Given triangle  ABC, find  min of :
\[ T \equal{} [1\plus{}cos^2(A)][1\plus{}cos^2(B)][1\plus{}cos^2(C)]\]

hello, here we get $A=B=C=\frac{\pi}{3}$ with the help of calculus.
Sonnhard.


Ring Theory
superior algebra
superior algebra theorems

Is it true that for all commutative rings $R$, non necessarily with $1$, there exists a commutative ring $A$ with $1$ and an injective ring homomorphism $\varphi_A: R\to A$ such that $\varphi_A(R)$ is an ideal of $A$?

-oo-'s proof in [url=http://www.mathlinks.ro/Forum/viewtopic.php?t=109996]this topic[/url], shows that for $K=\mathbb Z$ or any other ring that makes $R$ into an "algebra without unit", we can give the additive group $A=K\oplus R$ the structure of a $K$-algebra (with $1$), such that the natural map $R\to A$ identifies $R$ to an ideal in $A$.

$R$ doesn't even need to be commutative.


[quote="Interval"]A class of 43 students is to be divided into committees in such a way that each student serves on exactly one committee. Each committee must have at least 3 members and at most 5 members.  The possible number of committees that can be formed is

(1) at least 8 and at most 14

(2) at least 8 and at most 15

(3) at least 9 and at most 14

(4) at least 9 and at most 15

MY WORK:

43/3 = 14.3333333

43/5 = 8.6

My answer is at least 8 and at most 14.

Book's answer is choice 3 above.  Why?  How is right?

Thanks![/quote]
When you got 8.6 you dant have .6 left. You have to round it to 9.

Not round. You have to do $\lceil{8.6}\rceil$, which is $9$.

[quote="Interval"]

43/3 = 14.3333333

43/5 = 8.6

My answer is at least 8 and at most 14.
[/quote]

you were going well until this point, as explained above. when you do your work, check to make sure it makes sense. Can 43 people be divided into 8 groups without at least one group having more than 5 people? no!

So, basically I forgot to round 8.6 to the nearest unit, which, after doing so, becomes 9.

I clearly see what I forgot to do.

Thanks.

no, not just round to the nearest unit, but round up. It matters which way. If it had been 8.3 you wouldv'e had to round up. You are seeing how many committees 43 people can make with 5 people each- and 8*5 is only 40 so there are 3 people commitee-less. if you can't tell whether you need to round up or down, check your answer like that, and use common sense ;P

I see there is a difference if the decimal was 8.3 rather than 8.6.

Thanks!

i think you missed that too: there wouldn't be a difference if the decimal was 8.3, you would still have to round up! either way you need more than 8 comittees, so you need at least nine commitees!


Let $a,b,c,d$ be non-negative real numbers and $a+b+c+d=4$. Prove that
(a) (Sung Yoon Kim). \[a^{2}bc+b^{2}cd+c^{2}da+d^{2}ab \le 4.\]
(a) (Pham Kim Hung). \[ab^{2}c+bc^{2}d+cd^{2}a+da^{2}b \le 4.\]

[quote="Sung-yoon Kim"][quote="GacKiem"]Let $a,b,c,d$ be non-negative real numbers and $a+b+c+d=4$. Prove that
(a) (Sung Yoon Kim).
\[a^{2}bc+b^{2}cd+c^{2}da+d^{2}ab \le 4. \]
(a) (Pham Kim Hung).
\[ab^{2}c+bc^{2}d+cd^{2}a+da^{2}b \le 4. \]
[/quote]

That's not mine, actually.. that was proposed by Doosung Park. I just generalized it for n variable case.  :oops:[/quote]

You are really kind, Sung,

[quote="GacKiem"]Let $ a,b,c,d$ be non-negative real numbers and $ a \plus{} b \plus{} c \plus{} d \equal{} 4$. Prove that
(a) (Sung Yoon Kim).
\[ a^{2}bc \plus{} b^{2}cd \plus{} c^{2}da \plus{} d^{2}ab \le 4.
\]
[/quote]
See at http://www.mathlinks.ro/Forum/viewtopic.php?t=100960
Solution at this topic:
Suppose that  $ (x;y;z;t)$ is a permutation of  $ a;b;c;d$ such that $ x\geq y\geq z\geq t$
=>$ xyz\geq xyt\geq xzt\geq yzt$
and $ x.xyz \plus{} y.xyt \plus{} z.xzt \plus{} t.yzt\geq a^{2}bc \plus{} b^{2}cd \plus{} c^{2}da \plus{} d^{2}ab$
Then
$ x.xyz \plus{} y.xyt \plus{} z.xzt \plus{} t.yzt \equal{} (xy \plus{} zt)(xz \plus{} yt)\leq\frac {1}{4}(xz \plus{} xy \plus{} yt \plus{} zt)^{2}\leq 4$
Done!


I know how to do all of these problems, but I feel my method is either time consuming, confusing, or unorthodox. Therefore I want to see how you would do these problems.

If x is a positive integer, what is the value of x for the equation (x! - (x - 3)!) ÷ 23 = 1?

At a graduation ceremony, names were read at the rate of five names per minute. If the names had been read at the rate of seven names per minute, the ceremony would have taken 24 minutes less. How many names were read?

It might help to say what you've tried first. Then, people can comment on your methods and see if there is anything better.

Another way for #2:

If the cubes are not joined together, the surface area would be $ 7 \cdot 6\equal{}42$.

Every spot where two cubes are joined together, we subtract 2 (do you see why?)

There are $ 7$ places where the cubes are joined, so $ 42\minus{}14\equal{}28$.

um aleph0, did you mean to put the x in that function of yours? Also could you please elaborate on how you came up with that function?

For #4:

$ 1,5,13,25,41$ are the given $ (x,x)$ points. we want the $ (10,10)$ point.

We can easily find a recursion:

$ a_n\equal{}a_{n\minus{}1}\plus{}4n\minus{}4$

$ 2x^2\minus{}2x\plus{}1$ seems to give the points, but I'm not sure why.

Anyway, Plugging in $ 10$ gives $ 181$, or the recursion gives

$ 1,5,13,25,41,61,85,113,145,181$.

-------

$ x!\minus{}(x\minus{}3)!\equal{}23 \implies x!\equal{}23\plus{}(x\minus{}3)!$.

A little NT will help here. All factorials, except for $ 0!\equal{}1!\equal{}1$, are even. Thus, if $ x\minus{}3 > 1$, then one side is even, and the other is odd. Thus, $ x\equal{}4,3$. Clearly, 4 is the only solution.


quadratics
linear algebra
linear algebra unsolved

I don't get it, you give the question, you also give the result, and you post the actual problem in uzbek?  Something like find the normal is rather vague.   :huh:

[quote="rustam"]Please help.
I am tired of solving and not finding the right answer to this problem:
Given the quadratic form $\sum_{i=1}^{n}x_i^2 + \sum_{i<j}x_ix_j$.
[hide]Nayti normalniy vid i nevirojdennoye preobrazovaniye prevodyasheye k etomu vidu( i dont know how it should be in Enlish. Smth like find the normal ...[/hide]
The answer is:
$f(x_1,x_2,...,x_n)=\sum_{i=1}^{n}x_i^2 + \sum_{i<j}x_ix_j= y_i^2+\frac{3}{4}y_2^2+...+\frac{n+1}{2n}y_n^2 \text{ where }$
$y_1=x_1+\frac12(x_2+...+x_n);$
$y_2=x_2+\frac13(x_3+...+x_n);$
 ...
$y_n=x_n.$[/quote]

Ok, it is russian, though... I'll try to translate: Find the transformation which gives a diagonal form of this quadratic form.
Yes, rustam, you transformation seems to be ok.

Sorry, i'm not good in english. I just wanted to say that the answer in that book was that and i cant figure out this answer. :(

[quote="rustam"]Please help.
I am tired of solving and not finding the right answer to this problem:
Given the quadratic form $\sum_{i=1}^{n}x_i^2 + \sum_{i<j}x_ix_j$.
[hide]Nayti normalniy vid i nevirojdennoye preobrazovaniye prevodyasheye k etomu vidu( i dont know how it should be in Enlish. Smth like find the normal ...[/hide]
The answer is:
$f(x_1,x_2,...,x_n)=\sum_{i=1}^{n}x_i^2 + \sum_{i<j}x_ix_j= y_i^2+\frac{3}{4}y_2^2+...+\frac{n+1}{2n}y_n^2 \text{ where }$
$y_1=x_1+\frac12(x_2+...+x_n);$
$y_2=x_2+\frac13(x_3+...+x_n);$
 ...
$y_n=x_n.$[/quote]

Ah, i'll try to explain.
First you need as usually cancek out the term with $x_1$ in order to reduce you problem to the like one with variables $x_2,x_3,...,x_n$ and do it once again and so on... unitl you get the desired representation.
As a first step get together all the term of you quadratic form which contain $x_1$
So
$Q(x_1,x_2,...,x_n)=\sum_{i=1}^{n}x_i^2 + \sum_{i<j}x_ix_j= x_1+x_1x_2+...+x_1x_n+\sum_{i=2}^{n}x_i^2 + \sum_{2\leq i<j}x_ix_j = (x_1+\frac{x_2}{2}+...+\frac{x_n}{2})^2+\frac{3}{4}\sum_{i=2}^{n}x_i^2 + \frac{1}{2} \sum_{2\leq i<j}x_ix_j =y_1+\frac{3}{4}(x_2+\frac{x_3}{3}+...+\frac{x_n}{3})^2+ \sum_{i=3}^{n}x_i^2 + \sum_{3 \leq i<j}x_ix_j)=y_1+ \frac{3}{4}y_2+...$


[quote="Leonhard Euler"]prove that any triangle ABC incircle inscribe to triangle which vertex is the midpoint of side of triangle ABC.[/quote]

¿

  darij

[quote="Leonhard Euler"]prove that any triangle ABC incircle inscribe to triangle which vertex is the midpoint of side of triangle ABC.[/quote]

Can you restate the problem?
thanks.

Sorry for may bad English. I mean  nine-point circle of any triangle is tangent internally to the incircle(ie, Feuerbach's theorem). I already found proof.

I have also found a new purely synthetic proof of the Feuerbach's theorem and will be interested in your proof.
http://perso.orange.fr/jl.ayme/

look at : contenu, volume 1 (2007) Le theorème de Feuerbach    in french
or at Google : revistaoim
look at n° 26   in spain


modular arithmetic
induction
number theory unsolved
number theory

Let be given a prime number $p\ge 3$ and an even number $k$ such that $1<k<p-1$.Prove that:
$p^{k+1} | \sum_{i=1}^{p-1} \binom{p}{i}^k$

We will prove that if $p \geq 5$ is a prime ($p = 3$ is impossible) and $2 \leq k \leq p-3$ is an even natural number, then

$(1) \;\; \sum_{i=1}^{p-1}{\textstyle \binom{p}{i}}^{k}\equiv 0 \pmod{p^{k+1}}$

The fact that

$(2) \;\; x_{i}= \frac{\textstyle \binom{p}{i}}{p}= \frac{(p-1)!}{i!(p-i)!}$

is an integer when $1 \leq i \leq p-1,$ reduces the congruence (1) to

$(3) \;\; \sum_{i=1}^{p-1}x_{i}^{k}\equiv 0 \pmod{p}.$

Now

$(p-i)! = \prod_{j=i}^{p-1}p-j \equiv \prod_{j=i}^{p-1}-j = (-1)^{p-i}\: \frac{(p-1)!}{(i-1)!}\pmod{p},$

hence

$(4) \;\; i!(p-i)! \equiv (-1)^{p-i}\, i(p-1)! \pmod{p}.$

Then by (2)

$i!(p-i)!x_{k}= (p-1)!,$

which combined with (4) gives

$(-1)^{p-i}\, i(p-1)! \, x_{i}\equiv (p-1)! \pmod{p},$

From this congruence we obtain the following series of implications:

$(-1)^{p-i}\, i \, x_{i}\equiv 1 \pmod{p}$

$(-1)^{k(p-i)}\, i^{k}\, x_{i}^{k}\equiv 1^{k}\pmod{p}$

$i^{k}\, x_{i}^{k}\equiv 1^{k}\pmod{p}\;\; (k \;\: \mbox{is an even integer})$

$i^{p-1}\, x_{i}^{k}\equiv i^{p-k-1}\pmod{p}$

$x_{i}^{k}\equiv i^{p-k-1}\pmod{p}\;\; (\mbox{according to Fermats little theorem})$

$(5) \;\; \sum_{i=1}^{p-1}x_{i}^{k}\equiv \sum_{i=1}^{p-1}i^{n}\pmod{p},$

where $n = p-k-1 \in [2,p-3]$. Let $S_{m}= \sum_{i=1}^{p-1}i^{m}$ for all non-negative integers $m$. Then by the binomial formula

$p^{m}-1 = \sum_{i=1}^{p-1}(i+1)^{m}-i^{m}= \sum_{i=1}^{p-1}\sum_{j=0}^{m-1}{\textstyle \binom{m}{j}}i^{j}= \sum_{j=0}^{m-1}{\textstyle \binom{m}{j}}\Bigg[ \sum_{i=1}^{p-1}i^{j}\Bigg] = \sum_{j=0}^{m-1}{\textstyle \binom{m}{j}}S_{j}.$

Seeing $S_{0}= p-1$, this is equivalent to

$(6) \;\; mS_{m-1}=-p^{m}+p+\sum_{j=1}^{m-2}{\textstyle \binom{m}{j}}S_{j}$

for $m \geq 3$. Moreover $S_{1}={\textstyle \frac{p(p-1)}{2}}$ implies that $p\,|S_{1}$ because $p$ is an odd prime . Using this fact as the basis of induction on $m$ in the recurrence formula (6), it is easy to verify that $p\,|S_{i}$ when $1 \leq i \leq p-2$. Therefore $p \,| S_{n}$ since $n \in [2,p-3]$, hence by (5)

$\sum_{i=1}^{p-1}x_{i}^{k}\equiv \sum_{i=1}^{p-1}i^{n}= S_{n}\equiv 0 \pmod{p},$

which is in fact congruence (3).  [b]q.e.d. [/b]


For non-negative variables $ a$, $ b$, $ c$ and $ d$, show that if $ abcd\equal{}1$, then

$ \frac {ab}{1 \plus{} b} \plus{} \frac {bc}{1 \plus{} c} \plus{} \frac {cd}{1 \plus{} d} \plus{} \frac {da}{1 \plus{} a}\geq 2$.

[color=red]I've added a necessary condition.[/color]

take $ a\equal{}0 ,b\equal{}c\equal{}d\equal{}1$.

My mistake! I forgot the condition $ abcd\equal{}1$.

[quote="UlyssesGrant"]For non-negative variables $ a$, $ b$, $ c$ and $ d$, show that if $ abcd \equal{} 1$, then

$ \frac {ab}{1 \plus{} b} \plus{} \frac {bc}{1 \plus{} c} \plus{} \frac {cd}{1 \plus{} d} \plus{} \frac {da}{1 \plus{} a}\geq 2$.

[color=red]I've added a necessary condition.[/color][/quote]
Let $ a \equal{} \frac {x}{y},$ $ b \equal{} \frac {y}{z},$ $ c \equal{} \frac {z}{t}$ and $ d \equal{} \frac {t}{x},$ where $ x,$ $ y,$ $ z$ and $ t$ are positive numbers.
Then $ \sum_{cyc}\frac {ab}{1 \plus{} b} \equal{} \sum_{cyc}\frac {x}{y \plus{} z} \equal{}\sum_{cyc} \frac {x^2}{xy \plus{} xz}\geq\frac {(x \plus{} y \plus{} z \plus{} t)^2}{xy \plus{} yz \plus{} zt \plus{} tx \plus{} 2xz \plus{} 2yt}.$
But $ (x \plus{} y \plus{} z \plus{} t)^2\geq2(xy \plus{} yz \plus{} zt \plus{} tx \plus{} 2xz \plus{} 2yt)\Leftrightarrow(x \minus{} z)^2 \plus{} (y \minus{} t)^2\geq0.$


Show $ k,x\in\textbf f{R^+}, n\in\textbf f{N}\Rightarrow$

$ (x+1)^{n-1}(kx+1)\geq(\sqrt[n]{k}x+1)^n$

Is this just Holder on

$ (x\plus{}1)$ n-1 times and $ (kx\plus{}1)$ one time


topology
function
advanced fields
advanced fields unsolved

Let B(0,1) be the open ball with center 0, radius 1 and C complex plane.
Question: Is there an analytic f that is homeomorphism between B(0,1) and C?
I started thinking that there can´t be between B(0,1) and C\{0}, because B(0,1) is simply connected but C\{0} is not... but could there be between B(0,1) and C? There is homeomorphism between B(0,1) and R2 so it sounds possible, but I couldn´t find any... so maybe there is a proof that says there can´t be?

An analytic function which is a homeomophism has an analytic inverse. The classical name for such a thing is a "conformal mapping", and your question could be rephrased as asking whether the unit disk and the whole plane are "conformally equivalent."

They're not. Let $D=\{z : |z|<1\}$. Suppose there is an invertible surjective analytic mapping $f: D\mapsto\mathbb{C}.$ Then $f^{-1}$ is an analytic function from $\mathbb{C}$ to $D$, and that makes $f^{-1}$ a bounded entire function. By Liouville's Theorem, the only bounded entire functions are constants, so this just can't work.

However, this particular argument only came up when we tried to map $D$ onto all of $\mathbb{C}.$ It doesn't preclude a mapping onto an unbounded domain. In particular, let $E=\mathbb{C}\setminus\{x+0i : x\le 0\}.$ Can you find an invertible analytic function that maps $D$ onto $E$?

After we've dealt with this, for your next reading, you can look up
[hide="hidden key words."] the Riemann Mapping Theorem.[/hide]


MATHCOUNTS
geometry
analytic geometry
graphing lines
slope
rectangle

Triangle ABC has its vertices at A(0,4), B(0,9) and C(6,9). What is the area of the triangle, in square units?

If you draw out the triangle on the x-y coordinate plane, you'll see it has a base of 5 and a height of 9. So the area is just $ \frac{1}{2}(5)(9)\equal{}\boxed{22.5}$ square units.

Here is a much faster way to do the problem.
[hide="Solution"]
The difference between 0,9 and 0,4 is 5.
The difference between 6,9  and 0,9 is  6.
To get the area of this triangle you solve the expression
$ \frac{1}{2}(5)(6)\equal{} 15$
Note: I am a very unsmarticle person pm if I made a mistake.[/hide]

Monkeygirl - Your solution works just fine; however, it's important to graph it (or at least look at the slopes of the lines).  For all you know, they could have made it some irregular triangle and you would have needed to use Herron's Formula, Shoelace Formula, etc.

You realize that (0, 9) and (0, 4) share the same x value, and (6, 9) and (0. 9) share the same y-value, making the triangle right and half a rectangle that is "parallel" to the axes...

Well, You could plot the points out and use the distance formula for the base and height and multiply that by $ \frac {1}{2}$

So, it is $ \sqrt {(x_2 \minus{} x_1)^{2} \plus{} (y_1 \minus{} y_ 2)^{2}}$

Plug in what you know and you get the distances between the base: BC and the height: AB to get

$ \frac {1}{2}(5)(6) \equal{} 15$

The area is $ 15$ square units


The real number x between 0 and 1 has decimal representation [img]http://www.artofproblemsolving.com/Forum/latexrender/pictures/1c9f752cb01a4b76325c82d2f99742ff.gif[/img],with the following property:the number of distinct blocks of the form  [img]http://www.artofproblemsolving.com/Forum/latexrender/pictures/d129a66ddc9d0b333f71efb7c2a21e2a.gif[/img] as k ranges through all positive integers,is less than or equal to 2004.Prove that x is rational.Excuse me if it has been posted. :)

Yup, it's been posted: [url=http://www.mathlinks.ro/Forum/viewtopic.php?t=30601]www.mathlinks.ro/Forum/viewtopic.php?t=30601[/url]


Suppose that $f(x_1,\dots,x_n)$ is a polynomial in $\mathbb R[x]$ that with an order $\alpha x_1^{r_1}\dots x_n^{r_n}$ is the mononomial with the maximum degree. Suppose we have chosen $A_1,\dots,A_n\subset\mathbb C$ that $|A_i|>r_i$. Prove that there exist $(y_1,\dots,y_n)\in A_1\times\dots A_n$ that $f(y_1,\dots,y_n)\neq0$


symmetry
geometry
geometric transformation
reflection
inequalities
3D geometry
absolute value

I made the following proof and have two questions:

1) Are there any mistakes in it?

2) What could I do different to make it more rigorous?

Show that if $ x$ and $ y$ are such that if    $ \frac{x \plus{} y}{x}$ $ \equal{}$ $ \frac{y}{x \plus{} y}$

Then either $ x$ or $ y$ must be complex.

First I crossed multiplied giving $ x^2 \plus{} 2xy \plus{} y^2$ $ \equal{}$ $ xy$

If $ x$ and $ y$ are both real then $ x^2 \plus{} y^2$ $ > 0$ leaving us with $ 2xy \equal{} xy$ which is obviously false, thus either $ x$ or $ y$ must be complex.

Actually, when you get to the point about x^2 + y^2 > 0 a few things:

1) Not to be nitpicky, but that should be $ \ge$.
2) When you cancel them out you must change the equal sign to $ \le$

So in the end you get $ 2xy\le xy$, or $ xy\le 0$, which doesn't prove anything.

Instead, after cross multiplying, I would:

[hide="Hint"]Move all terms to one side, and solve for a variable.[/hide]

The last step: It doesn't leave it with $ 2xy \equal{} xy$, it leaves it with $ 2xy < xy$

Also, that statement isn't true: $ x^2 \plus{} y^2$ can also equal zero. Thus, the proper conclusion is $ 2xy \le xy$

EDIT: ahh, beat me to it.

[quote="dragon96"]

Instead, after cross multiplying, I would:

[hide="Hint"]Move all terms to one side, and solve for a variable.[/hide][/quote]

(The hint -- I knew this also before reading it.)

You can get to this formula, which is symmetric in x and y:

$ x^2 \plus{} xy \plus{} y^2 \equal{} 0.$

If you solve for y, you get:

$ y \equal{} \frac { \minus{} x \pm |x|\sqrt { \minus{} 3}}{2}$

For this expression to be equal to a real number, the only value
that x could be is 0 so that when its absolute value is multiplied
by the square root of negative three, that result would be zero.

However, x cannot be zero, because that would show division by zero
from the given problem.  If x can't equal zero, then y can't equal zero
because of it.  Otherwise, you're left with a complex number from
any other choice of x.

Because of the symmetry in the equation, solving for x, you get:

$ x \equal{} \frac { \minus{} y \pm |y|\sqrt { \minus{} 3}}{2}$

EDIT:

[QUOTE]
As with the x-values, here, you're going to have only complex values
with the y-values.

Then the statement in the original problem post "Then either $ x$ or $ y$
must be complex." can be made stronger:

Then both $ x$ and $ y$ must be complex.

[ENDQUOTE]

EDIT:  What if have in the QUOTED section is FALSE.  Only one of the variables
need be real, and this was pointed out after this post by user Yongyi781.

[hide="Just for fun"]
Interpret as sides of a triangle. If $ x,y$ are positive, then we have the third side of a triangle with one angle $ 2\pi/3$ (and hence nondegenerate) is zero; contradiction. If $ x,y$ are negative, then let $ a\equal{}\minus{}x,b\equal{}\minus{}y$ and use the same logic. If one of $ x,y$ is negative (wlog $ y$) let $ y'\equal{}\minus{}y$ and use the same logic but with a $ \pi/3$ angle. [/hide]

When you get to $ (x \plus{} y)^2 > xy$, the finish is simple, however, we have to prove a trivial claim first.

[hide="Neither x or y can be 0"]
Clearly $ x\neq 0$, and then if $ y\equal{}0$ we have $ 1\equal{}0$, a false statement. Therefore neither $ x$ or $ y$ is 0.[/hide]
[hide="Finish"]
The following holds for all real $ x$ and $ y$:
\[ (x \plus{} y)^2 > xy.\]
If exactly one of $ x$ or $ y$ is negative, then $ (x \plus{} y)^2$ is positive but $ xy$ is negative, and the result holds. If both $ x$ and $ y$ are positive, then the expression is equivalent to $ x^2 \plus{} xy \plus{} y^2 > 0$. All terms on the LHS are positive and so the result holds. If both are negative, then replacing $ x$ with $ \minus{} x$ and $ y$ with $ \minus{} y$ makes them both positive but does not change anything, therefore the result holds.

Therefore, either $ x$ or $ y$ must be complex.
[/hide]

Also, Arrange your tan, it is not true that both of them have to be complex. Just because solving for $ x$ yields a complex number for real $ y$ and solving for $ y$ yields a complex number for real $ x$ does not mean that both have to be complex. For example, let $ x \equal{} 1$ and solve for $ y$, and we get $ y \equal{} \frac { \minus{} 1\pm i\sqrt 3}2$, and we have $ x$ is real and $ y$ is complex.

Define $ r \equal{} \frac {x}{y}$

$ x^2 \plus{} 2xy \plus{} y^2 \equal{} xy \iff \frac {x}{y} \plus{} 2 \plus{} \frac {y}{x} \equal{} 1$
$ \implies r \plus{} 1 \plus{} \frac {1}{r} \equal{} 0 \iff 0 \equal{} r^2 \plus{} r \plus{} 1 > (r \plus{} 1/2)^2$

Since $ r$ is clearly nonreal, we cannot have $ x$ and $ y$ both reals.

[quote="Yongyi781"]

Also, Arrange your tan, it is not true that both of them have to be complex. Just because solving for $ x$ yields a complex number for real $ y$ and solving for $ y$ yields a complex number for real $ x$ does not mean that both have to be complex. For example, let $ x \equal{} 1$ and solve for $ y$, and we get $ y \equal{} \frac { \minus{} 1\pm i\sqrt 3}2$, and we have $ x$ is real and $ y$ is complex.[/quote]

I made an edit in my post to reflect your correction of my post.

[hide="my solution"]
Supose that exist real number $ (x,y)$ such that $ \frac{x\plus{}y}{x} \equal{}\frac{y}{x\plus{}y}$ .

Then :

$ x^2 \plus{}2xy \plus{}y^2 \equal{}xy \implies x^2 \plus{}y^2 \equal{}\minus{}xy$

Since the numbers $ x^2$ and $ y^2$ are positive we can apply AM-GM inequality ,

$ x^2 \plus{}y^2 \geq 2xy \implies \minus{}xy \geq 2xy \implies xy \leq 0$

Easily we see that they're both different from $ 0$ ,so one of the numbers $ x,y$ need to be negative.

WLOG $ x$ is negative and $ y$ is positive.

If $ |x| \geq |y|$ then $ LHS$ is positive and $ RHS$ is negative ,contradiction.

If $ |x| \leq |y|$ then $ LHS$ is negative and $ RHS$ is positive,contradiction.[/hide]

[hide="A different but a bit advanced perspective solution"]We already know that $ x\ne0$ and $ y\ne0$. Also, we may easily deduce that $ x\ne y$. Cross-multiply to obtain $ x^2 \plus{} xy \plus{} y^2 \equal{} 0$. Since $ x\ne y$, multiply both sides by $ x \minus{} y$ to get $ x^3 \minus{} y^3 \equal{} 0$. Thus, $ x \equal{} \omega y$ or $ x \equal{} \omega^2 y$, where $ \omega,\omega^2$ are the two primitive cube roots of unity. $ x$ is real only when $ y$ is the conjugate of the respective primitive roots; otherwise, $ x$ is complex for all other $ y\in\mathbb{C}$.[/hide]
edit: minor typos


polynomial
number theory proposed
number theory

Give the integer equation: [tex] a^2-ab+b^2=c^2 (a,b>1,a \neq b) [/tex] (1).
i)Prove that there exists $a,b,c$ satisfying (1).
ii)If $a$ is a prime,prove that $a<b$ :)

Okey,forget both that silly question :blush: ,we just need do one thing:
iii)Solve (1) :)

I'm not a expert in eisenstein integers  :( , but I'll try a solution:

let k = gcd(a ; b) so k | c . In this case we have: a = xk ; b=yk and c=zk

[tex]x^2 - xy + y^2 = (x + y\omega)(x + y \omega^2) = z^2[/tex] where [tex] \omega = -\frac{1}{2} + \frac{i\sqrt3}{2}[/tex]
now let [tex]gcd(x + y\omega ; x + y\omega^2) = d [/tex], then [tex]d | y(\omega^2 - \omega)[/tex] and
[tex]d | x(\omega - 1)[/tex] and how [tex]gcd(x, y) = 1[/tex] we have that d = 1 also!!
ence [tex]x + y\omega = (m + n\omega)^2 = m^2 - n^2 + (2mn - n^2)\omega[/tex] i.e the solution are:

[tex]a = (m^2 - n^2)k[/tex]
[tex]b = (2mn - n^2)k[/tex]
[tex]c = (m^2 - mn + n^2)k[/tex]
so, if a is prime is easy to see that b > a.


"Thanks for solving my problems. Would like to ask u if you would be interested to help me with some more quizzes. I have about 15 quizzes, about 5 questions each, this semester. Can pay u through paypal. Half before solving, half after.

my email: stefanstefan@hushmail.com

regards"

Is this sort of thing common?

I strongly recommend to find out why he offers money for doing this.

[quote="JBL"]I vote you take the first half, then tell him to do his own homework  :P[/quote]

Aw you stole my post.

These quizzes might even be competition related so you might help him cheat his way through.

I don't intend on doing it. But I was wondering whether it is common (nor do I intend on taking any money, so his anybody's (if they want him))

Is it common?

Once in a while I recieve e-mails from people who want me to collaborate on old contest problems. Personally, I have never experianced any monetary reward in solving problems.

[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=148707]Here[/url] is what I did the last time something like this happened to me (although the offender in that case was particularly dimwitted, and also posted his request on the forum rather than PMing me).


geometry
incenter
circumcircle
geometry unsolved

$ABCD$ is a quadrangle,and the lines $AC$ and $BD$ meet at the point $O$.The points $X,Y,Z$ and $W$ are the incenters of triangles $AOB,BOC,COD$ and $DOA$.

Prove that quadrangle $ABCD$ has a circumscribed circle $\iff$  quadrangle $XYZW$  is cyclic.


For $x,y > 0$, we have $\displaystyle \frac{1}{x} + \frac{1}{y} \geq  \frac{4}{x+y}$, so

(1) $\displaystyle \frac{a}{ab+1}+\frac{a}{ca+1} \geq \frac{4a}{(ab+1) + (ca+1)} = \frac{4a}{3-bc}$

Similarly, we have

(2) $\displaystyle \frac{b}{ab+1}+\frac{b}{bc+1} \geq  \frac{4b}{3-ca}$

(3) $\displaystyle \frac{c}{bc+1}+\frac{c}{ca+1} \geq  \frac{4c}{3-ab}$

Adding (1), (2) and (3) together and we are done.


calculus
derivative
integration
logarithms
inequalities
trigonometry
real analysis

Find asymptotical expression of the series when $ t\to 1-0$
$ S(t)=\sum\sigma (n)t^{n}$
where $ \sigma (n)$ is the sum of the division of the $ n$
I mean if
$ n=p_{1}^{n_{1}}p_{2}^{n_{2}}p_{3}^{n_{3}}...p_{k}^{n_{k}}$
then
$ \sigma (n) = \sum p_{j_{1}}^{n_{j_{1}}}p_{j_{2}}^{n_{j_{2}}}...p_{j_{k}}^{n_{j_{k}}}$

It seems very few are interested in this.
Here is my answer:
The original $ S(n)$ can easily be reduced to $ \sum_{n=1}^{\infty}\frac{nt^{n}}{1-t^{n}}$.
Using derivative to investigate the summand,we find that it's decreasing with $ n$.
So $ \int_{1}^\infty{\frac{yt^{y}}{1-t^{y}}}dy\leq \sum_{n=1}^{\infty}\frac{nt^{n}}{1-t^{n}}\leq \int_{1}^\infty{\frac{yt^{y}}{1-t^{y}}}dy+\frac{t}{1-t}$.
To evaluate $ \int_{1}^\infty{\frac{yt^{y}}{1-t^{y}}}dy$,we use this substitution:$ x=1-t^{y}$.Then it becomes $ -\frac{1}{(\log t)^{2}}\int_{1-t}^{1}{\frac{\log (1-x)}{x}}dx$.
Expanding the integand in powers of $ x$,we have $ \frac{\log (1-x)}{x}=-(1+\frac{1}{2}x+\frac{1}{3}x^{2}+\cdot\cdot\cdot)$ and it follows that the integral is approximately $ \frac{\pi^{2}}{6}\frac{1}{(\log t)^{2}}$ when $ t\to 1-$.Finally it is deduced from the above inequality that
$ S(t)=\frac{\pi^{2}}{6}\frac{1}{(\log t)^{2}}+O(\frac{1}{|\log t|})$.

Yes.
your first step was right
indeed you must find only
$ \sum\frac{n t^{n}}{1\minus{}t^{n}}$ when $ t\to\minus{}1$
or ...
 you must find
$ \int_{0}^{1}\frac{ln x}{1\plus{}x}dx$
now if you find this integral
$ \int_{0}^{1}\frac{ln x}{1\minus{}x}dx$ then by this substitution $ x\to x^{2}$
you can find this $ \int_{0}^{1}\frac{ln x}{1\plus{}x}dx$
you must represent $ (1\minus{}x)^{\minus{}1}$ in power series and then do inegration
and also use this
$ I_{N}\equal{}\sum_{n>N}\frac{1}{n^{2}}\equal{}\frac{4}{\pi}\frac{(2N)!!}{(2N\minus{}1)!!}\int_{0}^{\frac{\pi}{2}}x^{2}\cos^{2N}xdx$

some correction to me  :)
$ \sum\sigma (n) t^{n}\equal{}\sum\frac{nt^{n}}{1\minus{}t^{n}}$ (when $ |t| < 1$. because if we represent our sum as geometrical progression then $ \sum b_{n}t^{n}$, $ b_{n}$ will be $ \sigma (n)$)
$ \sum\frac{nt^{n}}{1\minus{}t^{n}}\to\int_{1}^{\plus{}oo}\frac{yt^{y}}{1\minus{}t^{y}}dy \equal{}\frac{1}{\ln^{2}t}\int_{0}^{1}\frac{\ln x}{1\minus{}x}dx$
(where $ x \equal{} t^{y}$)
$ \frac{1}{\ln^{2}t}\int_{0}^{1}\frac{\ln x}{1\minus{}x}dx \equal{}\frac{1}{\ln^{2}t}\int_{0}^{1}\sum_{i \equal{} 0}^{\plus{}oo}x^{i}\ln xdx$
last we can integration term by term because :
$ |\int_{0}^{1}(\sum_{n\geq N}x^{n}\ln x)dx| \equal{}\int_{0}^{1}\frac{x^{N}|\ln x|}{1\minus{}x}dx\leq\int_{0}^{1}x^{N\minus{}1}dx \equal{}\frac{1}{N}$


On the plane of triangle $ABC$ with $\angle BAC = 90^\circ$ we raise perpendicular lines in $A$ and $B$, on the same side of the plane. On these two perpendicular lines we consider the points $M$ and $N$ respectively such that $BN < AM$. Knowing that $AC = 2a$, $AB = a\sqrt 3$, $AM=a$ and that the plane $MNC$ makes an angle of $30^\circ$ with the plane $ABC$ find

a) the area of the triangle $MNC$;

b) the distance from $B$ to the plane $MNC$.

Before I post a solution, can you tell me if my answer is right?
[hide]a) $[MNC]=\frac{3a^2}{2}$[/hide]

Could anyone post a figure .I can't understand the problem . :maybe:

[hide="Part 1"]$\triangle ABC$ is the orthogonal projection of $\triangle MNC$, hence $[ABC]=[MNC]\cos\angle(ABC,MNC)$, which yields $[MNC]={(2a\cdot a\sqrt{3})/2\over\cos 30^\circ}=2a^{2}$[/hide]

Indeed, that's correct.

[hide="For the second part, we can use vectors."]
Let $A(0,0,0), \, B \left( 0, a \sqrt 3, 0 \right), \, C \left( 2a, 0, 0 \right), \, M \left( 0, 0, a \right), \, N \left( 0, a \sqrt 3, n \right)$.

Take $\overrightarrow v = x \overrightarrow i+y \overrightarrow j+z \overrightarrow k$ to be a vector $\perp$ to $(MNC)$, i.e. $\overrightarrow v \cdot \overrightarrow{CN}= 0, \, \overrightarrow v \cdot \overrightarrow{CM}= 0$. Solving this system we get a family of solutions, from which we pick one: $\overrightarrow v = \frac{a \sqrt 3}2 \cdot \overrightarrow i+(a-n) \overrightarrow j+a \sqrt 3 \cdot \overrightarrow k$.

The condition $\measuredangle \left( (MNC), (ABC) \right) = 30^\circ$ transforms into $\measuredangle \left( \overrightarrow v, \overrightarrow k \right) = 30^\circ$, i.e. $\cos 30^\circ = \frac{\overrightarrow v \cdot \overrightarrow k}{\left| \overrightarrow v \right| \cdot \left| \overrightarrow k \right|}$. Solving this, we get that either $NB = \frac{a}2$ or $NB = \frac{3a}2$. Since $NB < AM = a$, it follows that $NB = \frac{a}2$.

Now consider $\textrm{vol}\left( BMNC \right) = \frac{\left[ MNC \right] \cdot d \left( B, (MNC) \right)}3 = \frac{\left[ BMN \right] \cdot d \left( C, (BMN) \right)}3$.
It's easy to compute $d \left( B, (MNC) \right)$ from this relation.
[/hide]


The AIME II Math Jam is Friday, April 3 at 7 PM ET (4 PM PT).  We will cover all 15 problems.

how long will it be; somewhere around the lines of 2 hours?

As long as it takes; the last one was nearly 3 hours.

Will this math jam finish before 9:00 P.M?

Depends on what time zone you live in. It will probably end at about 9:00 if you live in the Central Time Zone.

That session was hard, but helpful/
 :spam:  :first:  :starwars:


What can you say about the void relation , it is :
[b](a)[/b] reflexive
[b](b)[/b] transitive
[b](c)[/b] symmetric.

Give reason with your answer. Thanks a lot for your help.

I suspect that it satisfies all three vacuously.

how so? i'm not getting this too  :(  :blush:


Five foreigners arrive in a restaurant. They are given a menu with nine choices. Each makes a different selection without knowing what he is going to eat (for example, on one particular night, they could select dish A, B, C, D, and E), and the waiter brings the five dishes to the table.

They are eventually able to deduce which dishes are which. What is their strategy, and what is the minimum number of meals they must eat at that restaurant to do so?

Generalize.

[hide="um is this reasonable?"] the food is served so it looks like the letter a, b, c, etc.?[/hide]

this was the only answer i couldve gotten. i didnt understand the question very much  :ninja:

[quote="D3m0n Shad0w"]the food is served so it looks like the letter a, b, c, etc.?[/quote]

Sorry. I used the letters A, B, C, etc. to represent menu selections that the foreigners cannot read. I could have called the foods asldjfksdlf, ewoityweiuoypbiodf, fiojuiofawyioey, iobdfuibopyweir, biopieorhjwielkrn, etc.

It's not a lateral thinking puzzle, it's a straight logic question.

[hide="Answer"]3
ADxxx
BDxxx
Cxxxx[/hide]

[hide] they could order ABCDE, and then the next day get ABCDF, so then they'd know which ones E and F are.  then they order EFGHA, so they know what A is (it's the only one that they order all three times).  then they get ABEFG to figure out B and G.  and then they'd also know what H is.  then they get ABCEI to figure out the other ones.

in summary:
ABCDE
ABCDF
EFGHA
ABEFG
ABCEI

so it takes 5 meals.  i don't know if this is the strategy with the lowest amount though.[/hide]

The actual problem in the book is a little different. It gives the minimum number of meals as [hide]three[/hide] and then asks you to find which dishes they ordered on each night.

[hide]
1) ABCDD
2) AEFGG
3) BEHII

How it is deduced:
A: Only dish ordered on both days 1 and 2.
B: Only dish ordered on both days 1 and 3
C: Only dish that was ordered on only day 1 once
D: Only dish that was ordered on day 1 twice
E: Only dish that was ordered on both days 2 and 3
F: Only dish that was ordered on only day 2 once
G: Only dish that was ordered on day 2 twice
H: Only dish that was ordered on only day 3 once
I: Only dish that was ordered on day 3 twice[/hide]


algebra
polynomial
superior algebra
superior algebra solved

A polinomial P(x) of degree $  \geq 2  $ has n real distinct roots,$ x_1, x_2,....,x_n $. Prove that:
\[
\displaystyle\sum_{k = 1}^n\frac{1} {x^{2}_k}= \displaystyle\frac{(\frac {d} {dx} P(0))^2 - P(0)\frac{d^2} {dx^2} [P(0)]} {P(0)^2}
\]

Or, write $P\left(  x\right)  =a\left(  x-x_{1}\right)  \left(  x-x_{2}
\right)  \ldots\left(  x-x_{n}\right)  .$ Then
\[
\frac{P^{\prime}\left(  x\right)  }{P\left(  x\right)  }=\sum\frac{1}{x-x_{k}}
\]
and
\[
\frac{P^{\prime\prime}\left(  x\right)  P\left(  x\right)  -\left(  P^{\prime
}\left(  x\right)  \right)  ^{2}}{P^{2}\left(  x\right)  }=-\sum\frac
{1}{\left(  x-x_{k}\right)  ^{2}}.
\]
Plugging $x=0$ gives the result.


Are the letters bound to you? (As in, if you get an o that came from Awesome's name, would it be counted for you?

[quote="toadoncart"]Toadoncart and Awesome are playing a game. Toadoncart places all of the letters of his name and all of the letters of Awesome's name (even if the same letter is used twice) into a bowl. If he randomly grabs four letters out of the bowl, what is the probability that he could be able to match the four letters such that two letters are part of his name and two letters are part of Awesome's name? (Notice that if he grabs A, O, A, O, he could give the AO to himself and the AO to Awesome but would only be counted once).  :)[/quote]

hmmm.......lets see...there are a total of 10+7=17 letters. 17 choose 4 = 7 x 5 x 4 x 17 ways.
The common letters are Toadoncart: o,a,o,a..AND..Awesome: a,o......Now the problem looks easier!
Since awesome has only 2 common letters, you only have to consider toadoncart. There are 2 x 2= 4 ways to get a,o from toadoncart's name. So the probability is 4/(7 x 5 x 4 x 17) =  1/ 595.
So..I guess thats the answer..correct me if i am wrong..


In a convex quadrilateral. eight segments are drawn, each of them connects a vertex with the midpoint of some opposite side. Seven of these segments have the same length $ a$. Prove that the eight one is also of length $ a$.

Let $ ABCD$ be the convex quadrilateral and $ X,Y,Z,W$ the midpoints of the segments $ AB, BC, CD, DA$, respectively.
Consider the lengths $ AB\equal{}2x$, $ BC\equal{}2y$, $ CD\equal{}2z$, $ DA\equal{}2w$, $ AC\equal{}2s$, $ BD\equal{}2t$.
By the median theorem for the median $ CX$ in triangle $ ABC$ we have
$ 2s^{2}\plus{}2y^{2}\equal{}x^{2}\plus{}(CX)^{2}$
By the same theorem for the medians $ DY,AZ,BW$ in triangles $ BCD,CDA,DAB$, respectively,
we obtain equations wich sum (with the previous one too) is
$ x^{2}\plus{}y^{2}\plus{}z^{2}\plus{}w^{2}\plus{}4s^{2}\plus{}4t^{2}\equal{}(CX)^{2}\plus{}(DY)^{2}\plus{}(AZ)^{2}\plus{}(BW)^{2}$
In the same way, we can obtain
$ x^{2}\plus{}y^{2}\plus{}z^{2}\plus{}w^{2}\plus{}4s^{2}\plus{}4t^{2}\equal{} (DX)^{2}\plus{}(AY)^{2}\plus{}(BZ)^{2}\plus{}(CW)^{2}$


Let x, y, z are real numbers and 3x+2y+z=1
What is the maximum value of  (1/1+lxl)  + (1/1+lyl)   +  (1/1+lzl)    ??
help!![/i][/quote]


[quote="didilica"]Evaluate the two limits where $a$ and $c$ are positive real numbers.

$\lim_{n\rightarrow \infty}\frac{(n+a)^{\frac{n+a}{n}}-n^{\frac{n+a}{n}}}{(n+c)^{\frac{n+c}{n}}-n^{\frac{n+c}{n}}}$

and

$\lim_{n\rightarrow \infty}\frac{(n+a)^{\frac{1}{n}}-n^{\frac{1}{n}}}{(n+c)^{\frac{1}{n}}-n^{\frac{1}{n}}}$[/quote]
transform by next: $e^{y}$ and then use the "beautifle limit". in the seccond you get: $a/c$
$\frac{e^{ln(n-a)/n}-e^{ln(n)/n}}{e^{ln(n-a)/n}-e^{ln(n)/n}}$=$\frac{ln(n+a)/n-ln(n)/n}{ln(n+c)/n-ln(n)/n}$=$\frac{a/n}{c/n}$=a/c.
analogysli in seccond


If Nonie could run a 7:55 mile at the beginning of the track season and she wanted to improve 10 seconds per meet and has had 4 meets, how fast should she be now?
[size=59](yeah simple question--I didnt exactly meet my goal cuz now I run a 7:34--yes I'm slow....still have tomorrows meet though)[/size]

[hide]she should be 7:15

$7: 55-: 10*4=7: 15$

BTW, i  ran the mile in 7:13 the latest fitness test and i was out of practice. :lol:[/hide]

I play cricket, and I don't run track...plus I'm only 12...so I run the mile at 8:26.

Anyway,

[hide]Nonie improves 10 seconds each time, and she originally had 55 seconds. She got 4 track meets, so 10*4=40. Subtract from 55. We get 7:15. Whoa, that's fast.[/hide]

my friend, who's 13 or 14 years old, can run it under 5 minutes.


A bottle water company has designed a new cup for its dispenser. The cup will be a right circular cone with a three-inch radius. How tall will the cup need to be to hold 93 cubic inches of water? Express your answer to the nearest whole number.

V = 1/3pi(r^2)(h)
  93 = 1/3pi (9)(h)
Simplify to get 31/pi = h. The answer is 10.


The number of distinct solutions of the equation $\big|x-|2x+1|\big| = 3$ is

$\textbf{(A) }0\qquad
\textbf{(B) }1\qquad
\textbf{(C) }2\qquad
\textbf{(D) }3\qquad
\textbf{(E) }4$

[hide]We break the equation up into cases. Either the equation equals 3 or -3. Then we consider the 2x-1 part if it's negative or positive.

If I didn't screw up (0.1% chance of that), $x=\frac{2}{3}, -4, -\frac{4}{3}, 2$. Checking each into the original equation, the last two numbers work. There are 2 solutions. [/hide]

[hide="Answer"]$|2x+1|$ is always greater than $x$, so $|x-|2x+1||<0$, and we have $x-|2x+1|=-3$, and therefore either $x-(2x+1)=-(x+1)=-3$ or $x+(2x+1)=3x+1=-3$, which gives two solutions, so our answer is $\boxed{C}$.[/hide]

[hide]Break the first absolute sign will give us two equations:

$x-|2x+1|=3$  and that gives $x\ge3$

$x-|2x+1|=-3$

Keep on breaking the second absoluate sigh and solve for $x$

The first equation has two roots: $x=-4, x=\frac{2}{3}$, both are omitted due to restriction $x\ge3$

The second equation has two roots: $x=2, x=\frac{-4}{3}$ which work

So the total number of solutions are $2$

$\boxed{C}$
[/hide]


calculus
integration
real analysis
real analysis unsolved

Hi again guys,

Working my way through some more little problems and i'm totally baffled with this one.

[url]http://putfile.com/pic.php?pic=main/8/21718293787.jpg&s=f10[/url]

Could anybody help me out with it? Give me some sort of idea what the graph will look like? - Are there any online graph plotting programmes?

This has nothing to do with graphs.  Simpson's rule is a method for numerical integration which can be found in any decent calculus textbook.  Plug in the numbers.

It says "compare this with the solution obtained analytically"...

I've had a look into it but as our maths teacher was infact a Media teacher.. i don't really understand it very well.

The analytic solution is what you get when you compute that definite integral using the fundamental theorem of calculus.


calculus
integration
LaTeX
logarithms
real analysis
real analysis unsolved

Hello,

I was woundering if anyone knew how to do the integral of:

(e^x)(ln x) dx[/code]

Does such an antiderivative exist?  I tried parts and substitution and didn't get anywhere; I doubt an elementary antiderivative exists.

Also, since you're new here, you may want to learn $\LaTeX$.  Your problem looks a little nicer when written as

\[ \int e^{x}\ln x\,dx. \]

I agree with you.

For $\int (x+1)e^x \ln x dx$ we can calculate.

If you put that integral into Mathematica's online Integrator (http://integrals.wolfram.com/index.jsp), it would spit out

$e^x\ln(x) - Ei(x)$

implying that your original integral was nonelementary.


function
logarithms
calculus
derivative
calculus computations

Expand the following function upto 4th power of x by using -
1. series expansion of logarithmic function,
2.Maclaurin's series formula.

 f(x)  =   log[(log(1+x))/x]

If any other method exist please post it .

Thanks.

This is pretty messy.

$ \log(1 \plus{} x) \equal{} x \minus{} \frac {x^2}2 \plus{} \frac {x^3}3 \minus{} \frac {x^4}4 \plus{} \frac {x^5}5 \plus{} \cdots,$ so

$ \frac {\log(1 \plus{} x)}x \equal{} 1 \minus{} \frac x2 \plus{} \frac {x^2}3 \minus{} \frac {x^3}4 \plus{} \frac {x^4}5 \plus{} \cdots$

$ \equal{} 1 \plus{} u,$ where $ u \equal{} \minus{} \frac x2 \plus{} \frac {x^2}3 \minus{} \frac {x^3}4 \plus{} \frac {x^4}5 \plus{} \cdots.$

The logarithm of that is $ u \minus{} \frac {u^2}2 \plus{} \frac {u^3}3 \minus{} \frac {u^4}4 \plus{} \cdots.$

$ u \equal{} \minus{} \frac x2 \plus{} \frac {x^2}3 \minus{} \frac {x^3}4 \plus{} \frac {x^4}5 \plus{} \cdots$

$ \minus{} \frac {u^2}2 \equal{} \minus{} \frac12\left( \minus{} \frac x2 \plus{} \frac {x^2}3 \minus{} \frac {x^3}4 \plus{} \cdots\right)^2 \equal{} \minus{} \frac {x^2}8 \plus{} \frac {x^3}6 \minus{} \frac {13x^4}{72} \plus{} \cdots$

$ \frac {u^3}3 \equal{} \frac13\left( \minus{} \frac x2 \plus{} \frac {x^2}3 \plus{} \cdots\right)^3 \equal{} \minus{}\frac {x^3}{24} \plus{} \frac {x^4}{12} \plus{} \cdots$

$ \minus{} \frac {u^4}4 \equal{} \minus{} \frac14\left( \minus{} \frac x2 \plus{} \cdots\right)^4 \equal{} \minus{} \frac {x^4}{64} \plus{} \cdots$

Adding those up, we get

$ \log\left(\frac {\log(1 \plus{} x)}x\right) \equal{} \minus{} \frac x2 \plus{} \left(\frac13 \minus{} \frac18\right)x^2 \plus{} \left( \minus{} \frac14 \plus{} \frac16 \minus{} \frac1{24}\right)x^3 \plus{} \left(\frac15 \minus{} \frac {13}{72} \plus{}\frac1{12}\minus{} \frac1{64}\right)x^4 \plus{} \cdots$

$ \equal{} \minus{} \frac x2 \plus{} \frac {5x^2}{24} \minus{} \frac {x^3}{8} \plus{} \frac {17x^4}{90} \plus{} \cdots$

... or something like that. There's probably only about a 20% chance that I got the arithmetic right. Someone who uses a CAS can check this.

And no, I don't want to compute the fourth derivative of that function.

In your second step there is (x^2)/3 and not (x^2)/2.
Also, in the ans coeff. of x^3 is  +1/8 and not -1/24.
Still thanks for the nice approach.
:coolspeak:

I fixed that $ \frac {x^3}3$ typo (which wasn't actually in the computation) and I fixed a couple of more mistakes. I now get $ \minus{} \frac {x^3}8.$


1. I am reading a book and it kept repeating the word arbirary. For example, let p be an arbitrary element of $A \times (B \cap C)$. What does arbitrary means? [b]Sorry, but English is just my second language, and I can't relate when I look at the dictionary.[/b]
Questions on Proofs:
I have answers to these proof questions but its still a little vague for now. I am just starting to do these things.

2. Suppose that A and B are sets, then prove that $A \times B = B \times A$, iff $A = \oslash$, $B = \oslash$ or $A = B$.

3. Suppose R is a partial order on a set $A$ and $B \subseteq A$, if $B$ has a smallest element say $b$, then prove that $b$ is unique.

[b]I am not sure if I am putting this on the right forum so please move it if isn't. Thanks.[/b]

1. "Let $p$ be an arbitrary element of $A\times (B\cap C)$" $\iff$ "$p\in A\times (B\cap C)$ and there are no restrictions for choosing $p$".
2. If $A$ or $B$ is empty, we have $A\times B=B\times A$. Let's suppose that $A,B$ aren't empty. We will prove that $A$ is included in $B$, proof of statement that $B$ is included in $A$ is analogous.
$A\subset B\iff \forall_{a\in A}a\in B$. Let's suppose that $a\in A$. Then there exists $b\in B$ that $(a,b)\in A\times B$. But $A\times B=B\times A$, so also $(a,b)\in B\times A$. It means that $a\in B, b\in A$, in particulary $a\in B$ which ends the proof.


Note that $ 2l \plus{} 2w \equal{} P$. Therefore, $ l \plus{} w \equal{} P \div 2$. Since the perimeter halved is 150, we know the length and width added together yields that result. Since the width is 1/6 of the perimeter, it is 50, and so the length is $ 150 \minus{} 50 \equal{} 100$.

Perimeter is the sum of all its sides.

So, 2(l+b) = 300
6b = 300. Therefore b = 50.
l+b = 300/2 = 150.
l = 150 - 50 = 100.

Length = 100m.

[quote="sri340"]The perimeter of a rectangular field is 6 times its width. If the perimeter is 300 m, what is the length of the field?[/quote]
all you need to do is find the width first. set x as the width and 6x=300 x=50 since the width is 50 to find the perimeter just multiply it by 2 because there are two sides of 50.  Subtract that from 300 then divide that by 2 because there are two lengths so you should get 100 as the length


[hide]a) coefficents respectively: 2, 7, 4, 6

b) coefficents respectively: 1, 3, 2, 3[/hide]

[url=http://www.chembuddy.com/?left=balancing-stoichiometry&right=balancing-reactions]Balancing chemical equations[/url] - lectures with examples.

Could someone show me how they got those answers?

Here's a hint, try to balance the C's and H's first, then the O's...

Nope, i couldn't get it working to the solutions you gave. Could you do one of them for me then i'll do the other on my own?

i'm not sure if this is correct, but this is where i got upto

C: 2x(1) = 2x(3)
H: 6x(1) = 2x(4)
O: 2x(2) - 2x(3) - x(4) = 0

Is that correct?

If that is, i'm having a problem then working that out with reducing to row echelon form and getting the values. Please help

Looks like you are trying [url=http://www.chembuddy.com/?left=balancing-stoichiometry&right=algebraic-method]algebraic method[/url].

I'll replace (1) with A, (2) with B, (3) with C and (4) with D.

Your equations take form:
2A = C
6A = 2D (or 3A = D)
2B - 2C - D = 0

Substituting C and D from the first two equations into the last one you get:

2B - 4A - 3A = 0

or

2B = 7A (or B=7/2 A)

There was not enough equations (which is a standard situation, as described in the lecture), so we have to assume some value for one of the coefficients. Let's start with A - from the last equation it is obvious that A must be even in order for B to be an integer. So
A = 2
B = 7
C = 2A = 4
D = 3A = 6

These are values already posted by g1sk8terz.

[quote="coolio2005"]Hi, could someone help me with the following please:

6. For each of the following chemical reactions, determine the balanced chemical equation:

(a) reactants C(2)H(6) and O(2); products CO(2) and H(2)O
(b) reactants C(2)H(5)OH and O(2); products CO(2) and H(2)O[/quote]

Balanced reactions are:
2C2H6 + 7O2 = 6H2O + 4CO2
C2H5OH + 3O2 = 3H2O + 2CO2
I have used this tool  [url=http://www.webqc.org/balance.php]www.webqc.org/balance.php[/url]
to balance them in one click:)

nah you shouldn't use online help because you won't have it when you're taking a test

For simple equations, a little trial and error works best. I'll do part a.
first, write out the unbalanced equation.
$C_2H_6+O_2{\rightarrow}CO_2+H_2O$
For this to be balanced, it needs the same number of each element on both sides.
try to deal with the molecules with bigger formulas first.
In $C_2H_6$, there are, of course, 2 carbons and 6 hydrogens. Make it the same on the other side.
$C_2H_6+O_2{\rightarrow}2CO_2+3H_2O$
Now count up the number of oxygens on the right side. There are 7. That means we need 3.5 $O_2$ molecules. So, express the equation like this.
$C_2H_6+3.5O_2{\rightarrow}2CO_2+3H_2O$
We cannot have a half a molecule, so just multiply the whole thing times two. This yields:
$2C_2H_6+7O_2{\rightarrow}4CO_2+6H_2O$
The more you do this, the less you have to think about the steps.

g1sk8terz is correct. try C and anything else, then H and at last O and then C etc


geometry
inequalities
search
trigonometry
geometry unsolved

Let $ ABC$ is a triangle. Let $ M$ is a change point in pland. Let $ AB\equal{}c, BC\equal{}a, CA\equal{}b$.

Prove that: $ \frac{MA}{a}\plus{}\frac{MB}{b}\plus{}\frac{MC}{c} \ge \sqrt{3}$

Use $ \frac {MB.MC}{b.c} \plus{} \frac {MC.MA}{c.a} \plus{} \frac {MA.MB}{a.b}\geq 1$. This inequality is somewhere in the forum but I couldn't find it. I'm sure you can :)

Can you find it for me. If it can be done by use this problem, so it's nice.

[quote="thanhnam2902"]Let $ ABC$ is a triangle. Let $ M$ is a change point in pland. Let $ AB \equal{} c, BC \equal{} a, CA \equal{} b$.

Prove that: $ \frac {MA}{a} \plus{} \frac {MB}{b} \plus{} \frac {MC}{c} \ge \sqrt {3}$[/quote]
Let : $ AM\equal{}p, BM\equal{}q,CM\equal{}r$ and $ \angle BMC\equal{}\psi_1,\angle CMA\equal{}\psi_2,\angle AMB\equal{}\psi_3$
$ \cos(\psi_1)\plus{}\cos(\psi_2)\plus{}\cos(\psi_3) \ge\minus{}3/2$ ;    $ (\psi_1\plus{}\psi_2\plus{}\psi_3\equal{} 2\pi)$

$ a\equal{} \sqrt{q^2\plus{}r^2\minus{}2qr\cos(\psi_1)}$ , $ b\equal{} \sqrt{r^2\plus{}p^2\minus{}2rp\cos(\psi_2)}$ , $ c\equal{} \sqrt{p^2\plus{}q^2\minus{}2pq\cos(\psi_3)}$

$ \frac {MA}{a} \plus{} \frac {MB}{b} \plus{} \frac {MC}{c}\equal{}\frac{p}{\sqrt{q^2\plus{}r^2\minus{}2qr\cos(\psi_1)}}$$ \plus{}\frac{q}{\sqrt{r^2\plus{}p^2\minus{}2rp\cos(\psi_2)}}\plus{}\frac{r}{\sqrt{p^2\plus{}q^2\minus{}2pq\cos(\psi_3)}}\ge$
$ (p\plus{}q\plus{}r)^{3/2}\frac{1}{\sqrt{pq^2\plus{}pr^2\plus{}qr^2\plus{}qp^2\plus{}rp^2\plus{}rq^2\minus{}2pqr(\cos(\psi_1)\plus{}\cos(\psi_2)\plus{}\cos(\psi_3))}} \ge$
$ (p\plus{}q\plus{}r)^{3/2}\frac{1}{\sqrt{pq^2\plus{}pr^2\plus{}qr^2\plus{}qp^2\plus{}rp^2\plus{}rq^2\plus{}3pqr}} \ge \sqrt{3}$ :)


Let's say there is an $n$-tiling of a square if it can be covered completely, with no overlap, by $n$ (smaller) squares, for $n\ge2$.

For which values of $n$ is there an $n$-tiling?

I think I seen this problem type somewhere before.
[hide]Any square number of small squares will obviously tile the large square.  Also, any number generated by this following process also works.  When you subtract x small squares and replace that with smaller squares that fill up the space, you create a n-tiling number.
In other words, x^(2n) where x and n are positive are tiling-numbers.
Also, x^(2n)-a+by^(2d)+cz^(2g)+... where a=b+c+...
There's infinitely many of both.[/hide]
But how many non-tiling #s are there?

[quote="dengmi"][hide]Any square number of small squares will obviously tile the large square.  Also, any number generated by this following process also works.  When you subtract x small squares and replace that with smaller squares that fill up the space, you create a n-tiling number.
In other words, x^(2n) where x and n are positive are tiling-numbers.
Also, x^(2n)-a+by^(2d)+cz^(2g)+... where a=b+c+...
There's infinitely many of both.[/hide][/quote][hide="hint"]It is more useful to note that if $k$ is a tiling-number, then so is $k+3$.[/hide]
[quote]But how many non-tiling #s are there?[/quote][hide="hint"]Very few. :D[/hide]

Why no more replies? If it's too hard it can be moved..


[quote="Moonmathpi496"]Find all polynomials $ f$ such that $ f(2x)$ can be written as a polynomial in $ f(x)$,
i.e. for which there exists a polynomial $ g$ such that

$ f(2x) \equal{} g(f(t)$[/quote]

Since $ f(2x)\equal{}g(f(x))$ and $ f(2x)$ and $ f(x)$ have same degree, $ g(x)\equal{}ax\plus{}b$
Then $ f(x)\equal{}\sum_{i\equal{}0}^{n}a_{i}x^{i}$ implies $ \sum_{i\equal{}1}^{n}a_{i}(2^{i}\minus{}a)x^{i}\plus{}a_{0}(1\minus{}a)\minus{}b\equal{}0$

And the only solutions are $ f(x)\equal{}cx^{n}\plus{}d$

Nice solution!
actually the main thing was to get that the degree r same and then we will be done after some calculation :wink:


can someone suggest name of a book, that should have [size=150]all[/size] Mathematic's theorems, from simple to advance. specially all the polynomial's theorems.no matter how much it costs

sorry i dont know if this is in the right place.

Bourbaki :wink:
There's much to much mathematic out there to fit into a book. For polynomials, you can try Barbeau: Polynomials, a problem book .
I posted this for ZetaX since he isn't able to do so because of his computer.

[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=163182]In case you haven't seen this yet, this is pretty amazing.[/url]


I need help on the Reed and Muench Method.

Background on what you're talking about is always nice.

It basically is a method to calculate the LD50 (lethal dosage to kill off 50%) of an experiment using the data.

And a specified form of help is nice.  What do you want people to help you with?

Why does the reed and muench method work?

Okay, so, from what I can gather from the internet (which is not very much), all it is is a statistical calculation.  So it works because statistics works.  I think it's some kind of "best-fit" idea -- it approximates a curve which goes through all the data points you give it, and uses that to figure out (with some degree of certainty) what the value is that is needed toresult in 50% death.


group theory
abstract algebra
superior algebra
superior algebra unsolved

A group G is such that the intersection of all subgroups of G is a subgroup distinct from (e). Prove that every element of G is of finite order.

Note that this makes little sense as stated:  the intersection of all subgroups is $ \{ e \}$

Anyways, if $ x \in G$ has infinite order consider the subgroups $ H_i \equal{} < x^i>$ for each positive integer $ i$
(cyclic subgroup generated by $ x^i$).  Do you know what the intersection is?

I would appreciate it if someone could give me a more elaborate solution. I will need it at this stage as I am trying to read up the subject without any tutor to guide me. As I get better I can work with just hints  :)

He did not give you a hint, he solved the problem for you. :)  Does it help to visualize the integers?  What is the intersection of the sets of integers divisible by $ 2$, those divisible by $ 3$, etc... If what you're looking for is a written out, formal proof, I suggest that you try to get the gist of what we're saying, then attempt to write out a proof and ask us where you think it is lacking.

And as dallasboy said, the problem does not really make sense the way you stated it, because there are no such groups with the given property!  Before you expect us to write out formal solutions, at least tell us what we're solving. :)

Oops! I somehow thought I had written it, but it should be:"intersection of subgroups distinct from (e)"

[quote="LydianRain"]What is the intersection of the sets of integers divisible by $ 2$, those divisible by $ 3$, etc... [/quote]
Once you've understood what LydianRain means here, you should understand dallasboy's remark.  What does an integer divisible by every other integer have to be?

Ok I got what this is leading to

Suppose there is an element $ x \in G$ of infinite order then the cyclic sub-groups $ <x^i>$, $ <(x^2)^i>$,...$ <(x^n)^i>$,...

are all of infinite order and do not have any non-identity elements in common.

Hence there is no element of infinite order.

I appreciate your help :)


[quote="henryyangrui"]I posted several math problems in a currently running contest. Now i feel truly regretful of doing that. Please accept my apology and understand that I am a new member and didn't fully understand the rules in a forum. I promise I would never do this again. Thanks.  :blush:[/quote]

Hi all, I am a new member of forum. Would a newcomer be warmly welcome here? Good day you guys!!!

__________________
[url=http://animewatching.com]Watch Anime Episodes[/url]

Of course! AoPS welcomes any people brave enough to join it's ranks.


$G$ an infinite group and $H$ a proper subgroup of finite index. Prove that $G$ is not simple.

Again, we use that problem posted so many times on ML, stating that if $[G:H]$ is finite, then there is a subgroup of $H$ which is normal in $G$ and which has finite index. I even remember it was posted twicer by the same user! :) (the username is Misha followed by some numbers which I forgot).

$core_G(H)$ is a normal subgroup of $G$, with $[G:core_G(H)]|[G:H]!$.

I suppose "core" is the intersection of all conjugates of the group? I recently derived and used a weaker version of this for a problem involving symmetric and alternating groups.

Apparently, I'm doomed to post and repost links towardss threadfs discussing that problem for the rest of my life :):

[url]http://www.artofproblemsolving.com/Forum/viewtopic.php?highlight=normal+interior&t=4295[/url]

[url]http://www.artofproblemsolving.com/Forum/viewtopic.php?highlight=normal&t=22796[/url]

[url]http://www.artofproblemsolving.com/Forum/viewtopic.php?highlight=normal&t=19724[/url]


Μήπως ξέρει κανείς τι είναι η συνάρτηση LCM;

Και αν ναι, ποιος είναι ο τύπος της;

Η απορία μου δημιουργήθηκε όπως διάβαζα το παρακάτω άρθρο για γεννήτριες πρώτων αριθμών:

Ευχαριστώ πολύ!


Circles $ \Gamma_1$ and $ \Gamma_2$ intersect at $ P$ and $ Q$. The common tangent of $ \Gamma_1$ and $ \Gamma_2$ touch the circles at $ A$ and $ B$ respectively. $ C$ is an arbitrary point on ray $ AB$. Let $ CP$ intersect $ \Gamma_1$ at $ D$. Point $ E$ is on $ CD$ such that $ BE$ is parallel to $ AD$. Let $ PQ$ intersect the common tangent at $ R$. Prove that $ \angle RED \equal{} 90^{\circ}$.

edit: sorry, the problem is wrong. additional information is needed: $ \angle RQC\equal{}90^{\circ}$.

Denote  H’  is the orthocenter of ⊿PRC , H is the orthocenter of ⊿PBA
K is the foot of altitude through B to PA , F is the foot of altitude through R to PC
Then we have B,H,K are collinear and R,H’,F are collinear
RA^2=RB^2=RQ*RP  =>  R is the midpoint of AB  =>  RK^2= RQ*RP  =>           ∠KRQ=∠RPK=∠QAR   => K,Q,R,A are concyclic   =>  PQ*PR=PK*PA=PH*PL=PH’*PL
 => H=H’  =>  BH*HK=PH*LH=RH*RF   =>  K,R,B,F are concyclic =>
∠ADP=∠KAR=∠AKR=90-∠RKB=90-∠RFB=∠BFC  =>  BF//AD  =>  F=E
Done!

Let $ E'$ and $ S$ be the projections of $ R$ and $ P$ on $ PC$ and $ AB.$ $ PS,RE'$ and $ CQ$ concur at the orthocenter $ H$ of $ \triangle PRC.$ Since $ RA^2 \equal{} RB^2 \equal{} RQ \cdot RP \equal{} RS \cdot RC,$ it follows by Newton's theorem that cross ratio $ (A,B,S,C)$ is harmonic $\Longrightarrow$ $ CB \cdot CA \equal{} CS \cdot CR,$ but since $ CS \cdot CR \equal{} CE' \cdot E'P$ $ \Longrightarrow$ $ CB \cdot CA \equal{} CE' \cdot CP$ $ \Longrightarrow$ $ PE'BA$ is cyclic. Then $\angle CE'B \equal{} \angle PAB \equal{} \angle CDA$ $\Longrightarrow$ $ BE' \parallel AD$ $ \Longrightarrow$ $ E \equiv E'.$

$ \bullet$ Another interesting result about this configuration is that, if  $ T$ is the second intersection of the ray $ DR$ with $ \Gamma_1,$  then circles $ \odot(RTA)$ and $ \odot(HRB)$ are congruent and externally tangent.


Sorry, if this question has been posted before. Is there a formula to find even positive divisors of a number. I know there's formula to determine number of divisors of a number.

I've to find number of even positive divisors of 22680
$22680 = 2^{3}* 3^{4}* 5 * 7$ and the number of divisors are $(3+1)*(4+1)*(1+1)*(1+1) = 80$

Could someone help me to find the even positive divisors of 22680, thanks in advance!

It's very similar; what is special about an even number? It has at least one power of $2$. So the exponent on $2$ is $\ge 1$. Hence there are $3 \cdot 5 \cdot 2 \cdot 2 = 60$ even divisors.

Alternately, you could figure out what the odd divisors are, which means there can't be any powers of $2$.  This is equivalent to finding the divisors of

$2835 = 3^{4}\cdot 5 \cdot 7$

And subtracting this number from your original number.

Edit:  This is a great example of an occasion where it helps you more to understand the concepts behind a rule (in this case, finding the number of divisors of an integer) than to remember the rule itself.  ;)

[quote="t0rajir0u"]Alternately, you could figure out what the odd divisors are, which means there can't be any powers of $2$.  This is equivalent to finding the divisors of

$2835 = 3^{4}\cdot 5 \cdot 7$

And subtracting this number from your original number.[/quote]

yep that's what i did, thanks for the help guys

[b]Ok..Lemme twist it a lil bit...Find all the divisors of 22680 of the form 1)4k 2)4k+1 3)4k+2 4)4k+3 and 5)5k 5)5k+1 6)5k+2 7)5k+3 8)5k+4 ???????? [/b]


I dunno what it was about today's episode, but it seemed kind of I didn't really like it that much.

I like the Simpsons too.... but would it kill you if you at least provided the name or the number and the season, or at least a description?

Do you really think that Mathlinkers outside of the USA will have watched this episode on the exact same day? :|

[quote="futuremontrealian"]Don't they only have like <400 episodes?  I can believe the number is that high for that episode, but why?[/quote]
I think hspotter has a weird provider that makes him pay for every character he types on mathlinks. :o

He meant
the 18th episode of season 17

Yes the current number is about 380, check the list out here :
[url]http://en.wikipedia.org/wiki/List_of_The_Simpsons_episodes[/url]

What?  I meant 1802.

Cause tv.yahoo.com was being stupid, and didn't tell me the right episode.

But yea, this one - [url]http://en.wikipedia.org/wiki/Jazzy_and_the_Pussycats[/url]


Explain the following steps in order to deduce the structure of this powerful tranquiliser.

1. Synthesize $ \text{4\minus{}chlorobenzoylchloride}$ from $ \text{4\minus{}chlorophenylmagnesium bromide}$

2. The product is treated with $ \text{vinylmagnesium bromide}$ in anhydrous ether followed by aqueous work up. This is then treated with $ \ce{HBr}$ in the presence of a peroxide followed by $ \ce{NH3}$.

Draw all the intermediate products. What is the structure of the final product ($ A$)?

3. $ \text{Fluorobenzene}$ reacts with $ \text{4\minus{}chlorobutanoyl chloride}$ in the presence of $ \ce{AlCl3}$. The product is predominantly para ($ B$).

4. Reacting $ A$ and $ B$ in an aqueous media. Gives Haloperidol. What is the structure of the product?

1. treat with formaldehyde and then oxidise to get 4- chloro ebnzoic acid... then treat with PCl5 to get the product

2.

[hide="Number 1"]Treat 4-chlorophenylmagnesium bromide with carbon dioxide, followed by acidic workup to get 4-chlorobenzoic acid. Then reflux this acid with thionyl chloride for about one hour to get 4-chlorobenzoyl chloride.[/hide]

That's great we can all solve the first question  :D . How about trying the other ones.

carcul, when  fluorobenzene reacts with  butanoyl chloride, why doesnt the product(b) cyclize or react with another molecule of fluorobenzene?
also isnt F a deactivativating group, so F & C reactions arent feasible rite?

1) Fridel-Crafts reactions are not feasible with [i]meta directing groups[/i], i.e, with groups more deactivating than halogens.

2) Cyclization? There are only meta positions available on the ring.

3) React with another molecule of fluorobenzene? One should use just one equivalent of the Lewis acid in order to prevent possible side reactions. Also, acylium cation is much more stable than a primary carbocation.

thanks. :)
but concerning your third point over here, isnt the lewis acid a catalyst, so even if you use one equivalent, it will still be regenrated. so why cant the product still react as the lewis acid is still present, or does alcl3 react with the hcl formed? :huh:

Structures look right to me (because they are the same ones I got) :D .

[quote="VARUNARASU"]isnt the lewis acid a catalyst, so even if you use one equivalent, it will still be regenrated. so why cant the product still react as the lewis acid is still present, or does alcl3 react with the hcl formed?[/quote]

The Lewis acid acts as a catalyst in all steps of the reaction except on the last one, where the final product is complexated on the oxygen by it. Then, to recover our product we must treat that complex with aqueous acid, yielding our product, 3 moles of HCl, and $ \ce{Al(OH)3}$. Therefore, the Lewis acid is necessarily destroyed.

I didnt get your last point please explain.

Acc to what is given here [url]http://en.wikipedia.org/wiki/Friedel-Crafts_reaction#Reaction_mechanism[/url]
Alcl3 gets regenerated automatically

That mechanism on wikipedia is incomplete: after the last step, aluminium chloride coordinates with ketone's oxygen (this is a natural reaction, right?).


geometry conics parabola ellipse hyperbola trigonometry circumcircle

I've already posted this problem in another Forum :blush:  but I really need an answer!

[url=http://www.artofproblemsolving.com/Forum/topic-84986.html]Look Here please[/url]

I've also proved a similar result, using Analytic Geometry again, for a parabola.
Let (K) be a parabola and (s) the podaria (ie, tangent to (K) in the vertex); (r) is a fixed straight line parallel to (s). P is a variable point in (r) and draw the tangents to (K) through P. This tangents cut (s) in T and U. Then OT·OU is a constant.

I didn't try this for an ellipse or hyperbola, maybe is a general result...

Thanx :)

The parabola tangent s at its vertex O is its pedal curve with respect to the focus F, i.e., the locus of the feet of normals from the focus to the parabola tangents. Thus the feet of normals from the focus F to the parabola tangents from the point P are the intersections T, U of the line s (the pedal curve) with a circle with diameter FP, bcause then the angles $\angle FTP = \angle FUP = 90^\circ$ are both right. Let Q be the intersection of the line r with the parabola axis OF (a normal to the line s from the focus F). Since the angle $\angle FQP = 90^\circ$ is also right, the circle with diameter FP passes through the point Q for any $P \in r.$ Power of the point O (the parabola vertex) to the circle with diameter FP is $OT \cdot OU = OF \cdot OQ = \text{const}.$

Cool yetti  :lol:

what about the problem with the circumference? I've  havent found a synthetic solution yet (yetti)

[quote="hpoincare"]I'd like to know, if possible, a way to solve the following problem, using  Synthetic Geometry (maybe Plane Inversion) instead of a standard Analytic Geometric approach.
Let $(K)$ be a circumference with diameter $OA$, $d(O,A)=a$. $P$ is a variable point in the straight line $(r)$, $r\perp OA$, $d(O,r)=b$. (s) is the perpendicular to $OA$ passing through $O$.
Let $t_1$ and $t_2$ be the tangent lines to $(K)$ passing through $P$. $t_1\cap s=T_1$, $t_2\cap s=T_2$. Prove that $OT_1\cdot OT_2$ is a constant and find it in terms of $a$ and $b$.[/quote]
The tangent theorem for a triangle $\triangle ABC$ is

$\frac{a - b}{a + b} = \frac{\tan \frac{A - B}{2}}{\tan \frac{A + B}{2}}$

Indeed, substituting $a = 2R \sin A = 4R \sin \frac A 2 \cos \frac A 2$ and $b = 2R \sin B = 4R \sin \frac B 2 \cos \frac B 2$ to the left fraction (where R is the triangle circumradius),

$\frac{a - b}{a + b} = \frac{\sin \frac A 2 \cos \frac A 2 - \sin \frac B 2 \cos \frac B 2}{\sin \frac A 2 \cos \frac A 2 + \sin \frac B 2 \cos \frac B 2} =$ $\left(\frac{\tan \frac A 2}{\cos^2 \frac B 2} - \frac{\tan \frac B 2}{\cos^2 \frac A 2}\right) : \left(\frac{\tan \frac A 2}{\cos^2 \frac B 2} + \frac{\tan \frac B 2}{\cos^2 \frac A 2}\right) =$

$= \frac{\tan \frac A 2 \left(1 + \tan^2 \frac B 2\right) - \tan \frac B 2 \left(1 + \tan^2 \frac A 2\right)}{\tan \frac A 2 \left(1 + \tan^2 \frac B 2\right) + \tan \frac B 2 \left(1 + \tan^2 \frac A 2\right)} =$ $\frac{\left(\tan \frac A 2 - \tan \frac B 2\right) \left(1 - \tan \frac A 2 \tan \frac B 2\right)}{\left(\tan \frac A 2 + \tan \frac B 2\right) \left(1 + \tan \frac A 2 \tan \frac B 2\right)} =$ $\frac{\tan \frac{A - B}{2}}{\tan \frac{A + B}{2}}$

O is the center of the circle $\mathcal K$ and I wil label it $\mathcal K \equiv (O)$. Radius of this circle is $r = OR$ (which you labeled a = OA). Let $l$ be an arbitrary line perpendicular to the arbitrary circle radius OR (you labeled this line r) and let N be the foot of a normal from the circle center O to the line $l$, so that the distance of this line from the circle center is $d = ON \neq r$ (you labeled this distance b). Let $P \in l$ be an arbitrary point on this line and let A, B be the tangency points of the tangents $a, b$ from the point P to the circle (O) (you labeled them $t_1, t_2$). Using the tangent theorem for the triangle $\triangle AON,$

$\frac{r - d}{r + d} = \frac{OA - ON}{OA + ON} = \frac{\tan \frac{\widehat{ONA} - \widehat{OAN}}{2}}{\tan \frac{\widehat{ONA} + \widehat{OAN}}{2}}$

The quadrilateral OAPN is cyclic, because the angles $\angle OAP = \angle ONP = 90^\circ$ are both right. Therefore,

$\angle OPA = \angle ONA,\ \ \angle OPN = \angle OAN$

$\frac{r - d}{r + d} = \frac{\tan \frac{\widehat{OPA} - \widehat{OPN}}{2}}{\tan \frac{\widehat{ONP} + \widehat{OAP}}{2}}$

Since the angles $\angle OPA = \angle OPB$ are equal, $\angle OPA - \angle OPN = \pm \angle NPB$ and $\angle OPA + \angle OPN = \angle NPA.$ The plus sign applies when the line $l$ intersects the circle (O) and the minus sign when it does not. Now give the circle (O) a direction, say counter-clockwise (counter-clockwise is positive, clockwise is negative). The tangents $a \equiv AP,\ b \equiv PB$ of the directed circle (O) are also directed, having the same direction as the directed circle at the tangency points A, B. Let the line $l$ be also directed, its direction being the same as that of the directed tangent at the point R. Two directed lines form a unique directed angle, not like undirected lines, where we have 2 possibilities. Choose the direction of the line $l$ so that $\widehat{lb} = \angle NPB$. Then $\widehat{la} = 180^\circ - \angle NPA$ and

$\frac{|r - d|}{r + d} = \frac{\tan \frac{\widehat{NPB}}{2}}{\tan \frac{\widehat{NPA}}{2}} = \tan \frac{\widehat {la}}{2} \tan \frac{\widehat{lb}}{2}$

The product $\tan \frac{\widehat {la}}{2} \tan \frac{\widehat{lb}}{2} = \frac{|r - d|}{r + d}$ is called power of the directed line $l$ to the directed circle (O) and it does not depend on the position of the point P on the directed line $l$ (because r, d do not). It is dual concept to power of a point to a circle, which does not depend on the line through this point. It is certainly possible to prove this in a different way, without using the tangent theorem.

Let the directed tangents a, b intersect the directed tangent $s \perp OR$ of the directed circle (O) at the point $R \equiv ON \cap (O)$ at points T, U, respectively. Since the radii OA, OB of the circle (O) are perpendicular to the tangents a, b, respectively, and since OT, OU bisect the angles $\angle ROA \equiv \angle NOA = \widehat{la},$ $\angle ROB \equiv \angle NOB = \widehat{lb}$, it follows that $RT = r \tan \frac{\widehat{la}}{2}$ and $RU = r \tan \frac{\widehat{lb}}{2}$, so that $\frac{RT \cdot RU}{r^2}$ is power of the directed line $l$ to the directed circle (O), i.e., independent of the point $P \in l.$ You could call power of the line $l$ to the circle (O) just the expression $\frac{RT \cdot RU}{r^2},$ but it would kind of obscure the fact that a directed angle is dual concept of a line segment.

Thanx for your reply! But you confused the letters. I attach a picture of the right problem I'm asking for. Thanx.

Note: OM=d(O,r)=b

Let I be the incenter of PTT' (I is also the center of (K))
It's easy to see that what I'm trying to prove is equivalent to prove that the circumference (T,I,T') (circumcircle, containing T',T and I) passes through another fixed point J because in that we would have constant=OJ·OI=OT·OT'

But everything is equivalent to prove taht the center L of cirnference (TIT') is straight line (l), perpendicular to OA. I should find d(l,O)...

Give arbitrary directions to the circle (K) and the line r. The directions of the tangents t, t' are determined by the direction of the circle (K). The purpose is to get the correct angles $\phi = \widehat{rt},\ \phi' = \widehat{rt'}.$ Power of the directed line r to the directed circle (K) is

$\tan \frac{\phi}{2} \tan \frac{\phi'}{2} = \tan \frac{\widehat{rt}}{2} \tan \frac{\widehat{rt'}}{2} = \frac{|OK - KM|}{OK + KM} = \frac{|OA - OM|}{OM} = \frac{|a - b|}{b}$

and it does not depend on the point $P \in r.$

$\frac{OK}{OT} = \tan \widehat{KTO} = \tan \frac{\phi}{2},\ \ \ \frac{OK}{OT'} = \tan \widehat{KT'O} = \tan \frac{\phi'}{2}$

$OT = \frac{OA}{2 \tan \frac{\phi}{2}},\ \ \ OT' = \frac{OA}{2 \tan \frac{\phi'}{2}}$

$OT \cdot OT' = \frac{OA^2}{4 \tan \frac{\phi}{2} \tan \frac{\phi'}{2}} = \frac{a^2b}{4|a - b|}$

BTW, the circumcenter L of the triangle $\triangle KTT'$ is the midpoint of the arc TT' of the circumcircle of the triangle $\triangle PTT'$ opposite to the vertex P. The excenter E of the triangle $\triangle PTT'$ against the vertex P also lies on the circumcircle (L) of the triangle $\triangle KTT'$. Since the inradius of the triangle $\triangle PTT'$ is constant and the parallel lines $r \parallel s \equiv TT'$ are fixed, the exradius of the excircle against the vertex $P \in r$ is also constant. The reflection J of this excenter E in the perpendicular bisector of the segment TT' lies on the circle (L) and on the line $OA \perp TT'$ through the incenter K, which is fixed, OJ is constant, so that this excenter reflection J is your fixed point.


If the sequence $\{a_n\}$ is defined by \begin{align*}a_1 &= 2,\\ a_{n+1} &= a_n + 2n\qquad (n\geq 1),\end{align*} then $a_{100}$ equals

$\textbf{(A) }9900\qquad
\textbf{(B) }9902\qquad
\textbf{(C) }9904\qquad
\textbf{(D) }10100\qquad
\textbf{(E) }10102$

What am I doing wrong here?

[hide="Answer"]Notice that the sequence goes $2+(2+2)+(2+2+4)+(2+2+4+6)+...+(2+2+4+6+...+198)$. Therefore, we have $2(\frac{99(100)}{2}+1)=2(4951)=9902\Rightarrow \boxed{B}$.[/hide]

[b]EDIT:[/b] Fixed.

[quote="JesusFreak197"]What am I doing wrong here?

[hide="Answer"]Notice that the sequence goes $2+4+(4+6)+(4+6+8)+(4+6+8+10)+...+(4+6+...+198)$. Therefore, we have $2(\frac{99(100)}{2}-1)=2(4949)=\boxed{9898}$.[/hide][/quote]

You make a mistake from your third term. It is $8$ not $10$ ($a_3=4+2*2=8$) and that's why you calculate it wrong.


function
combinatorics proposed
combinatorics

Prove The Following identity:
$ \sum_{j \equal{} 0}^n \left ({3n \plus{} 2 \minus{} j \choose j}2^j \minus{} {3n \plus{} 1 \minus{} j \choose j \minus{} 1}2^{j \minus{} 1}\right ) \equal{} 2^{3n}$.
The Second term on left hand side is to be regarded zero for j=0.

[quote="grfield"]I think that this is solvable by generating functions. I am not sure how to express the left hand side though. Does anyone know how to get the generating function for the left hand side?[/quote]

I think the following might work:

Consider coefficient of $ x^{M}$ in

$ \sum_{j\equal{}0}^{n} x^{2j}(a\plus{}x)^{M\minus{}j}$

Here is my proof using generating functions. First, I found the following lemma.

[b]Lemma[/b] Let $ P_{n}(x) \equal{} \sum_{j \equal{} 0}^{\infty}\binom{2j \plus{} n}{j}x^j$. Then we have
\[ \frac {1}{\sqrt {1 \minus{} 4x}} \left( \frac {1 \minus{} \sqrt {1 \minus{} 4x}}{2x} \right)^n.\]
[b]Proof of the lemma[/b]
$ P_c(x) \equal{} \binom{c}{0} \plus{} \binom{c \plus{} 2}{1}x \plus{} \binom{c \plus{} 4}{2}x^2 \plus{} \cdots$
$ \equal{} 1 \plus{} \left( \binom{c \plus{} 1}{0} \plus{} \binom{c \plus{} 1}{1} \right)x \plus{} \left( \binom{c \plus{} 3}{1} \plus{} \binom{c \plus{} 3}{2} \right)x^2 \plus{} \cdots$, which is equal to $ xP_{c \plus{} 1}(x) \plus{} P_{c \minus{} 1}(x)$.
Hence we have a recurrence formula $ P_{c \plus{} 1} \equal{} \frac {P_c \minus{} P_{c \minus{} 1}}{x}$.
Since we know that $ P_0 \equal{} \frac {1}{\sqrt {1 \minus{} 4x}}$ and $ P_1 \equal{} \binom{1}{0} \plus{} \binom{3}{1}x \plus{} \binom{5}{2}x^2 \plus{} \cdots \equal{} \frac {1}{2}(\binom{2}{1} \plus{} \binom{4}{2}x \plus{} \binom{6}{3}x^2 \plus{} \cdots$ which leads to $ P_1 \equal{} \frac {P_0 \minus{} 1}{2x} \equal{} \frac {1}{\sqrt {1 \minus{} 4x}}( \frac {1 \minus{} \sqrt {1 \minus{} 4x}}{2x})$, by the recurrence formula, we get the desired result. [b]End of the proof of the lemma[/b]

Now we prove the original problem. We are to prove that
\[ \sum_{n \equal{} 0}^{\infty} \sum_{j \equal{} 0}^{n} \left( \binom{3n \plus{} 2 \minus{} j}{j}2^j \minus{} \binom{3n \plus{} 1 \minus{} j}{j \minus{} 1}2^{j \minus{} 1} \right) x^n \equal{} \frac {1}{1 \minus{} 8x}.\]
Changing the order of the two summations, we have that the left hand side is equal to
$ \sum_{j \equal{} 0}^{\infty} \sum_{n \equal{} j}^{\infty} \left( \binom{3n \plus{} 2 \minus{} j}{j}2^j \minus{} \binom{3n \plus{} 1 \minus{} j}{j \minus{} 1}2^{j \minus{} 1} \right) x^n$
$ \equal{} \sum_{j \equal{} 0}^{\infty} \sum_{n \equal{} j}^{\infty} \binom{3n \plus{} 2 \minus{} j}{j}2^j x^n \minus{} \sum_{j \equal{} 1}^{\infty} \sum_{n \equal{} j}^{\infty} \binom{3n \plus{} 1 \minus{} j}{j \minus{} 1}2^{j \minus{} 1}x^n$
$ \equal{} \sum_{j \equal{} 0}^{\infty} 2^j \left( \binom{2j \plus{} 2}x^j \plus{} \binom{2j \plus{} 5}{j}x^{j \plus{} 1} \plus{} \cdots \right) \minus{} \sum_{j \equal{} 1}^{\infty} 2^{j \minus{} 1} \left( \binom{2j \plus{} 1}{j \minus{} 1}x^j \plus{} \binom{2j \plus{} 4}{j \minus{} 1}x^{j \plus{} 1} \plus{} \cdots \right)$
$ \equal{} \sum_{j \equal{} 0}^{\infty} 2^j \left( \binom{2j \plus{} 2}x^j \plus{} \binom{2j \plus{} 5}{j}x^{j \plus{} 1} \plus{} \cdots \right) \minus{} \sum_{j \equal{} 0}^{\infty} 2^{j} \left( \binom{2j \plus{} 3}{j}x^{j \plus{} 1} \plus{} \binom{2j \plus{} 6}{j}x^{j \plus{} 2} \plus{} \cdots \right)$
$ \equal{} P_2(2x) \plus{} xP_5(2x) \plus{} x^2P_8(2x) \plus{} \cdots \minus{} xP_3(2x) \minus{} x^2P_6(2x) \minus{} \cdots$
$ \equal{} \frac {1}{\sqrt {1 \minus{} 8x}}\frac {t^2 \minus{} xt^3}{1 \minus{} xt^3}$ where $ t \equal{} \frac {1 \minus{} \sqrt {1 \minus{} 8x}}{4x}$. We can calculate that this is equal to $ \frac {1}{1 \minus{} 8x}$, as desired.

Fabulous!  When I (failed) to solve this a while ago, I spent my time trying to construct a bijective proof.  (We should first change it to some subtraction-free form, probably; there are a couple of ways to do this.)  I would still be interested to see if this is possible.

For initial simplification we can use that ncr=(ncr-1)*(n/r).if that does not work out then we can employ generating functions or create a combinatorial problem and solve


number theory
least common multiple
pigeonhole principle
number theory proposed

Let $ N$ be a natural number and $ x_1, x_2, . . . , x_n$ further natural numbers less than $ N$ and such that the least common multiple of any two of these $ n$ numbers is greater than $ N$.

Prove that $ \frac {1}{x_1} \plus{} \frac {1}{x_2} \plus{}...\plus{} \frac {1}{x_n} < 2$

Notice that $ [\frac{N}{x_{i}}]$ gives the number of integers less or equal to N which are divisible by $ {x_{i}}$.
Suppose $ [\frac{N}{x_{1}}] \plus{} [\frac{N}{x_{2}}] \plus{} .. \plus{} [\frac{N}{x_{n}}]> N$. But then by pigeonhole principle some number <= N is divisible by at least two of $ x_{i}$. Contradiction.
So we have $ [\frac{N}{x_{1}}] \plus{} [\frac{N}{x_{2}}] \plus{} .. \plus{} [\frac{N}{x_{n}}] \leq N$ Since $ [a] < a\plus{}1$ we have $ \frac{1}{x_{1}} \plus{} \frac{1}{x_{2}} \plus{} .. \plus{} \frac{1}{x_{n}}\leq \frac{n\plus{}N}{N}$. From here the statement follows, since obviously n < N.

It is well known that if $ n>\frac{N}{2}$ then some of $ x_{i}$ divides another, which is impossible given the lcm condition. Using this observation we can replace the constant 2 with $ \frac{3}{2}$


It is given the isosceles triangle $ ABC$ ($ AC \equal{} BC$) where the angle $ \angle ACB$ is equal to $ 30^\circ$. The point $ M$ is symmetric to the vertex $ A$ with respect to the line $ BC$. $ N$ is symmetric to the $ M$ with respect to the vertex $ C$. If $ P$ is the intersecting point of the lines $ AC$ and $ BN$ and $ Q$ is the intersecting point of the lines $ AN$ and $ PM$ find the ratio $ AQ: QN$.

Since $ \angle{ACB} \equal{} 30^o$, we get $ \angle{ACM} \equal{} 60^o \Longrightarrow \triangle{ACN}$is equilateral.$ \angle{ACN} \equal{} 120^o$,$ AC \equal{} MC \equal{} CN$, so$ \angle{NAC} \equal{} 30^o$.

Hence$ BC \parallel AN$. So we get$ \frac {CP}{AP} \equal{} \frac {CB}{AN} \equal{} \frac {AC}{AN} \equal{} \frac {1}{\sqrt {3}}$.

By Menelaus on$ \triangle{ACN}$ and line $ MPQ$, we get $ \frac {AQ}{QN}\cdot \frac {NM}{CM} \cdot \frac {CP}{PA} \equal{} 1$.

Hence $ \frac {AQ}{QN} \equal{} \frac {1}{2} \cdot \frac {\sqrt {3}}{1} \equal{} \frac {\sqrt {3}}{2}$.

Very good.

Just for correctness, $ \triangle{ACN}$ is isoceles but not equilateral.

Can someone explain what a symmetric point is please? Or otherwise, can  the figure be provided?


algebra
function
domain
real analysis
real analysis theorems

[quote="Munkers, Topology 2nd"]If $A$ is a subspace of $X$, the inclusion function $f: A \rightarrow X$ is continuous

Proof: If $U$ is open in $X$, then $f^{-1}(U) = U \cap A$, which is open in $A$ by definition of the subspace topology.
[/quote]

Question: Why If $U$ is open in $X$, then $f^{-1}(U) = U \cap A$???? :blush:  :blush:

The elements of $A$ that map to elements in $U$ are... the elements that were already in $U$.

:maybe:  :maybe:
if $x \in f^{-1}(U)$ then $f(x) \in U$ we don't have any other information to argue that $x \in A \cap U$.

does the condition $U$ is open affect something??
thanks.

If $x \in f^{-1}(U)$ then in particular $x$ is in the domain of $f$.

You could think of the inclusion map as the identity map restricted to $A$. So $f(a)=a$ for all $a\in A$, so take an open subset of $X$ say $U$. Then $f^{-1}(U)=U$, but we have to get rid of stuff that was in $X$ but not in $A$ (since $A\subset X$). So we want $U\cap A$.


function
integration
real analysis
real analysis unsolved

Let $ f: [a,b]\minus{}>[0, \infty]$ be a integrable function.If $ \int_{a}^{b}\ f(x)dx\equal{}0$ then prove that in every interval $ (\alpha, \beta) \subseteq [a,b]$ exist's $ x_{0} \in (\alpha,\beta)$with the property $ f(x_{0})\equal{}0$.

Well you just write $ \left\{ x : f(x) > 0 \right\} \equal{} \bigcup_{n \equal{} 1}^\infty \left\{ x : f(x) > \frac1{n} \right\}$ and you arrive at a contradiction if one of the sets on the right has non-zero measure.


Source: Iberoamerican Olympiad, 1990

1. The function f is defined for the nonnegative integers.  f(2^n - 1) = 0 for n = {0,1,2,3,...}.  If m is not of the form 2^n - 1, then f(m) = f(m+1) + 1.  Show that for some k, f(n) + n = 2^k - 1, and find f(2^1990).

for all n, the k given in the problem can be computed by floor(log n base 2)+1. To see why, we look at the function for the first few values
f(0)=0
f(1)=0
f(2)=1 f(3)=0
f(4)=3 f(5)=2 f(6)=1 f(7)=0
f(8)=7 f(7)=6 etc.

to prove it, we note that f(2^k-1), where k is computed as I said, will equal 0 (by defenition). Now, f(2^k-2)=1+f(2^k-1)=1. f(2^k-3)=1+f(2^k-2)=2. etc. Thus, you can calculate the above expression.

f(n)=floor(log n base 2)+1-n. You can compute it from there.

yup that's right (psst...you [i]could[/i] use my formula, or notice that f(2^n) is always 2^n-1)

Yea, but i did the problem without looking at ur formula, since i wanted to do it myself, and i was too lazy to derive ur formula myself...

([i]In robotic voice[/i]) That's correct.


Find $ z_1,\ z_2,\ z_3\in\mathbb{C}$ such that $ |z_1|\equal{}|z_2|\equal{}|z_3|\equal{}a,\ z_1z_2z_3\equal{}a^3,\ a\in\mathbb{R},\ a>0$ .

$ z_1\equal{}z_2\equal{}z_3\equal{}1$
$ z_1\equal{}z_2\equal{}i ,z_3\equal{}i^2$
these are 2 trivial solutions

[quote="Mateescu Constantin"]Find $ z_1,\ z_2,\ z_3\in\mathbb{C}$ such that $ |z_1| \equal{} |z_2| \equal{} |z_3| \equal{} a,\ z_1z_2z_3 \equal{} a^3,\ a\in\mathbb{R},\ a > 0$ .[/quote]
 :play_ball:

What about any three points $ z_{j} \equal{} ae^{i\theta_{j}},  j\equal{}1, 2,3$, on the circle $ z\equal{}ae^{i\theta}$, such that the sum of their arguments is $ 2n\pi$ where $ n$ is an integer?

[quote="Mateescu Constantin"]Find $ z_1,\ z_2,\ z_3\in\mathbb{C}$ such that $ |z_1| \equal{} |z_2| \equal{} |z_3| \equal{} a,\ z_1z_2z_3 \equal{} a^3,\ a\in\mathbb{R},\ a > 0$ .[/quote]
Let $ z_1 \equal{} a \mbox{cis } \theta_1$, $ z_2 \equal{} a \mbox{cis } \theta_2$, $ z_3 \equal{} a \mbox{cis } \theta_3$. We have that $ a^3 \mbox{cis }(\theta_1 \plus{} \theta_2 \plus{} \theta_3) \equal{} a^3$ iff $ \mbox{cis }(\theta_1 \plus{} \theta_2 \plus{} \theta_3) \equal{} 1$ iff $ 2\pi | \theta_1 \plus{} \theta_2 \plus{} \theta_3$.

I probably overlooked something...


How do I make multiple-choice questions with LaTeX? I mean like the one AMC uses for its answer choices. When asking questions on the AMC marathon, it's weird to use normal text for those, it makes it a bit harder to read.

[code]
$\textbf{(A)}\ AnswerA \qquad
\textbf{(B)}\ AnswerB \qquad
\textbf{(C)}\ AnswerC \qquad
\textbf{(D)}\ AnswerD \qquad
\textbf{(E)}\ AnswerD \qquad$
[/code]

Use the enumitem package and multicols:
[code]
\usepackage{enumitem,multicols}

\setenumerate[2]{label=\textbf{(\Alph*)},ref=\Alph*}

\begin{enumerate}
\item Question 1
  \begin{multicols}{4}
  \begin{enumerate}
  \item Choice 1
  \item Choice 2
  \item Choice 3
  \item Choice 4
  \end{enumerate}
  \end{multicols}
\end{enumerate}
[/code]


2^n.a+b is a perfect square for each nonnegative integer n=0,1,2,3,4,.........
   show that a=0.(a and b are  integers.)(2^n means 2 of power n.) ;)

We've had this one before.
Assume $2^{n}a+b$ is always a square and a>0.  Then for all n there exist j,k such that:
$j^{2}=2^{n+2}+b$ and $k^{2}=2^{n}+b$ but then
$j^{2}=2^{n+2}+b=4(2^{n}+b)-3b=4(k^{2})-3b=(2k)^{2}-3b$

So $3b>2(2k)-1>4(2^{n/2})$ for all n, which is impossible.


What is the sum of all odd, positive multiples of 3 under 100?

How do you solve this quickly :?:

You can treat it as an arithmetic series and then use the formula for sum of an arithmetic series.


geometry
analytic geometry
radical axis
geometry proposed

Given triangle $ ABC$ with medians $ AA', BB', CC'. B'C'\cap (O)\equal{}\{A_1,A_2\}$, similar for $ B_1,B_2,C_1,C_2$. Prove that the radical center of $ (A'A_1A_2), (B'B_1B_2)$ and $ (C'C_1C_2)$ lies on [b]Brocard axis[/b] of triangle $ ABC$.

And $ AA'', BB'', CC''$ are the symmedians of triangle $ ABC, B''C''\cap (O)\equal{}\{A'_1, A'_2\}$, similar for $ B'_1, B'_2, C'_1, C'_2$ then the radical center of $ (A''A'_1A'_2), (B''B'_1B'_2), (C''C'_1C'_2)$ lies on [b]Brocard axis [/b]of triangle $ ABC$ too

Two nice problems  forgotten for too long!.

The general barycentric coordinates of the radical center $R$ of this triad of circles were given in the topic [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?f=47&t=297002]Radical center and line IG[/url] (see post #6). If $P$ coincides with the centroid $X_2(1:1:1),$ then $R$ is the Brocard midpoint (midpoint of the segment connecting the 2 Brocard points) $X_{39}(a^2(b^2+c^2):b^2(c^2+a^2):c^2(a^2+b^2)),$ which obviously lies on the Brocard axis $X_3X_6.$

If $P$ coincides with the symmedian point $X_6(a^2:b^2:c^2),$ then $R$ is given by

$R \ (a^2(b^2+c^2+2a^2):b^2(c^2+a^2+2b^2):c^2(a^2+b^2+2c^2))$

This isn't in the current edition of ETC, but it certainly lies on the Brocard axis $X_3X_6$

$\left [\begin {array}{ccc} a^2 & b^2 & c^2\\  a^2S_A&  b^2S_B& c^2S_C\\ a^2(b^2+c^2+2a^2) & b^2(c^2+a^2+2b^2)& c^2(a^2+b^2+2c^2)\end{array}\right]=0$


function
inequalities
real analysis
topology
triangle inequality
real analysis unsolved

1. Let (X,d) be a metric space and Y be a dense subset of X.  Suppose f:Y→R is uniformly continuous.  Show that f has an extension F:X→R that is continuous.

2. Let X denote the set of all bounded sequences of real numbers.  For x = xn ,
y = yn in X define the distance d(x,y) between x and y to be
d(x,y) = supn │xn-yn│.
(a)	Show (X,d) is a metric space.
(b)	Suppose A is contained in N and define xA in X by
Xn, A =   1  if n is in A
               0  if n is not in A.
           Show that if A,B are contained in N with A≠B the d(xA,xB) = 1
(c)	How many elements of X belong to {xA:A contained in N}
(d)	Show that (X,d) is complete

3. Let (X,d) be a metric space.  For a non-empty set A contained in X and and x in X, define d(x,A) by d(x,A) = inf {d(x,y):y is in A}
     (a)  Show that d(x,A) is a continuous real-valued function on X.
     (b)  Show that d(x,A) = 0 if and only if x is in A closure

These appear to be exercises from an analysis class- every single one of them is routine.
Hints:
1. [hide="hint"]Uniformly continuous functions map Cauchy sequences to Cauchy sequences[/hide]
2. [hide="hint"]There's really nothing much to it. The space is known as $\ell_\infty$, and the problem can be used to show that it is complete but not separable.[/hide]
3. [hide="hint"]Prove an extension of the triangle inequality: $d(x,A)+d(y,A)\ge d(x,y)$ and $d(x,A)+d(x,y)\ge d(y,A)$. This is already stronger than continuity- it's a Lipschitz condition.[/hide]

Yeah, the questions are for a practice test in my analysis class.  the only problem is the prof. isnt giving us the answers before the actual test, and even with you're hints i'm not getting very far.

1) For every $x\in X\setminus Y$, there exists a sequence of elements of $Y$ (say, $\{y_n\}$ which converges to $x$.  This is clearly a Cauchy sequence (every convergent sequence is Cauchy) and so $\{f(y_n)\}$ is Cauchy also (the proof is easy if you haven't already seen it).  Now, I think you're missing a condition - that $f$ is a uniformly continuous REAL function (this is how the question is posed in Rudin)

Anyways, $\{f(y_n)\}$ is a real Cauchy sequence and hence converges to some limit $q$.  Define $f(x)=q$ and we have a continuous function.

2) I don't really understand your notation

3) The second one is trivial, so I'll just do the first part.

Fix $z\in A$.  We have $d(x,A)\leq d(x,z)\leq d(x,y)+d(y,z)$ and hence, taking the infimum over $z$, we get that $d(x,A)-d(y,A)\leq d(x,y)$.  Switching $x,y$ gives $\vert d(x,A)-d(y,A)\vert \leq d(x,y)$.  This establishes more than continuity - it establishes uniform continuity, or a Lipschitz condition, or many other things.  And because I'm (kind of) nice, I'll do b) as well.

If $x\in A$, then $d(x,A)=d(x,x)=0$.  If $x\in A'$, then for every $\epsilon>0$, there exists $z$ in $A$ such that $d(x,z)<\epsilon$.  Taking the infimum over $z$ gives $d(x,A)=0$.  That's one direction.

Now, say that $x\not\in\overline{A}$.  Now, $\overline{A}$ is a closed set, so $\overline{A}^{c}$ is open, and thus there exists a $\epsilon$ neighborhood of $x$ such that $B_{\epsilon}(x)\cap \overline{A}=0$.  Hence $d(x,a)>\epsilon$.

Notation:

$A'$ - set of limit points of $A$
$A^c$ - (relative) complement of $A$
$\overline{A}$ - closure of $A$
$B_r(x)$ - ball around $x$ of radius $r$


I couldn't find this on the AoPS LaTeX pages, but how do you do the rotated { thingy where you can write x+x+x+...+x and under the rotated { you write "n x's" ? Thanks.


calculus
integration
combinatorics unsolved
combinatorics

Find total number of positive integral solutions of
$ xyz \equal{} 2^3 \cdot 3^4 \cdot 5^1$

A possible generalization would be: Finding the number of positive integer solutions for,
\[ \prod_{i \equal{} 1}^{n}x_i \equal{} \prod_{j \equal{} 1}^{m}p_j^{\alpha_j}\]
where $ p_j$'s are distinct primes with corresponding integer powers $ \alpha_j > 0$. This would be equivalent (q-analogue?) to finding the number of solutions of
\[ y_{j1} \plus{} \dots \plus{} y_{jn} \equal{} \alpha_j\]
for every $ j$ (Assuming $ \{x_1,\dots,x_n\}$ to be ordered. For the unordered case, I believe one brings in a partition instead of a composition (? How does one do this?)). Hence it is,
\[ \prod_{j}{\alpha_j \plus{} n \minus{} 1\choose n \minus{} 1}\]


integration
Putnam
calculus
calculus computations

Essentially this problem has been posted several times before. It's a Putnam problem, and it's in Zeitz's book as an exercise.