac{\sqrt3}{8}$ is the value of the term $y\left(1-y\right)\sqrt{1+y^2-y}$ for $y=\frac12$). And, in order to verify this inequality, we proceed as follows:

Let $u=y\left(1-y\right)$. Then, on one hand, $0\leq u\leq \frac14$ (in fact, $0\leq u$ is trivial since $y\in\left[0;\;\frac12\right]$ implies $y\geq 0$ and $1-y\geq 0$, and $u\leq \frac14$ is a rephrase of $y\left(l-y\right)\leq \left(\frac{y+\left(1-y\right)}{2}\right)^2$, what is trivial by the AM-GM inequality, applied to the nonnegative numbers y and 1 - y). On the other hand, $1+y^2-y=1-\left(y-y^2\right)=1-y\left(1-y\right)=1-u$, and thus

$y\left(1-y\right)\sqrt{1+y^2-y}=u\sqrt{1-u}$,

and the inequality that we have to prove, $y\left(1-y\right)\sqrt{1+y^2-y}\leq\frac{\sqrt3}{8}$, simplifies to $u\sqrt{1-u}\leq\frac{\sqrt3}{8}$. By squaring this inequality, we see that it is equivalent to $u^2\left(1-u\right)\leq\frac{3}{64}$. But this inequality is true, since it is the product of the inequality $u\leq\frac14$, which is known to hold, with the inequality $u\left(1-u\right)\leq\frac{3}{16}$, which follows from

$u\left(1-u\right)-\frac{3}{16}=-\frac{1}{16}\left(1-4u\right)\left(3-4u\right)\leq 0$

(since both numbers 1 - 4u and 3 - 4u are $\geq 0$, as $u\leq\frac14$). This completes the proof.

By the way, you could simplify the way you presented your solution if you had a priori assumed (WLOG) that the vertices of the equilateral triangle lie on the unit circle, so that a = 1 (this would have made needless the distinction between p and x).

  Darij

Here is my solution. Not so nice. But it doesn't take so many calculations.
Rewrite the original polynomial in the way
$
f(a,b,c)=(a^2+ab+b^2)(\frac{3}{4}c^2+(a+\frac{c}{2})^2)(\frac{3}{4}c^2+(b+\frac{c}{2})^2)
$
Now we want to increase a,b by e and decrease c by 2e(e=epsilon. You know what I mean). We know the polynomial increase when
$
\frac{3a+3b}{a^2+ab+b^2}\geq\frac{3c}{a^2+ac+c^2}+\frac{3c}{b^2+bc+c^2}
$
And this holds when
$
\frac{3b}{a^2+ab+b^2}\geq\frac{3c}{a^2+ac+c^2}
$
and
$
\frac{3a}{a^2+ab+b^2}\geq\frac{3c}{b^2+bc+c^2}
$
This happens when
$\frac{a}{b}+\frac{b}{a}$ is smaller than $\frac{a}{c}+\frac{c}{a}$ and $\frac{b}{c}+\frac{c}{b}$
Let $x\geq y\geq z$. If $\frac{x}{y}\leq\frac{y}{z}$, we have $\frac{x}{y}+\frac{y}{x}\leq \frac{y}{z}+\frac{z}{y}\leq \frac{x}{z}+\frac{z}{x}$ so
$f(x,y,z)\leq f(x+z/2,y+z/2,0)$. The rest can be easily proved.
Otherwise, we can reduce $x$ to $x'$, increase $y$, $z$ to $y'$, $z'$ such that $x'z'=(y')^2$ than go back to the first case.

By the way, thanks a lot to Fanfan for the information he gave. And I also wonder what Muirhead inequality is.


(M.Volchkevich, 9--11) Given two circles and point $ P$ not lying on them. Draw a line through $ P$ which cuts chords of equal length from these circles.

From P, take a random line that cut the smaller ciricle, say the chord's length is a.

Make a chord with length a at the bigger circle. Say the center of the bigger circle is O

Make another circle X, concentric with the bigger circle, with radius twice the length of the distance of the midpoint of the chord to O.

Let Q at X so PO= PQ
Take a line through P and the midpoint of the OQ

Consider the circles (K,r) and (L,R) with r<R. Consider also the line e which passes through P.
 Assume $ x = KK^'  \bot e\;and\;y = LL^'  \bot e.$ Basically we want: $ y^2  - R^2  = x^2  - r^2  \Rightarrow y^2  - x^2  = R^2  - r^2  = c^2 \;or\;\left( {y - x} \right)\left( {y + x} \right) = c^2$  (c constant).
If
 $ KK_1^'  \bot LL^' \;and\;K^{''}$ the symmetric point of
 $ K_1^'$    with axis symmetry the line e, which moves (the point  $ K^{''}$) along the constant circle (1).
If  $ K^{''}  \bot LK$  then $ LS \cdot LK = c^2$, that means S is constant. Therefore $ SK^{''}$  determines, finally, a constant line (2).
The common point of circle (1) with the line (2) are giving the point  $ K^{''} ....$

S.E.Louridas


Between 3:00 p.m. and 4:00 p.m., for what fractional part of the hour does the digit $ ^{\prime\prime}$2$ ^{\prime\prime}$ appear on a 12-hour digital clock that shows hour and minutes? Express your answer as a common fraction.


modular arithmetic
number theory
prime numbers
number theory proposed

Given two integers $ x\not\equal{} y$, prove that there are infinite many prime numbers $ p$ with $ p | x^{q\minus{}1}\plus{}y^{q\minus{}1}$ for some prime number $ q$.

we use the well-known method
we suppose that $ (x,y)\equal{}1$
suppose that $ p_1,...,p_k$ are primes such that for each of them there exists a prime $ q$ such that $ p_i | x^{q\minus{}1}\plus{}y^{q\minus{}1}$
then we know that there exists infinitely many primes of the form $ (p_1\minus{}1)..(p_k\minus{}1)t\plus{}1$ .
now assume that one of those primes is $ r$ then it's obvious that $ x^{r\minus{}1}\plus{}y^{r\minus{}1} \equiv 2 \pmod {p_1...p_k}$ which means that there exists another prime ,say $ s$, different from $ p_1,...,p_k$ and $ s | x^{r\minus{}1}\plus{}y^{r\minus{}1}$
do this infinitely many times to obtain infinitely many primes.
so we r done!

Good solution! There is only a very small thing ignored, when we get $ x^{r\minus{}1}\plus{}y^{r\minus{}1} \equiv 2 \pmod {p_1...p_k}$, there is still a possibility that one of the $ p_i$ happen to be $ 2$ and $ x^{r\minus{}1}\plus{}y^{r\minus{}1}$ may be equal to a power of $ 2$. This case can be easily ruled out by considering primes with the form $ 2(p_1\minus{}1)..(p_k\minus{}1)t\plus{}1$ which guarantees that $ x^{r\minus{}1}\plus{}y^{r\minus{}1} \equiv 1,2 \pmod{4}$.

yes of course but i have a bad habit and it is that i never consider trivial cases :)
another method to rule out the case mentioned is to consider the fact that $ (x,y)\equal{}1$ and if $ x^{r\minus{}1}\plus{}y^{r\minus{}1}$ is a power of two since $ r\minus{}1$ is even and $ x,y$ can't be both even so this number must be equal to two modulo $ 4$ which implies $ x^{r\minus{}1}\plus{}y^{r\minus{}1}\equal{}2$ and the rest is trivial