Find all positive $k$ for which the following holds :

$\sum_{\cyc}\frac1{k+x^2-2yz}\le\frac3{k-1}$ for all $x,y,z$ with $x^2+y^2+z^2=3$.
I think the answer is $k\ge6$ but I have no proof.

I think that your answer is correct..

first $x^2-2yz=3-(y+z)^2$
now whe must see what are the conditions in which is ${1\over{k-1}} \ge{1\over{k+3-(y+z)^2}}$
That is
${{4-(y+z)^2}\over{(k-1)(k+3-(y+z)^2)}}\ge0$
if $k>1$ then $4-(y+z)^2$ and $(k-1)(k+3-(y+z)^2)$ must have same sign and this can be done only for $k\ge6$ and it cannot be smaller because we can choose let say $(y+z)^2=4,01$
if $k<1$ then $4-(y+z)^2$ and $(k-1)(k+3-(y+z)^2)$ must have opposite sign and this can't be done because we can choose let's say $(y+z)^2=3,01$

It seems ok but I'm not sure about it....

[quote="Adalbert"]
now whe must see what are the conditions in which is ${1\over{k-1}} \ge{1\over{k+3-(y+z)^2}}$
[/quote]We can assume $k\ge1$. Then your condition is equivalent to $4\ge(y+z)^2$. But that is not always true as you state yourself. If it were, the originial ineq would be trivial. :)

[quote="Vasc"]The answer is $k \geq 7$. For $k=7$, equality occurs again when $x=y=2z$ or any other cyclic permutation.[/quote]Vasc, this is just incredible... an equality case in the middle of nowhere  :fool: Now I would be curious to see how you prove it from there  :roll:

For $k \geq 9$ we can use Cauchy-Schwarz Inequality.
For $k=7$, we have to show that
$x^2+y^2+z^2=3$ implies $\sum \frac 1{10-(y+z)^2} \geq \frac 1 {2}$.
The ineq. is equivalent to
$\sum \frac 1{10-(s-x)^2} \geq \frac 1 {2}$,
where $s=x+y+z$. Using (n-1)Equal Variable-Theorem (since $x^2+y^2+z^2=constant$ and $x+y+z=constant$), it suffices to consider $y=z$.
So, the inequality reduces to $(x-1)^2(3x^2-1)^2 \geq 0$.  :lol:

[quote="Vasc"]For $k \geq 9$ we can use Cauchy-Schwarz Inequality.
[/quote]Please give a hint, how? (I'd thought of that already, as the ineq occured to me first with $k=9$... but I can't see how  :huh:)

Write first the inequality as
$\sum \frac{3+x^2-2yz}{k+x^2-2yz} \geq \frac{6}{k-1}$,
then apply Cauchy-Schwarz.


$ ABCD$ is an isosceles trapezium.
Upper base $ \overline {AD}\equal{}2$  ,lower base $ \overline{BC}\equal{}8$.If $ M$is the mid-point of $ AB$ and $ MD \perp CD$.
Find the area of  isosceles trapezium $ ABCD$.

[quote="vinskman"]$ ABCD$ is an isosceles trapezium.
Upper base $ \overline {AD} \equal{} 2$  ,lower base $ \overline{BC} \equal{} 8$.If $ M$is the mid-point of $ AB$ and $ MD \perp CD$.
Find the area of  isosceles trapezium $ ABCD$.[/quote]
Let $ \overline{MD}\equal{}c,\overline{DA}\equal{}a,\overline{DC}\equal{}b$,so $ c\equal{}\frac{1}{2}(5a\plus{}b)$ and $ b*c\equal{}0$ so $ (5a\plus{}b)b\equal{}0$
$ 10|b|cos\angle ADC\plus{}b^2\equal{}0$ so $ |b|\equal{}10cos\angle ABC$ AH-altitude of trapezium so $ AH\equal{}10sin\angle ABCcos\angle ABC$ and
$ BH\equal{}10cos^2\angle ABC$, and $ BC\equal{}AD\plus{}20cos^2\angle ABC$ so $ cos\angle ABC\equal{}\frac{1}{\sqrt{5}}$ and $ AH\equal{}5$ and finaly
$ S_{ABCD}\equal{}AH(AD\plus{}BC)/2\equal{}25$ :)

$ |AM| \equal{} |BM| \equal{} \frac {b}{2}$

$ \cos \angle ABC \equal{} \frac {3}{b}$ $ \implies$ $ \frac {b^2}{4} \plus{} 10 \equal{} |MD|^2 \equal{} 25 \minus{} \frac {b^2}{4}$
Then $ \boxed{b \equal{} \sqrt {30}}$
$ S_{ABCD} \equal{} (8 \plus{} 2)/2.\sqrt {21}\equal{}5\sqrt{21}$

[quote="dima ukraine"][quote="vinskman"]$ ABCD$ is an isosceles trapezium.
Upper base $ \overline {AD} \equal{} 2$  ,lower base $ \overline{BC} \equal{} 8$.If $ M$is the mid-point of $ AB$ and $ MD \perp CD$.
Find the area of  isosceles trapezium $ ABCD$.[/quote]
Let $ \overline{MD} \equal{} c,\overline{DA} \equal{} a,\overline{DC} \equal{} b$,so $ c \equal{} \frac {1}{2}(5a \plus{} b)$ and $ b*c \equal{} 0$ so $ (5a \plus{} b)b \equal{} 0$
$ 10|b|cos\angle ADC \plus{} b^2 \equal{} 0$ so $ |b| \equal{} 10cos\angle ABC$ AH-altitude of trapezium so $ AH \equal{} 10sin\angle ABCcos\angle ABC$ and
$ BH \equal{} 10cos^2\angle ABC$, and $ BC \equal{} AD \plus{} 20cos^2\angle ABC$ so $ cos\angle ABC \equal{} \frac {1}{\sqrt {5}}$ and $ AH \equal{} 5$ and finaly
$ S_{ABCD} \equal{} AH(AD \plus{} BC)/2 \equal{} 25$ :)[/quote]

$ \boxed{BC \equal{} AD \plus{} 20cos^2\angle ABC}$ so $ cos\angle ABC$ [color=red]o[/color]$ \equal{} \frac {1}{\sqrt {5}}$ and $ AH \equal{} 5$
Sorry,I think you put wrong data to get the incorrect result!! :juggle:

Ptolemy!
Put F on DC so $ MF \parallel BC$
$ AF\equal{}MD\equal{}k$, $ AM\equal{}DF\equal{}m$
By Ptolemy on AMFD,
$ 10\plus{}m^2\equal{}f^2$
By Pythag on AMF,
$ m^2\plus{}f^2\equal{}25$
$ m\equal{} \frac{\sqrt{15}}{2}$
By Pythag, the height of the trapezoid is $ \frac{\sqrt{21}}{2}$, the area is $ 5\sqrt{21}$.

[quote="dgreenb801"]Ptolemy!
Put F on DC so $ MF \parallel BC$
$ AF \equal{} MD \equal{} k$, $ AM \equal{} DF \equal{} m$
By Ptolemy on AMFD,
$ 10 \plus{} m^2 \equal{} f^2$
By Pythag on AMF,
$ m^2 \plus{} f^2 \equal{} 25$
$ m \equal{} \frac {\sqrt {15}}{2}$
By Pythag, the height of the trapezoid is $ \frac {\sqrt {21}}{2}$, the area is $ 5\sqrt {21}$.[/quote]

beautiful and elegant work... :jump:

Let $N=BC\cap MD$; clearly $BN=AD\ (*)$, hence $CN=AD+BC=10\ (\ 1\ )$. Let $H$ be the projection of $D$ onto $BC$. Easily $CH=(BC-AD)/2=3$, so from the right-angled $\triangle CDN\implies DH^2=CH\cdot NH=3\cdot 7=21$. From $(*)$ we get $[ABCD]=[CDN]=CN\cdot DH/2=5\sqrt 21$.

Best regards,
sunken rock

Let $AB\cap CD=P$ and $F$ be the midpoint of $CD$. Then, $AP=PD=\frac{x}{3}$ by similar triangles. The right triangle $PDM$ gives us $\frac{25x^2}{36}-\frac{x^2}{9}=\frac{21x^2}{36}=MD^2$, so $MD=\frac{\sqrt{21}x}{6}$. Now, in right triangle $DMF$, $MD^2+DF^2=16$, so $\frac{21x^2}{36}+\frac{x^2}{4}=16$ or $\frac{30x^2}{36}=16$, hence $x=\frac{24}{\sqrt{30}}$. Then my calculations get off...


Hello!  I have a question about finding where a line crosses it's x-axis.

I have found an equation of my graphed line, and found the slope of the line also.  Now I'm trying to find it's x-axis intercept.  I only know how to find the y-axis intercept using the slope-intercept form.  Is there another form to find the x-axis?  Help.  Here are some numbers to work with=  (62, 157.5) and (64, 162.6)

Thanks.

[quote="heide.rumble"][quote="dogseatcheese"]In your equation of your line, if you make the y equal to 0, you will find the x intercept. Likewise, if you make the x zero, you will make the y intercept. Try to see why, in y intercept form, the last term is your y intercept[/quote]

Thank you so much!  Now, I think I'm having a mental block...My line equation goes like this:

-5.1x + 2y = -1.2

So, how do I make y=0?  Is there anything special I have to do to besides multiply y by 0?

This graphing is more confusing than I thought at first.

 :oops:[/quote]

To "make" y=0, you just substitute 0 for y. So for your equation:
$-5.1x+2\times0=-1.2, -5.1x=-1.2, \text{ and } x=\frac{4}{17}.$ So the x-intercept is at point $(\frac{4}{17},0)$.

Here's another way to think about it:
If you have a line in the form $Ax+By=C$, Then the slope is $\frac{-A}{B}$, x intercept is $\frac{A}{C}$, and the y intercept is $\frac{B}{C}$.

That's basically the same thing except the rearrangement is done in variables and you're not really thinking.  :P

*drops on couch*too lazy to think, shortcuts are better (if you understand them, so I guess it is thinking, ha). :)

I am still confused.  I can't see how we got 4/17?  I'm getting 4 & 1/4 (or 4/9).  I think I'm missing some logical step somewhere along the way.  That must be it.

Here's what I do after hearing the shortcut:

-5.1/-1.2 = 4.25 = 4 & 1/4 or 4/9.

Without hearing the shortcut, I get something completely different.  On the original equation, these are my steps:

-5.1x + (2*0) = -1.2
-5.1x = -1.2
divide both sides out by -5.1 to get x alone,
x=-1.2/-5.1
which equals pretty much nothing, and definitely not 4/17ths.

Help, what step am I missing?

I know, I know, I should be getting this by now.  Thanks for being patient with me.

Oops. :blush: major typo. sorry heide.rumble.

[quote="pianoforte"]The equations for x-intercept and y-intercept should of been $\frac{C}{A}$ and $\frac{C}{B}$.

$x=\frac{-1.2}{-5.1}$ does equal $\displaystyle\frac{4}{17}$ths.

$\frac{-1.2}{-5.1}=\frac{12}{51}=\frac{4}{17}$[/quote]

 :idea:  No one had ever told me to take the decimals out of a fraction before.  It makes sense (duh what is a fraction, but a decimal in the first place).

Thanks!  I knew it must have been something really simple that I was missing.  Here I was trying to divide the darn things.  LOL  (Sometimes I really think I should have been born a blond, cause I don't do justice to my fellow brunettes.)  But what I lack in math brains I make up for in other areas.  (I saved my house from burning down on Thursday everyone!!!   :lol: )  So, I thank you all for explaining me through my weak spots!!!

You are all WONDERFUL!!!!!


[hide="Maybe..."]
Is this the answer by any chance?  My solution is really long, and I don't want to tex it up until I'm sure this is right.

$6\log_6 (102+6\log_6 2)$[/hide]

Actually I remember the problem wrong. It should be $log_{\sqrt2}108^{3-r}-2r$  :D

:cursing:  :furious:

you mean i've got to start all over now? oh well, that happens sometimes. :roll:


Find H(orthocenter).Line GH cuts the circle at A and B with B in opositive to H .Bissector HB cuts the circle at C and D.Triangle ACD is the answer


geometry
rectangle
modular arithmetic
combinatorics unsolved
combinatorics

Given an nXn chessboard with the bottom right hand square removed, let S(n) be the set of numbers k such that this board can be tiled by kx1 rectangles, e.g. S(4) = (1, 3). Prove that x is an element of S(n) if and only if x divides (n-1).

We can easily show we can cover it if $ k \equal{} n \minus{} 1$, just cover the bottom right most square that isn't deleted horizontally, then the next bottom right most up from that and so on, we will be left with a $ n \minus{} 1$ by $ 1$ Vertical uncovered block which we can cover. To show this is true for all x such that x divides $ n \minus{} 1$, just split our kx1 block into m smaller blocks and arrange them to make our larger n-1 * 1 block, which we know we can tile.

To show we cannot cover it if x does not divide n-1, we have $ n^2 \minus{} 1$
squares = $ (N \plus{} 1)(N \minus{} 1)$ so one of k,1 divides N+1 and the other divides N-1. Now we just have to show it doesn't work if k divides only N+1 and not N-1

How do we prove that it doesn't work if k divides only n+1 and not n-1?

Couldn't figure out that bit, anyone know how to do it?

You are wrong about the condition you are trying to prove: that $ k \mid n^2 \minus{} 1$ and $ k \not \mid n \minus{} 1$ does [i]not[/i] imply $ k \mid n \plus{} 1$.  For example, $ 15 \mid 34^2 \minus{} 1 \equal{} 33 \cdot 35$ and $ 15 \not \mid 33$ and $ 15 \not \mid 35$.

Thanks JBL.

I can see what's happening by trying to tile when k does not divide n-1, but can't turn it into a proof.

Does anyone know how to solve this problem?

Arrange the board so that the "bottom" = "first" row is the one with the corner removed.  Let's say we're tiling with $ k$-blocks, and we begin by looking at our tiling from the first row.  We have $ n \minus{} 1 \equiv r \not\equiv 0 \pmod k$, so in the first row there must be a number of vertical blocks congruent to $ r \mod k$.  Then the second row has one more space, so the number of vertical blocks that start there is congruent to 1 mod $ k$.  Then for the 2nd through $ k$th rows, the number of vertical blocks that start there is divisible by $ k$.  But then in the $ k \plus{} 1$th row, we again need a number congruent to $ r$ mod $ k$ many vertical blocks to begin there.  Then in the next row we need $ \equiv 1 \pmod k$ new vertical blocks.  And so on.  Because $ r \not \equiv 0 \pmod k$ and $ 1 \not \equiv 0 \pmod k$ (or equivalently because $ k$ doesn't divide $ n \minus{} 1$), in [i]every[/i] row above the first, we have some vertical block that isn't terminating at that row.  This actually proves something a fair bit stronger than the original claim, I believe, assuming no errors on my part :)

You can do this in a completely algebraic way.  The if part is clear.  The only if part can be done by letting $ \varepsilon\equal{}e^{i\frac{2\pi}{k}}$.  By assigning the number $ \varepsilon^{i\plus{}j}$ to the square $ (i,j)$, where $ (1,1)$ is the bottom left corner and $ (n,n)$ is the top right corner, we must have $ \sum_{i,j}\varepsilon^{i\plus{}j}\equal{}\varepsilon^{n\plus{}1}\iff (\varepsilon^{n\plus{}1}\minus{}1)(\varepsilon^{n\minus{}1}\minus{}1)\equal{}0$.  Therefore, $ k|n\minus{}1$ or $ k|n\plus{}1$.  The latter case is impossible, and so $ k|n\minus{}1$, as required. $ \Box$

[quote="boxedexe"]You can do this in a completely algebraic way.  The if part is clear.  The only if part can be done by letting $ \varepsilon \equal{} e^{i\frac {2\pi}{k}}$.  By assigning the number $ \varepsilon^{i \plus{} j}$ to the square $ (i,j)$, where $ (1,1)$ is the bottom left corner and $ (n,n)$ is the top right corner, we must have $ \sum_{i,j}\varepsilon^{i \plus{} j} \equal{} \varepsilon^{n \plus{} 1}\iff (\varepsilon^{n \plus{} 1} \minus{} 1)(\varepsilon^{n \minus{} 1} \minus{} 1) \equal{} 0$.  Therefore, $ k|n \minus{} 1$ or $ k|n \plus{} 1$.  The latter case is impossible, and so $ k|n \minus{} 1$, as required. $ \Box$[/quote]
And why $ k|(n\plus{}1)$ is impossible?

Wow. A terribly unjustified statement.  I assumed $ n\plus{}1$ was a prime for some unknown reason and excluded the trivial case $ k\equal{}1$.

Thanks JBL. Now I know what I should have done.


logarithms
floor function
ceiling function
algebra unsolved
algebra

Sequence $ {a_n,n \equal{} 1,2,3,...}$.
We know that
$ a_1 \equal{} 1$,
$ a_{n \plus{} 1} \equal{} a_n \plus{} \sqrt {a_n} \plus{} \frac {1}{2},n \equal{} 0,1,2,...$.

Find the formula of $ {a_n}$.

I can prove only acimptotic formul $ \sqrt{a_n}\equal{}\frac n2 \plus{}\frac{\ln n}{4}\plus{}c\plus{}O(\frac{\ln n}{n})$ for some constant c.

[quote="t0rajir0u"]Do you have any reason at all to suspect that this sequence has a nice closed form?[/quote]

The problem is a really contest problem I heard that I can't solve it.

[quote="t0rajir0u"]Are you sure it isn't $ a_{n \plus{} 1} \equal{} \left\lfloor a_n \plus{} \sqrt {a_n} \plus{} \frac {1}{2} \right\rfloor$?  That one actually has a nice answer and it would explain the half.[/quote]

I can't be sure.I heard this problem from someone who also only heard from sombody!

By the way,how about(how to solve) your sequence?

$ a_n$ turns out to be of the form $ \left\lceil \frac {an^2 \plus{} bn \plus{} c}{d} \right\rceil$ and then it remains only to compute the relevant coefficients.


1)   We take the real positivs numbers x,y,z,t such that x+y+z+t=1.Prove that sqrt(x^2y^2+z^2t^2)+sqrt(x^2t^2+y^2z^2)>=(sin45)/4.
2)   Find the natural number n such that (2^n+4^n+...+2006^n)/(1^n+3^n+...+2005^n) is also natural.

Solution to problem 2:  n = 0.  Natural since 2005/2005 = 1.  Did you mean positive integer solution?

The previous posting should read:  1003/1003 = 1 is a natural number.

In number theory, natural numbers are usually taken to be ${1,2,3...}$. So natural numbers and positive integers are the same.

(And yes, it seems silly to me too, so let's not have an arguement over it)

And, in future, you may wish to the use the edit function rather than double posting  :)

I am really curious about the flaw which either exists in this argument or in the intentions of the problem setter:  Feedback appreciated!

The numerator can be factorised as 2^n(1^n + 2^n + ...+1003^n).
There are 1003 odd terms in the denominator (all terms are powers of odd numbers in the sequence 2k-1 for k = 1, 2, ..., 1003), so the denominator itself is odd.
Odd numbers do not have even factors.  Therefore the denominator has no proper factor in common with 2^n.  All factors of the denominator are factors of the numerator.  No proper factor of the denominator is a factor of 2^n.  Therefore all factors of the denominator are factors of (1^n + 2^n + ... + 1003^n).  Therefore the denominator, 1^n + 3^n + ...2005^n must divide (1^n + 2^n + ...1003^n).
But that is absurd unless n = 0 since both expressions have the same number of terms, and every term except the first is greater in the denominator, the first term being equal (to 1) in both cases.  The denominator is strictly greater than an expression it must divide exactly.
As this is clearly impossible, no such n > 0 exists.

In brief:
(i)  The denominator is odd so has nothing in common with the 2^n factor in the numerator.
(ii)  But the other factor in the numerator is smaller than the denominator.
(iii) There must therefore be odd primes left in the denominator after all
possible cancellation has been carried out, so the answer whatever it is is not an integer.

For problem 1, the inequality is false when t and y are really really tiny.

However, the real reason for my posting here was not to tell you that, but rather to say:

BRITISH!!!!


algebra
function
domain
number theory
prime factorization
Ring Theory
Galois Theory

Lef $ f$ in $ K[x]$ be irreducible and let $ L/K$ be a normal field extension. Show that if $ g, h$ are normalized prime factors of $ f$ in $ L[x]$, there exist $ \sigma \in G(K/L)$ such that $ \sigma g \equal{} h$

Deduce that in the prime factorization of $ f$ in $ L[x]$ all prime factors have the same exponent.

Reduce to the case where $ L/K$ is separable. Then $ L/K$ can be assumed to be Galois with group $ G\equal{}Gal(L/K)$. Hence $ G$ operates $ K[X]$-linearly on the Dedekind domain $ L[X]$ with $ Fix_G(L[X])\equal{}K[X]$. By assumption, $ gL[X]$ and $ hL[X]$ are prime ideals of $ L[X]$ lying above the prime ideal $ fK[x]$ of $ K[X]$. By Galois theory of Dedekind domains, the prime ideals $ gL[X]$ and $ hL[X]$ are conjugated by some $ \sigma\in G$. Because $ g$ and $ h$ are normalized it follows that they are conjugated by $ \sigma$ as elements in $ L[X]$.

The additional claim follows trivially.

What is going on with this problem?

[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=244275]just 2 days ago[/url]

Interestingly enough, this wants you to deduce that the prime factors are of the same degree, whereas I wanted to use that fact to prove this.

Ouups, I was not aware of the fact of this double posting.

@ son of Skanderbeg: This should be avoided in any case.  :mad:


LaTeX
inequalities unsolved
inequalities

For a,b,c,d>0 and abcd=1 we have
$ \frac{a}{b}$+$ \frac{b}{c}$+$ \frac{c}{d}$+$ \frac{d}{a}$<a+b+c+d
Prove that $ \frac{b}{a}$+$ \frac{c}{b}$+$ \frac{d}{c}$+$ \frac{a}{d}$>a+b+c+d

My Latex in the last version no good :lol: I write new
[b][i]We have[/i][/b] $ a,b,c,d>0$ [i]and[/i] $ abcd\equal{}1$
[b]Prove that[/b]
If $ a/b\plus{}b/c\plus{}c/d\plus{}d/a<a\plus{}b\plus{}c\plus{}d$ then
$ b/a\plus{}c/b\plus{}d/c\plus{}a/d>a\plus{}b\plus{}c\plus{}d$


inequalities
geometry
search
inequalities unsolved

$ a,b,c$ are the side lengths of $ \triangle ABC$, $ S$ is the area of $ \triangle ABC$

prove:

$ \frac{ab\plus{}bc\plus{}ca}{4S}\geq\sqrt{3}$

Similar with the well known Weiscnbock ineq


I noticed some participants are already registrated, but I can only see the number of participants but not the link that lead to the pics and names etc...
   when we can see the participants? we must wait until all register??

Click on the country code? (at http://www.imo-official.org/year_reg.aspx )

thanks... i was trying to do this on the http://www.imo-2008.es haha


Prove that for every integer $ n\geq 3$, the equation $ x^{n}+y^{n}=z^{n+1}$ has infinitely many solutions in positive integers $ x,y,z$

From the eye you can see that $ (2,2,2)$ is a solution. Afterwards if $ \alpha \in \mathbb N$ then $ (2\cdot \alpha^{n+1},2\cdot \alpha^{n+1},2\cdot \alpha^{n})$ si also a solution.


From a point within a triangle, line segments are drawn to the vertices. A necessary and sufficient condition that the three triangles thus formed have equal areas is that the point be:

$ \textbf{(A)}\ \text{the center of the inscribed circle} \qquad \\
\textbf{(B)}\ \text{the center of the circumscribed circle}\qquad \\
\textbf{(C)}\ \text{such that the three angles fromed at the point each be }{120^\circ}\qquad \\
\textbf{(D)}\ \text{the intersection of the altitudes of the triangle}\qquad \\
\textbf{(E)}\ \text{the intersection of the medians of the triangle}$

[hide="My Solution"]
It is known that medians form equal areas so the answer is $ \fbox{(E)}$.
[/hide]

Answering the questions you post doesn't really give anyone else a chance to respond.

Are all these problems for the resources page or something?  :maybe:

@grn_trtle: Every ten problems I posted, I went through problems that weren't answered and added the solution. So it's not like I completely didn't give anybody a chance to post.

@ThinkFlow: Yeah.

the rihgt answer is last one. because median bisects an area of triangle.

Ok the right answer is letter A because the centroid of a triangle divides the triangle in three triangles with equal areas, but the ''if and only if'' is not that easy, so  a rigorous proof to this problem (as geometry requests) has not been posted yet on this topic.

I think you mean letter E, because letter A involves the incenter.

Let the triangle be ABC, with medians AP,BQ, and CR and centroid N. I can prove the centroid is sufficient.
Proof 1: [BRC]=[ABP]=1/2[ABC], so subtracting [BRNP] from both sides, [ARN]=[NPC]. Similarly for the other three. But [BNP]=[NPC] since they have the same base and height, so all six triangles BNP, PNC, PNQ, QNA, ANR, and RNB have the same area. So [BNC]=[CNA]=[ANB].
Proof 2: It is well-known that AN/NP=2. This can be proved either by Menelaus or by the fact that the medial triangle, POR, is similar to the original triangle with ratio 1/2. So dropping perpendiculars from A and N to BC, we find the perp from A is three times the perpendicular from N, so [NBC]=1/3[ABC]. Similarly for the other three triangles.

Now to prove it is necessary.
Drop perpendiculars U and V from N to BC and AC, respectively. We must have $ NU \cdot BC\equal{}NV \cdot AC, or \frac{NU}{NV}\equal{}\frac{BC}{AC}$= a constant for the areas to be equal. From here it follows that N must lie on CR. Similarly it must lie on BQ and AP, so we are done.


linear algebra
matrix
function
linear algebra unsolved

$(a(i,j))\in M(n,R)$ such that $a(i,j)+a(j,k)+a(k,i)=0$ for any i,j,k

Prove that there exist a sequence $t(1),...,t(n)$ such that $a(i,j)=t(i)-t(j)$ for
all i,j

I think it is something related to graph theory. For example, if one consider
an electronique circuit, with $n$ vertices, each vertice connected to every others.
An oriented edge from $i$ to $j$ has a number that represents a difference of potential. All
we know here is that, on every $3$-cycle, the sum of the differences of potential is null. And what we
want to prove is that there actually exists a potential function, that is a number $t(i)$ for each vertices, such that
the number on the edge from $i$ to $j$ is precisely $t(j)-t(i)$.

so one can check that:
$t(1)=0$ and $t(i)=a(1,i)$ for $i \geq 2$ works ...


I can't figure out this one on the 1996 AIME

Find the average value of |a1 - a2| + |a3 - a4| + |a5 - a6| + |a7 - a8| + |a9 - a10| for all permutations a1, a2, ... , a10 of 1, 2, ... , 10.


calculus
inequalities
inequalities proposed

And I think it's harder than the old result: $ \sum\ \frac{a^2\plus{}b^2}{a\plus{}b} \geq\ \sqrt{3(a^2\plus{}b^2\plus{}c^2)}$
 :)

this inequality is equivalent to :
$ \frac{3}{\sqrt{2}} \geq \sum\frac{cos^2x}{siny\plus{}sinz}$

[quote="Evariste-Galois"]this inequality is equivalent to :
$ \frac {3}{\sqrt {2}} \geq \sum\frac {cos^2x}{siny \plus{} sinz}$[/quote]
Can you post your full proof? Thank you  :)

[quote="nguoivn"]Given $ a, b, c \geq\ 0$. Prove that:
$ \frac {3}{\sqrt {2}}.\sqrt {a^2 \plus{} b^2 \plus{} c^2} \geq\ \frac {a^2 \plus{} b^2}{a \plus{} b} \plus{} \frac {b^2 \plus{} c^2}{b \plus{} c} \plus{} \frac {c^2 \plus{} a^2}{c \plus{} a}$

PS: Please don't use calculus  :)[/quote]

My solution :)
Assume:
$ a \ge b \ge c$
We have:
$ RHS \le \frac{a^2\plus{}b^2}{a\plus{}b}\plus{}a\plus{}b \le 3\sqrt{\frac{a^2\plus{}b^2}{2}} \le LHS$
Done! :)

[quote="caubetoanhoc94"][quote="nguoivn"]Given $ a, b, c \geq\ 0$. Prove that:
$ \frac {3}{\sqrt {2}}.\sqrt {a^2 \plus{} b^2 \plus{} c^2} \geq\ \frac {a^2 \plus{} b^2}{a \plus{} b} \plus{} \frac {b^2 \plus{} c^2}{b \plus{} c} \plus{} \frac {c^2 \plus{} a^2}{c \plus{} a}$

PS: Please don't use calculus  :)[/quote]

My solution :)
Assume:
$ a \ge b \ge c$
We have:
$ RHS \le \frac {a^2 \plus{} b^2}{a \plus{} b} \plus{} a \plus{} b \le 3\sqrt {\frac {a^2 \plus{} b^2}{2}} \le LHS$
Done! :)[/quote]
Nice proof, caubetoanhoc94  :)
My proof is for the stronger:
$ \frac{6(a^2\plus{}b^2\plus{}c^2)(a\plus{}b\plus{}c)}{3\sum\ a^2\plus{}2\sum\ ab} \geq\ \sum\ \frac{a^2\plus{}b^2}{a\plus{}b}$
 :)


A circle with center O is tangent to the sides AB, BC, CA and DA of a convex quadrilateral ABCD at K, L, M, and N respectively. The lines KL and MN intersect at point S. Prove that BD is perpendicular to OS.

I didn't consider the problem to be so difficult. However, I hope you'll enjoy solving it and hopefully help me out with the solution. Thank you in advance!

[quote="Tiger100"]just see that the line BD is the polar of S in the circle (O) then the result comes .  :wink: , You can read here for further questions about polars http://mathworld.wolfram.com/Polar.html[/quote]

I have heard something about poles and polars but I would never think of using them. The site You recommended seems really useful. Thank you very much!


[quote="ramox"]let  $x ,y$ such that  $1 \le x^{2}+y^{2}-xy \le 2$

prove that : $x^{4}+y^{4}\ge \frac{2}{9}$ .[/quote]
Why $x^{2}+y^{2}-xy \le 2$ $?$
I think, $x^{4}+y^{4}\ge \frac{2}{9}$ is true for all $x$ and $y$ such that $1 \le x^{2}+y^{2}-xy$ only. :wink:

The problem was to prove that  $\frac{2}{9}\le x^{4}+y^{4}\le 8$ .
$x, y$ real numbers such that $1\le x^{2}+y^{2}-xy \le 2$ .
the other inequality is easy to prove ...


calculus
derivative
calculus computations

Find the 1000th derivative of f(x) = x*e^(-x). I did the problem but i just want to check my answer.

i found the 1st two derivatives and i got:

f '(x) = e^(-x) - [x*e^(-x)]

f ''(x) = -2e^(-x) + x*e^(-x)

so i noticed the coefficient of the first term is equal to the order of the derivative and that the signs of both terms alternate and concluded that the 1000th derivative would have 1000 as the coefficient in the first term and that since 1000 is an even number, the signs are the same as what appears in the 2nd derivative and got:

1000th derivative of f(x) = -1000e^(x) + x*e^(-x)

Is this correct?

Yes. Prove it by induction.
$f^{(n)}(x)=(-1)^{n}(x-n)e^{-x}$

There's Leibnitz's rule
\[(f(x)g(x))^{(n)}=\sum_{k=0}^{n}\binom nkf^{(k)}(x)g^{(n-k)}(x)\]
You have f(x)=x, hence $f^{(k)}(x)=0$, k>1. This is a general method for such problems.


Let $ a,b,c \in R^\plus{}$ and $ ab\plus{}bc\plus{}ca \equal{} 3$
Prove that \[ \sum_{cyclic}\frac{a^3}{bc(b\plus{}c)^3} \geq \frac{3}{8}\]

[hide]Hint:
Using the Chebyshev inequality after assuming that a>=b>=c[/hide]

Multiplying the LHS by $ ab\plus{}bc\plus{}ca$ and the RHS by $ 3$ homogenizes the inequality, right?

[quote="iwishiwasgoodatmath"]Multiplying the LHS by $ ab \plus{} bc \plus{} ca$ and the RHS by $ 3$ homogenizes the inequality, right?[/quote]

Right.

I would just do the problem with the AM-GM inequality...

[hide="Solution"]
By AM-GM:
$ \sum_{cyclic}\frac {a^3}{bc(b \plus{} c)^3} \geq \frac{3abc}{(a\plus{}b)(a\plus{}c)(b\plus{}c) \sqrt[3]{a^2b^2c^2}}$

Again by AM-GM, we have:
$ 1\equal{}\frac{ab\plus{}ac\plus{}bc}{3}\geq \sqrt[3]{a^2b^2c^2}$

Thus,
$ \frac{3abc}{(a\plus{}b)(a\plus{}c)(b\plus{}c) \sqrt[3]{a^2b^2c^2}} \geq \frac{3abc}{(a\plus{}b)(a\plus{}c)(b\plus{}c)}$

Clearly, we must prove that $ \frac{abc}{(a\plus{}b)(a\plus{}c)(b\plus{}c)}\geq \frac{1}{8}$, which is shown, yet again, through AM-GM...

$ (a\plus{}b)(a\plus{}c)(b\plus{}c)\equal{}a^2b\plus{}a^2c\plus{}b^2a\plus{}b^2c\plus{}c^2a\plus{}c^2b\plus{}abc\plus{}abc \geq 8abc$.

That completes the proof...[/hide]

I think this is my first post in the Pre-Olympiad Section  :)

[quote="krsattack"]I would just do the problem with the AM-GM inequality...

[hide="Solution"]
By AM-GM:
$ \sum_{cyclic}\frac {a^3}{bc(b \plus{} c)^3} \geq \frac {3abc}{(a \plus{} b)(a \plus{} c)(b \plus{} c) \sqrt [3]{a^2b^2c^2}}$

Again by AM-GM, we have:
$ 1 \equal{} \frac {ab \plus{} ac \plus{} bc}{3}\geq \sqrt [3]{a^2b^2c^2}$

Thus,
$ \frac {3abc}{(a \plus{} b)(a \plus{} c)(b \plus{} c) \sqrt [3]{a^2b^2c^2}} \geq \frac {3abc}{(a \plus{} b)(a \plus{} c)(b \plus{} c)}$

Clearly, we must prove that $ \frac {abc}{(a \plus{} b)(a \plus{} c)(b \plus{} c)}\geq \frac {1}{8}$, which is shown, yet again, through AM-GM...

$ (a \plus{} b)(a \plus{} c)(b \plus{} c) \equal{} a^2b \plus{} a^2c \plus{} b^2a \plus{} b^2c \plus{} c^2a \plus{} c^2b \plus{} abc \plus{} abc \geq 8abc$.

That completes the proof...[/hide]

I think this is my first post in the Pre-Olympiad Section  :)[/quote]
but $ \frac{abc}{(a\plus{}b)(b\plus{}c)(c\plus{}a)} \leq \frac{1}{8}$  :wink:

[quote="apollo"]Let $ a,b,c \in R^ \plus{}$ and $ ab \plus{} bc \plus{} ca \equal{} 3$
Prove that
\[ \sum_{cyclic}\frac {a^3}{bc(b \plus{} c)^3} \geq \frac {3}{8}
\]
[/quote]
I'm surprised that this way works  :)

by AM-GM we have
$ \frac {a^3}{bc(b \plus{} c)^3} \plus{} \frac {b \plus{} c}{16abc} \plus{} \frac {b \plus{} c}{16abc} \plus{} \frac {b \plus{} c}{16abc} \geq \frac {1}{2bc}$
creating 2 other similar inequality, we have :
$ LHS \plus{} \frac {3}{8}(\frac {1}{ab} \plus{} \frac {1}{bc} \plus{} \frac {1}{ac}) \geq \frac {1}{2}(\frac {1}{ab} \plus{} \frac {1}{bc} \plus{} \frac {1}{ac})$ , so that $ LHS \geq \frac{1}{8}(\frac {1}{ab} \plus{} \frac {1}{bc} \plus{} \frac {1}{ac})$
and by Cauchy-S we have $ \frac {1}{8}(\frac {1}{ab} \plus{} \frac {1}{bc} \plus{} \frac {1}{ac}) \geq \frac {1}{8}\frac {9}{ab \plus{} bc \plus{} ac} \equal{} \frac {3}{8}$

[quote="HTA"][quote="apollo"]Let $ a,b,c \in R^ \plus{}$ and $ ab \plus{} bc \plus{} ca \equal{} 3$
Prove that
\[ \sum_{cyclic}\frac {a^3}{bc(b \plus{} c)^3} \geq \frac {3}{8}
\]
[/quote]
I'm surprised that this way works  :)

by AM-GM we have
$ \frac {a^3}{bc(b \plus{} c)^3} \plus{} \frac {b \plus{} c}{16abc} \plus{} \frac {b \plus{} c}{16abc} \plus{} \frac {b \plus{} c}{16abc} \geq \frac {1}{2bc}$
creating 2 other similar inequality, we have :
$ LHS \plus{} \frac {3}{8}(\frac {1}{ab} \plus{} \frac {1}{bc} \plus{} \frac {1}{ac}) \geq \frac {1}{2}(\frac {1}{ab} \plus{} \frac {1}{bc} \plus{} \frac {1}{ac})$ , so that $ LHS \geq \frac {1}{8}(\frac {1}{ab} \plus{} \frac {1}{bc} \plus{} \frac {1}{ac})$
and by Cauchy-S we have $ \frac {1}{8}(\frac {1}{ab} \plus{} \frac {1}{bc} \plus{} \frac {1}{ac}) \geq \frac {1}{8}\frac {9}{ab \plus{} bc \plus{} ac} \equal{} \frac {3}{8}$[/quote]
Nice solution  :coolspeak:
There is another Nice solution with Holder's Inequality.

Well using Holder was a little more suggested by the problem than AM-GM

[hide]
\[ (1\plus{}1\plus{}1)(bc\plus{}ca\plus{}ab)\left( \sum{\frac{a^3}{bc(b\plus{}c)^3}}\right) \ge \left (\sum{\frac{a}{b\plus{}c}}\right) ^3\ge \frac{27}{8}\]
[/hide]

Nice inequality by the way
:)


superior algebra
superior algebra unsolved

here

[img]http://www.geocities.com/zettz1/alg.jpg[/img]

thanks...

BLANK = prime. Now try to prove it.

For the first part, assume that an algebraic $ \alpha$ is a root of $ a_n x^n \plus{} ... \plus{} a_0$ for rational $ a_i$. Multiply this with a common denominator of all $ a_i$, thus we may assume that all $ a_i$ are integers. Then multiply again, this time by $ a_n^{n\minus{}1}$ and show that $ a_n\alpha$ is an algebraic integer.

Suppose  a=bc
i
then N(a)= N(bc)= N(b) N(c)= prime

which implies that N(b) or N(c) is a 1 which implies that b or c is a unit which implies that a is irreducible.

does that work?


Let $f(n)$ be the number of lattice points in the region $2|x|+|y|\le n$. Find the formula for $f(n)$.

my answer...
[hide]
Assuming that the lattice points on the borders count, $f(n)=n^2+n+1$[/hide]

[quote="kunny"]We are to consider $ f(n+1)-f(n) $.[/quote]
Yes, that's the correct approach.
Does anyone have a complete solution to share?


The polynomials $ P(x)$ and $ Q(x)$ are given. It is known that for a certain polynomial $ R(x, y)$ the identity $ P(x) \minus{} P(y) \equal{} R(x, y) (Q(x) \minus{} Q(y))$ applies. Prove that there is a polynomial $ S(x)$ so that $ P(x) \equal{} S(Q(x)) \quad \forall x.$

I  solve it firstly in case q has no double roots.
(1) Let p(x)=q(x)c(x)+r(x).
Then writing the equation for z[1]..z[deg[q]] solutions of q(x)=0 we get
r(x) is constant.
Performing (1) for c(x) and so on, we get the desired result
So suppose q has double root.
Now we substitute q(x)=q'(x)+c to have q'(x) no double roots.
The condition follows/
But I can't prove at the moment such q' exists ( though it seems obvious)

http://www.bath.ac.uk/~masgcs/ukimo2004/cc2sol.pdf here is a solution

Is it really true that for any polynomial Q(x) there is some c such that Q(x)+c doesn't have double roots?
It sounds natural but how to prove it? (Look above at iura's solution)

Of course it's true: every double root is also a root of the derivative. Since the derivative remains unchanged if you add a constant $c$ to the polynomial, simply add a constant $c$ s.t. $P(x)+c$ is non-zero in all the roots of the derivative $P'$.

Does this work?
Let $P(x)=Q(x)T(x)+U(x)$, where $\deg U < \deg Q$.
Then, $Q(x)-Q(y) | Q(x)T(x)+U(x)-Q(y)T(y)-U(y)$ i.e. $Q(x)-Q(y) | Q(y)(T(x)-T(y))+U(x)-U(y)$. Let $r$ be a root of $Q$. Substitute $y=r$, implying that $Q(x)-Q(r) | Q(r)(T(x)-T(r))+U(x)-U(r)$ i.e. $Q(x) | U(x)-U(r)$. However, $\deg U < \deg Q$, which means that $U$ is a constant.
Thus, $Q(x)-Q(y) | Q(y)(T(x)-T(y))$. However, since $\gcd (Q(x)-Q(y), Q(y)) = \gcd(Q(x), Q(y)) = 1$, $Q(x)-Q(y) | (T(x)-T(y))$, which means that we are back where we started, so we can repeat our process until $T=$ constant, in which case we are done.

Let $P(x)=\sum_{i=0}^np_ix^i$ ,$Q(x)=\sum_{i=0}^mq_ix^i$, W.L.O.G $p_n,q_n>0$. Let $\omega$ be a primitive $m-th$ root of unity. Substitute $y=\omega x$. The degree of $Q(\omega x)-Q(x)$ is strictly less than $m$, therefore the degree of $P(\omega x)-P(x)$ is less than $n$ ( $R(x,\omega x)$ may be the zero polynomial ), therefore  $p_n \omega^n -p_n=0$, which implies $m|n$. Let $P_1(x)=P(x)-\frac{p_n}{q_n^{n/m}}(Q(x))^{n/m}$, $R_1(x,y)=\left(R(x,y)-\frac{p_n}{q_n^{n/m}}\frac{(Q(x))^{n/m}-(Q(y))^{n/m}}{Q(x)-Q(y)}\right)$. We have $P_1(x)-P_1(y)=R_1(x,y)(Q(x)-Q(y))$ and $\deg(P_1)<\deg(P)$ and hence by induction hypothesis we are done !

Does anyone know if my solution to this problem is correct? Thank you so much for all your help!

Cool, thank you very much for the confirmation, r3mark! I really appreciate it!

We have $Q(x)-Q(y) \mid P(x)-P(y)$. Use the Division Algorithm to write $P(x)=A(x)Q(x) + B(x)$ for some $A,B\in \mathbb{R}[x]$ with $\deg B < \deg Q$. We have
\[ Q(x)-Q(y) \mid \big[A(x)Q(x)-A(y)Q(y) \big]+\big[B(x)-B(y)\big] \qquad (\clubsuit). \]
Let $\{r_1,\ldots,r_k\}$ be the set of distinct roots of $Q$, and let $Q(x) = (x-r_1)^{a_1}\cdots (x-r_k)^{a_k}$ for some $(a_i) \in \mathbb{Z}_{\ge 1}$. If $r$ and $s$ are roots of $Q$, then plugging in $x=r$ and $y=s$ above implies $B(r)=B(s)$. Hence $B(r_1)=\cdots=B(r_k)$. Therefore,
\[ B(x)-c = (x-r_1)^{b_1}\cdots (x-r_k)^{b_k} \qquad (\spadesuit)\]
for some constant $c\in \mathbb{R}$ and for some $(b_i)\in \mathbb{Z}_{\ge 1}$.
-----------------------
From $(\clubsuit)$, we get
\[ Q(x)-Q(y) \mid \big[A(x)-A(y) \big]Q(x) +\big[B(x)-B(y)\big]. \]
We know $x-y\mid P(x)-P(y)$ for any polynomial $P$. Hence we can write -- as a polynomial statement -- the following:
\[ \frac{Q(x)-Q(y)}{x-y} \Biggr| \left[ \frac{A(x)-A(y)}{x-y} \right ]Q(x) +\left[\frac{B(x)-B(y)}{x-y}\right]. \]
Taking the limit as $y$ approaches $x$, the above implies
\[ Q'(x) \mid A'(x)Q(x) + B'(x) \implies A'(x)Q(x)+B'(x) = Q'(x)S(x,y) \qquad (\heartsuit)\]
for some $S(x,y) \in \mathbb{R}[x,y]$.
----------------------
[color=#00f][b]Lemma: [/b][/color]If $Q$ has a root $r$ with multiplicity $n$, then $r$ is also a root of $B$, and it has multiplicity at least $n$.

[color=#00f][i]Proof:[/i][/color] Replace $n$ with $n+1$. We know $Q(r)=0$ and $Q'(r)=0$, and so on till $Q^{(n)}(r)=0$.  Taking $n-1$ derivatives of $(\heartsuit)$, every term on the LHS (except $B^{(n)}(x)$) and every term on the RHS will have a term $Q^{(m)}(x)$ for some $m\in \{0,\ldots,n\}$, and is hence $0$. Therefore, $B^{(n)}(x)=0$, and also by the same logic $B^{(m)}(x)=0$ for any $m \le n$. Therefore, $B$ has a root at $r$ with multiplicity $n$ as well. $\blacksquare$

[color=#00f][b]Claim: [/b][/color] $B(x)$ is a constant polynomial.

[color=#00f][i]Proof:[/i][/color] Suppose $Q$ has a root $r_i$ with multiplicity $a_i\ge 2$. By the Lemma, so does $b$, so $(x-r_i)^{a_i} \mid B(x)$. Hence $B'(x)$ has a root at $r_i$ with multiplicity at least $a_i-1$. By $(\spadesuit)$, $B'(x)$ has a root at $r_i$ with multiplicity $b_i-1$. Therefore, $b_i-1\ge a_i-1$, so $b_i\ge a_i$. Summing over all $i$, we get $\deg B \ge \deg Q$, contradiction. Therefore, we must have $B(x)=c$ in fact, since the $0$ polynomial has any number as a root, and any root has infinite multiplicity. $\blacksquare$
--------------------
Now we are almost done. We claim $A(x)$ also works in the place of $P(x)$ in the original statement. Indeed, since $P(x)=A(x)Q(x)+c$, and $Q(x)-Q(y) \mid P(x)-P(y)$, we have $Q(x)-Q(y) \mid A(x)Q(x)-A(y)Q(y)$, which implies $Q(x)-Q(y) \mid [A(x)-A(y)]Q(x)$. But $\gcd(Q(x),Q(x)-Q(y)) = \gcd(Q(x),Q(y))=1$ since they are polynomials purely in different variables. Therefore, $Q(x)-Q(y)\mid A(x)-A(y)$, as claimed. Now we can continue the process, and backtracking once the process is finished, we get that $P(x)$ is a polynomial in $Q(x)$ (the process essentially builds up the polynomial in $Q$, increasing the degree by $1$ at each step).

[hide=Remarks]
The main ideas and motivation of the solution are:
[list]
[*] Writing $P=AQ+B$ using Division Algorithm. This is motivated since the final result of the problem is that we want to show $P$ is a polynomial in $Q$, so we want to ``build up'' $P$ in terms of $Q$, motivating using Division Algorithm. We want to show $B$ is a constant.
[*] After manipulating the above into $(\clubsuit)$, the idea is to plug in $r$ and $s$ if they are two roots of $Q$ -- this gets rid of everything except $B(r)-B(s)$, so it is also $0$. From here, we have that the set of roots of $B$ (translated by a constant) and $Q$ is the same. This is a pretty key step, and a trick the problem uses a lot -- vanish a lot of stuff so the remaining stuff is also $0$.
[*] Now that the set of roots is the same, the hard part of the problem is dealing with multiplicity. The best way to deal with multiplicity in general is through derivatives. We get $Q' \mid A'Q+B'$ after some trickery. We need to use the fact that $\deg Q > \deg B$, and using $(\heartsuit)$, we can prove that the multiplicities of the roots of $B$ and $Q$ are the same; not just that they have the same set of roots. Now we are basically done.
[/list]
[/hide]

Here is a solution based on looking at coefficients. Please let me know if it doesn't work.

We induct on $\deg P$, keeping $Q$ constant. WLOG suppose that both $P$ and $Q$ are monic.
Suppose that $\deg Q=d$. The main claim is that $d \mid \deg P$. Suppose the terms of $R(x,y)$ with maximum degree are $r(x,y)$. Then the maximum degree terms of $R(x,y)(Q(x)-Q(y))$ are $(x^d-y^d)r(x,y)$, which should also be $x^r-y^r$ for some $r$ by considering $P(x)-P(y)$. Evidently there must be an $x^k$ term in $r(x,y)$, else $y \mid (x^d-y^d)r(x,y)$ in the polynomial sense—which is untrue. Then there is a $-x^ky^d$ term in the (unsimplified) representation of $P(x)-P(y)$, but this is illegal so it must be cancelled out by an $x^{k-d}y^d$ term in $r(x,y)$ (since $x^d(x^{k-d}y^d)-y^d(x^k)=0$). Then there is a $-x^{k-d}y^{2d}$ term in $P(x)-P(y)$, so it must be cancelled out by an $x^{k-2d}y^{2d}$ term in $r(x,y)$, and so on. This process only terminates when $k-di=0$; for example, if $k-di=d-1>0$ at some point, then we still need an $x^{k-d(i+1)}y^{d(i+1)}$ term in $r(x,y)$, which is illegal. This readily implies that $d \mid k$. Since $\deg P = d+k$, our claim is true.
We are now ready to induct. Suppose that some (monic) $P_0$ exists with $\deg P_0=ad$ and $P_0(x)-P_0(y)=R_0(x,y)(Q(x)-Q(y))$. Let $P(x)=P_0(x)-Q(x)^a$, which has degree at most $ad-1<\deg P_0$. It is clear that $Q(x)-Q(y) \mid Q(x)^a-Q(y)^a$, so $P(x)-P(y)=R(x,y)(Q(x)-Q(y))$ for some suitable choice of $R$ as well, completing the inductive step.
It remains to take care of the base case of $\deg P = 0$, but this is trivial; thus we are done. $\blacksquare$

Firstly , note that for each arbitrary complex number $\alpha \in \mathbb{C}$ we have $Q(x)-Q(\alpha)|P(x)-P(\alpha)$ and for each two polynomials with this property , we'll show that we have $deg(Q)|deg(P)$. So suppose that for polynomial $P_{0}(x) \in \mathbb{C}[X]$ we have :
$$P(x)=P_{0}(x)(Q(x)-Q(\alpha))+P(\alpha)$$
Now we'll show that for each number $\beta \in \mathbb{C}$ that $Q(\beta) \neq Q(\alpha)$ , we have $Q(x)-Q(\beta)|P_{0}(x)-P_{0}(\beta)$ and thus , one can see that :
$$Q(x)-Q(\beta)|P(x)-P(\beta)=P_{0}(x)(Q(x)-Q(\alpha))+P(\alpha)-P(\beta) \implies P_{0}(x)(Q(x)-Q(\alpha)) \equiv P_{0}(\beta)(Q(x)-Q(\alpha)) \pmod{Q(x)-Q(\beta)}$$
So while polynomials $Q(x)-Q(\alpha)$ and $Q(x)-Q(\beta)$ are coprime , as the result we can get :
$$P_{0}(x) \equiv P_{0}(\beta) \pmod {Q(x)-Q(\beta)}$$
Note that the equality $deg(P_{0})=deg(P)-deg(Q)$ holds , then by defining polynomials $P_1(x) , P_2(x) , ...$ in the same way , for a number $k \in \mathbb{N}$ we have $deg(P_k) < deg(Q)$ and while for a number $\omega \in \mathbb{C}$ we have $Q(x)-Q(\omega)|P_k(x)-P_k(\omega)$ , thus polynomial $P_k(x)$ is constant which implies $deg(Q)|deg(P)$ obviously.

Now suppose that we have $P(x)=a_{nk}x^{nk}+...+a_1x+a_0$ and $Q(x)=b_nx^n+...+b_1x+b_0$. Then we define polynomial $P_0(x)$ such as below :
$$P_0(x)=P(x)-\frac{a_nk}{b_{n}^{k}}Q(x)^{k}$$
And according to this defining , note that for each number $\alpha \in \mathbb{C}$ , polynomial $Q(x)-Q(\alpha)$ divides $P_0(x)-P_0(\alpha)$ too , because one can see that :
$$P(x)-P(\alpha) \equiv \frac{a_{nk}}{b_{n}^{k}}(Q(x)^k-Q(\alpha)^k) \equiv 0 \pmod {Q(x)-Q(\alpha)}$$
So we have $deg(Q)|deg(P_0)$ and by defining polynomials $P_1 , P_2 , ...$ in the same way , while $deg(P_{t+1}) < deg(P_t)$ there exist a number $t \in \mathbb{N}$ such that $P_t(x) \equiv 0$ and for some complex numbers $c_k , ... , c_0 \in \mathbb{C}$ one can see that :
$$P_t(x)=P(x)-c_kQ(x)^k-c_{k-1}Q(x)^{k-1}-...-c_0=0 \implies S(x)=c_kx^k+c_{k-1}x^{k-1}+...+c_0 , P(x)=S(Q(x))$$
So we're done.

Solved with [b]everythingpi3141592[/b]
By the divisibility condition we have that when $Q(x) = Q(y) \implies P(x) = P(y)$, which essentially helps us prove that all turning points of $Q(x)$ are of $P(x)$, now if we construct a polynomial $S(x)$ such that when $P'(x) = 0 \implies S'(Q(x)) = 0$ (that is the turning points of $P(x)$ which aren't of $Q(x)$ are of $S(Q(x))$), which is just a matter of langrage interpolation, so we're done.

This tells us so because like from here we get that $P'(x) = k \cdot S'(Q(x)) \cdot Q'(x)$, where k is just a constant, now note that if some constant was vanished in taking the derivative of $P(x)$, we can just add that up to $S(x)$ accordingly.

We use Lagrange interpolation polynomial to do this.

First we take $ m $ points over function $ Q$ such that:$$Q(x_1)=y_1,Q(x_2)=y_2,\cdots Q(x_m)=y_m.$$
where $m>\frac{\deg P}{\deg Q}$.
Then by Lagrange interpolation polynomial we deduce that there exists a function $S:\mathbb{R} \rightarrow \mathbb{R}$ such that
$$P(x_k)=S(Q(x_k))$$
holds for $k=1,2,\cdots m$.Now what we need to prove is that actually $P=S \circ Q$.
Using the fact that $Q(x)-Q(y)|P(x)-P(y)$ and $x-y|f(x)-f(y)$ for any $f\in R[x]$,then we have
$$Q(x)-y_k \bigg| P(x)-S(Q(x))$$
holds for any $k=1,2,\cdots m$.Then it is easy to get that $\gcd (Q(x)-y_i,Q(x)-y_j)=1$ for $1\leq i < j \leq m $.Thus we get
$$\prod_{k=1}^{m}\bigg( Q(x)-y_k \bigg) \bigg| P(x)-S(Q(x))$$.But look at the expotent we must have $\deg P-S\circ Q=0$,thus we get $P=S\circ Q$ holds in $ \mathbb{R}$.$\blacksquare$

It suffice to show the problem with real replaced by complex, for if we get that to work, then $f(x)=P(g(x))$ for some $P\in \mathbb{C}[x]$. Now, taking conjugates imply that $f(x) = \overline{P}(g(x)) = P(g(x))$ (since $g\in \mathbb{R}[x]$), which implies $\overline{P}=P$, so $P\in \mathbb{R}[x]$.

Fix $g$; we induct on $\deg f$. The base case, $f\equiv 0$, is clear.

It suffices to show that $\deg g \mid \deg f$, because if that is the case, say $\deg f = k\deg g$, then we can apply inductive hypothesis on $f(x)-cg(x)^k$ where $\deg(f(x)-cg(x)^k) < \deg (f(x))$.

Let $n=\deg g$. The main intuition is that if $x\in \mathbb{R}$ is a large real, then there is a root around $x\omega_n$, where $\omega_n=e^{\frac{2\pi i}{n}}.$

Let's formalize that intuition. Let $h(x)=g(x+b)$ for some constant $b$ such that $[x^{n-1}]h(x)=0$. Write $$h(x) = h_nx^n + h_{n-2}x^{n-2} + \cdots + h_0$$

Now we can see that

$$|h(x\omega_n) - h(x)| = O(x^{n-2}) $$

Let $z$ be a unit complex number. Then

$$\frac{\partial h}{\partial z}(x) = \lim_{\substack{d\to 0^+ \\ d\in \mathbb{R}^+}} \frac{h(x+dz)-h(x)}{d}= zh'(x)$$

We use a Newton-style approximation method to find $x'$ such that $|x\omega_n-x'| = O(x^{-1})$ and $h(x')=h(x)$. Consider a sequence of points $x_0,x_1,\cdots$ defined by $x_0=x\omega_n$ and

$$x_k-x_{k-1} = \frac{h(x) - h(x_{k-1})}{h'(x)}$$

We induct on $k$ to show that $x_k-x_{k-1} = O(x^{-k})$. For the base case $k=1$, check that

$$h(x_1) = h\left(x_0 + \underbrace{\frac{h(x)-h(x_0)}{h'(x)}}_{O(x^{-1})}\right)=h(x_0)+nx_0^{n-1} \frac{h(x)-h(x_0)}{h'(x)} + O(x^{n-3})$$

Thus, $$h(x)-h(x_1) = (h(x)-h(x_0))\left( 1-\frac{nx_0^{n-1}}{h'(x)}\right) = (h(x)-h(x_0)) O(1/x)$$

Inductive step is similar: keep in mind that $x_{k-1}=x\omega_n + O(1/x)$ is a consequence of inductive hypothesis.

$$h(x_k) = h\left(x_{k-1} + \underbrace{\frac{h(x)-h(x_{k-1})}{h'(x)}}_{O(x^{-k})}\right)=h(x_{k-1})+nx_{k-1}^{n-1} \frac{h(x)-h(x_{k-1})}{h'(x)} + O(x^{n-k-2})$$

Thus, $$h(x)-h(x_k) = (h(x)-h(x_{k-1}))\left( 1-\frac{nx_{k-1}^{n-1}}{h'(x)}\right) = (h(x)-h(x_{k-1})) O(1/x)$$

Since the sequence of points $x_0,\cdots,$ is Cauchy and $\mathbb{C}$ is a complete metric space, it must converge to $x'$. We can see that $h(x)=h(x')$. This means that for any large $x$, there exists a constant $a$ such that $g(x) = g(x\omega_n + a + O(1/x))$. This means that $f(x) = f(x\omega_n + a + O(1/x)) \rightarrow n\mid \deg f$

If $Q(x)$ is constant then $P(x)$ is constant, so assume $\deg Q(x)\ge 1$. Let us fix $Q(x)$. We prove the problem statement by strong induction on $\deg(P(x))$. We can shift $P(x),Q(x)$ by a constant, hence we can assume that $P(0)=Q(0)=0$. Substitute $y=0$, implying that $Q(x)\mid P(x)$. Hence $P(x)=T(x)Q(x)$ for a polynomial $T(x)\in\mathbb R(x)$. Hence
$$Q(x)-Q(y)\mid T(x)Q(x)-T(y)Q(y)$$
so
\begin{align*}
Q(x)-Q(y)&\mid T(x)Q(x)-T(y)Q(y)+T(y)(Q(x)-Q(y))\\
&=(T(x)-T(y))Q(x).
\end{align*}
Now
$$\gcd(Q(x)-Q(y),Q(x))=\gcd(Q(x),Q(y))=1$$
as these are polynomials of different variables, so $Q(x)-Q(y)\mid T(x)-T(y)$. Since $\deg T(x)<\deg P(x)$, we are done by induction hypothesis.

Here is another idea. If $Q(x)$ is constant, then $P(x)$ is constant as well, so assume $Q(x)$ is nonconstant. Again, we fix $Q(x)$ and induct on $\deg P(x)$. WLOG assume $P(x)$ and $Q(x)$ are monic. Assume $\deg Q(x)=m$ and $\deg P(x)=n$. By taking terms of makimal degree we see that $x^n-y^n=r(x,y)(x^m-y^m)$ for a polynomial $r(x,y)\in\mathbb R[x,y]$. Substituting $y=1$ implies that $x^m-1\mid x^n-1$ in $\mathbb R[x]$. It is well-known that $\gcd(x^m-1,x^n-1)=x^{\gcd(m,n)}-1$, hence $\gcd(m,n)=m$ so $m\mid n$. Now let $d\doteqdot\frac{n}{m}$. Since $Q(x)-Q(y)\mid Q(x)^d-Q(y)^d$, we can replace $P(x)$ by $P(x)-Q(x)^d$ and continue by induction hypothesis.

We will strong induct on $\deg P$. Note obviously $\deg Q < \deg P$.

Note that we can shift $P$ and $Q$ by any constant and the condition will still be satisfied; thus, set $P(x) = Q(x)R(x)$ for some polynomial $R$. Then $$Q(x)-Q(y) \mid (Q(x)R(x) - Q(y)R(y)) - (Q(x)R(x) - Q(y)R(x)) = Q(y)(R(x) - R(y))$$ in $\mathbb R[x, y]$. Note that as $Q$ is nonconstant, $\gcd(Q(x) - Q(y), Q(y)) = 1$, hence $$Q(x) - Q(y) \mid R(x) - R(y).$$ But by the inductive hypothesis, as $\deg R < \deg P$, we have $R(x) = S(Q(x))$ for some $S$. Then $$P(x) = S(Q(x))Q(x) = (S+1)(Q(x)).$$

blud what is this question

Write
\[P=A_dQ^d+\dots+A_0Q^0\]
for polynomials $A_0\dots,A_d$ with degree less than $\text{deg }Q$. Our aim is to show that all $A_i$ are constant. Define $\equiv$ for polynomials in the obvious way (dunno if this is legit)
Notice that
\[\sum A_i(y)Q(y)^i=P(y)\equiv P(x)=\sum A_i(x)Q(x)^i\equiv \sum A_i(x)Q(y)^i\pmod {Q(x)-Q(y)}.\]
Hence $Q(x)-Q(y)$ divides
\[\sum (A_i(x)-A_i(y))Q(y)^i\]
Since the maximal degree of $x$ in this is $\text{deg }Q-1$ and yet $Q(x)-Q(y)$ has a maximal degree in $x$ of $\text{deg }Q$ it follows that the entire expression is equal to $0$. Now we just induct from the top down. FTSOC suppose at some point $A_i$ isn't constant. Consider the maximal $m$ where this holds, i.e.,
\[A_m(x)-A_m(y)\ne 0.\]
Clearly then the degree of $y$ in the original expression at least $m\cdot \text{deg }Q$. But for all $n<m$ the maximal degree of $y$ in
\[(A_n(x)-A_n(y))Q(y)^n\]
is at most
\[n\cdot \text{deg }Q + \text{deg }Q-1<m\cdot \text{deg }Q\]
hence something doesn't cancel and the expression isn't zero. So all $A_i$ are constant, done.

(Also this is essentially equivalent to the solution with remainders, I don't know why I didn't try that. Oh well.)


Let $P$ be the intersection point of the diagonals $A$C and $BD$ in a cyclic quadrilateral.
A circle through $P$ touches the side$CD$ in the midpoint $M$ of this side and intersects
the segments $BD$ and $AC$ in the points$Q$ and$R$respectively. Let$S$ be a point on the
segment $BD$ such that $BS = DQ$. The parallel to$AB$ through $S$ intersects $AC$ at $T$ .
Prove that $AT = RC$.

$BP \cdot PD = AP \cdot CP \Rightarrow  \frac{BP}{AP} = \frac{CP}{PD}$

$\triangle PBA : ST \parallel AB \Rightarrow \frac{BP}{AP} = \frac{BS}{AT}$

Hence,  $\frac{CP}{PD} = \frac{BS}{AT}$

$CM^2 = CP \cdot CR$
$DM^2 = DP \cdot DQ$

$CM^2 = DM^2 \Rightarrow  CP \cdot CR = DP \cdot DQ \Rightarrow  \frac{CP}{DP} = \frac{DQ}{CR} = \frac{BS}{CR}$

$\Rightarrow  \frac{BS}{AT} = \frac{BS}{CR} \Rightarrow  AT=CR$


prove that the equation
x+y+1/x+1/y=3*n
has no solution in positive rational numbers.

we write x=a/b, y=c/d with (a,b)=(c,d)=1. then the eqn is: (a^2+b^2)*cd  +(c^2+d^2)*ab=3nabcd. We notice that cd/(c^2+d^2)*ab. Because (c,d)=1 we find that (cd, c^2+d^2)=1, so cd/ab. By the same manner we find ab/cd, so ab=cd. We can now divide the equality by ab=cd, so we have a^2+b^2+c^2+d^2=3nab=3ncd. Because 3/a^2+b^2+c^2+d^2 and x^2=1 or 0 (mod 3), we either have 3/a^2, b^2, c^2 and d^2, which is impossible, or a^2=0(mod 3) and b^2=c^2=d^2=1 (mod 3). We can put b, c, or d instead of a, but let's choose a. Therefore 3/a and 3 does not divide any of b, c, d. But because ab=cd, 3/cd, so 3/c or d, so we get a contradiction. We found that there are no solutions in positive rational numbers.


Go to the resources section at the top of the forum:

[url]http://www.artofproblemsolving.com/Forum/resources.php?c=182[/url]

EDIT: ah, my bad.. the 10 B test.. you should be able to find some posted by users on the forum here..? don't have the packet with me, though..

[quote="aznjonny24"]I need to know questions 8 and 20. Could I have both the problem and the answer if possible? Thanks.[/quote]

EDIT: They're both up in the Contests section.

[quote="aznjonny24"]thanks for number 20, worthawholebean. but i still cant find numero ocho.[/quote]

Go to the link that undefined posted.


For which prime numbers $p$ and $q$ is $(p + 1)^q$ a perfect square?

i think only for $q=2$ and any $p$ and for any $q$ and $p+1$ a square.

Naphthalin

[hide]

if $q=2$ then $\forall p$ prime the condition is satisfied. If $q \neq 2$ then $p+1$ has to be a perfect square, because if it isn't, $(p+1)^q$, since $q$ is odd, contain at least one prime factor repeated an odd number of times, so it can't be a square. So $p+1=k^2 \rightarrow p=(k+1)(k-1)$ that has solution only $p=3$
[/hide]

year, this is the only prime $p$ where $p+1$ is a square.

Naphthalin


imi puteti da si mie un indiciu sau rezolvare(dar as prefera indiciu)
sa se determine nr nat n cu proprietatea ca exista nr intregi a si b a.i
n^2 =a+b    si         n^3=a^2+b^2     :)     multumesc

Ecuatia redactata frumos:$n^3=a^2+b^2$ si $n^2=a+b $
O mica indicatie poate a=b=n=2. :)

de cand am postat problema i have been working hard si mi-a iesit in cele din urma ca 2n^3 >= n^4 si prin urmare n <= 2 si am incercat cazurile...

10x anyway...

as mai avea oproblema care mi-a dat bataie de cap ziua asta...si ma voi mai gandi la ea dar cred ca oricum am nevoie de ajutor :blush: :
sa se demonstreze ca ecuatia x 2 +y 2 +z 2 +t 2 =2^2004,
unde 0 <= x <= y <= z <= t,are exact doua solutii in multimea nr intregi :blush:  sper sa nu ma fac de ras

merci


Find all the functions $f$ such that  $f^2(x)\cdot f(\frac{1-x}{1+x})=64x$

for each $x\in R-\left\{ 0,-1,1 \right\}$

Substitute $x$ by $\frac{1-x}{1+x}$ then:
$f(x).f^2(\frac{1-x}{1+x})=64\frac{1-x}{1+x}$ with the original equation we have:
$f^3(x)=64x^2\frac{1-x}{1+x}$
$f(x)=4\sqrt[3]{x^2\frac{1-x}{1+x}}$
Easy to verify its indeed solution.


Order the numbers 1, 2, ... , n any way you want. Add up the differences between successive numbers, with all differences counted positively: for example, if you start with the ordering 5, 3, 1, 4, 2, you get 2 + 2 + 3 + 2 = 9. What is the largest number you can obtain in this way?

Not sure if this is right but if you order them like this

$n, 1, n-1, 2, n-2, 3...$ then the sum is
$\frac{n(n-1)}{2}$

Well thats not right here is something that gives a higher sum if you order them like this

$2,n,1,n-1,3,n-2...$ then the sum is
$\frac{n(n-1)}{2}+1$...hmm is there something higher??


Let $ a,b,c,d,e,f $ be positive real numbers.

Prove that:

(a^3+d^3)(b^3+e^3)(c^3+ f^3 ) \geq  (abc+def)^3

Speaking about Holder and Huygens inequality...

just by brute force, we may take x = a/d, y = b/e, z = c/f, and obtain
x+y+z + (xy)+(yz)+(zx) >= 3xyz + 3xyz, which is trivial.

Or just by Cauchy(Hey I mean AM-GM :D ).


Just like other forums, this will be a marathon thread that is restricted to ONLY PROBLEMS TAUGHT AT A REGULAR MIDDLE SCHOOL LEVEL! Please do not go overboard with problem difficulty. Thanks!

Simply answer the problem given by the poster above you, showing all work. Then post a new problem yourself, so as to keep it going.

I'll start:

New Problem:

What is the area of a trapezoid with bases 4 and 6 and height 12?

$ A \equal{} \frac12 \times b_1b_2 \times h$

$ A \equal{} \frac12 \times 4 \times 6 \times 12$

$ A \equal{} 5 \times 12$

$ A \equal{} \boxed{60}$

[b]Problem 2:[/b]

Find the number of centimeters in the hypotenuse with legs 9 and 40.

For some reason, I think this is useless because there are only a few users who cannot solve questions like those, and half of them probably never come online.  And, easy problems are too easy.

Oh, BTW the answer to your problem is 41.

[quote="gauss1181"]Sum of squares of legs is 1600+81=1681, so hypotenuse is square root of 1681 or 41.

New problem:

Olly is 12 years older than Izzy. Two years later, Olly will be twice as old as Izzy. Find their current ages.[/quote]

olly is 22, izzy is 10.

new: find the area of a square with the same perimeter as the right triangle with legs 11 and 60.

so evaluating 7*3 is waaayyy to hard for classroom math.
in that case
NEW PROBLEM
evaluate: 3+7

[quote="qwertythecucumber"][quote="D3m0n Shad0w"]Please do not create any more marathons in Classroom Math - they're never going to last and it's really a waste of effort.[/quote]

yeah...[/quote]

DOT DOT DOT

$ 2^4\equal{}\boxed{16}$ ways.

New Problem:

A truck travels 30 miles per hour. How far does it travel in 150 minutes?

[quote="AIME15"][quote="qwertythecucumber"][quote="D3m0n Shad0w"][size=200][b]Please do not create any more marathons in Classroom Math - they're never going to last and it's really a waste of effort.[/b][/size][/quote]

yeah...[/quote]

DOT DOT DOT[/quote]

Is anyone listening to these posts??

[quote="athunder"]uhh define anyone... gauss1181 probably has and she is fine with... soooo we will make it continue :)

LOL WHAT AM I SAYING wow... does anyone in aops really find this marathon to be useful? i would say its used to increase post count which is lame.[/quote]

why not continue this marathon

Also, if you're going to start a marathon, use hide tags.

[code] [hide=Problem # or Solution #] Text [/hide] [/code]

[hide=" Optional"]

To box in your answers use

[code] \boxed {ANSWER} [/code]

[/hide]

Please do not revive marathons after this amount of inactivity. There are many other marathons in this forum that are essentially the same as this one (except this covers all the MS classroom math subjects in one marathon). Locked.


AMC USA(J)MO USAMO geometry probability AIME Putnam

Hello AoPS users. I am beginning a new game, and in it, people sign up, and the first person to sign up will be the contestant. Anybody else that signs up is a mob member. The contestant is going to go against the mob. The contestant's goal is to beat everybody else. The mob's goal is to beat the contestant. Basically, I will ask everybody questions. The questions will be on this post. You will have to pm me your answers. Anybody that answers the questions wrong will be eliminated. The contestant has life lines that he can use. I will tell them their lifelines later.

At the end of each competition there are two possibilities;

A.) The contestant wins.

If this happens, then he can be the contestant has four choices.

         A.A.) He can be the contestant again.
         A.B.) He can choose somebody from the mob to be the next contestant.
         A.C.) He can leave it to me to decide randomly.
         A.D.) He can let the mob compete.

B.) The mob wins.

If this happens then they will all have a vote. These are the three choices they have.

         A.A.) They can compete against each other.
         A.B.) They can let me pick who becomes the new contestant.
         A.C.) They vote on who becomes the new contestant.

Anybody that chooses to compete after somebody has won will become a mob member.

Sign-ups begin now.

Hmmm, this looks like my game from months ago... I join.

umm... wait. math questions. right??

><
i'm in.

I'll join, but your games never end and take way tooo long.

Is lotsofmath the first contestant?

I join.

ARe these math questions?

[quote="ProtestanT"]I'll join too.

And I don't think they're math questions. If they're math questions this competition will take too long, since most people here are used to answering math questions [b]correctly[/b]. (like in the last 1vs100, it went on for a long time)[/quote]

*cough* add USAMO level questions *cough*

idiot want join! idiot want join! idiot want join! idiot want join! idiot want join! idiot want join! idiot want join! idiot want join! idiot want join! idiot want join! idiot want join! idiot want join! idiot want join! idiot want join! idiot want join! idiot want join! idiot want join! idiot want join! idiot want join! idiot want join! idiot want join! idiot want join! idiot want join! idiot want join! idiot want join! idiot want join! idiot want join! idiot want join! idiot want join! idiot want join! idiot want join! idiot want join! idiot want join! idiot want join! idiot want join! idiot want join! idiot want join! idiot want join! idiot want join! idiot want join! idiot want join! idiot want join! idiot want join! idiot want join! idiot want join! idiot want join! idiot want join! idiot want join! idiot want join! idiot want join! idiot want join! idiot want join! idiot want join! idiot want join! idiot want join! idiot want join! idiot want join! idiot want join! idiot want join! idiot want join! idiot want join! idiot want join! idiot want join! idiot want join! idiot want join! idiot want join! idiot want join! idiot want join! idiot want join! idiot want join! idiot want join! idiot want join! idiot want join! idiot want join! idiot want join! idiot want join! idiot want join! idiot want join! idiot want join! idiot want join! idiot want join! idiot want join! idiot want join! idiot want join! idiot want join! idiot want join! idiot want join! idiot want join! idiot want join! idiot want join! idiot want join! idiot want join! idiot want join! idiot want join! idiot want join! idiot want join! idiot want join! idiot want join! idiot want join! idiot want join! idiot want join! idiot want join! idiot want join! idiot want join! idiot want join!

[quote="1=2"]If pythag011 joints I quit.[/quote]

I quit

I join. [b]Mewto55555 chooses to be first contestant[/b] Pwnt. I got there first.

I'm positive that I sent you a PM with my answer.

Oh, yes sorry crazypianist answered the question, but pythag didn't. Only about 5 hours left to pm me.

Still waiting on;
mewto55555
BOGTRO
pythag011

Alright, mewto and BOGTRO are too late, and they are eliminated from this competition.

james4l, you can either use one of your lifelines, or you can answer the question. If you would like to use a lifeline, then just say so, and I will show you the three lifelines.

james4l got the question correct, so now there are only five mob members left. Unfortunately, one of these five people got the question incorrect.

It was one of these five;

crazypianist116
pythag011
azjps
SMB
nick42

[hide]pythag011 got the question correct

[hide]crazypianist1116 got the question correct

[hide]azjps got the question correct

nick42 got the question correct

SuperMathBoy was incorrect. [/hide][/hide][/hide]

james4l has 75,000 up for grabs, so he can take the money, or try to earn more by reducing the mob.

Go for more (I won't settle for less than 300k!)

lol, if you want more than 300k, you'll have to narrow the mob down to two people.

Which topic do you pick;

Arithmetic
Algebra
Geometry
Probability

Okay, the question will be posted tomorrow.

Here is the question;

What is the probability that and integer between 1 and 1000 inclusive is divisible by 3, but not 4, and 5, but not 7?

james4l, please wait to solve this problem until the mob has answered.

BTW - The answer to the previous problem was 112, in case I haven't said already.

Do you mean, "is a multiple of 3 and 5 but not a multiple or 4 and/or 7" or "is a multiple of 3 and 5 but not 4 and 7?"

It is divisble by 3 and 5, but it is not divisible by 4 or 7.


tamame tavabe haghighi $f$ ra tori peida konid ke :
$f(x^{2}+f(f(x)-y))=2f(f(x))+2y^{2}, x,y \in \mathbb R$

tamame tavabe haghighi $f$ ra tori biabid ke
$(x+y)(f(x)-f(y))=(x-y)f(x+y) ,x,y \in\mathbb R$


geometry
rectangle
ratio
similar triangles
geometry unsolved

Let two circles $ (O; R)$ and $ (O; r)$ with $ R > r$. Let $ P$ is a fixed point on $ (O; r)$ and let $ B$ is a change point on $ (O;R)$. Let $ BP$ meet $ (O;R)$ at $ C$, let $ (d)$ is a line pass point $ P$ and $ (d) \perp BP$, let $ (d)$ meet $ (O;r)$ at point $ A$.

a) Find $ BC^2\plus{}CA^2\plus{}AB^2$ by $ R$ and $ r$.

b) Let $ I$ is midpoint of $ AB$. Find locus of point $ I$ ?

[quote="thanhnam2902"]Let two circles $ (O; R)$ and $ (O; r)$ with $ R > r$. Let $ P$ is a fixed point on $ (O; r)$ and let $ B$ is a change point on $ (O;R)$. Let $ BP$ meet $ (O;R)$ at $ C$, let $ (d)$ is a line pass point $ P$ and $ (d) \perp BP$, let $ (d)$ meet $ (O;r)$ at point $ A$.

a) Find $ BC^2 \plus{} CA^2 \plus{} AB^2$ by $ R$ and $ r$.

[/quote]

[hide="Solution a)"] Let's circle $ (O;r)$ and line that pass through $ PB$  intersects second time in $ K$. Then we have :
$ CK\equal{}BP$, $ AK\equal{}2r$
Determine $ CK\equal{}BP\equal{}a$, $ KP\equal{}x$, $ AP\equal{}b$.
$ AB^2\plus{}AC^2\plus{}BC^2\equal{}2AP^2\plus{}BP^2\plus{}PC^2\plus{}BC^2\equal{}2b^2\plus{}a^2\plus{}(a\plus{}x)^2\plus{}(2a\plus{}x)^2\equal{}2(b^2\plus{}x^2)\plus{}6a(a\plus{}x)\equal{}2\cdot 4r^2\plus{}6(R\minus{}r)(R\plus{}r)\equal{}8r^2\plus{}6R^2\minus{}6r^2\equal{}6R^2\plus{}2r^2$[/hide]

[color=darkblue]Let $ M$ be foot of the perpendicular from $ O$ onto $ BC$ and let $ BC$ cuts $ (O,r)$ at $ T$. Then $ T$, $ O$ and $ A$ are collinear and $ OM$ is midline of the triangle $ TPA$. So $ OM \equal{} \dfrac{1}{2}PA$. Also note that $ M$ is midpoint of $ BC$. Denote $ \{G\} \equal{} OP \cap AM$. Then from similar triangles $ OMG$ and $ PGA$ we obtain that

$ \frac {MG}{GA} \equal{} \frac {OG}{GP} \equal{} \frac {OM}{AP} \equal{} \frac {1}{2}$

Thus $ G$ is centroid of the triangle $ BAC$ and also $ OG \equal{} \frac {1}{3}OP \equal{} \frac {1}{3}r$.

Now apply Leibnitz theorem to the triangle $ BAC$ and point $ O$ and obtain

$ AB^2 \plus{} BC^2 \plus{} CA^2 \equal{} 3OA^2 \plus{} 3OB^2 \plus{} 3OC^2 \minus{} 9OG^2 \equal{} 3r^2 \plus{} 6R^2 \minus{} r^2 \equal{} 2r^2 \plus{} 6R^2$.

For $ b)$ part let us prove the value $ IQ$ is constant where $ Q$ is midpoint of $ OP$. Then, the required locus will be circle with center $ Q$ and radius $ QI$.

By median theorem we have

$ 8QI^2 \equal{} 4AQ^2 \plus{} 4BQ^2 \minus{} 2AB^2$
$ 4BQ^2 \equal{} 2BO^2 \plus{} 2BP^2 \minus{} PO^2 \equal{} 2R^2 \plus{} 2BP^2 \minus{} r^2$
$ 4AQ^2 \equal{} 2AO^2 \plus{} 2AP^2 \minus{} PO^2 \equal{} 2R^2 \plus{} 2AP^2 \minus{} r^2$

$ 8QI^2 \equal{} 4R^2 \minus{} 4r^2 \plus{} 2BP^2 \plus{} 2AP^2 \minus{} 2AB^2 \equal{} 4R^2 \minus{} 4r^2$[/color]

[img]http://pic.ipicture.ru/uploads/090112/b7cU3tBq7d.png[/img]

[color=darkblue]Your solution is nice. I can find locus of point by use homothetic transformation.

+) Easy we get $ PBMA$ is a rectangle $ \Longrightarrow$ $ I$ is midpoint of PM.

+) Hence we get $ \overrightarrow {PI} \equal{} \frac{1}{2} \overrightarrow {PM}$

$ \Longrightarrow$ point $ I$ be image of point $ M$ of homothetic center $ O$, ratio $ k\equal{}\frac{1}{2}$

But $ M \in (O) \Longrightarrow I \in (O')$ with $ (O')$ be image of $ (O)$ of homothetic center $ O$, ratio $ k\equal{}\frac{1}{2}$.[/color]


group theory
abstract algebra
function
LaTeX
superior algebra
superior algebra unsolved

Let $G$ be a group with $m$ elements and let $H$ be a proper subgroup of $G$ with $n$ elements. For each $x\in G$ we denote $H^x = \{ xhx^{-1} \mid h \in H \}$ and we suppose that $H^x \cap H = \{e\}$, for all $x\in G - H$ (where by $e$ we denoted the neutral element of the group $G$).

a) Prove that $H^x=H^y$ if and only if $x^{-1}y \in H$;
b) Find the number of elements of the set $\bigcup_{x\in G} H^x$ as a function of $m$ and $n$.

[i]Calin Popescu[/i]

Consider the rel P on G by the following aPb iff a(b^-1) belongs to G.We can show that it is an equivalence relation.
Now work with the equivalence classes generated by P.
I think we can get H_x=H_y or id I do anything wrong. :?

a) $H^x=H^y\Leftrightarrow xHx^{-1}=yHy^{-1}\Leftrightarrow y^{-1}xHx^{-1}y=H\Leftrightarrow H^{y^{-1}x}=H\Leftrightarrow y^{-1}x\in H$.
The last equivalence follows from the condition $H^x\cap H=\{e\}$ for $x\in G-H$.

b) First we show that $H^x$ and $H^y$ can only have trivial intersection ($\{e\}$ or $H^x$). If $x^{-1}y\in H$ we know with a) that $H^x=H^y$. Now consider the case $x^{-1}y\not\in H$. Then we cannot have $h_1,h_2\in H\backslash\{e\}$ with $xh_1x^{-1}=yh_2y^{-1}$ since otherwise $y^{-1}xh_1x^{-1}y=h_2$ and $H\cap H^{x^{-1}y}\neq\{e\}$ which is a contradiction to one of the conditions. Now we have exactly $[G:H]=\frac{m}{n}$ different sets of the form $H^x$ with pairwise intersection $\{e\}$. In total we get
$n+\left(\frac{m}{n}-1\right)\cdot(n-1)=\frac{mn-m+n}{n}$
elements in $\cup_{x\in G} H^x$.

I hope I didn't miscalculate in the end of b)

Misha

I dont think so because there is a problem in Herstein where one has to give an example of group G such that for a subgroup H of G  :cup: H_x x belonging to G  where ( :cup: H_x x belonging to G
is defined in the same way)is not the whole of G.I have constructed a similar example.

[quote="Misha123"]I don't see a contradiction to my solution. :?
Could you please explain what you mean (and use LATEX) ?

Misha[/quote]
What contradiction!!!I said theres a similar problem in Herstein (the 2nd part) & that two I have done similarly.
Sorry I ll try to learn latex.I know its very clumsy& its equally difficult for me to express myself.. :(

All right! I just misunderstood you, sorry :blush:

Misha

(a) If $x^{-1}y\in H$, then $\{(y^{-1}x)h(x^{-1}y)\mid h\in H\}=H$ (note that $(y^{-1}x)^{-1}=x^{-1}y$): for each $h'\in H$, we can take $h:=(x^{-1}y)h'(y^{-1}x)$ which satisfies $(y^{-1}x)h(x^{-1}y)=h'$. It follows that $H^x=H^y$.

Now assume $H^x=H^y$. Clearly either both $x$ and $y$ are in $H$ or none of them are in $H$. Since $|H^x|=|H^y|=n$, there is a permutation $\sigma$ of $H$ such that $xhx^{-1}=y\sigma(h)y^{-1}$ for all $h\in H$. It follows that $H^{y^{-1}x}=\{(y^{-1}x)h(x^{-1}y)\mid h\in H\}=H$. But this can only happen when $y^{-1}x\in H$, which implies $x^{-1}y\in H$, as desired.

(b) We claim that for all $x,y\in G,x^{-1}y\not\in H$, we have $H^x\cap H^y=\{e\}$. Assume otherwise. Then there exists $h_1,h_2\in H$ with $h_1,h_2\neq e$ such that $xh_1x^{-1}=yh_2y^{-1}$ so $(y^{-1}x)h_1(x^{-1}y)=h_2$. But now $H^{y^{-1}x}$ contains $h_2\neq e$, a contradiction.

It follows that for all $x,y\in G$, either $H^x=H^y$ or $H^x\cap H^y=\{e\}$. For each $x\in G$, the set $G_x:=\{y\in G\mid H^x=H^y\}$ has cardinality $n$ (there are exactly $n$ $y\in G$ such that $H^x=H^y$: one for each element in $H$). Thus the $G_x$ define equivalence classes: $G_x\cap G_y=\{e\}$ if and only if $y\not\in G_x$. There are $\frac{m}{n}$ such equivalence classes so $\left|\bigcup_{x\in G}H^x\right|=1+\frac{m}{n}(n-1)$. $\square$


If the major and minor axes of an ellipse are denoted by 2a and 2b respectively, why is the distance between the diagonals $ \sqrt{a^{2}-b^{2}}$? What other properties of ellipses are worth taking note? Any help/information is appreciated  :)

when you say diagonal do you mean chord??

---
you should be able to prove tangent/normal formula, chord formula

foci directrices etc

also you should be able to prove that PS+PS'=2a

Sorry I meant the distance between the foci, not the diagonals..  :blush: my mistake. I read somewhere that the distance the foci is $ \sqrt{a^{2}-b^{2}}$..

hmm that is an interesting property I have neevr come across

$ \sqrt{a^{2}-b^{2}}$
$ \sqrt{a^{2}-a^{2}(1-e^{2})}$ using $ b^{2}$=$ a^{2}(1-e^{2})$
$ \sqrt{a^{2}(1-(1-e^{2})}$
$ \sqrt{a^{2}e^{2}}$
$ ae$

and distance from origin to foci is ae therefore
$ \sqrt{a^{2}-b^{2}}$=$ ae$


calculus
integration
limit
function
Putnam
logarithms
calculus computations

If $0<a<b$, find $\lim_{t\rightarrow 0}\left\{ \int_{0}^{1}\left[ bx+a(1-x)\right]^{t}dx\right\}^{\frac{1}{t}}$.

Any relation to $L^{p}(a,b)$ norms?

  -Ben

[quote="didilica"]Hi,
 I think this problem is classical. It is a simple application of a more general result.

Let $f$ be a positive and continuous function on $[a,b]$ and let $M=\sup_{x\in[a,b]}{f(x)}$.
 Prove that

$\lim_{n\rightarrow \infty}\sqrt[n]{\int_{a}^{b}(f(x))^{n}dx}=M$.
Thus, in your case the limit is $b$.
Didi[/quote]

But our limit is at zero.

I told you it was fun, Firey!

By the way, this problem is also Putnam 1979 B2.

[quote="FieryHydra"]A simple substitution $y = (b-a)x+a$ transforms the integral into one that can be evaluated easily. So, the given limit is equal to $\lim_{t \to 0}\left( \frac{b^{t+1}-a^{t+1}}{(t+1)(b-a)}\right)^\frac1{t},$ which is of $1^{\infty}.$ Now you know what is to be done.[/quote]

Let \[J=\lim_{t\to 0}\left(\frac{b^{t+1}-a^{t+1}}{(t+1)(b-a)}\right)^\frac{1}{t}\] so that \[\log J=\lim_{t\to 0}\frac{\log\left(\frac{b^{t+1}-a^{t+1}}{(t+1)(b-a)}\right)}{t}\] which is an indeterminent form of the type $-\frac{\infty}{\infty}$ so apply l'Hospital's Rule to get \[\log J=\lim_{t\to 0}\frac{(t+1)(b-a)}{b^{t+1}-a^{t+1}}\left(\frac{(b^{t+1}\log b-a^{t+1}\log a)(t+1)(b-a)-(b^{t+1}-a^{t+1})(b-a)}{(t+1)^{2}(b-a)^{2}}\right)\]  \[{=\lim_{t\to 0}\left( \frac{b^{t+1}\log b-a^{t+1}\log a}{b^{t+1}-a^{t+1}}-\frac{1}{t+1}\right) = b\log b-a\log a}{b-a}-1\qquad\qquad\qquad ,\]  \[= \frac{1}{b-a}\log\left( \frac{b^{b}}{a^{a}}\right)-1 \qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad ,\] and hence \[J=e^{\frac{1}{b-a}\log\left( \frac{b^{b}}{a^{a}}\right)-1 }=\frac{1}{e}\left( \frac{b^{b}}{a^{a}}\right)^{\frac{1}{b-a}}=\frac{b}{e}\left( \frac{b}{a}\right)^{\frac{a}{b-a}}=\frac{a}{e}\left( \frac{b}{a}\right)^{\frac{b}{b-a}}\]


Hey Kids!  [url=http://www.research.ibm.com/ponder/]IBM Research[/url] has posted their monthly puzzle on their [url=http://www.research.ibm.com/ponder/]Ponder This[/url] Web Site.  These puzzles are notoriously difficult.

This month's problem reminds me a little of two of this year's favorite AIME problems!  (Remember those number 15's?)

There is a 64-disk "Tower of Hanoi" puzzle.  You are to determine the specific configuration of the disks, subject to certain conditions.  One such condition is that the minimum number of moves to complete the puzzle is exactly  [tex]\lfloor\pi\times10^{18}\rfloor[/tex] moves.

Since these problems are expected to be your own work, please don't post any hints or solutions until after June 1.  [I suppose you could tell us if you've correctly solved the puzzle.]

Disclaimer:  I have no connection with [url=http://www.research.ibm.com/ponder/]IBM's "Ponder This"[/url], except as an occasional puzzle solver.

the Tower of Hanoi is coolio...there's a A LOT that we can learn about it and the patterns used in solving it...


[quote="Leonhard Euler"]$ Let$ $ a,b,c > 0$. $ Show$ $ that$
$ \sum_{cyc}\sqrt{a^{4}\plus{}a^{2}b^{2}\plus{}b^{4}}\geq\sum_{cyc}a\sqrt{2a^{2}\plus{}bc}$[/quote]
$ \sum_{cyc}\sqrt{a^{4}\plus{}a^{2}b^{2}\plus{}b^{4}}\geq\sum_{cyc}a\sqrt{2a^{2}\plus{}bc}\Leftrightarrow$
$ \Leftrightarrow\sum_{cyc}(a^{2}b^{2}\minus{}a^{2}bc)\plus{}2\sum_{cyc}\sqrt{(a^{4}\plus{}a^{2}b^{2}\plus{}b^{4})(a^{4}\plus{}a^{2}c^{2}\plus{}c^{4})}\geq2\sum_{cyc}ab\sqrt{(2a^{2}\plus{}bc)(2b^{2}\plus{}ac)}.$
But $ (a^{4}\plus{}a^{2}b^{2}\plus{}b^{4})(a^{4}\plus{}a^{2}c^{2}\plus{}c^{4})\geq\left(a^{4}\plus{}\frac{a^{2}(b^{2}\plus{}c^{2})}{2}\plus{}b^{2}c^{2}\right)^{2}$ and
$ 2a^{2}\plus{}bc\plus{}2b^{2}\plus{}ac\geq2\sqrt{(2a^{2}\plus{}bc)(2b^{2}\plus{}ac)}.$
Thus, it remains to prove that
$ \sum_{cyc}(a^{2}b^{2}\minus{}a^{2}bc\plus{}2a^{4}\plus{}4a^{2}b^{2})\geq\sum_{cyc}(2a^{3}b\plus{}2a^{3}c\plus{}2a^{2}bc),$ which obviously true.
See here the stronger inequality:
http://www.artofproblemsolving.com/Forum/viewtopic.php?t=165319


Solve the equation
$ max \limits_{b}(min \limits_{a}(a^2-b^2+ab+ax-bx))=3$


Is this an olympiad problem?............ :|

I don't think, but our algebra teacher that  teaches us algebra for olympiad said this problem. ;)

And he told you that it has a solution?!! :huh:

$|\sin x -\sin y|\le |x-y|$. At first, prove $f\nearrow$. Maybe $f^{-1}(x_0)$, where $x_0$ is known. For example, $f^{-1}(\pi )=\pi$. $\boxed {f^{-1}(y)=x\Longleftrightarrow f(x)=y}$.


geometric transformation
geometry
power of a point
radical axis
geometry proposed

(A.Zaslavsky, 9--11) Given two circles. Their common external tangent is tangent to them at points $ A$ and $ B$. Points $ X$, $ Y$ on these circles are such that some circle is tangent to the given two circles at these points, and in similar way (external or internal). Determine the locus of intersections of lines $ AX$ and $ BY$.

Assume without loss of generality that the third circle $ \Gamma$ is tangent internally to the first two given ones (the other case can be treated analogously). Denote by $ X^{\prime}$, and by $ Y^{\prime}$ the intersections of the line $ XY$ with the first, and second circle respectively, and let $ O_{1}$, $ O_{2}$ be the centers of these first two given circles $ \gamma_{1}$, and $ \gamma_{2}$; finally let $ Q$ be the center of $ \Gamma$. Since $ \angle{O_{1}XX^{\prime}}\equal{}\angle{O_{1}X^{\prime}X}\equal{}\angle{QXY}\equal{}\angle{QYX}\equal{}\angle{O_{2}YY^{\prime}}\equal{}\angle{O_{2}Y^{\prime}Y},$ we have that $ O_{1}X \| O_{2}Y^{\prime}$, and $ O_{1}Y \| O_{1}X^{\prime}$. Thus, $ \angle{AXX^{\prime}}\equal{}\angle{BY^{\prime}Y}$, and since $ AB$ is tangent to $ \gamma_{2}$, we conclude that $ \angle{AXX^{\prime}}\equal{}\angle{BY^{\prime}Y}\equal{}\angle{ABY}$. This shows that the points $ A$, $ B$, $ X$, and $ Y$ lie on a same circle.

Let $ \omega_1 ,\omega_2$ denote the two given circles and $ \omega$ denotes the third circle internally/externally tangent to $ \omega_1 ,\omega_2$ at $ X,Y.$ Let $ O$ be the exsimilicenter of $ \omega_1 \sim \omega_2,$ $ X,Y$ are the exsimilicenters/insimilicenters of $ \omega \sim \omega_1$ and $ \omega \sim \omega_2.$ Therefore, by Monge & d'Alembert theorem $ XY$ passes through $ O$ $\Longrightarrow$ $ X,Y$ and $ A,B$ are inverse points under the direct inversion through pole $ O$ that takes $ \omega_1$ and $ \omega_2$ into each other $ \Longrightarrow$ $ A,B,X,Y$ are concyclic $ \Longrightarrow$  $ P \equiv AX \cap BY$ moves on the radical axis of $ \omega_1 ,\omega_2.$

First, let's see that $\{T \} \equiv AX \cap BY$ belongs to the third circle.

[b]Proof:[/b] if $\omega_1$ and $\omega_2$ are the first two circles and $\Omega$ - the third one, then $X$ is the similicenter (in- or ex-) of $\omega_1$ and $\Omega$ and, obviously, if $\{X, T' \} \equiv AX \cap \Omega$, then the tangent to $\Omega$ at $T'$ is parallel to $AB$, the same for $B$ and $T"$, $T"$ being the intersection of $BY$ with $\Omega$, hence $T" \equiv T' \equiv T$.

Next, if $(TZ$ is the direction $(BA$ of this tangent, we have $\widehat{BAT}=\widehat{XTZ}$ $(1)$, but from the circle $\Omega $: $\widehat{XTZ}=\widehat{XYT}$, $(2)$, from $(1)$ and $(2)$ getting that $ABYX$ is cyclic, i.e. $T$ belongs to the radical axis of $\omega_1$ and $\omega_2$.

Best regards,
sunken rock


factorial
superior algebra
superior algebra unsolved

The problem reads:
"Prove that if a>1, then there is a prime p with a<p<=a!+1.
I understand that in a!+1 will have a prime in it  but I have absolutely no idea how to even get started!
Please help!
Thanks

Let $ p\le a$, then $ p|a!$ so $ p\nmid a!\plus{}1$. Therefore any prime factor of $ a!\plus{}1$ must be greater than $ a$.

I'm not sure if i understand why any prime factor of a!+1 has to be greater than a?!


inequalities
Cauchy Inequality
inequalities unsolved

Let $ a,b,c > 0$. Prove the inequality:

$ abc(a\plus{}2)(b\plus{}2)(c\plus{}2) \le \left( 1 \plus{} \frac{2}{3}(ab\plus{}bc\plus{}ca) \right)^3$

[quote="Mashimaru"]Let $ a,b,c > 0$. Prove the inequality:

$ abc(a \plus{} 2)(b \plus{} 2)(c \plus{} 2) \le \left( 1 \plus{} \frac {2}{3}(ab \plus{} bc \plus{} ca) \right)^3$[/quote]

This inequality is from KMO(Korean Mathematical Olympiad) 2006, Round 2.

[Proof]
Transform the RHS into this form;
$ RHS \equal{} ( \frac {(1 \plus{} ab \plus{} bc) \plus{} (1 \plus{} bc \plus{} ca) \plus{} (1 \plus{} ca \plus{} ab)}{3})^3$
$ \geq (1 \plus{} ab \plus{} bc)(1 \plus{} bc \plus{} ca)(1 \plus{} ca \plus{} ab)$ due to AM-GM.

Moreover, by Cauchy Inequality, the following holds;
$ (1 \plus{} ab \plus{} bc)(bc \plus{} ca \plus{} 1) \geq (\sqrt {bc} \plus{} a \sqrt {bc} \plus{} \sqrt {bc})^2 \equal{} bc(a \plus{} 2)^2$
$ (1 \plus{} bc \plus{} ca)(ca \plus{} ab \plus{} 1) \geq (\sqrt {ca} \plus{} b \sqrt {ca} \plus{} \sqrt {ca})^2 \equal{} ca(b \plus{} 2)^2$
$ (1 \plus{} ca \plus{} ab)(ab \plus{} bc \plus{} 1) \geq (\sqrt {ab} \plus{} b \sqrt {ab} \plus{} \sqrt {ab})^2 \equal{} ab(c \plus{} 2)^2$

Multiplying the inequalities and taking square roots of both inequalities kills the problem =)

Great solution, [b]qwerty414[/b] :D

[b]P/S:[/b] non-homogenous inequalities are always as hard as its beauty :blush:


Let $E$ be a real normed vectorial space and $H$ an hyperplan of $E$.Show that $E\setminus H$ is connexe par arcs iff $H$ is not closed.

Dear Killer, I remember you that the last RMS issue should not be posted on Mathlinks till later. And moreover, I remember you that the same problem was already posted by you and solved by grobber.  ;)

Actually ...I posted this problems in order for someone to give me a solution for them,since I dont know how to solve some of them and I find them interesting.And that's the purpose of mathlinks,isn't it Harazi?I dont see why  shouldnt I post them on this site(you know very well that they do not belong to that RMS isssue.They are there simply because they were given in last year(s) at the ENS oral exams.So everyone is free to think over them.I dont understand your remark at all ;) )

And sorry if i posted this pb for a second time.I simply forgot.In that case..some moderator should delete it.If not ,enjoy it.

Here is this link to problem http://www.mathlinks.ro/Forum/viewtopic.php?highlight=arc&t=60058#


algebra
difference of squares
special factorizations

What is the value of $ 513^2 \minus{} 487^2$?

Sorry, accidentally pressed "report error" on this problem.


Let $ P$ be a polynomial with integer coefficients.Suppose that
(i)$ P(n)>n$  $ \forall n>0$
(ii)For each positive integer $ m$,there exists an integer $ k>0$ such that $ P_k(1)$ is divisible by $ m$.
Prove that $ P(x)\equal{}x\plus{}1$.

Note: $ P_k(x)\equal{}P(P(P(...P(x)))$

Mathlinkers,could you help me with this.Thanks.

because $ P(n)>n$ so we take $ m\equal{}P(1)\minus{}1$
there exists a positive integer k such that $ P_{k}(1)$ is divisible by $ P(1)\minus{}1$ it follows that $ P(1)$ is divisible $ P(1)\minus{}1$ and we must have P(1)=2;
we prove that the degree of P equals 1, if not :
we take m=$ P_{l}(2)\minus{}P{l\minus{}1}(2)$ ,with a value of $ l$ we have a value of k which $ P_{k}(2)$ is divisible by m;
 so with a large $ l$ we must have k$ \ge$l
but we have with $ k \ge l$ we have $ P_{k}(2)\minus{}P_{k\minus{}1}(2)$ is divisible m so, we must have $ P_{l\minus{}1}(2)$ is divisible m with a large l; so $ P_{l\minus{}1}(2) \ge P_{l}(2)\minus{}P{l\minus{}1}(2)$
but the degree of P is greater than 1 so |$ P_{l}(2)\minus{}2P_{l\minus{}1}(2)$| is very large when l is enough large so the degree of P equals 1;
and, now it is easy to find that P(x)=x+1

You got a mistake in typing . I think you mean "We take $ m = P_l(2) - P_{l - 1}(2)$" when you wrote $ m = P_l(2) - P{l - 1)(2)}$"^^ and I don't know the last step .
Where is the relation between proving   " $ 2P_{l - 1}(2) \ge P_l(2)$ is imposible when $ l$ large enough"(1) and  "|$ 2P_{l - 1)(2) - P_l(2)|}$ is very large when $ l$ large enough " ?

To prove (1) ,when can do like this:
* $ P_{l - 1}(2)$ is very large when $ l$ large enough
* if $ degP(x) > 1$  then $ P(x) > 2x$ when  $ x$ large enough
=> $ P( P_{l - 1}(2)) > 2P_{l - 1}(2)$ when $ l$ large enough


factorial
inequalities
absolute value
complex numbers
number theory unsolved
number theory

Please someone help me with this i got stuck somwhere in the middle:
find all integer solutions to:

$ y^{3}\equal{}x^{2}\plus{}x\plus{}1$

Thanks. very nice solution. it turns out to be these are the only solutions $ (x,y)\equal{}(\minus{}1,1),(0,1),(18,7),(\minus{}19,7)$

[quote="tdl"]We have: $ (y \minus{} 1)(y^2 \plus{} y \plus{} 1) \equal{} x(x \plus{} 1)\Rightarrow x(x \plus{} 1)\vdots (y \minus{} 1)$
Because $ GCM(x,x \plus{} 1) \equal{} 1$ then we have two case:
1) $ x \equal{} k(y \minus{} 1)$ and $ y^2 \plus{} y \plus{} 1 \equal{} k(x \plus{} 1)$, in which $ k$ be a natural number. (Easy to check that $ k > 1$)
$ y^2 \plus{} y \plus{} 1 \equal{} k(ky \minus{} k \plus{} 1)\Leftrightarrow y^2 \plus{} (1 \minus{} k^2)y \plus{} k^2 \minus{} k \plus{} 1 \equal{} 0$
2) $ x \plus{} 1 \equal{} k(y \minus{} 1)$ and $ y^2 \plus{} y \plus{} 1 \equal{} kx$: Similarly![/quote]
It is wrong. For exfmple $ 28|20*21$, but $ 28\not |20$ and $ 28\not |21$.

[quote="Jure the frEEEk"]well... lol yes its obviously wrong. :blush:
ur absolutly right. u can't asume anything like that from $ (y \minus{} 1)(y^2 \plus{} y \plus{} 1) \equal{} x(x \plus{} 1)$

Anyway I approached it simmilar to that fermat problem $ x^2 \plus{} 2 \equal{} y^3$:

$ y^3 \equal{} x^2 \plus{} x \plus{} 1\Leftrightarrow4y^3 \equal{} (2x \plus{} 1 \plus{} i\sqrt3)(2x \plus{} 1 \minus{} i\sqrt3)$
from what u can deduce $ (1 \plus{} i\sqrt3)(a \plus{} bi\sqrt3)^3 \equal{} (2x \plus{} 1\pm\sqrt3i)$ and from this $ a^3 \plus{} 3a^2b \minus{} 9ab^2 \minus{} 3b^3 \equal{} \pm1$ which could be said unleast looks harder than the original equasion. At this point i gave up.

Can anyone prove it? and can please someone tell me if my thinking above is correct as an idea, even if it doesn't lead anywhere at this particular problem, because i don't quite handle this kind of aproach yet.[/quote]
$ Z[i\sqrt 3 ]$ is not factorial, for example $ 2*2 \equal{} (1 \plus{} i\sqrt 3 )(1 \minus{} i\sqrt 3)$, but you can use $ Z[e^{\frac {2\pi i}{3}}]$.

Here's my solution:(3o minutes for it) :blush:
I easily prove that y is positive integer!
Then I have 2 cases considered:
1st case:x>=2
Then I have:8y^3=8x^2+8x+8
And I use this inequality: (2x)^3<8x^2+8x+8<(2x+1)^3
So this equation doesn't have any roots for this case!
When x=1 then y^3=3(unreasonable)
When x=0 then y^3=1.Therefore,(0,1) is one root of this equation!
x=-1 then y=0(another root)
x=-2 then y^3=3(unreasonable)
2nd case:x=<-3
Since x(x+1) is divisible by 2, then I have:y must be an odd integer
Let y=(2k+1)(k>=0 and k is integer)
Then replacing it into the equation I have:2k(3+4k^2+6k)=x^2+x
Since x<=-3 then x must divide by k or x must divide by 3+4k^2+6k
Let x=m.k then I have:k(-m^2+8k+12)=-4(m,k is integer and k>=0)
Similar to the 2nd equation(using 2 variable) I have:3=k(2kl-7-4k)(k, l is integer and k>=0)
Finally I'm done! :lol:

[quote="ghjk"]
And I use this inequality: (2x)^3<8x^2+8x+8<(2x+1)^3
[/quote]

Why $ (2x)^3<8x^2\plus{}8x\plus{}8<(2x\plus{}1)^3$, where is root $ x\equal{}18,y\equal{}7$?
You are wrong!

[quote="tdl"][quote="ghjk"]
And I use this inequality: (2x)^3<8x^2+8x+8<(2x+1)^3
[/quote]

Why $ (2x)^3 < 8x^2 \plus{} 8x \plus{} 8 < (2x \plus{} 1)^3$, where is root $ x \equal{} 18,y \equal{} 7$?
You are wrong![/quote]

Yeah, you're right :blush: . So i decide to take off the 1st case and againI have to consider the case when: /x/>=3(since x=2 is not the value of this equation).Then you can continue with the rest of my solution!  :)

[quote="ghjk"]
Since x<=-3 then x must divide by k or x must divide by 3+4k^2+6k
[/quote]
sorry but this is not true. same mistake as above. check rust's post

@Rust thanks for the hint

$ Z[i\sqrt3]$ isn't factorial but $ |x\parallel{}y| \equal{} |z|$ for comlex $ x,y,z$ limits the number of "choises for factorisation". I mean $ 4 \equal{} 4\cdot1 \equal{} 2\cdot2 \equal{} (1 \minus{} i\sqrt3)(1 \plus{} i\sqrt3)...$ but there are only a few of this possible factorisations. now please tell me if this is right or wrong:

we're operating with numbers of form $ a \plus{} bi\sqrt3$ where $ a,b$ are integers.
now $ 4y^3 \equal{} (2x \plus{} 1 \minus{} i\sqrt3)(2x \plus{} 1 \plus{} i\sqrt3)$. if there is such number of form $ a \plus{} bi\sqrt3$, other than $ 1$, that devides $ 2x \plus{} 1 \minus{} i\sqrt3$ and $ 2x \plus{} 1 \plus{} i\sqrt3$ than it has to devide $ 2i\sqrt3$. noticing 2 doesn't devide $ 2x \plus{} 1\pm\sqrt3i$ and with a limited selection of such numbers with absolute value of less or equal to $ 2i\sqrt3$ we come to the conclusion that the only such number could be $ i\sqrt3$.

if that were the case than $ y^3$ would be devisable by $ 27$, but $ x^2 \plus{} x \plus{} 1$ can't be devisable by $ 9$. therefore there is no such number other than $ 1$ that would devide both $ 2x \plus{} 1 \minus{} i\sqrt3$ and $ 2x \plus{} 1 \plus{} i\sqrt3$

$ 4$ has to devide $ (2x \plus{} 1 \minus{} i\sqrt3)(2x \plus{} 1 \plus{} i\sqrt3)$ and since $ 2$ doesn't devide $ 2x \plus{} 1\pm\sqrt3i$ we come to the only possible conclusion that $ (1 \plus{} i\sqrt3)$ devides $ 2x \plus{} 1 \plus{} i\sqrt3$ and $ (1 \minus{} i\sqrt3)$ devides $ 2x \plus{} 1 \minus{} i\sqrt3$ or the other way around.

therefore we come to the conclusion that $ 2x \plus{} 1 \plus{} i\sqrt3$ can be written in te form $ (1\pm\sqrt3i)(n \plus{} mi\sqrt3)^3$ with $ n,m$ integers.

From this, with regarding the imaginary part, we now get the equasion:
$ n^3 \plus{} 3n^2m \minus{} 9nm^2 \minus{} 3m^3 \equal{} \pm1$ which we unfortunatelly don't know how to solve either

can anyone solve this one? it's equivalent. i give up :(

$ y^3\minus{}3x^2y\pm1\equal{}x^3$


group theory
abstract algebra
superior algebra
superior algebra unsolved

If p, q are primes then a group G of order $ p^2q$ can not be simple. I have a solution but i feel it contains a mistake...
If q<p then there exists a subgroup of order q and as far as q is the smallest prime dividing order of G it must be normal(there is a theorem that claims this). If p<q then there is a subgroup of order $ p^2$, and in this subgroup there is an element of order p, which generates a subgroup of G, and by the same argument it must be normal. Is it correct?

No, there is no theorem stating that $ U\leq G$ is simple if $ |U|$ is the least prime dividing $ |G|$. If this would be true, there would be no simple groups at all. (Okay, $ \mathbb{Z}_p$ would still be simple...)

There is a similar theorem: If $ |G: U|$ is the least prime dividing $ |G|$ then U is a normal subgroup of G. So one case can be handled that way: If $ q<p$ then a p-sylowsubgroup of G has index q and is therefore normal.

The (standard?) proof is different:

We can assume that no p-sylowgroup is normal. Hence there are q p-sylowgroups at least.

If $ S_1$ and $ S_2$ are distinct p-sylowgroups $ D: \equal{}S_1\cap S_2$ can only have order 1 or order p.

1.Case: There are $ S_i$ s.t. $ |S_1\cap S_2|\equal{}p$
Then $ N_G(D)$ contains $ S_1$ and $ S_2$ as both groups are of order $ p^2$ and hence abelian. So $ p^2$ divides $ |N_G(D)|$ and $ |N_G(D)|$ is stricly greater than $ p^2$. So $ |N_G(D)|\equal{}p^2q\equal{}|G| \Rightarrow N_G(D)\equal{}G$, so D is normal in G.

2.Case: $ S_1\cap S_2\equal{}\{1\}$ for all distinct p-sylowgroups
The p-sylowgroups contain $ q(p^2\minus{}1)$ elements of order p or $ p^2$. There are only $ |G|\minus{}q(p^2\minus{}1)\equal{}q$ elements left, so there can only be one single q-sylowgroup which must be normal.

Gockel, thanks for pointing out my mistake. I think i can prove that it is solvable moreover: there must be a normal p-Sylow or q-Sylow subgroup, which implies solvability in this particular case. I used 2 statements of Sylow theorems: number or p-Sylow subgroups divides order of group G and #{p-Sylow subgroups}$ \equiv 1(mod \ p)$. Let $ q > p$; the # of p-Sylow subgroup is 1 or  q. If it is 1 then we are done, suppose it is $ q$. # of q-Sylow subgroups is 1, $ p$ or $ p^2$. If it is 1 we are done; $ q>p$ so it can not be p, then it is $ p^2$ and $ p^2 \minus{} 1\equiv 1 (mod \ q)$. q is prime, so we must have $ p \plus{} 1$ divides $ q$ and thus $ q \equal{} p \plus{} 1$, so q=3 and p=2, we can check this particular case easily.
Are there any mistakes?

[quote="Gockel"]Solvability can be proved easier: We already proved that there is a nontrivial normal subgroup. N and G/N are of order p,q or p^2. Therefore N and G/N are solvable and so is G.[/quote]
How about $ |G/N|\equal{}pq$?

Oh, yeah, i forget this case. But this case is known to be solvable too for it has a normal syslowgroup.


Herzlichen Glückwunsch allen deutschen IMO-Teilnehmern!
;)  :first:  ;)  :first:  ;)

Immerhin habt ihr 2,5 von drei großen Gronau-Zielen erfüllt (Für Nicht-Insider: Die Ziele lauten: mindestens eine Gold-Medaille, eine Medaille für jeden und TOP 10). Hier zum Vergleich die Bilanzen der Vorjahre:

2004: kein Ziel
2003: zwei Ziele
2002: ein Ziel
2001: ein Ziel

Nochmal herzlichen Glückwunsch!

Herzlichen Glückwunsch an Alle!

wenn die Ziele damals die gleichen waren hast du 2003 mit 2002 vertauscht oder ist ne Anerkennung seit neustem eine Medalie ;-) ? (Friedrich und Alex hatten beide nur ne A.) dafür waren wir 2002 aber auf platz 10

Strimmt, du hast natürlich Recht. Das gute Jahr 2002 liegt ja schon drei Jahre zurück!  ;)

Und Christians Ziel, vor der Mongolei zu landen, haben wir auch erfüllt. Landen im wörtlichen Sinne...

  dg

Mein Glückwunsch an alle IMO-Teilnehmer. :)


function
abstract algebra
group theory
superior algebra
superior algebra solved

a)(H.B.Mann)Let $G$ be a finite group,and let $S$ and $T$ be(not necessarily distinct) noempty subsets.Prove that either $G=ST$ or $|G|\geq |S|+|T|$.
b)Prove that every element in a finite field is a sum of two squares.

[i]Rotman[/i]

b/ find the number of squares. The conclusion will follows.

a) Assume $|G|<|S|+|T|$.
Let $x\in G$. The sets $S^{-1}\cdot x$ and $T$ must intersect because $|S^{-1}\cdot c|+|T|=|S|+|T|>|G|$. Let $s^{-1}x=t$. Then $x=st$ and we are done. I denoted $S^{-1}\cdot x=\{s^{-1}x|s\in S\}$. $S^{-1}\cdot x$ and $S$ have the same number of elements because the application $G\to G: g\to g^{-1}\cdot x$ is bijective (be careful about the distinction bijective vs. isomorphism).

b)
If $K$ is of characteristic 2, then all the elements of $K$ are squares and we have nothing to prove. This is because the function $K\to K: x\to x^2$ is Frobenius's automorphism, hence surjective.

If $K$ has an odd number of elements $n$, then it has odd characteristic. The kernel of the group homomorphism $K^*\to K^*: x\to x^2$ is $\pm 1$, hence it's image has $\frac{n-1}2$ elements. So $K$ has $\frac{n+1}2$ squares (the element 0 which is a square has to be counted also in $K$).
Let $S$ be the set of squares. Then $|S|+|S|=n+1>n=|K|$ and we can apply (a) for $S=T$ and $G=(K,+)$.

This followed alekk's idea.

[quote="amfulger"]
If $K$ is of characteristic 2, then all the elements of $K$ are squares and we have nothing to prove. This is because the function $K\to K: x\to x^2$ is Frobenius's automorphism, hence surjective.

[/quote]

I no understand,help me!

[quote="jmerry"]We have $(x+y)^2=x^2+y^2$ and $(xy)^2=x^2y^2$, so squaring is a (ring) homomorphism. A field has no nontrivial ideals, so a nonzero homomorphism from a field is injective. An injective map from a finite set to itself is surjective, so squaring is an automorphism.

The same applies to the $p$th power map in a finite field of characteristic $p$. This is known as the Frobenius automorphism, and it actually generates the automorphism group of the field.

These maps do not need to be surjective if the field is infinite.[/quote]

Thank you :)


limit
logarithms
Euler
real analysis
real analysis unsolved

1.Find all $\alpha \in \mathbb{R}$ such that $\exists \lim_{n\to+\infty}x_{n}\in \mathbb{R}$
$x_{n}=1^{\alpha}+2^{\alpha}+...+n^{\alpha}+\alpha .lnn$
2.the same question with:
$y_{n}=1^{\alpha}+2^{\alpha}+...+n^{\alpha}+\frac{1}{\alpha .n^{\alpha}}$

[quote="didilica"]I will discuss only the first problem.

Definitely that if $\alpha \geq 0$ then the limit is $\infty$.

let us consider the case when $\alpha=-\frac{1}{\beta}<0$, where $\beta >0$.

Then $x_{n}=1+\frac{1}{2^{\beta}}+\cdots+\frac{1}{n^{\beta}}-\frac{\ln{n}}{\beta}$.

We distinguish here three cases:

1) $\beta =1$, in this case the sequence converges to $\gamma$.

2) $If \beta >1$ then the sequence diverges.

3) If $\beta \in (0, 1)$ then we have that

$x_{n}=\left(1+\frac{1}{2^{\beta}}+\cdots+\frac{1}{n^{\beta}}-\frac{n^{1-\beta}}{1-\beta}\right)+\frac{n^{1-\beta}}{1-\beta}-\frac{\ln n}{\beta}$
Now, since the paranthesis converges and $\lim \left(\frac{n^{1-\beta}}{1-\beta}-\frac{\ln n}{1-\beta}\right)=\infty$ it simply follows that your sequence diverges, hence the only real values for which the initial sequence converges is $\alpha=1$ for which the limit is nothing else but the famous Euler-Mascheroni constant.[/quote]
Yes,that's right.I have a similar solution
part 1:$\alpha =-1$
and part 2:$\alpha =-\frac{1}{2}$


geometry
3D geometry
tetrahedron
inequalities
triangle inequality

[quote="moldlee"]How many different tetrahedrons are there with whole number-long

edges, and with 17 as the sum of the lengths of the edges?[/quote]
[hide]$x+y+z+a=17$
balls walls urns sticks dots barriers laters
$\binom{16}{3}=560$[/hide]

[quote="bpms"]
[hide]$x+y+z+a=17$
balls walls urns sticks dots barriers laters
$\binom{16}{3}=560$[/hide][/quote]

[hide]1.   Tetrahedrons have more than four edges.
2.   Triangle Inequality[/hide]


inequalities
Cauchy Inequality
inequalities proposed

Let $ a,b,c$ be sidelengths of a triangle. If $ A$ and $ B$ are like below, prove that $ AB\ge9$.

$ A \equal{} \frac {a^2 \plus{} bc}{b \plus{} c} \plus{} \frac {b^2 \plus{} ca}{c \plus{} a} \plus{} \frac {c^2 \plus{} ab}{a \plus{} b}$

$ B \equal{} \frac {1}{\sqrt {(a \plus{} b \minus{} c)(b \plus{} c \minus{} a)}} \plus{} \frac {1}{\sqrt {(b \plus{} c \minus{} a)(c \plus{} a \minus{} b)}} \plus{} \frac {1}{\sqrt {(c \plus{} a \minus{} b)(a \plus{} b \minus{} c)}}$

We will prove that
$ A \ge a \plus{} b \plus{} c$ and $ B \ge \frac {1}{a} \plus{} \frac {1}{b} \plus{} \frac {1}{c}$

Therefore, $ AB \ge (a \plus{} b \plus{} c)\left( \frac {1}{a} \plus{} \frac {1}{b} \plus{} \frac {1}{c} \right) \ge 9$

[b][u]First part[/u][/b]:

$ \sum_{cyc} \frac {a^2 \plus{} bc}{b \plus{} c} \plus{} \sum_{cyc}a \equal{} \sum_{cyc}\frac {(a \plus{} b)(a \plus{} c)}{b \plus{} c}$

Applying cauchy inequality, we derive

$ \frac {(a \plus{} b)(a \plus{} c)}{b \plus{} c} \plus{} \frac {(b \plus{} a)(b \plus{} c)}{a \plus{} c} \ge 2(a \plus{} b)$

[b][u]Second part[/u][/b]:

$ \sum_{cyc}\frac {1}{\sqrt{(a \plus{} b \minus{} c)(b \plus{} c \minus{} a)}} \ge \sum_{cyc}\frac {2}{(a \plus{} b \minus{} c) \plus{} (b \plus{} c \minus{} a)} \equal{} \sum_{cyc}\frac {1}{b}$

We are done!

[quote="great math"]We will prove that
$ A \ge a \plus{} b \plus{} c$ and $ B \ge \frac {1}{a} \plus{} \frac {1}{b} \plus{} \frac {1}{c}$

Therefore, $ AB \ge (a \plus{} b \plus{} c)\left( \frac {1}{a} \plus{} \frac {1}{b} \plus{} \frac {1}{c} \right) \ge 9$

[b][u]First part[/u][/b]:

$ \sum_{cyc} \frac {a^2 \plus{} bc}{b \plus{} c} \plus{} \sum_{cyc}a \equal{} \sum_{cyc}\frac {(a \plus{} b)(a \plus{} c)}{b \plus{} c}$

Applying cauchy inequality, we derive

$ \frac {(a \plus{} b)(a \plus{} c)}{b \plus{} c} \plus{} \frac {(b \plus{} a)(b \plus{} c)}{a \plus{} c} \ge 2(a \plus{} b)$

[b][u]Second part[/u][/b]:

$ \sum_{cyc}\frac {1}{\sqrt {(a \plus{} b \minus{} c)(b \plus{} c \minus{} a)}} \ge \sum_{cyc}\frac {2}{(a \plus{} b \minus{} c) \plus{} (b \plus{} c \minus{} a)} \equal{} \sum_{cyc}\frac {1}{b}$

We are done![/quote]

Nice solution, great math!