algorithm
function
quadratics
vector
linear algebra
matrix

hi every body
[url]http://acm.sgu.ru/problem.php?contest=0&problem=200[/url]
plZ help me to solve this problem?
thX

[quote="moogra"]wtf. This is a contest problem as it says so on the link[/quote]

Resource:	Petrozavodsk Winter Trainings 2003
Date:	2003-02-06

Yeah, from 4 years ago.

Anyway, I have a sort of brute force algorithm that won't run in time but may (or may not) be insightful. Represent each number $ b_m$ as a binary string of length $ t$, where the $ i$th digit is a $ 1$ iff the largest power of $ p_i$ that divides $ b_m$ is odd. We want to find sets of numbers so that the total number of $ 1$'s in any digit place is even.

Given the constraints, I think we can recursively find sets that work by considering it digit-by-digit but in a sort of clever way. Start the the digit that has the least number of $ 1$'s occurring in it (over all $ b_j$'s), and continue, always choosing the digit that has the next least number of $ 1$'s. That way you limit the number of early calls of your recursive function and hopefully cut down on the time. There's no way that this runs in anywhere near two seconds (it won't ever finish for well-designed cases), but each binary string has at most nine $ 1$'s in it so maybe there is some way of modifying this to not need to find every working possibility.

So the link seems to be finally working, that's great. An interesting problem with a simple solution - I happen to know it because we're doing the quadratic sieve in school right now :D - it's nothing more than linear algebra in $ \mathbb{Z}_2$.

First, take the vectors as paladin8 explained. We're trying to find all such subsets of these vectors that add up to the zero vector in $ \mathbb{Z}_2$, or in other words minimal linearly dependent subsets. This can be done by solving a system of linear equations - the variables $ x_1$, $ x_2$, ... represent the presence or absence of i-th vector in the resulting linear combination and the equations are obtained from the vectors themselves, each original vector producing one column in the matrix of the system.

The number of solutions of a system of linear equations in $ \mathbb{Z}_2$ is easily calculated from the number of independent rows in the system. ($ 2^{n \minus{} m}$.) You have to substract one for the trivial $ (0, 0, ...)$ solution.

EDIT: I'm not sure if anybody will be able to understand what I'm saying, sorry if I use some odd english :lol:, so here's an example for the sample input {9, 20, 500, 3}:

$ 9\rightarrow (0,2,0)\rightarrow (0,0,0)$
$ 20\rightarrow (2,0,1)\rightarrow (0,0,1)$
$ 500\rightarrow (2,0,3)\rightarrow (0,0,1)$
$ 3\rightarrow (0,1,0)\rightarrow (0,1,0)$

The system of linear equations has the following matrix (I transposed the above vectors and used them as columns in the matrix):

$ \left( \begin{array}{cccc}
0 & 0 & 0 & 0 \\
0 & 0 & 0 & 1 \\
0 & 1 & 1 & 0 \end{array} \right)$

You can easily see that the number of independent rows is 2 and the number of columns is 4, therefore the answer is $ 2^{4\minus{}2}\minus{}1\equal{}3$.


Prove that for an integer $ n>\equal{}1$ we have $ n(1\plus{}\frac{1}{2}\plus{}\frac{1}{3}\plus{}\dots\plus{}\frac{1}{n})\geq (n\plus{}1)(\frac{1}{2}\plus{}\frac{1}{3}\plus{}\dots\frac{1}{n\plus{}1})$

Expanding the LHS and RHS we get:

$ n\plus{}\frac{n}{2}\plus{}\frac{n}{3}\plus{}...\plus{}1\geq  \frac{n}{2}\plus{}\frac{n}{3}\plus{}...\plus{}1\plus{}\frac{n}{n\plus{}1}\plus{}\frac{1}{2}\plus{}\frac{1}{3}\plus{}...\plus{}\frac{1}{n\plus{}1}$

canceling out what we can, we get:

$ n \geq \frac{n}{n\plus{}1}\plus{}\frac{1}{2}\plus{}\frac{1}{3}\plus{}...\plus{}\frac{1}{n\plus{}1}$

since $ n \geq 1$ we can multiply with $ n\plus{}1$ and get:

$ n^2\geq (n\plus{}1)\left(\frac{1}{2}\plus{}\frac{1}{3}\plus{}...\plus{}\frac{1}{n}\right)\plus{}1$

$ n^2 \minus{}1\geq (n\plus{}1)\left(\frac{1}{2}\plus{}\frac{1}{3}\plus{}...\plus{}\frac{1}{n}\right)$

divide by $ n\plus{}1$:

$ n \geq 1\plus{}\frac{1}{2}\plus{}\frac{1}{3}\plus{}...\plus{}\frac{1}{n}$

$ n \geq H(n)$

where $ H(n)$ is the harmonic number. Since:

$ \ln(n\plus{}1) \plus{} C \geq H(n)$

where $ 0<C<1$ we can say:

$ n \geq \ln(n\plus{}1)\plus{}C$

And that's true for all $ n \geq 1$.

We need to prove that
$ \frac {n^2}{n \plus{} 1}\geq\sum_{i \equal{} 2}^{n \plus{} 1}\frac {1}{i}$.
But $ \frac {n^2}{n \plus{} 1} > n \minus{} 1 > \sum_{i \equal{} 2}^{n \plus{} 1}\frac {1}{i}$.
The last inequality is true because every number in this sum is less than $ 1$.
I think that all of your problems must be moved to pre-olympiad zone.

Let $ 1\plus{}\frac 12\plus{}\dots\plus{}\frac 1n\equal{}H$. Then we need to prove that

$ nH\ge (n\plus{}1)\left(H\minus{}1\plus{}\frac{1}{n\plus{}1}\right)$

$ \iff n(H\plus{}1)\ge (n\plus{}1)H$

$ \iff n\ge H$

which is obviously true. Equality holds iff $ n\equal{}1$.


Sum of all digits of an integer $n$ is $100$, while the sum of all digits of $44n$ is $800$
Determine the sum of all digits of $3n$

hmm.. ;)
it looks like you need to have one more condition....

Well one possible solutions seems apparent...

[hide]
I came across this problem, and I had the right solution in mind right away.

$n = 1111....................1111$, where there are 100 1's.
Notice that when we multiply by 44, we have
$48888888........8888888888884$, where there are 99 8's and 2 4's, adding up to 800.

so $3n = 3333333333333333.............33333$, where there are 100 3's. and the sum is $\boxed{300}$
[/hide]

Let $S(n)$ be the sum of the digits of $m$.

Theorem: $S(mn)\le S(m)S(n)$, with equality if and only if we never have to carry when multiplying $m$ and $n$.
Proof: Multiply $m$ and $n$ as polynomials in the variable 10. $S(m)S(n)$ is the value of this product evaluated at $10=1$. Now we reduce to standard form by carrying; each step clearly reduces $S$ by a multiple of $10-1$.

Now we merely need to show that if the product $44n$ can be done without carrying, the product $3n$ can also be done without carrying.

[quote="jb05"]That proof isn't very satsifying. If I ran out of time, I may put that merely because I know that there is only 1 answer, but what if there are other solutions to these conditions?[/quote]

well.... i was in a hurry as well. Sorry about that. I will make more elegant proof. I was just giving one of the answer because I was doing something else and I just saw that it was a problem "to solve", not "to prove".


[quote="Anaxerzia"]Since AoPS have discontinued the Olympiad classes, I suggest that AoPS eventually compile these problems/ideas into books. I think people here would totally buy them, even for a high price...[/quote]If you were a good candidate for the Olympia classes, you should take WOOT. It has replaced the Olympiad classes altogether.

[quote="Anaxerzia"]I did take WOOT, but didn't the Olympiad classes have 12 classes or so for each subject? WOOT has about 4.
I'm not saying to bring them back, and I'm not complaining that they are gone. I'm just suggesting they are put into a book :).[/quote]I doubt we'll ever do that. It's just not worth the financial effort (not many people would buy a high-level book, and there are already tons and tons of Olympiad books out there which would make it ever more difficult to sell yet another one). IMO besides WOOT you just need to get a lot of practice. WOOT should be used as a guide in your preparation, and shouldn't be your [b]only[/b] olympiad preparation (ie to get to the IMO you would have to work roughly 3 times more stuff than what we gave you at WOOT :) ).


function
induction
number theory solved
number theory

find all function f :
f: N->N
f(x^2+y^2)=x*f(x)+y*f(y)

hey ,duzn't this thing hold
f(n) = f(k) + f(n-k).this i think finishes off the job ,duznt it.though i havent tried this full thing yet
cheers


The reason Im posting this is because it seems too good to be true; that Ive found a way to trisect an angle using only a compass and a straightedge. However, this has worked with every angle Ive tried, so verification from others would be nice.

The steps will assume an acute angle, however the construction for obtuse angles is fundamentally the same, the only difference being the positioning of certain lines. Anyway, on with the steps.

1) Construct the angle ABC
2) Put a point somewhere on line AB, call it D
3) Construct a line parallel to line BC through point D
4) Construct a line perpendicular to line BC through point D, the point where it meets line BC will be point E
5) Split line DE into three equal segments
Note: Construction for number five has already been proven so I will not go into detail. An explanation can be found [url=http://www.jimloy.com/geometry/trisect0.htm]here[/url].
6) Construct point F on the line parallel to BC the same length as 2/3 of line DE
7) Construct line BF, it will trisect angle ABC

Follow those steps and meansure the angles. Angle ABF will be 1/3 of angle ABC.

-m3

Nice approximate construction  ;)
It is even wrong for angles that can be trisected, e.g. $90^\circ$.

This trisects the angle $\theta$ if and only if $\frac{3\sin\theta}{3\cos\theta+2\sin\theta}=\tan\frac23\theta$.
That only happens for a few values; the only nonzero acute angle for which it is true is approximately 51.75 degrees.

The angle you construct is off by about 2-3 degrees for most acute angles; try it for a $30^\circ$ angle to see how bad it is.


Let $A_{n}$ be the set of all real numbers of the form $1+\frac{\alpha_{1}}{\sqrt 2}+\frac{\alpha_{2}}{(\sqrt 2)^{2}}+\cdots+\frac{\alpha_{n}}{(\sqrt 2)^{n}}$, where ${\alpha_{j}\in{-1, 1}}$ for each $j$. Find the number of elements of $A_{n}$, and find the sum of all products of two distinct elements of $A_{n}$

[quote="Rust"]It is obviosly number of elements $A_{n}$ is $2^{n}$. \[\sigma_{2}=\sum_{x\not=y,x,y\in A_{n}}xy=\frac{1}{2}(s_{1}^{2}-s_{2}),\] were \[s_{i}=\sum_{x\in A_{n}}x^{i}.\] It is easy to calculate $s_{1}=2^{n},s_{2}=2^{n+1}-1.$
Therefore $\sigma_{2}=\frac{1}{2}(2^{n}-1)^{2}.$[/quote]

Well, is that obvious?
I don't think so.


Hello everyone,
I am writing a document and i want my subsections to appear in italics. i am using the report style file. can anyone help me how to modify that? Thank a lot.

At the beginning of your document add[code]\let\oldsubsection\subsection
\renewcommand{\subsection}[1]{\oldsubsection{\textit{#1}}}[/code]\oldsubsection is used to prevent recursion otherwise the new version of \subsection would keep referring to itself ad infitum.


geometry
perimeter
ratio
area of a triangle
Heron\u0027s formula

I need help on this question:

A triangle has three altitudes, 12,15, and 20. What is its perimeter?

Hint:
[hide]Let the side that has the altitude 12, be x.  Now figure out the other three sides using 1/2bh.[/hide]

hmm i have to think about this. I didn't know you could use 1/2bh just to do that.

ok, i have the equation:

$\25y^{2}-25xy=18x^{2}$

I cant find a way to solve for y or x.

nicely done. I couldn't find the equation either.

[quote="indianamath"]In terms of x, the three sides would be x, 3/5x, and 4/5x.  Bbut I still cant find an equation to solve for x[/quote]
[hide]Just use Heron's Formula to find the area of the triangle in terms of $x$, and then divide this by one of the side lengths to find the corresponding altitude, and you have the equation you need.[/hide]

After finding the lengths of the sides you could use heron's formula.

Thanks so much! I really never thought about Heron's formula. it was drifting in the back of my mind.

[hide]
The altitudes are $12$, $15$, and $20$, so the bases opposite the altitudes are in the ratio of  $\frac{1}{12}: \frac{1}{15}: \frac{1}{20}$, or $25: 20: 15$. Since this is a pythagorean triplet, the triangle is a right triangle. The legs are the two longest altitudes ($15$ and $20$), so the third side of 25 and the perimeter is $60$.
[/hide]

I know where you got this.

Rose Hulman competition.

I just took it today. And I missed this problem!


[quote="joejia"]What positive integer is 56 less than its square?[/quote]

[hide]$x+56=x^2$

$x^2-x-56=0$

$x=\frac{1\pm\sqrt{1+224}}{2}$

$x=\frac{1\pm\sqrt{225}}{2}$

$x=\frac{1\pm15}{2}$

$x=\frac{16}{2}=8$

$x=\frac{-14}{2}=-7$

$\text{Therefore the anwer must be 8.}$[/hide]

[hide]

The equation is $  x + 56 = x^2  $
The closest square from 56 is 64.

$ 64 - 56 = 8; 8^2 = 64  $

Therefore, the answer is $  \framebox{8}  $[/hide]

[quote="prototypex24"][quote="joejia"]What positive integer is 56 less than its square?[/quote]

[hide=""Solution""]$x=x^2-56$
$x^2-x-56=0$
$(x-8)(x+7)=0$
$x=8$[/hide][/quote]

There is a minor mistake in your answer. It would be $x^2-x+56$ instead of $x^2-x-56$


Let $a, b, n$ be positive integers with $\gcd(a, b)=1$. Prove that \[\sum_{k}\left\{ \frac{ak+b}{n}\right\}=\frac{n-1}{2},\] where $k$ runs through a complete system of residues modulo $m$.

[quote="Peter"]Let $ a, b, n$ be positive integers with $ \gcd(a, b) = 1$. Prove that
\[ \sum_{k}\left\{\frac {ak + b\}{n}\right\} = \frac {n - 1}{2},
\]
where $ k$ runs through a complete system of residues modulo $ m$.[/quote]
I think it not true.
For example $ a = n$ then
$ \{\frac {ak + b}{n}\} = \{\frac {b}{n}\}$
If we chose $ b = 1$ then
$ \sum_{k}\{\frac {ak + b}{n}\} = 1$
It is $ \gcd(a,n) = 1$ but it is quite easy.
Because $ \{ak + b\}$ is a complete reside mod n when $ k$ run through an complete reside mod n.

And there is a typo too: the last $ m$ should be $ n$ in the statement. I will correct both for the next release. Thanks for reporting.


limit
combinatorics unsolved
combinatorics

let \[ (x_n) \] be a sequence, determined by the following recursive formula

\[ x_1  \equal{} \frac{1}{2}\]
and

\[ x_{n \plus{} 1}  \equal{} x_n  \minus{} x_n ^2  \plus{} x_n ^3  \minus{} x_n ^4  \plus{} ... \plus{} x_n ^{2007}  \minus{} x_n ^{2008} (n \in N*)
\]
Find the limit \[ \mathop {\lim }\limits_{n \to \infty } nx_n
\]

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


Many times I have wanted to import code from an external file into a $ \text{\LaTeX}$ document. Usually I would want code in a block like

[code]
\begin{verbatim}
  MY CODE
  ...
\end{verbatim}
[/code]

Now, if i wanted to import code in plaintext, it would be easy with something like

[code]
\input{codefile.cpp}
[/code]

Yet horrible things happen with

[code]
\begin{verbatim}
\input{codefile.cpp}
\end{verbatim}
[/code]

As that will obviously just put in verbatim the line "\input{codefile.cpp}". I had solved this problem by hacking together a simple latex preprocessor which I signaled with code hidden in comments, but that would be less desirable than a native solution. Is it possible to escape the verbatim environment?[/code]

Use the [url=http://tug.ctan.org/cgi-bin/ctanPackageInformation.py?id=listings]listings package[/url]. At its simplest:
[code]\usepackage{listings}
...
\lstinputlisting{codefile.cpp}[/code]
but do read the manual as it has a vast number of options.


The diagram below shows 3 circles,A,B and C,overlapping one another. The total are of B and C is 4/5 that of A. Given that 1/6 is shaded 1/5 of B is shaded and 1/3 of c is shaded, find the ratio of the aren of A tot he area of B to the area of C. :huh:

Is the diagram necessary for the problem?

and is this what you mean?

The diagram below shows 3 circles, A, B and C,overlapping one another. The total areas of B and C is 4/5 that of A. Given that 1/6 of A is shaded 1/5 of B is shaded and 1/3 of C is shaded, find the ratio of the area of A to the area of B to the area of C.

[quote="theflame9"]B is over lapping A.A is overlapping B and C and C is overlapping A[/quote]
Could you scan it? Or upload a picture?

[hide]Let
$ a=\text{Area of circle A}$

$ b=\text{Area of circle B}$

$ c=\text{Area of circle C}$

This is all the information we need:
$ b+c = 4a/5$
$ b/5+c/3=a/6$

We want a:b:c.
Eliminate $ b$:
$ b+5c/3=5b/6$
$ 2c/3=a/30$
$ a=20c$

Substitute into the 1st equation:
$ b+c=16c$
$ b=15c$

So $ a: b: c = \boxed{20: 15: 1}$[/hide]

[quote="abacadaea"]i think i made a mistake.

let x, y, and z be there Area of the circles A, B, and C respectivley. The shaded areas together is equal to $ \frac{1}{6}x$. its is also equal to the left shaded area, $ \frac{1}{5}y$, plus the right shaded area, $ \frac{1}{3}z$. now we have the equation 1 equation: $ \frac{1}{6}x=\frac{1}{5}y+\frac{1}{3}z$. we also know $ x=\frac{4}{5}(y+z)$. so i have 2 equations

$ x=\frac{4}{5}(y+z)$
and
$ \frac{1}{6}x=\frac{1}{5}y+\frac{1}{3}z$

....and im stuck.[/quote]

Why don't you just read my solution?

Anyways, since we want $ x: y: z$, set $ z=1$ or something and solve for $ x, y$.

i try to solve things on my own. if i read solutions its no help to me. i have to actually try it to learn it

Generally, unless you have an original way of solving it, which you didn't it's best to leave the question. I can't stand it when an answer is posted a hundred times... once is enough. Even two makes my hands twitch.

dont post a reply unless you have alteast the slightest clue about what we r talking about and if you dont, read other posts

What you could do is just I think express B in terms of A, and C in terms of A. Create 3 variables and solve. So then you get

(b/5)+(c/3)=(a/6) where a,b,c are the areas of the three circles.
=>b+c=(4a/5)
=>b+5c=(5b)/6
=>b=15c
=>a:b :: 20:15

b+c=16c
=>15a
=>a:b:c :: 20:15:1

Uh, the solution has been posted already.


What is the angle formed by the hour and the minute hands of an analog clock at 3:54 PM.

meh, use that one formula angle = |5.5m - 30h| and you get 207 degrees which is equal to 153 degrees.

Psh, that doesn't show that you like using math!

You're supposed to draw it, then actually calculate using fractions and other fun math stuff. xD

Sorry but how you can calculate it? I mean where is the hour pointer is anyway? It's not fixed at 3 or 4 its something in between  :wink:

From one hour to another, the degree is $ \frac{360}{12}\equal{}30$.  For each minute, the hour hand moves $ \frac{30}{60}\equal{}\frac{1}{2}$ a degree.  We have $ 54$ minutes, so it is $ 3\times 30\plus{}\frac{54}{2}\equal{}117$ from the $ 12$, or $ 3$ degrees from the $ 4$.  In the meanwhile, the minute hand goes $ \frac{360}{60}\equal{}6$ degrees every minute.  So it is $ 54\times 6\equal{}324$ degrees away from the $ 12$ when going the same way as the hour hand (counterclockwise).  We do $ 324\minus{}117\equal{}207$, or $ 153$, just like cognos599 said.


modular arithmetic
combinatorics unsolved
combinatorics

prove that if n is a natural number such that 1+(2^n)+(4^n) is prime then n=3^k (k=0 or k is natural)

If $ n\equal{}3^k.l , (l,3)\equal{}1$

$ i) l\equal{}3t\plus{}1 \Longrightarrow$ $ 1\plus{}2^n\plus{}4^n \equiv 1\plus{}2^{3^k.l}\plus{}4^{3^k.l} \equiv 1\plus{}x^{3t\plus{}1}\plus{}x^{6t\plus{}2} \equiv 1\plus{}x\plus{}x^2 \equiv 0 \pmod {1\plus{}x\plus{}x^2}$ $ (x\equal{}2^{3^k})$

$ ii) l\equal{}3t\plus{}2 \Longrightarrow$ $ 1\plus{}2^n\plus{}4^n \equiv 1\plus{}2^{3^k.l}\plus{}4^{3^k.l} \equiv 1\plus{}x^{3t\plus{}2}\plus{}x^{6t\plus{}4} \equiv 1\plus{}x^2\plus{}x \equiv 0 \pmod {1\plus{}x\plus{}x^2}$ :wink:


For a,b,c,d positive real numbers such that a+b+c=1 prove that

a^3+b^3+c^3 +abcd >= min(1/4,1/9+d/27)

I have a different solution:
  Suppose it's not the case. Then d< 1/abc*[1/4-sum a^3]. We will prove that sum a^3+abcd>=d/27+1/9. This is d(1/27-abc)+1/9<=sum a^3. But it's enough to prove that [1/4-sum a^3]/abc* [1/27-abc]<=sum a^3-1/9, or sum a^3+15/4abc>=1/4. Since sum a^3=3abc+1-3sum ab, this comes down to sum ab<=(1+9acb)/4. This is really a magnificient problem.


geometry
incenter
circumcircle
rhombus
symmetry
geometry unsolved

For some reason I am not getting this concept at all.

Here is a couple sample questions:

1. Show how to construct a triangle given the 3 midpoints of its sides.

2. Show how to construct a traingle given the length of one side, the size of an adjacent angle and the length of the median from that angle

3. Show how to construct a triangle given the length of one side, the distance from an adjacent vertex to the incenter and the radius of the incircle.

These are my attempts:

1. Given 3 midpoints of 3 sides we can erect perpendiculars from these midpoints. Then where these 3 perpendicular lines meet is the incenter. So I think we are done. Is this right?? Seems too ez.

2. Let X be the angle we are given. Let L be the length of the adjacent side, call it AB. Just copy the angle over to the other side of the median with the same lenght L, say AC. Then we simply join B to C, and we are done. Again looks too simple, and probably wrong. I never even use the length of the median.

3. Form the incircle with radius r. Copy your line L (call it AB) to the other side of the line segment going through the incenter. This line (call it AC) and our original line L must hit the incircle and form a 90degree with the incenter. Now form the circle at center A, with radius L. Your point C must line on this circle. So we can simply draw a line from B to C. Again doesn't even take into account the distance to the incenter.

So I know I am definately doing stuff completely wrong and probably missing some fundamental concepts behind this, but I just don't know what I am doing wrong or how to fix it.

1: It's the circumcenter, not the incenter, and you don't know the circumradius. Here's how I'd do it:

Let the midpoints be $A,B,C$. Draw the parallel to $AB$ through $C$, the parallel to $BC$ through $A$, and the parallel to $AC$ through $B$. These three lines are the sides of your new triangle.
That's not the only way, and there are probably more efficient variations on this idea.

2: You've just constructed an isosceles triangle with given side and angle. The median length could be anything. My idea:

Start with points $A$ and $B$, with a ray $\ell$ through $B$ at the given angle $\theta$ with $AB$. Let $C$ be the third vertex, and $M$ be the midpoint of $AC$. What is the locus of $M$ as $C$ varies along $\ell$? Find the point on this locus at the right distance from $B$ (intersect with a circle) and the rest is easy.

3: I can't even follow what you're trying to do.

Let's start by naming a few points: $I$ is the incenter, and we have a circle $\Gamma$ of radius $r$ around $I$. Choose an arbitrary point $P$ on $\Gamma$; we want this point to be on the side $AB$ of the triangle (with given length). Also, we know the length of $AI$. Use the tangency; $AP$ and $IP$ are perpendicular, so we have a known line through $P$ and a known circle centered at $I$ both containing $A$.
Once you find vertex $A$, the rest is easy.

Excellent, thank you.  Here is a couple more if you are up for it.

4.  Show how to construct triangle ABC given the length of side BC and the lengths of the altitude from B and C

5. Show how to construct a traingle given the measure of one angle, the length of the internal bisector of that angle and the radius of the incircle.

6. Prove
a) the diagonals of a rhombus bisect each other at right angles
b) the disagonals of a kite or dart (possibly extended) intersect at right angles.

I'd rather not do your homework for you.

A few hints:
#4: A side and an adjacent altitude determine an angle of the triangle.

#5: Start with the angle and its bisector. Find the incenter and the incircle, then use the fact that the incircle is tangent to the third side.

#6: Use symmetry and the obvious isosceles triangles.

I think I got it (well the first 3 at least).  Will look at the rest later.  Thx for all the help.


Find all number triples $(x,y,z)$ such that when any one of these numbers is added to the product of the other two, the result is 2.

[hide="My attempt.."]

By the problem, we know $x+yz = y+xz = z+xy = 2$. Ignoring 2 for now, we can get several equations:

$x+yz = y+xz$
$y+xz = z+xy$
$z+xy = x+yz$

Solving this gives $x = y = z = 1$. So there is only one triple and that is $(1,1,1)$. [/hide]

Right?  :|

Okay, given:

\[ ab+c=2 \]
\[ ac+b=2 \]
\[ bc+a=2 \]

Seeing as every equation is equal to the other, choose the first two equations to get $ab+c = ac+b \Longleftrightarrow a(c-b)-1(c-b) = 0 \Longleftrightarrow (a-1)(c-b)=0.$  Applying this for every two equations, we see that $a=b=c$, and also $a=b=c=1.$  Specifically, if one sets all variables equal to one sole variable, then we get all answers:  $a^2+a = 2 \Longleftrightarrow (a+2)(a-1)=0 \Longleftrightarrow a=1,-2$.

So your solutions are $(1,1,1)$ and $(-2,-2,-2).$

Good solution.

I should've known.. CMO problems often have two answers. I should be careful next time when I solve problem like this.  :wallbash:


integration
function
limit
linear algebra
matrix
group theory
abstract algebra

My team took 1st place! I congratulate all my team-mates! :first:

These are selected problems.

1. http://www.mathlinks.ro/Forum/viewtopic.php?t=31689
Given two piles of stones. An automatic machine choose the pile, which contain an even number of stones (if both piles contains even numbers of stones then machine selects an arbitrary pile), and put half of these stones to another pile. If both piles contains an odd number of stones then machine stops. Find all pairs of initial numbers of stones in the piles s.t. the machine will necessarily stop.

2. http://www.mathlinks.ro/Forum/viewtopic.php?t=31690
Is it possible that the equation $y'=y^2+f(x)$ has two distinct periodic solutions $y_1$ and $y_2$ with the same period $T$ s.t. \[\int_0^T(y_1(x)+y_2(x))dx\neq 0\,?\]

3. http://www.mathlinks.ro/Forum/viewtopic.php?t=31691
Let $\displaystyle f(x)=\frac{1}{1+2x+3x^2+...+2005x^{2004}}$. Find $f^{(2005)}(0)$.

4. http://www.mathlinks.ro/Forum/viewtopic.php?t=31704
Let $S$ be finite commutative semigroup wrt multiplication consisting of integer-valued non-invertable matrices of order 2. Prove that if $|S|>5$ then $AB=0$ for all $A,B\in S$.

5. http://www.mathlinks.ro/Forum/viewtopic.php?t=31697
Is there a word $x=x_1x_2...x_{2005}$ ($x_i\in\{a,b\}$) s.t. it is impossible to partition $x$ to at most
a) 811 palindromes?
b) 669 palindromes?
c) 668 palindromes?

6. http://www.mathlinks.ro/Forum/viewtopic.php?t=31692
Let $\{a_n\}$ be a sequence of positive numbers. Is it possible that \[\limsup_{n\to\infty}\left(\frac{1+a_{n+1}}{a_n}\right)^n<e.\]

7. http://www.mathlinks.ro/Forum/viewtopic.php?t=31698
Given a simple graph of $n$ vertices, a degree of each vetrex is at most 8. Is it true that we can color all vertices into 3 colors in such a way that the number of edges with monochromatic vertices will be at most $n$?

8. http://www.mathlinks.ro/Forum/viewtopic.php?t=31694
Prove or disprove that each function $f:\mathbb{R}\to\mathbb{R}$ has at most countable number of discontinuos points of first kind ($x_0$ is a point of discontinuos of first kind iff there exist $\lim_{x\to x_0-0}$ and $\lim_{x\to x_0+}$).

9. http://www.mathlinks.ro/Forum/viewtopic.php?t=31700
A square matrix $2005\times 2005$ is filled with 1 and 0. It is known that for each unit the quantity of units in the same column OR in the same row is at most 40 percents. What is maximal number of units in the matrix?

10. http://www.mathlinks.ro/Forum/viewtopic.php?t=31705
Is it true that there is a subgroup, which is isomorphic to $S_3$, in the multiplicative group of invertable matrices from $M_2(\mathbb Z)$ ?

11. http://www.mathlinks.ro/Forum/viewtopic.php?t=31701
Myth calculated for $n$ real numbers all $n(n-1)/2$ pairwise sums. Pierre wants to find initial numbers from these sums. He is allowed to ask questions, which assume only answers [b]Yes[/b] or [b]No[/b]. How many questions does he need?
a) $n=4$;
b) $n=5$;
c) $n=6$;
d) $n$ is an arbitrary natural number.

12. http://www.mathlinks.ro/Forum/viewtopic.php?t=31702
A cube $2005\times 2005\times 2005$ consists of unite cubes. Some of unit cubes are colored in black color in such a way that for each black cube there is at least one row, which contains this cube and doesn't contain other black cubes (there are horizontal, vertical and perpendicular rows). What is the maximal number of black cubes?

13. http://www.mathlinks.ro/Forum/viewtopic.php?t=31703
There are 250 hobbits in Shir. Each hobbit lives in his own house. Every evening some hobbits pay a visit to all hobbits which sit in their houses this evening.
a) How many evenings do they need in order each hobbit pays a visit to each other hobbit?
b) How many evenings do they need in order for each two hobbits $A$ and $B$ either $A$ pays a visit to $B$ OR $B$ pays a visit to $A$?

14. http://www.mathlinks.ro/Forum/viewtopic.php?t=31695
Let $X\subset \mathbb{R}$ posses the following property: any function $f:\mathbb{R}\to \mathbb{R}$, satisfying the condition that its restriction to any segment $[a,b]\subset X$ is a continuous fucntion, is continuous on $X$.

Prove that $X$ is union of at most countable number of intervals and points. (Note: point $\{x\}$ is a segment $[x,x]$.)

15. http://www.mathlinks.ro/Forum/viewtopic.php?t=31696
The function $f:\mathbb{R}\to\mathbb{R}$ satisfies \[\forall x\in\mathbb{R}\,\exists\delta>0\,\forall y\in\mathbb{R}:\ |y-x|<\delta\implies f(y)\geq f(x).\]
What are possible values for cardinality of $f(\mathbb{R})$?

[i]P.S. Most of problems are not original, unfortunately.[/i]

Congratulations, Myth! Could you tell the rules of the contest (how many members per team, how much time for solving problems, etc.)?

It is a regular math fight (математический бой), but a preparation time was significatly decresed to 1.5 hours for 8 problems. A team consists of 6 members. It can include university students, at most 1 school student and at most 1 PhD studentd (it was me :) ).

[quote="Myth"]A team consists of 6 members. It can include university students, at most 1 school student and at most 1 PhD studentd (it was me :) ).[/quote]It is now clear how you obtained first place :D :D :D Congratulations :P

[quote="Myth"]It is a regular math fight (математический бой), [/quote]

What does this mean?   Is it a written or oral competition?

[quote="fleeting_guest"]
What does this mean?  [/quote]

Below is a short  description of math. fights you can find on the web. It concerns a two-team version. There is also a three-team version where the rules are even more complicated: there are regulations about who can challenge who during each turn and special rules for the team that is neither challenging nor challenged; basically they say that this third team can interfere if the first two teams have nothing more to say but the third team can add something valuable to the discussion (for instance, if there was an incorrect challenge, the third team gets right to present its solution or, if the opponent agreed with the solution, the third team gets the right to try to refute it).
Another thing I should add is that, at least in Leningrad, we also had math. fights when the problems were given several days before the fight rather than several hours. This practice was especially common for student math. fights and allowed to include really hard problems.

Now the promised description:

"Mathematical fights have been conducted in Soviet mathematical schools for more than three decades. I. Ya. Verebeychik, a mathematics teacher of school No. 30 in Leningrad, invented and organized them in his school since 1965. They became popular all over Russia since the beginning of 70-s, when the report about the mathematical fight that took place in Chelyabinsk during the All-Union Mathematical Olympiad, appeared in the “Kvant” magazine (Fedotov, 1972). Almost two decades later, a more detailed account of the rules was published in the journal “Matematika v Shkole” (Deryagin et al., 1990).
A mathematical fight is a competition between two teams in solving non-standard difficult problems selected by the jury, in oral explaining the solutions, in checking the solutions of the other team. It consumes usually one day’s time (from the morning till the evening). Mathematical fights have well-developed and rather complicated rules.
2. The rules.
Two teams consisting of equal number of members (usually from 6 to 12), receive from the jury a set of problems (the number of problems is even and usually approximately the same as the number of members in a team) and try to solve them sitting in different rooms for some time (usually from 2 to 5 hours, sometimes much less). After the lunch, the main fight begins. First, two captains of the teams compete in answering a mathematical puzzle posed by the jury. The winner of the captains’ contest decides which team will make the first move. Then the chosen team makes the first move (moves in mathematical fights are called “challenges”), choosing one problem and challenging the second team to present a solution of it. Two cases are possible: 1) The second team accepts the challenge; 2) The second team rejects the challenge.
If the second team accepts the challenge, it must set one of its members to present the solution. In this case, the first team must set somebody to refute the solution found by the second team.
If the second team rejects the challenge, the teams exchange their roles, the first team must present the solution of the problem chosen, and the second can refute the solution. In this case, if the first team does not itself have a solution and so refuses to present it, the challenge of the first team is declared incorrect. However, if he first team presents the solution and the second team proves it to be false, the challenge is also declared incorrect. Accepted challenge is always considered "correct" whether the first team has a solution or not.
Each problem “costs” 12 points (thus, the difficulty of the problems should not be indicated). During one turn, the challenged team gets all 12 points if the solution presented is correct and full. On the other hand, the challenging team gets 6 points for the full and correct refutation. Moreover, in this case the roles also (as in the case of the rejection of the challenge by the second team) become inverted, and the first team, presenting its own solution, can get up to 6 additional points for their solution (the second team becoming the opponent). If the first team can not present own solution, the jury gets the rest of points (or the part of the rest if the solution is partial). During one turn, the roles can be inverted only once (in cases of the rejection or false solution by the second team).
After the first turn, if the challenge was admitted to be correct, the second team makes the move with another problem and everything is repeated in similar way. If the challenge is admitted "incorrect", the first team has to challenge again.
In some moment, one of the teams may admit that it does not have solutions of the problems left. In this case, it may give up challenging. In case of such refusal, the other team has the right to present all other solutions it has got. The team that gave up challenging has the right only to refute.
The fight finishes when there are no more undiscussed problems, or one of the teams gives up challenging in some moment and the other refuses to present the solutions of the rest of the problems.  The jury sums the points of each team and names the winner. If there is time left, the jury demonstrates the solutions of problems not solved by the participants".


vector
abstract algebra
Ring Theory
superior algebra
superior algebra unsolved

Let $ E_1,E_2,$ and $ F$ be three quasi-finite fields and $ E_1$ relatively algebraically closed in $ F$. Let $ \tilde{E}_1$ be the algebraic closure of $ E_1$. Prove that
\[ \left (\tilde{\mathbb{E}}_1\bigotimes_{\mathbb{E}_1}\mathbb{E}_2\right )\bigotimes_{\mathbb{E}_2}\mathbb{F}\approx_\mathbb{F}\tilde{\mathbb{E}}_1\bigotimes_{\mathbb{E}_1}\mathbb{F}.
\]
Looks like it follows from the universal property of tensor products but I'm not quite experienced to use the universal properties.

Edit: $ E_1\subset E_2\subset F$ and a quasi-finite field = a perfect field with precisely one extension of each degree.

First, I'm almost positive that you've named some unnecessary conditions in this problem.  Where did it come from?

Second, what does $ \tilde{\mathbb{E}}_1\bigotimes_{\mathbb{E}_1}\mathbb{E}_2$ mean?  How are you identifying $ \mathbb{E}_2$ as a vector space over $ \mathbb{E}_1$ without the assumption $ \mathbb{E}_1 \subset \mathbb{E}_2$?

Third, if you define terms such as "quasi-finite", you may get more responses.

let $ R \to S, R \to A, S \to B$ be (commutative, unital) algebras. then

$ (A \otimes_R S) \otimes_S B \equal{} A \otimes_R B_{/R}$

where $ B_{/R}$ is the algebra $ R \to S \to B$. this is because for every test $ S$-algebra $ T$ we have natural bijections

$ Hom_S((A \otimes_R S) \otimes_S B, T) \equal{} Hom_S(A \otimes_R S,T) \times Hom_S(B,T)$
$ \equal{} Hom_R(A,T_{/R}) \times Hom_S(B,T) \equal{} Hom_S(A \otimes_R B,T)$

for the last step, remark that $ S \to B \to A \otimes_R B_{/R}$ defines a $ S$-algebra structure on $ A \otimes_R B_{/R}$ so that the natural bijection

$ Hom_R(A \otimes_R B_{/R},T_{/R}) \equal{} Hom_R(A,T_{/R}) \times Hom_R(B_{/R},T_{/R})$

restricts by construction to the desired bijection. finally, the yoneda-lemma yields the desired natural isomorphism. this arguments works for pushouts in arbitrary categories (here rings). if you want to do it more explicitely, you will repeate these arguments.

This is from J. Ax's paper "The elementary theory of finite fields". I was wondering what do you -oo- mean by "algebra $ R\to S\to B$. I have been taught that an algebra is a set with a module structure and a bilinear mapping. But in $ R\to S\to B$ there are two maps.

if $ f : R \to A$ is a ring homomorphism, $ ra : \equal{} f(r)a$ yields a $ R$-algebra structure on $ A$. conversely, if $ A$ is a $ R$-algebra, the structure comes from the ring homomorphism $ R \to A, r \to r \cdot 1$. thus the category of $ R$-algebras is isomorphic to the category of ring homomorphisms on $ R$.


your right i forgot to check if the answers were valid. I guess there is no solution.

[quote="10000th User"]No one checked his solution. It has an error at the fifth line.[/quote]

yeah it should be divide 12 because of divide
[quote]
\log_{2}(x+1)+\log_{2}(x-3)-\log_{2}(12)=0
\log_{2}(12(x+1)(x-3))=0[/quote]


Asymptote
geometry
geometric transformation
LaTeX
AMC
USA(J)MO
USAMO

Hi guys,

It's been 3 years now since we have the [url=http://www.artofproblemsolving.com/resources.php]Resources (Contests)[/url] pages up and since we're backstaging a small revamp of the site right now, and I will be recoding this section anyway, I'm opening this topic for you guys to share your suggestions about the resources pages.

I would like you to be concise, short and clear in what you are suggesting. Think of most of the people that could use that, not just yourself. Also some features might be technically not fesible at this point, so I will not take everything into consideration. Also some of the stuff might not arrive in the Contests 2.0.

I will reserve the second post in this topic to see what I decide is worth adding and not.

These the features that are to be included:

- multiple nested navigation (ie right now it's kind of hard-coded, it's hard to add new competitions, new countries, etc. -- this will be modified in the future)
- wiki style problem entering (ie anyone can enter problems, but they have to be vetted by either the community or the resource managers)
- problems will be held in their own database
- tagging
- problem rating (this is to be further discussed -- does it mean difficulty rating or beauty rating, or both?)
- adding solutions wiki style with vetting by resource managers required
- automatic reports on missing problems / wrong solutions / typos / mistags
- ability to create problem sets
- images in PDF files (asymptote only, sorry guys don't need jpgs or stuff like that)
- translations of problems / solutions

- "tagging": What do you want to tag? Topic area? Difficulty rating?

- "ability to create problem sets": You should be able to define the topic areas and therein the difficulty ranges you want to have included in your sample paper.

[quote="orl"]- "ability to create problem sets": You should be able to define the topic areas and therein the difficulty ranges you want to have included in your sample paper. Besides the classes AoPS/ML could generate profit from it by providing a service where submitted student papers ($ \text{\LaTeX},$ scan etc.) are scored and clients are given detailed feedback on their attempts/solutions.[/quote]We do provide services like that only that they are inside an actual course (ie you take a class, and for that class you get assignments which are graded, etc.)

This is rather of low importance, but perhaps a source-code option for an entire contest page? It would make uploading problems onto the wiki much quicker, whereas I would usually open up each problem on a separate window and quote, etc.

azjps can you develop on that idea a little bit please? I have a vague understanding of what you are trying to say.

Hopefully this is what he meant. Similar to how you can click on $ \text{longer blocks of LaTeX in a post}$ to see what the code used for it was, have some button in a contest page pop up a large block of code that [url=http://www.artofproblemsolving.com/Forum/resources.php?c=182&cid=27&year=2009]for this page[/url] would look something like this.

[hide="Code"][code]USA USAMO 2009

Day 1 - 28 April 2009

1
Given circles $\omega_1$ and $\omega_2$ intersecting at points $X$ and $Y$, let $\ell_1$ be a line through the center of $\omega_1$ intersecting $\omega_2$ at points $P$ and $Q$ and let $\ell_2$ be a line through the center of $\omega_2$ intersecting $\omega_1$ at points $R$ and $S$.  Prove that if $P, Q, R$ and $S$ lie on a circle then the center of this circle lies on line $XY$.

2
Let $n$ be a positive integer.  Determine the size of the largest subset of $\{ - n, - n + 1, \ldots , n - 1, n\}$ which does not contain three elements $a, b, c$ (not necessarily distinct) satisfying $a + b + c = 0$.

3
We define a [i]chessboard polygon[/i] to be a polygon whose sides are situated along lines of the form $x = a$ or $y = b$, where $a$ and $b$ are integers. These lines divide the interior into unit squares, which are shaded alternately grey and white so that adjacent squares have different colors. To tile a chessboard polygon by dominoes is to exactly cover the polygon by non-overlapping $1 \times 2$ rectangles. Finally, a [i]tasteful tiling[/i] is one which avoids the two configurations of dominoes shown on the left below. Two tilings of a $3 \times 4$ rectangle are shown; the first one is tasteful, while the second is not, due to the vertical dominoes in the upper right corner.

[asy]
size(100); pathpen = linewidth(2.5);
void chessboard(int a, int b, pair P){
 for(int i = 0; i < a; ++i) for(int j = 0; j < b; ++j)
  if((i+j) % 2 == 1) fill(shift(P.x+i,P.y+j)*unitsquare,rgb(0.6,0.6,0.6));
 D(P--P+(a,0)--P+(a,b)--P+(0,b)--cycle);
}
chessboard(2,2,(2.5,0));fill(unitsquare,rgb(0.6,0.6,0.6));fill(shift(1,1)*unitsquare,rgb(0.6,0.6,0.6)); chessboard(4,3,(6,0)); chessboard(4,3,(11,0)); MP("\mathrm{Distasteful\ tilings}",(2.25,3),fontsize(12));

/* draw lines */
D((0,0)--(2,0)--(2,2)--(0,2)--cycle); D((1,0)--(1,2)); D((2.5,1)--(4.5,1)); D((7,0)--(7,2)--(6,2)--(10,2)--(9,2)--(9,0)--(9,1)--(7,1)); D((8,2)--(8,3)); D((12,0)--(12,2)--(11,2)--(13,2)); D((13,1)--(15,1)--(14,1)--(14,3)); D((13,0)--(13,3));
[/asy] a) Prove that if a chessboard polygon can be tiled by dominoes, then it can be done so tastefully.

b) Prove that such a tasteful tiling is unique.

Day 2 - 29 April 2009

4
For $n\geq2$ let $a_1, a_2, \ldots a_n$ be positive real numbers such that
\[
(a_1 + a_2 + \cdots + a_n)\left(\frac {1}{a_1} + \frac {1}{a_2} + \cdots + \frac {1}{a_n}\right) \leq \left(n + \frac {1}{2}\right)^2.
\]
Prove that $\max(a_1, a_2, \ldots, a_n)\leq 4\min(a_1, a_2, \ldots, a_n)$.

5
Trapezoid $ABCD$, with $\overline{AB}||\overline{CD}$, is inscribed in circle $\omega$ and point $G$ lies inside triangle $BCD$.  Rays $AG$ and $BG$ meet $\omega$ again at points $P$ and $Q$, respectively.  Let the line through $G$ parallel to $\overline{AB}$ intersects $\overline{BD}$ and $\overline{BC}$ at points $R$ and $S$, respectively.  Prove that quadrilateral $PQRS$ is cyclic if and only if $\overline{BG}$ bisects $\angle CBD$.

6
Let $s_1, s_2, s_3, \ldots$ be an infinite, nonconstant sequence of rational numbers, meaning it is not the case that $s_1 = s_2 = s_3 = \cdots.$  Suppose that $t_1, t_2, t_3, \ldots$ is also an infinite, nonconstant sequence of rational numbers with the property that $(s_i - s_j)(t_i - t_j)$ is an integer for all $i$ and $j$.  Prove that there exists a rational number $r$ such that $(s_i - s_j)r$ and $(t_i - t_j)/r$ are integers for all $i$ and $j$.[/code][/hide]

The main thing I want out of this would be a quick one-click-popup to get the BBCode used for a problem. Right now it's click link and quote first post, which is a lot of page loading when you're trying to grab a lot of these.

1.) It would be good to have some synchronization with the wiki so people can edit solutions.

2.) There should be an option to download all the problems in one file from all the years. Downloading files year by year can be very tedious.

3.) There should be an unifying theory which is to be retrieved from the resources section and to be included into the actual post. Sometimes I shortened the source description because a problem appeared in several contests, e.g. sometimes I wrote ISL than IMO Shortlist again, or AIMO and then again German TST.

4.) Wrt loading there should be the option just to give the thread URL and it automatically extracts the post-ID from the first post to be used for the resources section.

5.) Tagging by topic area similar to [url=http://cran.r-project.org/web/classifications/MSC.html]MSC.[/url] This should have several layers, e.g.:

Number theory -> Diophantine equations -> Congruences in many variables.

But there should also be the option to tag the problem with the different classifications because a problem may be a hybrid of several areas, or different people consider the intended the solution to follow ideas from different fields.

6.) Tagging of difficulty should be two-ways. One score where everybody can vote and a score by trusted community members. Based on the problem-specific rating there should be something like a moving average to be aggregated such that you know what the average difficulty level of say IMO problem 4 was for the last 5 or 10 years. Surely it cannot hurt to provide a quick option where people see suggested difficulty descriptions as [url=http://www.artofproblemsolving.com/Wiki/index.php/Competition_Rating#Scale]here.[/url]

[quote="MellowMelon"]The main thing I want out of this would be a quick one-click-popup to get the BBCode used for a problem. Right now it's click link and quote first post, which is a lot of page loading when you're trying to grab a lot of these.[/quote]Once again, the Wiki should not be a problem repository (except maybe for those particularly useful, meaningful problems that teach something important -- so called examples for different techniques, and so on.)

[quote="orl"]1.) It would be good to have some synchronization with the wiki so people can edit solutions.
[/quote]
[quote="orl"]In the wiki there should be a separation between problem text and solution because it is hard to avoid to take a glance at it while reading the problem statement.[/quote]Refer to my post above.

[quote="orl"]2.) There should be an option to download all the problems in one file from all the years. Downloading files year by year can be very tedious.
[/quote]
I doubt we will make that available. Some people, countries, etc. don't want their problems easily download-able like that.

[quote="orl"]
3.) There should be an unifying theory which is to be retrieved from the resources section and to be included into the actual post. Sometimes I shortened the source description because a problem appeared in several contests, e.g. sometimes I wrote ISL than IMO Shortlist again, or AIMO and then again German TST.
[/quote]

What do you mean by a unifying theory? The forum post will not be connected to the resources problem, but rather the other way. You will be able to add multiple sources (also known as tags) to the problems.

[quote="Nukular"][quote="Valentin Vornicu"]
I doubt we will make that available. Some people, countries, etc. don't want their problems easily download-able like that.
[/quote]

On the other hand, the PDF file names differ /only/ in the year number, so it would be trivial to write a one-line wget query
in a for loop to get all of the files for a particular competition. Maybe this is a level of sophistication above a simple "get all" option, but it's certainly not a big one. (If the admins want this to be less possible, the file names really need to be
a bit more obfuscated than they are right now.)[/quote]

I know that I can write a short wget statement. But I wanted to have all the problems in one file, and yes I do not mean downloading separate files and then concatenating those files by some other tool. But Val just said anyway he does not want to do it.

[quote="Valentin Vornicu"][quote="orl"]
3.) There should be an unifying theory which is to be retrieved from the resources section and to be included into the actual post. Sometimes I shortened the source description because a problem appeared in several contests, e.g. sometimes I wrote ISL than IMO Shortlist again, or AIMO and then again German TST.
[/quote]

What do you mean by a unifying theory? The forum post will not be connected to the resources problem, but rather the other way. You will be able to add multiple sources (also known as tags) to the problems.[/quote]

I mean sometimes I look at a problem and wonder what the source is. So I am looking at the source field but do not know what the abbreviation stands for. If you are coming from the resources' section it is obvious but less so if you are coming from the forum.

Either you take the information from the resources section and give the source field in a posting a regular consistent source field description, or you make a second source field because the old source field may have a source reference in the source field but was not added to the resources section yet. So imagine the old source field has A and B whereas the problem was only added as problem as part of competition A so it may get overwritten by substituting through the regular source description A'. So it is better to introduce a new source field with A' and keep the old source field with A and B.

That will be fixed (the ambiguities of the sources in the 2.0 version)

In the current resources, there are many errors and mistakes and gaps.

PEN is not a national/regional competition

Under PEN, there are A-S categories, and [deleteme] category (yet not deleted)

USAMO 2009/3 in PDF is some code error, no picture (there are probably more problem errors, but it's hard to find)
But, as IMO/ISL/ILL are important:

ISL'08 G6: I think the proposer is Dusan Djukic (proposer of 2006/2,2006/6), not Duskan Dukic (the names are very similar+I never heard of the second). Same with N4 ISL'08

And, the solution project (clicking on S) should be done for all problems, or for none. Not partial as it is. I think that there are enough people to do that, and that way users can see the best solutions, and not spam which appears in some threads, or 5 new problems (inequalities - it never stops on 1 problem in a thread).


It seems to me to be an extended form of Nesbitt's inequality! Is it?

Is k integer here or not? Are a, b, c positive?

Specify man ....specify??????/What are $ a,b,c,k$?

For specification, I think :
1. $ abc\equal{}1$
2. $ a,b,c,k \in \mathbb{R}^\plus{}$


Find all $ n,k\in N$ such that $ \binom{n}{k}=2002$

I don't really understand what you said!

Please tell clearly what you're doing, and what's the answer?


My immediate thought was "What are the 96 senses of 'slow'?"

Then I thought, "Who would have the time to tell me that I'm slow in all 96?"

Hmm, for me, something strange kind of happens in my brain. Kind of like a circle...

In a way, it makes me feel the way I feel when I hear something ironic.

If you've ever read Hitchhiker's Guide to the Galaxy, that's a good book with a lot of examples of statements that kind of confuse you in an ironic way...

[quote="banblink182"]Describe what goes through your mind when you hear the phrase, "I've been called slow, in all 96 senses of the word."[/quote]

First, "What are the 96 senses of slow? I don't know that many..." then "Oh, I get it."

First, "Huh?"
Then, "Wow you must be slow."
Then, "Wait let me read that again, it makes no sense."
Finally, "?????"

Yep that's basically what happened to me too :D.

The quote is kind of an "dead-ended" or "non-cyclic/repeating" logical irony.

An example of a "circular" or "closed" logical irony would be something like, "this statement is false".

What is even more interesting is if I said, "I've been called slow in all three sense of the word." People don't dismiss the possibility of there being three definitions as quickly, and it gives more possible logical pathways.

Think about this one: "My response to the next thing you think is, 'I don't know, I was just being random'."

Or what about George W. Bush's quote:

[i]"This may sound a little West Texan to you, but I like it. When I'm talking about.. when I'm talking about myself, and when he's talking about myself, all of us are talking about me."[/i]

If you meant "no" in a sarcastic way, then yes.

No, the response is supposed to respond to your response of something like, "What? I don't get it..."

or something like  :huh:

 :)

Here's a better one.

"In my judgment, when the United States says there will be serious consequences, and if there isn't serious consequences, it creates adverse consequences."

--Meet the Press, February 8,2004.

And here's a another cyclical thingy--aren't I the articulate one?

I always lie.

Well , to be precise , it's a strange kind of a feeling....At first , nothing happens , and then , the I start thinking of whatever you might probably mean.

[quote="ambierona"]"i always lie" isn't circular. If you were lying there, then you could only sometimes lie, and that works. more like if it was given that you always tell the truth, and then you say "i am lying."[/quote]

Yes, that is true.

Your signature is hilarious  :lol:


1. how do u find the answer to:

3($ 122^o$ 26' 18") - 4($ 91^o$ 12' 14") = ??

and

2. A 10 inch chord of a larger of two concentric cricles is tangent to the smaller circle.  Find the square of the area of the region that is not shaded.

is there a formula for this?

[quote="DaReaper"]
2. A 10 inch chord of a larger of two concentric cricles is tangent to the smaller circle.  Find the square of the area of the region that is not shaded.
[/quote]

It would help if you post where the shaded area is.
[asy]fill(Circle((0,0),25/3),rgb(0.8,0.8,0.8));
fill(Circle((0,0),20/3),white);
dot((0,0));
draw((0,0)--(0,20/3)--(15/3,20/3));
label("\(5\)",((0,20/3)+(15/3,20/3))/2,N);[/asy]
Did I guess right? If so use the Pythagorean Theorem. (by the way, if these are homework problems, it would help if you posted your work/specific questions)

[quote="DaReaper"]3($ 122^o$ 26' 18") - 4($ 91^o$ 12' 14") = ??[/quote]

You can convert this to degrees with decimals and perform the subtraction there.

Perhaps converting to fractions is quicker if calculators are not allowed.


when a coin is tossed four coins consecutively:

how many sequences are possible?

how many sequences have exactly one tail?

Assuming ahead or tail is equally likely, what is the probability of getting exactly one head in four tosses?

[hide="click to reveal..."]
On each toss $2$ things can happen: there are $4$ tosses and hence $2^{4}=16$ possible sequences.
There are $4$ 'positions' (first, second etc.) that our sole tail can be in and for each of these, if the others are all heads we get a sequence with only one tail. Hence there are $4$ sequences with exactly one tail.

Similarly the probability of getting one head is $\frac{4}{16}=\frac14$.
[/hide]


:)  Two Inequality can be solve in case $n-1$ variable equal.

[quote="hungkhtn"]Do you have solution? Zhaobin?[/quote]
Sure,
Use General Popoviciu's inequality we can get:
$a_1^{n-1}+a_2^{n-1}+...+a_n^{n-1}+n(n-2) \ge (n-1)a_1a_2...a_{n}(\frac{1}{a_1}+\frac{1}{a_2}+...+\frac 1 {a_n})$
and use AM-GM:We can get:
$a_1^{n-1}+a_2^{n-1}+...+a_n^{n-1}+n(n-2) \ge (n-1)(a_1+a_2+...+a_n)$
Right :lol: ?


linear algebra
matrix
linear algebra unsolved

Dear friends,

I just want to know for any matrix $A$ of size $n\times n$ and with $\mbox{rank}(A)<n$, is it correct to say that the column space of $A$ is same as space spanned by eigenvectors corresponding to nonzero eigenvalues? If yes how can we prove it?

Regards

Bhupala.

This statement is false.
We may consider a matrix A of the form a_i_j=1 if j=i+1, else a_i_j=0
The eigenvectors space of A is zero-space is not equal to the non-zero column-space.

Mr Slon,

can you be more eloborate on this. I cant understand what you are saying. What do you mean by eigenvector space is a zero space?

regards

bhupala

That example was a nilpotent matrix- there are no eigenvectors associated to nonzero eigenvalues, but the column space is nontrivial.
That's not the only way this could fail- it's easy to construct matrices that don't have enough eigenvectors.

The rank<n condition is meaningless- we can just pad a smaller matrix with zeros if we want to.

@jmerry,

So do you mean that the argument given is wrong? or you are supprting it?

regards

bhupala


function
integration
calculus
limit
real analysis
real analysis unsolved

As known, in 1881 while pupil of Ulisse Dini, Vito Volterra (1860-1940) constructed the first example of a differentiable real function $ f$ defined over $ [0,1],$ [u] with bounded $ f',$[/u] such that $ \int_a^b f'(x) dx$ does not exist as Riemann integral for any $ 0 \le a < b \le 1$ ([i]
Giornale di Matematica diretto dal Prof. G.Battaglini} Vol.XIX
1881, pp.76--86  and
333--372)[/i]. By the way, as far as I know, this example togheter with the problem of passing to the limit under integral sign generated the Lebesgue's researches about his measure.

In the first of the two papers there is also the proof of the nonexistence of a function continuous on the rationals and discontinuous on the irrationals of a given interval (a short proof usually employs Baire's category theorem which came later than 1881).

[u]The problem proposed is to show  that an analogous function does not exist if the integral is intended as Lebersgue's.[/u]

Consider $ f_h(x)\equal{}\frac{f(x\plus{}h)\minus{}f(x)}h$, $ x\in[a,b\minus{}h]$, $ f_h(x)\equal{}0$, $ x\in[b\minus{}h,b]$. We have $ f_h$ is bounded and summable, $ f_h(x)\to f'(x)$ as $ h\to 0$, so $ \lim_{h\to 0}\int_a^bf_h(x)dx\equal{}  \int_{a}^{b}f'(x) dx$ due to Lebesgue's theorem.

What can we say if we drop the condition that $ f'$ be bounded? Can still we affirm that $ f'$ must be integrable within an
interval? I say yes.

[quote="Myth"]Consider $ f_h(x)$.. [/quote]


geometry AMC USA(J)MO USAMO Euler MATHCOUNTS ARML

[quote="r15s11z55y89w21"]Do you guys believe Pleasanton's schools would eventually catch up to, or even (hopefully not) overtake Mission and Lynbrook? If so, when?[/quote]

This is exactly our goal :)

[quote="junggi"]this is a really dumb topic tbh[/quote]
can only be expected :(

[quote="DataBox"]IMhO Palo Alto is bigger threat to Cupertino and Fremont than Pleasanton.[/quote]
What? Palo Alto has *already* achieved parity, at the very least.

Imagine my surprise when I saw a topic with the name of my hometown...

Hi, I'm William and I'm the president of the Math Team at Amador Valley High School in Pleasanton. I have been working hard to organize our math team here in our school and city, and like AIME15 said, we actually have been wanting to get as much recognition as certain schools in Fremont and San Jose.

[b]Pleasanton Math League[/b]
Amador Valley High School
Foothill High School
Harvest Park Middle School
Hart Middle School

It's great to know that we're being noticed in the bay area - in the past, we've never really left an impression at all. It makes me feel like what I'm doing is beneficial to our Math League, and I'm actually helping make a name of our team beyond our city...

It is my hope that someday we will catch up to Mission San Jose and Lynbrook. To do that we'll need to attract more mathy students in our middle schools. It's a really slow process, but I'm trying out everything I can to help.

But still, Pleasanton doesn't even have any USAMO qualifiers, so we've got some ways to go.

Keep watching our Pleasanton Math League. We're going places.

Anyway, you just helped make my years on the math team much more worth it. Can I ask what school you're from?  :lol:

What Lynbrook's lack in number of Moppers is made up by our general numbers.  :P
Quantity over Quality!!

EDIT: Y's Euler going against Lynbrook? (Aren't u going here next yr??)   :huh:

[quote="Ubemaya"]Also, I don't understand why everyone is talking about MSJ/Lynbrook when Gunn has been historically the best school, and Saratoga is currently the best school.  :huh:[/quote]
qft

[quote="Eulers_Apprentice"]In response to the original topic:
Yes I think it might be possible, especially since Lynbrook hasn't had any moppers in forever.

Bay Area needs more blue/black moppers if it wants more recognition.[/quote]

Bay Area had 2/6 IMO people. That's enough for now... you don't need 6/6 IMO people right?

i am, of course, totally at a loss to understand how someone can throw out an assertion like "Saratoga is currently the best school" without any data to back it up.  of course, that then begs the question of what would constitute acceptable data.  if you want to find the best *math team* in the area, i've created the only systematic and objective method in the Bay Area to test that (at least at the high school level), and Lynbrook is that best math team this year.  if you define the best school based on some individual performance metric, you have a lot of different systems (AMC, mathleague.com, AP tests, CAHSEE scores, whatever) to choose from, with what i expect will be a bewildering degree of inconsistency.  i'm interested to hear what metrics people use to answer this kind of question; let me know your thoughts..

Saratoga has the highest number of people who can score decently on USAMO, the highest number of people who are at MOP level, the highest number of people who qualified for USAMO, and probably has the highest number of people who have a chance at USAMO. For what nearly everyone who does math cares, that is all that matters.

Hmm okay, I don't know why I threw SCVMA in there. Not smart, I admit.

Yeah, turns out the Gunn kids were the wimps? I'm not in the class so I don't actually know how good it is in and of itself (though it must be awesome if Ms. Antink is in charge) but it's very convenient for keeping math contest participation afloat.

I'm in the class and I have to say that it's pretty awesome. Our textbooks are AOPS 1 and 2 and Ms. Antink is thinking of making a usamo class later.

Huh? I thought they said they would have 1/2 at Paly and 3/4 at Gunn, and we had 15 from Gunn sign up, but 0 from Paly. This year it looks like we might have enough. I think I might get enrolled if they have it since I signed some sheet of paper? It'll basically become extended math club or something.

=-O that is probably true
hmm
you, thief, hot person, indian girl
I am not sure who else
also did you all make red MOP?

That's about it I believe

also
I heard not all of you made red mop

that's pretty unfortunate.

[quote="CircleSquared"][quote="Ubemaya"][quote="r15s11z55y89w21"]For quite a while now, Fremont (Mission) and Cupertino (Lynbrook) have been prominent in the Bay Area, and in the whole of California, at math. However, the schools of Pleasanton are getting better at quite a progressive rate. Do you guys believe Pleasanton's schools would eventually catch up to, or even (hopefully not) overtake Mission and Lynbrook? If so, when?[/quote]

Which schools are from Pleasanton?

Also, for the past two years Saratoga has been the "best" in the Bay Area, and since all of our best members are underclassmen, that'll probably hold for the next two years at least.[/quote]

SARATOGA FTW
for once it seems Ubey is not being fat

EDIT: my school's class of 2012 is totally a beast factory[/quote]
Dude yes it appears that aside from you 4, there are like several other people who have a chance at USAMO who don't even do math :O

[quote="junggi"]first of all, i feel like that will's post is implying that other teams [i]dont[/i] have the enthusiasm, attitude, belief, dedication, which slightly annoys me, but that was probably not the intention since he was just listing the good points of amador math team.

moving on to the main part of this post:

there seems to have been a slight confusion in defining what exactly "math team" was supposed to be

most of tim/silas's posts just talked about results/attendence of certain team contests, hence we thought you guys meant best math team as in working to solve problems together. by this definition, gunn/saratoga [i]clearly are[/i] the best math teams in the bay area, since we just outperformed everyone else at smt (which i will now assume to be the only important team contest around here, since tim doesnt deny that mathleague is unimportant [as of now]), it wasn't even close, we were ahead by quite a large margin.
[/quote]
The only reason you guys are miles ahead is because everyone at Lynbrook prioritized the science bowl over SMT :(.

Dude what we had like only one person on our team who got below 7 on the individual stuff, and most people got pretty much all the questions correct.

[quote="junggi"]what okay i am willing accept that lynbrook may be better than gunn
but the point still holds that we are clearly ahead of amador in team contests[/quote]
Well actually I'm not sure if Lynbrook has a team persay, seeing as some of our top members decide not to show up to certain team events (:P) and we do our own thing more than actually contributing to the team i think, and you guys also owned us at team mandlebrot.

BackHo should be on the power rank this month, yes?

[quote="Scrambled"][quote="junggi"]what okay i am willing accept that lynbrook may be better than gunn
but the point still holds that we are clearly ahead of amador in team contests[/quote]
Well actually I'm not sure if Lynbrook has a team persay, seeing as some of our top members decide not to show up to certain team events (:P) and we do our own thing more than actually contributing to the team i think, and you guys also owned us at team mandlebrot.
[/quote]
lol
one of the team mandelbrots
i did/wrote up like 5/6 problems
the geo one
[quote]
BackHo should be on the power rank this month, yes?[/quote]
hell no

[quote="not_trig"][quote="junggi"][b]team contest[/b][/quote]

that is all[/quote]
i definitely saw this coming

[quote="junggi"]....other teams [i]dont[/i] have the enthusiasm, attitude, belief, dedication, which slightly annoys me, but that was probably not the intention since he was just listing the good points of amador math team.[/quote]
From where I'm standing, I'm pretty sure my school hasn't had most of that for years.


limit
real analysis
real analysis unsolved

Prove that $\forall x \in [0,\infty)$ there exists two sequences of postitive integers $(a_n)_{n\geq },(b_n)_{n\geq 1}$ such that :
   $\lim_{n\to\infty}\left (\frac{1}{a_n+1}+\frac{1}{a_n+2}+...+\frac{1}{a_n+b_n}\right )=x$.

[hide]
Please check and correct this :
I supposed from the start that $a_n\to\infty,b_n\to\infty$.So $\frac{1}{a_n+1}+\frac{1}{a_n+2}+...+\frac{1}{a_n+b_n}=x_{a_n+b_n}-x_{a_n}-\ln(a_n)+\ln(a_n+b_n)$where $x_n$ is the sequence that gives us $c$ (Euler`s constant) so $\lim_{n\to\infty}\left (\frac{1}{a_n+1}+\frac{1}{a_n+2}+...+\frac{1}{a_n+b_n}\right )=\lim_{n\to\infty}\ln(1+\frac{b_n}{a_n})$. So we must choose $a_n,b_n$ such that $\frac{a_n}{b_n}\to e^x-1$ which is possible because $Q$ is dense in $R$ and also we can choose $a_n=na'_n$ such that $a_n\to \infty$ [/hide]

[quote="spix"]Prove that $\forall x \in [0,\infty)$ there exists two sequences of postitive integers $(a_n)_{n\geq },(b_n)_{n\geq 1}$ such that :
   $\lim_{n\to\infty}\left (\frac{1}{a_n+1}+\frac{1}{a_n+2}+...+\frac{1}{a_n+b_n}\right )=x$.
[/quote]
If $x =0$, choose $a_n = n, \, b_n = 1$.
If $x > 0$, choose $a_n = 2^{n+K}$ for sufficiently large $K$ and $b_n = \max\{ k \, | \, \sum_{j=1}^k \frac{1}{a_n + j} < x \}$. This is well defined, since $\sum_{j=1}^\infty \frac{1}{a_n + j} = \infty$.
Let $c_n = \sum_{j=1}^{b_n} \frac{1}{a_n+ j}$, then $c_n < x \le c_n + \frac{1}{a_n+b_n+1} < c_n + 2^{-n-K}$, and thus $c_n \to x$.


inequalities
function
calculus
derivative
inequalities proposed

Let $ a,b,c$ be nonnegative real numbers, no two of them are zero. Prove that

$ (a) \ \ \ \ \ \ \ \ \  \sum\frac {ab\minus{}2bc\plus{}ca}{(3a\plus{}b\plus{}c)(b\plus{}c)} \ge 0$;

$ (b) \ \ \ \ \ \ \ \ \  \sum\frac {3a(b\plus{}c)\minus{}2bc}{(2a\plus{}b\plus{}c)(b\plus{}c)} \ge \frac 3{2}$.

:P Can I use $ R,r,p$, let $ a\to p \minus{} a,b\to p \minus{} b,c\to p \minus{} c$

$ (a)$ is equivalent to
\[ \sum \frac {(p \minus{} b)(p \minus{} c) \plus{} (p \minus{} c)(p \minus{} a) \plus{} (p \minus{} a)(p \minus{} b) \minus{} 3(p \minus{} b)(p \minus{} c)}{(3p \minus{} 2a)a}\ge 0
\]

\[ \Leftrightarrow \minus{} 384r^4R^2 \minus{} 512R^3r^3 \minus{} 96r^5R \minus{} 8r^6 \plus{} ( \minus{} 10r^4 \plus{} 96r^3R \minus{} 32R^2r^2)p^2 \plus{} (4Rr \minus{} 2r^2)p^4\ge 0
\]

\[ \Leftrightarrow 4r^4(r \plus{} 40R)(R \minus{} 2r) \plus{} 2r^2(189r^2 \plus{} 188r(R \minus{} 2r) \plus{} 48(R \minus{} 2r)^2)(p^2 \minus{} 16Rr \plus{} 5r^2) \plus{} 2r(2R \minus{} r)(p^2 \minus{} 16Rr \plus{} 5r^2)^2\ge 0
\]
Which is true.

$ (b)$ is equivalent to
\[ \Leftrightarrow \sum\frac {3(p \minus{} a)a \minus{} (p \minus{} b)(p \minus{} c)}{(2p \plus{} a)a}\ge 0
\]

\[ \Leftrightarrow \minus{} 2r^6 \minus{} 24r^5R \minus{} 96r^4R^2 \minus{} 128R^3r^3) \plus{} (32R^2r^2 \plus{} 52r^3R \minus{} 4r^4)p^2 \plus{} (12Rr \minus{} 2r^2)p^4\ge 0
\]

\[ \Leftrightarrow 16r^3(1296r^3 \plus{} 2145r^2(R \minus{} 2r) \plus{} 1180r(R \minus{} 2r)^2 \plus{} 216(R \minus{} 2r)^3) \plus{} 4r^2(354r^2 \plus{} 383r(R \minus{} 2r) \plus{} 104(R \minus{} 2r)^2)(p^2 \minus{} 16Rr \plus{} 5r^2) \plus{} 2r(6R \minus{} r)(p^2 \minus{} 16Rr \plus{} 5r^2)^2\ge 0
\]
Which is true.

There is a mistake at (b). You must replace $ 2p\plus{}a$ with $ 2p\minus{}a$.

That's right, thanks, but no problem with my solution because I worked with $ 2p\minus{}a$ :)

[quote="encyclopedia"]That's right, thanks, but no problem with my solution because I worked with $ 2p \minus{} a$ :)[/quote]
Can you remake your proof ? Because the last inequality is wrong (it is not equivalent to the original inequality).

oh thank you I saw  :blush: I will repost it, my big mistake  :( , now I used $ p^2\le 4R^2\plus{}4Rr\plus{}3r^2$

$ (b)$ is equivalent to
\[ \Leftrightarrow \sum\frac {3(p \minus{} a)a \minus{} (p \minus{} b)(p \minus{} c)}{(2p \minus{} a)a}\ge 0
\]

\[ \Leftrightarrow \minus{}4r^2[(48R^2r^2\plus{}64rR^3\plus{}12r^3R\plus{}r^4)\plus{}(2r^2\minus{}4R^2\minus{}20Rr)p^2\plus{}p^4]\le 0
\]

\[ \Leftrightarrow \minus{}16r^4(2r\minus{}R)^2\plus{}16r^2(2r\minus{}R)(r\minus{}R)(p^2\minus{}4R^2\minus{}4Rr\minus{}3r^2)\minus{}4r^2(p^2\minus{}4R^2\minus{}4Rr\minus{}3r^2)^2\le 0\]

Sorry for my mistake !

[quote="Vasc"]Let $ a,b,c$ be nonnegative real numbers, no two of them are zero. Prove that

$ (a) \ \ \ \ \ \ \ \ \ \sum\frac {ab \minus{} 2bc \plus{} ca}{(3a \plus{} b \plus{} c)(b \plus{} c)} \ge 0$;

$ (b) \ \ \ \ \ \ \ \ \ \sum\frac {3a(b \plus{} c) \minus{} 2bc}{(2a \plus{} b \plus{} c)(b \plus{} c)} \ge \frac 3{2}$.[/quote]
Here is my solution :)
For a), we have the following trivial lemma (the readers can check it easily):
Lemma. If $ x \ge y \ge z,x \plus{} y \plus{} z \equal{} 0, m \ge n$ and $ m \plus{} n \ge 2p$, then $ xm \plus{} yn \plus{} zp \ge 0.$
Now, we assume that $ a \ge b \ge c$. With this hypothesis, we will show that
\[ \frac {1}{(3a \plus{} b \plus{} c)(b \plus{} c)} \ge \frac {1}{(3b \plus{} c \plus{} a)(c \plus{} a)},
\]
and
\[ \frac {1}{(3a \plus{} b \plus{} c)(b \plus{} c)} \plus{} \frac {1}{(3b \plus{} c \plus{} a)(c \plus{} a)} \ge \frac {2}{(3c \plus{} a \plus{} b)(a \plus{} b)}.
\]
Indeed, we have
\[ \frac {1}{(3a \plus{} b \plus{} c)(b \plus{} c)} \minus{} \frac {1}{(3b \plus{} c \plus{} a)(c \plus{} a)} \equal{} \frac {(a \minus{} b)(a \plus{} b \minus{} c)}{(a \plus{} c)(b \plus{} c)(3a \plus{} b \plus{} c)(3b \plus{} c \plus{} a)} \ge 0.
\]
For the right inequality, after using Cauchy Schwarz, we can deduce the inequality to
\[ 2(3c \plus{} a \plus{} b)(a \plus{} b) \ge (3a \plus{} b \plus{} c)(b \plus{} c) \plus{} (3b \plus{} c \plus{} a)(c \plus{} a)
\]
which is true because
\[ LHS \minus{} RHS \equal{} (a \minus{} b)^2 \plus{} c(a \plus{} b \minus{} 2c) \ge 0.
\]
Now, applying the above lemma with $ x \equal{} a(b \plus{} c) \minus{} 2bc,y \equal{} b(c \plus{} a) \minus{} 2ca,z \equal{} c(a \plus{} b) \minus{} 2ab$ and $ m \equal{} \frac {1}{(3a \plus{} b \plus{} c)(b \plus{} c)},n \equal{} \frac {1}{(3b \plus{} c \plus{} a)(c \plus{} a)},p \equal{} \frac {1}{(3c \plus{} a \plus{} b)(a \plus{} b)}$, we can get the result. :)
For b), to have simple calculation, let us subsitute $ x \equal{} \frac {b \plus{} c}{2},y \equal{} \frac {c \plus{} a}{2},z \equal{} \frac {a \plus{} b}{2}$, then
\[ \frac {3a(b \plus{} c) \minus{} 2bc}{(2a \plus{} b \plus{} c)(b \plus{} c)} \equal{} \frac {6x(y \plus{} z \minus{} x) \minus{} 2(z \plus{} x \minus{} y)(x \plus{} y \minus{} z)}{4x(y \plus{} z)} \equal{} \frac {3x(y \plus{} z) \plus{} (y \minus{} z)^2 \minus{} 4x^2}{2x(y \plus{} z)} \equal{} \frac {(y \minus{} z)^2 \minus{} 4x^2}{2x(y \plus{} z)} \plus{} \frac {3}{2}.
\]
Hence, we can rewrite the inequality as
\[ \sum \frac {(y \minus{} z)^2}{x(y \plus{} z)} \ge 4\sum \frac {x}{y \plus{} z} \minus{} 6
\]
Since
\[ 4\sum \frac {x}{y \plus{} z} \minus{} 6 \equal{} 2\sum \frac {(y \minus{} z)^2}{(x \plus{} y)(x \plus{} z)}
\]
Our inequality is equivalent to
\[ \sum (y \minus{} z)^2\left( \frac {1}{x(y \plus{} z)} \minus{} \frac {2}{(x \plus{} y)(x \plus{} z)}\right) \ge 0
\]

\[ \sum \frac {(y \minus{} z)^2(x \minus{} y)(x \minus{} z)}{x(y \plus{} z)(z \plus{} x)(x \plus{} y)} \ge 0
\]

\[ \frac {(x \minus{} y)(y \minus{} z)(z \minus{} x)}{(x \plus{} y)(y \plus{} z)(z \plus{} x)} \sum \frac {z \minus{} y}{x} \ge 0
\]
Again, we see that
\[ \sum \frac {z \minus{} y}{x} \equal{} \frac {(x \minus{} y)(y \minus{} z)(z \minus{} x)}{xyz}
\]
Therefore, the last inequality can be rewritten as
\[ \frac {(x \minus{} y)^2(y \minus{} z)^2(z \minus{} x)^2}{xyz(x \plus{} y)(y \plus{} z)(z \plus{} x)} \ge 0,
\]
which is obviously true.

Equality in a) occurs when $ a \equal{} b \equal{} c$ or $ a \equal{} b,c \equal{} 0$ and its cyclic permutations. Equality in b) occurs when $ a \equal{} b$ or $ b \equal{} c$ or $ c \equal{} a$.

The proof is finished. :)

[quote="Vasc"]I think the last your inequality is not yet equivalent to my original inequality.
You must replace $ 3(p \minus{} a)a \minus{} (p \minus{} b)(p \minus{} c)$ with  $ 3(p \minus{} a)a \minus{} 2(p \minus{} b)(p \minus{} c)$.[/quote]

You are right Vasc, follow solution of can_hang2007, your inequality is equivalent to
\[ \frac {(x \minus{} y)^2(y \minus{} z)^2(z \minus{} x)^2}{xyz(x \plus{} y)(y \plus{} z)(z \plus{} x)} \ge 0
\]
this mean it is equivalent to Bludon's inequality of $ R,r,p$ and actually your inequalities is stronger than the inequalities I used in my second solution so, it is wrong  :( . Sorry.

[quote="Vasc"]Nice your solutions, Can-hang.  :lol:
We can also prove (a) by SOS method.
The following inequality is also nice.

$ (c) \ \ \ \ \ \ \sum \frac {9a(b \plus{} c) \minus{} 8bc}{(3a \plus{} b \plus{} c)(b \plus{} c)} \le 3$.[/quote]

Inequality is equivalent to:

$ 4 \cdot \left(\frac{(a\minus{}b)^2(4c^3a\plus{}4c^3b\plus{}abc(a\plus{}b))\plus{}(b\minus{}c)^2(4a^3b\plus{}4a^3c\plus{}abc(b\plus{}c))\plus{}(c\minus{}a)^2(4b^3c\plus{}4b^3a\plus{}abc(a\plus{}c))}{(a\plus{}b)(b\plus{}c)(c\plus{}a)(3a\plus{}b\plus{}c)(3b\plus{}c\plus{}a)(3c\plus{}a\plus{}b)}\right)$

[quote="Vasc"]Very nice your proof, Shaaam.  :lol:
Can you post it ?[/quote]

I am sorry  :blush: . The inequality I gave above is result of expansion. I also wait for nice solution.
So, lets add one more problem to this series:
(My favorite number is 19  :) )

Given that $ a,b,c \geq 0$, Prove that:
$ (d)$
\[ \sum_{cyc}\frac {19a(b \plus{} c) \minus{} 13(b^2 \plus{} c^2)}{(4a \plus{} b \plus{} c)(b \plus{} c)} \leq 3
\]

[quote="Vasc"]Let $ a,b,c$ be nonnegative real numbers, no two of them are zero. Prove that

$ (a) \ \ \ \ \ \ \ \ \ \sum\frac {ab \minus{} 2bc \plus{} ca}{(3a \plus{} b \plus{} c)(b \plus{} c)} \ge 0$;

[/quote]

The following stronger, ugly and hard inequality holds for $ a , b, c \geq 0$
e)
$ \boxed{\sum_{cyc}\frac {ab \minus{} 2bc \plus{} ca}{(3a \plus{} b \plus{} c)(b \plus{} c)} \geq \frac {(a \minus{} b)^2(b \minus{} c)^2(c \minus{} a)^2}{(a \plus{} b)(b \plus{} c)(b \plus{} c)(3a \plus{} b \plus{} c)(3b \plus{} c \plus{} a)(3c \plus{} a \plus{} b)}}$

Who can defeat this [u]monster[/u] by elementary way?

Thanks   :)

[quote="encyclopedia"]$ (b)$ is equivalent to
\[ \Leftrightarrow \sum\frac {3(p \minus{} a)a \minus{} (p \minus{} b)(p \minus{} c)}{(2p \minus{} a)a}\ge 0
\]

\[ \Leftrightarrow 4r^2[(48R^2r^2 \plus{} 64rR^3 \plus{} 12r^3R \plus{} r^4) \minus{} 2 (2R^2 \plus{} 10Rr \minus{} r^2)p^2 \plus{} p^4]\le 0
\]
[/quote]
This result and the proof by Can_hang to the inequality (b) is a complete proof of Blundon's Inequality, with establishes the upper and lower bounds of $ p$ for given $ R$ and $ r$:

$ 2R^2 \plus{} 1 \minus{} Rr \minus{} r^2 \minus{} 2(R \minus{} 2r)\sqrt {R(R \minus{} 2r)} \le p^2 \le 2R^2 \plus{} 1 \minus{} Rr \minus{} r^2 \plus{} 2(R \minus{} 2r)\sqrt {R(R \minus{} 2r)}$,

with equality for an isosceles triangle.

Can somebody prove the inequality (c) without expanding ?

[quote="Vasc"]
For $ k \equal{} 4$, there is a nice proof by SOS method.[/quote]

Can you please post it?
Thank you  :)

I think this kind of inequalities cannot have [u]a brilliant[/u] solution, because denominator is already ugly.
It takes a long time to expand but it is the most effective method.

I am really looking forward to see your solution for $ k\equal{}4$.   :)

[quote="Vasc"]Nice your solutions, Can-hang.  :lol:
We can also prove (a) by SOS method.
The following inequality is also nice.

$ (c) \ \ \ \ \ \ \sum \frac {9a(b \plus{} c) \minus{} 8bc}{(3a \plus{} b \plus{} c)(b \plus{} c)} \le 3$.[/quote]
This inequality can be also proved by SOS method as follow:
It is easy for us to rewrite the inequality as
\[ \sum \frac{(a\minus{}b)^2(3c^2\minus{}ac\minus{}bc\plus{}ab)}{(a\plus{}c)(b\plus{}c)(3a\plus{}b\plus{}c)(3b\plus{}c\plus{}a)} \ge 0\]
\[ 2\sum \frac{c^2(a\minus{}b)^2}{(a\plus{}c)(b\plus{}c)(3a\plus{}b\plus{}c)(3b\plus{}c\plus{}a)} \plus{}\frac{(a\minus{}b)(b\minus{}c)(c\minus{}a)}{(a\plus{}b)(b\plus{}c)(c\plus{}a)(3a\plus{}b\plus{}c)(3b\plus{}c\plus{}a)(3c\plus{}a\plus{}b)} \sum (b^2\minus{}a^2)(a\plus{}b\plus{}3c) \ge 0\]
\[ \sum \frac{c^2(a\minus{}b)^2}{(a\plus{}c)(b\plus{}c)(3a\plus{}b\plus{}c)(3b\plus{}c\plus{}a)} \minus{}\frac{(a\minus{}b)^2(b\minus{}c)^2(c\minus{}a)^2}{(a\plus{}b)(b\plus{}c)(c\plus{}a)(3a\plus{}b\plus{}c)(3b\plus{}c\plus{}a)(3c\plus{}a\plus{}b)} \ge 0\]
Without loss of generality, we can suppose that $ a \ge b \ge c.$ Then, we will show the slightly stronger inequality is
\[ \frac{b^2(a\minus{}c)^2}{(b\plus{}a)(b\plus{}c)(3a\plus{}b\plus{}c)(3c\plus{}a\plus{}b)} \ge \frac{(a\minus{}b)^2(b\minus{}c)^2(c\minus{}a)^2}{(a\plus{}b)(b\plus{}c)(c\plus{}a)(3a\plus{}b\plus{}c)(3b\plus{}c\plus{}a)(3c\plus{}a\plus{}b)} \ge 0\]
\[ b^2(a\plus{}c)(a\plus{}c\plus{}3b) \ge (a\minus{}b)^2(b\minus{}c)^2\]
which is obviously true because
\[ b^2 \ge (b\minus{}c)^2\]
and
\[ (a\plus{}c)(a\plus{}c\plus{}3b) \ge a^2 \ge (a\minus{}b)^2\]
That's all. :)

[quote="Vasc"][quote="shaaam"]
The following stronger, ugly and hard inequality holds for $ a , b, c \geq 0$
e)
$ \boxed{\sum_{cyc}\frac {ab \minus{} 2bc \plus{} ca}{(3a \plus{} b \plus{} c)(b \plus{} c)} \geq \frac {(a \minus{} b)^2(b \minus{} c)^2(c \minus{} a)^2}{(a \plus{} b)(b \plus{} c)(b \plus{} c)(3a \plus{} b \plus{} c)(3b \plus{} c \plus{} a)(3c \plus{} a \plus{} b)}}$

Who can defeat this [u]monster[/u] by elementary way?
Thanks   :)[/quote]
Using computer, we have
$ \boxed{\sum_{cyc}\frac {ab \minus{} 2bc \plus{} ca}{(3a \plus{} b \plus{} c)(b \plus{} c)} \geq \frac {k(a \minus{} b)^2(b \minus{} c)^2(c \minus{} a)^2}{(a \plus{} b)(b \plus{} c)(b \plus{} c)(3a \plus{} b \plus{} c)(3b \plus{} c \plus{} a)(3c \plus{} a \plus{} b)}}$
for $ k\le 6$.
For $ k \equal{} 4$, there is a nice proof by SOS method.[/quote]
The inequality
\[ \boxed{\sum_{cyc}\frac {ab \minus{} 2bc \plus{} ca}{(3a \plus{} b \plus{} c)(b \plus{} c)} \geq \frac {k(a \minus{} b)^2(b \minus{} c)^2(c \minus{} a)^2}{(a \plus{} b)(b \plus{} c)(b \plus{} c)(3a \plus{} b \plus{} c)(3b \plus{} c \plus{} a)(3c \plus{} a \plus{} b)}}\]
is no need computer to solve it, VasC ;)
Rewrite it as
\[ \sum c(a\minus{}b)^2(a\plus{}b)(a\plus{}b\minus{}c)(a\plus{}b\plus{}3c) \ge 6(a\minus{}b)^2(b\minus{}c)^2(c\minus{}a)^2\]
Now, we can assume that $ a \ge b \ge c$. Then we have $ b(a\minus{}c)^2 \ge \frac{a^2}{b} (b\minus{}c)^2$, using this inequality, we can deduce our inequality to
\[ \frac{a(b\minus{}c)^2}{b} [a(a\plus{}c)(a\plus{}c\minus{}b)(a\plus{}c\plus{}3b) \plus{}b(b\plus{}c)(b\plus{}c\minus{}a)(b\plus{}c\plus{}3a)] \ge 6(a\minus{}b)^2(b\minus{}c)^2(c\minus{}a)^2\]
Consider the function
\[ f(c)\equal{}a(a\plus{}c)(a\plus{}c\minus{}b)(a\plus{}c\plus{}3b) \plus{}b(b\plus{}c)(b\plus{}c\minus{}a)(b\plus{}c\plus{}3a) \equal{}a(a\plus{}c)^3\plus{}2ab(a\plus{}c)^2\minus{}3ab^2(a\plus{}c)\plus{}b(b\plus{}c)^3\plus{}2ab(b\plus{}c)^2\minus{}3a^2b(b\plus{}c)\]
We have
\[ f'(c)\equal{}3a(a\plus{}c)^2\plus{}4ab(a\plus{}c)\plus{}3b(b\plus{}c)^2\plus{}4ab(b\plus{}c)\minus{}3a^2b\minus{}3ab^2 \ge 0\]
Therefore, $ f(c)$ is increasing and we obtain
\[ f(c) \ge f(0)\equal{}a^2(a\minus{}b)(a\plus{}3b)\plus{}b^2(b\minus{}a)(b\plus{}3a)\equal{}(a\minus{}b)^2(a^2\plus{}b^2\plus{}4ab)\ge 6ab(a\minus{}b)^2\]
(actually, we can prove this inequality without using derivatives but it will need more calculations which make our proof become more complicated)
Therefore
\[ \frac{a(b\minus{}c)^2}{b} [a(a\plus{}c)(a\plus{}c\minus{}b)(a\plus{}c\plus{}3b) \plus{}b(b\plus{}c)(b\plus{}c\minus{}a)(b\plus{}c\plus{}3a)] \ge 6a^2(a\minus{}b)^2(b\minus{}c)^2 \ge 6(a\minus{}b)^2(b\minus{}c)^2(c\minus{}a)^2\]
:)

[quote="can_hang2007"][quote="Vasc"]Nice your solutions, Can-hang.  :lol:
We can also prove (a) by SOS method.
The following inequality is also nice.

$ (c) \ \ \ \ \ \ \sum \frac {9a(b \plus{} c) \minus{} 8bc}{(3a \plus{} b \plus{} c)(b \plus{} c)} \le 3$.[/quote]
This inequality can be also proved by SOS method as follow:
It is easy for us to rewrite the inequality as
\[ \sum \frac {(a \minus{} b)^2(3c^2 \minus{} ac \minus{} bc \plus{} ab)}{(a \plus{} c)(b \plus{} c)(3a \plus{} b \plus{} c)(3b \plus{} c \plus{} a)} \ge 0
\]

\[ 2\sum \frac {c^2(a \minus{} b)^2}{(a \plus{} c)(b \plus{} c)(3a \plus{} b \plus{} c)(3b \plus{} c \plus{} a)} \plus{} \frac {(a \minus{} b)(b \minus{} c)(c \minus{} a)}{(a \plus{} b)(b \plus{} c)(c \plus{} a)(3a \plus{} b \plus{} c)(3b \plus{} c \plus{} a)(3c \plus{} a \plus{} b)} \sum (b^2 \minus{} a^2)(a \plus{} b \plus{} 3c) \ge 0
\]

\[ \sum \frac {c^2(a \minus{} b)^2}{(a \plus{} c)(b \plus{} c)(3a \plus{} b \plus{} c)(3b \plus{} c \plus{} a)} \minus{} \frac {(a \minus{} b)^2(b \minus{} c)^2(c \minus{} a)^2}{(a \plus{} b)(b \plus{} c)(c \plus{} a)(3a \plus{} b \plus{} c)(3b \plus{} c \plus{} a)(3c \plus{} a \plus{} b)} \ge 0
\]
Without loss of generality, we can suppose that $ a \ge b \ge c.$ Then, we will show the slightly stronger inequality is
\[ \frac {b^2(a \minus{} c)^2}{(b \plus{} a)(b \plus{} c)(3a \plus{} b \plus{} c)(3c \plus{} a \plus{} b)} \ge \frac {(a \minus{} b)^2(b \minus{} c)^2(c \minus{} a)^2}{(a \plus{} b)(b \plus{} c)(c \plus{} a)(3a \plus{} b \plus{} c)(3b \plus{} c \plus{} a)(3c \plus{} a \plus{} b)} \ge 0
\]

\[ b^2(a \plus{} c)(a \plus{} c \plus{} 3b) \ge (a \minus{} b)^2(b \minus{} c)^2
\]
which is obviously true because
\[ b^2 \ge (b \minus{} c)^2
\]
and
\[ (a \plus{} c)(a \plus{} c \plus{} 3b) \ge a^2 \ge (a \minus{} b)^2
\]
That's all. :)[/quote]
Nice solution.
An other similar solution is to write the inequality as
$ \sum S_A(b\minus{}c)^2 \ge 4(a\minus{}b)^2(b\minus{}c)^2(c\minus{}a)^2$,
where
$ S_A\equal{}a(a\plus{}b)(a\plus{}c)(3a\plus{}b\plus{}c)$.

[quote="can_hang2007"][quote="Vasc"][quote="shaaam"]
The following stronger, ugly and hard inequality holds for $ a , b, c \geq 0$
e)
$ \boxed{\sum_{cyc}\frac {ab \minus{} 2bc \plus{} ca}{(3a \plus{} b \plus{} c)(b \plus{} c)} \geq \frac {(a \minus{} b)^2(b \minus{} c)^2(c \minus{} a)^2}{(a \plus{} b)(b \plus{} c)(b \plus{} c)(3a \plus{} b \plus{} c)(3b \plus{} c \plus{} a)(3c \plus{} a \plus{} b)}}$

Who can defeat this [u]monster[/u] by elementary way?
Thanks   :)[/quote]
Using computer, we have
$ \boxed{\sum_{cyc}\frac {ab \minus{} 2bc \plus{} ca}{(3a \plus{} b \plus{} c)(b \plus{} c)} \geq \frac {k(a \minus{} b)^2(b \minus{} c)^2(c \minus{} a)^2}{(a \plus{} b)(b \plus{} c)(b \plus{} c)(3a \plus{} b \plus{} c)(3b \plus{} c \plus{} a)(3c \plus{} a \plus{} b)}}$
for $ k\le 6$.
For $ k \equal{} 4$, there is a nice proof by SOS method.[/quote]
The inequality
\[ \boxed{\sum_{cyc}\frac {ab \minus{} 2bc \plus{} ca}{(3a \plus{} b \plus{} c)(b \plus{} c)} \geq \frac {k(a \minus{} b)^2(b \minus{} c)^2(c \minus{} a)^2}{(a \plus{} b)(b \plus{} c)(b \plus{} c)(3a \plus{} b \plus{} c)(3b \plus{} c \plus{} a)(3c \plus{} a \plus{} b)}}
\]
is no need computer to solve it, VasC ;)
Rewrite it as
\[ \sum c(a \minus{} b)^2(a \plus{} b)(a \plus{} b \minus{} c)(a \plus{} b \plus{} 3c) \ge 6(a \minus{} b)^2(b \minus{} c)^2(c \minus{} a)^2
\]
Now, we can assume that $ a \ge b \ge c$. Then we have $ b(a \minus{} c)^2 \ge \frac {a^2}{b} (b \minus{} c)^2$, using this inequality, we can deduce our inequality to
\[ \frac {a(b \minus{} c)^2}{b} [a(a \plus{} c)(a \plus{} c \minus{} b)(a \plus{} c \plus{} 3b) \plus{} b(b \plus{} c)(b \plus{} c \minus{} a)(b \plus{} c \plus{} 3a)] \ge 6(a \minus{} b)^2(b \minus{} c)^2(c \minus{} a)^2
\]
Consider the function
\[ f(c) \equal{} a(a \plus{} c)(a \plus{} c \minus{} b)(a \plus{} c \plus{} 3b) \plus{} b(b \plus{} c)(b \plus{} c \minus{} a)(b \plus{} c \plus{} 3a) \equal{} a(a \plus{} c)^3 \plus{} 2ab(a \plus{} c)^2 \minus{} 3ab^2(a \plus{} c) \plus{} b(b \plus{} c)^3 \plus{} 2ab(b \plus{} c)^2 \minus{} 3a^2b(b \plus{} c)
\]
We have
\[ f'(c) \equal{} 3a(a \plus{} c)^2 \plus{} 4ab(a \plus{} c) \plus{} 3b(b \plus{} c)^2 \plus{} 4ab(b \plus{} c) \minus{} 3a^2b \minus{} 3ab^2 \ge 0
\]
Therefore, $ f(c)$ is increasing and we obtain
\[ f(c) \ge f(0) \equal{} a^2(a \minus{} b)(a \plus{} 3b) \plus{} b^2(b \minus{} a)(b \plus{} 3a) \equal{} (a \minus{} b)^2(a^2 \plus{} b^2 \plus{} 4ab)\ge 6ab(a \minus{} b)^2
\]
(actually, we can prove this inequality without using derivatives but it will need more calculations which make our proof become more complicated)
Therefore
\[ \frac {a(b \minus{} c)^2}{b} [a(a \plus{} c)(a \plus{} c \minus{} b)(a \plus{} c \plus{} 3b) \plus{} b(b \plus{} c)(b \plus{} c \minus{} a)(b \plus{} c \plus{} 3a)] \ge 6a^2(a \minus{} b)^2(b \minus{} c)^2 \ge 6(a \minus{} b)^2(b \minus{} c)^2(c \minus{} a)^2
\]
:)[/quote]
Nice your solution for $ k\equal{}6$.
Below is my solution for $ k\equal{}4$.

[b]Remark.[/b] Let $ \triangle ABC\ .$ Prove that $ \sum\frac {ab \minus{} 2bc \plus{} ca}{(3a \plus{} b \plus{} c)(b \plus{} c)} \ge 0\ .$

Denote $ a \plus{} b \plus{} c \equal{} 2p\ \ \wedge\ \ \begin{array}{c} x_1 \equal{} a(b \plus{} c) \minus{} 2bc \\
 \\
x_2 \equal{} b(c \plus{} a) \minus{} 2ca \\
 \\
x_3 \equal{} c(a \plus{} b) \minus{} 2ab\end{array}\ \ \wedge\ \ \begin{array}{c} y_1 \equal{} \frac {1}{2(p \plus{} a)(2p \minus{} a)} \\
 \\
y_2 \equal{} \frac {1}{2(p \plus{} b)(2p \minus{} b)} \\
 \\
y_3 \equal{} \frac {1}{2(p \plus{} c)(2p \minus{} c)}\end{array}\ .$

Observe that $ \sum_{k \equal{} 1}^3 x_k \equal{} 0$ and $ a\le b\le c\ \Longleftrightarrow\ x_1\le x_2\le x_3\ \Longleftrightarrow\ y_1\le y_2\le y_3\ .$

From the Tcebyshev's inequality obtain easilly

$ \sum\frac {ab \minus{} 2bc \plus{} ca}{(3a \plus{} b \plus{} c)(b \plus{} c)} \equal{} \sum_{k \equal{} 1}^3x_ky_k\ge \frac 13\cdot\sum x_k\cdot\sum y_k \equal{} 0\ .$

[b][u]Here is 1:50 AM[/u]. Good night/good morning ![/b]

[quote="shaaam"]
I think this kind of inequalities cannot have [u]a brilliant[/u] solution, because denominator is already ugly.
It takes a long time to expand but it is the most effective method.
[/quote]

Indeed, i am mistaken  :oops: . Can_Hang and Vasc, your solutions are more than brilliant.  :)
Thanks for sharing your knowledge with us.

A generalization of Iran 1996 Inequality to above direction is the following:

Let $ a,b,c$ be nonnegative real numbers, no two of them are zero. If $ ab\plus{}bc\plus{}ca\equal{}3$, then

$ \frac 1{(a\plus{}b)^2}\plus{}\frac 1{(b\plus{}c)^2}\plus{}\frac 1{(c\plus{}a)^2}\ge \frac 3{4}\plus{}\frac {15(a\minus{}b)^2(b\minus{}c)^2(c\minus{}a)^2}{4(a\plus{}b)^2(b\plus{}c)^2(c\plus{}a)^2}$.

[quote="Vasc"]A generalization of Iran 1996 Inequality to above direction is the following:

Let $ a,b,c$ be nonnegative real numbers, no two of them are zero. If $ ab \plus{} bc \plus{} ca \equal{} 3$, then

$ \frac 1{(a \plus{} b)^2} \plus{} \frac 1{(b \plus{} c)^2} \plus{} \frac 1{(c \plus{} a)^2}\ge \frac 3{4} \plus{} \frac {15(a \minus{} b)^2(b \minus{} c)^2(c \minus{} a)^2}{4(a \plus{} b)^2(b \plus{} c)^2(c \plus{} a)^2}$.[/quote]

I proved the following inequality;

$ (ab \plus{} bc \plus{} ca)\left(\frac {1}{(b \plus{} c)^2} \plus{} \frac {1}{(c \plus{} a)^2} \plus{} \frac {1}{(a \plus{} b)^2}\right) \geq \frac {9}{4} \plus{} \frac {15(a \minus{} b)^2(b \minus{} c)^2(c \minus{} a)^2}{4(a \plus{} b)^2(b \plus{} c)^2(c \plus{} a)^2}$

After the expansion, (took a long time, and i have a physics test tomorrow  :( , what i will do with this inequality ?  :| )  i found interesting identity:

Let

$ A \equal{} (ab \plus{} bc \plus{} ca)\left(\frac {1}{(b \plus{} c)^2} \plus{} \frac {1}{(c \plus{} a)^2} \plus{} \frac {1}{(a \plus{} b)^2}\right)$

and
$ B \equal{} \frac {9}{4} \plus{} \frac {15(a \minus{} b)^2(b \minus{} c)^2(c \minus{} a)^2}{4(a \plus{} b)^2(b \plus{} c)^2(c \plus{} a)^2}$

Then we have:

$ A \minus{} B \equal{} \frac {ab(a \minus{} b)^4 \plus{} bc(b \minus{} c)^4 \plus{} ca(c \minus{} a)^4 \plus{} 8abc(a^3 \plus{} b^3 \plus{} c^3 \plus{} 3abc \minus{} a^2(b \plus{} c) \minus{} b^2(c \plus{} a) \minus{} c^2(a \plus{} b))}{(a \plus{} b)^2(b \plus{} c)^2(c \plus{} a)^2}$

which is obviously nonnegative.

 :)

P.S: Notice the denominator doesn't contain the 4 as the original, because at last, this 4 cancels out nicely.
However, i just  now noticed that, the inequality Vasc posted above is different inequality and stronger than mine.... Because if we multiply both sides by 3, that inequality will become:

$ (ab \plus{} bc \plus{} ca)\left(\frac {1}{(b \plus{} c)^2} \plus{} \frac {1}{(c \plus{} a)^2} \plus{} \frac {1}{(a \plus{} b)^2}\right) \geq \frac {9}{4} \plus{} \frac {45(a \minus{} b)^2(b \minus{} c)^2(c \minus{} a)^2}{4(a \plus{} b)^2(b \plus{} c)^2(c \plus{} a)^2}$

which is definitely stronger than the inequality i proved above...   :|

[quote="shaaam"][quote="Vasc"]A generalization of Iran 1996 Inequality to above direction is the following:

Let $ a,b,c$ be nonnegative real numbers, no two of them are zero. If $ ab \plus{} bc \plus{} ca \equal{} 3$, then

$ \frac 1{(a \plus{} b)^2} \plus{} \frac 1{(b \plus{} c)^2} \plus{} \frac 1{(c \plus{} a)^2}\ge \frac 3{4} \plus{} \frac {15(a \minus{} b)^2(b \minus{} c)^2(c \minus{} a)^2}{4(a \plus{} b)^2(b \plus{} c)^2(c \plus{} a)^2}$.[/quote]

I proved the following inequality;

$ (ab \plus{} bc \plus{} ca)\left(\frac {1}{(b \plus{} c)^2} \plus{} \frac {1}{(c \plus{} a)^2} \plus{} \frac {1}{(a \plus{} b)^2}\right) \geq \frac {9}{4} \plus{} \frac {15(a \minus{} b)^2(b \minus{} c)^2(c \minus{} a)^2}{4(a \plus{} b)^2(b \plus{} c)^2(c \plus{} a)^2}$
[/quote]

Actually, I made a mistake. :oops:  The inequality you proved is the best of this type. :lol:

[quote=Vasc]Let $ a,b,c$ be nonnegative real numbers, no two of them are zero. Prove that

$ (a) \ \ \ \ \ \ \ \ \  \sum\frac {ab\minus{}2bc\plus{}ca}{(3a\plus{}b\plus{}c)(b\plus{}c)} \ge 0$;
[/quote]
It's just $\sum_{sym}(a^5b+2a^4b^2-3a^3b^3+a^4bc+a^3b^2c-2a^2b^2c^2)\geq0$, which is  Muirhead.


algebra
polynomial
superior algebra
superior algebra unsolved

Prove that the polynomial
\[ f(x) \equal{} \frac {x^n \plus{} x^m \minus{} 2}{x^{\gcd(m,n)} \minus{} 1}\]
is irreducible over $ \mathbb{Q}$ for all integers $ n > m > 0$.

Solved with [b]mira74, AIME12345[/b]:

It is sufficient to prove irreducibility over $\mathbb{Z}$. By direct computation, it is easy to see that $f(x)$ is monic and has constant term $2$.

We claim that all roots of $f(x)$ have magnitude greater than $1$.

Let $x_0$ be a root of $f(x)$. If $|x_0| < 1$, then $0 < |x_0^n + x_0^m| \leq |x_0^n| + |x_0^m| = |x_0|^n + |x_0|^m = 1^n+1^m = 2$, contradiction. If $|x_0| = 1$, then since $x_0^n + x_0^m = 2$ and $|x_0| = 1$, and as \[D(x^n+x^m) = nx^{n-1} + mx^{m-1}\] is relatively prime to $f$, $f$ has no double roots, so $x_0^n = x_0^m = 1$. Thus, $x_0^{\gcd(m, n)}-1 = 0$, and $f(x)$ is undefined. Therefore, $|x_0| > 1$.

If $f(x) = g(x)h(x)$ is a nontrivial factorization into monic polynomials of strictly lower degree than $f(x)$, then without loss of generality $g(x)$ has constant term $\pm 1$. Since all complex roots of $g(x)$ are roots of $f(x)$ and in particular have magnitude greater than $1$, we have a contradiction. $\square$


AMC AIME USA(J)MO USAMO FTW probability geometry

In light of the Olympics, I have decided to host a Math Olympics. It will proceed as follows:

Everyone will be broken up into 3 or 4 person "countries".(I haven't decided as of yet.) You may make requests as to whom you would like to be with, and I will do my best to make sure everyone is happy, but I will ultimately decide. You can choose your team name together, but I have to approve of it. (So that it's not too long or inappropriate, etc.)

We will have different "sports"(such as Counting, etc.) The scores of your country in each sport will be added up in a total score. The sum of these total scores of each sport will be your final score. The team with the highest score will win.

I myself will not be participating, but I hope many people will. :)

Sign Up List(colors indicate people who have requested to be on a team together):

1. [color=indigo]PI-Dimension[/color]
2. [color=green]Urc[/color]
3. [color=red]Yongyi781[/color]
4. [color=red]5849206328x[/color]
5. [color=olive]herefishyfishy1[/color]
6. [color=blue]FantasyLover[/color]
7. [color=blue]AIME_is_hard[/color]
8. [color=orange]kl2836[/color]
9. [color=red]Bachukas[/color]
10. [color=green]Poincare[/color]
11. [color=indigo]27r[/color]
12. [color=green]suma_milli[/color]
13. [color=indigo]pythag011[/color]
14. [color=orange]qwertythecucumber[/color]
15. [color=olive]dannyhamtx[/color]
16. [color=olive]shentang[/color]
17. [color=olive]vallon22[/color]
18. [color=blue]superllamel[/color]
19. [color=orange]ghjk[/color]
20. [color=indigo]xpmath[/color]
21. [color=red]tinytim[/color]
22. [color=orange]mewto55555[/color]

[color=indigo]Galazawalagooki[/color]
[color=green]Waffle Union[/color]
[color=red]1337 Beasts[/color]
[color=olive]MATH13375[/color]
[color=orange]Mewtoland[/color]
                          [img]http://w3.byuh.edu/library/curriculum/Olympics/olympics1.gif[/img]

Do you two want to be represent a country together?

The real Olympics should do this too! :D

But unfortunately, the real Olympics say that a sport "must require physical exertion". *Sigh* Isn't scratching your head physical exertion?

Haha. They really should have a Nerd Olympics...I was actually considering calling this the "Nerd Olympics" but decided against it.

Are you joining, Yongyi? You would definitely pwn everyone. :wink:

Sure. :D

About the pwning everyone part... I'm not so sure... :maybe:
 :D

ill join. uh, doesnt matter who im with.

I will join to [color=red]red team[/color]

[quote="Poincare"]i wanna be with Urc!  :D[/quote]

Yeah, we can be together :D

Oops, I forgor to submit.

CAn you extend the deadline due to iTest?

I believe the entire competition has ended... right? :maybe:

This is what you sent me

[quote=isabella2296]Sorry, I didn't see them.

#1) Yes, greater than 2.

#7) I'll get back to you on that. [/quote]

#7- you never got back to me, and I have all the answers filled in except for number 7 which I am still waiting on.  :(

[quote="Wickedestjr"]Are you going to begin another one?[/quote]

plz don't post the same thing 30 minutes after you posted the first one

and BTW, i have the answers ready, i just didn't have time to pm them to you =)

which is an extremely sad way to lose points..... :(

shentang: I think I did get back to you, but I must have sent it to the wrong person by accident - sorry. :oops:

Wicked: I may do another, but not anytime soon. Maybe in the summer.

You mean next summer? Because this summer has ended :P

Well, yes, that's what I meant. I wish it was still summer though...  :(

Hey Izzy after this coudl you set up another tournament like this or the other one.

I will in the summer or later this year - I'm way too busy in the school year.

5849206328x and I will make a really hard tournament,(it will be able to be solved with mathcounts techniques) We plan to start in a month or so, depending on how much questions and their solutions we get done.....

There already exists a really hard one...

See if you can solve all the ones on mine....

Some of them are IMO difficulty...but they can all be solved using MathCounts Techniques!

In fact, I think I went a little too far in making my problems hard...but a lot of people said that the previous round with an IMO Shortlist problem and a ToT problem was easy....


let Sn=1/C(0,n)+1/C(1,n)+1/C(2,n)+..+1/C(n,n)
what is the maximum of (Sn) over N ??

The maximum of the sequence S_n is 8/3 = S_3 = S_4.
It is easy to compute S_1 = 1, S_2 = 5/2.

Now, we prove that the sequence (S_n) is decreasing for n >= 4 :
The following sums are over k = 0,..,n.
We have S_(n+1) - S_n = -1 + sum( 1/C(k,n) - 1/C(k,n+1) ) = -1 + sum( k/((n+1)*C(k,n)) ) = -1 + (1/(n+1))*sum( k/C(k,n) ) > -1 + (1/(n+1))*(0/C(0,n) + 1/C(1,n) + (n-1)/C(n-1,n) + n/C(n,n) ) = 0.
And we are done.

Pierre.