calculus
integration
function
Galois Theory
calculus computations

Can someone explain me the following tips of calculate this form of integral?
$\int x^{m}(a+bx^{n})^{p}dx$(a,b are real numbers,and m,n,p are fractions)
a,If p is integer,Put $t^{N}=x$,N is the denominator of m and n.
b, If $\frac{m+1}{n}$ is integer.Then put $a+bx^{n}=t^{N}$,where N is the denominator of p.
c,If $\frac{m+1}{n}+p$ is integer.Then put $ax^{-n}+b=t^{N}$,where N is the denominator of p.

That type of function is called "differential binomial", and those tips are the Chebichev (Integrability) Conditions. They provide you a mean of finding a substitution that will greatly simplify the job. As an exercise, calculate the following:

\[\int \frac{1}{\sqrt{x^{3}}\sqrt[3]{1+\sqrt[4]{x^{3}}}}dx\]

On the other hand, if none of the Chebichev Conditions is met, then you cannot express your antiderivative as a finite sum of elementary functions. A simple example is the following:

\[\int \sqrt{1+x^{4}}dx\]

Hope this helps.

Carcul

[quote="Carcul"]
On the other hand, if none of the Chebichev Conditions is met, then you cannot express your antiderivative as a finite sum of elementary functions. A simple example is the following:
\[\int \sqrt{1+x^{4}}dx \]
Hope this helps.

Carcul[/quote]
Thanks Carcul,can you show me more about the statement above?

[quote="greengrass"]With de substitution $x^{n}=y$,

${I= \int x^{m}\left( a+bx^{n}\right)^{p}dx = \frac{1}{n}\int y^{\frac{m+1}{n}}( a+by)^{p}dy = \frac{1}{n}\int \left( \frac{a+by}{y}\right)^{p}y^{p+\frac{m+1}{n}-1}}dy$

[color=blue][b]A.-[/b][/color] If $p$ is an integer, introduce de new substitution $y=v^{s}$, where $\frac{m+1}{n}-1 = \frac{r}{s}$, an irreductible fraction, so

$I = \frac{s}{n}\int v^{r+s-1}\left(a+b^{s}\right)^{p}\,dv$, a rational integral.

[color=blue][b]B.-[/b][/color] If $\frac{m+1}{n}$ is an integer and $p = \frac{\rho}{\sigma}$, an irreductible fraction, with de new sustition $a+by = w^{\sigma}$,

$I = \frac{\sigma}{nb^{(m+1)/n}}\int \left( w^{\sigma}-a \right)^{(m+1)/n-1}dw$, another rational one.

[color=blue][b]C.-[/b][/color] If $\frac{m+1}{n}+p$ is an integer, with de new substition $\frac{a+by}{y}= z^{\sigma}$,

$I =-\frac{a^{p+(m+1)/n}}{n}\int \frac{z^{\rho+\sigma+1 dz}}{ \left( z^{\sigma}-b \right)^{1+p+(m+1)/n}}$ a new rational integral.

Nesting two substitions you get

[color=blue][b]A.-[/b][/color] $x^{n}= y = v^{s}$

[color=blue][b]B.-[/b][/color] $w^{\sigma}= a+by = a+bx^{n}$

[color=blue][b]C.-[/b][/color] $z^{\sigma}= \frac{a+by}{y}= \frac{a+bx^{n}}{x^{n}}= ax^{-n}+b$[/quote]
I know,but can anyone explain the following statement?
We can't calculate the integral $\int \sqrt{1+x^{4}}dx$ as a elementary function.

Read [url=http://math.hunter.cuny.edu/ksda/papers/churchill.pdf]this[/url].  If you succeed in understanding the article, you should be able to prove it yourself after finishing it; if any questions arise, just ask; if you find it too difficult overall, wait a year and try again :) . The article is almost self-contained, though some knowledge of advanced algebra (more precisely, of elements of Galois theory) is assumed.

Otherwise the Chevysev integral becomes the incomplete beta function


Determine the best constant $ M$

$ \frac{x^6\minus{}2x^5\minus{}7x^4\minus{}4x^2\minus{}8x\plus{}4}{(x*(x\plus{}1))^2}\ge M$

where $ x$ is real number. Find when equality occurs.

[easy with calculus...can you find a well-motivated solution w/o calculus?]

[quote="Altheman"]Determine the best constant $ M$

$ \frac {x^6 \minus{} 2x^5 \minus{} 7x^4 \minus{} 4x^2 \minus{} 8x \plus{} 4}{(x*(x \plus{} 1))^2}\ge M$

where $ x$ is real number. Find when equality occurs.

[easy with calculus...can you find a well-motivated solution w/o calculus?][/quote]
$ \frac {x^6 \minus{} 2x^5 \minus{} 7x^4 \minus{} 4x^2 \minus{} 8x \plus{} 4}{x^2(x \plus{} 1)^2}\equal{}$
$ \equal{}\left(\frac{x^3\minus{}x^2\minus{}2x\plus{}2}{x^2\plus{}x}\right)^2\minus{}4\geq\minus{}4.$
The equality occurs when $ x\in\{1,\sqrt2,\minus{}\sqrt2\}.$

just wondering... arqady, how do you randomly pull perfect squares out of thin air??

Or did you think "oh, it's going to need these terms" and guess-and-check or something?

The denominator is a perfect square, so it makes sense to look for perfect squares in the numerator.

Just looking at the first few terms of the numerator, you can tell it's $ x^3 \minus{} x^2 \plus{} ax\plus{}b$. From there, it's not too bad.

Nice question, nice solution.


Δίνονται τα μη μηδενικά διανύσματα $ \vec{a_1}, \vec{a_2},...,\vec{a_n}$, με άθροισμα $ \vec{s}\neq 0$. Έστω $ m \equal{} min{\frac {|\vec{s}|}{|\vec{a_i}|}, i \equal{} 1,2,...,n }$. Θεωρούμε το καρτεσιανό επίπεδο συντεταγμένων με αρχή Ο, τον κύκλο (Ο,m) και μία εφαπτομένη του ε: Αx+By+Γ=0. Να βρεθεί η ελάχιστη τιμή της παράστασης:
$ \sqrt [2^{l \plus{} 1}]{\frac {2^{2^l \minus{} 1}(A^{2^{l \plus{} 1}} \plus{} B^{2^{l \plus{} 1}})}{\Gamma^{2^{l \plus{} 1}}}}, l \in N$.

απ την την ιδιοτητα με την αποσταση παίρνουμε $ G^2\equal{}m^2(A^2\plus{}B^2)$ υψωνουμε στην $ 2^l$ και επειτα με $ B\minus{}C\minus{}S$ κατω απο την ριζα παίρνουμε το ελαχιστο συναρτησει του ..$ m$....(Mε LATEX λογω ριζας-δύναμης-$ 2^l$ θέλω καμμια εβδομαδα να γραψω αναλυτικα την λυση μου)
Η κατι δεν κανω καλα ή τα διανυσματα ειναι για να ειναι...
Επισης το $ l$ πως βοήθησε στην πλοκη της ασκησης..
Καλη χρονια σε όλους

Δυστυχώς, η ανισότητα είναι μερικώς σωστή, ενώ το αποτέλεσμα είναι λάθος. Πρέπει να βρεθεί η ελάχιστη τιμή συναρτήσει σταθερών ποσοτήτων και μόνο. Το m μεταβάλλεται, όπως και το l. To l δεν είναι σταθερό. Το αποτέλεσμα πρέπει να είναι ανεξάρτητο της τιμής του. Αν και στη λύση σας το l δεν παρεμβαίνει, γενικά πρέπει να προσεχθεί.
Καλή Χρονιά εύχομαι επίσης! Είθε το νέο έτος να δεκαπλασιάσει τα καλά!

1o)Τι σημαίνει οτι το αποτελεσμα ειναι ανεξαρτητο της τιμης του ; Δεν το εχω ξανασυναντησει . Mηπως εννοείς της μεταβλητης του;
2o) Eχω συναντησει απειρα ακροτατα συναρτησει παραμετρων ...
3ο) Ποια θεωρείς σταθερα και ποια μεταβλητα στην ασκηση; (βλεπεις οι αγωστοι δεν ειναι και λιγοι Α, Β, Γ, l ,m, s, a1, a2,......, an, n)
4ο) Μηπως θελεις την αριθμητικη τιμη του ελαχίστου; (δεν το αναγράφεις...)
5ο) Υποθετω οτι στην συνεχεια θελεις και ελαχιστοποίηση του $ m$...

Φιλικα X.P.

Η ελάχιστη τιμή είναι $ \frac{1}{n}$, καθώς το n δεν είναι άγνωστος!
1)Εννοώ ότι δεν πρέπει να βρεθεί αποτέλεσμα που να περιέχει το l.
3)Στην άσκηση μόνο το n θεωρείται σταθερά.
4)Όχι.
5)Όχι, μεγιστοποίηση χρειάζεται!


geometry
3D geometry
rotation
geometric transformation
combinatorics
IMO Shortlist
IMO Longlist

Is it possible to find a set of $100$ (or $200$) points on the boundary of a cube such that this set remains fixed under all rotations which leave the cube fixed ?

Please post your solutions. This is just a solution template to write up your solutions in a nice way and formatted in LaTeX. But maybe your solution is so well written that this is not required finally. For more information and instructions regarding the ISL/ILL problems please look here: [url=http://www.mathlinks.ro/Forum/viewtopic.php?t=15623]introduction for the IMO ShortList/LongList project[/url] and regarding[url=http://www.mathlinks.ro/Forum/viewtopic.php?t=15623]solutions[/url]  :)

For $200$, just put $18$ equally spaced points on each side. You get $12\cdot 18-16=200$ points. I subtracted $16$ because the $8$ corners belong to three sides each.

For $100$ I don't think it's possible. All corners are the same (they all have a point or none of them has one). All  interiors of sides have the same number of points, and all interiors of faces have the same number of points, meaning that we must either have $6|100$ or $6|100-8=92$, and none of these is true.

Is it Ok?


calculus
derivative
inequalities
inequalities theorems

[u][b]The author of this posting is : Megus[/b][/u]
____________________________________________________________________

Ok, because some people ask me about it, I post it , here it goes:

-----------------------------------------------------------

[b]Sum of derivatives[/b]

If $f\left(x_1,x_2,...,x_n\right)$ fulfills two conditions:

[b]1.[/b] $f$ is non-negative with assumption that one variable is equals zero, and others are non-negative also:

$f\left(x_1,x_2,...,x_{i-1},0,x_{i+1},...,x_n\right) \geq 0$

for non-negative numbers $x_1$, ..., $x_n$;

[b]2.[/b] for every set of non-negative numbers $x_1$, ..., $x_n$ following inequality holds

$f^{\prime}_{x_1}\left(x_1,...,x_n\right) + ... + f^{\prime}_{x_n}\left(x_1,...,x_n\right) \geq 0$

Then

$f\left(x_1,...,x_n\right) \geq 0$

-------------------------------------------------------

The proof of this is not long and not too hard so I leave it to you  8)

Any comments welcome.

I guess billzhao gave a short proof for this but it was lost due to attack on server

This comes up in my mind, is this valid?
Assume the contrary, i.e. there exists a set of $(y_1,...,y_n)$ such that $f(y_1,...,y_n) \le 0$. Since $f(0,y_2,...,y_n)$,...,$f(y_1,...,0)$ are all non-negative, there must be some intervals $[0,z_i]$ for $f(y_1,..,y_i,..,y_n)$ to decrease (when we are considering y_i, assume other variables to be constant).
Take z to be $min(z_i)$, then at the interval [0,z], $f'_{y_i}(y_1,...,y_n) \le 0$ for all $i=1,...,n$. Then $\sum f'_{y_i}(y_1,...,y_n) \le 0$ which contradicts with the second condition.

Here's proof taken from book "Powrót do krainy nierówności" by Lev Kourliandtchik (I've shorten it a bit)

$a=min(x_1,...,x_n)$ then $y_i=x_i -a$ are nonnegative and so $f(y_1,...,y_n) \geq 0$. Consider $g(t)=f(y_1+t,...,y_n+t)$. Then $g(0) \geq 0$ and $g'(t)=f' _{y_1}+...+f' _{y_n} \geq 0$ and so for each $t\geq 0$ , $g(t) \geq 0$ in particular $f(x_1,...,x_n)=g(a) \geq 0$

[quote="arqady"]But it does not help for hard inequalities.
Let $a>0$, $b>0$  and $c>0$. Prove that:
\[\frac{a^2}{b}+\frac{b^2}{c}+\frac{c^2}{a}\geq\frac{37(a^2+b^2+c^2)-19(ab+ac+bc)}{6(a+b+c)}\]
Nothing! :wink:[/quote]

Well, your inequality is strong. Stronger method is needed to solve your problem, of course.

For the general problem,
$\forall x,y,z \geqslant 0 \wedge x+y+z=3$
\[ \frac{x^2}{y}+\frac{y^2}{z}+\frac{z^2}{x} \geqslant 3k - \left( k-1 \right) \cdot \left( xy+yz+zx \right)\]

The best constant is
\[k_0= 6+\cfrac{1}{5+\cfrac{1}{1+\cfrac{1}{5+\cfrac{1}{1+...}}}}\]
Second approximation to $k_0$ is $k_2=6+\frac{1}{5+1}=\frac{37}{6}$, which is your inequality.

[quote="arqady"]
Let $a>0$, $b>0$  and $c>0$. Prove that:
\[\frac{a^2}{b}+\frac{b^2}{c}+\frac{c^2}{a}\geq\frac{37(a^2+b^2+c^2)-19(ab+ac+bc)}{6(a+b+c)}\]
Nothing! :wink:[/quote]

how to prove this one ? thanks !


Find all positive integers a,b,c such that
((a+b)(b+c)(c+a))/2 + (a+b+c)^3 = 1-abc

There are must be typo, since LHS>=1 and RHS<=0.

I think this is $+abc$.
Therefore the problem is easily solved with the identity :
$(a+b)(b+c)(c+a)+2(a+b+c)^3+2abc=(2a+b+c)(a+2b+c)(a+b+2c)$

Pierre.

Sorry for my typo. It should have been "Find all integers".


analytic geometry
graphing lines
slope
linear algebra
matrix
algebra
system of equations

I appericate all the help and want to thank you all in advance..
Im just a little stumped on how to figure these out and show the work for it..Espsically questions #3-6

p.s - i apologize if this is the wrong forums. I just came here because it said college playground, and this is from my college class..

1.  Find the slope of the line that passes through the points (2, 3) and (5, 8).

2.  Find the equation of the line that passes through the points (3, -2) and (4, -2).

3.  Find the equation, in standard form, with all integer coefficients, of the line perpendicular to x + 3y = 6 and passing through (-3, 5).

4.  Solve the system of equations using the substitution method.
If the answer is a unique solution, present it as an ordered pair: (x, y). If not, specify whether the answer is “no solution” or “infinitely many solutions.”
-3x + y = 1
5x + 2y = -4

5.  Solve the system of equations using the addition (elimination) method.
If the answer is a unique solution, present it as an ordered pair: (x, y). If not, specify whether the answer is “no solution” or “infinitely many solutions.”
-7x + y = 8
2x – y = 2

6.  Solve the system of equations using the addition (elimination) method.
If the answer is a unique solution, present it as an ordered pair: (x, y). If not, specify whether the answer is “no solution” or “infinitely many solutions.”
3x – 2y = -7
-9x + 6y = 21

[Moderator: moved]

[hide="hints"]
1. The slope of the line passing through the points $ M_{1}(x_{1},y_{1})$ and $ M_{2}(x_{2},y_{2})$ is
$  m=\frac{y_{1}-y_{2}}{x_{1}-x_{2}}$

2. The equation of the line passing through the points $ M_{1}(x_{1},y_{1})$ and $ M_{2}(x_{2},y_{2})$ is
$  \frac{x-x_{1}}{x_{2}-x_{1}}=\frac{y-y_{1}}{y_{2}-y_{1}}$ or $ \begin{vmatrix}x & y & 1\\x_{1}& y_{1}& 1\\x_{2}& y_{2}& 1\end{vmatrix}=0$

3. If $ d_{1}, \ d_{2}$ have the slopes $ m_{1}, \ m_{2}$ and $ d_{1}\perp d_{2}$ then $ m_{1}\cdot m_{2}=-1$
The line $ x+3y=6$ has the slope $  m_{1}=-\frac{1}{3}$, then the slope of the perpendicular is $ m_{2}=3$.
Now, the equation of the line with the slope $ m$ and passing through the point $ M_{0}(x_{0},y_{0})$ is
$ y-y_{0}=m(x-x_{0})$.

4. From the first equation, $ y=1+3x$ and substitute in the second and solve for $ x$.

5. Adding the two equations $ y$ will be eliminated and solve for $ x$. Then replace $ x$ in one equation and find $ y$.

6. Multiplying the first equation by 3, and adding you'll get $ 0=0$, then the system has infinetly many solutions.
[/hide]


There are approximately 7 months to the AMC 12. Right now, I can just get about a 100 on the test. I am doing Volume 2 and I am done with the whole intro series. I have never gotten above a 5 on an AIME. Is it still practical for me to set USAMO as a goal? I know this a pretty vague question and seems really stupid to ask a question like this.

Yes. I know someone who started math and got a score that would have made red MOP in the same. You're already ahead of him.

You have to make sure to work hard though. In my opinion, the best way to get better at AIMEs/AMCs is to do Olympiad math. After mucking around with that, when you look at AIMEs, you'll go... woah this stuff is so easy o_O. That's what happened to me.

Yeah, thats what happened to me too. I definitely second that advice. However, make sure you dont completely ignore aimes, cuz youll end up making tons of really stupid mistakes...

You should also not neglect MathCounts/other speed tests, or else you will be too slow/careless on the AMCs and AIME.

I was in a similar situation as you last year (108 amc 12, 6 aime) and I made usamo by substantial margin this year by working on olympiad problems and practicing a few amcs and aimes before the test.

So, what olympiads should I start with?

BTW, I have olympiads from around the world (i think 3 of em).

According to pythag011, the best Olympiads to start with are Canadian Olympiads. He says they're more like AIME level. Once you've mastered these (or at least get comfortable with Olympiads), you can move onto American Olympiads and maybe Chinese Olympiads.

[quote]After mucking around with that, when you look at AIMEs, you'll go... woah this stuff is so easy o_O.[/quote]

How much time do you need to spend "mukking around" with this. Are you guys suggesting that he should always practice Olympiad problems until the competition?

[quote="ritwik_anand"]from Intro to olympiads, isn't that a BIG leap?[/quote]

I'm doing volume 2 also (on chapter 6) and i've done some AIME problems so yeah...


A train leaves Newcastle for Sydney at 7.45am while another train leaves from Sydney for Newcastle at 8.15am. The first train arrives in Sydney 40 minutes after the trains pass while the train to Newcastle arrives 1 hour 40 minutes after they pass. Assuming that the trains travelled at constant speeds (not necessarily the same), at what time did they pass?

Source: Australian Math Competition, S24, 2001


What kind of program is used to run math jams? Is there something similar where one of us can host our own private math jams?  It sounds like a great online tutoring tool.

I'm not sure if this is what you're asking, but they use a programming language, Java.


topology
geometry
3D geometry
sphere
real analysis
real analysis solved

hi evrone,

how could we prove that there is no homeomorphism between R^2 and R^3 ..
(i've just seen a proof for the case of R and R^2. It uses connexity .. )

any idea ..

really .. so I suppose it's a hard question .. could you give some more information moubinool ?

Doesn't the Borsuk-Ulam theorem state that any continuous mapping from the sphere to the plane maps a certain pair of antipodes antipodes to the same point? In particular, a continuous mapping from a sphere to a plane cannot be injective. This means that we can do a lot less than find a homeomorphism.


Find all pairs of nonnegative integers $ (x, y)$ satisfying $ (14y)^x \plus{} y^{x\plus{}y} \equal{} 2007$.

$ x\equal{}0$ or $ y\equal{}0$ are not solutions

$ y^x(14^x\plus{}y^y)\equal{}2007\equal{}3^2\cdot 223 \Rightarrow y\in \{ 1,3,3^2,223,223\cdot 3,223 \cdot 3^2 \}$

Finally, the only possible solution is $ (x,y)\equal{}(2,3)$

$ y \equal{} 0,x \equal{} 0$ are not solutions and $ 14^3 \equal{} 2744$ so $ x \equal{} \{1,2\}$ and $ x \equal{} 1$ not solution so $ x \equal{} 2$ and $ y \equal{} 3$

[quote="dima ukraine"]$ y = 0,x = 0$ are not solutions and $ 14^3 = 2744$ so $ x = \{1,2\}$ and $ x = 1$ not solution so $ x = 2$ and $ y = 3$[/quote]  I solved it the same way using that $ y$ is odd and $ y^{y + 1}\le2007 , y^{y + 2}\le2007$ for $ y = 1,3$


\[
\begin{gathered}
  \sum {\frac{{4a^3  + 5b^3  - 3a^2 b + 10ab^2 }}
{{3a + b}} - \left( {a^2  + 4b^2  - ab} \right)}  \geqslant 0 \hfill \\
   \Leftrightarrow \sum {\frac{{4a^3  + 5b^3  - 3a^2 b + 10ab^2  - \left( {a^2  + 4b^2  - ab} \right)\left( {3a + b} \right)}}
{{3a + b}}}  \geqslant 0 \hfill \\
   \Leftrightarrow \sum {\frac{{a^3  + b^3  - ab^2  - a^2 b}}
{{3a + b}}}  \geqslant 0 \hfill \\
   \Leftrightarrow \sum {\frac{{(a + b)(a - b)^2 }}
{{3a + b}}}  \geqslant 0 \hfill \\
\end{gathered}
\]


geometry
parallelogram
geometry unsolved

Let $ ABCD$ be a quadrilateral such that $ AB \equal{} CD$. Let $ I$ and $ J$ be the midpoints of the segments $ AD$ and $ BC$. Prove that $ \measuredangle\left(\overrightarrow{AB},\overrightarrow{IJ}\right) \equal{} \measuredangle\left(\overrightarrow{IJ},\overrightarrow{DC}\right)\text{ mod }2\pi$.

I have a solution but it's not a general solution .
  We erects parallelogram AKLD with I, J are the midpoints of AD , KL . Hence , tri ABK = tri DLC .---> <BAK = <LDC  .

    I think this solution depends on pictures . In case ABCD is a unconvex quadrilateral  , is that true?

See http://www.mathlinks.ro/Forum/viewtopic.php?t=193499 for a related problem.


complex analysis
complex analysis unsolved

this question is of two parts:

1)if zo is real and G is a region symmetric with respect to the real axis
show that f satisfies the symmetric relation f(z bar)=f(z) bar.

2) if zo is real and G is a region symmetric with respect to zo then find a symmetric
that f satisfies in G.
 thx

These questions make no sense. Proofread yourself.

maybe u are right but i had final exam on monday and i have toooooo many to study and solve and i do not have time.
for this reason iam ASKING FOR HELP BECAUSE I RUN OUT OF TIME

People need to know what you're asking if you want them to help you.  ;)


number theory
combinatorics unsolved
combinatorics

Find all primes p such that each nonzero element of Z(p) is its own multiplicative inverse.  You must prove that the primes you find work and are the only ones that work.

Don't multipost (this was posted in Number Theory, Advanced Fields and Superior Algebra too).

But I think some moderator should simply delete that one (the one in Superior Algebra was already deleted)...

PS: I just copied this post from http://www.mathlinks.ro/Forum/viewtopic.php?t=85134 where one more of these posts is ;)


Find the least positive number $k$ such that the inequality holds for all nonegative real numbers $a,b,c,d$ \[|(a-b)(b-c)(c-d)(d-a)(a-c)(b-d)|\leq k(a+b+c+d)^{6}.\]

Have you solved this problem?I don't think this problem is interesting.But the solution may be very long ,I don't have the time to solve it . :P

What a pity that you do not like it. I like it.

Cuong, can you draft your ideas for this problem here?

[quote="paladin8"]Not sure if this gives the minimum $k$ at all, but here's an idea. WLOG assume $a < b < c < d$ (if any two are equal, the LHS is $0$ and uninteresting). Set $b = a+x_{1}$, $c = a+x_{1}+x_{2}$, and $d = a+x_{1}+x_{2}+x_{3}$ where $x_{1}, x_{2}, x_{3}$ are positive reals. The inequality becomes

$x_{1}x_{2}x_{3}(x_{1}+x_{2}+x_{3})(x_{1}+x_{2})(x_{2}+x_{3}) \le k(4a+3x_{1}+2x_{2}+x_{3})^{6}$.

Clearly setting $a = 0$ forces $k$ to be higher, so it remains to find the smallest $k$ satisfying

$x_{1}x_{2}x_{3}(x_{1}+x_{2}+x_{3})(x_{1}+x_{2})(x_{2}+x_{3}) \le k(3x_{1}+2x_{2}+x_{3})^{6}$,

which may just hold and have equality at $x_{1}= x_{2}= x_{3}\Rightarrow k = \frac{1}{3 \cdot 6^{4}}$.[/quote]
I think we can just deduce that$k\geq \frac{1}{3 \cdot 6^{4}}$,And I believe that that is not the best $k$

Sure sir, indeed now we can assume that the equality holds when $x=ny=nmz$ and use AM-GM. The equation to find $m$ and $n$ however is at big degree, but no problem with a computer, right  :D


$ 1.1.6$ Beginning with 2 and 7, the sequence 2,7,1,4,7,4,2,8,... is constructed by multiplying successive pairs of its members and adjoining the result as the next one or two members of the sequence, depending on whether the product is a one- or a two-digit number. Prove that the digit 6 appears an infinite number of times in the sequence.

$ 1.1.7$ Let $ S_1$ denote the sequence of positive integers $ 1,2,3,4,5,6,.....,$ and define the sequence $ S_{n\plus{}1}$ in terms of $ S_{n}$ by adding 1 to those integers in $ S_n$ which are divisible by n. Thus, for example, $ S_2$ is $ 2,3,4,5,6,7,...$, $ S_3$ is $ 3,3,5,5,7,7,...$. Determine those integers $ n$ with the property that the first $ n\minus{}1$ integers in $ S_n$ are $ n$.

[hide="My Solutions"] 1.1.6 First, we continue writing out some more elements of this sequence: 2,7,1,4,7,4,2,8,2,8,8,1,6,1,6,1,6,6,4,8,6,6,6,6,6,3,6,2,4,.....
Next we notice, that there are a chain of 6's which will form a sequence like: 3,6,3,6,3,6...
This in turn will form something like: 1,8,1,8,1,8...
Which will lead to 8,8,8,8,8,8...
And then 6,4,6,4,6,4,6,4....
$ \Longrightarrow$ 2,4,2,4,2,4,2,4,2,4,
$ \Longrightarrow$ 8,8,8,8,8,8,8,8
And the cycle repeats which gurantees that 6 appears an infinite amount of times in this sequence.

1.1.7 We write out the first few elements of the first few sequences:
$ S_1$=1,2,3,4,5,6,...
$ S_2$=2,3,4,5,6,7,...
$ S_3$=3,3,5,5,7,7,9,9...
$ S_4$=4,4,5,5,7,7,10,10...
$ S_5$=5,5,5,5,7,7,10,10...
$ S_6$=6,6,6,6,7,7,11,11,...
$ S_7$=7,7,7,7,7,7,11,11,...
Apparently, this is true for all the primes.
How do we prove this though?
[/hide]

Hey, can you look over my proof for 1.1.6 and give me a hint for 1.1.7?
My proof writing skills are really bad-sorry.

P.S. This chapter is called "Search for a Pattern". Thanks!

Here's a hint for 1.1.7.
Try to prove the following things:

1) For all $ n$, $ S_n$ is nondecreasing.
2) For all $ n$, the first term of $ S_n$ is $ n$
3) For all primes $ p$ and for all $ n$ such that $ 1 < n \leq p$, the $ (p \minus{} 1)$st term of $ S_n$ is $ p$.
4) For all composite numbers $ c$, the $ (p \minus{} 1)$st term of $ S_c$ is greater than $ c$.

How can I make my proof "better" for 1.1.6?

bump.
Question: Do people not like Larson's problems??

BUMP! again can someone tell me if my proof for the first problem is correct?

Proving that the cycle size gets larger would be nice. Please don't bump that many times, by the way.


1. A stage course is attended by $ n$ students ($ n>4$). The day before the final test, each group of three students conspires against another student, to throw him or her out of the competition. Prove that there is a student against whom there are at least $ \sqrt[3]{(n\minus{}1)(n\minus{}2)}$ conspirators.

2. Seated at a round table are 100 representatives of 25 countries (four persons from each country). Prove that they can be partitioned into four groups satisfying the following conditions: (i) each group contains a representative of each country, (ii) each person has no neighbors among the members of his group.

[hide="wrong solution"]There are $ C^n_3\equal{}\frac{n(n\minus{}1)(n\minus{}2)}{6}$ groups.

So there are $ \frac{n(n\minus{}1)(n\minus{}2)}{6}$ conspiracies going on.

There are $ n$ students so there is at least one of them against whom there are at least $ \frac{(n\minus{}1)(n\minus{}2)}{2}$ conspirators.

Why is this not right?[/hide]


quadratics
Ring Theory
geometry
3D geometry
number theory
relatively prime

Prove elementary, that the equation $x^{2}-2=y^{3}$ has no the solutions in the set of positive integers

Let $x=3k+i$ and $y=3k'+j$ where $i,j=0,1,2.$ By $(3k+i)^{2}=(3k'+j)^{3}+2$
we conclude that $i^{2}\equiv j^{3}+2{\text{ }}(\bmod 3).$ So $(i,j)=(0,1),(1,2),(2,2).$
If $(i,j)=(0,1)$, we have $9|3$; contradiction.
If $(i,j)=(1,2)$ or $(i,j)=(2,2)$ we have $y=3k+2$ and $y\equiv 2(\bmod 3).$ So $3\nmid y^{3}-1.$ Now we have $x^{2}\equiv y^{3}-1(\bmod 3)$. Therefore by Legender symbol $1 = (\frac{{y^{3}-1}}{3}) = (\frac{{y-1}}{3})(\frac{{y^{2}-y+1}}{3}) = 0$, because $3 | y^{2}-y+1$; contradiction.
For Legender symbol refer to [url]http://planetmath.org/encyclopedia/LegendreSymbol.html[/url]

Umm, in the last step you factored wrong.  $y^{3}-1 \neq (y-1)(y^{2}-y+1)$

Last remark is very interesting. You state that $y^{3}-1$ - quadratic reduce to mod 3 but then there is given contradiction itself.
I made the mistake that this proposition is given here. For the equation $x^{2}-2=y^{3}$ there is not elementary solution.
I thought that it is the consequence of Mordell's theorem -  for all $k$ - integer the equation $x^{2}+k=y^{3}$ have final number of integer solutions.
For an example, the equations $x^{2}+3=y^{3}$, $x^{2}-7=y^{3}$ is enable the elementary solutions.

I heard somewhere that this can be solved by factoring out the complex roots or something (presumably of the $y^{3}$), but I haven't managed anything yet. It turned up on a sheet of exercises at a British camp this week (although it was penultimate, and the question following it was $x^{n}+y^{n}=z^{n}$ with $n>2$ :)).

$\mathbb{Z}(\sqrt{2})$ is an UFD so what are we waiting for?  :wink:. Just see that if ($p$ is prime) $p|x+\sqrt{2}$ and $p|x-\sqrt{2}$ then $p=\sqrt{2}$ so $2|x^{2}-2$ hence $2|x$ so $2|y$ then $8|y^{3}$ but $8\not |x^{2}-2$ contradiction. Hence $x+\sqrt{2}=(a+b\sqrt{2})^{3}$ for some natural $a,b$ hence.

$1=2b^{2}+3a^{2}b$ - so $b=1$ or $b=-1$. Both cases lead to a contradiction. Hence no solution.

Certainly this equation is easier than Fermat's theorem.
But for demonstration it is necessary to construct arithmetics of ring $\mathbb{Z}(\sqrt{2})$.
$(x-\sqrt{2})(x+\sqrt{2})=y^{3}$ hence ${x+\sqrt2}=(a_{1}+b_{1}\sqrt2)^{3}$, ${x-\sqrt2}=(a_{2}+b_{2}\sqrt2)^{3}$
It is correct if $x-\sqrt2$, $x+\sqrt2$ - relatively prime and the main theorem of arithmetics is saved for $\mathbb{Z}(\sqrt{2})$.

I've never seen UFDs before. Is it essentially taking numbers of the form \[a+b\sqrt{2}\] and pretending they are integers in a complex-numbery sort of way?

@Megus: what about multiplication with units¿

@Ilthigore: there is more behind it than that, but that's the first idea. I don't think it's possible to explain it here, so I suggest to get some book about such stuff.

You're right ZetaX - I forgot.

It can be proved that the only units of $\mathbb{Z}(\sqrt{2})$ are $\pm w^{n}$ where $w=1+\sqrt{2}$ and $n \in \mathbb{Z}$. So we have some cases to deal with (four if I'm not mistaken), but I don't feel like doing it now  :)

In fact, you have reduced Mordell equation $x^{2}-2=y^{3}$ to the Thue equation $y^{3}+3x^{2}y+2x^{3}=1$, so probably you have to prove that this last equation has no integer solutions but $(x,y)=(0,1)$, or something similar.
But this DOES NOT SEEM TO BE A BIG IMPROVEMENT. In fact, this equation appears to be at least as hard to solve than the initial one.
Anyway, what is evident is that the last equation has the rational solution $(x,y)=(-\frac{3}{2},1)$ and as $2(-\frac{3}{2})^{2}-(-\frac{3}{2}+1)^{2}=\frac{17}4$ we obtain a rational solution to the initial equation $(\frac{71}8)^{2}-2=(\frac{17}4)^{3}$.

The rank of the elliptical curve  $x^{2}-2=y^{3}$ is one, therefore there is indefinite much rational solutions  and it is propagate from one generatrix. But it has poor relation to the solution of this equation in integers.

Apparently, there is an elementary solution to  the diophantine equation $x^{2}-2=y^{3}$.
The article [url=http://mathworld.wolfram.com/MordellCurve.html]Mordell Curve[/url] of MathWorld states that Uspensky and Heaslet, in his 1939's book "Elementary Number Theory", give elementary solutions to the Mordell equations $x^{2}-n=y^{3}$ for $n=-4,-2,2$, and then give $n=-1,-5,-6,1$ as exercises.
So, please, if anyone out there has access to this book, post the elementary solution.

The equation is solved by means of a simple device due to A. Brauer -said Uspensky and Heaslet-.
First, rewrite it as $(y+2)^{3}=x^{2}+6(y+1)^{2}$, from where it is clear that $2\nmid y+2$, $3\nmid y+2$ and $x$ and $y+1$ are relatively prime. In the ring $\mathbb Z[\sqrt{-6}]$ of integers of $\mathbb Q(\sqrt{-6})$ we have the ideals $\langle x+(y+1)\sqrt{-6}\rangle$ and $\langle x-(y+1)\sqrt{-6}\rangle$ relatively prime to each other because any prime ideal $\mathfrak p$ dividing both should divide $\langle 2x\rangle$, $\langle 2\sqrt{-6}(y+1)\rangle$ and $\langle y+2\rangle$, so both ideals should be the cube of another ideal, but the class-number of $\mathbb Q(\sqrt{-6})$ is $2$, so that both ideals are the cube of another principal ideal, and as the only units of $\mathbb Z[\sqrt{-6}]$ are $\pm 1$ we have the identity \[x+(y+1)\sqrt{-6}=(r+s\sqrt{-6})^{3}\] where $r$, $s$ are integers.
NOTE: In their book, Uspensky and Heaslet prove this identity using elementary means, no ideal theory nor class-number.
Using the last identity it is trivial to derive the two equations \[y+1=3s(r^{2}-2s^{2})\]  \[y+2=r^{2}+6s^{2}\] yielding $9r^{2}=18s^{2}+24s+8-\frac 1{3s-1}$, from where it follows inmediately that $s=0$, $r=\pm 1$, $y=-1$, $x=\pm 1$ are the only solutions of the diophantine equation.


geometry
power of a point
radical axis
geometry unsolved

Does anyone know how to prove brianchon's theorem, which states that the diagonals of a hexagon which can be circumscribed around a circle are concurrent, using radical axis theorem, which states that the pairwise radical axes of three circles are concurrrent? Thanks.

You can find the solution in page 77 -[b]Problems in plane and solid geometry[/b], Vol.1-[b]Viktor Prasolov[/b].


In a convex polygon, the degree measures of the interior angles form an arithmetic progression.  If the smallest angle is 159 degrees and the largest is 177 degrees, how many sides does the polygon have?

[hide="i hope this is sufficient solution"]
let the number of sides be n.
the average of the angles is $\frac{177+159}{2}=168$.
using the formula for the total angle in a regular n-gon,
$\frac{180(n-2)}{n}=168$ (total angle measure/number of angles)
$n=30$
[/hide]

That can't work.  That formula is only for regular polygons, and this can't be regular if the angle measures are changing.  Also, there is a maximum of 18 numbers in that series, so there can't be more than 18 sides...

[quote="DanK"]Also, there is a maximum of 18 numbers in that series, so there can't be more than 18 sides...[/quote]

And there's a law against angles having a measure in minutes and seconds?

His solution is fine.

We can find the sum of the interior angles by taking the average of the smallest and the largest angle and multiplying that by the number of vertices (=number of sides). This works because we know that the angles form an arithmetic sequence.

So $180(n-2)=168n$, which is essentially the same as his equation.

Actually, I did it exactly like that at first, but then reread the problem and changed my mind.  Stupid overthinking, that got me into trouble on the AIME this year too...

Thanks for clarifying.


Given $ a, b, c > 0$. Prove that: $ \frac{a^2\plus{}b^2\plus{}c^2}{ab\plus{}bc\plus{}ca} \geq\ \frac{ab}{a^2\plus{}bc\plus{}ca}\plus{}\frac{bc}{b^2\plus{}ca\plus{}ab}\plus{}\frac{ca}{c^2\plus{}ab\plus{}bc}$
 :)

It is similar to another inequality proposed by nguoivn on this forum

$ \frac{\sum a^2}{\sum ab} \geq \sum \frac{a^2}{a^2\plus{}ab\plus{}bc}$

and again unsolved :(

I cannot find a neat solution for it. :oops:  Has anyone a nice hint ?

Thank you very much. :)

[quote="nguoivn"]Given $ a, b, c > 0$. Prove that: $ \frac {a^2 \plus{} b^2 \plus{} c^2}{ab \plus{} bc \plus{} ca} \geq\ \frac {ab}{a^2 \plus{} bc \plus{} ca} \plus{} \frac {bc}{b^2 \plus{} ca \plus{} ab} \plus{} \frac {ca}{c^2 \plus{} ab \plus{} bc}$
 :)[/quote]
From the Cauchy Schwarz Inequality, we find that
$ \frac {ab}{a^2 \plus{} bc \plus{} ca} \equal{} \frac {ab(b^2 \plus{} bc \plus{} ca)}{(a^2 \plus{} bc \plus{} ca)(b^2 \plus{} bc \plus{} ca)} \le \frac {ab(b^2 \plus{} bc \plus{} ca)}{(ab \plus{} bc \plus{} ca)^2}.$
Adding this to the two analogous inequalities, we infer that the right hand side of the original inequality does not exceed
\[ \frac {\sum ab(b^2 \plus{} bc \plus{} ca)}{(ab \plus{} bc \plus{} ca)^2}.\]
Therefore, it suffices to prove that
\[ (a^2 \plus{} b^2 \plus{} c^2)(ab \plus{} bc \plus{} ca) \ge \sum ab(b^2 \plus{} bc \plus{} ca).\]
After some simple computations, we find that it is equivalent to
\[ a^3b \plus{} b^3c \plus{} c^3a \ge abc(a \plus{} b \plus{} c).\]
It is easy to see that this is a direct consequence of the Cauchy Schwarz Inequality. Therefore, our proof is completed. Note that the equality holds if and only if $ a \equal{} b \equal{} c.$

Manlio, I think  you get $ \sum_{cyc} a^3 c \ge \sum_{cyc} a^2 b^2$ which is true by Rearrangement inequality.

No, definitely not. Take for example $ a\equal{}0$, $ b\equal{}2$, $ c\equal{}1$.

Yeah sorry, wrote the variables in a wrong way while writing it in that form.

Can Hang's method doesnt work here but an approach similar to his may solve the problem. I am still trying one.

[quote="manlio"]It is similar to another inequality proposed by nguoivn on this forum

$ \frac {\sum a^2}{\sum ab} \geq \sum \frac {a^2}{a^2 \plus{} ab \plus{} bc}$

and again unsolved :(

I cannot find a neat solution for it. :oops:  Has anyone a nice hint ?

Thank you very much. :)[/quote]

for you my friend  :)

write the inequality as :  $ \frac {\sum a^2}{\sum ab}  \plus{} \sum{ \frac{b(a\plus{}c)}{a^2\plus{}b(a\plus{}c)}} \geq  3$ and then apply cauchy shwarz , to make it symmetric I suggest to use Vasc ineq : $ (\sum{a^2})^2\geq 3\sum{a^3b}$ :wink:

[quote="manlio"]Nice proof, can_hang2007.  :)

Thank you very much.

Using your method for

$ \frac {\sum a^2}{\sum ab} \geq \sum \frac {a^2}{a^2 \plus{} ab \plus{} bc}$

we obtain $ \sum ab^3 \geq \sum a^2b^2$ which is not true. :([/quote]
My dear friend, for this inequality, you can use the Cauchy Schwarz Inequality as follows:

Note that $ 2(a^2\plus{}ab\plus{}bc)\equal{}(a^2\plus{}2bc)\plus{}(a^2\plus{}2ab),$ the Cauchy Schwarz Inequality implies that
$ \frac{a^2}{a^2\plus{}ab\plus{}bc} \equal{}\frac{2a^2}{(a^2\plus{}2bc)\plus{}(a^2\plus{}2ab)} \le \frac{a^2}{2(a^2\plus{}2bc)}\plus{}\frac{a^2}{2(a^2\plus{}2ab)}.$
Using this inequality and the two similar ones, we can reduce the inequality to proving that
$ \frac{2(a^2\plus{}b^2\plus{}c^2)}{ab\plus{}bc\plus{}ca} \ge \sum \frac{a^2}{a^2\plus{}2bc}\plus{}\sum \frac{a}{a\plus{}2b}.$
On the other hand, it is easy to see that
$ \frac{a^2\plus{}b^2\plus{}c^2}{ab\plus{}bc\plus{}ca} \ge \sum \frac{a^2}{a^2\plus{}2bc}$
using the Vornicu Schur Inequality. Therefore, it remains to check that
$ \frac{a^2\plus{}b^2\plus{}c^2}{ab\plus{}bc\plus{}ca} \ge \sum \frac{a}{a\plus{}2b},$
which is equivalent to
$ \frac{a^2\plus{}b^2\plus{}c^2}{ab\plus{}bc\plus{}ca}\plus{}2\sum \frac{b}{a\plus{}2b} \ge 3.$
Now, we apply the Cauchy Schwarz Inequality again so that the inequality comes down to
$ \frac{a^2\plus{}b^2\plus{}c^2}{ab\plus{}bc\plus{}ca}\plus{}\frac{2(a\plus{}b\plus{}c)^2}{2(a^2\plus{}b^2\plus{}c^2)\plus{}(ab\plus{}bc\plus{}ca)} \ge 3,$
or
$ \frac{a^2\plus{}b^2\plus{}c^2}{ab\plus{}bc\plus{}ca}\minus{}1 \ge 1\minus{}\frac{3(ab\plus{}bc\plus{}ca)}{2(a^2\plus{}b^2\plus{}c^2)\plus{}(ab\plus{}bc\plus{}ca)}.$
However, this is obviously true since
$ \frac{1}{ab\plus{}bc\plus{}ca} \ge \frac{2}{2(a^2\plus{}b^2\plus{}c^2)\plus{}(ab\plus{}bc\plus{}ca)},$
and
$ a^2\plus{}b^2\plus{}c^2\minus{}ab\minus{}bc\minus{}ca \ge 0.$
Our proof is completed. It is easy to see that the equality holds iff $ a\equal{}b\equal{}c.$

Very nice proofs, can_hang  :lol:
[quote="manlio"]

$ \frac {\sum a^2}{\sum ab} \geq \sum \frac {a^2}{a^2 \plus{} ab \plus{} bc}$

[/quote]
With this ineq, I also had nice solution by only CS and Am-Gm, manlio  :)
My way is use CS and Am-Gm to prove:
$ \frac {a^2 \plus{} b^2 \plus{} c^2}{ab \plus{} bc \plus{} ca} \plus{} \sum\ \frac {ab}{a^2 \plus{} ab \plus{} bc} \geq\ 2$
 :)

[quote="nguoivn"]Given $ a, b, c > 0$. Prove that: $ \frac {a^2 \plus{} b^2 \plus{} c^2}{ab \plus{} bc \plus{} ca} \geq\ \frac {ab}{a^2 \plus{} bc \plus{} ca} \plus{} \frac {bc}{b^2 \plus{} ca \plus{} ab} \plus{} \frac {ca}{c^2 \plus{} ab \plus{} bc}$
 :)[/quote]
From this result, we obtain another nice problem:
$ \frac{1}{a^2\plus{}ab\plus{}bc}\plus{}\frac{1}{b^2\plus{}bc\plus{}ca}\plus{}\frac{1}{c^2\plus{}ca\plus{}ab} \leq\ \frac{(a\plus{}b\plus{}c)^2}{(ab\plus{}bc\plus{}ca)^2}$  :)


How can the schools decide whether to take AIME1 or AIME2?

It's just whatever the school decides, though schools are definitely encouraged to take the AIME I.  My school, however, is on spring break this week so we are doing the AIME II.


AMC USA(J)MO USAMO induction inequalities Harvard college

So did anyone from NC make it to MOP other than Arnav?

I am fairly sure Yasha did not, I know Misha did not; maybe Jeremy did if he scrapes in somehow.  I (Vitek) did not, I have not heard from Berman or Damien.  Vivek is going to RSI, although he seems to have done decently well (probably 10+) on USAMO.  Obviously not enough for MOP.

 - Daniel

It's just Arnav and Damien, unless Pardon or one of the unknowns made it.

EDIT: Wait, Pardon can't make it; he's a senior...

[quote="MysticTerminator"]DUDE HOW CAN YOU BE SURE I THOUGHT THEY HAD FOUND CRIPPLING ERRORS IN YOUR 2 4 AND 6[/quote]
OH MAN, THAT IS TRUE I GUESS, BUT I COULD STILL SCRAPE UP ENOUGH POINTS FOR MOP

and i will probably get 15 and miss blue by the same amount i missed red by last year.  \sad

oh well ZHAI DO YOU SEE THE TITLE NC MOPPERS?!?!  I KNOW THAT THERE WAS SOME CONFUSION ON MY PART EARLIER BUT IT IS ALL SORTED OUT NOW (thanks to jb05/b-flat) AND WE HAVE EMPIRICALLY AND LOGICALLY DETERMINED THAT YOU ARE NOT A NORTH CAROLINIAN.  sorry - it had to be said.

although we do appreciate your posts.  i think i'll give one of them a 6 now.

jb05 - how do you know Damien made red?

Umm... he told me. Plus I have a list of mospers.

Au contraire, as long as brains are machines, there will always be some doubt as to whether Zhai lives in North Carolina.

so i could start a new topic, but what did everybody get?  i got 14, and arnav said elsewhere that he got 25.  (I told you he underestimated - what now, jb05?)

vivek got 12, misha got 14 also.  zhai got 28 (if we consider him a north carolinian).

how did you get a list of MOPpers?

I got 14. Cool! We're all clustered around in the top 100, so that's impressive. Next year, we'll be looking for 4 winners :wink:, one a perfect scorer (zhai).

I guess I'm the outlier with 3.

#1:  I read over my proof and basically, it is the same outline/style as their 3rd solution, but I had a lemma that I thought to be obvious and so I basically said because we have the assumption, what we want is proven.  Most of the proof was based on it, so I doubt I got any points on it.  I probably only needed 2 more lines (see below) to get an actual score on this one.

If $S_{x}=x^{2}$, then $S_{x+1}=x(x+1)$ because $(x+1)(x-1)=x^{2}-1<x^{2}$ and $(x+1)(x+1)=x^{2}+2x+1>x^{2}$.

#4:  I probably got 3 points here, because my proof was correct, but as my dad pointed out, the wording on one part could be interpreted in 2 ways; the easier not giving what I said afterward.  I also didn't format it very well; I had a paragraph that said what I was going to prove, but I didn't say because or anything and the proof was in the next paragraph.

Damien got 12.

I probably could have gotten both correct if I'd written them more clearly.

umm...yeah for some reason damien can't post (probably parental restrictions) but yeah he got a 12 like bryce said.

jb05, how [i]did[/i] you get that list of MOPpers? :?:

bryce - don't worry about it - you're only in 8th grade.  you will get MUCH better by the time you're a sophomore/junior.

EDIT: anybody (other than jeremy) know what jeremy got?  yasha?

oh ha ha now i get your joke, berman.  thanks for pointing that out arnav.  wow - i know a 1994 mopper.

to pander to your use of the official moniker for the camp, how [i]did[/i] you get that list of MOSPpers?

i still don't know where i got the other 7 points in my score.  i know i got #1 and i did a little bit of cool extremal +contradiction on #2 that didn't get me anywhere; #3 i made a silly mistake and forgot turan's theorem; #4 maybe - it seems rigorous but w/e; #5 i just got the induction but no beast factorization, and #6 i just did equality case and then used triangle inequality but stupidly flipped a sign.

so that's like 7/1/1/1/1/1 which leaves me 2 points short or 7/0/0/7/0/0 which gets me to 14 but which i think is wrong since my #4 sucked (imo).

[quote="jb05"]Jeremy got 15, if you really must know. I don't know about Yasha.

@b-flat: it is vastly more likely that you got 1/0/0/2/0/0 than 0/0/0/3/0/0, but of course I wouldn't know because I haven't seen your solutions. What you said about #1 would not cause any point deduction - it is clear that that case works and the problem says that the sequence is unique. I suspect there is a different flaw in the proof. Perhaps if you post it, we could determine what that is. Also, are you sure #4 was completely rigorous? I only ask because SO many people were wrong about that being rigorous. On the other hand, a lot of the people with bad proofs seem to have gotten lots of points on it. You're lucky though - in middle school, the USAMO doesn't matter at all.

@teki-teki: Here's the most recent list of MOPpers, but you could have easily found this online.

[u][size=150]MOP 1994[/size][/u]
Jeremy Bem, Christopher Chang, Li-Chung Chen, Jay Chyung, Nathan Curtis, Samit Dasgupta, Craig DeAlmeida, Andrew Dittmer, Todd Geldon, Andrei Gnepp, Tao He, Anders Hendrickson, Theodore Hwa, Daniel Johnston, Aleksandr Khazanov, Jacob Lurie, Davesh Maulik, Johanna Miller, Josh Nichols-Barrer, Jacob Rasmussen, Noam Shazeer, Frank Thorne, Aaron Ucko, Stephen Wang, Jonathan Weinstein, Donald Ying[/quote]

it's really bizarre that this broken list keeps circulating. i can assure you that andrew dittmer was not there in 94 (too late) and neither was donald ying (too early)

donald ying really should be some sort of unofficial mop mascot by the way. i seem to remember him giving up on a mop test early and going to sleep at his desk, then waking up with a solution to one of the problems he hadn't gotten and writing it out

[quote="MysticTerminator"][quote="Johnclyde"][quote="MysticTerminator"]to the contrary, the summer program in years after 1994 still continues to be known as "MOP" except by (1) official people, (2) stupid people, and (3) official stupid people[/quote]

arguably once they started taking more than 24 people it became a mosp. and if not, 300 or so people in whatever year that was HAD to be a mosp[/quote]

???

that ... did not make sense[/quote]

one can argue that "mop" refers to the program that was in place up through 1994 that only allowed 24 of the best students to participate. in 1995 there were still 24 students and people resisted the name change, but every year since then the summer program has let in 32 students or more (including one year that there were hundreds of students), thus diluting the "MOP" brand and creating something that, perhaps, should be called a mosp.

did that make sense

[quote="jb05"]no...[/quote]

sigh, let's try this.

MOP: [n] a summer program for math students that, in addition to clearing other hurdles, also succeeded in solving multiple problems on the USAMO. Was abolished after 1994, although many people try to pretend that it still exists.

MOSP: [n] a slightly weaker form of MOP that often lets in students who scored fewer than 10 points on the usamo, in one case even inviting students who scored zero (0) points on the USAMO.

i'm being tongue-in-cheek here, but it frustrates me that this concept is difficult to grasp ... personally I think it's fine if people want to keep calling it MOP. It's just that it isn't as hard to get into as it was when the official name was MOP. Like if Harvard changed its name to "Slightly dumbed down harvard" next year and accepted 100,000 students the year after that, but ten years later people kept insisting "it's called Harvard, not SDDH", you might imagine that some people would say "no, actually this new school IS sddh .. I don't think you can still call it Harvard", and I don't think that concept would be so difficult to grasp.

[quote="MysticTerminator"]oh you're talking about the hundreds of people going to MOP

well that's easy

we just have a real MOP (blue + black) and fake MOP (red + yellow + cgmo + whatnot)[/quote]

yeah yeah

well this has gotten blown out of proportion but i'll just finish with my (long-winded) anecdote.

in 1996 in addition to letting 32 people attend instead of 24, they also instituted the "three year rule" (does this still exist?) which affected exactly two underclassmen by telling them they couldn't come to MOP because they weren't in the top 8 and they had already gone twice. additionally, eligible members of the previous year's team opted to skip the IMO that year for various reasons and a disproportionate number of the previous year's underclassmen went straight to college (as I write this I'm blown away to see how many factors went into this).

I actually got curious just now, so I ran the numbers -- the result of what happened was that you went from the previous year having 12 veterans and 12 rookies, to now having a MOP with 8 veterans and 11 combined years of previous mop experience (this 11 figure must be an all-time low to the extent that relevant comparisons can be made), who were surrounded by 24 rookies. Meanwhile, at RSI you had 4 students who combined for 8 years of mop experience..

Hence, the first question I was asked when I arrived at RSI after leaving MOP a couple days early: "[b]So John, was it a MOSP?[/b]" followed by snickers

(incidentally the result at IMO that year was that China bombed and finished like 10th (ok maybe 6th), and the US team finished 2 points out of 1st place. also probably warrants mentioning how crazy the team selection was that year: they were in the 3rd year of experimenting with inviting the top 6 USAMO winners straight to IMO, and it was the first time they tried emulating the IMO with a 6 question test out of 42 instead of 5 questions out of 100, and what they got was a 10-way tie for 3rd place at 28 points. literally 5 students were able to say that they tied for 3rd on the final round of selection, yet were not invited to IMO (although none of those 5 were relevant to what i was saying up above) )


geometry
rectangle
pigeonhole principle
combinatorics unsolved
combinatorics

Every point of the plane is colored red or blue. Show that there exits a rectangle with vertices of the same color.

I think there is a solution in Arthur Engel... Extremal Principe (chapter 1), if I ain't wrong...  :maybe:  Do u have one ?

This is true in more general case: Every point in the plane is
colored with one of $n$ colors, then there is a rectangle with
monochromatic vertexes.

We will observe only points with integer coordinates. Let $L_{i}$ be
the array of points $(i,0),(i,1),....,(i,n)$. Such array consist
of $n+1$ points and every point is colored with one of $n$ colors,
so there are certainly two of them with the same color.

There are $n^{n+1}$ ways to color $n+1$ points with $n$ colors. If
we look at arrays $L_{0},L_{1},L_{2},...,L_{n^{n+1}}$ (there are
$n^{n+1}+1$ of them) then, between them, there are two arrays
$L_{x}$ and $L_{y}$, such that color of $(x,i)$ and $(y,i)$ are the
same for every $i=0,1,...,n$. As we conclude earlier, in every
array $L_{i}$ there are two points with the same color, so lets say
that, in $L_{x}$ those two points are $(x,a)$ and $(x,b)$. Then
$(y,a)$ and $(y,b)$ have the same point as $(x,a)$ and $(x,b)$, so
those four points are vertexes of wanted rectangle.

Bye

The initial problem is easy with Pigeonhole Principle.
   But what about the "SQUARE version"?
I am meaning:
SHow that there exist a monochromatic square!
  Follow my mind it was discused here ago,posted by Lhvietbao,I think

[quote="HUY VAN"]There are $n^{n+1}$ ways to color $n+1$ points with $n$ colors.
Why?[/quote]

There are $n$ ways to color one point (because we have $n$ colors). If we want to color two points both of them we can color in $n$ diferent ways, so there are $n\cdot n=n^{2}$ ways to color two points (f.i. if there are two colors, $1$ and $2$, possible colorings are $(1,1),(1,2),(2,1),(2,2)$). So, if we have $n+1$ points, we can color first of them in $n$ ways, second also in $n$ ways,....,$n+1$-st point in $n$ ways, so there are $\underbrace{n\cdot n\cdot ... \cdot n}_{n+1}=n^{n+1}$ to color $n+1$ points with $n$ colors.

I hope that you will understand this, if not, just ask, and I will try to explane it more clearly.


Duke
college
AMC
AIME
FTW
summer program
Mathcamp

Guys I would totally recommend you go to Duke TIP (East Campus is 133457). It was fun, I finally learned what in the world ultimate frisbee is, I learned stuff[size=9]They don't have AC[/size]related to math (game theory), and yeh. So yeh, go there.

BLEH DUDES I TOTALLY WOULDN'T

HOW MUCH DOES TIP COST ANYWAY

TOTALLY NOT WORTH IT

GO TO MOP

DUDE IM GOING TOMORROW FOR DUKE TIP. WEST CAMPUS PWNS EAST

ANY YALL DURHAM KIDS WANNA STALK ME IM RIGHT THERE FOR A WHOLE 3 WEEKS

GOOD LUCK THOUGH IM VERY SCHNEAKY

DANGIT WHY ARE Y'ALL WASTING YOUR TIME AT TIP

:-/

DAGNABBIT

I DID NOT DO ANY OF THESE TIME-WASTERS BACK IN THE DAY, NO SIR I DID NOT

hmm

my 2 cents

DYWC ftw

duke tip camps are pretty silly

do not go to duke TIP.
i applied, and then after i got accepted, withdrew (with good reason).  :P
go to something besides duke TIP.

well i was a [hide]bevan scholar[/hide] which means [hide]no tuition[/hide] so I thought[hide]might as well use it[/hide]

OK FINE THAT IS NOT *COMPLETELY* USELESS THEN

I AM STILL NOT FOND OF DUKE TIP IN GENERAL THOUGH

There were no camps that I was interested in for my age.

[quote="dnkywin"]im at tip now, same reason as Pi dimension[/quote]

davidson? what course.

[quote="#H34N1"]GUYS WHY DID YOU CREATE LIKE 300 ACCOUNTS[/quote] lol thats the problem solving class XD

calvin go tell them to delete their accounts

OF COURSE YOU ARE AT CAMP! :alien:  :gathering:  :ddr:

THE MATH CLASS WAS BULL THE CLASSES WERE BULL.

The food was ok. I just lived on Lucky Charms and sandwiches.

I SHOULDA GONE TO MATHCAMP OR AWESOMEMATH OR ATTEMPTED PROMYS!

go to ross. duke tip is completely useless unless youre learning language arts because i bet most of us suck at that.

I just came back from a course called Biology of Cancer at Duke West. It really teaches you a lot of biochemistry and cell signaling. You go to Duke TIP not to "train" for math contests, etc. or to learn more of what you know, but instead to learn something new. Obviously you should not go to a math class then, but for other subjects that you want to learn about, I really recommend it.

i just got back from davidson...

@ PI Dimension it was International relations

i have infiltrated your forum...i took engineering

[quote="Teki-Teki"]Dude guys just friggin' go to RSI, okay?  Be done with it.  The only thing that RSI is slightly bad at is Frisbee, which most of you are too small to play anyway.[/quote]

Or go to MOP... :wink:

(no hidden meanings... lol... making RSI is hard, at least as hard as MOP)

Dude I suck so bad at frisbee

like its not even funny

i once clocked a guy in the head with a frisbee at ARML by accident

he was on my team too

so ya

i sucked

What's the point of frisbee anyways.

I mean, all you do is throw a freaking plastic plate around.

Not worth my time.  :rotfl:

LOL...a vid of us at tip..[youtube]xjTY70BzNQs[/youtube]

Hmm guys watch out for Duke tip. Trivial classes, very very strict rules, and gross food. They just try to trick people into thinking its decent. Find something useful to do.

LOL i see dnkywin :D

hm which one are you Aime? (don't say the azn one)

the person who goes in the closet right after dnkywin :|

In front of dnkywin is asiangenius314 and behind me is crazyasianvampire


rotation
geometry
geometric transformation
articles
LaTeX

hi all
i ve joined this forum recently.
i got a problem in editing tables in LaTeX.
i wanna tilt a table/figure, 90 degrees so that it can fit into the page.
can u help me on this? is there any command for dat?

regards
gmc

to rotate a figure, just do:

\includegraphics[angle=90]{./filename.png}

the rotation is counterclockwise

for tables, I think you need to put the page in landscape. I didn't try it but it is suppose to work with:

\usepackage{lscape}
....
\begin{document}
...
\begin{landscape}
...
\end{landscape}
...
\end{document}

if you want to whole document in landscape just add the option landscape in the documentclass

\documentclass[landscape]{article}

ledurt

hi ledurt

i need to tilt a table 90 degrees so that it can fit into the width of the page.
the solution u gave did not work.
can anybody help me out from this?\

regards
gmc

ok

try this:

\rotatebox{90}{
\begin{tabular}...
...
\end{tabular}
}

I didn't try to add the caption in the box but it problely works but i'm not sure you should rotate the caption too.

ledurt

hi

It also didn't work :(  .
It is always going out of the margins.
When I put \rotatebox{90}, the table is going to the right bottom corner (not tilting) and \rotatebox{270}, to the right top corner.

I am not getting the table in the center of the page even though I am using "\centering".

regards
gmc

ok, well that's weird because I tried it and it worked...
is your table really large? I've tried it with large tables and I have like 300 errors...

try the same thing but with

\usepackage{graphicx}

ledurt


[quote="guard2"]Consider the sequence $(a_{n})_{n=1}^{\infty}: a_{n}>0\forall n\in N^{*}~;\varlimsup_{n\rightarrow \infty}a_{n}^{\frac{1}{n}}=1$ and $\varliminf_{n\rightarrow \infty}a_{n}^{\frac{1}{n}}<1$.Prove that there is a subsequence $(a_{n_{k}})_{k=1}^{\infty}$ of $(a_{n})_{n=1}^{\infty}$ such that :
$\lim_{k\rightarrow \infty}a_{n_{k}}^{\frac{1}{n_{k}}}=1$ and$\lim_{k\rightarrow \infty}|a_{n_{k}}^{2}-a_{n_{k}+1}.a_{n_{k}-1}|^{\frac{1}{n_{k}}}=1$[/quote]
nobody?why? :(

[quote="guard2"]Consider the sequence $(a_{n})_{n=1}^{\infty}: a_{n}>0\forall n\in N^{*}~;\varlimsup_{n\rightarrow \infty}a_{n}^{\frac{1}{n}}=1$ and $\varliminf_{n\rightarrow \infty}a_{n}^{\frac{1}{n}}<1$.Prove that there is a subsequence $(a_{n_{k}})_{k=1}^{\infty}$ of $(a_{n})_{n=1}^{\infty}$ such that :
$\lim_{k\rightarrow \infty}a_{n_{k}}^{\frac{1}{n_{k}}}=1$ and$\lim_{k\rightarrow \infty}|a_{n_{k}}^{2}-a_{n_{k}+1}.a_{n_{k}-1}|^{\frac{1}{n_{k}}}=1$[/quote]
I don't know what is the difference between $\varlimsup_{n\rightarrow \infty}a_{n}^{\frac{1}{n}}$ and $\varliminf_{n\rightarrow \infty}a_{n}^{\frac{1}{n}}$.?


[quote="fjwxcsl"]Let $x,y,z\ge 0$,$xy+yz+zx=1$,$a,b\ge 0$, prove that
$a(x+y+z)+b(\frac{1}{x}+\frac{1}{y}+\frac{1}{z})\ge 3(a+b)$.[/quote]

No, $xy+yz+zx=1$. I guess that he had some mistake in type. But if $xy+yz+zx=3$, it's too easy

If $xy+yz+zx=1$, the correct inequality is
\[a(x+y+z)+b\left( \frac{1}{x}+\frac{1}{y}+\frac{1}{z}\right) \geq a\sqrt{3}+3b\sqrt{3}\]
which is also "too easy". The one he posted is just wrong when $a=1,b=0,x=y=z=\frac{1}{\sqrt{3}}$.


Let $ABC$ be an equilateral triangle, and $D,E,F$ be points on $BC,CA,AB$ respectively.Let $\angle BAD=\alpha$,$\angle CBE=\beta$, $\angle ACF=\gamma$. Suppose $\alpha+\beta+\gamma \geq 120$.
Prove that the union $\Sigma$ of triangular regions $BAD,CBE,ACF$ covers the triangle $ABC$.

smbdy pleees help. its kinda urgent. i sort of need to send thee solutions.
sorry.
thanx.


Find all functional equations satisfying $ f(xy)\equal{}f(x)f(y)$ for all real $ x$ and $ y$

[quote="enndb0x"]Put $ y\equal{}0 \implies f(x)\equal{}1$[/quote]

But that is only if $ f(0)$ is not equal to zero.

This reduces, with one caveat, to the [url=http://en.wikipedia.org/wiki/Cauchy_functional_equation]Cauchy functional equation[/url] by setting $ \ln f(x) \equal{} g(\ln x)$.


Find b>0 such that
$ a_{1}\equal{}1$
$ a_{n\plus{}1}^{b}\equal{}a_{1}\plus{}...\plus{}a_{n}$
satisfy the sequence $ \frac{a_{n}}{n}$ converge to a point different from 0

Obviosly $ b \equal{} 2$ and $ \\ lim_{n\to\infty}\frac{a_{n}}{n}\equal{}\frac{1}{2}.$

$ a_{n\plus{}1}^{b}\equal{}a_{n}^{b}\plus{}a_{n}$
Use the same method as p8


Find the number of times between 1 o'clock and 4 o'clock when the hour hand is perpendicular to the minute hand.

At $ 1$ o'clock, the hour hand is at $ 30^\circ$, and the minute hand is at $ 0^\circ$. Every minute, the minute hand moves $ 6^\circ \minus{} 0.5^\circ \equal{} 5.5^\circ$ relative to the hour hand. The first meeting will be after the minute hand moves $ 120^\circ$ relative to the hour hand, which occurs at $ \frac {120^\circ}{5.5^\circ} \equal{} \frac{900}{11}\implies1: \frac{900}{11}$ o'clock. After this, every $ 180^\circ$ the minute hand moves relative to the hour hand, the hour and minute hands will be perpendicular. Since it takes $ \frac {180^\circ}{5.5^\circ} \equal{} \frac {360}{11}$ minutes per $ 180^\circ$, and there are $ 4\cdot60 \minus{} \frac {900}{11} \equal{} \frac {1740}{11}$ minutes left, there are $ 1 \plus{} \left\lfloor\frac {\frac {1740}{11}}{\frac {360}{11}}\right\rfloor \equal{} \boxed5$ times total.

The idea in this solution is great (I probably would not have come up with something this simple), but I think there are some errors.

$ \frac {120^\circ}{5.5^\circ} \equal{} \frac{900}{11}$

I think this should be

$ \frac {120^\circ}{5.5^\circ} \equal{} \frac{240}{11}$

Also, at the end, you said $ 4\cdot60 \minus{} \frac {900}{11} \equal{} \frac {1740}{11}$

Same thing as before, but also between 1:00 and 4:00 is only 3 hours so I think that should be:

$ 3\cdot60 \minus{} \frac {240}{11} \equal{} \frac {1740}{11}$

Oddly, it still comes out to the same thing, so the answer is fine either way.

Yes, thanks for the corrections! :oops:  :)


Hello,

we are preparing a meeting of young mathematican in Rostock Germany.
We want to invite some IMO participants and other interested in mathematican.

Did anybody know an organization where kids who are interested in math were supported?

Have a look here :
http://www.animath.fr

Pierre.


geometry
circumcircle
trigonometry
number theory
relatively prime
trig identities
Law of Cosines

Triangle $ABC$ is a right triangle with $AC=7,$ $BC=24,$ and right angle at $C.$ Point $M$ is the midpoint of $AB,$ and $D$ is on the same side of line $AB$ as $C$ so that $AD=BD=15.$ Given that the area of triangle $CDM$ may be expressed as $\frac{m\sqrt{n}}{p},$ where $m,$ $n,$ and $p$ are positive integers, $m$ and $p$ are relatively prime, and $n$ is not divisible by the square of any prime, find $m+n+p.$

$AM=MC=MB$ because M is circumcircle of ABC.

let $\angle AMC = \alpha$

$49=2\cdot (12.5)^{2}-2\cdot (12.5)^{2}\cdot \cos \alpha$ by law of cosines in triangle AMC

$\cos \alpha = \frac{263.5}{312.5}= \frac{527}{625}$

since $\angle CMD = 90-\alpha$

$\sin \angle CMD = \frac{527}{625}$

hence $\text{area of CMD}= \frac{1}{2}\cdot CM \cdot MD \cdot \frac{527}{625}$

using pythag in $\triangle AMD$, $MD=\frac{5\sqrt{11}}{2}$

hence the desired area is $\frac{1}{2}\cdot (12.5) \cdot (\frac{5\sqrt{11}}{2}) \cdot \frac{527}{625}$

$=\frac{527sqrt{11}}{8}$

$\implies m+n+p=546$

I did the same thing except I set $\ \angle BAC=\theta$ making $\ \angle CMD =2\theta-90$  So $\ \sin \angle CMD =-\cos 2\theta$.  Since we know both $\ \sin \theta$ and $\ \cos\theta$ we can calculate that with the double angle formula.  However I believe you messed up in the very last step and the denominator should be $\ 40$.  Making the final answer $\ 11+40+527 =\boxed{578}$

Drawing out our diagram, we have $MD = 5\sqrt{11}/2$ by the Pythagorean Theorem applied to $\Delta MBD$ (DM and BM are perpendicular because BMD and AMD are congruent triangles). We know that $CM = 25/2$ since $M$ acts as the circumcentre to $\Delta ABC.$ If we let $\angle CMD = \theta$, from $\Delta ACM$, we observe that $\sin\theta = 527/625$ by the Law of Cosines. Thus, $[CMD] = 5\sqrt{11}/2 \cdot 25/2 \cdot 527/625 = 527\sqrt{11}/40 \Rightarrow m+n+p=578.$

[quote="SM4RT"]u also get
$\frac{527\sqrt{11}}{40}$ so
$m+n+p=527+11+40=\boxed{578}$..how do u get $546$ is your fraction incorrect???[/quote]

WOW...  :rotfl:

I so can't add. Sorry about that  :D Edited.

[hide="Really easy for an 11 :wink:"]
Use pyth. thm. to find DM; it is just like $\sqrt{15^{2}-12.5^{2}}=\frac{5\sqrt{11}}{2}$

Then since circumcircle is midpoint of AB, we have $CM=12.5$.

It remains to find $\sin a$. Consider that: $90-a=180-2\cos^{-1}\frac{7}{25}$. The rest is standard and very simple, no messy calculations, as the sin becomes a cosine through the 90 and we end up:$\sin a = \cos 2\cos^{-1}\frac{7}{25}= 2(\frac{7}{25}^{2}-1$
[/hide]
oh an of course, area of triangle is half sin a bc

[hide="Trig is bad"]

Extend $MD$ to $BC$ and call where they intersect $X$. By area lemma (or common sense) we have that $CX:XB = [CXM]:[XMB] = [CDX]:[XDB] = [CXM] - [CDX]:[XMB] - [XDB] = [CDM]:[MDB]$

Next notice that $\triangle{MXB} \simeq \triangle{ABC}$. So we have that $XD:25 = MD:24 = 12.5:24 \rightarrow XB = \frac{312.5}{24}$

This means that $CX = BC - \frac{312.5}{24} = 24-\frac{312.5}{24} = \frac{263.5}{24}$

Thus $[CDM]:[DBM] = CX:XB = 263.5:312.5 = 527:625$.

Next observe that $\triangle{ADB}$ is isosceles. Thus, $DM$ is a height of $\triangle{ADB}$ since $M$ is the midpoint and $D$ is the vertex opposite of the base. Thus we have that $DM = \sqrt{15^2 - 12.5^2} = \sqrt{68.75}$

Thus, the area of $\triangle{MDB} = \frac{12.5\sqrt{68.75}}{2}$

Thus the area of

$\triangle{CDM} = \frac{527}{625} * \frac{12.5\sqrt{68.75}}{2} = \frac{527}{50} * \frac{\sqrt{68.75}}{2} = \frac{527\sqrt{68.75}}{100}  = \frac{527\sqrt{275}}{200}= \frac{527*5\sqrt{11}}{200} = \boxed{\frac{527\sqrt{11}}{40}}$

Thus the final answer is $527+11+40 = \boxed{578}$

[/hide]

[hide=Pretty Elegant when you drop some altitudes]Let $E$ be the projection of $C$ onto $MD$, and $F$ be the foot of the altitude from $C$ to $AB$. Then $FMEC$ is a rectangle, so $CE=FM$. You can find $MD=\frac{5}{2}\sqrt{11}$ by Pythagorean and $CE=FM=AM-AF=\frac{25}{2}-\frac{49}{25}=\frac{527}{50}$ and you have $[CMD]=\frac{MD\cdot CE}{2}$, giving $\frac{527}{40}\sqrt{11}\longrightarrow \boxed{578}$.[/hide]

[b]Solution:[/b] It is quite easy to calculate the value of $DM$ with the Pythagorean theorem. From here, we observe that the height of $\triangle{CDM}$ must be a "nice" number due to the answer extraction. Hence, we try to draw a segment from $C$ parallel to $\overline{DM}$ until it intersects the triangle again at $\overline{AB}$. Letting that $X$ be the foot of that altitude, we can assign $MX=x$ and $XA=12.5-x$. If we also let $CX=y$, noticing that $CM = 12.5$, we have a nice system of equations that we can easily solve for and get the height $MX=x$ of the triangle.


If $a,b,c,d > 0$, is it true that
$a^4+b^4+c^4+d^4+12abcd \ge (a+b+c+d)(abc+bcd+cda+dab)$?

Yes, it is.

I will use theorem from here:

http://www.mathlinks.ro/Forum/viewtopic.php?t=20851

Denote $f(a, b, c, d) = \sum a^4 + 8abcd - \sum a^2bc$
We see that $f(a, b, c, 0) = \sum a^3 - \sum a^2bc$ which is non-negative by AM-GM/Muirhead or anything similar.

Now, $\sum \frac{\partial f}{\partial a} = 4\sum a^3 + 2\sum abc - 2\sum a^2b$

So, we have to prove that:
$2\sum a^3 + \sum abc \geq \sum a^2b$

To prove this one we will prove the following 3-variable inequality:
$\frac{2}{3} \sum a^3 + abc \geq \frac{1}{2} \sum a^2b$

After proving this we can write similar 3 inequalities and just add them up to obtain the result.

For the proof of this inequality rewrite it as:
$4\sum a^3 + 6abc \geq 3 \sum a^2b$

Now, use the easy inequality:
$2\sum a^3 \geq \sum a^2b$
and
$2\sum a^3b + 6abc \geq 2 \sum a^2b$
which is just Schur.
After adding them up we get desired inequality, QED.

Does the general inequality hold?
i.e. $a_1^n + \cdots +a_n^n + n(n-1)a_1 \cdots a_n \ge \left( \sum a_1 \right)\left(\sum a_1 \cdots a_{n-1}\right)$
If it is true, we can prove http://www.mathlinks.ro/Forum/viewtopic.php?t=69084

Edit: Yes By Tomicio's method we can prove this inequality. Thanks. :D

[quote="Vasc"][quote="Soarer"]Does the general inequality hold?
i.e. $a_1^n + \cdots +a_n^n + n(n-1)a_1 \cdots a_n \ge \left( \sum a_1 \right)\left(\sum a_1 \cdots a_{n-1}\right)$[/quote]
I posted this inequality in this Forum in 2004 and Harazi given a nice solution.
See
http://www.mathlinks.ro/Forum/viewtopic.php?p=105910#p105910[/quote]

Thanks for the beautiful solution!  :ninja: Mine's too ugly...


quadratics
algebra
polynomial
number theory unsolved
number theory

Find all integers $ x,y$ so that $ x^3 \plus{} 10x \minus{} 1 \equal{} y^3 \plus{} 6y^2$.
Thanks in advanced.

I think it 's not hard.We denote z=y+2
and wil have solution

using substitution $ x \equal{} y \plus{} a$ where $ a$ is some integer will solve this type of equation
by doing so we get the equation:
$ y^2(3a \minus{} 6) \plus{} y(3a^2 \plus{} 10) \plus{} (a^3 \plus{} 10a \minus{} 1)$ which is a quadratic equation in $ y$. if we want $ y$ to be an integer first off all this expression's discriminant must be positive. therefore:
$ (3a^2 \plus{} 10)^2 \minus{} 4(3a \minus{} 6)(a^3 \plus{} 10a \minus{} 1)$ which is a degree four polynomial in $ a$ with leading coeficient $ \minus{} 3$ and is therefore positive for only finitely many $ a$
from here on it is just calculations to get all the solutions


The number of values of $x$ satisfying the equation \[ \frac{2x^2-10x}{x^2-5x}=x-3 \]is:

$\text{(A)} \ \text{zero} \qquad \text{(B)} \ \text{one} \qquad \text{(C)} \ \text{two} \qquad \text{(D)} \ \text{three} \qquad \text{(E)} \ \text{an integer greater than 3}$

[hide]Simplify the fraction and get $2 = x - 3$. So there's only one solution.[/hide]

[hide="A twist"]For $x=5$ the fraction isn't defined. Hence there's no solution.[/hide]


世界杯，大家支持哪个队？
我==〉阿根廷

你多好呀！可以天天看。我们只能今天晚上看个开幕式了。。。。。。

我支持偉大的祖國,即使不能進入決賽

开幕式完了， 感觉怎么样？

Frings的远射太牛屄了！看着真过瘾。

By the way，金龙你现在是在北京还是在东北？

[quote="jin"]开幕式没怎么看 不过看了两场球 厄瓜多尔不错 在A组有可能出线
在死亡之组谁出线都不是冷门[/quote]

居然嬴了波蘭，太令人意外了

我周三晚上火车去北京 周四报到

[quote="dingdongdog"]来猜一下小组出线队伍吧。
个人意见：
A组：德国   哥斯达黎加
B组：英格兰 巴拉圭
C组：阿根廷 荷兰
D组：葡萄牙 伊朗
E组：捷克 意大利
F组：巴西 日本
G组：法国 瑞士
H组：西班牙 乌克兰[/quote]

Group D should be Portugal and Mexico, consider they rank top 10 in the world. Iran got killed mexico today 3-1

哥斯达黎加 已经出局了。

to Jin: 你来了北京还能看世界杯么？

[quote="dingdongdog"]来猜一下小组出线队伍吧。
个人意见：
A组：德国   哥斯达黎加
B组：英格兰 巴拉圭
C组：阿根廷 荷兰
D组：葡萄牙 伊朗
E组：捷克 意大利
F组：巴西 日本
G组：法国 瑞士
H组：西班牙 乌克兰[/quote]
我愈发地觉得我的预测水平能赶上贝利了... :blush:

死亡之组名不符实，塞黑和科特迪瓦一点还击之力都没了。

黄建翔怎么了？意大利赢球至于那么激动么？居然还高呼意大利万岁！！难道他忘了他还是一个解说员么？！

[quote="Soarer"]巴西輸了...[/quote]
巴西昨天踢得太臭了，要是正常发挥，绝不至于连一脚像样的射门都没有。。。

我预测：　冠军德国。

[quote="ifai"]施丹, 唉............ :([/quote]
他三届世界杯，一次小组未出线，剩下的两次都因为恶意犯规被直接红牌罚下。居然还能得金球奖！？

不是世界"盃"嗎? :D

[quote="VIPeter"]不是世界"盃"嗎? :D[/quote]
 :rotfl:
盃 = 杯


How many ways are there to make change for 67 cents using pennies, nickels, dimes, and quarters?

Whats the simplest way to do it?  Without listing?

Well theres some random formula that you can use which you probably wont remember a couple of days from now, so listing is the best way to go (imo). But remember, dont do random listing. Make it systematic.

IMO, unfortunately, the best way is to list:

2 quarters, 1 dime, 1 nickel, 2 pennies
2, 0, 3, 2
2, 0, 2, 7
2, 0, 1, 12
2, 0, 0, 17
1, 4, 0, 2
etc. Slowly convert the big coins into the little ones.

[quote="darkdieuguerre"]If you wanted the answer...

[hide="Answer"]$ 87$ combinations.[/hide][/quote]

So basically all you can do is list?  You went and listed all 87?   :o

[quote="Ihatepie"]You don't really have to list them. For example, once you set the quarters and dimes, if you find a combo with x nickels and y pennies (with y less than 5), there are x+1 combinations with that set number of quarters and dimes.[/quote]

Yeah, thats sort of what I was getting at.

Start from the highest number of the highest type of coin you have.
ie the most quarters you can have is 2. From there, you can have at most 2 dimes, and from there at most 1 nickle. The pennies are set once we choose the others, so lets not worry about that.

From the two quarters, 1 dime thing, we can have 1 nickle or no nickles, so that gives us 2 ways
Now what if we have two quarters and no dimes? The most nickles we can have is 2, and the least we can have is 0, so thats a total of 3 ways.
1 quarter, the most dimes we can have is 4, which gives 1 way
1 quarter, 3 dimes, 2 nickles......1 quarter, 3 dimes, 0 nickles=3 ways
1 quarter, 2 dimes, 4 nickles.......1 quarter, 2 dimes, 0 nickles=5 ways
.............
And we can keep doing that.

From [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=292950]here[/url]:
[quote="Mewto55555"]Mewto's infamous coin formula:

The number of ways to make n cents (n is an integer, 11-49) out of pennies, nickels, dimes, and quarters is $ 3 \plus{} \frac {(\lfloor\frac {n}{5}\rfloor)(\lfloor\frac {n}{5}\rfloor \minus{} 1)}{2}$[/quote]
I'm not sure about 67, though. I agree with Thunder.

[quote="AdithyaGanesh"]So basically all you can do is list?  You went and listed all 87?   :o[/quote]

Nah. I wrote an algorithm to do it for me.  :D

Is this algorithm you speak of simple to derive?  Rather than listing, it would be faster to derive it wouldn't it?

Sort of; the algorithm falls in the category of dynamic programming. It's a bit difficult to justify and explain. However, it's very troublesome to do on paper, which is very listing would be more faster.


Given $ a, b, c$ satisfy $ a\plus{}b\plus{}c\equal{}3$. Prove that:
$ (a\plus{}bc^3)(b\plus{}ca^3)(c\plus{}ab^3) \leq\ 16$

This is a nice inequaliy. Equality occurs (a,b,c)=(0,1,2)
But I haven't find the solution for it. I tried to use pqr, but it isn't effect.
So I need help  :wink:

I think it is have a connection with the following result:
\[ \sum a^2b\le 4\minus{}abc,\forall a,b,c\ge 0,a\plus{}b\plus{}c\equal{}3\]

Another problem have same form with this one. Maybe it is easier.
Let $ a,b,c\ge 0$ and $ a\plus{}b\plus{}c\equal{}3$. For each $ k\ge 0$ find the maximum of:
\[ H\equal{}(a^2b\plus{}k)(b^2c\plus{}k)(c^2a\plus{}k)\]

It was created by hungkhtn  :wink:

My hint is the following \[ 8\sum a^4(b^2\plus{}c^2) \le (\sum a)^3 (\sum a^2(b\plus{}c) \minus{}4abc)\] :)

I think:
If $ a \geq\ b \geq\ c$. Use AM-GM:
$ (a \plus{} bc^3)(b \plus{} ca^3)(c \plus{} ab^3) \leq\ (\frac {a \plus{} b \plus{} c \plus{} ab^3 \plus{} bc^3 \plus{} ca^3}{3})^3 \leq\ (\frac {a \plus{} b \plus{} c \plus{} \frac {\sum\ ab(a^2 \plus{} b^2)}{2} }{3})^3 < 16$
We only need to prove in cases: $ a \leq\ b \leq\ c$.
Use my old result:$ (ab \plus{} bc \plus{} ca)(ab^3 \plus{} bc^3 \plus{} ca^3) \leq\ 16$with $ a \plus{} b \plus{} c \equal{} 3$.
We will prove: $ (a \plus{} bc^3)(b \plus{} ca^3)(c \plus{} ab^3) \leq\ (ab \plus{} bc \plus{} ca)(ab^3 \plus{} bc^3 \plus{} ca^3)$
$ \Leftrightarrow 1 \plus{} a^3b^3c^3 \plus{} \sum\ a^4b^2 \leq\ \sum\ ab^2 \plus{} a^3 \plus{} b^3 \plus{} c^3$
Is it true with  $ a \leq\ b \leq\ c$ ?  :maybe:

[quote="nguoivn"]Use my old result:$ (ab \plus{} bc \plus{} ca)(ab^3 \plus{} bc^3 \plus{} ca^3) \leq\ 16$with $ a \plus{} b \plus{} c \equal{} 3$.
We will prove: $ (a \plus{} bc^3)(b \plus{} ca^3)(c \plus{} ab^3) \leq\ (ab \plus{} bc \plus{} ca)(ab^3 \plus{} bc^3 \plus{} ca^3)$
[/quote]
This inequality is not true, try $ a\equal{}\frac{1}{3},b\equal{}\frac{5}{3},c\equal{}1$.
By the way, the following is stronger for your old inequality
\[ (ab\plus{}bc\plus{}ca)(ab^3\plus{}bc^3\plus{}ca^3\plus{}abc) \le16\]
:)

[quote="can_hang2007"]
This inequality is not true, try $ a \equal{} \frac {1}{3},b \equal{} \frac {5}{3},c \equal{} 1$.
By the way, the following is stronger for your old inequality
\[ (ab \plus{} bc \plus{} ca)(ab^3 \plus{} bc^3 \plus{} ca^3 \plus{} abc) \le16
\]
:)[/quote]
No, I asked that: "Is it true with $ a \leq\ b \leq\ c$ ?  :maybe:

[quote="nguoivn"][quote="can_hang2007"]
This inequality is not true, try $ a \equal{} \frac {1}{3},b \equal{} \frac {5}{3},c \equal{} 1$.
By the way, the following is stronger for your old inequality
\[ (ab \plus{} bc \plus{} ca)(ab^3 \plus{} bc^3 \plus{} ca^3 \plus{} abc) \le16
\]
[/quote]
Can you post your solution for this stronger result? I can't solve it  :wink:[/quote]
Try to prove the following by your own
\[ 4\sum ab \plus{}\sum ab^3 \plus{}abc \le 16\]

Yes, if it true, it can be solved easily by your lenma.
So, come back my problem, we only need to prove that:
$ 1 \plus{} a^3b^3c^3 \plus{} \sum\ a^4b^2 \leq\ \sum\ ab^2 \plus{} a^3 \plus{} b^3 \plus{} c^3 \plus{} ab\plus{}bc\plus{}ca$
with  $ a \leq\ b \leq\ c$ ?
I  think it's not hard  :wink:

[quote="can_hang2007"]My hint is the following
\[ 8\sum a^4(b^2 \plus{} c^2) \le (\sum a)^3 (\sum a^2(b \plus{} c) \minus{} 4abc)
\]
:)[/quote]
I think you should try with my lemma ;)
It is very easy to prove and it is really helpful for your problem.
By the way, I haven't check your lemma yet, since I dont think it is nice.

[quote="nguoivn"]Yes, if it true, it can be solved easily by your lenma.
So, come back my problem, we only need to prove that:
$ 1 \plus{} a^3b^3c^3 \plus{} \sum\ a^4b^2 \leq\ \sum\ ab^2 \plus{} a^3 \plus{} b^3 \plus{} c^3 \plus{} ab \plus{} bc \plus{} ca$
with  $ a \leq\ b \leq\ c$ ?
I  think it's not hard  :wink:[/quote]
Now, I have checked your inequality, and it is also wrong, try $ (a,b,c)\equal{}(0,\frac{10}{7},\frac{11}{7}).$

Can U post your sulotion, please ?
I can't solve it yet.  :wink:

[quote="nguoivn"]Given $ a, b, c$ satisfy $ a \plus{} b \plus{} c \equal{} 3$. Prove that: $ (a \plus{} bc^3)(b \plus{} ca^3)(c \plus{} ab^3) \leq\ 16$[/quote]
Any another idea for it ? For a long time, I still can't solved it yet  :(

[quote="nguoivn"]Given $ a, b, c \ge 0 $ satisfy $ a\plus{}b\plus{}c\equal{}3$. Prove that:
$ (a\plus{}bc^3)(b\plus{}ca^3)(c\plus{}ab^3) \leq\ 16$[/quote]
It is also my inequality,see here http://forum.mathscope.org/showthread.php?t=30445
I think it is not hard inequality.
We can use two inequalities:

\[ \max(a^2b+b^2c+c^2a;ab^2+bc^2+ca^2) \le 4-abc \]

\[ \min(a^2b+b^2c+c^2a;ab^2+bc^2+ca^2) \le \frac{27-3abc}{8} \]

[quote="nguoivn"]Given $ a, b, c \ge 0 $ satisfy $ a+b+c=3$. Prove that:
$ (a+bc^3)(b+ca^3)(c+ab^3) \leq\ 16$[/quote]
The first inequality of lemma is well known:Vasc’s inequality.
The second inequality of lemma follows from Schur inequality:
\[ (a+b+c)^3-3abc \ge 4(a^2b+b^2c+c^2a+ab^2+bc^2+ca^2). \]

Now,rewrite the inequality as the follows:
\[ abc+a^4b^4c^4+a^2b^4+b^2c^4+c^2a^4+abc(a^4b^2+b^4c^2+c^4a^2) \le 16. \]
Since
\[ a^4b^4c^4 \le abc.\]
\[ a^2b^4+b^2c^4+c^2a^4=(ab^2+bc^2+ca^2)^2-2abc(a^2b+b^2c+c^2a). \]
\[ a^4b^2+b^4c^2+c^4a^2 \le (a^2b+b^2c+c^2a)^2. \]
It suffices to show that:
\[ 2abc+(ab^2+bc^2+ca^2 )^2+abc((a^2b+b^2c+c^2a )^2-2(a^2b+b^2c+c^2a )) \le 16 .\]
We consider two cases:
[b] Case 1 [/b] . $ c \ge b \ge a \ge 0 $.
Since $ ab^2+bc^2+ca^2 \ge a^2b+b^2c+c^2a $ , by lemma:
\[ ab^2+bc^2+ca^2 \le 4-abc. \]
\[  a^2b+b^2c+c^2a \le \frac{27-3abc}{8} \le \frac{28-4abc}{8} \le \frac{7-abc}{2}\]
The last inequality deduces:
\[  (a^2b+b^2c+c^2a )^2-2(a^2b+b^2c+c^2a ) \le \frac{(7-r)^2}{4}-(7-r). \]
Put $ abc=r $,it suffices to show that:
\[ 2r+(4-r)^2+\frac{r(7-r)^2}{4}-r(7-r) \le 16.\]
Or
\[  r(3+6r-r^2) \ge 0.\]
Which is obvious true for $ 0 \le r \le 1 $.
[b] Case 2 [/b] .$ a \ge b \ge c \ge 0 $.
Since
\[ a^2b^4+b^2c^4+c^2a^4+abc(a^4b^2+b^4c^2+c^4a^2)
\le a^4b^2+b^4c^2+c^4a^2+abc(a^2b^4+b^2c^4+c^2a^4).\]
So,similar case 1,the inequality is proved.Our proof are completed.
Equality holds if and only if  $ a=2;b=0;c=1 $ and any cyclic permutations.

[quote="quykhtn-qa1"]
It is also my inequality,see here http://forum.mathscope.org/showthread.php?t=30445
[/quote]
I found it in 2008, not 2012, Quy  :wink:.
[quote="quykhtn-qa1"]
I think it is not hard inequality.
[/quote]
Of course, this problem is only hard in four years ago. Now, we can solve it by many ways. Example, I'll complete my old hint.
I proved the case $a \ge b \ge c$. So, we only need to consider the cases $a \le b \le c$.
We need to show that \[2abc+\sum a^2b^4+abc(a^4b^2+b^4c^2+c^4a^2) \le 16 \]
Or \[2abc+(ab^2+bc^2+ca^2)^2+abc(\sum a^4b^2-2 \sum a^2b) \le 16 (1)\]
By contradict method and AM-GM, we can easily to show that: \[a^4b^2+b^4c^2+c^4a^2 -2(a^2b+b^2c+c^2a) \le 5\]  with $a \le b \le c$. And use this lenma, we have \[LHS(1) \le 2abc+(4-abc)^2+5abc=abc(abc-1)+16 \le 16=RHS(1)\] The proof completed :).
@Quy:
[hide]Làm toán cũng cần trung thực, không nên nhận vơ bài của người khác là của mình. Đến đứa trẻ con cũng biết nếu là bài của em thì trong topic kia em đã khoe "Bài 1 là của Toursish, còn bài thứ 2 là của tôi, anh QA có thể thử sức", phải không nào?

Anh đoán em vô tình lượm được bài này ở một số diễn đàn toán VN nên đưa lên MS cho vui. Nhưng cái mà anh thấy thất vọng là sau đã biết bài toán này của anh rồi mà còn cố tình nhận vơ một cách vô duyên như vậy. Hay em không nhìn thấy cái mốc thời gian 2008 trong bài post đầu của topic này?

Nói như kiểu "trùng ý tưởng" của em thì bài nào anh cũng nói là của anh được. Vô tình trùng thôi mà??? Nếu em tìm được bất đẳng thức này ở 1 topic nào trước năm 2008 thì anh rất vui lòng đính chính lại. Còn nếu không thì đừng nên nhận vơ kiểu như thế. Chỉ có em biết rõ nhất là bài này có phải do em nghĩ ra không? Cá nhân anh thì chưa bao giờ thấy em sáng tác được một bất đẳng thức đại số nào kiểu thế này.

Gần đây trên MS có rất nhiều đứa mặt dày thích ăn cắp ý tưởng của người khác, khỏi nói em cũng biết là những ai rồi. Hi vọng em không trở thành một kẻ như vậy. Tính anh quen nói thẳng. Anh ghét cay ghét đắng những thằng háo danh mà không dựa trên thực lực của mình.[/hide]

Your proof is similar with my proof and we do not need use contradict method
\[ (a^2b+b^2c+c^2a)^2-2(a^2b+b^2c+c^2a) \le \left(\frac{27}{8}\right)^2-2.\frac{27}{8} < 5 \]
@ nguoivn
[hide] Anh nên xem lại cách nói của mình.Em tham gia mạng từ năm 2009 thì lấy đâu ra topic nào như vậy.Em mới mượn được đứa bạn mấy cuốn sách do anh viết.Thú thực là phần nhiều những bài toán của anh không phải là bài toán có trước mà là lời giải có trước.Em học được nhiều thứ từ những bài của anh Cẩn hơn.Còn cái chuyện sáng tác những bài dạng này thì không phải là quá khó như anh nghĩ .Nói như vậy nhưng em rất tôn trọng anh,hi vọng không vì bài toán này mà lại cãi nhau ở đây.[/hide]

[quote="nguoivn"]
By contradict method and AM-GM, we can easily to show that: \[a^4b^2+b^4c^2+c^4a^2 -2(a^2b+b^2c+c^2a) \le 5\]  with $a \le b \le c$. [/quote]
I also think it's similar hint. With this lenma, here is my way.
Assume that \[a^4b^2+b^4c^2+c^4a^2 -2(a^2b+b^2c+c^2a) \ge 5\] Hence, easily to show that \[a^4b^2+b^4c^2+c^4a^2 \ge 7+2\sqrt{6} (*)\] In the other hand, by AM-GM we have again \[a^4b^2+b^4c^2+c^4a^2 \le \frac{1}{2}\sum ab(a^2+b^2) \le \frac{729}{64}\] It's contradict with $(*)$. So, the assumption was wrong. And we have q.e.d
@Quy.[hide] Anh Cẩn cũng là người anh mà anh kính trọng và học hỏi được nhiều nhất. Tất nhiên, mỗi người mỗi sở thích, mỗi hướng đi. Em cũng thế mà anh cũng vậy. Anh chỉ khuyên em là ở một sân chơi quốc tế thì chúng ta nên chơi theo luật quốc tế. Bất cứ bài bđt nào hay được mọi người post lên đa phần đều được các lão làng như Vasc, Araqady, Jichen xác nhận cũ hay mới (nếu cũ sẽ dẫn link đi kèm).... Anh chưa thấy ai xử sự như em ở trên. Cứ coi như là trùng ý tưởng đi, đúng hay sai anh chưa biết nhưng thế có lẽ hơi vô duyên. Những gì cần nói anh nói cả rồi.[/hide]


number theory
greatest common divisor
number theory unsolved

Let $ (a_n)$ be a sequence of natural numbers such that $ (a_i,a_j)\equal{}(i,j)$ for any $ i,j\in \mathbb{N}$ ($ (...)$ denotes greatest common divisor). Prove that $ a_n\equal{}n$ for all $ n$.

Take $ i \equal{} n,j \equal{} 2n$ then
$ n|f(n)$
Suppose exist n such that $ x_n \equal{} kn$ and $ k > 1$
From condition we have
$ \gcd (x_n,x_{kn}) \equal{} \gcd(n,kn) \equal{} n$
But we have $ kn|x_n,kn|x_{kn}$
So $ kn|n \Rightarrow k \equal{} 1$ .
It give tradition .
So $ f(n) \equal{} n,\forall n\in N$


[quote="RC-7th"]I thought guests couldn't see profile information. Yet, on the "who is online" list, it sometimes says that a guest is looking at profile information.[/quote]We don't make a difference from "trying" and "actually seeing" profile information.
When you access the profile page, it automatically registers you as using that page. The simplest way to check this is just by browsing as a guest ;)


Let a,b,c are positive numbers. Prove that
$ \frac {a}{b}\plus{}\frac {b}{c}\plus{}\frac{c}{a}\plus{}\frac{8(a\minus{}b)(b\minus{}c)(c\minus{}a)}{(a\plus{}b)(b\plus{}c)(c\plus{}a)}\geq 3$
 :D

This is interesting! I will prove this for cases k=1 of the inequality:
$ \frac {a}{b}\plus{}\frac{b}{c}\plus{}\frac{c}{a}\plus{}\frac{k(a\minus{}b)(b\minus{}c)(c\minus{}a)}{(a\plus{}b)(b\plus{}c)(c\plus{}a)}\geq 3$
We have
$ \frac{a}{b}\plus{}\frac{b}{c}\plus{}\frac{c}{a}\minus{}3\equal{}\sum \frac{a\minus{}b}{b}\equal{}\frac{a\minus{}b}{b}\minus{}\frac {a\minus{}b}{a}\plus{}\frac{b\minus{}c}{c}\minus{}\frac {b\minus{}c}{a}\equal{}\frac{(a\minus{}b)^2}{ac}\plus{}\frac{(b\minus{}c)(a\minus{}c)}{ac}$.
Thus
$ LHS\minus{}RHS\equal{}\frac{(a\minus{}b)^2}{ac}\plus{}(b\minus{}c)(a\minus{}c)(\frac{1}{ac}\minus{}\frac{a\minus{}b}{(a\plus{}b)(b\plus{}c)(c\plus{}a)})\geq 0$
Assume that c=min{a,b,c}, we have that inequality is true. :)
For k=2 and k=8, can anyone prove?

Oh, it isn't very hard. So, who can prove for k=8? A post bored in these days. :(

hey.can anyone post a solution for it???

not beautiful solution.

let's suppose $ a$ is the largest.
it is obviously true when $ a \ge c \ge b$
so it is enough to show the inequality when $ a \ge b \ge c$
let's put $ x\equal{}b\minus{}c, y\equal{}a\minus{}b$
if we substitute $ b\equal{}c\plus{}x, a\equal{}c\plus{}x\plus{}y$ into the inequality,
the last expression is like below,

$ (8x^2\plus{}8xy\plus{}8y^2)c^4\plus{}(24x^3\plus{}24x^2y\plus{}16xy^2\plus{}8y^3)c^3\plus{}(26x^4\plus{}28x^3y\plus{}6x^2y^2\plus{}4xy^3\plus{}2y^4)c^2\plus{}(12x^5\plus{}17x^4y\plus{}2x^3y^2\minus{}2x^2y^3\plus{}xy^4)c
\plus{}x^3(x\plus{}y)^2(2x\plus{}y) \ge 0$

trivial.(because $ x^3y^2\plus{}xy^4 \ge 2x^2y^3$)

equality holds when a=b=c


Will I get points for finding an answer and then proving that it is the answer, rather than writing a normal proof.

Yes, but you might have to prove that your answer(s) is unique if the problem requires it.

so if i kind of guess an answer-then prove that it is definitely the unique and correct answer i will get full points?

If it's mathematically correct, you will get points. That's how it usually works, right? Who cares if you guess it, that's how a lot of mathematicians did and do things.


How do I graph points in the (x, y, z) form?

EXAMPLE:

Graph the following points on the xyz-plane?

(0, 0, 0), (1, -2, 3), (2, -3, -3) and (4, 1, -1)

You graph them in $ \mathbb{R}^{3}$ as you said.  If you can't visualize what this looks like, just go to google images and type in xyz-plane or something to get a picture.  Then you plot the points just as you would on an xy-plane.


$ 1.)$ Fie $ z_1,z_2,z_3,z_4 \in \mathbb{C^*}$ cu proprietatile :

$ (i)$ $ |z_1|\equal{}|z_2|\equal{}|z_3|\equal{}|z_4|$ ;

$ (ii)$ $ z_1\plus{}z_2\plus{}z_3\plus{}z_4 \equal{} 0$ ;

Demonstrati ca : $ \frac{1}{z_{1}^n}\plus{}\frac{1}{z_{2}^n}\plus{}\frac{1}{z_{3}^n}\plus{}\frac{1}{z_{4}^n} \equal{}0$ , unde $ n$ este un numar impar .

Fie $ M_1, M_2, M_3, M_4$ punctele din planul complex cu originea in $ O$ care au afixele $ z_1, z_2, z_3, z_4$. Este destul sa se arate ca cele patru puncte sunt varfurile unui dreptughi. Astfel ar rezulta ca $ z_1\equal{}\minus{}z_3$ si $ z_2\equal{}\minus{}z_4$ de unde rezulta ca suma respectiva este zero. Revenim la demonstratia ca cele patru puncte sunt varfurile unui dreptunghi.
$ z_1\plus{}z_2\equal{}\minus{}(z_3\plus{}z_4)$ adica $ \vec{OM_1}\plus{}\vec{OM_2}\equal{}\minus{}(\vec{OM_3}\plus{}\vec{OM_4})$ (vectorii suma sunt opusi). Cum cei patru vectori au modulele egale rezulta ca cele patru puncte sunt intersectiile dintre cercul unitate centrat in $ O$ cu mediatoarele vectorilor suma. Rezulta ca $ M_1M_2M_3M_4$ este dreptunghi.

[quote="turcas_c"][color=darkred]Fie $ z_1,z_2,z_3,z_4 \in \mathbb{C}$ cu proprietatile $ \{\begin{array}{c} |z_1| = |z_2| = |z_3| = |z_4| = r > 0 \\
 \\
z_1 + z_2 + z_3 + z_4 = 0\end{array}$ .

Demonstrati ca $ \frac {1}{z_{1}^n} + \frac {1}{z_{2}^n} + \frac {1}{z_{3}^n} + \frac {1}{z_{4}^n} = 0$ , unde $ n$ este un numar natural impar.[/color] [/quote]
[color=darkblue][b][u]Demonstratie algebrica.[/u][/b] Fie ecuatia $ z^4 - s_1\cdot z^3 + s_2\cdot z^2 - s_3\cdot z + s_4 = 0$ ale carei radacini sunt $ z_k\ ,\ k\in\overline {1,4}$ .

$ \boxed {s_1\equiv z_1 + z_2 + z_3 + z_4 = 0}$ $ \implies$ $ \overline {\sum_{k = 1}^4}z_k = 0$ $ \implies$ ${ \sum_{k = 1}^4}\overline z_k = 0$ $ \implies$ $ \sum_{k = 1}^4\frac {r^2}{z} = 0$ $ \implies$
$ \boxed {s_3\equiv z_1z_2z_3 + z_1z_2z_4 + z_1z_3z_4 + z_2z_3z_4 = 0}$ . Asadar, ecuatia ale carei radacini sunt $ z_k\ ,\ k\in\overline {1,4}$ este

$ \boxed {z^4 + s_2\cdot z^2 + s_4} = 0$ , unde $ \{s_2,s_4\}\subset\mathbb C$ . Notam $ S_n = \sum_{k = 1}^4z_k^n$ , $ n\in\mathbb N$ . Folosind relatia de recurenta

$ \{\begin{array}{c} \boxed {S_1 = 0}\ \ \ ;\ \ \ \boxed {S_3 = 0} \\
 \\
\boxed {S_{2n + 3} + s_2\cdot S_{2n + 1} + s_4\cdot S_{2n - 1} = 0}\end{array}$ , unde $ n\in\mathbb N$ se obtine inductiv concluzia problemei propuse.

[b]Remarca.[/b] Am folosit relatia cunoscuta $ \boxed {S_3 = s_1^3 - 3\cdot s_1\cdot s_2 + 3\cdot s_3}$ .[/color]

[quote="turcas_c"]$ 1.)$ Fie $ z_1,z_2,z_3,z_4 \in \mathbb{C^*}$ cu proprietatile :

$ (i)$ $ |z_1| \equal{} |z_2| \equal{} |z_3| \equal{} |z_4|$ ;

$ (ii)$ $ z_1 \plus{} z_2 \plus{} z_3 \plus{} z_4 \equal{} 0$ ;

Demonstrati ca : $ \frac {1}{z_{1}^n} \plus{} \frac {1}{z_{2}^n} \plus{} \frac {1}{z_{3}^n} \plus{} \frac {1}{z_{4}^n} \equal{} 0$ , unde $ n$ este un numar impar .[/quote]
Can you translate in english the last line please?
Thanks!

[quote="DIMITRIS T"]Can you translate in english the last line please? Thanks![/quote]
Prove that $ \sum_{k\equal{}1}^4\frac {1}{z_k^n}\equal{}0$ , where $ n$ is an odd natural number.

Se poate dezvolta si generalizarea.In conditiile date sa demonstram ca \[ \displaystyle\frac{1}{z_1^k} \plus{}\displaystyle\frac{1}{z_2^k}\plus{}...\plus{}\displaystyle\frac{1}{z_n^k}\equal{}0\]
Putem presupune ca $ |z_1|\equal{}...\equal{}|z_n|\equal{}r>0$ si $ z_1^k\plus{}...\plus{}z_2^k\equal{}0$ avem:
\[ \displaystyle\frac{1}{z_1^k}\plus{}\displaystyle\frac{1}{z_2^k}\plus{}...\plus{}\displaystyle\frac{1}{z_n^k}\equal{}\displaystyle\frac{\overline{z_1^k}}{r^{2k}}\plus{}...\plus{}\displaystyle\frac{\overline{z_n^k}}{r^{2k}}\equal{}\displaystyle\frac{1}{r^{2k}}(\overline{z_1^k\plus{}z_2^k\plus{}...\plus{}z_n^k})\equal{}0\]


induction
vector
group theory
abstract algebra
superior algebra
superior algebra unsolved

Let $\mathbb{G}$ be a finite group such that $x^{2}=e$ $\forall x \in \mathbb{G}$. Prove, without using Cauchy/Sylow Theorems/similar results, that $ord(G)=2^{k}$:
a. By induction
b. Using vectorial spaces

Nobody? c'mon :P  I know it may seem boring to solve this problem without using Cauchy but I assure you that the ideas are nice :)

[quote="ciprian"]
b. Using vectorial spaces[/quote]
Rewrite as $2x=0\forall x\in G$ then $G$ is finite dimesion vector space on $\mathbb{F}_{2}$. Done ! :D

[quote="ciprian"]You are right N.T.TUAN :)  What about part a.?[/quote]
At this time , I can't. Hmm, can you post solution?  :D

We already had this before and I asked the same question:

How can you even use Cauchy¿

its a nice problem here you are my solution;
1: use lagrange and the fact that $A=(e,x)$ is a supgroup of $G$.
2:prove that $G$ is isomorphe (Z/2Z)^{k} and done

@ZetaX: Maybe I am completely wrong... Assume that ord(G) has a prime divisor p>2. Then, using Cauchy, G has a subgroup with p elements i.e it has an element of order p. But all the elements in G have order 2 (except e), contradiction.


topology
real analysis
real analysis unsolved

How could I show that the Borel $\sigma$-algebra has same cardinal as $%Error. "mathdd" is a bad command.
{R}$ ? ( it is very intuitive since the set of the open sets in $%Error. "mathdd" is a bad command.
{R}$ has same cardinal than $%Error. "mathdd" is a bad command.
{R}$, and every Borelian is obtained through a finite number of operations on these open sets. But how to show it rigorously ?)

The way I know is to show that the set of all Suslin's sets is of cardinality $c$ (which is trivial, because a Suslin set
is determined by countably many closed sets) and to note that every Borel set is Suslin.

I didnt know what a Suslin set was, and the definition that appears in Wikipedia confused me a bit.

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

Given that intersection of closed sets is a closed set, i dont really understand that definiton (it seems that a Suslin set is just a union of closed sets)

Am I right? Is that the definition of Suslin set, or is there something missing? I'd like to know how to show that the Borel $ \sigma$-algebra has cardinality $ c$. Is there an proof wich doesn't involve Suslin sets? Or where can I read more about Suslin sets and their relation with the Borel sets? Thanks a lot.

:oops:  I think that the interesting thing about Suslin sets must be that they arent just 'any' union of closed sets.. Anyway, does anybody know where can I read more about them?


Let $a,b,c$ be real positive numbers, prove that:
\[ (1+ \frac{a}{b})(1+ \frac{b}{c})(1+ \frac{c}{a}) \geq 2(1+ \frac{a+b+c}{\sqrt[3]{abc}}) \]

Thanks in advanced!!

Fabycv  :rotfl:

make the substitution
\[ \sqrt[3]{\frac{a^2}{bc}}=x \]  and others similarly

then \[ xyz=1 \]

and u get \[ (x+y)(y+z)(z+x)\geq2(1+x+y+z) \]
which is true by Murheads


geometry
trigonometry
ratio
angle bisector
geometry proposed

In an attempt to create a more or less simple geometry problem for ML, I came up with the following:

If A', B', C' are the midpoints of the sides BC, CA, AB of a triangle ABC, and if the incircle of the triangle A'B'C' touches its sides B'C' and C'A' at the points X and Y, then prove that the lines AY and BX meet on the angle bisector of the angle BCA.

  Darij

This looks like a straightforward application of Menelaus and Ceva, but I wanted to find a nicer proof. I failed, so.. :D

I use slightly different notations, trying to show that $BZ,CY$ concur on the $a$-bisector of $ABC$.

Let $BZ\cap CA=N,\ CY\cap AB=P$. Then $\frac{C'P}{PB}=\frac 12\cdot\frac{p-c}{p-a},\ \frac{B'N}{NC}=\frac 12\cdot\frac{p-b}{p-a}$, so $\frac{AP}{PB}=\frac{b}{p-a},\ \frac{AN}{NC}=\frac{c}{p-a}$, etc. ($p$ is the semiperimeter).

Let AY and BX intersects at P. From Ceva sine theorem it is enough to prove
sinCAP/sinPAB * sinABP/sinPBC = 1 , but  <CAP = <AYC' , <PBC =<BXC'.
Hence sinAYC'/sinPAB * sinBXC'/sinPBC = AC'/C'Y *XC'/BC' = 1. c.t.d.

Actually, my solution is quite similar to what both of you did, although it doesn't explicitely use Menelaos.

Let the parallel to BC through A meet the line BX at U, and let the parallel to CA through B meet the line AY at V. Then, since C'X || AU (this is because the line C'X, being a midparallel of triangle ABC, is parallel to BC, and BC is parallel to AU), we have by Thales

  AU : C'X = AB : C'B.

Since the point C' is the midpoint of the segment AB, we have AB : C'B = 2 and thus AU : C'X = 2 and hence AU = 2 C'X. Similarly, BV = 2 C'Y. But C'X = C'Y (since the two tangents from a point to a circle are always equal in length). Thus, AU = BV.

Now, let the lines BX and AY intersect the lines CA and BC at the points S and T, respectively. And let the angle bisector of the angle BCA meet the side AB at the point R. Then we have to prove that the lines AT, BS and CR concur. After Ceva, this is equivalent to showing that

$\frac{BT}{TC}\cdot \frac{CS}{SA}\cdot \frac{AR}{RB}=1$

(we use non-directed lengths here, since everything lies inside :-) ).

Now, since an angle bisector in a triangle divides the opposite side in the ratio of the adjacent sides, we have

$\frac{AR}{RB}=\frac{CA}{BC}$;

also, since BV || CA and AU || BC, Thales yields

$\frac{BT}{TC}=\frac{BV}{CA}$ and $\frac{CS}{SA}=\frac{BC}{AU}$.

Thus,

$\frac{BT}{TC}\cdot \frac{CS}{SA}\cdot \frac{AR}{RB}=\frac{BV}{CA}\cdot \frac{BC}{AU}\cdot \frac{CA}{BC}=\frac{BV}{AU}=1$,

since AU = BV. Proof complete.

I think this all was a neat Ceva exercise. First I thought about proposing it for a periodical (for proposing olympiad problems I'm still a bit too young now :D ), but then I thought: In a periodical, I'll probably have to wait a year until some solutions get published, while here there's a good chance to see the first one after two hours. And yeah, it was as I expected. Thanks!

  Darij


calculus
integration
trigonometry
complex analysis
real analysis
real analysis unsolved

this [i]might [/i]be hard...  :maybe:  [u][b]without[/b][/u] the aid of complex analysis, show

$ \textbf{(i)}\;\int_{0}^\infty\;\sin \;(x^{2})\;\;\textbf dx\;=\;\int_{0}^\infty\;\cos \;(x^{2})\;\;\textbf dx\;=\;\boxed{\frac{1}{2\sqrt 2}\;\Gamma\left(\frac{1}{2}\right)}$

$ \textbf{(ii)}\;\int_{0}^\infty\;\sin \;(x^{3})\;\;\textbf dx\;=\;\boxed{\frac{1}{6}\;\Gamma\left(\frac{1}{3}\right)}$

$ \textbf{(iii)}\;\int_{0}^\infty\;\cos \;(x^{3})\;\;\textbf dx\;=\;\boxed{\frac{1}{2\sqrt 3}\;\Gamma\left(\frac{1}{3}\right)}$

can you now generalize it to finding

$ \textbf{(iv)}\;\int_{0}^\infty\;\sin \;(x^{k})\;\;\textbf dx\;\;\;\quad \text{and}\;\;\;\quad \textbf{(v)}\;\int_{0}^\infty\;\cos \;(x^{k})\;\;\textbf dx\;\;\;\quad ?$
( where $ k\;=\;2,3,4,5,6\cdots$ )

\begin{eqnarray*}\int_{0}^{\infty}\sin (x^{k}) \; dx & = & \frac{1}{k}\int_{0}^{\infty}\frac{\sin t}{t^{1-1/k}}\; dt \quad \quad (\text{where }x^{k}= t)\\ & = & \frac{1}{k \Gamma(1-1/k)}\int_{0}^{\infty}\sin t \int_{0}^{\infty}u^{-1/k}e^{-tu}\; du \; dt\\ & = & \frac{1}{k \Gamma(1-1/k)}\int_{0}^{\infty}\frac{u^{-1/k}}{u^{2}+1}\; du\\ & = & \frac{1}{k \Gamma(1-1/k)}\int_{0}^{\infty}\frac{v^{k-2}}{v^{2k}+1}\; dv \quad \quad (\text{where }u = v^{k})\\ \end{eqnarray*}

Similarly, $ \int_{0}^{\infty}\cos (x^{k}) \; dx \; = \; \frac{1}{k \Gamma(1-1/k)}\int_{0}^{\infty}\frac{v^{2k-2}}{v^{2k}+1}\; dv$

This works well for $ k=2$.

But for other values, computing $ \int_{0}^{\infty}\frac{v^{k-2}}{v^{2k}+1}\; dv$ with no aid of complex analysis is much harder, and it may need the reflexion formula $ \Gamma(z)\Gamma(1-z) = \pi \csc \pi z$.