Learn $ \text{\TeX}$.  I have no idea what you're trying to say.

hello, do you mean $ 4^{2x}\plus{}\left(\frac{1}{2}\right)^{2x}\equal{}8$?
Sonnhard.

I'm guessing he means $ 4^{2x\plus{}1}(\frac{1}{2})^{2x}\equal{}8$

I'm going to go with Ihatepie's guess.

Note that $ 4\equal{}2^{2}$, $ \frac{1}{2}\equal{}2^{\minus{}1}$, and $ 8\equal{}2^{3}$.  Then $ 2^{4x\plus{}2\minus{}2x}\equal{}2^{3}$, so $ 2x\plus{}2\equal{}3\Rightarrow x\equal{}\frac{1}{2}$.


[quote="Singular"]A hexagon has sides 1,2,3,4,5,6 and equal angles.  What is its area?[/quote]

The problem is not well-defined unless you specify in which order the sides 1, 2, 3, 4, 5, 6 occur (e. g. in the order 1, 6, 2, 4, 3, 5 or in the order 1, 6, 3, 2, 5, 4 - this makes a difference). Once this is specified, the area can be obtained from http://www.mathlinks.ro/Forum/viewtopic.php?t=16758 post #4.

  darij

By the way, a beautiful problem: prove that there exists an equiangular n-gon whose side lengths are $1,2,\ldots ,n$ (in some order) iff $n$ is not a prime power.

wow, that's a beautiful problem enescu! Any hint for proving it? :)

Use the lemma in post #4 of [url=http://www.mathlinks.ro/Forum/viewtopic.php?t=33922]this topic[/url].


abstract algebra
vector
geometry
3D geometry
sphere
advanced fields
advanced fields unsolved

If $f$ is a linear functional on $X$, $f \neq 0$, $N(f) = \{ x \in X: f(x) = 0 \}$ and $x_{0} \in X \setminus N(f)$, then any $x \in X$ can be uniquely represented as $\alpha x_{0} + y$ where $\alpha \in \mathbb{R}$ and $y \in N(f)$.

How can this be generalized to the case where there are $n$ linear functionals $f_{1},\ldots,f_{n}$? What should be the analogous statement?

It is because $\mathrm{codim}\,\ker f: =\dim\,(X/\ker f)=1$.

Yes that explains the first statement. But I am thinking about how it can be generalized to the case of n linear functionals.

Denote $Y: =\cap_{i=1}^n\ker f_i$. All I can say is that there exists a decomposition $X=X_0\oplus X'\oplus Y$ with $\dim X_0=1$, so $x\in X\Rightarrow x=\alpha x_0+x'+y$ where $x_0\in X$, $x'\in X'$ and $(\forall\,i=1,\ldots,n)\;f_i(y)=0$. In the case of $n=1$ $X'=\{0\}$.

I have managed to show that if $f_{1},\ldots,f_{n}$ are nonzero linear functionals on $X$, $Y_{m} = \cap_{i=1}^{m} Ker (f_{i})$ and $x_{i} \in Y_{i-1} \setminus Ker(f_{i})$ for $i=1,\ldots,n$ (where $Y_{0} : = X$), then any $x \in X$ has a unique representation as $a_{1}x_{1} + \cdots + a_{n}x_{n} + y$ where $y \in Y_{n}$.
I hope I didn't make a mistake.

Hopefully Vladimir (or anyone else) can help me check the statement in my previous post. The next thing I would like to ask is if I have n linear functionals  $f_{1},\ldots,f_{n}$ on X, $Y=\cap_{i=1}^{n} Ker(f_{i})$ and $x_{1},\ldots,x_{n} \in X$ are linearly independent and not in $Y$, do I get some similar result? Or what other conditions can I impose on $x_{1},\ldots,x_{n}$ so that $x_{1},\dots,x_{n}$ together with $Y$ will span $X$?

Yes, you are right but there is inaccuracy in your previous statement. It would be more correctly to say that there exits family of vector subspaces $(X_i)_{0\le i\le n-1}$ with $Y_i=X_i\oplus Y_{i+1}$ and $X=X_0\oplus\ldots\oplus X_{n-1}\oplus Y_n$ because if $(f_i)_{1\le i\le n}$ are linearly dependent, for example $f_n=\sum\limits_{i=1}^{n-1}\lambda_if_i$, then $Y_{n-1}\subset\ker f_n$ and hence $Y_{n-1}\setminus\ker f_n=\varnothing$ i.e. $x_n=\varnothing$ that is absurdity while in the decomposition above some of $X_i$ are allowed to be trivial (the sequence $(Y_i)_{0\le i\le n-1}$ is decreasing but not obligatory strictly decreasing hence in the decomposotion $Y_i=X_i\oplus Y_{i+1}$ $X_i$ is allowed to be $\{0\}$).

I have another question that seems to be related. Now $X$ is a Banach space. Take $F \subset S(X^{*})$ such that $|F|=n$ and let $X_{F} = \{ x \in X: x^{*}(x) = 0 \: \forall x^{*} \in F \}$. Let $x_{1},\ldots,x_{n+1} \in X$ be linearly independent. Then there exists $x \in S([x_{i}]_{1}^{n+1})$ such that $x \in X_{F}$. Why is that so?

:?: What do you mean under $S$? Does $|F|$ means cardinal number of $F$?

$F$ is a set of n bounded linear functionals with norm 1. In the last line, $S([x_{i}]_{1}^{n+1})$ refers to the unit sphere of the (closed) linear span of $x_{1},\ldots,x_{n+1}$.

As to your next question formulated in the terms of intersection of kernels. The responsibility for spanning of $X$ lies not on $x_1,\ldots,x_n$ but on all the kernels. For example in the case of 2 functionals, linearly independent $\{x_1,\ldots,x_k\}\subset X\setminus Y$ together with $Y$ will span $X$ iff $\dim\,\ker f_1/Y=k-1$.

As to the last statement formulated in the terms of Banach spaces. It is not true in general. You actually state that given a finite set of linear functionals $F=\{x_i^*\}_{1\le i\le n}$ and arbitrary vector subspace $X_0$ with $\dim X_0=n+1$, $X_F=\cap_{i=1}^n\ker x_i^*$ and $X_0$ have nontrivial intersection. It is not genarally true bacause $X_0$ may lie in a complement of $X_F$ which has infinite dimension (if we choose $F$ properly). Again all depends not on $X_0$ but on the kernels...

Anaway, I could certainly make a mistake.

The questions actually come from proofs that I did not understand in a paper so I could have missed out some details or misunderstood something. I still cannot figure them out but many thanks to Vladimir for spending time to reply.


[hide]The man was a millionaire when he walked in![/hide]

Good answer, but there is a situation where the man could've walked into the building not being a millionaire.

[hide]He collected lottery prize money? [/hide]

[hide]It was a game show? (i.e. Deal or No Deal, Apprentice, Power of 10)[/hide]

Neither of the above are what I had in mind.  Besides, that would be earning money.

[hide="Hint"]He entered the building with money.  He knew that he was going to be a millionaire.
Keep in mind the definition of "millionaire": a person whose wealth amounts to a million or more in some unit of currency[/hide]

[hide]did he walk out a building CALLEd a millionaire?[/hide]

[quote="t0rajir0u"][quote="hunter34"][hide]The Building he walked into was an Exchange Currency Place.[/hide][/quote]

I like this answer :)

[hide="Alternately..."] Hyperinflation occurred while he was in the building.  Don't laugh, it really can happen!  :D [/hide][/quote]
He was living in post-WWI Germany with many non-monetary assets?

[quote="t0rajir0u"][quote="hunter34"][hide]The Building he walked into was an Exchange Currency Place.[/hide][/quote]

I like this answer :)

[hide="Alternately..."] Hyperinflation occurred while he was in the building.  Don't laugh, it really can happen!  :D [/hide][/quote]

True, but mine seems a bit more reasonable.

[quote="lingomaniac88"]A man walks into a building.  Ten minutes later, he walks out of the building a millionaire.  He did not do anything illegal, nor did he spend or earn any money.  What happened?[/quote]
[hide]
He could have found a million dollars on the ground.
or he could have money poofed into his pocket the moment before walking out of the building
or he could have been a millionaire before entering the building
or he could have legally got a million dollars without earning it.
There are many possibilities.

[/hide]

[quote="t0rajir0u"][quote="hunter34"][hide]The Building he walked into was an Exchange Currency Place.[/hide][/quote]

I like this answer :)

[hide="Alternately..."] Hyperinflation occurred while he was in the building.  Don't laugh, it really can happen!  :D [/hide][/quote]
[hide]
Hyperinflation doesn't increase the amount of money you own unless its in the bank or in stocks or anything, which would mean that it's not your money, it's your assets.
Hyperinflation just decreases the value of your money because other people have more money.
[/hide]

[quote="ProtestanT"][hide]
Hyperinflation doesn't increase the amount of money you own unless its in the bank or in stocks or anything, which would mean that it's not your money, it's your assets.
Hyperinflation just decreases the value of your money because other people have more money.
[/hide][/quote]

[hide="Well..."] I wass, in fact, referring to his assets under the loosest definition of "millionaire," which is a person whose net worth is at least a million (in some currency). [/hide]

[quote="Hamster1800"][hide]
Technically, exchanging currency could be considered spending money, since you're buying a different currency with your currency.
[/hide][/quote]

[hide="Yeah!"] A very good point!  And if we take this technicality into consideration then my answer actually ends up being better  :P [/hide]

Another possibility that I'm surprised no one has mentioned is that he could have [hide]inherited a large sum of money.[/hide]

After reading the responses above already,

I see that dropping off money in the building would not violate "nor did he
spend ...any money" because he would be just giving away a certain amount.

And to be different from Hamster's reply, the man was a BILLIONAIRE at the start, dropped off enough money into the building until he was down into the millions of dollars, and left.

< It was already stated that some other answer was the 'desired' answer. >

[quote="box of rocks"]the man was a BILLIONAIRE at the start, dropped off enough money into the building until he was down into the millions of dollars, and left.[/quote]
Millionaire $ \in$ billionaire?

What you mean is $ \{\textrm{millionaires}\}\subset\{\textrm{billionaires}\}$.

[quote]nor did he spend or earn any money.[/quote]
Key statement


geometry
3D geometry
sphere
vector
function
analytic geometry
Pythagorean Theorem

It is geometrically obvious that any circle on a sphere (in R^3) is equal to the intersection points between the sphere and a suitable plane. I am sure that it is possible to prove this by just writing general equations, changing basis to an ON basis where one base vector is a normal vector to the plane and then just finding the new equations. However, this method seems to be very messy. Can anyone think of an easy/elegant proof?

Spherical distance is a function of Euclidean distance, so a circle on a sphere is the intersection of two spheres. Choose cylindrical coordinates around the axis between the centers of these spheres; they have equations $(z-c_{1})^{2}+r^{2}=R_{1}^{2}$ and $(z-c_{2})^{2}+r^{2}=R_{2}^{2}$ respectively. Subtracting the two equations, $2(c_{1}-c_{2})z=R_{2}^{2}-R_{1}^{2}+c_{1}^{2}-c_{2}^{2}$, and the circle is the intersection of the plane $z=\frac{R_{2}^{2}-R_{1}^{2}+c_{1}^{2}-c_{2}^{2}}{2c_{1}-2c_{2}}$ with the original sphere.

[quote="jmerry"]Spherical distance is a function of Euclidean distance, so a circle on a sphere is the intersection of two spheres.[/quote]
Sorry for being daft, but I don't get this! I've looked up what spherical distance is, but I'm still not sure. Could you please elaborate a little? Thanks

The spherical distance between two points $A$ and $B$ on a sphere with center $O$ is either the angle $AOB$ or the radius times that angle (the distance along a great circle path). We can relate the spherical distance and the Euclidean distance with trigonometry.

I still don't understand how this means that a circle on a sphere is the intersection of two spheres. Sorry.

It's the intersection of our original sphere and of another sphere centered at a point on the original sphere; that point is the "spherical center" of the circle.

[quote="Kalle"]Can anyone think of an easy/elegant proof?[/quote]
What is a circle anyway? I define it as the set of all points in a plane that are at some fixed distance from some fixed point (the center). So, any circle $C$ is contained in some plane $P$ simply by definition. If we also assume that $C$ lies in a sphere $S$, then.$C\subseteq S\cap P$. It remains to prove that $S\cap P$ is actually a circle and not its superset. Let $A$ be the orthogonal projection of the center of $S$ onto $P$; by the Pythagorean theorem all points of $S\cap P$ are at the same distance from $A$.

[quote="mlok"]What is a circle anyway? [/quote]

I also would like to know what is a "circle on a sphere".

Maybe the intention was "all points at a fixed spherical distance from a given point on the sphere".  Due to rotational invariance this is a union of (Euclidean) circles in $R^{3}$, but one can easily show that there is at most one circle for any given distance.

Question:  is there a local definition of "circle on a sphere"?


calculus
integration
function
geometry
complex numbers
complex analysis
complex analysis unsolved

hello

I need some help with the following proof:

$ Re (\int_{\gamma}\overline{z}dz) \equal{} 0$

$ \gamma$ is a closed contour
nothing is given about $ z$ i guess we can assume it's analytic

thank you  :)

$ z$ is the variable, and $ \overline{z}$ is the function we're integrating.

I'd go back to real variables here: we have
$ \int_{\gamma}(x\minus{}iy)(dx\plus{}i\,dy) \equal{}\int_{\gamma}(x\,dx\plus{}y\,dy)\plus{}i(x\,dy\minus{}y\,dx)$
We only care about the real part. Applying Green's theorem, it becomes $ \iint_{R}(0\minus{}0)dA \equal{} 0$, where $ R$ is the region enclosed.

The imaginary part is twice the area if we move in a counterclockwise direction.


Note the following:

Brackets when used as m[n]! do not mean multiplication.  Instead they are to be read as "m of n factorial."

Definitions
-------------

m[1]! = m!

Let m >= n

Then m[n]!  = m[n-1]!*(m-1)[n-1]!*(m-2)[n-1]!*...*2[n-1]!*1[n-1]!

Example)

Find the value for m[n]! when m = 4 and n = 3.

       4[3]!  =

                 = 4[2]!*3[2]!*2[2]!

                 =  4[1]!*3[1]!*2[1]!*1[1]!*3[1]!*2[1]!*1[1]!*2[1]!*1[1]!

                 = 4!*3!*2!*1!*3!*2!*1!*2!*1!

                 = 24*6*2*6*2*2

                 = 6912

Question:  What magnitude (or approximately how many digits)

                  does 9[9]! have when written out in base 10 form?

First,

9[9]! = 9[8]!*8[8]!

Second,

9[8]! = 9[7]!*8[7]!*7[7]!  and 8[8]! = 8[7]!*7[7]!

So, up to this point (I'm using braces here for multiplication),

9[9]! = 9[7]!{8[7]!*7[7]!}^2

After many steps I got <here I am using brackets back to the usual way for grouping and multiplying> :

9![(8!)^8][(7!)^35][(6!)^110][(5!)^275][(4!)^572][(3!)^1001][(2!)^1430]

Then I took the log (base 10) of this last result to see about the number of digits in 9[9]! and got 3056.9+

I rounded down to 3056 (digits).


algebra
polynomial
modular arithmetic
quadratics
number theory unsolved
number theory

Let $p(X) \in \mathbb{Z}_{p}[X]$ be an odd polynomial, where $p$ is prime and $p \equiv 3 \pmod{4}$.

How many solutions $(X,Y)$ does the equation $Y^{2}= p(X)$ have in $\mathbb{Z}_{p}^{2}$?

It seems to be too easy. Since $p(-x)=-p(x)$ and $-1$ is not a quadratic residue $\mod{p}$ (because $p \equiv 3 \mod{4}$) we have $(\frac{x}{p})(\frac{-x}{p})=(\frac{-1}{p})(\frac{x^{2}}{p})=-1$. It means that exactly one of $x$ and $-x$ is a quadratic residue $\mod{p}$. Therefore, we have $p-1$ solutions for $x \not \equiv 0 \mod{p}$ (since $y \equiv (-y)^{2}$). $p$ is odd so it has no constant term and of course we have exactly one solution for $x \equiv 0$. Therefore, the number of solution is $p$.


You can make it however you want.

I would like to see hehehehehe because it's fun.

   I will post my own later.

Again something that definitely doesn't belong here!!!

This poem is about me.
What isn't? Haikus own all.
All including you.

I have not any
friends because I am in
love with mathematics.

Most people on this web
site or a forum are in
the same case as me.

ZetaX hates me
for some odd reason. I do
not know the reason.

Perhaps it is because
cyberbullying me makes
his self esteem raise.

[quote="Ignite168"]ZetaX hates me
for some odd reason. I do
not know the reason.

Perhaps it is because
cyberbullying me makes
his self esteem raise.[/quote]
(1)
How about this one:
you [i]try[/i] to annoy, never he;
he's respected, not you.

---
(2)
I just got pwned hard!
Stupid olympiad stuff
shouldn't be on Putnam.

---
(3)
As ZetaX just said,
this thread should be moved away
down to GFF.

[quote="Xevarion"][quote="Ignite168"]ZetaX hates me
for some odd reason. I do
not know the reason.

Perhaps it is because
cyberbullying me makes
his self esteem raise.[/quote]
(1)
How about this one:
you [i]try[/i] to annoy, never he;
he's respected, not you.[/quote]

He always tries to
annoy in my topics and
never on topic

No one respects him
outside of people who don't
go outside ever

ZetaX has a
very awesome-unique name
too bad I couldn't

take it and give it
positive connotations
(nobody likes him)

The Only ME

loves to do anything that spells math.

to take the exact one precise path

always thinking

always pondering

all about one thing.

Math


[u]A immediate solution.[/u]

$a_1=b_1=55$; for $k\in N^*$, $a_k=a_1+(k-1)r_a$, $b_k=b_1+(k-1)r_b$.
I note $x=r_a+r_b$, $y=r_a-r_b$. Therefore, for any $k\ge 2$
$a_k+b_k=110+(k-1)x;\ a_k-b_k=(k-1)y$ and

$\displaystyle \begin {array}{ccc} a^2_{k-1}-b^2_{k-1}=78 & \Longleftrightarrow & (k-2)^2xy+110(k-2)y=78;\\ a^2_k-b^2_k=53 & \Longleftrightarrow & (k-1)^2xy+110(k-1)y=53;\\ a^2_{k+1}-b^2_{k+1}=27 & \Longleftrightarrow & k^2xy+110ky=27.\\ \end {array}$

[u]Answer:[/u] $k=54;\ x=-2,\ y=\frac 14\Longrightarrow \overline {\underline {\left| k=54,\ r_a=-\frac 78,\ r_b=-\frac 98\right| }}.$

Thanks for the solution Ph-An. I solved this problem a long time ago but your solution is just beautiful.
But maybe your answer is wrong  ;)  check again and post a little of your solution, becouse i found different from your answer numbers

Verify you the your solution ? I verified the my solution ans is O.K.


[quote="ehsan2004"]Then we know that
$x^2+y^2+z^2\geq \frac{(x+y+z)^2}{3}\Longrightarrow
\frac{(x^2+y^2+z^2)^4}{(x+y+z)^4}+\frac{3}{x+y+z}\geq \frac{(x+y+z)^4}{3^4}+\frac{3}{x+y+z}\geq \frac{x+y+z}{3}+\frac{3}{x+y+z}\geq 2$ ;)[/quote]

Hi, on the last inequality you have
\[
\frac{(x+y+z)^4}{3^4}+\frac{3}{x+y+z}\geq \frac{x+y+z}{3}+\frac{3}{x+y+z}
\]

which is not true, for example when $x+y+z=1$.

Yes oldman you have right.For example if x=y=z=-1 A=0

[quote="siuhochung"][quote="silouan"]Find the min price of [(x^5)+(y^5)+(z^5)+3]/ (x+y+z)[/quote]
I suppose x,y,z are positive, then $x^5+ 1 \ge 5(1/4)^{1/5} x $. Do the same to others we get the min = $5(1/4)^{1/5}$[/quote]

Very nice one line solution :)


Given quadrilateral $ ABCD$. Its sidelines $ AB$ and $ CD$ intersect in point $ K$. Its diagonals intersect in point $ L$. It is known that line $ KL$ pass through the centroid of $ ABCD$. Prove that $ ABCD$ is trapezoid.

[i]Author: F.Nilov[/i]

Assuming that $ AD\cap BC\equal{}\{G\}, KL\cap AD, BC\equal{}\{E,F\}$ then $ (ADEG)\equal{}(BCFG)\equal{}\minus{}1$ so $ \frac{AE}{ED}$ and $ \frac{BF}{FC}\neq 1$, which follow KL doesn't pass through midpoints of AD, BC therefore does'nt pass through the centroid of ABCD.
Thus $ ABCD$ is a trapezoid.


function
inequalities
algebra
functional equation
algebra proposed

My generalization of a Romanian contest problem. It is black beacause you can't see it's beauty from the beginning...:
$f:(0,\infty)\rightarrow (0,\infty)$. Find $f$ if ,for every $x_1,x_2...x_n\in(0,\infty)$ we have:
$f(x_1)f(x_2f(x_1))f(x_3f(x_2+x_1))f(x_4f(x_1+x_2+x_3))...f(x_nf(x_1+x_2+...x_{n-1}))=f(x_1+x_2+...+x_n)$.
Add the condition that $f$ is non-constant on any interval $(k,\infty)$, although it may work without this condition too.

i found it in a romanian contest (Dan Barbilian), for n=3. But I don't know if the second condition is really necesarry.
On a course held in the romanian-ungarian imo preparation, mr. Pelikan (the leader of the ungarian team and also the chairman of the imo) told us that most problems in functions are based on 1 idea.  He mentioned one problem with functions given in the imo in 1989 or something, which is extremelly beautiful and needed 5 ideas.
Well I think this one is a 3-ideas problem.

hello (salut) dospi (gabi), I only realized now who you really are by looking in  valentin's pdf version of romanian selection tests. As about me...if you didn't figured out by yourself....hint: izarah is harazi.
So who is that friend? Are we talking about the same contest? On my paper says Radu Remus proposed it, but I don't know who is he.
Now I see that you and dzeta also had a problem there. Maybe you also know if anyone solved the functional equation during the contest.
Ok. Now I also took a look on imc2000. I found case n=2 there. My proof for the injectivity used the second condition. But we can get rid of it.  I realise that your friend (Radu Remus) had other solution then mine for n=3 since he didn't gave the problem for an arbitrary n.

Hi, pachi! I hope I didnt make any mistake, but from your post regarding the very nice problem of Murray Klamkin, I deduced that you must be jkap. Radu Remus is one of my colleagues from University. He proposed that problem and of course, nobody solved it. Knowing that this will happen, me and Adrian Iuga proposed an easy problem so that the maximum score shouldnt be 14p. Unfortunately, there werent solution with more than 4 points for that problem either. About this problem, I consider it very hard, especially if you dont know that problem from IMC. The answer is not that natural and the solution either. Radu Remus gave a quite complicated and long solution, but I must say that I couldnt find an easy solution either. So, could you please post your solution? Regarding the IMO, dont be that sad, you wont die if you didnt participated at IMO. I know this from my experience. But something bothers me: from where do you know about izarah?

[quote="harazi"]That's nice, because my friend gave that problem in the contest as a generalization of a problem from IMC. Cute problem![/quote]which problem from IMC was he generalizing?

That cool problem with $ f(x)f(xf(y))=f(x+y) $.

we had some form of this problem in our imo-selection tests:
"find all functions $f: (0, \infty)->R$ such that
a) $f(2)=0$
b) $f(x)f(xf(y))=f(x+y)$
c) $f(x)\neq 0 \forall 0<x<2$

having solved that problem i know that there are much more than just the natural solutions.

Peter

That problem seems to have a bit different approach.
As for the solution, Radu Remus must see the lines below. So, for n we:
-[b]first idea[/b]: we will show that it exists $g:(0,\infty)\rightarrow (0,\infty)$ such that $g(x+y)=f(x)f(yf(x))$. To do this fix $x_1+x_2$ and $x_3,x_4...x_n$ in our equation. We get that $f(x_1)f(x_2f(x_1))$ is fixed for $x_1+x_2$ constant. And this implies that there exists $g$.
-[b]second idea[/b] Multiply the original functional equation with $f(x_1+x_2),f(x_1+x_2+x_3)....f(x_1+x_2+....x_{n-1})$. After proper grouping of the terms, we have $\ds g(x_1+x_2)g(x_1+x_2+x_3)....g(x_1+x_2+...x_n)=f(x_1+x_2)f(x_1+x_2+x_3)...f(x_1+x_2+...x_n)$. Now define $h(x)=f(x)/g(x)\Rightarrow$ for $y_1,y_2...y_n$ all different $h(y_1)...h(y_n)=1$. Now fix the first $n-1$ terms. We get that for any other values of $y_n$,  $h(y_n)$ is the same. Now fix other $n-1$values than those and we get that $h$ is a constant and more than that $h(x)=1$ for any x, so $g=f\Rightarrow f(x+y)=f(x)f(yf(x))$.
-[b]third idea [/b] look on the imc site for the third idea:). In fact this third idea also contains more subideas.
The first 2 ideas make the word "beauty", and the third one makes the word black, allthough it has some contribution to "beauty", too.
As about harazi-izarah. It was just a hint man, for you  to discover the correct meaning of $ikap$.  But it seems I really hit you nice:). As a matter of fact one of my friends had the yahoo id armanddnamra and this is where I met the idea. What about the solution??? Sorry to have remembered you about that. [/b]

Dear paki-ipak, it seems that you have a real pleasure in hiting me nice. I dont understand the reason, but anyway, I asked that question with izarah since for me it is not just the reversed of harazi. Anyway, I realized the meaning of the hint one our later. It seems that my brain doesnt work quickly. Well, thats me:old, tired, boring. Why are you sorry to have remembered me that? Was the problem proposed for me to solve or what? Sincerely, I didnt think about this general case since I knew how to reduce the problem proposed by Radu Remus to that from the IMC and even if you dont believe me, its just like yours. And maybe Radu Remus is not obliged to see your solution, is he? Finally, congratulations for your solution to that nice inequality of Klamkin. I saw that you have already noticed very well (two times) what I have said about the one who finds a nice geometric solution. Youre damn good. But you know, in inequalities Im not satisfied only with that. Amaze me with some other problems posted and unsolved for months. Sorry to have remembered you about that.

Dear Gabi, it seems you misunderstood my tone. It's really hard to be precise in these mesagges. Anyway. I said that i'm sorry for remembering you about your miss to the imo.
And I think, about that reducing to the imc problem, that is nice, and probably you were expecting me to solve precisely the imc problem in a nicer way .
And well, i was happy for finding this and the tone that I used when saying "Radu Remus must see this" was like "Radu Remus would want to see this thing, cause he gave a long solution to the problem", and I thought you were refering to the n=3 case when you said that he gave a long solution, and I also thought that Radu Remus is a mathlinker and it would only be a mouse click for him to get here.
I didn't wanted to make you angry, and I didn't insinuated nothing about you.
It is the second time when someone misunderstood me here and i only am here since 4 august. Hopefully last time the misunderstood was a positive one (one of my sentences was taken as a compliment). I should probably change my signature in "Don't get me wrong".
Ohh. And the word "hit". I used it as "nimerit" in romanian, I couldn't find a good one in english. And I don't want to get into izarah's intimity.
Ohh.. and ok I will take your challenge and try to get reabilitated.

Now i see that i didn't press enter before my last sentence in the last post, and I must recognize that I used the word "nice" in the most unproper context.

[quote]
(one of my sentences was taken as a compliment)
[/quote]

So, it wasn't and you are laughing at me!!! :(  :(  :(   ;)

Pierre.[/quote]

Oh  God.
I already said there that my sentence was only a confirmation, and only a confirmation. No other hidden meanings for the sentence.
I think I'll take a break with this site. Maybe I should take an english course and then start posting 100-lines messages in which i should explain exactly what i want to say.

:D :D
Yet, I added a last emoticon....

Pierre.

Ok, Pachi, I apologise for not understanding what you really meant. You dont need to be reabilitated, I was just a little bit angry. But it is clear that I didnt understand the right meaning of your words. Peace, ok? No, you dont need an english course.

I can see now why you guys call this thread [b]black[/b] beauty :D:D:D

[quote="Peter VDD"]He stated it right in the first post ;)
[quote="ikap"]It is black beacause you can't see it's beauty from the beginning...[/quote][/quote]I was pointing out the little flame there :P

I wouldn't call that a flame. It was just a misunderstanding.

There's nothing burning here neither. Only some smoke but it will be gone in 60 seconds.

Hmm.. no that's pierre smoking in the jacks. :)


1)Given $ m;n$ be positive integers. Solve the system of equation in $ \mathbb Z^{ \plus{} }$:
\[ x \plus{} y \plus{} z \equal{} m
\]

\[ x^2 \plus{} y^2 \plus{} z^2 \equal{} n
\]
2) what will it  happen when I assume we solve the equation in $ \mathbb Z$

[quote="314 Einstien"]There are just two equations and three variables thus there are infinitely many solution. Consider one variable to be free variable .
Let us say we assume z =0.
then we have x+y =m , x^2 + y^2 = n.
I hope this is trivial to solve :)[/quote]

Not so simple, Einstein. You should better read the statement : we are looking for variables in $ \mathbb Z^\plus{}$, not in $ \mathbb R$

Sorry i just thought it was trivial .
Of course even now we can consider z as the free variable .
The necessary condition fo the solution to exist is 2n - (m-z)^2 - 2 z^2 should be a perfect number .
So given m and n one always choose z in such a way that this is made possible .
However i am not very sure of the cases when no solutions exist . In most cases solution do exist .
We choose z based on the above condition if the value of this z< m then of course solution lies in Z+.
Or else no solution in Z+.

[quote="314 Einstien"] So given m and n one always choose z in such a way that this is made possible .[/quote]
This is the required problem :)

[quote]This is the required problem [/quote].
So it is also basically a perfect square problem right.
U wanna know when this is perfect square. As of now i don't have a proof i will let you know once i am done with this.

No one helps me. It is a problem proposed by professor Khoai Ha Huy. But I can't understand his proof (he still did not show any triples $ (x,y,z)$).
I am waiting for your nice solutions. ZetaX, Rust or pco,.... Can you?.

I would be surprised if a simple solution exists for this problem :

1) not all $ n$ can be written as sum of three squares (all $ n\equal{}4^p(8q\plus{}7)$ cant)

2) when a number $ n$ can be written as such a sum, the number of representations is limited (see for example http://mathworld.wolfram.com/SumofSquaresFunction.html).
So there is a limited number of values of $ m$.

So the set of $ (m,n)$ for which a solution exists is rather "low density" in $ \mathbb N^2$


geometry
3D geometry
number theory unsolved
number theory

Suppose $a,b,c\in Q$ and $a+b+c=a^{2}+b^{2}+c^{2}$ are integers. Prove that $abc$ can be written as a quotient of a perfect cube and a perfect square which are relatively prime.

It is Bulgarian National olympiad 2003 Problem5.

Hint: If $ t = a+b+c=a^{2}+b^{2}+c^{2}$, then $ t\leq 3$ and from here there are 2 cases to consider(because t = 3 is trivial).


Euler
geometry
circumcircle
geometric transformation
homothety
ratio

Prove that  a triangle has an angle with the measure 90 degrees if his circumcircle is tangent to the Euler's circle of that triangle

[quote="not_trig"]Is the "Euler's Circle" the nine-point circle?[/quote] Yep...the 9 point circle

Let $ ABC$ be a right triangle with circumcenter $ O$ on $ BC$ (that is $ \angle A\equal{}90$). Let $ AO$ intersect the circumcircle again at $ A'$. Consider a homothety center $ A$ with ratio $ \frac{1}{2}$. The image of the circumcircle of $ \triangle ABC$ is also a circle. It must pass through the images of $ B$, $ C$, and $ A'$, which are the midpoints of the sides of $ ABC$. So this homothety takes the circumcircle of $ \triangle ABC$ to the nine-point circle of $ ABC$. However, a homothety with a center on a circle maps the circle to another circle to a circle internally tangent there (since the centers and the center of homothety are collinear), so we are done.


i got selected in RMO (additional) can anyone please tell me what`s the difference between additional and actual one

what do you mean by additional, i have never heard of it before


A sequence of positive real numbers is defined as:

Y(0) = 1
Y(n+2) = 2*Y(n) - Y(n+1), for n = 0, 1, 2, ...

What is the value of Y(2007) ?

[quote="Rust"]$Y_{n}=\frac{1-2Y_{1}}{3}(-2)^{n}+\frac{Y_{1}+2}{3}.$[/quote]

[hide="Explanation"]
The linear recurrence you give us has a closed form of $Y_{n}= A \cdot \alpha^{n}+B \cdot \beta^{n}$, where $A,B, \alpha, \beta \in \mathbb{R}$. We can find these via what are called [u]characteristic equations[/u]. For your recurrence, we have:
\[Y_{n+2}+Y_{n+1}-2 Y_{n}=0 \Longrightarrow x^{2}+x-2=0 \]
where the roots of the polynomial will be our $\alpha$ and $\beta$. Solving, we get that
\[\alpha=1, \ \ \beta=-2 \]
so our recurrence will be of the form $Y_{n}= A+B \cdot (-2)^{n}$.

If we had a second term, $Y_{1}$, we could set up a system of linear equations in $A$ and $B$ to solve for these constants; but you haven't given us one, and [b]Rust[/b] has solved the system anyway.
[/hide]


geometry
circumcircle
geometric transformation
geometry proposed

Let $ (O) $ be the circumcircle of the triangle $ ABC $ . A circle $ (O_1) $ internally tangents $ (O) $ at $ D $ and tangents $ AB ,AC $ .  A circle $ (O_2) $ internally tangents $ (O) $ at $ E $ and tangents $ BA ,BC $ .  A circle $ (O_3) $ internally tangents $ (O) $ at $ F $ and tangents $ CA ,BC $ . Prove that : $ AD, BE , CF $ are concurrent

Of course, romano. The concurrent point is the external center of similitude of the circumcircle and incircle.

Can you show that : the concurrent point lies on OI ?

 PS : What is the external (internal) center of similitude of the circumcircle and incircle?

Dear romano:
$A$ is the exsimilicenter of circle $(O1)$ and the incircle $(I)$ further $D$ is the exsimilicenter of Circumcircle $(O)$ and circle $(O1)$. Now the line $AD$ passes trough the exsimilicenter of circles $(O)$ and $(I)$ by the Monge's theorem.Similiar reasoning for the lines $BE$ and $CF$, hence $AD,BE,CF$ pass trough the exsimilicenter of $(O)$ and $(I)$. Also we can to find only the concurrency from the theorems of Casey and Ceva due to:$AE.BF.CD=AF.BD.CE$

Here is some property I note here
     1) The polar of $A$ with respect to $(O_1)$ is the line through $I$ and perpendicular to the A_ bisector.
     2) Let $C', B'$ be the intersection of that polar with $AB, AC$. Then $C', B'$ are the tangency points of the circle $O_1$ with $AB, AC$.
     3) $DBMI$ and $DCNI$ are cyclic.
     4) $DI$ is the bisector of $\measuredangle{BDC}$
 Property 4 is a special case of a very nice problem from Iran Olympiad  (I think 1996).


group theory
abstract algebra
superior algebra
superior algebra unsolved

$n \geq 5$
Suppose that $G$ is a subgroup of $S_n$ of index $n$.
Prove that $G$ is isomorphic to $S_{n-1}$

if $n<5$ a simple verification works so assume $n\ge5$
the action on the cosets gives a gp homomorphism from $S_n$ to $S_n$. This will be injective because the kernel is surely not included in $A_n$. therfore it is an isomorphism and the subgroup is mapped into the stabilizer of a point in $S_n$ solving the problem. Note that in the case $n=6$ the "strange" automorphism of $S_6$ gives an "strange" $S_5$ inside $S_6$.


function
logarithms
inequalities
inequalities proposed

I don't know if anyone post this before, it shouldn't be new but interesting:
Prove that $x^yy^x \le \left(\frac{x+y}{2}\right)^{x+y}\/$ for $x,y>0\/$.

[quote="pipi"]I don't know if anyone post this before, it shouldn't be new but interesting:
Prove that $x^yy^x \le \left(\frac{x+y}{2}\right)^{x+y}\/$ for any real $x,y\/$.[/quote]
It's nice ! However my opinion is that you must  suppose $ x,y\in D:=(0,\infty) . $ Use the fact that the harmonic mean is less than arithmetic mean , as well as following properties of the function  $ \ln:D\to {\mathbb R} $:
i) $ \ln $ is increasing on $  D $ , for instance   $\; (*)\; \; \; \ln\left(\frac{2xy}{x+y}\right) < \ln\left(\frac{x+y}{2}\right) $, for $ x,y \in D ,$
 ii) $ -\; \ln $ is convex on $ D $, e.g. according to Jensen inequality
\[ \ln\left(\varepsilon x +(1-\varepsilon)y\right)\ge \varepsilon \ln{x} +(1-\varepsilon ) \ln {y} \; \; \; ,\; \; \forall
x,y \in D\; \; \mbox{and}\; \; \varepsilon \in [0,1] .\]
If you select $  \; \varepsilon =\frac{y}{x+y} $ and use  (*) you find   $ \frac{y}{x+y}\ln{x} +\frac{x}{x+y}\ln{y} \le  \ln\left(\frac{2xy}{x+y}\right)\le  \ln\left(\frac{x+y}{2}\right) $ which  proves your  inequality.

[quote="flip2004"]It's nice ! However my opinion is that you must  suppose $ x,y\in D:=(0,\infty) . $ [/quote]
Thanks, flip2004, I edit it already.
And your answer is correct.
I use another method, but actually similar.

[quote="vedran6"]If we write this as
              (2x/(x+y))^y<=((x+y)/2y)^x
         and supose that y<=x we get AG inequality (x+y)^2>=4xy
          in other case where is x<=y we write ineq. as
          (2y/(x+y))^x<=((x+y)/2x)^y and again this is AG (x+y)^2>=4xy
            the profe is done.Equaility holds for x=y[/quote]
I like this proof, easy to follow and less calculation :D  :D  :D
Thanks a lot...


In this multiplication problem, replace each letter  with a digit chosen from 0 through 9, inclusive.  M not equals to 0     Replace identical letters with the same number  Different letters represent different numbers

      MEGSL
           x 4
   _________
      LSGEM

The sum    M+E+G+S+L    equals

A.20   B.	22   C.23     D.26     E.27

[hide="Solution"]Let $ n$ be the number, and $ s$ be the sum of the digits of the number, and let $ 4n$ have the sum of digits be $ t$. We note that $ 4s$ modulo $ 9$ is equivalent $ t$ because the digits are the same, just in a different permutation, so among the multiple choice answers, we seek potential answers such that multiplying the answer by $ 4$ leaves the same remainder modulo $ 9$. The only valid answer is $ 27$, so our answer is $ \boxed{E}$.

[/hide]


geometry trigonometry calculus MATHCOUNTS \/closed

Well, you can take geometry, interm. alg, but there's no pre-calc class.

Wait, isn't precalc just more algebra and trig?

Yeah, it's supposed to be an epic review, or something.

Did you ask your counselors at school if they would allow that?

I haven't asked, but why not? If I can skip geometry, why not skip them all since learning from AoPS is better? I stil get math credit by taking a final test...

[quote="modx07"]yes but you might not get Honors or Pre-AP credit (if your school offers that), which can effect your weighted GPA, and in turn, your class rank...which is not a good thing....colleges want you to be ranked very high in your classes[/quote]
Most colleges (even the not-so-good ones) unweight grades before looking at them.  So the weighted GPAs are meaningless.  I know this fact personally.

By the end of this year, we expect to have the full range from beginning algebra through calculus available, including a thorough exploration of discrete math.  So, yes, you can get all your core math education for middle and high school through us.  I, of course, think that's a fine idea, provided you can work well independently and put in the time doing the readings and the homework.

[quote="mathswimmer"]What are the benefits (if any) of taking math classes in high school as a supplement for AoPS learning?[/quote]

Not much, but I do find that math class is a good place to practice mental math and do homework--this probably won't help much when you're out of MATHCOUNTS, though. By the way, good job at chapter! You should have tied for 1st with us three, but I guess not? Or they could've given you the 2nd, 3rd, and 4th place trophies. :)

[quote="mathswimmer"]Hey, I'm thinking about totally skipping math in high school and just learning from AoPS and taking a final test whenever I'm ready so I can take more classes at school that I can't learn myself. Does this seem like a good or bad idea? What are the benefits (if any) of taking math classes in high school as a supplement for AoPS learning?[/quote]

If your high school gives you credit for this, that's fine. Otherwise, you may have a problem with getting enough credits for graduation (I believe most schools require math credits?).

Sorry to revive, but I have another question.

[quote="rrusczyk"]By the end of this year, we expect to have the full range from beginning algebra through calculus available[/quote]

Would statistics be included in that range?

No -- that's on a longer-term todo list.


Find all $ n\in N$ such that you can divide the group of divisors of $ 30^n$ in groups of $ 3$ elements, and the product of the $ 3$ numbers of each group is equal

we can write $ 30^n$ as $ (2^n)*(5^n)*(3^n)$
so the number of factors of this number is $ (n \plus{} 1)^3$
Since u want to divide the factors into groups of 3, 3 divides  $ (n \plus{} 1)^3$
Case 1:
If $ 30^n$ is not a perfect square
u know that the product of divisors of $ 30^n$ is$ 30^{((n \plus{} 1)^3)/2}$
so if the product of each group is x u get $ x^{((n \plus{} 1)^3)/3}$=$ 30^{((n \plus{} 1)^3)/2}$
So we get $ x^2$=$ 30^3$
then x is not a natural number so in this case no solution

Case 2:
If $ 30^n$ is  a perfect square
we get that  $ x^{((n \plus{} 1)^3)/3}$ = $ 30^({(n \plus{} 1)^3} \minus{} 1)/2 * 30^n/2$
Since $ {((n \plus{} 1)^3)/3}$ is always lesser than $ ({(n \plus{} 1)^3} \minus{} 1)$/$ 2$ +$ n/2$
we get x is a power of 30.
So let $ x\equal{}30^l$
But x=$ 30$ is not a solution.
For $ l>1$
we get $ l{((n \plus{} 1)^3)/3}$=$ ({(n \plus{} 1)^3} \minus{} 1)$/$ 2$ +$ n/2$
which simplifies to $ 2l\minus{}1{((n \plus{} 1)^3)}$=$ 3n\minus{}3$
which has no solutions in natural numbers

P.SI changed my wording of the second case a bit , cud anyone check my solution.

I think your solution is wrong,please check it
I know the answer because I have read it and your solution doesn´t make any sense
Daniel

could anyone :blush: point out the errors in my solution.

P.S. i checked and changed the wording of my solution a bit, and i could not find any mistake :blush:

[quote="befuddlers"]u know that the product of divisors of $ 30^n$ is $ 30^{((n \plus{} 1)^3)/2}$[/quote]
I think this is not true and should be $ 30^{\frac {n(n \plus{} 1)^3}{2}}$

Hi akashnil,
we observe that if $ 30^n$ is not a perfect square then for each factor a there is a factor b such that $ ab$ is $ 30^n$. So the number of pairs of $ 30^n$ is (number of factors)/2
if $ 30^n$ is a perfect square all factors have their complement such that their product is $ 30^n$ except $ 30^{n/2}$.
OMG yes i realized my mistake
the divisors can be grouped into pairs of $ 30^n$, in my hurry i took the factors group as $ 30$ :blush:

here is my revised solution
as akashnil pointed out and as i proved above
take x as product of each group, i proved in my earlier post that x is a power of 30
So let x=$ 30^l$
we get $ 30^{(l(n\plus{}1)^3)/3}$=$ 30^{n((n\plus{}1)^3)/2}$
so we get 2l=3n
so there exist solutions if n is even,
but since we assumed $ 30^n$ is not a perfect square there are no solutions in this case.

Case 2:
If $ 30^n$ is a perfect square then
$ 30^{l(n\plus{}1)^3/3}$= $ 30^{n(n\plus{}1)^3\minus{}1/2}$*$ 30^{n/2}$
so simplifying we get $ l\equal{}3n/2$
but since n is even if $ 30^n$ is a perfect square so there exists a solution for all even n.
Lambrusckoid is my solution right now

Hi Krishna,
I'm sorry but to be frank I find your solution very difficult to understand. Although I think the approach is correct, but to me the latexing, explaining and detailing is not understable easily :( .

I find some of the things a bit difficult to understand:

[quote="befuddlers"] ... no solutions in this case.[/quote]
Which case is this? You didn't write the case no. and the assumption-statement of the case above the paragraph.

[quote="befuddlers"]i proved in my earlier post that x is a power of 30[/quote]
I don't find it. And there is something in case 2 in the previous post but the latexing and explaining is messed up in the beginning of that case. So I can't follow upto that.

[quote="befuddlers"]If $ 30^n$ is a perfect square then
$ 30^{l(n \plus{} 1)^3/3}$= $ 30^{n(n \plus{} 1)^3 \minus{} 1/2}$*$ 30^{n/2}$[/quote]
I don't understand this part at all where do you get the RHS?

[quote="befuddlers"]so simplifying we get $ l \equal{} 3n/2$
but since n is even if $ 30^n$ is a perfect square so there exists a solution for all even n.[/quote]
You just failed to draw a contradiction on the existence of $ l$. But that doesn't imply that such grouping is possible. You'll have to show a strategy of grouping possible or something.

Hm... I didn´t solve the problem, but I had [hide="an idea."] Suppose we can divide the divisors of $ p$ and $ q$, with $ gcd(p, q) \equal{} 1$, in groups of three elements with constant product, as the problem says. Then we have the partitions

$ D(p) \equal{} \{a_{11}, a_{12}, a_{13}\} \bigcup \{a_{21}, a_{22}, a_{23}\} \bigcup ... \{a_{m1}, a_{m2}, a_{m3}\}$

and

$ D(q) \equal{} \{b_{11}, b_{12}, b_{13}\} \bigcup \{b_{21}, b_{22}, b_{23}\} \bigcup ... \{b_{n1}, b_{n2}, b_{n3}\}$

where $ D(k)$ is the set of divisors of $ k$. Clearly $ D(pq)$ is partitioned like this:

$ D(pq) \equal{} \bigcup_{i \equal{} 1}^m \bigcup_{j \equal{} 1}^n \{a_{i1}b_{j1}, a_{i1}b_{j2}, a_{i1}b_{j3}, a_{i2}b_{j1}, a_{i2}b_{j2}, a_{i2}b_{j3}, a_{i3}b_{j1}, a_{i3}b_{j2}, a_{i3}b_{j3}\}$

Therefore we can divide the divisors of $ pq$ in groups like $ \{a_{i1}b_{j1}, a_{i2}b_{j2}, a_{i3}b_{j3}\}$, $ \{a_{i1}b_{j2}, a_{i2}b_{j3}, a_{i3}b_{j1}\}$ and $ \{a_{i1}b_{j3}, a_{i2}b_{j1}, a_{i3}b_{j2}\}$. We can observe that all this groups have the same product. So we can divide the divisors of $ pq$ in groups of three elements with constant product.

Now it suffices to prove that the divisors of $ 2^n$ can be divided in such groups for $ n \equal{} 6k \plus{} 2$ (since Befuddlers proved that $ n \plus{} 1 \equiv 0 mod 3$ and $ n \equiv 0 mod 2$). It appears to be true, because I have already proved this is true for $ 0 \le k \le 10$. I hope this is helpful.[/hide]

I don´t know if I can bring this topic to life, but I am very curious...

No solutions?


induction
abstract algebra
number theory proposed
number theory

Today, I've invented [during biology lesson  :P] this problem.

Take Mersenne's number $2^p-1$, where $p \in \mathbb{P}$ and $q \in \mathbb{P}$ such that $q|2^p-1$. Prove that for every non-negative integer $n$ it is true that $p^{n+1}|(q-2)^{p^n}+1$

[quote="Myth"]Avtually, you are WRONG!  :lol:

You need to specify that $p$ is odd. For odd prime number we are to use induction and the fact that $p\mid q-1$, since $q\mid 2^p-1$.[/quote]

But you are obviously right Myth  :D, indeed $p$ must be odd. Your way of thinking is the one I used, but how do you prove that $p|q-1$? Because to get that I used Lagrange's Theorem (that abstract algebra's one) and I wonder whether there is any simple argument?

Indeed, there is a very simple argument. We know that $q\mid 2^{q-1}-1$ and $q\mid 2^p-1$. It follows that $q\mid 2^{(q-1,p)}-1$, but then it is clear $(q-1,p)=p$.


algebra
polynomial
induction
number theory
algebra unsolved

Let be given a polynomial $ P(x)=x^3+153x^2-111x+38$
Prove that there are at least 9 natural numbers $ n<3^{2000}$ satisfying $ \frac{P(n)}{3^{2000}} \in N$

Well, when reading the statement I think this one is "nice and easy", but now I realize that it's "nice but not easy". Can anyone give me a hint?

Here's a hint: this is a number theory problem.

If $x\equiv 1 (\mod 3)$ and $3|k$, $9k|f(x+k)-f(x)$ (If there are zeros mod $3^{2000}$, they are equivalent to 1 mod 4). Therefore, if $x$ is a zero mod $3^{2000}$, $x+a\cdot 3^{1998}$ is also a zero mod $3^{2000}$, so if there is one zero, there are at least nine.

To finish the problem, we just need to prove that there is a zero.

Yes, thanks. I know the way to prove the existence of a zero of $ 3^{2000}$.

This problem is a consequence of  theorem:If$P(b)\equiv 0(mod 3^{s+2});P(d)\equiv 0(mod 3^{s+2})$;then $b\equiv d(mod3^s),s\geq3$

pigfly , can you post the solution of the theorem you mentioned ?

Proof by induction on $s\geq 3$
For s=3 ,it's easy to verify that: $P(n)\equiv 0(mod243)$ if and only if $n\equiv 19(mod27)$.We are done in this case.
 Assume that it holds for some $s\geq 3$.Suppose $P(d)\equiv P(b)(mod3^{s+3})$.
We have:
$d\equiv b(mod 3^s)\rightarrow b=d+h3^s\rightarrow 0\equiv P(d+h3^s)\equiv P(d)+h3^sP'(d)(mod3^{s+3})$(1)
Because:
 $d\equiv 1(mod9)\rightarrow P'(d)$ is divisible by 9 but is not divisible 27 $\rightarrow 3^sP'(d)$ is divisible by $3^{s+2}$ but is not divisible $3^{s+3}$(2)
Using (1) and (2) we dedu ce that : $h=3t\rightarrow b=d+3t3^s=d+3^{s+1}t\rightarrow b\equiv d(mod3^{s+1})$
Thus, we are done $\forall s\geq 3$.Or at least I hope so :P


Three schools have a chess tournament. Four players come from each school. Each player plays three games against each player from the other schools, and plays one game against the each other player from his or her own school. How many games of chess are played?

If izzy was in a school, then she would have to play each of her 3 friends once and each of the 8 other people 3 times. So each person plays 3(1)+8(3)=27 different people. There are 12 people, but there is an overcount, so there are $ \frac{27\times 12}{2}\equal{}162$ games in total.


[b]1.[/b] $ \lim_{x\to 1}\frac{nx^{n\plus{}1}\minus{}(n\plus{}1)x^n\plus{}1}{(x\minus{}1)^2}$

[b]2.[/b] $ \lim_{x\to 0}\frac{(1\plus{}mx)^n\minus{}(1\plus{}nx)^m}{x^2}$ where $ m,n$ are positive integers

By applying the L' Hospital's rule we have:

1)$ {n(n\plus{}1)}/2$
2)$ {nm(n\minus{}m)}/2$  :)

[quote="vinskman"][quote="hamidreza"]By applying the L' Hospital's rule we have:

1)$ {n(n \plus{} 1)}/2$
2)$ {nm(n \minus{} m)}/2$  :)[/quote]

Yes,these two may applied in the L' Hospital's rule .

But, we might remain some space of thought for the others to solve without using calculus.Agree?
Thanks.
 :lol:[/quote]
Ok; good idea.  :)


Sa se determine f,g,h: $\mathbb{R}$ -> $\mathbb{R}$, stiind ca au primitivele F,G,H si f=G+H, g=F+H, h=F+G.

Banuiesc ca domnul Enescu iti va raspunde imediat ce vede problema asta.  ;)

Nu chiar imediat  ;) . Acum plec la aeroport. Daca am conexiune internet, peste o zi doua postez solutia. Cred, totusi, ca va  gasi cineva timp sa o rezolve pana atunci.

Ok, atunci sa dam cateva indicii. Adunam ecuatiile si obtinem ca $ F+G+H$ verifica o ecuatie de forma $ y'=2y$ de unde $ F+G+H= c e^{2x}$ cu c constanta. Atunci $ F'+F=c e^{2x}$ ecuatie diferentiala ce se rezolva fara nici o dificultate. Aflam astfel F si la fel G, H, deci avem f,g,h.

$f,g,h$ is derivable

$f'=g+h$
$g'=f+h$
$h'=f+g$

again $f,g,h$ are derivable twice

$f\"=g'+h'=2f+f'$

By symetry $g,h$ verify the same differential equation $y\"-y'-2y=0$

Moubi, can you solve the general case of a system of n functions with this method?  :)

Easy

Let $g=f_1+...+f_n$

$f'_1=g-f_1$
$f'_2=g-f_2$
.................
$f'_n=g-f_n$

Sum all these equations

$g'=(n-1)g$ solutions are  $g(x)=ce^{(n-1)x}$

And to finish solve $f'_k+f_k = ce^{(n-1)x}$ with variations of constants

I spoke about your method for three functions. Itt was clear that with this method it is obvious.  ;)

You can aslo use matrix and differential equation

Y the column

Y=
(f_1)
(f_2)
......
(f_n)

$A$ is a matrix with $0$ on the diagonal and $1$ every where

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

And solve $Y'=AY$

Probably, our teacher is ill now and we are in holiday. Probably after.  ;)  :D


[quote="8parks11"]Jan can mow her lawn in 60 minutes by herself. If she hires Wally to do it, he takes 30 minutes, while Peter can do it in 40 minutes. If Jan starts mowing the lawn for 15 minutes, then decides to hire Wally and Peter to help her finish, how long will it take (in hours), from when Jan started to when the three finsihed together?[/quote]

It'll take them x minutes working together:

$\frac{1}{x}= \frac{1}{60}+\frac{1}{30}+\frac{1}{40}$
$\frac{1}{x}= \frac{9}{120}$
$x = \frac{40}{3}$

However, $\frac14$ of the job has already been completed by Jan, meaning $\frac34$ of the job is left.

$(\frac34)(\frac{40}{3}) = 10$

10 minutes is $\boxed{\frac16 \text{ hours.}}$

It says "from when Jan started", hence the correct answer is $25\ \mathrm{min}={5\over 12}\ \mathrm{hrs}$


induction
floor function
superior algebra
superior algebra unsolved

Show that $ ord_{p}n ! \equal{} \sum_{i\equal{}1}^{\infty}[\frac{n}{p^{i}}]$, where [] represents the greatest integer function.

Apply induction on [b]n[/b]. The formula is true for [b]n=1[/b]. Assume it is also true for [b]n[/b] and write $ n \plus{} 1 \equal{} p^{M} \cdot m$ where $ p^{M}$ is the greatest power of $ p$ that divides $ n \plus{} 1$. Then, by the induction hypothesis we have that

$ \displaystyle \mathrm{ord}_{p}(n \plus{} 1)! \equal{} \mathrm{ord}_{p}n! \plus{} M \equal{} \sum_{k \equal{} 1}^{M} \left( \left \lfloor \frac {n}{p^{k}} \right \rfloor \plus{} 1 \right) \plus{} \sum_{k > M} \left \lfloor \frac {n}{p^{k}} \right \rfloor.$

The result follows now after noticing that

$ \left \lfloor \frac {n \plus{} 1}{p^{k}} \right \rfloor \equal{} \left \lfloor \frac {n}{p^{k}} \right \rfloor \plus{} 1$

if $ 1 \leq k \leq M$ and

$ \left \lfloor \frac {n \plus{} 1}{p^{k}} \right \rfloor \equal{} \left \lfloor \frac {n}{p^{k}} \right \rfloor$

if $ k > M.$

Or set $ n' \equal{} \left \lfloor \frac {n}{p} \right \rfloor$.
Since we are looking for the highest power of p, we only consider factors of $ n!$ that are divisible by $ p$.
Such divisors are of the form: $ \{pt: 1\leq t \leq n' \}$.

We are after the highest power of p that divides the product: $ 1\cdot 2\cdot ... \cdot n' \cdot p^{n'}$.

Consequently, such a power equals the sum of the $ n'$ and the exponent of the highest power of $ p$ that divides $ n'!$.

Hence, the desired power equals $ n' \plus{} n'' \plus{} n'' \plus{} ...$, where $ n'' \equal{} \left \lfloor \frac {n'}{p} \right \rfloor$, etc.

Note that the set $ \{n', n'', n''', ... \}$ is a decreasing sequence and so it will terminate.

Also, $ n' \equal{} \left \lfloor \frac {n}{p} \right \rfloor$, $ n'' \equal{} \left \lfloor \frac {n}{p^2} \right \rfloor$, $ n''' \equal{} \left \lfloor \frac {n}{p^3} \right \rfloor$, etc because $ k \equal{} \left \lfloor \frac {n}{p} \right \rfloor$ and $ j \equal{} \left \lfloor \frac {k}{q} \right \rfloor$ imply that $ j \equal{} \left \lfloor \frac {n}{pq} \right \rfloor$

[quote="akech"]Hey Coquitao,
Can you prove those last two equalities in your proof please?[/quote]

[b]1[/b]) On the one hand, we have that $ \left \lfloor \frac {n \plus{} 1}{p^{k}} \right \rfloor \equal{} p^{M \minus{} k} \cdot m$. On the other, $ p^{M \minus{} k}\cdot m \minus{} 1 \leq p^{M \minus{} k} \cdot m \minus{} \frac {1}{p^{k}} < p^{M \minus{} k}$, and we're done.

[b]2[/b]) Repeat the reasoning in [b]1[/b], [i]mutatis mutandis[/i].

Quite an evolved discussion on this folklore Legendre formula  :)


what do you think? post your own creations!

oh, nice picture, the effect of the water is very realistic. ^^

well, this is my best coloured picture  :P it's from a manga book called naruto

[url=http://img474.imageshack.us/my.php?image=misasuke31zf.jpg][img]http://img474.imageshack.us/img474/2189/misasuke31zf.th.jpg[/img][/url]

[quote="thatin"]Thanks! You colored it or you designed and colored it? Anyway, the snake is really frightening![/quote]

Snake?

*Look around*

What snake???

I just coloured it, and yes, it is a snake, not a dragon  :lol:

ID card that I made for fun on another website. :D

[img]http://i256.photobucket.com/albums/hh175/jeffrey529/7.png[/img]

this was a concept of windows 7 i made... before the new screenies came out...

lol its too big but i don't know how to resize...

http://en.wikipedia.org/wiki/Windows_7

next version of windows


Let $\{n_{k}\}_{k \ge 1}$ be a sequence of natural numbers such that for $i<j$, the decimal representation of $n_{i}$ does not occur as the leftmost digits of the decimal representation of $n_{j}$. Prove that \[\sum^{\infty}_{k=1}\frac{1}{n_{k}}\le \frac{1}{1}+\frac{1}{2}+\cdots+\frac{1}{9}.\]

Here is solution :
Enough to solve with finite sequence .
Now call the largest number is $ 10N + d$ where $ d\in\{1,..,9\}$
From the condition we must have $ N$ not occur in this sequence .
Now if the Number $ 10N + i,\forall i = 0,..,9$ occur then remove it by N.
We has :
$ \sum_{i = 1}\frac {1}{n_1} + \frac {1}{N} - \sum_{d = 0}^9\frac {1}{10N + d}\geq\sum_{i}\frac {1}{n_i}$
Then repeat this with the sequence ${ \{n_1,...,N}$ then we have result:
$ \sum_{i}\frac {1}{n_i}\leq \sum_{k = 1}^9\frac {1}{k}$


PSAT's are coming everyone!!! Is everyone ready?  How did you prepare?

I m as ready as I will be although I m not as ready as I wish I could be.  School and extracurriculars has taken up so much time.   I still haven't even registered for the PSAT... the person who takes the money is always not in his office!!!  O well, I WILL give him the money monday.

Good luck to every who is taking it.

I don't even know if I'm taking it or not.  I'm not really preparing that much since I'm only in 8th grade, so it shouldn't even be in the high-school transcript (I know the SATs aren't if you take them in 8th grade).

As far as I remember, no one actually sees your PSAT scores, unless you're applying to a summer program or something. The only thing that PSAT scores really matter for is national merit scholarships. The only thing that the PSATs might give you an indication of is if you need to start studying really hard for the SATs. But don't be alarmed if you screw up on the PSATs. Like I said it doesn't really affect anything, so there should be no pressure when taking them. I just viewed it as a test to let you know where you stand in terms of the SAT. In fact, I sort of did screw up on the PSATs a little, but I did not screw up on the SATs, so it worked out fine.

Just my two cents on the PSAT.

Yea... I'd suggest anyone under 11th grade to take it as many times as possible to they can track their progress.  When you get closer to 11th grade, you can see what you need to improve and you're not left with like 2 months to prepare for it... You've got years  :D

[quote="lunmu"]took em this morning..  :?

and got a math question wrong..

im such a freaking idiot.. -.-[/quote]
How do you know what you got if you yook the test this morning? I took the PSAT this morning, too, but I won't see my score until December. :?

p.s.(a.t.)  I prepared by taking the PSAT last year nad by doing 4 SATs in the last month.

A group of kids I know took it today as well. For the main part, we are taking it , of course, on Saturday. Did everyone take the test that comes with the reading material after registration? How did you do?

The PSAT is [i]supposed[/i] to give you an inclination on how much you need to study for the SAT, but don't a goodly number of people take the SAT before their junior year?

[quote="MagnusRex"]I MISSED A MATH QUESTION!!![/quote]

You could still get an 80 if you're lucky.  Well, it sometimes works that way on the SAT, so I'm just assuming.

[quote="MagnusRex"]DAMMIT!!! I MISSED A MATH QUESTION!!![/quote]

Yeah, so did I.  Forgot that the green marbles couldn't share a vertex with the red ones for the second to last grid-in.  Oh well, it's like it's the end of the world, I'm only in 8th grade.

[quote="Ragingg"]Which questions.... Btw, why do your schools take them today? Mine are on Sat![/quote]

Lol.. yea.. we're really going to tell you the questions when you're taking them on saturday..  :) nice try..  ;)

And to answer someone else's question earlier, I basically figured out right after the test that I had done something wrong.. -.- arghh..

hopefully, since it is the first year they've done the new version of the PSAT, they'll attach a nice curve to it, and maybe, just maybe, I'll still get a 80 on math.. *prays*  :?

[quote="lunmu"]Lol.. yea.. we're really going to tell you the questions when you're taking them on saturday..  :) nice try..  ;) [/quote]

I'm pretty sure that the people taking the test on Saturday will have another version.  The CollegeBoard may be stupid, but they aren't that stupid.

[quote="Simfish"]Did anyone of you remember the equilateral triangle problem? (two equilateral triangles). I couldn't solve that one; it was the only one I had trouble with. I ultimately just subtracted and made a guess: 9. Probably fails but anyways...
[/quote]

Are you referring to the one where the two eq. triangles overlap and they gave u the lengths of the 6 edges of the hexagon formed by the overlap?...I think it asked you to calculate the difference in  the perimeter of triangle ABC and DEF or something like that...

the answer to that was 9...

Yes, I also got nine.  The six little triangles are equilateral because of some parallel thingy.

wow frank, seeing you on two forums in one day is quite a coincidence.  thats crazy.

god, i was sick the day of the PSAT

predicted scores:
math: 80
critcial reading: 78-80
writing: 76-78

i got a 224 last year. too bad im in CA :( the cut off's like 215 here

Great! :) I have a chance of getting 80!

(Unless I get some very stupid questions wrong for some reason). I once took a Kaplan practice test and got the third math question wrong. LoL...

I think (or at least hope) that I'll be within a few points of 80 on each of the three sections.  I can't think of any math questions I had any trouble on, and I found the Criticial Reading and Writing sections to be relatively easy.

Here's hoping December comes fast.

But not TOO fast...

I'm not very confident in my scores.  I might be a commended student if i'm lucky.  Every single sat or psat i have ever taken, i got a score of a 54 on the verbal.  that's not good.  i'm pretty sure i got an 80 math, and between a 60 and 70 writing.  not good scores.  My scores havn't improved any since middle school.  I guess i should've prepared myself.

Does anyone know when we get our scores?  I forgot.

I had an interesting (aka extremely irritating) experience with the PSAT, particularly the math section. I did all the problems without too much difficulty, and then I checked my answers carefully before bubbling in my answers. The trouble is, I probably checked a little too carefully, since I wasn't able to finish transfering one set of my math answers on a section- I ended up with five answers that were circled in the test booklet and not marked on the answer sheet. Consequently, my math score is probably at best a 68 or a 69...

Note that this is the same mistake JBL once made on an AMC- do not bubble at the last second, even if you think you have enough time. My moderator was going by the clock and I was looking at my watch, so I think he actually chopped 30 seconds off of the time allotted. Don't take the risk of bubbling all at the last moment-something like this really can happen, even on the math section. Even if bubbling as you go along is slower, it carries far less risk.

The other two sections were alright though, and I thought writing in particular was rather easy. My problems with the math section make national merit somewhat problematic, however, and it'll be tough to make the cutoff with the blunder(213 in my state).

I think we get our scores sometime in December.

I've always bubbled in scores as I went along, and I've always gotten high scores on multiple-guess tests published by ETS. I don't see the point in waiting until the last minute to mark answers on the one-and-only official answer sheet.

never, ever, EVER bubble in at the end. this can cost you SO much time and SO many points, even though you know the answers. We took all our bio tests on scantron sheets and it has really improved my bubbling speed. If you get really good, you can bubble while reading the next question. Also, if you are desperately trying to bubble at the last second, there is one thing you can try- just make a mark through the answer bubble. Our school scantron picks these up- I'm not sure about the big guys, but if you are running outo of time fast, its definitely worth a shot to just mark through the answers really quickly (and accurately, one hopes).

[quote="bleumoose"]If you get really good, you can bubble while reading the next question.[/quote]

I thought everyone can do that. If you CAN'T read the next question while bubbling the answer to the question you've just answered, you definitely need to practice taking actual tests under actual test conditions more.

[quote="tokenadult"]I thought everyone can do that. If you CAN'T read the next question while bubbling the answer to the question you've just answered, you definitely need to practice taking actual tests under actual test conditions more.[/quote]

Yea, no need to concentrate on bubbling...especially if you think there's even a slight possibility of not having enough time...

Although I really shouldn't be talking :-D I spend like 10 minutes bubbling my math sections :-)

haha...random question, but anyone know the best way to bubble?

[quote="MithsApprentice"]haha...random question, but anyone know the best way to bubble?[/quote]

I've never thought it made much difference HOW I bubbled, and I just made sure to get a good black coverage of the entire bubble space. I've read, in a book, that one can just make a quick circle of black pencil in the space, saving a fraction of a second per space, and still have all the answers get scored correctly.


3 points are given on a circumference. You can move one of the 3 points on the midpoint of one of two arches delimited by the other 2 points. Which inizial states let you have:

a) a isosceles triangle with angles <= 1/2004 degrees;

b) an equilater triangle

i think the question a) is quite simple...but i can't do the second part.. it's to simple to be correct

enjoy!

$ b)$ The only initial state is when $ 2$ of the $ 3$ initial points form a central angle of $ 120$ degrees

Let us work backwards from the final state of the equilateral triangle,
the above solution is when there only $ 1$ move.
if there is at least an intermediate state then the previous one is with two $ 60$ degrees consecutive angles
with previous state the equilateral triangle or the above solution.

are you sure of that? it sounds too simple...i'll check it better..see u


prove that  the equation :
$ x^2\minus{}y^3\equal{}7$  doesn't have  any solutions  $ (x,y) \in  N^2$


function
logarithms
algebra
polynomial
calculus
calculus computations

I have a situation where I need to approximate a data set with a specific function. I know that the function is of the form

$ \Psi(x) \equal{} \alpha \minus{} \beta \, e^{\minus{}\delta x}$

but that's it. I tried the least mean square method, but it didn't help. So, if you have any suggestions - I'd be more than thankful.

How about just a straightforward Taylor series approximation?  Maybe first or second order?

What do you mean?  He's trying to approximate a data set with a function, not the other way around.

If the $ x$-values in your data set $ (x_k, y_k)$ are evenly spaced, use least squares on $ \ln (y_{k \plus{} 1} \minus{} y_k)$.  Otherwise, any idea I can think of requires guessing either the value of $ \alpha$ or the value of $ \delta$.

I understand what JRav said. I could take a few element's of the Taylor series, but that would be approximating the function with a polynomial, and since I know what function I have, I find this a bit degrading. And yes, they are evenly spaced and now that you've mentioned it, I'm quite surprised I didn't think of this my self! :) Thanks for the help!  Still, I was thinking of approximating the $ \delta$ with the mean of $ \frac{\Psi''(\vec{x})}{\Psi'(\vec{x})}$ since it represents the same value. I thought of using $ \Psi'(x_{i\plus{}1}) \equal{} x_{i\plus{}1}\minus{}x_i$ and so on, but I think the error got the best of my method. But, I can do yours! :)


given triangle ABC , prove the following :
$ \frac{1}{(cotA\plus{}cotB)^2} \plus{} \frac{1}{(cotA\plus{}cotC)^2}\plus{}\frac{1}{(cotC\plus{}cotB)^2} \geq \frac{9}{4}$

[hide="hint"]

the equation reduces
to $ \frac {sin^2Asin^2B}{sin^2C} \plus{} \frac {sin^2Asin^2C}{sin^2B} \plus{} \frac {sin^2Csin^2B}{sin^2A}$

now we can use AM-GM
so we get
$ \frac {sin^2Asin^2B}{sin^2C} \plus{} \frac {sin^2Asin^2C}{sin^2B} \plus{} \frac {sin^2Csin^2B}{sin^2A} \geq 3 \sqrt [3]{sin^2Asin^2Bsin^2C} \equal{} \frac{3.3}{4} \equal{}\frac{9}{4}$ [/hide]

[quote="implex"][hide]$ \sqrt [3]{\sin^2A\sin^2B\sin^2C} \equal{} \frac {3.3}{4}$[/hide][/quote]
That would mean that $ \sin A \cdot \sin B \cdot \sin C \equal{} \frac {3 \sqrt3}{8}$ but this is generally not true (try $ A \equal{} 30^{\circ},B \equal{} 60^{\circ},C \equal{} 90^{\circ}$).
Am I wrong?

Bye
GM

[quote="implex"][hide="hint"]

the equation reduces
to $ \frac {sin^2Asin^2B}{sin^2C} \plus{} \frac {sin^2Asin^2C}{sin^2B} \plus{} \frac {sin^2Csin^2B}{sin^2A}$

now we can use AM-GM
so we get
$ \frac {sin^2Asin^2B}{sin^2C} \plus{} \frac {sin^2Asin^2C}{sin^2B} \plus{} \frac {sin^2Csin^2B}{sin^2A} \geq 3 \sqrt [3]{sin^2Asin^2Bsin^2C} \equal{} \frac {3.3}{4} \equal{} \frac {9}{4}$ [/hide][/quote]
actually , we have $ sinA.sinB.sinC \leq \frac {3\sqrt {3}}{8}$ ,so your solution is not true , the AM-GM part is not enough to prove  :)


Define $E(a,b,c)=a\times{b^c}$. What is the value of r if it satisfies $E(r,r,3)=625$?

You have the correct answer, and it's fairly easy to get by intuition, but that isn't a rigorous proof of the answer. Can you do that?

indeed it is

[hide]

By definition:

$ r \cdot r^3 = 625$

$r^4=625$

Taking each to the 4th root yields:

$\boxed{r=5}$ [/hide]

Are you guys sure you have found all values of $r$?

[hide="Notice that your answer is not exactly complete."]-5 also works, as (-5)^4=625

There are also other solutions involving $i=\sqrt{-1}$[/hide]

So there are technically four roots, but I think the question is referring to the positive real root, although that should be specified.

The actual problem in the handbook did ask for "positive value", but it was left out of the first post.

Oh yeah  :blush: I copied wrong  :P
Ravi B is right about the original question... the question I asked isn't really at the level of this forum.


linear algebra
matrix
geometry
geometric transformation
rotation
linear algebra unsolved

Let $A \in M_2(\mathbb{Z})$ such that there exists $n \in \mathbb{N}$ s.t. $A^n=I$.Prove that $A^{12}=I$.

You didn't mean $\mathbb{R}$ there...
As written, the matrix could have any order; just consider a rotation by $\frac{2\pi}n$.

Yes, i meant $\mathbb{Z}$ instead of $\mathbb{R}$.
It becomesa an annoying rule for me to do typos lastly.

such a matrix is diagonalizable over $\mathbb{C}$ and its eigenvalues $a,b$ satify:
$|a|=|b|=1$, $a+b \in \{-2,-1,0,1,2\}$, $ab \in {-1,+1}$. From here conclusion follows easily. (just solve the equations)

Yeah such a matrix is diagonalisable and i guess we can just put $6$ instead of $12$

with alekk denotation since $a,b$ are roots of on of those equations:
 $x^2+\epsilon_1x+\epsilon_2=0$,$\vert \epsilon_1\vert \in \{0,1,2\},\vert \epsilon_2\vert=1$
we deduce that $a^3=1$ and $b^3=1$ so i thank mohamed that we can put $3$ instead $6$

yeah right it was an error of my computations ;) coz i resolved the exercice in an other way sorry :blush:


superior algebra
superior algebra unsolved

Let $ K$ be a finite field with $ q$ elements. Prove that $ x^q\equal{}x$ for all $ x\in K$.

For $ x \neq 0$, use Lagranges theorem for the group $ G\equal{}\mathbb F_q^\times$ and multiplication. For $ x\equal{}0$, use the ultimative power of eternal $ 0$.


search
inequalities unsolved
inequalities

[quote="metdzoi"]3/ If a,b,c,d > 0 and abcd = 1 then prove that :$ \frac {1}{(a \plus{} 1)^2} \plus{} \frac {1}{(b \plus{} 1)^2} \plus{} \frac {1}{(c \plus{} 1)^2} \plus{} \frac {1}{(d \plus{} 1)^2} \geq {1}$

[/quote]
China TST 2005 :wink:
[hide]http://www.mathlinks.ro/viewtopic.php?search_id=539604098&t=98967
http://www.mathlinks.ro/viewtopic.php?p=112435#p112435
http://www.mathlinks.ro/viewtopic.php?p=190407#p190407
http://www.mathlinks.ro/Forum/viewtopic.php?t=98967
http://www.mathlinks.ro/Forum/viewtopic.php?t=66028
http://www.mathlinks.ro/Forum/viewtopic.php?t=83784
http://www.mathlinks.ro/Forum/viewtopic.php?t=32031
http://www.mathlinks.ro/Forum/viewtopic.php?t=30724
http://www.mathlinks.ro/Forum/viewtopic.php?t=16027[/hide]


[quote="guest"]x,y,z > 0
so that
1/2(x+y) + 1/2(y+z) + 1/2(x+z) = x + y + z
[/quote]

is equivalent to saying 0=0!
loool
there must be a mistake!

I assume it is 1/(2*(x+y)), ......

Setting a=x+y,b=y+z,c=z+x

We have s = Sum 1/a = Sum a so by Cauchy-Schwarz s >= 3

Ineq to prove is Sum 1/(s+a) < 1/4

Sum 1/(s+a) <= Sum 1/(3+a)

It's enough to prove Sum 1/(3+a) <= 1/4 for s >= 3

It's the same as f(a,b,c)=Sum 1/(1+a) <= 9/4 for s >= 1.
It's enough to consider the maximum for s = 1.

It's easy to see that f(a-u,b+u,c) >= f(a,b,c) so max is
attained for example for a=b=0 and c=1.

So Sum 1/(s+a) <= 1/4 with equality for a,b, or c=0 which is impossible.

ofcourse it was 1/[2(x+y)]...
my solution was:
set a = x + y, b = x + z and c = z + y
then we get:
ab+ac+bc=abc(a+b+c)
and the ineq becomes:
1/(2a+b+c)^2 + 1/(2b+a+c)^2 + 1/(2c+a+b)^2 < 1/4

Using (2a + b + c)^2 >= 4a(a+b+c) by AM-GM with equality if a = a + b + c which is for positive reals impossible, we get that 1/(2a+b+c)^2 + 1(2b+a+c)^2 + 1/(2c+a+b)^2 < 1/[4a(a+b+c)] + 1/[4b(a+b+c)] + 1/[4c(a+b+c)] = (bc + ac + ab)/[4abc(a+b+c)] = 1/4

hi dear guest!
although there are no limitations as to post as a guest, it is however better to register :D


What is the sum of all integer multiples of $ 3$ between $ \minus{}118$ and $ 128$?

Sum of all integer multiples of $ 3$ between $ \minus{} 118$ and $ 118$ is $ 0$.
So, "sum of all integer multiples of $ 3$ between $ \minus{} 118$ and $ 128$" $ \equal{}$ "sum of all integer multiples of $ 3$ between $ 119$ and $ 128$".
Thus, $ 120 \plus{} 123 \plus{} 126 \equal{} \boxed{369}$.


Let  a and  b be positive integers with a>b. It is known that   (a^n+b^n) is divisible by  n  for infinitely many positive integers  n.
Let  n1<n2<... be all positive integers for which  ni|a^ni+b^ni
(i.e. for all i ni|a^ni+b^ni and for all n st n|a^n+b^n there exists a natural number i st n=ni).
Prove that   there exist natural numbers i and j (i<j) st gcd(ni,nj)>1. .

We may assume that at most one of the $ n_i$ is even. Let $ n_i$ and $ n_j$ be odd and suppose that they are coprime. In that case, it is easy to see that $ n_in_j$ divides $ a^{n_in_j}\plus{} b^{n_in_j}$.


conics
ellipse
parabola
calculus
analytic geometry
graphing lines
slope

how can i get shortest distance between line 3x+4y+5=0 and curve (27/28)*x^2+(9/7)*y^2=1 ???
Thank You!

[hide]Use the fact that the distance between the point $ (x,y)$ and your line is $ \frac{|3x\plus{}4y\plus{}5|}{\sqrt{3^2\plus{}4^2}}\equal{}\frac{|3x\plus{}4y\plus{}5|}{5}$.  Then, solve for one variable in the ellipse, and plug that into the distance.  Then it should be the form of a parabola which you can easily find the minimum value of.  This looks like the arithmetic would be horrible, so their should be a better way, but this should work.[/hide]

If we could use calculus, I would say implicitly differentiate the curve and find where it has the same slope as the line.
But since this forum doesn't allow it, I think WATR's method is the best.

Alternatively, you could try scaling the x-y axes to turn that ellipse into a circle (which shouldn't be too hard), then finding the nearest point (because it becomes much easier), then un-scaling.

please tell me how it will be with using calculus !

[quote="grn_trtle"]If you want a calculus solution, find the distance between any point $ (x,y)$ on your line and the curve, then take the derivative and set it equal to zero.
It would be ugly, though.[/quote]

yeah ... but i need shortest distance

that DOES find the shortest distance
but the best way is to use implicit differentiation with respect to x and find when the derivative is equal to the slope of the line
That finds the point on the curve closest to the line (next question: Why?)

The line of shortest distance between two non-intersecting curves always lies along the line which is normal to both the curves.
Slope of line normal to $ 3x\plus{}4y\plus{}5\equal{}0$ is obviously $ \frac{4}{3}$.
Slope of normal to the curve at any point (x,y) on it
$ \equal{}\minus{}\frac{1}{\frac{dy}{dx}}\equal{}\frac{4y}{3x}$
Therefore,$ \frac{4y}{3x}\equal{}\frac{4}{3}$
i.e.  $ y\equal{}x$
On plugging $ y\equal{}x$ in the given curve we get the point as $ (\minus{}\frac{2}{3},\minus{}\frac{2}{3})$ and also
$ (\frac{2}{3},\frac{2}{3})$.
Use the formula $ \frac{|3x\plus{}4y\plus{}5|}{\sqrt{3^2\plus{}4^2}}\equal{}\frac{|3x\plus{}4y\plus{}5|}{5}$ to find distance between these points and the given line.The lesser value is obviously the answer

Before you embark on all those calculations, do check whether the line intersects or touches the ellipse...


ratio geometry parallelogram trigonometry

Let ABCD be a parallelogram. let P and Q be points on the side BC such that AB = BP, P is between B and Q and angleBPA + angleBQA + angleBCA = angleBAD. If AP = PQ.sqrt 2, prove that PQ = QC

excuse me but you double posted, please do not do that,  :spam:
[url=http://www.mathlinks.ro/Forum/viewtopic.php?t=41127]same problem posted here[/url]

Sorry! I didn't know that I can't do that...Terribly sorry... :(

[color=blue][b]Lemma:[/b][/color] Let $\triangle APQ$ be a triangle with an arbitrary angle $\angle APQ$ and with the sides $PQ, AP$ in the ratio $\frac{PQ}{AP} = \sqrt 2$. Let $C$ be a point on the side $PQ$. Then the angles $\angle PQA, \angle PAC$ are equal if and only if $C$ is the midpoint of the side $PQ$.

[color=blue][b]Proof:[/b][/color] Assume that $C$ is the midpoint of the side $PQ$. Then $AC$ is the median from the vertex $A$. Using the parallelogram rule, the median length is given by

$2AP^2 + 2AQ^2 = PQ^2 + 4AC^2,\ \ \ AC^2 = \frac{2AP^2 + 2AQ^2 - PQ^2}{4}$

Since $\frac{PQ}{AP} = \sqrt 2,\ PQ^2 = 2AP^2$, we obtain

$AC^2 = \frac{2AP^2 + 2AQ^2 - 2AP^2}{4} = \frac{AQ^2}{2},\ \ \ \frac{AQ}{AC} = \sqrt 2$

Using the sine theorem for the triangles $\triangle APC,\ \triangle AQC$, we have

$\frac{\sin{\widehat{ACQ}}}{\sin{\widehat{AQC}}} = \frac{AQ}{AC} = \sqrt 2$

$\frac{\sin{\widehat{ACP}}}{\sin{\widehat{PAC}}} = \frac{AP}{PC} = \frac{2AP}{PQ} = \frac{2}{\sqrt 2} = \sqrt 2$

Consequently,

$\frac{\sin{\widehat{ACQ}}}{\sin{\widehat{AQC}}} = \frac{\sin{\widehat{ACP}}}{\sin{\widehat{PAC}}}$

Since the angles $\angle ACQ + \angle ACP = 180^o$ add up to a straight angle, $\sin{\widehat{ACQ} = \sin{\widehat{ACP}}}$, which means that $\sin{\widehat{AQC}} = \sin{\widehat{PAC}}$ as well. Since the angle $\angle AQC \equiv \angle AQP$ is against the side $AP = \frac{PQ}{\sqrt 2} < PQ$ of the triangle $\triangle APQ$, which definitely is not its longest side, $\angle AQC < 90^o$. Likewise, since the angle $\angle PAC$ is against the side $PC = \frac{PQ}{2} = \frac{\sqrt 2}{2} AP$ of the triangle $\triangle APC$, which definitely is not its longest side, $\angle PAC < 90^o$. Then from equality of the sines follows equality of the angles, $\angle PAC = \angle AQC \equiv \angle AQP$.

Assume that $C$ is not the midpoint of the side $PQ$. Let $C'$ be the midpoint of this side. As we have just demonstrated, the angles $\angle PAC' = \angle AQC'$ are then equal. If the point $C$ is on the segment $PC'$, then $\angle PAC < \angle PAC' = \angle AQP$ and if the point $C$ is on the segment $QC'$, then $\angle PAC > \angle PAC' = \angle AQP$. Consequently, the angles $\angle PAC \neq \angle AQP$ cannot be equal.

[color=red][b]Proof[/b][/color] of the problem proposition: The triangle $\triangle ABP$ is iscosceles with $AB = BP$, hence, the angles $\angle BPA = \angle BAP$ are equal. The angles $\angle BCA = \angle CAD$ are equal, as the alternate angles formed by the transversal $AC$ cutting the parallel lines $BC \parallel AD$. Applying the above lemma to the defined configuration with the parallelogram $ABCD$: The angles $\angle BQA \equiv \angle CAQ = \angle PAC$ if and only if $C$ is the midpoint of the segment $PQ$. Consequently,

$\angle BAD = \angle BAP + \angle PAC + \angle CAQ = \angle BPA + \angle BCA + \angle BQA$

if and only if $C$ is the midpoint of $PQ$, i.e., iff $PC = QC$.

Thank you very much for your detailed explanation. But I still have one more qn: does that mean for this qn to be true, C must be the midpoint of PQ?

Certainly. The proof demonstrates that if $C$ is the midpoint of $PQ$, then

$\angle BPA + \angle BCA + \angle BQA = \angle BAD$

and if $C$ is not the midpoint of $PQ$, then

$\angle BPA + \angle BCA + \angle BQA \neq \angle BAD$

Hence, if the equality of angles is true, $C$ must be the midpoint of $PQ$.

I don't get that. How come in the question they ask to prove that PQ=QC, but according to ur drawing Yetti, it turns our that PQ = 2 QC (if C is in the middle of PQ).  :?:  :|

Besides, the question said that AP = PQ.sqrt2 => the ratio of AP/PQ = sqrt2, not PQ/AP = sqrt 2 as in ur answer.  :|

Sorry for creating this confusion, I misread the problem. This is what I proved:

[color=blue]Let $ABCD$ be a parallelogram. Let $P$ and $Q$ be points on the side $BC$ such that $AB = BP$, $P$ is between $B$ and $Q$. If $\angle BPA + \angle BCA + \angle BQA = \angle BAD$ and $PQ = AP \sqrt 2$, prove that $PC = QC$ ($C$ id the midpoint of $PQ$).[/color]

And this is what I was supposed to prove:

[color=blue]Let $ABCD$ be a parallelogram. Let $P$ and $Q$ be points on the side $BC$ such that $AB = BP$, $P$ is between $B$ and $Q$. If $\angle BPA + \angle BQA + \angle BCA = \angle BAD$ and $AP = PQ \sqrt 2$, prove that $PQ = QC$ ($Q$ is the midpoint of $PC$).[/color]

The proofs of both propositions are very similar. In the proof of the supporting lemma and in the proof of the problem proposition, the only change consists of exchanging the points $Q, C$ and using the midpoint $Q'$ of the segment $PC$ in the contradiction part, with one exception: The point $C$ is still one of the vertices of the parallelogram $ABCD$.

[color=blue][b]Lemma:[/b][/color] Let $\triangle APC$ be a triangle with an arbitrary angle $\angle APC$ and with the sides $PC, AP$ in the ratio $\frac{PC}{AP} = \sqrt 2$. Let $Q$ be a point on the side $PC$, i.e., the segments $AP, PQ$ are in the ratio $\frac{AP}{PQ} = \frac{2AP}{PC} = \frac{2}{\sqrt 2} = \sqrt 2$ . Then the angles $\angle PQA, \angle PAC$ are equal if and only if $Q$ is the midpoint of the side $PC$.

[color=blue][b]Proof:[/b][/color] Assume that $Q$ is the midpoint of the side $PC$. Then $AQ$ is the median from the vertex $A$. Using the parallelogram rule, the median length is given by

$2AP^2 + 2AC^2 = PC^2 + 4AQ^2,\ \ \ AQ^2 = \frac{2AP^2 + 2AC^2 - PC^2}{4}$

Since $\frac{PC}{AP} = \sqrt 2,\ PC^2 = 2AP^2$, we obtain

$AQ^2 = \frac{2AP^2 + 2AC^2 - 2AP^2}{4} = \frac{AC^2}{2},\ \ \ \frac{AC}{AQ} = \sqrt 2$

Using the sine theorem for the triangles $\triangle APQ,\ \triangle ACQ$, we have

$\frac{\sin{\widehat{AQC}}}{\sin{\widehat{ACQ}}} = \frac{AC}{AQ} = \sqrt 2$

$\frac{\sin{\widehat{AQP}}}{\sin{\widehat{PAQ}}} = \frac{AP}{PQ} = \frac{2AP}{PC} = \frac{2}{\sqrt 2} = \sqrt 2$

Consequently,

$\frac{\sin{\widehat{AQC}}}{\sin{\widehat{ACQ}}} = \frac{\sin{\widehat{AQP}}}{\sin{\widehat{PAQ}}}$

Since the angles $\angle AQC + \angle AQP = 180^o$ add up to a straight angle, $\sin{\widehat{AQC} = \sin{\widehat{AQP}}}$, which means that $\sin{\widehat{ACQ}} = \sin{\widehat{PAQ}}$ as well. Since the angle $\angle ACQ \equiv \angle ACP$ is against the side $AP = \frac{PC}{\sqrt 2} < PC$ of the triangle $\triangle APC$, which definitely is not its longest side, $\angle ACQ < 90^o$. Likewise, since the angle $\angle PAQ$ is against the side $PQ = \frac{PC}{2} = \frac{\sqrt 2}{2} AP$ of the triangle $\triangle APQ$, which definitely is not its longest side, $\angle PAQ < 90^o$. Then from equality of the sines follows equality of the angles, $\angle PAQ = \angle ACQ \equiv \angle ACP$.

Assume that $Q$ is not the midpoint of the side $PC$. Let $Q'$ be the midpoint of this side. As we have just demonstrated, the angles $\angle PAQ' = \angle ACQ'$ are then equal. If the point $Q$ is on the segment $PQ'$, then $\angle PAQ < \angle PAQ' = \angle ACP$ and if the point $Q$ is on the segment $CQ'$, then $\angle PAQ > \angle PAQ' = \angle ACP$. Consequently, the angles $\angle PAQ \neq \angle ACP$ cannot be equal.

[color=red][b]Proof[/b][/color] of the problem proposition: The triangle $\triangle ABP$ is iscosceles with $AB = BP$, hence, the angles $\angle BPA = \angle BAP$ are equal. The angles $\angle BQA = \angle QAD$ are equal, as the alternate angles formed by the transversal $AQ$ cutting the parallel lines $BC \parallel AD$. Applying the above lemma to the defined configuration with the parallelogram $ABCD$: The angles $\angle BCA \equiv \angle ACP = \angle PAQ$ if and only if $Q$ is the midpoint of the segment $PC$. Consequently,

$\angle BAD = \angle BAP + \angle QAD + \angle PAQ = \angle BPA + \angle BQA + \angle BCA$

if and only if $Q$ is the midpoint of $PC$, i.e., iff $PQ = QC$.

Yetti

[quote="YooSam"][quote="yetti"]Sorry for creating this confusion, I misread the problem. This is what I proved:

[color=blue]Let $ABCD$ be a parallelogram. Let $P$ and $Q$ be points on the side $BC$ such that $AB = BP$, $P$ is between $B$ and $Q$. If $\angle BPA + \angle BCA + \angle BQA = \angle BAD$ and $PQ = AP \sqrt 2$, prove that $PC = QC$ ($C$ id the midpoint of $PQ$).[/color] [/quote]

Are we supposed to change the question of dogtrees? You actually did change the question a bit.

This is the real question: [b]Let ABCD be a parallelogram. let P and Q be points on the side BC such that AB = BP, P is between B and Q and angleBPA + angleBQA + angleBCA = angleBAD. If [color=red]AP = PQ.sqrt 2[/color], prove that PQ = QC[/b]

And in the question they don't say that C is the midpoint of PQ though. I still think that my drawing is logicall.

Does anyone have any idea?[/quote]
i dont understand. she has the drawing exactly the same as yetti's corrected drawing. and the highlighted in red [b]does[/b] appear in yetti's corrected solution. what's the complain? i think yetti's solution is good...  :?

[quote="yetti"]YooSam, would you mind to read more than the first 3 lines of the last post ?

Yetti[/quote]

Yup sorry, your first drawing is wrong, and then you corrected it. I did not see that.  :blush:

Yes, your solution is good  :oops:   :thumbup: .


calculus
derivative
algebra
polynomial
superior algebra
superior algebra unsolved

Please refer to attachment. I didn't know how to post the problem so I simply attached it in a MS word document.  I would appreciate any help though!

What don't you understand?

If $p(x)=a_0+a_1x+\ldots+a_nx^n$ and $q(x)=b_0+b_1x+\ldots+b_mx^m$

then
\begin{eqnarray*} p'(x)+q'(x)&=&a_1+2a_2x+\ldots na_nx^{n-1}+b_1+2b_2x+mb_mx^{m-1} \\ &=&(a_1+b_1)+2(a_2+b_2)x+3(a_3+b_3)x^2+\ldots \\ &=&(p+q)'(x) \end{eqnarray*}
The other one is similar.


geometry
circumcircle
geometric transformation
homothety
trapezoid
trigonometry
cyclic quadrilateral

Let $ABCD$ be a cyclic quadrilateral, and $P$ a point in its interior such that $\angle BPC= \angle PAB+\angle PDC$. Let $E$, $F$ and $G$ be the orthogonal projections of $P$ on the sides $AB$, $DA$ and $CD$, respectively. Show that triangles $BPC$ and $EFG$ are similar.

[i]Alternative formulation.[/i] A point $P$ is in the interior of the cyclic quadrilateral $ABCD$ and has the property $\angle BPC = \angle BAP+\angle PDC$. The feet of the perpendiculars from $P$ to $AB$, $AD$ and $DC$ are respectively denoted $E$, $F$ and $G$. Show that $\triangle FEG$ and $\triangle PBC$ are similar.

IM  pretty sure ive seen this problem before its from yugoslavia or chekoslovakia somewhere like that  :?

If I didn't remember wrongly, this one is from Austrian-Polish Match.
For the problem, construct the circumcircles of $\triangle ABP$ and $\triangle CDP$. The crucial point is to make use of the radical center and radical axis of the three circles.

We need to prove that $\frac{AP \cdot CP}{BP \cdot DP} = \frac{AC}{BD}$.

(It is very easy to see that the statement which we need to prove is equivalent to this equality.)

I still need to decide how I'm going to do that :lol: seems to be not so easy.

Starting with 2 arbitrary circles (K), (L) intersecting at points P, Q and their internal bisector circle (X) centered at their internal homothety center X and with radius XP = XQ, draw 2 arbitrary lines through the point X, the 1st one in the angular sector $\angle PXL$, the other one in the angular sector $\angle PXK$. Let the 1st line meet the circle (K) at points $A, A_0$ and the circle (L) at points $B, B_0$, the points A, B lying on the same side of the center line KL. Likewise, let the 2st line meet the circle (K) at points $C, C_0$ and the circle (L) at points $D, D_0$, the points A, B, C, D lying on the same side of the center line KL. The power of the point X to the circles (K), (L) is $XA \cdot XA_0 = XC \cdot XC_0,$ $XB \cdot XB_0 = XC \cdot XC_0,$ respectively. Since X is the internal homothety center of the circles (K), (L), $\frac{XA_0}{XB} = \frac{XA}{XB_0} = \frac{XC_0}{XD} = \frac{XC}{XD_0}.$ Combining, $XA \cdot XB = XC \cdot XD = XA_0 \cdot XB_0 = XC_0 \cdot XD_0$, hence the quadrilaterals $ABCD,\ A_0B_0C_0D_0$ are both cyclic and generally not similar. Since (X) is the common P-Apollonius circle of the triangles $\triangle APB,\ \triangle CPD$ and the common Q-Apollonius circle of the triangles $\triangle A_0QB_0,\ \triangle C_0QD_0$, WE HAVE $\angle BPC = \angle PAB + \angle PDC$ and $\angle D_0QA_0 = \angle QB_0A_0 + \angle QC_0D_0.$

Invert the cyclic quadrilateral ABCD with a circumcircle (O) and the circles (K), (L), (X) in a circle (P) centered at the point P and with radius PX. The pencil of circles (K), (L), (X) is carried into a pencil of concurrent lines $k, l, x$, such that the line x bisects the angle formed by the lines $k, l$ (inversion preserves angles). The quadrilateral vertices A, B, C, D are inverted into the intersections A', B', C', D' of the lines $k, l$ with the inverted circumcircle (O'). Since the circle (X) is congruent to the inversion circle (P), the line x is the perpendicular bisector of the segment PX. By the basic properties of inversion, the triangle pairs $\triangle B'PA' \sim \triangle APB,$ $\triangle D'PC' \sim \triangle CPD,$ $\triangle C'PB' \sim \triangle BPC,$ $\triangle A'PD' \sim \triangle DPA$ are similar. Hence, $\angle PB'A' = \angle PAB,$ $\angle PC'D' = \angle PDC,$ $\angle C'PB' \equiv \angle BPC$, and from the equality $\angle PAB + \angle PCD = \angle BPC$, it follows that the lines A'B', C'D' form an angle

$\angle (A'B', C'D') = \angle A'B'P + \angle PB'C' + \angle B'C'P + \angle P'C'D' =$

$= \angle PAB + \angle PDC + (180^\circ - \angle CPB) = 180^\circ$

which means that these lines $A'B' \parallel C'D'$ are parallel. Thus A'B'C'D' is a cyclic trapezoid, isosceles, and B'C' = A'D'. Consequently, and the line x bisecting the angle formed by its diagonals $k \equiv A'C',\ l \equiv B'D'$ is also the perpendicular bisector of the sides A'B', C'D', so that the triangles $\triangle C'PB' \cong \triangle D'XA'$ are congruent. As a result, the triangles $\triangle BPC \sim \triangle D'XA'$ are similar. But the pedal triangle $\triangle EFG$ of the triangle $\triangle AXD$ with respect to the point P is similar to the image $\triangle D'XA'$ of this triangle in the inversion in the circle (P), so that the triangles $\triangle BPC \sim \triangle EFG$ are similar as well.

Nice, Yetti. I think my solution is a bit shorter, but I'm not sure whether it's actually correct :)

Invert $ABCD$ with respect to $P$ (with factor $r^2$). It is easy to see that $A'B'C'D'$ is a trapezoid (using the condition given in the problem). It must also be cyclic since $ABCD$ is cyclic itself (I think? :|), so it must be an isosceles trapezoid and $A'C' = B'D'$. Now, since $A'C' = r^2 \cdot \frac{AC}{AP \cdot CP}$ and $B'D' = r^2 \cdot \frac{BD}{BP \cdot DP}$, we get that $\frac{AP \cdot CP}{BP \cdot DP} = \frac{AC}{BD}$. And that is, of course, what we wanted to show. Indeed, $\angle EFG = \angle EFP + \angle PFG = \angle BAP + \angle CDP = \angle BPC$ and $\frac{EF}{FG} = \frac{AP \sin \angle BAD}{DP \sin \angle ADC} = \frac{AP \cdot BD}{DP \cdot AC} = \frac{BP}{CP}.$ So we're done. $\blacksquare$

It is correct?

Both of you seem to have nice solutions to this question by inversion.

(Arne: Inversion preserves angles. If your cyclic quad is inverted, it either stays cyclic or the four points turn collinear.)

But there was a non-inversion solution that no one in the UK squad could find:

[hide="Reveal Solution"]The key is to draw a line from $P$ which cuts $BC$ at $T$ such that $\angle BPT = \angle BAP$. Now $\angle CPT = \angle CDP$ and by the converse of the Alternate Segment Theorem, $PT$ is the common tangent of the circles through $PAB$ and $PCD$. By the Radical Axis Theorem on these two circumcicles and the circle through $ABC$, $AB$, $CD$ and $PT$ meet at a point $Q$.

Assume w.l.o.g. that $Q$ lies on the opposite side of $A$ to $B$. $AEPF$ is cyclic so $\angle FEP = \angle DAP$. $QEPG$ is cyclic so $\angle GEP = \angle GQP$. So $\angle FEG = \angle FEP - \angle GEP = \angle DAP - \angle DQP = \angle QDA - \angle QPA$ (This follows from $\angle DAP + \angle QPA = \angle QDA + \angle DQP$). As $\angle QPA = \angle PBA$ by the Alternate Segment Theorem, and $\angle QDA = \angle CBA$, $\angle FEG = \angle PBC$

Simple angle chasing instantly shows $\angle BPC = \angle EFG$.[/hide]

And we're done.  :lol:

[quote="Centy"]
(Arne: Inversion preserves angles. If your cyclic quad is inverted, it either stays cyclic or the four points turn collinear.)
[/quote]

Indeed, pretty embarrassing :blush:

[quote="Centy"]The key is to draw a line from $P$ which cuts $BC$ at $T$ such that $\angle BPT = \angle BAP$. Now $\angle CPT = \angle CDP$ and by the converse of the Alternate Segment Theorem, $PT$ is the common tangent of the circles through $PAB$ and $PCD$. By the Radical Axis Theorem on these two circumcicles and the circle through $ABC$, $AB$, $CD$ and $PT$ meet at a point $Q$.[/quote]

I drew that line... (I drew it like 100 times, in fact :D) but I didn't manage to complete the proof. In fact, I tried considering the second intersection of the circumcircles of $ABP$ and $CDP$ as well, but that didn't really work - probably because there is no second intersection :( Anyways, the idea of that construction is not thát unnatural...

[b]Very nice problem, but it isn't hardly ! Must you have a little idea.[/b] [hide="Here is a proof without inversion."][color=darkblue][b]Proof.[/b] Suppose w.l.o.g. that $L\in AB\cap CD$ and the line $BC$ separates the points $L$, $P$. Denote the circumcircle $w$ of the quadrilateral $ABCD$ and the circumcircles $w_1$, $w_2$ for the triangles $ABP$, $CDP$ respectively.The quadrilaterals $AEPF$ and $DFPG$ are cyclically and $m(\widehat {BPC})=m(\widehat {EAP})+m(\widehat {GDP})$ $\Longrightarrow$ $\widehat {EFP}\equiv \widehat {EAP}$, $\widehat {GFP}\equiv \widehat {GDP}$ $\Longrightarrow$ $m(\widehat {EFG})=m(\widehat {EFP})+m(\widehat {GFP})$  $\Longrightarrow$ $m(\widehat {EFG})=m(\widehat {EAP})+m(\widehat {GDP})$ $\Longrightarrow$ $\boxed {\ \widehat {EFG}\equiv \widehat {BPC}\ }\ \ (\alpha )\ .$[/color]
[color=darkred]At this time here is [b]the little idea[/b] : there is one point $S\in (BC)$ for which $\widehat {SPB}\equiv \widehat {BAP}$ and $\widehat {SPC}\equiv \widehat {CDP}$. Thus, the line $PS$ is the common tangent in the point $P$ to the circumcircles $w_1$ and $w_2$, i.e. the line $PS$ is the radical axis of the circles $w_1$, $w_2$. But the line $AB$ is the radical axis between the circles $w$, $w_1$ and the line $CD$ is the radical axis between the circles $w$, $w_2$. Therefore, the lines $AB$, $CD$ and $PS$ are concurrently (in the point $P$).[/color]
$\triangle LAP\sim \triangle LPB$ and $\triangle LDP\sim\triangle LPC\Longrightarrow$ $\frac{PB}{PA}=\frac{LP}{LA}$, $\frac{PD}{PC}=\frac{LD}{LP}$. [color=darkblue]But $\frac{LD}{LA}=\frac{\sin A}{\sin D}$. From the product of the last three relations obtain $\frac{PB}{PA}\cdot \frac{PD}{PC}=\frac{\sin A}{\sin D}\ .$ Therefore, $EF=PA\sin A$ and $FG=PD\sin D$ $\Longrightarrow$ $\frac{FE}{FG}=$ $\frac{PA}{PD}\cdot \frac{\sin A}{\sin D}=$ $\frac{PA}{PD}\cdot \frac{PB}{PA}\cdot \frac{PD}{PC}=$ $\frac{PB}{PC}$ $\Longrightarrow$  $\boxed {\ \frac{FE}{FG}=\frac{PB}{PC}\ }\ \ (\beta)\ .$

$(\alpha)\ \wedge\ (\beta)\Longrightarrow \boxed {\ \triangle BPC\sim\triangle EFG\ }\ .$[/color][/hide]

have seen  these two geometry problem in the journal in ky country

[b]Indication.[/b] Given are a fixed line $d$ and two fixed points $A$, $B$ so that $[AB]\cap d =\emptyset$ and $AB\not\parallel d$. Construct the points $L_1\in d$, $L_2\in d$ for which the circumcircles of the triangles $ABL_1$, $ABL_2$ are tangent to the line $d$. Then the point $L$ for which $m(\widehat {ALB})$ is maximum belongs to the set $\{L_1,L_2\}$.

[quote="bilarev"]This beautiful problem is like G4 from IMO2003 shortlist![/quote]

Indeed, I noticed the similarities too :)


Here is a very easy inequality problem:

Let p, q be positive real numbers.
Suppose p, A', A", q are in arithmetic, and p, G', G", q are in geometric progression.
Show that A'A" \geq G'G".

[i]Canadian Mathematical Olympiad 1979[/i]

q=p+3d=pr^3 , so p^2+3pd=p^2*r^3
We prove (p+d)(p+2d)  \geq pr*pr^2 => p^2+3dp+2d^2 \geq p^2+3pd => 2d^2 >= 0 which is true.


Hi, I play both the piano and violin, but have always had trouble reading rhythms and playing to the beat of a metronome. Anyone have any ideas on how I can become better at this? My teacher tell me to count, but that's not really helping, because sometimes I have no idea what the song sounds like until someone else plays it. And I just go along with that until i've memorized the rhythm and mastered the notes in the piece and it has really slowed my progress. Just looking for some suggestions, thanks!

I guess the idea is to think about subdividing. Subdivide it as much as you need to figure out the relative lengths of things. At the same time, look for the down beats in each measure.

maybe try playing 1/2 speed, keep the metronome the same, and have it count for an eight note, instead of a quarter note, it will make subdividing easier, it will help u get a feel for it, then u could try playing it full speed, or gradually speeding up?

My Jazz band had the same problem.

What helps to get the rhythms is use words, my teacher always used Animal names.

Hip-po hip-po al-li-gat-tor pig.

Each word is one count, and each syllable is one subdivision.  Like in the example I posted above, Hippos are eighth notes, alligators are sixteenths, and pigs are quarter notes.  Emphasize is on the beginning of each word.

It's really fun saying them all...on a beat...then try saying it with a metronome.

We had other words as well, for triplets, but I don't remember them.  If I do I'll post them!  You can easily find other words that work for yourself too.  I personally like the animals, very catchy =D


how many permutation {$ {a_1,a_2,...,a_n}$}of set {$ {1,2,3,...,n}$} such that $ a_i$  and $ a_{i\plus{}1}$ not equal $ i$ for all $ i\equal{}1,2,...,n$.

[quote="hoangclub"]how many permutation {$ {a_1,a_2,...,a_n}$}of set {$ {1,2,3,...,n}$} such that $ a_i$  and $ a_{i \plus{} 1}$ not equal $ i$ for all $ i \equal{} 1,2,...,n$.[/quote]
after spending 5 minutes thinking:
I guess the majority of mathlinkers know how to determine the number of sequences with $ a_i\neq i$. it can be expressed in both recurrent and explicit way.
now let $ d_n$ is the number of sequences such that $ a_i\neq i$.
therefore:
$ d_n \equal{} 1 \plus{} \sum_{i \equal{} 0}^{n}\binom{n}{0}\cdot b_i$,where $ b_n$ is the value we are looking for.

See for example [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=1158]this thread[/url], and sequences [url=http://www.research.att.com/~njas/sequences/A000179]A000179[/url], [url=http://www.research.att.com/~njas/sequences/A000271]A000271[/url] in the [url=http://www.research.att.com/~njas/sequences]OEIS[/url].  This is known as the "ménage problem."