A bear walks 10 miles south, 10 miles east and 10 miles north, and he is back on his starting point. What is the color of the bear?

[quote="alan"][hide]white[/hide][/quote][hide="Actually..."]The question is ambiguous, because the hair is white, but the bear itself (its skin) is dark dark black.[/hide]

And of course there are no bears in Antarctica of any color.


A two-person game begins with the random selection of an integer
from the set {1, 2, 3}, each choice being equally likely. Then two
players, not knowing the value of x, simultaneously select integers
from {1, 2, 3}. . Each player’s objective is to select an integer x. If
only one player is successful, he wins. If both players are successful,
and their choices are not the same, the winner is the player who
chose the smaller integer. In all other cases, the game is a draw.
The winner gets $ \$$1 from the loser. Describe the pure strategies
for each player, and the payoffs. . Show that the game has a
unique equilibrium strategy solution (this means that if one player
consistently plays this strategy, there is no advantage to the other
in changing their strategy).

hmmm...
YOU'RE CLOSE!
CLOSE!
AW SO VERY VERY CLOSE!
CLOSE!
AW YOU'RE CLOSE, SO CLOSE, BUT NO CIGAR!
wrong forum.
mods plz move to Intermediate/pre-olympiad.


If $ x$ is rational and non-integer, $ x>1$, is the sequence: $ a_{n} \equal{} frac(x^n)$ dense on $ [0,1]$? (where frac means the fractional part).
I understand that Hardy and Littlewood proved that for almost all $ x$, the same sequence is equally distributed over $ [0,1]$ for almost all $ x$, with exceptions being e.g. $ x \equal{} \frac {n \plus{} \sqrt{n^2\plus{}4}}{2}$. But for the exceptions to this fact, are any of them rational? If not, this problem is a direct consequence of Hardy and Littlewood's result.
I tried using Weyl's criterion ( [url]http://planetmath.org/encyclopedia/WeylsCriterion.html[/url] ).
However, I have no idea how to evaluate that sum.

Are you asking if $ {n\plus{}\sqrt{n^2\plus{}4}\over 2}\in\mathbb{Q}$ for some $ n\in\mathbb{N}$,
because that is easy.

Maybe you could outline the proof of Hardy and Littlewood?
If it's not to long.

[quote="olorin"]Are you asking if $ {n \plus{} \sqrt {n^2 \plus{} 4}\over 2}\in\mathbb{Q}$ for some $ n\in\mathbb{N}$,
because that is easy.

Maybe you could outline the proof of Hardy and Littlewood?
If it's not to long.[/quote]

The solution is only $ n\equal{}0$ :huh:
$ x\equal{}{n \plus{} \sqrt {n^2 \plus{} 4}\over 2}$ is only an example of the exceptions, there should be other.

[quote="olorin"]Are you asking if $ {n \plus{} \sqrt {n^2 \plus{} 4}\over 2}\in\mathbb{Q}$ for some $ n\in\mathbb{N}$,
because that is easy.

Maybe you could outline the proof of Hardy and Littlewood?
If it's not to long.[/quote]

No, that's not what i'm asking, clearly that number is irrational for all $ n > 0$: i'm asking if the sequence $ frac (x^n)$ is dense on $ [0,1]$ for all $ x$, such that $ x > 1$, $ x$ is non-integer and rational.
Hardy and Littlewood proved that this is true for almost all $ x$, and SOME of the exceptions are of that form (the surd) - but I'm not sure if ALL of the exceptions are of that form.
As for the proof, I do not know it. I was reading an article about the problem on mathworld wolfram site, and since it said that it was only proved by Hardy and Littlewood in the last century; so it is probably very long and difficult, and uses heavy machinery.


Let $ a_n$ be a sequence such that $ a_{n} \equal{} \frac {a_{n \minus{} 1}^{2} \plus{} T}{a_{n \minus{} 2}}\; \forall n\geq 3$, where $ T,a_1,a_2$ are   coprime positive integers (only $ 1$ is  common divisor ) .
Prove  that if $ a_3,a_4$ are integers  then all $ a_n$ are integers..  :D ..   (Related to http://www.mathlinks.ro/viewtopic.php?t=201188 )

 :cool:

[quote="TTsphn"]Posted before . ....  Problem claimed .[/quote]  .
Have you got a link ?
 :cool:


A work crew of 3 people requires 3 weeks and 2 days to do a certain job.  How long would it take a work crew of 4 people to do the same job if each person of both crews works at the same rate as each of the others?  Note: each week contains six work days.

[hide]3 people*20 days = 4 people*x days (60 = work)
x = 15, or 2weeks, 3 days[/hide]

[hide]  With 3 people working it takes, 3 * [( 6*3)  + 2]= 60-manpowered days of work.  With 4 people it will take 60/4= 15-days of manpowered work.  This is equivlant to 2 weeks and 3 days.[/hide]


geometry
function
3D geometry
conics
ellipse

What Round 3 problem was your favorite? Mine was number 5, because it had phi in it.

5 was incredibly cool ... seemingly difficult until you played with it and everything dropped kind of dropped away (and the source of some awesome TeXing).

4 was fun but tedious, 2 was fun in that it couldn't be solved outright but only via use of some manipulation and then some case solving.

1 was tedious.  3 was messy and the answer worried me (what kind of triangle has area $12\sqrt{5}-12\sqrt{3}$?)

JB

I LOVE functional equations.  So five all the way :)

#5: First functional equation I'd ever solved. :D

[quote="Xantos C. Guin"]#5: I found $f(x)$ and wrote a 1/2 page solution.[/quote]
Watch out - finding a specific solution to the equation doesn't prove that $f(1)$ is equal to a specific value for all possible solutions.  Unless you can prove that $f(x) = \frac{\phi}{x}$ (which I assume is the solution you came up with; I encountered it as well when playing with the problem) is the only monotonically increasing solution, then you can't say that $f(1) = \phi$ for all possible functions $f(x)$.

Edit: I looked at your solution, and you appear to do that; cool!

[quote="besttate"][quote="zanttrang"][quote="Xantos C. Guin"]Unless you can prove that $f(x) = \frac{\phi}{x}$[/quote]

I thought it was $\frac{-1}{\phi x}$[/quote][/quote]

Yes you are right.

$f(x)=\frac{1-\sqrt{5}}{2x}=\frac{-1}{\phi x}$ for all $x>0$

BTW: your post makes it look like I said that quote.

Mine was 1, I came up with a pretty cool solution for it. If you color a formation looking like |_| with three different colors, the rest of the cube is forced. There are 3 distinct colorings for this. If the |'s are the same color, then the _'s have to be the same color, or else it would be the same case as the first. So if we continue to color the second case, there is one point where the coloring isn't forced, but again, if there's only one way to color it differently than in case 1. So total 4.

That's the basic outline.

#2 was all right, but it wasn't too interesting, since I got the solution so easily. #3 was cool, I saw the ellipse part, but it took me 3 days to get off my laziness and actually do the algebra. I got it anyway. #4 was pretty tedious, and too long, I didn't want to read it. And #5 seemed pretty cool, but I didn't get it.

[quote="besttate"][quote="zanttrang"]Unless you can prove that $f(x) = \frac{\phi}{x}$[/quote]

I thought it was $\frac{-1}{\phi x}$[/quote]

Sorry - I was thinking of $\phi = \frac{1 \pm \sqrt{5}}{2}$, both of which work for the function (though only one is monotonically increasing).

[quote="Xantos C. Guin"]#5: I found $f(x)$ and wrote a 1/2 page solution.[/quote]

Oh, I did that too :D

It was very nice!!!


geometry
3D geometry
trigonometry
function

If n is the smallest positive integer for which positive a and b exist that make (a + bi)^n = (a - bi)^n , compute the numerical value of b/a.

[hide]By DeMoivre, we can see that n=3, and we find the imaginary cube roots of -1.

Therefore, b =  :rt3:*a.  b/a = :rt3:[/hide]

Edit: oops, it's b/a

[b][color=red]Alex, your solution is correct but they were asking for b/a not a/b so the answer would be :rt3:.[/color][/b]

[quote]By DeMoivre, we can see that n=3[/quote]

Can you explain how you used DeMoivre's theorem here?

You dont really know what quadrant a+bi and a-bi are in, so thats confusing me about finding  :theta: .

You have:

( :sqrt: a^2+b^2)^n * cis(n*arctan(b/a)) = ( :sqrt: a^2+b^2)^n *cis(n*arctan(-b/a))

cis(n*arctan(b/a)) = cis(n*arctan(-b/a))
cos(n*arctan(b/a)) + i*sin(n*arctan(b/a)) = cos(n*arctan(-b/a)) + i*sin(n*arctan(-b/a))

From there I am confused.

[quote="probability1.01"]They're in quadrant I because they're positive...[/quote]

You don't know that.  Furthermore, even if you did, they wouldn't both be in the same quadrant.

[quote="mallachaitanya"]If n is the smallest integer for which [b]positive [/b]a and b exist that make (a + bi)^n = (a - bi)^n , compute the numerical value of b/a.[/quote]

btw I meant a + bi.  Obviously a - bi would be in quadrant IV.  Sorry for the confusion.

Yes, okay, thanks, but how do you apply demoivre's theorem to get n=3?

[b][color=red]My Solution using DeMoivre's Theorem:[/color][/b]
[color=red]
a+bi = r cis  :theta:
a-bi = r cis  -:theta:

It states that a and b are positive numbers so 0 < :theta: < 90 :^o:.

(r cis  :theta:)^n = (r cis - :theta:)^n
r^n cis n:theta: = r^n cis -n:theta:
cis n:theta: = cis -n:theta:
n :theta: = 360 :^o:-n:theta:
2n :theta: = 360 :^o:

0 < 2n :theta: < 180 :^o:n
0 < 360 :^o: < 180 :^o:n
0 < 2 < n

The least possible value of n is 3.

n = 3
2n :theta: = 360 :^o:
6:theta: = 360 :^o:
:theta: = 60 :^o:

b/a = tan(:theta:) = tan(60 :^o:) = :rt3:[/color]

[quote]n  = 360 -n
[/quote]

This technically should be

n :theta: = k*360 - n :theta: where k is an (positive due to constraints on :theta:) integer

but it's basically the same, since the end result is
:theta: = 180*k/n

and k might as well equal 1 because the term k/n and as a result :theta: is smallest when k = 1.

gj warpedklown

Thanx.  I was going to use 360 :^o:z, but I should've just stated that and said that z = 1 to minimize n.  Either way still works.

[quote]cis n = cis -n
n  = 360 -n [/quote]

Why?

Also, how do you know that there is a one-to-one property w/cis?  Cause trig functions are not one-to-one.  Like if tanx=tany that dosent mean x=y.

[quote="JS1527"][quote]cis n :theta: = cis -n :theta:
n :theta: = 360 - n :theta:[/quote]

Why?
[/quote]

We want n :theta: to be equivalent to -n :theta: (in other words, n :theta: = k*360 + -n :theta: as discussed somewhere earlier).

[quote]
Also, how do you know that there is a one-to-one property w/cis?  Cause trig functions are not one-to-one.  Like if tanx=tany that dosent mean x=y.[/quote]

While it is true that tan x = tan y doesn't mean x = y, here we have cis, which has an inverse function (think of a unit circle).


conics ellipse trigonometry parabola hyperbola

at what time is the equation r=L/[1-ecos(a)] an elipse?
e and L are constants,(the eccentricity and semilatus rectum)
also transform this polor co-ordinates equation into a cartesian co-ordinate equation.

I havent got a very good background in conics... please help thx :roll:

r= x^2 + y^2?....i thought r^2 = x^2+y^2.. can u show me how did u do that?

err wait. i am wrong. you are right.. i lose.. :| :blush:

\begin{eqnarray*}r=\frac{L}{1-e\cos\theta}&\implies& r-er\cos\theta=L\\ &\implies& r=L+er\cos\theta\\ &\implies& \sqrt{x^{2}+y^{2}}=L+ex\\ &\implies& x^{2}+y^{2}=L^{2}+2eLx+e^{2}x^{2}\\ &\implies& (1-e^{2})x^{2}-2eLx+y^{2}=L^{2}\end{eqnarray*}

If $|e|<1$ then it's an ellipse (with the special case of circle for $e=0$);
If $|e|=1$ then it's a parabola;
If $|e|>1$ then it's a hyperbola.

It doesn't really matter if $e$ is negative, since $1+e\cos\theta=1-e\cos(\theta+\pi)$, hence it's still the same curve with the different angular offset.

or think of it this way...
to be an ellipse, r must be bounded and well defined for all angles, hence |e|<1

May be it is noted before, but  he basic definition of any conic curve (ellipse/parabola/hyperbola) etc .

[quote]A conic section may more formally be defined as the locus of a point P that moves in the plane of a fixed point F called the focus and a fixed line d called the conic section directrix (with F not on d) such that the ratio of the distance of P from F to its distance from d is a constant e called the eccentricity. If e==0, the conic is a circle, if 0<e<1, the conic is an ellipse, if e==1, the conic is a parabola, and if e>1, it is a hyperbola.
 [/quote]

Here write $r=\frac{L}{1-e\cos\theta}$

as $r = e (r \cos \theta+L/e)$

If you take focus at the origin and directrix as a vertical line ($x=-a \ constant$  which you will see as $L/e$

Then distance from the focus of any point  = $r$
Distance from the directrix = $x+(L/e) = r \cos \theta+(L/e)$

And according to the definition of conic section :

$e<1 \implies ellipse$  ($e=0 \implies circle$)
$e=1 \implies parabola$
$e>1 \implies hyperbola$


Hey guys,

I'm definitely sure that I want to be a hedge fund analyst, preferably a quant, eventually, but I'm not sure what level of education would be necessary to get a quant job. Are there many hedge funds that frequently hire those with only undergraduate degees for quant jobs, outside of DE Shaw and Jane Street Capital? And would a MS in financial engineering/financial mathematics be beneficial for getting hired at these same firms?

Thanks for helping out.

[quote="brycha"]Are there many hedge funds that frequently hire those with only undergraduate degees for quant jobs, outside of DE Shaw and Jane Street Capital? [/quote]

Not as many as there used to be!

I don't think anyone will know the answer to this question for another year or so -- it's not even clear who will be left in 6 months.  There will still be people hiring quants for trading, but not nearly as many as were hiring 2 years ago.  As for what everything will look like 3-5 years from now, who knows?

[quote="brycha"]
And would a MS in financial engineering/financial mathematics be beneficial for getting hired at these same firms?[/quote]

These things vary a lot from firm to firm.  The few that I know a lot about are more interested in people with proven high-level quantitative knowledge (CS, physics, math, applied math) than with specific financial knowledge.  Obviously, having both would be a big win.

The current edition of the economist has the following story: [url=http://www.economist.com/finance/displaystory.cfm?story_id=12778069]The hedge-fund industry imprisons its clients and weakens its appeal.[/url] It also mentions funds like Fortress Investment Group, Tudor Investment, Citadel, D.E. Shaw and Farallon Capital Management.

To play devil's advocate, I'll note that more than a few people blame the quickness of capital flight for playing a large role in the current crisis.  Not only that, but many countries place similar barriers to expatriating funds once they've been invested.  By agreeing to not pull your money out very quickly, you are also buying the protection from *others* pulling their money out quickly and thereby damaging your investment.  As with most issues, this one's much more complicated than meets the eye.

(That said, I still find the fee structure for investing in hedge funds simply slobberknocking.  But then, I find the prices people pay for Rolexes pretty nuts, too.)

Sure what is the preferred strategy on the micro level may not be good on the macro level. May we interpret this money retention stop as "behaviorally oriented regulation" as Stephen J. Dubner discusses in [url=http://freakonomics.blogs.nytimes.com/2009/01/09/this-is-your-brain-on-prosperity-andrew-lo-on-fear-greed-and-crisis-management/]his exciting blog entry?[/url] We can view it plainly as a hedge fund's strategy to stay into business but I think the behavioral finance view adds important insights.


Please find out all the numbers "a" which makes possible that there are five nonnegative numbers going as follows and satisfying the following three equations.

LaTeX:

$x_1,x_2,x_3,x_4,x_5$
such that
$\displaystyle\sum_{k=1}^{5}kx_k=a$
$\displaystyle\sum_{k=1}^{5}k^3x_k=a^2$
$\displaystyle\sum_{k=1}^{5}k^5x_k=a^3$

It's from IMO 1979. See:
http://www.mathlinks.ro/Forum/viewtopic.php?p=367346#p367346


Since $ \sinh (x)$ is an odd function and $ \cosh (x)$ is an even function, then $ \tanh(\minus{}x) \equal{} \frac{\sinh (\minus{}x)}{\cosh(\minus{}x)} \equal{} \frac{\minus{}\sinh(x)}{\cosh(x)} \equal{} \minus{}\tanh(x)$. So $ \tanh(\minus{}3) \equal{} \minus{}\tanh(3)$.

$ \frac{e^{\minus{}3} \minus{} e^3}{e^{\minus{}3} \plus{} e^3} \equal{} \frac{\minus{}(\minus{}e^{\minus{}3}\plus{}e^3)}{e^{\minus{}3}\plus{}e^3}$


geometry
3D geometry
tetrahedron
combinatorial geometry
combinatorics proposed
combinatorics

Given a shape $X \subset \mathbb{R}^n$ prove that if $X$ is connected then the convex hull of $X$ is the union of convex hulls of its subsets with size $n$
${ \displaystyle \\conv(X)=\bigcup \{{\sum_{i=1}^n{\lambda_ix_i} \mid \lambda_i\geq 0 , \sum{\lambda_i}=1,x_i\in X}}\}$

Wow! Cool! :) It's a generalization of [url=http://www.mathlinks.ro/Forum/viewtopic.php?t=22792]this one[/url], where we had $n=2$.

This is something stronger than Caratheodory theorem for connected sets.
Me and Ali proved this 5 days ago

And also Caratheodory theorem is that:
$ \displaystyle \\conv(X)=\cup {\sum_{i=1}^{n+1}{\lambda_ix_i}}$ such that $\lambda_i\geq 0$ and ${{\sum{\lambda_i}=1,x_i\in X}}$

Hint:
We prove it for $n=2$
if $P \in conv(X)$ then by Caratheodory theorem is in the convex hull of three points of X.so there is $A,B,C\in X$ so that P is in triangle $ABC$.If P is on one of the sides then the assertion is true(we want to prove that $P$ is between two points of $X$).If not, Then a connected path between $A,B$ cuts one of halflines showed below.If this path cuts for example half line $L$ at $M$ then $P\in AM$.

I think I proved it for $n=2$ in that thread I mentioned in my first mssage on this topic. Besides, I think you considered your set to be path connected, while the problem asks to prove it for connected sets. Please don't post any more hints, because I don't think people like solving problems after being given copious hints... :)

Can anybody definitely say what kind of connectivity we should use?

I think it's connectedness (i.e. we don't need path connectedness). I can visualize it in three dimensions and I think I can solve it in the general case, but I don't know if I can explain it properly.

I'll solve it for $n=3$, and I'm sure it will be obvious what to do for higher dimensions.

By Caratheodory (I didn't know this theorem when I solved the problem in the other thread, for $n=2$ :)), we know that the point we're working with, $P$, lies in a tetrahedron $ABCD$ with $A,B,C,D\in\mathcal X$. We may assume $P$ is strictly inside this tetrahedron and does not belong to $\mathcal X$, otherwise we're done. Consider now the (open) region $\mathcal R$ of space bounded by the plane angles $\angle B'PC',\angle C'PD',\angle D'PB'$, where $B',C',D'$ are on $(BP,(CP,(DP$ respectively ($B,B'$ are on opposite sides of $P$, adn the same holds for $C,C'$ and $D,D'$).

If none of these angles contains a point from $\mathcal X$, the partition $\mathcal X=(\mathcal X\cap\mathcal R)\cup (\mathcal X\setminus \mathcal R)$ shows that $\mathcal X$ is not connected, because both $\mathcal X \cap \mathcal R$ and $\mathcal X\setminus\mathcal R$ are non-void, the first one containing $A$, and the second one containing $B,C,D$, for example.
Think it's Ok? It seems too short.. :?


An equilateral triangle of side length $ n$ is divided into unit triangles. Let $ f(n)$ be the number of paths from the triangle in the top row to the middle triangle in the bottom row, such that adjacent triangles in a path share a common edge and the path never travels up (from a lower row to a higher row) or revisits a triangle. An example is shown on the picture for $ n \equal{} 5$. Determine the value of $ f(2005)$.

The answer is $ 2004!$. Let's consider only upward directed triangles. The number of triangles on the first level is $ 1$, on the second level there are  $ 2$ triangles and so on. Let's call the current position of the path a [b]point.[/b] If the point is on the $ k$-th level, it may pass through the base of any triangle on the $ k$-th level. And notice, if you choose randomly any bases on each level, there will be a unique path, which goes through those bases.

The number of such choices is $ 2004!$, as there are $ 1$ base on the first level, $ 2$ bases on the second,...,and $ 2004$ bases on the $ 2004$-level.

This can also be seen -
To go from $n$ th level to $n+1$ th level there are exactly $n$ ways.
So $f(n+1)= nf(n)$.
Hence problem can be solved by noticing that  $f(1)=1$.


Evaluate $ \frac{\sqrt[3]{108}}{\sqrt[3]{32}}$. Express your answer as a common fraction.


If Arccos(x) = Arctan(f(x)), determine the exact value of f(1/3).

Don't use a computer/calculator/slide rule!  Use your brain! :)

By the way, is the capitalization of the "A" at the beginning of "Arc" a standard method for indicating that we only want the principal values?  What are the principle values for cosine -- 0 to :pi:?

[hide]Arccos(x) = Arctan(f(x))

tan(Arccos(x)) = f(x) = sin(Arccos(x))/cos(Arccos(x)) = sin(Arccos(x))/x

so f(1/3) = 3sin(Arccos(1/3))

cos(Arccos(1/3)) = 1/3, so sin(Arccos(x)) = :pm: :sqrt: 8/9 =  :pm: 2 :rt2: /3

hence 3sin(Arccos(1/3)) =  :pm: 2 :rt2: [/hide]

JBL wrote:By the way, is the capitalization of the "A" at the beginning of "Arc" a standard method for indicating that we only want the principal values?  What are the principle values for cosine -- 0 to :pi:?

Yes, the capital A means it's a function. Thus the Arccos can only take on values from 0 to  :pi:

ThAzN1 wrote:[hide]Arccos(x) = Arctan(f(x))
tan(Arccos(x)) = f(x) = sin(Arccos(x))/cos(Arccos(x)) = sin(Arccos(x))/x

so f(1/3) = 3sin(Arccos(1/3))
cos(Arccos(1/3)) = 1/3, so sin(Arccos(x)) =   8/9 =  2  /3
hence 3sin(Arccos(1/3)) =   :pm: 2 :rt2: [/hide]

Since Arccos can only range from 0 to  :pi: , sin(Arccos(x)) is always positive. So the answer is just [hide]2sqrt(2)[/hide].


geometry
geometric transformation
dilation
angle bisector

In convex quadrilateral $ ABCD$, the rays $ BA,CD$ meet at $ P$, and the rays $ BC,AD$ meet at $ Q$. $ H$ is the projection of $ D$ on $ PQ$. Prove that there is a circle inscribed in $ ABCD$ if and only if the incircles of triangles $ ADP,CDQ$ are visible from $ H$ under the same angle.

Let $ K$ and $ L$ be the centers of incircles $ k$ and $ l$ of triangles $ ADP$ and $ CDQ$ respectively. Denote their inradii by $ r_k$ and $ r_l$ respectively. Let lines $ PQ$ and $ KL$ meet at $ M$ (on the projective plane). The incircles are seen from $ H$ under the same angle $ \Leftrightarrow$ $ HK/HL \equal{} r_k/r_l \equal{} VD/DL$ $ \Leftrightarrow$ $ HD$ bisects angle $ KHL$ $ \Leftrightarrow$ $ DHM$ is an Apollonius circle of triangle $ KHL$ $ \Leftrightarrow$ $ (K,L,D,M)$ is harmonic $ \Leftrightarrow$ the external common tangents to $ k$ and $ l$ meet at $ M.$ Now, if a circle $ w$ is inscribed into $ ABCD,$ then $ k$ and $ l$ meet at $ M$ simply by the Three Circles Theorem applied to $ k,l,w.$ Conversely, if $ k$ and $ l$ meet at $ M,$ then let $ v$ be the excircle of triangle $ ADP$ opposite to $ P.$ Applying the Three Circles Theorem to $ k,l,v$ yields that the external common tangents to $ v$ and $ l$ meet on $ MP,$ so they have to meet at $ Q$ and thus $ BC$ is tangent to $ v$ as well.

2discredit. Can you also specify which class this problem belongs? It has some sense for ARO, as its number also.

[quote="lego"] $ \Leftrightarrow$ $ HK/HL \equal{} r_k/r_l \equal{} VD/DL$ $ \Leftrightarrow$ $ HD$ bisects angle $ KHL$ $ \Leftrightarrow$ $ DHM$ is an Apollonius circle of triangle $ KHL$ $ \Leftrightarrow$ $ (K,L,D,M)$ is harmonic .[/quote]

Please clarify more... what is point $ V$?

And no matter what $ V$ is, I can't see how the harmonic comes...

[b]lego[/b]'s solution is fine: it just has the minor typo: $ V \equal{} K$.

It was proven that $ \frac{KH}{HL} \equal{} \frac {r_k}{r_l} \equal{} \frac {KD}{DL}$, meaning that $ HD$ bisects $ \angle KHL$.  Furthermore, since $ \angle DHM \equal{} 90$ and the external and internal bisectors of an angle are perpendicular, it is seen that $ HM$ is the exterior bisector of $ \angle KHL$.  Thus, $ \frac {MK}{ML} \equal{} \frac {HK}{HL} \equal{} \frac {KD}{DL}$ and $ (M, K, D, L)$ is harmonic.

Using lego's notation, and define $ I$ be center of circle which inscribed in quadrilateral $ ABCd$
if we use menelaus theorem at $ \triangle IKL$ and line $ {MPQ}$, we can show without using harmonic quadruples.

Let $w_1,w_2$ be the incircles of $\triangle ADP, \triangle CDQ$, and $I_1,I_2$ the incenters. Let $I=PI_1 \cap QI_2$.  Consider the dilation about $D$ (with negative factor) which takes $w_1$ into $w_2$.  Then $I_1 \mapsto I_2$, and define $R,H'$ as the images of $P,H$.  Let $S=RI_2 \cap PQ$.

By the conditions of the problem, $H,H'$ subtend equal angles from $w_2$, which means $HI_2=H'I_2$, or $I_2$ lies equidistant from parallel lines $PQ,RH'$.  Thus $I_2$ is the midpoint of $RS$.  If $X$ is the point at infinity of $RS$, then $X,I_2;S,R$ are harmonic.  Projecting from $P$ onto $IQ$ yields that $I,I_2;Q,T$ are harmonic as well ($PI \parallel RI_2$), where $QT$ is the angle bisector in $\triangle CDQ$.  This means that $I$ is the excenter of $\triangle CDQ$ opposite $Q$.  Similarly it is the excenter of $\triangle ADP$ opposite $P$, and we are done.  To prove the converse simply reverse the steps.

It is fairly well known that $I,I_2;Q,T$ are harmonic, but can be easily shown by projecting the points from $C$ onto the line through $Q$ parallel to $CI_2$.

[img]https://hostr.co/file/970/BOojGUuE7xVC/2008_5.PNG[/img]

Define $K$ to be the incenter of $ADP$ and $M$ to be the incenter of $CDQ$. Let $F$ be the intersection of $KDM$ (external bisector of $\angle{PDQ}$). Clearly, the ratio of the inradii of $ADP$ and $CDQ$ is $KD$ to $DM$, so the condition is equivalent to $\frac{KH}{MH} = \frac{KD}{DM}$. This is equivalent to $HD$ being the angle bisector of $\angle{KHM}$, which due to $DH \perp HF$ is equivalent to $(K, M; D, F) = -1$.

If $ABCD$ has an inscribed circle, call it $I$. Now let $ID \cap PQ = R$. Since $R$ is the foot of the interior angle bisector of $PDQ$, it can be seen that $(P, Q; R, F) = -1$. Since $RD$, $PK$, and $QM$ all pass through $I$, observe that

\[ (K, M; D, F) \stackrel{I}{\doublebarwedge} (P, Q; R, F) = -1 \] as desired.

The reverse implication is not hard either: let $PK$ and $MQ$ intersect at $I'$, then let $R' = I'D \cap BC$. Clearly $(P, Q; R', F) = -1$ by a perspecitivity from $I$, so $R' = R$, and the angle bisectors of $\angle{ADC}$, $\angle{BAD}$, and $\angle{BCD}$ concur at a point $I$. Observe that this implies that $I$ is the incenter; indeed, \[ \delta(I, AB) = \delta(I, AD) = \delta(I, DC) = \delta(I, BC) \] as desired.

Probably similar to the above solutions. Posting anyway :P

Solution: We assume that both Incircles are visible from $H$ under the same angle. Let $K$ be the incenter of triangle $APD$ and $J$ be the incenter of triangle $CQD$. Clearly, $KJ$ is the external bisector of angle $ADC$. Let $I$ be the intersection of the lines $PK$ and $QJ$. It is clear that $I$ has equal distance from both pairs of opposite sides of the quadrilateral $ABCD$. Therefore, it is sufficient to show that $I$ lies on the internal bisector of $\angle ADC$. Let $T=KJ \cap PQ$ and let $L=ID \cap PQ$. Since the angle of visibility is the same, by the sine rule in corresponding triangle, on the half of the visibility angle yields that $\frac{d(K,AD)}{d(J,DC)}=\frac{HK}{HJ}=\frac{KD}{DJ}$. Since $\angle THD=90^{\circ}$ we conclude that $(K,L;D,T)=-1$. Projecting from $I$ we get that $I(K,L;D,T)=(P,Q;L,T)=-1$. Since $KJ$ is the external bisector of angle $ADC$ we get that $D(P,Q;L,T)=-1=(DP,DQ;DT,DI)=-1$ and so $DI$ bisects angle $ADC$. It is clear that all these steps can be reversed and the only if implication follows as well. The result clearly follows.

Really nice problem! :)

[asy]
unitsize(50);
pen n_purple = rgb(0.7,0.4,1),
n_blue = rgb(0,0.6,1),
n_green = rgb(0,0.4,0),
n_orange = rgb(1,0.4,0.1),
n_red = rgb(1,0.2,0.4);
pair pole(pair a, pair b) {return extension(a, rotate(90, a) * (0, 0), b, rotate(90, b) * (0, 0));}
pair S1, S2, S3, S4, A, B, C, D, P, Q, H, I, I1, I2;
S1 = dir(173);
S2 = dir( 53);
S3 = dir(313);
S4 = dir(253);
A = pole(S4, S1);
B = pole(S1, S2);
C = pole(S2, S3);
D = pole(S3, S4);
P = extension(B, A, C, D);
Q = extension(B, C, A, D);
H = foot(D, P, Q);
I1 = incenter(A, D, P);
I2 = incenter(C, D, Q);
I = extension(P, I1, Q, I2);
draw(unitcircle, n_blue);
draw(A--B--C--D--cycle);
draw(P--A^^P--D^^Q--C^^Q--D, gray(0.6));
draw(P--Q, n_green);
draw(D--H, n_green + dashed);
draw(P--I^^Q--I, gray(0.6));
draw(I1--I2, gray(0.6));
draw(H--I1^^H--I2, n_green);
draw(incircle(A, D, P)^^incircle(C, D, Q), n_purple);
dot(A^^B^^C^^D^^P^^Q^^H^^I1^^I2^^I);

label("$A$", A, dir(160));
label("$B$", B, dir(110));
label("$C$", C, dir(40));
label("$D$", D, dir(300));
label("$P$", P, dir(230));
label("$Q$", Q, dir(320));
label("$H$", H, dir(270));
label("$I$", I, dir(90));
label("$I_1$", I1, dir(140));
label("$I_2$", I2, dir(60));
[/asy]
Let $\omega_1, \omega_2$ be the incircles of $\triangle ADP, \triangle CDQ$, and let $I_1, I_2$ be their centers. Let $r_1, r_2$ be their radii, and lines $PQ$ and $I_1I_2$ intersect at $X$.

First, suppose that the incircle $\omega$ of quadrilateral $ABCD$ exists. Then by Monge on $\omega, \omega_1, \omega_2$ we deduce that the exsimilicenter of $\omega_1, \omega_2$ is on line $PQ$. This exsimilicenter also must lie on line $I_1I_2$, hence it is the point $X$.

Since $D$ is the insimilicenter of $\omega_1, \omega_2$, $(I_1I_2; XD) = -1$. Thus $H(I_1I_2; XD) = -1$; as $\overline{HX} \perp \overline{HD}$ we obtain $\overline{HD}$ bisects $\angle I_1HI_2$. Finally by Angle Bisector Theorem we deduce $\tfrac{HI_1}{HI_2} = \tfrac{DI_1}{DI_2} = \tfrac{r_1}{r_2}$; thus $\omega_1 \cup \{H\} \sim \omega_2 \cup \{H\}$ and $\omega_1, \omega_2$ are visible from $H$ under the same angle.

Now suppose that $\omega_1, \omega_2$ are visible from $H$ under the same angle. Then $\omega_1 \cup \{H\} \sim \omega_2 \cup \{H\}$, so $\tfrac{HI_1}{HI_2} = \tfrac{r_1}{r_2} = \tfrac{DI_1}{DI_2}$, thus $\overline{HD}$ is the internal bisector of $\angle I_1HI_2$.

Since $\overline{HD} \perp \overline{HX}$ we obtain $H(I_1I_2; DX) = -1$, thus $(I_1I_2; DX) = -1$. Since $D$ is the insimilicenter of $\omega_1, \omega_2$, $X$ must be the exsimilicenter, so the exsimilicenter lies on $\overline{PQ}$.

Let $\omega$ be the $P$-excircle of $\triangle ADP$. Then the exsimilicenter of $\omega, \omega_1$ is $P$, so by Monge on $\omega, \omega_1, \omega_2$ we deduce that the exsimilicenter of $\omega, \omega_2$ is on $\overline{PQ}$. Since line $AQ$ is an external common tangent, this exsimilicenter must be point $Q$. Thus, $\omega$ is tangent to line $BQ$, so $\omega$ is an incircle for quadrilateral $ABCD$.

I'll prove the first direction the second one is done the same way with "fantoming" the exsimillicenter of the incircles $\triangle ADP,CDQ$ (call them $\omega_2,\omega_1$ from now on.)

Let $R$ be the exsimilicenter of $\omega_1,\omega_2$ and let $R'$ be the projection of $R$ onto $QA$.Note that by Monge-de Alambert we have that $R\in\overline{QP}$.We have:$\tfrac{P_{\omega_1}(R)}{P_{\omega_2}(R)}=\tfrac{P_{\omega_1}(D)}{P_{\omega_2}(D)}=\tfrac{r_1^2}{r_2^2}$.Also form similarity we have $\tfrac{P_{\omega_1}(R')}{RO_C^2}=\tfrac{P_{\omega_2}(D)}{DO_A^2}$analogously for $\omega_2$ we have that $\tfrac{P_{\omega_1}(R')}{P_{\omega_2}(R')}=\tfrac{r_1^2}{r_2^2}$ and as  $H\in \odot DRR'$,taking in account that $\odot DRR'$ is coaxial with $\omega_1,\omega_2$ we're done.

neat problem
[hide = solution]
Let $I_1, I_2$ denote the incentres of $DAP$ and $DCQ$ respectively. Denote by $\omega_1, \omega_2$ their corresponding incircles. Finally let $r_1, r_2$ be their respective radii. Remark that $I_1, D, I_2$ are collinear since $I_1D$ and $I_2D$ both externally bisect $\angle CDA$. Now since $H$ subtends the same angle with respect to the two incircles, it follows that $HI_1/r_1 = HI_2/r_2$. Since $D$ subtends equal angles wrt $\omega_1, \omega_2$ as well, it follows similarily that $DI_1/r_1 = DI_2/r_2$. Now we have $HI_1/HI_2 = DI_1/DI_2$. Observe that this metric condition is equivalent to the equivalence of angles subtended by $H$ wrt $\omega_1, \omega_2$.

Let us prove the forward direction first. Since $HI_1/HI_2 = DI_1/DI_2$ it follows that $HD$ internally bisects $\angle I_1HI_2$. In particular, since $PQ \perp HD$ it follows that $PQ$ is the external bisector of $\angle I_1HI_2$. Define $X = I_1I_2 \cap PQ$. We now have $(I_1I_2;DX) = -1$. Since $D$ is the insimiliencenter of $\omega_1, \omega_2$ which are centred at $I_1, I_2$ respectively, it follows that $X$ is the exsimilicenter of $\omega_1, \omega_2$. Let $\omega$ denote the $Q$-excircle of $\triangle QDC$. From Monge on $\omega_1, \omega_2, \omega$ we conclude that the exsimilicenter of $\omega_1, \omega$ lies on $XQ$. Since $PC$ is tangent to both $\omega_1, \omega$ we conclude that the exsimilicenter must be $P$ itself. Since $PA$ is tangent to $\omega_1$ it must be tangent to $\omega$ as well. Since $\omega$ already touches sides $AD, DC, CB$ by definition, it follows that $\omega$ is an incircle to $ABCD$.

Now for the backward direction. Retracing the Monge we find that $X = I_1I_2 \cap PQ$ is the exsimilicentre of $\omega_1, \omega_2$. Since $\angle XHD = 90^{\circ}$ and $(I_1I_2;DX) = -1$ it follows from the Apollonian circle config that $HI_1/HI_2 = DI_1/DI_2$; exactly what we wanted to retrieve.
[/hide]


Solve in the set of the real numbers this equation :

$ ab\cdot x^2\plus{}(a^2\plus{}b^2)\cdot x\plus{}1\equal{}0\ ,\ a,b\in \mathbb{Z}$

[quote="Carcul"]Excuse my ignorance, but what exactly is the purpose here?[/quote]

First a solution , i have a long solution .

[quote="alex2008"]Solve in the set of the real numbers this equation :

$ ab\cdot x^2 \plus{} (a^2 \plus{} b^2)\cdot x \plus{} 1 \equal{} 0\ ,\ a,b\in \mathbb{Z}$[/quote]

Hemmmm, if you want solution, I think college forums would give you :

if $ a\equal{}b\equal{}0$ : no solution

if $ a\equal{}0$ and $ b\neq 0$, $ x\equal{}\minus{}\frac{1}{b^2}$

if $ b\equal{}0$ and $ a\neq 0$, $ x\equal{}\minus{}\frac{1}{a^2}$

if $ ab\neq 0$ then $ (a^2\plus{}b^2)^2\geq 4a^2b^2\geq 4ab$ and two solutions :

$ x_1\equal{}\frac{\minus{}a^2\minus{}b^2\minus{}\sqrt{(a^2\plus{}b^2)^2\minus{}4ab}}{2ab}$

$ x_2\equal{}\frac{\minus{}a^2\minus{}b^2\plus{}\sqrt{(a^2\plus{}b^2)^2\minus{}4ab}}{2ab}$

:)

But maybe you made a mistake in the problem statement and you want we solve in $ \mathbb Z$ and not in $ \mathbb R$

[quote="pco"]But maybe you made a mistake in the problem statement and you want we solve in $ \mathbb Z$ and not in $ \mathbb R$[/quote]
But an analysis over $ \mathbb Z$ would boil down to finding rational values of $ \sqrt {(a^2 \plus{} b^2)^2 \minus{} 4ab}$.
For $ a\ne b$, the value $ (a^2 \plus{} b^2)^2 \minus{} 4ab$ lies strictly between $ (a^2 \plus{} b^2\minus{}1)^2$ and $ (a^2 \plus{} b^2)^2$.
For $ a\equal{}b$, the value $ (a^2 \plus{} b^2)^2 \minus{} 4ab$ is $ 4a^2(a^2\minus{}1)$, which implies $ a\equal{}b\equal{}0$ or $ a\equal{}b\equal{}1$ or $ a\equal{}b\equal{}\minus{}1$.

[quote="pco"][quote="alex2008"]Solve in the set of the real numbers this equation :

$ ab\cdot x^2 + (a^2 + b^2)\cdot x + 1 = 0\ ,\ a,b\in \mathbb{Z}$[/quote]

Hemmmm, if you want solution, I think college forums would give you :

if $ a = b = 0$ : no solution

if $ a = 0$ and $ b\neq 0$, $ x = - \frac {1}{b^2}$

if $ b = 0$ and $ a\neq 0$, $ x = - \frac {1}{a^2}$

if $ ab\neq 0$ then $ (a^2 + b^2)^2\geq 4a^2b^2\geq 4ab$ and two solutions :

$ x_1 = \frac { - a^2 - b^2 - \sqrt {(a^2 + b^2)^2 - 4ab}}{2ab}$

$ x_2 = \frac { - a^2 - b^2 + \sqrt {(a^2 + b^2)^2 - 4ab}}{2ab}$

:)

But maybe you made a mistake in the problem statement and you want we solve in $ \mathbb Z$ and not in $ \mathbb R$[/quote]

If $ ab\neq 0$ let be $ x_1,x_2$ the roots of the equation : $ abx^2+(a^2+b^2)x+1=0\ (1)$, then $ x_1\neq 0,x_2\neq 0$, and let $ y_1=\frac{1}{x_1},y_2=\frac{1}{x_2}$ the roots os the equation $ y^2+(a^2+b^2)y ,\ (2)$ .

For $ k\in \{1,2\}$ we have $ x_k\in \mathbb{Q}\Leftrightarrow y_{k}\in \mathbb{Q}\Leftrightarrow y_k\in \mathbb{Z}$ . Now we must solve $ (2)\in \mathbb{Z}$ .

Assume the equation $ (2)$ has the root $ y_1\in \mathbb{Z}$ , but because $ y_1+y_2=-(a^2+b^2)\in \mathbb{Z}\Rightarrow y_2\in \mathbb{Z}$ and

$ |y_1|+|y_2|\ge |y_1+y_2|=a^2+b^2\ge 2|ab|=2|y_1y_2|\Rightarrow \frac{|y_1|+|y_2|}{|y_1y_2|}\ge 2\Rightarrow \frac{1}{|y_1|}+\frac{1}{|y_2|}\ge 2\Rightarrow |y_1|=|y_2|=1$ .

Since $ y_1+y_2<0\Rightarrow y_1=y_2=-1\Rrightarrow a^2+b^2=2$ and $ ab=1\Rightarrow a=b=\pm 1$ .

So for $ a=b=\pm 1$ the equation $ (1)$ has the rational roots $ x_1=x_2=-1$ . Otherwise the equation $ (1)$ has no rational roots .

[quote="alex2008"] ... Otherwise the equation $ (1)$ has no rational roots .[/quote]

Whoever asked for rational roots ?


geometry
3D geometry
modular arithmetic
number theory unsolved
number theory

Hi, can anyone help me with this math problem,

I'm thinking of a 6-digit number. The sum of the digits is 43. And only two of the following three statements about the number are true: (1) it's a square number. (2) it's a cube number, and (3) the number is under 500000."

i'll really appreciate it if anyone can help me, thanks.

I'm moving this to "number theory" where I'm sure one of the regulars will explain to you why no such number can exist.

Well, there is such a number. If the sum of the digits of $n$ is 43, then $n\equiv 43 \equiv 7\pmod{9}.$ All cubes $\pmod{9}$ are $0,1,$ or $-1$, so $n$ is not a cube and statement (2) is false. That means that $n<500000$ and $n$ is a square.

The only square $\pmod{9}$ is $7,$ and the numbers that square to $7$ are $4$ and $5$. That narrows the set of possibilities down to the following:

$n=k^2,\,k\equiv 4 \text{ or } 5\pmod{9}$, and $100000\le k^2<500000$ which means that $319\le k \le 707.$ This leaves us with about 90 possibilities, and brute force seems possible.

There are more restrictions. The first digit of $n$ cannot be greater than 4, so the remaining 5 digits must add up to 39, which means we need some 8's and 9's in our number.

It turns out that there is only one such number, the largest number on the "brute force" list: $707^2=499849.$


[quote="radac_mail"]Aflati $I=\int^{3}_{0}\frac{\sqrt{x}}{\sqrt{x}+\sqrt{3-x}}dx$[/quote]
[Prea simpla !]
Utilizati egalitatea $(*)\; \; \; \int_{a}^{b}E(x)\; dx =\int_{a}^{b}E(a+b-x)\; dx .$ !
Fie $I: =\int_{a}^{b}\frac{f(x)}{f(x)+f(a+b-x)}\; dx$ . In cazul dumneavoastra $[a,b]: =[0,3]$ si $f(x)=\sqrt{x}.$ Din $(*)$
$I=\int_{a}^{b}\frac{f(a+b-x)}{f(a+b-x)+f(x)}\; dx$, deci $I+I=\int_{a}^{b}1\; dx$, sau $I=\frac{b-a}{2}.$

O problema asemanatoare cu cea expusa de subsemnatul, a aparut in Gazeta Matematica, acum vreo 25 ani (??).
De asemenea, la unul din concursurile de admitere la Politehnica [perioada 1980-1989 ?] s-a cerut sa se calculeze
$I_{n}: =\int_{0}^{\frac{\pi}{2}}\frac{\sin^{n}x}{\sin^{n}x+\cos^{n}x}\; dx$ pentru o anumita valoare a lui $n$ [nu mai retin !]. Observati ca rezultatul este independent de $n$ !

In cazul unei sume, folositi "trick"-ul $\; \; (*)'\; \; \; \sum_{k=p}^{q}A(k)=\sum_{k=p}^{q}A(q+p-k) .$

Probabil,prin abuz de limbaj, am putea spune ca $(*)$ resp. $(*)'$ constituie "scrierea de la dreapta la stanga" a unei integrale (resp. a unei sume) ?.

Cred ca ar fi suficienta discutia asupra $\ast$,integrala fiind o suma($Riemann$).
Presupunand ca avem $f-integrabila \: pe\: [a,b]$,putem alege un sir de diviziuni $\Delta_{n}=(a,x_{1},x_{2},...,x_{i},...,b) , (a=x_{0},b=x_{n},n\rightarrow \infty)$ si un sir de puncte intermediare $\Sigma_{i}\in [x_{i},x_{i-1}] ,\forall i,0\leq i\leq n$ astfel incat $\int^{b}_{a}f(x) dx=\sum_{i=1}^{n}f(\Sigma_{i})(x_{i}-x_{i-1})$.
Scrierea de la dreapta la stanga s-ar face cu un nou sir de diviziuni $\Delta'_{n}=(b,x'_{1},x'_{2},...,x'_{i},...,a) , (b=x'_{0},a=x'_{n})$ si bineinteles cu un alt sir de puncte intermediare $\Sigma'_{i}\in [x_{i},x_{i-1}] ,\forall i,0\leq i\leq n$
Observam ca sirul primei diviziuni este crescator si sirul celei de-a doua diviziuni este descrescator.
Daca $f$ este integrabila atunci putem alege $p= x_{i}-x_{i-1}=x'_{i-1}-x'_{i}, \forall i,1\leq i\leq n$cu$p\rightarrow 0$

In acest caz este verificabil ca $x_{i}=a+pi$ si $x'_{i}=b-pi ,\forall i,0\leq i\leq n$ deci $x'_{i}=b+a-x_{i}$.Similar pentru $\Sigma'_{i}=a+b-\Sigma_{i}$.

Problema pe care ati pus-o dumneavoastra este daca $\int_{a}^{b}f(a+b-x)\; dx =\sum_{i=1}^{n}f(\Sigma'_{i})(x'_{i}-x'_{i-1})$?

Cum $\sum_{i=1}^{n}f(\Sigma'_{i})(x'_{i}-x'_{i-1})=\sum_{i=1}^{n}f(a+b-\Sigma_{i})(x_{i-1}-x_{i})=\int_{b}^{a}f(a+b-x)\; (-dx)=\int_{a}^{b}f(a+b-x)\; dx \rightarrow q.e.d$

[quote="radac_mail"]Cred ca ar fi suficienta discutia asupra $\ast$,....[b]integrala fiind o suma($Riemann$)[/b]....
[/quote]
Nu este exact  ! Nu intervine si un "proces" de limita ?

[quote="radac_mail"]Ai putea face,te rog, o referire mai exacta?[/quote]
Citeste definitia integralei (definite) Riemann !
Integrala definita Riemann ,[b]nu[/b] este o suma !

Ai dreptate.Scuzele mele....($(\ast)'$ iese dintr-o simpla examinare a extinderii sumei)


Is that for every $x,y,a,b\in\mathbb{R}$, or for every $x,y\in\mathbb{R}$ with fixed $a,b\in\mathbb{R}$?

[quote="doicu"]hallo,
find alll functions $f: R\to R$ such that:
$(f(x)+f(y))(f(a)+f(b))=f(xa+yb)+f(xb-ya)$
 :) :ninja:[/quote]
Why this problem reminse me something :blush:  :D
Guys this is very famous problem from IMO 2002 ;) ,

hallo,
yes its for all $x, y, a, b \in R$

Here is the solution :)

[quote="Tiks"]
Guys this is very famous problem from IMO 2002 ;) ,[/quote]


calculus
integration
logarithms
trigonometry
calculus computations

Hi, I was wondering if anyone knows how to solve this integral.  Thank you.

Question:  $ \displaystyle\int_0^{\frac{\pi}{2}}\sec x\ln\left(\frac{1\plus{}b\cos x}{1\plus{}a\cos x}\right)dx$

Your integral equals to the double integral $ \int_{0}^{\pi /2}\!{\int_{a}^{b}{\frac {dy\,dx}{(1 \plus{} y\cos x)}}},$ now reverse integration order.

Then we are left with

$ \int_a^b { \frac{2\arctan\left( \frac{\sqrt{y^2 \minus{} 1}}{y \plus{} 1} \right) } {\sqrt{y^2 \minus{} 1}} dy }$

$ I \equal{} \int_0^{\pi/2} \int_a^b \frac {1}{1 \plus{} y \cos x} \,dy \,dx \equal{} \int_a^b \int_0^{\pi/2} \frac {1}{1 \plus{} y \cos x} \,dx \,dy$.

But $ \int_0^{\pi/2} \frac {1}{1 \plus{} y \cos x} \,dx \equal{} \int_0^1 \frac {1}{1 \plus{} y \frac {1 \minus{} t^2}{1 \plus{} t^2}} \cdot \frac {2}{1 \plus{} t^2}\,dt$

$ \equal{} 2 \int_0^1 \frac {1}{(1 \plus{} y) \plus{} (1 \minus{} y)t^2}\,dt \equal{} \frac {2}{\sqrt {1 \minus{} y^2}} \int_0^1 \frac {\sqrt {\frac {1 \minus{} y}{1 \plus{} y}}}{1 \plus{} \left[\sqrt {\frac {1 \minus{} y}{1 \plus{} y}} \cdot t\right]^2}\,dt$

$ \equal{} \frac {2}{\sqrt {1 \minus{} y^2}} \arctan \left(\sqrt {\frac {1 \minus{} y}{1 \plus{} y}}\right)$.

Now observe that

$ \left[\arctan \left(\sqrt {\frac {1 \minus{} y}{1 \plus{} y}}\right)\right]' \equal{} \minus{} \frac12 \cdot \frac {1}{\sqrt {1 \minus{} y^2}}$. Therefore,

$ I \equal{} \int_a^b \frac {2}{\sqrt {1 \minus{} y^2}} \arctan \left(\sqrt {\frac {1 \minus{} y}{1 \plus{} y}}\right) \,dy$

$ \equal{} \minus{} 4 \int_a^b \minus{} \frac12 \cdot \frac {1}{\sqrt {1 \minus{} y^2}} \arctan \left(\sqrt {\frac {1 \minus{} y}{1 \plus{} y}}\right)\,dy$

$ \equal{} \minus{} 4 \left[\left( \arctan \left(\sqrt {\frac {1 \minus{} y}{1 \plus{} y}}\right)\right)^2 \right]_a^b$

$ \equal{} 4 \left[\left(\arctan \sqrt {\frac {1 \minus{} a}{1 \plus{} a}}\right)^2 \minus{} \left(\arctan \sqrt {\frac {1 \minus{} b}{1 \plus{} b}}\right)^2\right]$

$ \equal{} (\arccos a)^2 \minus{} (\arccos b)^2$.

[quote="Carcul"]$ \equal{} \minus{} 4 \int_a^b \minus{} \frac12 \cdot \frac {1}{\sqrt {1 \minus{} y^2}} \arctan \left(\sqrt {\frac {1 \minus{} y}{1 \plus{} y}}\right)\,dy$

$ \equal{} \minus{} 4 \left[\left( \arctan \left(\sqrt {\frac {1 \minus{} y}{1 \plus{} y}}\right)\right)^2 \right]_a^b$[/quote]
I think it should be $ \equal{} \minus{} 4 \left[\frac{1}{2} \left( \arctan \left(\sqrt {\frac {1 \minus{} y}{1 \plus{} y}}\right)\right)^2 \right]_a^b$.
Am I wrong?

Goodbye
GM

[quote="GioMott"][I think it should be $ \equal{} \minus{} 4 \left[\frac {1}{2} \left( \arctan \left(\sqrt {\frac {1 \minus{} y}{1 \plus{} y}}\right)\right)^2 \right]_a^b$.
Am I wrong?

Goodbye
GM[/quote]
Yes, you are right.  The final answer is half of what Carcul wrote, but the important thing is that I understood Carcul's technique and the general idea.  Thank you very much again to everyone who had helped me with this.


How can I generate something like

[img]http://img33.picoodle.com/img/img33/4/1/31/f_b8fcabed18am_1be04bb.png[/img]

?

\usepackage{fancybox}
\usepackage{graphicx}
...
\setlength{\shadowsize}{1pt}
\rotatebox{30}{\shadowbox{Hint}}

Thanks Steve, one more question:

I was using

\begin{pspicture}(-0.6,-0.2)(0.8,1)
\rput{30}{
\psshadowbox[framearc=0.5,linewidth=0.3pt,framesep=5pt,shadowcolor=Blue,linecolor=Blue]{\color{Blue}\text{Hint}}}
\end{pspicture}

but it tells me that pspicture's environment is undefined (I'm using the pspicture package), also it does not recognize \psshadowbox and \rput. Do I need another package to make this work?

I don't know where you got this code from as there are a lot of errors:
1. pspicture is an ancient package superseded by pstricks;
2. there is no pspicture environment in the pspicture package, hence pspicture is undefined;
4. psshadowbox and rput are not in the pspicture package;
3. unless you've defined it yourself there is no Blue.
[b]It is important that you give a [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=157945]minimal example[/url] to show packages loaded and definitions made.[/b]
Corrected code as a [i]minimal example[/i]
[code]\documentclass{article}
\usepackage{amsmath}
\usepackage{amssymb}
\usepackage{pstricks}
\begin{document}
\begin{picture}(-0.6,-0.2)(0.8,1)
\rput{30}{
\psshadowbox[framearc=0.5,linewidth=0.3pt,framesep=5pt,shadowcolor=blue,linecolor=blue]{\color{blue}\text{Hint}}}
\end{picture}
\end{document}[/code]

Thank you again.

Now I get how to give the minimal example, sorry.

Okay, last question:

How would you generate this

[img]http://img209.imageshack.us/img209/9793/69a23fa84cf7b2db9ff8220jv9.png[/img]

Thanks for your time.

That looks like a gradient fill which PSTricks can do. Examples and code are [url=http://tug.org/PSTricks/main.cgi?file=Examples/gradient]here[/url].


[size=117]
記得四年前我剛回國時，
第一個想到要買的就是一部車。

經過一段時間的評估後，我決定買一部墨綠色的中型轎車。

當時我的印象是一般人的車都買白色或黑色，

所以認為自己的選擇很獨特，而且又很有品味。

正在為自己能買到一部與眾不同的車而沾沾自喜時，

我突然發現不論是在高速公路上、小巷子裡，甚至於我住的大樓停車場中，

都看到許多與我同型而且是墨綠色的轎車。我開始覺得很奇怪，

為什麼大家突然間都開始買墨綠色的車，所以我就把我的觀察與同事們分享。

有一位女同事當時正好懷孕，
聽我講完後就說：「我倒是沒有看到很多墨綠色的車。

可是最近我發現，無論在那裡都會看到孕婦。我記得上個星期天在逛百貨公司時，

短短二小時就看到六個孕婦，台灣的人口出生率最近是不是有提高呢？」

我與其他同事異口同聲地都說沒發現孕婦有增加的現象，
她看到那麼多大概是很湊巧。

後來我有一次在國外聽演講，才了解到這種現象在心理學上叫做「視網膜效應」。

簡單地說，這種效應的意思就是當我們自己擁有一件東西或一項特徵時，

我就會比平常人更會注意到別人是否跟我們一樣具備這種特徵。

這個發現對我自有什麼影響呢？卡內基先生很久以前就提出一個論點，

那就是每個人的特質中大約有百分之八十是長處或優點，
而百分之二十左右是我們的缺點。

當一個人只知道自己缺點是什麼，而不知發掘優點時，

「視網膜效應」就會促使這個人發現他身邊也有許多人擁有類似的缺點，

進而使得他人際關係無法改善，生活也不快樂。

有沒有發現那些常罵別人很兇的人，其實自己脾氣也不太好？

這就是「視網膜效應」的影響力。

一個人要人緣好、受人歡迎，一定要養成欣賞自己與肯定自己的能力。

[color=blue]因為在「視網膜效應」的運作下，

一個看到自己優點的人，才有能力看到他人的可取之處。

能用積極態度看待他人，往往是良好人際關係的必備條件。

所以，從現在起，學習欣賞自己的優點和長處吧！[/color][/size]

[quote="Dreamer"]古书上记载：有个宋国人都东西了，怀疑他的邻居的小孩。于是乎每次看到那个小孩怎么看都像个贼；
后来，在自己家找到了那个东西，于是乎每次看到那个小孩怎么看都像个好孩子。

 :D 道理都一样。[/quote]
怎么感觉跟某一年的全国高考作文题一样……

[quote="dingdongdog"][quote="Dreamer"]古书上记载：有个宋国人都东西了，怀疑他的邻居的小孩。于是乎每次看到那个怎么感觉跟某一年的全国高考作文题一样……[/quote][/quote]

你指的可能是韩非子的智子疑邻，确实考过，也是宋国的故事。情节差不多。


Two parallel chords in a circle have lengths 6 and 8. The distance between them is 1. What is the diameter of this circle?

The diameter of the circle is 10.
Without the loss of generality, let's assume the circle is centered at the origin and that both chords lie in the upper semicircle.
The equation for this semicircle is $y=\sqrt{r^2-x^2}$.
The chord of length eight then intersects the circle at $(4,y)$, and the chord of length six intersects the circle at $(3,y+1)$.
Plugging these points into the equation for the semicircle yields the following system of equations
$y+1 = \sqrt{r^2-9}$
$y=\sqrt{r^2-16}$
The only positive solution to this system is $r=5$ and $y=3$.

Jimmy - [hide]The answer is correct, something you may like to look at is:  how did you solve for  $r$ and $y$ ? Was it a guess, or can you    solve it for any other given value.? For  example, suppose the lengths of the chords were, say 7, and 8 (instead of 6 and 8) would you still be able to solve it?  [/hide]

Good point, the solution was not a guess, but I did not solve it by hand; I used a TI-89 (I do know how to solve it by hand though).
As for the lengths of 7 and 8, the equations are
$y=\sqrt{r^2-16}$
$y+1=\sqrt{r^2-12.25}$
The solution is $r=\frac{\sqrt{1145}}{8}$ and $y=\frac{11}{8}$ I think we can both be fairly certain that I didn't guess this :lol:.

I didn't use co-ordinate geometry, just plain circle geometry. (I don't know how to upload a diagram, so i'll just have to describe it). Anyways, draw a line through the centre of the circle, perpendicular to the two parallel chords. Since it's the diameter, it must bisect the two chords as well. connect the centre to the endpts. you end up with two right-angle triangles, both with the radius as the hypotenuse. Let the radius be r. Therefore one of the triangle has sides: r, 4, and sqrt(r^2-16). The other: r, 3, sqrt(r^2-9). Since the two sqrt sides add to give the distance between the two chords, which is 1, therefore:

sqrt(r^2-9)+sqrt(r^2-16)=1
moving one sqrt term to the other side and squaring and then simplifying, we get:
sqrt(r^2-9)=4
therefore: r=5.
And the diameter is 10.

Cheers!


integration
complex analysis
complex analysis solved

Let f be analytic on {IzI <1}.
Suppose I f ' (z)-f ' (0)I< I f ' (0)I for all IzI<1.
Prove f is injective.

Thanks for your help.

Consider two points $z_0$ and $z_1$ in $D=\{|z|<1\}.$ Let $\gamma$ be the line segment connecting $z_0$ and $z_1,$ parameterized in the usual way. Since $D$ is convex, $\gamma$ is contained in $D.$

$f(z_1)-f(z_0)=\int_{\gamma}f'(z)\,dz$

$=\int_{\gamma}f'(0)\,dz+\int_{\gamma}(f'(z)-f'(0))\,dz$

$=f'(0)(z_1-z_0)+\int_{\gamma}(f'(z)-f'(0))\,dz$

Now estimate:

$\left|\int_{\gamma}(f'(z)-f'(0))\,dz\right|\le \int_{\gamma}|f'(z)-f'(0)|\,|dz|$

$<|f'(0)|\int_{\gamma}|dz|=|f'(0)||z_1-z_0|.$

Thus, $|f(z_1)-f(z_0)|>|f'(0)||z_1-z_0|-|f'(0)||z_1-z_0|=0.$

$|f(z_1)-f(z_0)|>0$ implies $f(z_1)\ne f(z_0).$ Since this holds for all pairs of points in $D,$ we have that $f$ is injective.

You'll notice that the only property of $D$ that we used was that it was convex; the same theorem should hold on any convex domain.


geometry
3D geometry
combinatorics unsolved
combinatorics

[quote="April"]Is it possible to build a $ 4\times 4\times4$ cube from blocks of the following shape consisting of $ 4$ unit cubes?[/quote]
It can be built, but it is a bit hard to explain how without opportunity to draw it. Just try it out yourself :P

Who can post the official solution ? Thanks.

Solution will goes this way: cover the whole cube in a chessboard fashion now as the tetramino covers up 2 black and 2 white cubes it must coverup the whole cube

but i made a sketchup file showing that how 8 tetraminos fill up 2*4*4 cube .now it is easy to double that i hop eyou have google sketchup! what you have to do is: open my file and then view>animation>play
my browser says"The extension skp is not allowed."so you have to download it from http://www.4shared.com/file/PxYZUl2Z/cube_revisted.html

but i have a doubt that what the condition that the tetraminoes will fill up a m*n*p cube? can anybuddy generalise>? :)

ACTUALLY i am too in doubt can any one please have a look
http://www.artofproblemsolving.com/Forum/viewtopic.php?f=41&t=394305&p=2192229#p2192229

I suck at visualization but this seems to build a $2 \times 4 \times 4$.
[code]
1122
3114
3554
5566
----
7224
7784
3788
3668
[/code]

EDIT in response: these are two layers of cubes, so you should imagine one 4x4 square on top of the other in 3D, and the cubes with the same number are part of the same tetromino.

[quote="math_explorer"]I suck at visualization but this seems to build a $2 \times 4 \times 4$.
[code]
1122
3114
3554
5566
----
7224
7784
3788
3668
[/code][/quote]
 srrry dude can u please explain me these code arrangement? how u create this?

Same numbers are parts of the same tetromino. They are two adjacent layers.

How did he create that? Use the [ code ][/ code ] tags (without the spaces).

[quote="chaotic_iak"]Same numbers are parts of the same tetromino. They are two adjacent layers.

[/quote]

thanks dude i was looking for this!! :)


geometry
trapezoid
3D geometry
function
algebra
floor function
puzzles

\............/   10
.\........../
..\......../
...\....../       0 <-- This is the bottom & the cup is not drawn to scale.

Imagine you are looking at this as a sideways view of a cup.  This view is in the shape of an _isosceles trapezoid_ [EDIT].  Let the diameter of the top be equal to twice the diameter of the bottom.  Also, let the bottom have height 0 units and let the top have height 10 units.

What vertical height number to the nearest tenth (between 0 & 10), would you reach if you poured a liquid into the cup so that it would become half-full?
 (Ignore any thickness of the cup; that is, make it negligible.)

Well, if I understand you correctly, the "inside" is:
[hide]5 times the cube root of 2.  Are those brackets
supposed to be the "greatest integer function" of it?
I don't require the exact origination of the value, just
the decimal to the nearest tenths.
You tell me the value as a decimal number (to the nearest
tenths) so I am not presuming something from you.
If you wish, your next, if any, reply can be by hiding
it in your post (or not).  I will reply later; maybe others
will reply before I do with a decimal number candidate.[/hide]

There is no one answer.  It's ambiguous.

The diameter of the top being twice the diameter of the bottom is why.


Find the limit
$ \lim   x\rightarrow\infty x ([ \dfrac{1}{x}] +[ \dfrac{2}{x}] +...+[ \dfrac{2009}{x}] )$

_______________________________________
Azerbaijan Land of Fire
$ \Rightarrow\cup\Leftarrow$

[quote="Pirkuliyev Rovsen"]Find the limit
$ \lim x\rightarrow\infty x ([ \dfrac{1}{x}] + [ \dfrac{2}{x}] + ... + [ \dfrac{2009}{x}] )$[/quote]

When $ x>2009$, the parenthesis is exactly $ 0$ and so the limit is $ 0$


A convex quadrilateral $ ABCD$ with area $ 2002$ contains a point $ P$ in its interior such that $ PA=24$, $ PB=32$, $ PC=28$, and $ PD=45$.  Find the perimeter of $ ABCD$.

[hide]
Let's calculate the maximum possible area for each of the four triangles $ PAB,PBC,PCD,PDA$

$ (PAB) = \frac{PA\cdot PB}{2}\cdot \sin APB$

So $ (PAB)_{max}= \frac{24\cdot 32}{2}=384$

Similarly

$ (PBC)_{max}= \frac{32\cdot 28}{2}=448$
$ (PCD)_{max}= \frac{28\cdot 45}{2}=630$
$ (PDA)_{max}= \frac{45\cdot 24}{2}=540$

It is $ 384+448+630+540=2002$, so it is necessary that all the triangles are right at $ P$ (or else the total area is less than $ 2002$)

Now it's easy

$ AB=\sqrt{24^{2}+32^{2}}=\sqrt{576+1024}=\sqrt{1600}=40$
$ BC=\sqrt{32^{2}+28^{2}}=\sqrt{1024+784}=\sqrt{1808}=4\sqrt{113}$
$ CD=\sqrt{28^{2}+45^{2}}=\sqrt{784+2025}=\sqrt{2809}=53$
$ DA=\sqrt{45^{2}+24^{2}}=\sqrt{2025+576}=\sqrt{2601}=51$

and the perimeter is $ 144+4\sqrt{113}$

.. and I hope I didn't do any mistakes[/hide]

[quote="ordinary boy"]yea your answer is correct, nice solution but how did you know to try it with max areas because once you realize that its pretty trivial, but its kinda hard to realize that.[/quote]

I just had this idea  :idea:

Well, let's assume that the problem said 'Area=2000'
Then I would not be able to find a solution, because there would be a lot of possible cases

I always have this in my mind: [b]If some problem appears to have few clues, [u]maybe[/u] the solution is the only[/b]. I can't explain it in english, well, like this problem, in which there is only one possibility

Actually I ... hoped that the answer would be something like that. I searched about the possible values of the area and I found that its maximum is $ 2002$, so I felt lucky!!!  :D  If the maximum was larger, I would not solve this problem (however, if it was smaller, I'd say that the problem is incorect)

I hope you understand what I'm trying to say


quadratics
calculus
integration
linear algebra
matrix
algebra
polynomial

[size=150]Here is a problem in elementary number theory.

1. Prove that every integer $ n$ can be expressed in the form $ n \equal{} x^2 \plus{} y^2 \minus{} z^2.$

2. Prove that $ x^2 \plus{} y^2 \equal{} z^4$ has infinitely many solutions with $ (x, y, z) \equal{} 1.$ $ i.e.,$ $ x,y,z$ are relatively prime in pairs.[/size]

[hide="1"] This is a cop-out, but problems of this type are trivialized by the [url=http://en.wikipedia.org/wiki/15_and_290_theorems]15 and 290[/url] theorems.

An actual solution:  a number is representible in the form $ y^2 \minus{} z^2$ if and only if it is not congruent to $ 2 \bmod 4$.  Hence if $ n \not \equiv 2 \bmod 4$ let $ x \equal{} 0$ and if $ n \equiv 2 \bmod 4$ let $ x \equal{} 1$.[/hide]
[hide="2"] Let $ (a, b, c)$ be a primitive Pythagorean triple.  Then $ (2ab, a^2 \minus{} b^2, c)$ is a primitive solution to the desired Diophantine. [/hide]

To t0rajir0u,

What do you mean by the word "cop-out?" I'm interested in learning new words.

[url=http://dictionary.reference.com/browse/cop-out]Try dictionary.com[/url].

I mean that using such a strong theorem to answer a question like this is citing the full proof of Fermat's Last Theorem to solve the cases $ n \equal{} 3, 4$; it avoids the point of the exercise.

[quote]The 15 theorem of John H. Conway and W. A. Schneeberger (Conway-Schneeberger Fifteen Theorem), proved in 1993, states that if an integral quadratic form with integral matrix represents all positive integers up to 15, then it represents all positive integers.[/quote]

Also I was reading wikipedia on the 15 theorem. I don't know what it means to call a quadratic form with integral matrix. What does matrix have to do with quadratic forms?

A quadratic form in $ n$ variables can be represented in the form $ \mathbf{XAX}^T$, where $ \mathbf{X} \equal{} \left < x_1, x_2, ... x_n \right >$ and $ \mathbf{A}$ is an $ n \times n$ symmetric matrix that encodes the coefficients of $ \mathbf{X}$.  For example,

$ ax^2 \plus{} bxy \plus{} cy^2 \equal{} \left[ \begin{array}{cc} x & y \end{array} \right] \left[ \begin{array}{cc} a & \frac {b}{2} \\
\frac {b}{2} & c \\
\end{array} \right] \left[ \begin{array}{c} x \\
y \end{array} \right]$.

Note that the condition that $ a, b, c$ be integer is weaker than the condition that the entries in the corresponding array be integer.  The matrix representation turns out to be the natural way to analyze quadratic forms; the determinant of the matrix of a $ 2$-variable quadratic form, for example, is (the negative of one-fourth of) the familiar discriminant.

[b]Correction:[/b]  the matrix corresponding to $ x^2 \plus{} y^2 \minus{} z^2$ is $ \left[ \begin{array}{ccc} 1 & 0 & 0 \\
0 & 1 & 0 \\
0 & 0 & \minus{} 1 \end{array} \right]$, which is integral but not [url=http://en.wikipedia.org/wiki/Positive-definite_matrix]positive-definite[/url], so the 15 Theorem does not apply.  My apologies.

To t0rajir0u,

I see. But does the matrix being not positive-definite imply that the quadratic form not positive-definite?

Also I have some other problems I couldn't solve.

[size=134]1. Find all integral solutions of the equation $ x^4 \plus{} 2x^3 \plus{} 2x^2 \plus{} 2x \plus{} 5 \equal{} y^2$

2. Show that the equation $ x^2 \equal{} y^3 \plus{} 23$ has no solution in integers.  [/size]

[quote="lenny"]I see. But does the matrix being not positive-definite imply that the quadratic form not positive-definite?[/quote]

I believe that's the motivation behind the definitions, yes.

[hide="1: technique"] Bound the polynomial on the left between two polynomials of the form $ (x^2 \plus{} x \plus{} a)$. [/hide]

Thanks!! I think I see the solution now.

[hide]Let $ f(x) \equal{} x^4 \plus{} 2x^3 \plus{} 2x^2 \plus{} 2x \plus{} 5.$
Then we can rewrite $ f(x)$ as $ f(x) \equal{} (x^2 \plus{} x \plus{} 1)^2 \plus{} (x\plus{}2)(x\minus{}2).$
Now the last term of the right hand side is positive except for the values when $ x$ is in the interval $ I \equal{} [\minus{}2,2].$
Hence $ f(x) > (x^2 \plus{} x \plus{} 1)^2$ if $ x \notin I.$
We can also write $ f(x)$ as $ f(x) \equal{} (x^2 \plus{} x \plus{} 2)^2 \minus{} (x\plus{}1)(3x\minus{}1).$
The last term of the right hand side is positive except for the values when $ x$ is in the interval $ J\equal{}[\minus{}1, \frac{1}{3}].$
Hence $ f(x) < (x^2 \plus{} x \plus{} 2)^2$ if $ x \notin J.$
We have just shown that if $ x \in \mathbb{Z},$ $ x \notin I,$ and $ x \notin J,$ then $ (x^2 \plus{} x \plus{} 1)^2 < f(x) < (x^2 \plus{} x \plus{} 2)^2.$
This means that $ f(x)$ can not be a perfect square except possibly for $ x \in I$ or $ x \in J.$
We check these values individually to find that all the integral solutions of $ f(x)$ are
$ (\minus{}2,3), (\minus{}1,2), (2,7)$ and no others.[/hide]

By the way, what's your background and how long did it take you to be so good at these?

I'm a senior in high school.  I guess I started doing math seriously freshman year, so four years?   :?:

As for your second question, I'm not sure how easy that particular question is, but see [url=http://mathworld.wolfram.com/MordellCurve.html]Mordell curve[/url].


simplify the following into lowest terms.

$\frac{a^2+b^3}{a^4+b^5}$ + $\frac{c^3}{a^2+b^4}$

4everwise, you really shouldn't double post, just edit your first one. also, wouldn't saying "good job" or testing latex be spaming?

[quote="nat mc"]how is that spamming, hye just said good job[/quote]
nothing against you, 4everwise, but you didnt post the original question and did the same thing in several threads.


Let $p$ be an odd prime. Find all functions defined on the integers (taking integer values) such that
1) $f(n + p) = f(n)$ for all integers $n$,
2) $f(mn) = f(m)f(n)$ for all integers $m,\;n$.

For simplicity, I'll take $f:\mathbb Z_p\to Z$. We also know that $f$ is multiplicative. One solution is $f(n)=0,\ \forall n\in\mathbb Z_p$, so let's assume the function isn't this one. In this case we see that $f(1)=1$. Since all non-zero $n\in \mathbb Z_p$ are invertible, it means that $f(m)\in \{\pm 1\},\ \forall m\in \mathbb Z_p^*$. Since $f(0)=f^2(0)\Rightarrow f(0)\in \{0,1\}$.

Since $Z_p^*$ is cyclic ($p$ has a primitive root), assume $a$ is such a primitive root. $f$ is uniquely determined by $f(a)\in \{\pm 1\},\ f(0)\in \{0,1\}$. Conversely, we can see that all four choices are ok (they satisfy the condition of multiplicity).

one little correction: if we choose $f(a)=-1$, then we must have $f(0)=0$(take m=0,n=a). thus there are three solutions: either everything 0, everything 1 or the Legendre symbol.

Peter


Đặt  :  $ a \plus{} b \minus{} c \equal{} z , b \plus{} c \minus{} a \equal{} x,c \plus{} a \minus{} b \equal{} y$
  Ta có :  $ 2a \equal{} y \plus{} z, 2b \equal{} x \plus{} z, 2c \equal{} x \plus{} y$
  BĐT  tương đương với :
 $ \frac {2}{x} \plus{} \frac {2}{y} \plus{} \frac {2}{z} \ge \frac {4}{x \plus{} y} \plus{} \frac {4}{z \plus{} y} \plus{} \frac {4}{x \plus{} z}$
 Hiển nhiên  đúng khi ta  áp dụng 3 lần BĐT : $ \frac {1}{x} \plus{} \frac {1}{y} \geq \frac {4}{x \plus{} y}$

[size=200][color=darkblue]Bạn giải cho bài toán nào thế?[/color][/size]


algebra
polynomial
complex analysis
function
topology

Let $k$ be a positive integer.  find all polynomials $P(x)$ with real coefficients s.t.
$P(P(x))=[P(x)]^{k}$

[hide="hint"]
identity thm[/hide]

Let $y=P(x)$
Then $P(y)=y^{k}$
For any $y$ in the range of $P$

Since P is a polynomial, its holomorphic on $\mathbb{C}$
For any $I$ an open set in $\mathbb{C}$ we know that $P(I)$ is an open set. By the identity theorem (and holomorphic properties) we can extend $P(I)$ to the whole complex plane to get $P(x)=x^{k}$

Can someone please verify my complex analysis steps? Im not to good on that kinda stuff

i think that answer is correct.
although dont forget about constant polynomials (p(x)=0,1 or -1 for odd k.)

holomorphic??

uhhhhhh

it's either constant so P(x) = 0,1 or y=P(x) has infinitely many values, so P(y) = y^k has infinitely many values, ie. the finite degree polynomial P(y)-y^k has infinite roots, so it's 0 everywhere and P(y) = y^k.

holomorphic man come on

Are you asking what holomorphic is? Basically its a useful property defined over functions over $\mathbb{C}$, but it boils down to if theres an infinite sequence $x_{n}$ that converges to $x$ then $f(x_{n})$ converges to $f(x)$. But the point is that unlike the real number line, where it can approach the limit in only two directions, the sequence can approach it in any direction it likes. So we know all polynomials, exponential and pretty much all the elementary functions are holomorphic.

Now, one of the properties of a holomorphic function is that if you know the values in some neibhbourhood of a point $x$ then you can extend the function to the whole complex plane, and theres only one function that remains holomorphic over the whole plane.

So in this problem, i extended the polynomial to be defined over the whole complex plane instead of just the reals. Nothing new there. then i just picked a closed set in $\mathbb{C}$ (which is a set not including boundry). Since we have a polynomial i knew this got mapped to another closed set  in $\mathbb{C}$ then i could extend this set to the whole complex plane to get $P(x)=x^{k}$ which gives the constant functions mentioned when $k=0$, $k=-\infty$ im not too sure about the $-1$ solution when $k$ is odd, ill check it out later, im off somewhere now.


How can $\binom{2n+2}{n+1}$ be simplyfed?

I don't think it can. The first thing I would do is to represent it as $\binom{2(n+1)}{n+1}$, but that gets me nowhere.

[quote="Ezbakhe Yassin"]If I am not wrong, it could get simplified to $2\cdot\binom{2n+1}{n+1}$[/quote]

That's right, but you can also simplify it as
$\frac{2(2n+1)}{n+1}\cdot\binom{2n}{n}$

[quote]That's right, but you can also simplify it as
\frac{2(2n+1)}{n+1}\cdot\binom{2n}{n}[/quote]

Isn't that a little bit more complicated?

Depends on what you want to do simplifying it! I was just trying to put it in terms of $\binom {2n}{n}$

Yes, and since : $\frac{\binom{2n}{n}}{n+1}$ is the n-th Catalan number, you could say that, it is equal to :
$(2(2n+1)) \cdot C_n$ (moreover, this is equal to $(n+2) \cdot C_{n+1}$)

Thus, I have to say that it was a nice problem ;)

What is a Catalan number? :?:
How can we use them?

Wiki has the answer ;)  : http://en.wikipedia.org/wiki/Catalan_number

Thanks... now i know what $C_n$ means.
I didn't knew it. :D

Alright, and you'll see how they often appear in combinatorics ;)


I can hardly believe that this is number theory...


inequalities
number theory
prime numbers
number theory proposed

Prove that $\frac{n}{\pi(n^2)}>\frac{n+m}{\pi((n+m)^2)}$ for any natural $n$ greater than 2 and for any $m$ -  natural number  greater than 0 where $\pi(x)$ represents the number of prime numbers  before $x$.

[i]Edited by Myth[/i]

I was wondering why there were so many views yet no replies yet.

[quote="ADDY____"]Prove that $\frac{n}{\pi(n^2)}>\frac{n+m}{\pi((n+m)^2)}$ for any natural $n$ greater than 2 and for any $m$ -  natural number  greater than 0 where $\pi(x)$ represents the number of prime numbers  before $x$.[/quote]
Are you sure? At least it means that there is at least one prime number between $n^2$ and $(n+1)^2$. I thought it is an open question. Maybe I am wrong, but the best known estimate is there is a prime between $n$ and $n+c\cdot n^{5/8+\epsilon}$. Right?

[quote]
Maybe I am wrong, but the best known estimate is there is a prime between  and . Right?
[/quote]

Right I am not so sure about

[quote]
Prove that  for any natural  greater than 2 and for any  - natural number greater than 0 where  represents the number of prime numbers before .
[/quote]

either. Need clarification.

[quote="ADDY____"]I found a way to prove this inequality ... i`m just waiting another point of view, another way to resolve the problem[/quote]Did you read post #3 in this thread?

I have to agree with Myth. Where does the estimate come from though? What method was used?


For any non-empty subset $A$ of $\{1, 2, \ldots , n\}$ define $f(A)$ as the largest element of $A$ minus the smallest element of $A$. Find $\sum f(A)$ where the sum is taken over all non-empty subsets of $\{1, 2, \ldots , n\}$.

Element $k$ is the largest element for $2^{k-1}$ subsets of $\{1,2,\dots,n\}$ and is the smallest element for $2^{n-k}$ subsets. Hence,
\[ \sum_A f(A) = \sum_{k=1}^n k(2^{k-1} - 2^{n-k}) = (n-3)2^n + n + 3. \]


13) Zachary has a superball which rebounds to 7/8 of its previous height. On the third bounce, the ball lands on a piece of gum and only rebounds to 1/8 of its previous height. Find the total distance the superball has traveled, in feet, if it starts at a height of 4 feet and is caught at its highest point after the third bounce.

What I did, was see that when you dropped it (1st bounce), that is 4 feet, and it goes up (7/8) (4), and down the same distance (7/8) (4) for the 2nd bounce. It goes up (1/8)(7/8)(4) and then down the same distance (1/8)(7/8)(4), and up (7/8)(1/8)(7/8)(4), and that's where you catch it. I get 1569/128, but that's incorrect, and I don't know what I'm thinking wrong.

14) If the probability of event A is 0.6, the probability of event B is 0.4, and the probability of A union B is 0.7, find the probability of A given that B has already occured.

I didn't know whether to throw that last part about B has already occured out, and whether it was irrelevant information. What does A union B have to do with this? I've seen union with inequalities, but not with problems like these.

15) Nathan has a 1 dollar bill, a 2 dollar bill, a 5 dollar bill, a 10 dollar bill, a 20 dollar bill, a 50 dollar bill, and a 100 dollar bill. He wants to give Stephanie and Tyler all of his money so that the difference of the amounts he gives them differes by 25 dollars or less. In how many different ways can Nathan give Stephanie and Tyler his money?

Okay. What I did here was, pretend one person is going to get the 100 dollar bill, and other person will get the 50, since those two bills have to be on opposite sides of the hand. Therefore, I knew that the remaining money summed up to 38, and the combinations 37 and 1, 36 and 2, 35 and 3, 33 and 5, and 32 and 6 were possible. Then, I think either person could get the 100 dollar bill, so my answer was 2 (5) = 10. On this one, like the others, I don't know what I did wrong.

13.  You have the right idea, but it lands on the gum on the third bounce, so on the second bounce it will still rebound 7/8 of its original height.  If you are still confused, sketch a graph of the height of the ball as a function of time.

14.  The probability of $ A$ given that $ B$ has occured is denoted as $ P(A|B)$ and can be calculated by $ P(A|B) \equal{} \frac{P(A \cap B)}{P(B)}$.  The probability of $ A$ union $ B$ ($ P(A \cup B)$) can be used with $ P(A)$ and $ P(B)$ to find $ P(A \cap B)$.  If you're not sure how, try drawing a Venn Diagram.

15.  Again a nice start.  I believe you are just forgetting the "38 and 0" case.
Another way to simplify the casework is to note that the total amount of money is 188, so the average is 94.  In order to differ by less than 25, they must be on the interval bounded by $ 94 \pm 12.5$, so one person has at most 106 and the other has at least 82.  Thus one person gets the 100 dollar bill and the other gets the 50,20, and 10.  Then some simple casework can determine how the 1,2, and 5 can be distributed.

[hide="bouncy!"]The equation for the height of the bounce $ b_n$ is $ h_n\equal{}k^{\minus{}n}h_o$ where $ k\equal{}\frac{7}{8}$ and $ h_o\equal{}4$
Accounting for gum $ h_3\equal{}\frac{1}{8}h_2\Longrightarrow\boxed{h_3\equal{}\frac{32}{49}}$  :jump: [/hide]That right?


[quote="joml88"]Prove that if $n$ is an odd integer greater than 1 then the sequence
\[\binom{n}{1},\binom{n}{2},\ldots,\binom{n}{\frac{n-1}{2}}.\]
has an odd number of odd terms.[/quote]

I would like to see if someone finds a combinatorical argument for it.

[hide="My Solution"]As $\binom{n}{k}=\binom{n}{n-k}$, and $\sum_{k=0}^{b}\binom{n}{k}=2^n$, then $\binom{n}{0}+\binom{n}{1}+\binom{n}{2}+\cdots+\binom{n}{\frac{n-1}{2}}=2^{n-1}$, then, there are a even number of odd terms in the sum, but $\binom{n}{0}=1$ is odd.[/hide]

Best,


For intersecting two circles ( the points are [b]E[/b], [b]H[/b]) their two common external tangents meet at [b]A[/b].
One of two tangents touches one of two circles at [b]D[/b]. Let [b]DE[/b] meet the other circle at [b]F[/b].
[b]AF[/b] meets the circle at [b]G[/b]. Prove that [b]A[/b], [b]H[/b], [b]G[/b], [b]D[/b] are cyclic.

Is this a well-known property? Thanks in advance for reading my post.

Dear Mathlinkers,
it is not a converse of the pivot theorem applied to the triangle AFD, H being the pivot?
Sincerely
Jean-Louis


real analysis
function
integration
rudin
real analysis unsolved

I am having trouble with this "big"  Rudin problem:

Suppose A and В are measurable subsets of R^1, having finite positive measure. Show that the
convolution \chi_a * \chi_b is continuous and not identically 0. Use this to prove that A + В contains a
segment.
 Thanks.

The convolution is continuous because the convolution of an $ L^1$ function with an $ L^{\infty}$ function is continuous (or an $ L^2$ function with an $ L^2$ function, or several other choices.) You can prove that by approximating the $ L^1$ function in the $ L^1$ sense by a continuous compactly supported function.

The convolution is not identically zero because $ \int_{\mathbb{R}}\chi_a*\chi_b \equal{} \int \chi_a\cdot\int\chi_b \equal{} m(A)m(B).$

A continuous function that is not identically zero is nonzero everywhere on some interval.

If $ \chi_a*\chi_b(x)\ne 0,$ then $ x\in A \plus{} B.$

Do you understand all of these assertions or is there one you still have trouble with?

[quote="sempre"]Thanks a lot for your reply. I understand most of the claims. The only thing I still can't see is why we get a segment A+B... maybe I don't have a clear understanding of what segment means ... Thanks again.[/quote]

Suppose $ \int \chi_A(x)\chi_B(y \minus{} x) dx > 0$ for $ y$ in a segment. Then for every such $ y$ , the integrand must be positive for some $ x$, which is the same as saying  $ x\in A$ and $ y \minus{} x\in B$. That implies $ x \plus{} (y \minus{} x)\in A \plus{} B$.

In this context, "segment" means "interval of positive length." It's an odd word to use, and I think I was reading it as "interval" all along.


[quote="Kalle"]The polynomial $p(z)$ gives the remainder $z + 3$ when divided by $z^2 - 1$ and the remainder $z - 1$ when divided by $z^2 + 1$. Decide the remainder of $p(z)$ when divided by $z^4 - 1$.[/quote]
[hide]
By the conditions of the problem,

p(z) = (z2 - 1)Q(z) + z + 3 where Q(z) is some polynomial of degree n-1 , n being the degree of p(z)

Similarly

p(z) = (z2 + 1)G(z) + z - 1 where G(z) is some polynomial of degree n-1

Also Let

 p(z) =  (z4 - 1)q(z) + r(z)

 We have to find r(z) which should be a cubic polynomial as the degree of z4 - 1 is 4

  Let r(z) = az3 + b2  + cz + d

  Then p(1) = a + b + c + d = 1+3 =4

          p(-1) = -a + b - c + d  = -1+3 = 2

          p(i) = -ai - b +ci + d = i-1

           p(-i) = ai -b - ci + d = -i-1 where i= square root of -1

Solving the above 4 eqns, we get the values of a,b,c,d and hence we can define the remainder polynomial r(z)[/hide]

[quote="fourierseries"]There's a problem with your solution.  If $p(z)$ has degree n, then your defined $q(z)$ should have degree $n-2$.[/quote]

That's perfectly right..Sorry for the mistake!!

the degreee of Q should be n-2

[quote="fourierseries"]What I recommend is:

Write $p(z)=(z^4-1)q(z)+r(z)$.

We know that $z^2+1$ and $z^2-1$ are solutions, so plug those into p, and use the remainder theorem to try to find your function![/quote]

That's exactly what I've done if u closely observe my solution  ;)


Hey i got this challenge set and tehre are some probs taht i jus cant seem to get the right answer for. I got some ideas though.

1. How many ways are there to distribute 15 distinct oranges into three different boxes with at most 8 oranges in a box?

ideas: i first thought this was a partition type problem but then realized that all oranges were distince so iw as thinking of doing cases but that would take too long.

2. How many arrangements are tehre of seven a's, eight b's, three c's, and six d's witn no occurrence of consecutice pairs ca or cc?

ideas: i found total number of arrangements but then finding the number wiht ca's or cc's is prooving to be very difficult.

I woudl really appreciate it if you could use permutations or combinations to solve these problems. Thanks a lot in advance.

[hide]
For number 1, order matters right?  Since it asks for the number of ways to distribute 15 distinct oranges.
If so I was thinking you could do cases:
15 nPr 8 + 7 nPr 6 + 1 nPr 1
15 nPr 8 + 7 nPr 5 + 2 nPr 2
... and so on ...

Would this work?

[/hide]

ya i was thinin of doin cases but wat happens is that since oranges and boxes are distinct i geuss you end up wiht too many cases. I geuss that is teh best way of doin it. K thanks fr the tip im gna try it like that.

Yea your right.  Where did these problems come from though?  There is probably an easier way we are not thinking of.

these are teh challenging problems that our teacher gave us I dont kno what he was thinking off some of the other ones are very complicated.


inequalities
calculus
integration
search
induction
limit
inequalities unsolved

Maybe induction can help ...

Anyway - it's really magnificent

Sorry very much, Harazi. I didn't realize that, as I have deduced it from the SUPER-HARD inequality of Pecaric

Let $a^{(n)}=(a_1,a_2,.....,a_n)$ be a positive sequence $(n=1,2,....)$. We define the powermeans of $a^{(n)}$ in the following way

$M_{r,n} = M_{r,n}(a^{(n)})=(\frac{\sum_{k=1}^{n}(a_k)^r }{n})^{\frac{1}{r}} $ if $r \neq 0$

and
$M_{r,n} = M_{r,n}(a^{(n)})=( \prod_{k=1}^{n} a_k)^{\frac{1}{n}} $ if $r = 0$

Now we define the following sequence of power means

$M^{r,n} =(M_{r,1},M_{r,2},...,M_{r,n})$

And now finally the SUPER HARD ineq

$  M_{s,n}(M^{r,n}) \leq M_{r,n}(M^{s,n})$

for every $r \leq s$.

I found it in this very interesting DOCTORAL THESIS file

epubl.luth.se/1402-1544/2003/17/LTU-DT-0317-SE.pdf

Hi Manlio!
Thank you for link!
Unfortunately, I didn't know Hardy's inequality five years ago, when I invented this inequality and it was my course work to prove disrete form. And I did it by proving integral form and after that disrete form.

[quote="manlio"]For any positive real numbers $a_i$ with $i=1,...,n$ define

$A_k=\frac{\sum_{i=1}^{k}a_i }{k}  $

$G_k=( \prod_{i=1}^{k} a_i)^{\frac{1}{k}} $

for $k=1,..,n$.

Prove that

$\frac{\sum_{k=1}^{n} G_k}{n} \leq  ( \prod_{k=1}^{n} A_k)^{\frac{1}{n}} $[/quote]

Indeed, if somebody doesn't know yet, two papers by Kedlaya himself on this inequality are online at http://www-math.mit.edu/~kedlaya/papers/ . A question I have been asking myself for a long time is whether there is a cyclic or a symmetric variant of this inequality. In fact, Problem 2 of the MathLinks Contest 3rd edition, 1st round asked to prove the inequality

$\sqrt[3]{\frac{a+b}{2}\cdot\frac{b+c}{2}\cdot\frac{c+a}{2}}\geq\frac{\sqrt{ab}+\sqrt{bc}+\sqrt{ca}}{3}$

for any three positive reals a, b, c. This is already hard enough. Now, one could try to generalize this problem to n positive variables $a_1$, $a_2$, ..., $a_n$ in two ways: the symmetric version,

$\sqrt[\binom{n}{2}]{\frac{a_{1}+a_{2}}{2}\cdot\frac{a_{1}+a_{3}}{2}\cdot...\cdot\frac{a_{2}+a_{3}}{2}\cdot...\cdot\frac{a_{n-1}+a_{n}}{2}}\geq\frac{\sqrt{a_{1}a_{2}}+\sqrt{a_{1}a_{3}}+...+\sqrt{a_{2}a_{3}}+...+\sqrt{a_{n-1}a_{n}}}{\binom{n}{2}}$,

where we take all $\binom{n}{2}$ pairs of variables, and the cyclic version,

$\sqrt[n]{\frac{a_{1}+a_{2}}{2}\cdot\frac{a_{2}+a_{3}}{2}\cdot...\cdot \frac{a_{n-1}+a_{n}}{2}\cdot\frac{a_{n}+a_{1}}{2}}\geq\frac{\sqrt{a_{1}a_{2}}+\sqrt{a_{2}a_{3}}+...+\sqrt{a_{n-1}a_{n}}+\sqrt{a_{n}a_{1}}}{n}$,

where only the n pairs of cyclically neighboring variables are taken. If any of these versions turns out to be true for every n, or, at least, for some n > 3, I will be very happy. I remember having had a short talk on this with Christian Reiher in March 2004, and he said he had an approach to questions like this using combinatorics, but I am not really sure whether he got my question right or he supposed that I was meaning Kedlaya's inequality (whose proof actually uses combinatorics). Anyway, I will be glad to see any advance on this problem.

  Darij

Can somebody show me the proof of Kiran Kedlaya's inequality because I'm not able to open this file on Kedlaya's website [wrong format]. I'll be grateful.

[quote="harazi"]The symmetric inequality you have posted here and which generalizes the problem from Mathlinks contest  is true, Darij. I saw it in a romanian journal, proposed by Mihaly Bencze and Szilard Andras, but the problem is that I do not know a solution for this magnificient problem. I suppose it can be proved by a very smart series of applications of Holder's inequality, but I couldn't find anything till now. Keep searching.[/quote]

Harazi, that's great! I won't give up the hope that there is a nice solution for this inequality.

And, just for the case that somebody is losing his time for nothing: The cyclic (i. e., non-symmetric) version I posted is trash. It fails already for n = 4, $a_1=a_2=1$ and $a_3=a_4=2$. Sorry! :blush:

@Megus: see the attached ZIP file with the two PDFs concerning the inequality.

  Darij

at last i was right :)...darij told me some time in march about that problem(at imo training, at about 10 o'clock in the evening) and i just guessed that the cyclic version is wrong and the symmetric one would be correct :D
however...i guess one might slaughter it by factoring everything out and using muirhead and similar brutalities, but that's actually combinatorics. maybe that's the approach christian meant ;) (but i guess his approaches are much nicer)

Indeed you were right at Bad Homburg...

I believe that Christian meant combinatorics in the way Kedlaya uses it in his proof of his inequality. The combinatorial results he uses come from a book by Donald Knuth, and I believe Christian is a big fan of Knuth, so I guess his approach is similar. Maybe he will thematize this approach in the next IMO training.

  Darij

i just read the beuatiful proofs of kedlaya. that forces me to conjecture that if $w_{ij1}+w_{ij2}=1$ for all $i,j=1,...,n, i<j$, then
\[\sqrt[n\choose 2]{\left(w_{121}a_1+w_{122}a_2\right)\cdot \left(w_{131}a_1+w_{132}a_2\right)\cdots \left(w_{(n-1)n1}a_{n-1}+w_{(n-1)n2}a_n\right)}\geq\]
\[\frac{a_1^{w_{121}}a_2^{w_{122}}+a_1^{w_{131}}a_3^{w_{132}}+... +a_{n-1}^{w_{(n-1)n1}}a_n^{w_{(n-1)n2}}}{\left(n\atop 2\right)}\],

thus just a weighted version of it. sometimes the more general ones are easier to prove, but i didn't yet think about how to prove this one.

Peter

Your conjecture is wrong: take $w_{ij1}=1$ and $w_{ij2}=0$, then you will obtain anty-Cauchy.

Wow, proof of Kedlaya is really nice - I seldom see combinatorics in inequalities.

Thanks Darij for these files  :D

@myth: that's what i recognized 10 seconds after posting it  :blush:

[quote="darij grinberg"]Now, one could try to generalize this problem to n positive variables $a_1$, $a_2$, ..., $a_n$ in two ways: the symmetric version,

$\sqrt[\binom{n}{2}]{\frac{a_{1}+a_{2}}{2}\cdot\frac{a_{1}+a_{3}}{2}\cdot...\cdot\frac{a_{2}+a_{3}}{2}\cdot...\cdot\frac{a_{n-1}+a_{n}}{2}}\geq\frac{\sqrt{a_{1}a_{2}}+\sqrt{a_{1}a_{3}}+...+\sqrt{a_{2}a_{3}}+...+\sqrt{a_{n-1}a_{n}}}{\binom{n}{2}}$,

where we take all $\binom{n}{2}$ pairs of variables, and the cyclic version,

$\sqrt[n]{\frac{a_{1}+a_{2}}{2}\cdot\frac{a_{2}+a_{3}}{2}\cdot...\cdot \frac{a_{n-1}+a_{n}}{2}\cdot\frac{a_{n}+a_{1}}{2}}\geq\frac{\sqrt{a_{1}a_{2}}+\sqrt{a_{2}a_{3}}+...+\sqrt{a_{n-1}a_{n}}+\sqrt{a_{n}a_{1}}}{n}$,

where only the n pairs of cyclically neighboring variables are taken.[/quote]

I have already explained that the cyclic version is wrong, and now Vasc kindly pointed out that the symmetric version is also wrong for any $n\geq 4$. Why? Just take $a_1=a_2=0$.

So I am sorry for posting the above nonsense here. When I talked with Christian Reiher, we probably meant different things.  :blush:

  darij

[quote="harazi"]Manlio, this is Kiran Kedlaya's inequality, already discussed in this forum. Unfortunately, I do not remember to have seen a solution for the general case.[/quote]

 Because the general case is false. :D

Cezar you continue to amaze me each day. First, if it was a revenge, then you should probably quoted the other part, in which I was saying stupid things about an inequality that turned out to be false and not this part, which is correct, since believe it or not there are hard and correct inequalities that cannot be solved using Cauchy-and this is not a reason for them to be wrong. Secondly, this remark is extremely useful for us: of course I realised the OTHER inequality was wrong after Vasc indicated the counter-example, it is useless for you to point it now (but I see you started to search for my mistakes on the forum- very well, enjoy it, there are many) and not to mention the enormous contribution that you brought to this topic. Continue like this, it's really useful.

I'm sorry, but I will not spend my time in searching your mistakes. Very soon, I will show that this Kedlaya's inequality is wrong. ;)

Yes, obviously, like all other promesses. Anyway, this means that Kedlaya and all those people from AMM are really dumb. Congratulations, please show us these extraordinary discoveries and then we will believe you. But if I were you, I would be less confident.  ;)

[quote="manlio"]For any positive real numbers $a_i$ with $i=1,...,n$ define

$A_k=\frac{\sum_{i=1}^{k}a_i }{k}  $

$G_k=( \prod_{i=1}^{k} a_i)^{\frac{1}{k}} $

for $k=1,..,n$.

Prove that

$\frac{\sum_{k=1}^{n} G_k}{n} \leq  ( \prod_{k=1}^{n} A_k)^{\frac{1}{n}} $[/quote]

 This inequality is false. We shall use Carleman inequality which states that $\sum_{k=1}^{n}  G_{k} < e\cdot\sum(\prod_{k=1}^{n} A_{k})^{\frac{1}{n}}$. This inequality can be also written as:$\frac{1}{n}\cdot\sum_{k=1}^{n} G_{k} < e\cdot A_{n} $.Now, if Kedlaya's inequality holds true (we suppose that) if $a_{1}\leq a_{2}\leq... \leq a_{n}  $. From here we deduce that $A_{1}\leqq_{2}\leq....\leq A_{n}$ so Kedlaya's inequality would imply $\frac{1}{n}\cdot\sum_{k=1}^{n} G_{k}\leq A_{n}$. But we can deduce that $\lim\frac{\sum_{k=1}^{n} G_{k}}{\sum_{k=1}^[n} a_{k}=e$,where $e$ is the best constant possible.From this, for any $\epsilon >0$ there is $n(\epsilon)=n_{0}$ such that for $n\geq n_{0}$ we have $\frac{1}{n}\cdot\sum G_{k}/A_{n}> e-\epsilon $. Putting $\epsilon=0,71...$ then for $n$ large enough for $a_{k}=\frac{1}{k}$, ordered in increasing order, we get a contradiction. ;)  :D

[quote="cezar lupu"]Do you agree with me? :?[/quote]
I think I don't. I think that e is the best possible constant only when there is no other condition. When you add the condition $a_1 \le a_2 \le \cdots \le a_n$, things may change much. I doubt e is still the best possible constant unless you can convince me that's the truth by proving it.  ;)

With the condition $a_1 \le a_2 \le \cdots \le a_n$ you ''destroy'' the generalisation of the problem I think.

[quote="cezar lupu"]Silouan, I don't quite think so. ;)[/quote]

have you read my comment? Prove it if you don't think so.  Remember this is not a symmetric inequality (in fact not even cyclic) :lol:

[quote="silouan"][quote="harazi"]Let him sleep. Congratulations, Cezar, your proof works perfectly. I suggest you to send it immediately to AMM. Too bad it is totally wrong. By the way, have you ever tried at least to read the proof?  :P[/quote]
Harazi I dont understand you? you mean that the ineq of Kiran Keldaya is wrong? :?[/quote]
it was an irony


[quote="N.T.TUAN"][quote="FOURRIER"]Find all  p,q,r £ P with:
[img]http://www.gnux.be/latex/data/e6f5a656ec0ac9746afe1f7e2c88ad1a.png[/img]
P={2,3,5,7,11,13...}
And thanks  :lol:[/quote]
$q^{2}+r^{2}=p^{2}(p-1)$ is even. Done ! :P[/quote]

why is it done?

[quote="aviateurpilot"]if $p=q$ then $p|r$ then $p=r=q$
so $p^{3}=3p^{2}$ => $p=r=q=3$

if $p\neq q$ then $there\ exist\ u,such\ that\ uq\equiv 1[p]$
we have $r^{2}\equiv-q^{2}[p]$, then $(ur)^{2}\equiv-(qu)^{3}[p]\equiv-1[p]$
[/quote]
Until here I got (c bon)
But why
[quote="aviateurpilot"]so $p\equiv 1[4]$[/quote]?
Thank you enlighting me.

[quote="aviateurpilot"]$r^{2}\equiv-1[p]$
$1\equiv r^{p-1}\equiv (r^{2})^{\frac{p-1}{2}}[p]$
then ${\frac{p-1}{2}}$ is even
so, $p\equiv 1[p]$[/quote]
Do we have $u^{2}r^{2}\equiv-1[p] \Rightarrow  r^{2}=-1[p]$?

[quote="N.T.TUAN"][quote="FOURRIER"]Find all  p,q,r £ P with:
[img]http://www.gnux.be/latex/data/e6f5a656ec0ac9746afe1f7e2c88ad1a.png[/img]
P={2,3,5,7,11,13...}
And thanks  :lol:[/quote]
$q^{2}+r^{2}=p^{2}(p-1)$ is even. Done ! :P[/quote]
Can you show us your quick method please
 :maybe:


inequalities
trigonometry
combinatorial geometry
geometry unsolved
geometry

Let $M$ be a set of 6 points in the plane such that the distance between any 2 points is at least 1. Prove that there are two points $A$, $B$ in $M$ such that
\[ AB \geq \sqrt { \frac { 5 + \sqrt 5 } 2 } . \]

[hide="First ideas"]
If there exist 3 points $A$, $B$, $C$ such that $\angle ABC \geq \dfrac {4\pi } 5$, then by the Cosine rule we have
\[ AC^2 = BA^2+BC^2 - 2 AB \cdot BC \cdot \cos {\dfrac {4\pi }5 } \geq 1+1+ 2 \cos { \dfrac {\pi }5 } = \sqrt { \frac {5 +\sqrt 5} 2  }. \]
Also observe that the equality takes place when $ABCDE$ is a regular pentagon and $F$ is the center of the circumscribed circle.

It remains to study the cases when the convex hull of $M$ is a) a hexagon; b) a pentagon; c) a quadrilateral; d) a triangle.
I suspect that in case a) we can actually prove that there is a distance in $M$ greater or equal with 2. However simple Dirichlet in b), c), d) doesn't seem to work.
[/hide]

Any other ideas? :)

Maybe this can be ironed out into a proof: translate and scale (if necessary) so that $A = (0,0)$ and $B = (1,0)$. Prove a contradiction, by arguing that

1. The regions in which the four other points must be are the two congruent, kite shaped regions (four vertices connected by circular arcs) formed by $1 \le AX, BX < \sqrt{\frac{5+\sqrt{5}}{2}}$.
2. Of the four remaining points $C, D, E,$ and $F,$ all four must lie in the same kite region.

3. Contradiction stemming from 2.


[quote="Vasc"]If $a,b,c$ a non-negative numbers such that
$a^2+b^2+c^2=a+b+c$,
then
$ab+bc+ca \geq a^2b^2+b^2c^2+c^2a^2$.[/quote]
$a^4+b^4+c^4-(a^2+b^2+c^2)=2(ab+bc+ca )-2(a^2b^2+b^2c^2+c^2a^2).$
Hence, $ab+bc+ca \geq a^2b^2+b^2c^2+c^2a^2\Leftrightarrow$
$\Leftrightarrow a^4+b^4+c^4\geq a^2+b^2+c^2\Leftrightarrow a^4+b^4+c^4\geq\frac{(a^2+b^2+c^2)^3}{(a+b+c)^2}\Leftrightarrow$
$\Leftrightarrow(a+b+c)^2(a^4+b^4+c^4)\geq(a^2+b^2+c^2)^3,$  which true by Holder.

[quote="Arqady"]$\Leftrightarrow a^4+b^4+c^4\geq a^2+b^2+c^2$[/quote]
It is easy to show more generally $a^{n+1}+b^{n+1}+c^{n+1}\ge a^n+b^n+c^n$ for all $n\ge2$ :)

Very nice solution, Arqady.
And Spanferkel is right. ;)

A nice related inequality is the following:

If $n \geq 2$ is an integer and $a,b$ are non-negative numbers
such that $a+b=a^n+b^n$, then $a^{n+1}+b^{n+1} \leq 2$.


I believe the rezults of the Iranian team will be
IRN1 7 7 7 7 7 7
IRN2 7 7 7 7 7 7
IRN3 7 7 7 7 7 7
IRN4 7 7 7 7 7 7
IRN5 7 7 7 7 7 7
IRN6 7 7 7 7 7 7
 :)
I think Iran is going to win this olympiad.
And what do you think about it?

only one iranian solved problem 3
so i dont think they will be 1 st

And you really think German team will take 2nd place?

the results werent so good,

they could be the first, but they couldnt.

i hope we will win the nexst imo :wink:

I'm ashamed of this result ,very bad result for Iran   :(

[quote="José"]Searching, I found that you were 8º with 145 points, but can anyone post the individual results? Thanks.[/quote]

Well, you could've done more searching and you could find what you want at the official IMO 2006 site:
http://imo2006.dmfa.si/results_itd.html
or at http://www.imo.org.yu/index.php?options=gl|imotres&zemljarez=IRN&p=34:31

[quote="José"]Thanks.

Well, I don't see why Iran had a bad result: 3 golds and 3 silvers!!! That's preety nice eh! Congratulations![/quote]
 I just think that we can do much better next year,because our team was younger than the average.next year we ar experienced
[/quote]

[quote="rouzbeh_k"] I just think that we can do much better next year,because our team was younger than the average.next year we ar experienced
[/quote]
How do you know your team is younger than the average?

[quote="José"]Thanks.

Well, I don't see why Iran had a bad result: 3 golds and 3 silvers!!! That's preety nice eh! Congratulations![/quote]

because iran have won the imo in 1999.

and it is a kind of decrising.

i should say that all of the boys in iran team2006, are so clever. :D


geometry
circumcircle
rectangle
parallelogram
geometry proposed

Let $ O$ be the circumcenter of an acute triangle $ ABC$ and let $ k$ be the circle with center $ P$ that is tangent to $ O$ at $ A$ and tangent to side $ BC$ at $ D$. Circle $ k$ meets $ AB$ and $ AC$ again at $ E$ and $ F$ respectively. The lines $ OP$ and $ EP$ meet $ k$ again at $ I$ and $ G$. Lines $ BO$ and $ IG$ intersect at $ H$. Prove that $ \frac{{DF}^2}{AF}\equal{}GH$.

It is actually an easy problem: $ P$ lies on $ OA$, while $ D$ on the bisector of the angle $ \angle BAC$, so $ DF \equal{} DE$; next, see that $ AEIG$ is a rectangle and $ EBHG$ a parallelogram, so $ GH \equal{} BE$, consequently, the relation to be proved becomes
$ DE \cdot DF \equal{} AF \cdot BE$ ( 1 ). It's also easy to see that $ EF \parallel BC$ and $ DF \parallel CK$ and $ DE \parallel KB$.
If $ r$ and $ R$ are the radii of the circles $ (AEF)$ and $ (ABC)$.
Now: $ \frac {BK \cdot CK}{AB \cdot AC} \equal{} \frac {KD}{AD}$ ( 2 ), $ AD\cdot AK \equal{} AB \cdot AC$ ( 3 )
But: $ \frac {AE}{BE}\equal{}\frac{AF}{CF}$, or $ AF\cdot BE\equal{}AE\cdot CF$, so we have to prove the equality $ AE\cdot CF\equal{}DE\cdot DF$ ( 4 ).
Since $ \frac{DE\cdot DF}{KB\cdot KC}\equal{}\left(\frac{r}{R}\right)^2$, with (2) and (3) we get an identity in (4), q.e.d.

Best regards,
sunken rock

[quote="sunken rock"]It is actually an easy problem: $ P$ lies on $ OA$, while $ D$ on the bisector of the angle $ \angle BAC$, so $ DF \equal{} DE$; next, see that $ AEIG$ is a rectangle and $ EBHG$ a parallelogram, so $ GH \equal{} BE$, consequently, the relation to be proved becomes
$ DE \cdot DF \equal{} AF \cdot BE$ ( 1 ). It's also easy to see that $ EF \parallel BC$
[/quote]
since $\angle DAF=\angle DEF=\angle BDE$ and $\angle ADF=\angle AEF=\angle ABC=\angle DBE$ ie $\triangle BDE\sim \triangle AFD$ so $DE/EB=AF/FD$

First we observe that ($A,P,O$ are collinear) $EGHB$ is a parllellogrm.$\Longrightarrow GH=EB$
Next $EP||BO$ and $CP||CO \Longrightarrow \frac{AE}{EB}=\frac{AP}{PO}=\frac{AF}{FC} \Longrightarrow FE||BC \Longrightarrow FD=ED$ we suffice to prove that $ED^2=EB.AF$. Since $ED=DF, AD$ is the angular bisector of $\angle BAC$ Let $DA$ intersect $EF$ at $L \Longrightarrow DE^2=DL.DA$  .But since $\Delta EAL \sim \Delta  ADF$ we have $\frac{EB}{LD}=\frac{AE}{AL}=\frac{AD}{AF} \Longrightarrow EB.AF=DL.DA$ and we are done

Please ,

is there an idea how to construct with ruler and compass the shape?

Thanks


Tim has three times as many coins as Mike. If Tim gives one coin to Mike, Mike will
have a total number of coins equal to half the number of coins Tim started with. How
many total coins do Tim and Mike have together?

let's say that Tim has 3x coins (so Mike has x coins)

$ x\plus{}1\equal{}\frac{3x}{2}$

and $ 1\equal{}\frac{x}{2}$ so x is 2.

so there are total 6+2=8 coins.

answer : 8