algorithm
function
trigonometry
parameterization
combinatorics unsolved
combinatorics

consider all permutations of the first n integers.take a particular permutation.then the no of runs of the particular permutation is calculated as shown below:
example n=6
then let 1 3 2 4 6 5 be a permutation.
then 1->3 is increasing
3->2 is decreasing
2->4->6 is increasing
6->5 is decreasing
therefore the different monotone sequences are counted.and the no of runs of the above permutation is 4.similarly for others
now we are supposed to find the no of permutations of first n integers  with given k runs.
please help
regards

The only idea I have is the following. A permutation with $k$ runs whose first run is increasing, can be obtained like this: color the numbers from $1$ to $n$ in $k$ colors and then write the numbers colored by color $1$ in the increasing order; then write the numbers of color $2$ in the decreasing order, and so on. The problem is that some groups may be empty or contain just $1$ element, in which case you may get fewer than $k$ runs and also each permutation can be obtained this way from several colorings ( (1,3,2) may be viewed as (1)(32) or as (13)(2). If you try to take all that accurately into account, you arrive at formulae whose complexity grows fast with $k$, so I'm not sure there exists a good answer to the question whether the sought number can be given by a relatively simple formula for arbitrary $n$ and $k$. Meanwhile, this idea allows to prove that for fixed $k$ and $n\to\infty$ the number of permutations with $k$ runs is $\frac {k^n}{2^{k-2}}(1+o(1))$. I don't know whether this is of any use though... :(

Many years ago my students told one trick related to this problem, but, unfortunately, I don't remember details.
Anyway, the closed form for Catalan numbers is also far from any immediate insight. So let's hope we will find something nice here too.

Another idea is to insert $n+1$ into a permutation of first $n$ numbers and see what happens. If I haven't mistaken, this results in the following nice recurrent relation:
$U(n+1,k)=kU(n,k)+2U(n,k-1)+(n-k+1)U(n,k-2)$
(someone better check me on this), where $U(n,k)$ is the number of permutations with $k$ runs and the first run going up (so the total number of permutations with $k$ runs is $2U(n,k)$). This allows to compute the $U(n,k)$ for small values of $k$ by hand and gives a fairly fast algorithm for a computer (with running time about $nk$) to compute $U(n,k)$. In particular,

$U(n,1)=1$ ($n\geq 1$)

$U(n,2)=\frac12 2^{n}-2$ ($n\geq 2$)

$U(n,3)=\frac14 3^n-2^n-\frac12 n+\frac{11}4$ ($n\geq 3$)

$U(n,4)=\frac18 4^n-\frac12 3^n-\frac14 (n-6) 2^n+n-\frac 72$ ($n\geq 4)$.

Do these numbers look in any way familiar to anyone?

[quote="Myth"]  Anyway, the closed form for Catalan numbers is also far from any immediate insight. So let's hope we will find something nice here too.   [/quote]

Using generating functions, it is possible to show that (assuming for convenience $U(1,0)=1$)

$\sum_{n\geq 0}\frac {U(2n+1,2n)}{(2n+1)!}x^{2n+1}=\tan x$

I don't include the proof because this observation  is too far from the answer to the problem and I mention it only to make the following point: while it still may be feasible to find  the general generating function for $U(n,k)$ (or, maybe a better idea is to write them as $U(k+\ell,k)$ and fix $\ell$ :?), there is no hope for a simple explicit formula for $U(n,k)$ just because there is no known simple formula for [url=http://mathworld.wolfram.com/BernoulliNumber.html]Bernoulli numbers[/url] that appear in the coefficients of $\tan x$.

At last, the answer to a close but somewhat different problem can be found on [url]http://mathworld.wolfram.com/PermutationRun.html[/url]

Here is my (rather pitiful) attempt to find a generating function for $U(n,k)$. Since $\sum_{k\geq 1}U(n,k)=\frac {n!}2$, it makes sense to define the generating function as $f(x,y)=\sum_{n,k}\frac{U(n,k)}{n!}x^n y^k$ (other possible natural choices result either in divergent series or in even more complicated differential equations). Putting $U(n,k)=0$ for all pairs of $n,k$ except those with $n>k\geq 1$, it is possible to check that the recurrence relation in my previous post holds for all pairs $n,k$ except $n=k=1$. Thus, the differential equation for $f$ is
$f_x=yf_y+2yf+x(y^2 f)_x - y(y^2 f)_y+y^2 f + xy$
or, after some simplification,
$(1-xy^2)f_x+(y^3-y)f_y=(2y-y^2)f + xy$
Now, this is a first order linear partial differential equation, so it is solvable, which is a good news. The bad news is that the solution seems to be not an elementary function. Indeed, to solve this equation, one needs first to solve the auxiliary system of ordinary differential equations $X'(t)=1-X(t)Y(t)^2$, $Y'(t)=Y(t)^3-Y(t)$, whose solution, if I don't mistake, is $Y(t)=\pm\sqrt\frac{1}{1-e^{2t}}$, $X(t)=\sqrt{e^{-2t}-1}(\arcsin e^t-\arcsin e^T)$ where $T$ is a free parameter, varying which we get a system of curves filling the plane. So far, we can also express $t$ and $T$ in terms of $x$ and $y$, which is good. But the next step is to solve the following  initial value problem for the function $g(t)=f(X(t),Y(t))$:
$g'(t)=(2Y(t)-Y(t)^2)g(t) + X(t)Y(t)$ ,  $g(T)=0$
and here one has to find an antiderivative that doesn't seem to be an elementary function. This ruined my hope for a nice analytic solution though, maybe, I overlooked something  :(

I found a silly mistake in my computations! :lol: The generating function is elementary after all, so Myth was right! Unfortunately, I cannot call it nice: the answer is

$f(x,y)=\frac1{1+y}\left[-1-xy+\sqrt{\frac{1-y}{1+y}}\,\cdot\,\frac {y+\cosh(x\sqrt{1-y^2})-\sqrt{1-y^2}\sinh(x\sqrt{1-y^2})} {\sqrt{1-y^2}\cosh(x\sqrt{1-y^2})-\sinh(x\sqrt{1-y^2})}\right]$

Nothing is clear when looking at this monster, not even the fact that $U(n,k)=0$ for $n\leq k$ :( The derivation is 3 pages of rather boring computations: you do exactly what I tried to do in the previous post except for one thing: it is better to use the curves $X(t)=\sqrt{e^{-2t}+1}(\text{arcsinh}\, e^t-C)$, $Y(t)=\sqrt{\frac 1{1+e^{2t}}}$, which immediately gives you a parametrization for small $x$ and $y$ rather than for $y>1$ as it was in my previous post. If someone wants to see the complete reasoning, just let me know and I'll post it.  Sorry for posting 3 messages in a row.


geometry
trigonometry
rhombus
summer program
MathPath
probability
logarithms

This is just a contest for fun.  you are given a few problems.  you HAVE to find the answer to the first problem first because you need that answer in the following question.  say the answer to 1) is 101, this is called the "AFIN1"(Answer found in number 1).

1)In the figure, Point A is the center of a 100 by 100 square.  Find x such that the shaded region has an area that is 1/5 of the area of the square.  (see figure 1)

2) what is 1+2+3...+(AFIN1)?

3) What is the greatest factor of AFIN2?

4) what is the sum of the digits in AFIN3?

5) how many diagonals does an AFIN4-sided polygon have?

6) find the sum of all the factors of AFIN5.

7) Aproximate the cosine of AFIN6 to the nearest hundreth.

PM me the answers by Friday (this shouldnt take you very long)...you dont have to show your work.

all questions are worth 1 poit except 7 which is 2 points

first person to tern it in gets a free point

if you don't get it...just ask

Just wondering, in the figure, is A the center of the square?

[quote="JBL"][quote="jli"]1)In the figure, Point A is the center of a 100 by 100 square... (see figure 1)[/quote][/quote]

Oops...Yeah! Now I know I got the answer right  :ddr: unless I made a careless mistake  :mad:

[quote="Iversonfan2005"][quote="jli"] PM me the answers by Friday (this shouldnt take you very long)...you dont have to show your work.[/quote]

 :o  :o  :o  :o  :o  :o  :mad:  :mad:  :mad:  :mad:

I feel stupid now  :( I'm just gonna go crawl in a hole... :([/quote]

its ok...i deleted the posts...hope nobody saw

[quote="jli"]i just wanted to say...BE CAREFUL...the one the you both missed is a freebie.[/quote]

I see what you're saying. You have to read those questions carefuly. I'm an idiot for making those mistakes...now I know why I missed those two  :blush:  :mad:

yeah... so the rakings are

:winner_first:
[hide="first place"]Xantos C. Guin[/hide]

:winner_second:
[hide="second place"]Iversonfan2005[/hide]

 :winner_third:
[hide="third place"]break314dancer[/hide]

there aren't any trophies for the rest

Only 2nd place?  :(  :?  :wallbash:

I hate math...

oh well... im going to post another one on the weekend

[quote="samath"]I agree.  I think another problem is that your questions are a lot different from the older ones.  For example, I think that all of the old questions had integer answers.  Maybe it might help if you made your questions more  like the old ones fromm the beginning of AFIN.[/quote]

How can that affect the people joining though?

I'm not talking about people joining, I'm talking about people returning.  They might have left because they didn't like how it was being run, or something like that.

Okay, now we have an experiment  :P  I'll post something tomorrow or as soon as I can.  :)

Oh well, no one replied. The rounds are no longer cumulative....

Last round:

chess64: 6.5
Melissa: 5.7

[size=200]AFIN 18[/size]

1. Given the system of equations:

$\dfrac{1}{x}+\dfrac{2}{y}+\dfrac{3}{z}=0$

$\dfrac{1}{x}-\dfrac{6}{y}-\dfrac{5}{z}=0$,

Find $\dfrac{x}{y}+\dfrac{y}{z}+\dfrac{z}{x}$.

2. What is |AFIN 1|?

3. How many different sets {a,b,c} are there such that a+b+c=AFIN 2, and a, b, and c are non-negative integers?

4. What is sum of the first AFIN 3 integers?

5. What is greatest number of sides in a polygon with AFIN 4 obtuse angles?

These are easy! Due next Sunday.

[quote="samath"]Was that me?  I think I gave solutions...[/quote]

Yeah it was you.... anyone else want to try?  :(

[quote="Melissa"]Sorry, I can't do nr. 1 :blush:[/quote]

Oh well, that can be a problem  :D . I'm still waiting for one other person's response...

OK, I want this to move on, so I will give the easiest method I can think of right now.

[hide]

Guess and Check  :lol:  :rotfl: [/hide]

Okay someone else submitted answers.

The scores for this round are:

samath: 5
Melissa: 4

The answers were:

1. -1
2. 1
3. 3
4. 6
5. 9

For number 5, make the obtuse angles as big as possible (180 degrees), and make the acute angles (the remaining ones) as big as possible (90 degrees). The sum of these angles must be greater than the sum of the angles in this polygon. Let this polygon have sides n. We wanted to find the greatest number of sides a polygon could have if it had 6 obtuse angles (as you found previously).

$180(6)+90(n-6)>n-2(180)$

Divide by 90 to get

$12+n-6>2n-4$

$n<10$.

Thus, the largest n = 9.

Problems hopefully coming soon  :D

[size=200]AFIN 19[/size]

1. What is the sum of the coeffients of the quotient $\frac{x^4-15x^2-10x+24}{x^2-x-12}$?

2. How many numbers are between AFIN 1 and 101?

3. What is the largest integer value of $n$ such that $8^n$ evenly divides $(AFIN 2)!$?

4. What is the units digits of $(AFIN3)^{44}$?

5. Two marbles are randomly selected without replacement from a bag containing blue and green marbles. The probability they are both blue is $\frac{1}{AFIN4}$. If three marbles are randomly selected, the probability they are all blue is $\frac{1}{21}$. What is the fewest number of marbles that must have been in the bag before any were drawn?

Some questions:

2.  Does that mean inclusive?
3.  That means that $8^n$ is a divisor of $(AFIN2)!$, right?
5.  By "both" in the last part, do you mean that all three were blue?

[quote="236factorial"]
 If three marbles are randomly selected, the probability they are both blue is $\frac{1}{21}$.
[/quote]

2. not inclusive
3. yes
5.  :oops:  all three I meant

Everyone, we can't have one person answer each round  :roll:  samath got a nice score... anyone else?  :surrender:


logarithms
calculus
calculus computations

Verify that the following notations of big-Oh are true:

(i) $ k^{300}\equal{}O(2^k)$

(ii) $ (\ln k)^{375}\equal{}O(k^{0.001})$

(iii) $ N^{10}2^N\equal{}O(e^N)$[/code]

All of these are in fact little-o, and little-o implies big-O. (As long as it is understood that $ k\to \infty$ and $ N\to\infty$,)

The third one is a little clearer if we rearrange it to

$ N^{10}\equal{}o\left(\left(\frac{e}{2}\right)^N\right)$


linear algebra
matrix
linear algebra unsolved

dear friends,

Let us consider a matrix $A$ with size $n\times n$ with $rank(A)=k<n$ and its $k$ eigenvalues are all $1$. What can we say about the corresponding eigenvectors? Are they also all equal? Or else how to prove that they are not equal?

please help me in this regard.

bhupala

Of course the corresponding eigenvectors are not all equal.
And the don't all belong to the same subspace of dimension 1.
Try this example \[\begin{pmatrix}1 & 0 & 0 & 0\\ 0 & 1 & a & b\\ 0 & 0 & 1 & 0\\ 0 & 0 & 0 & 1 \end{pmatrix}\] and the following eignevectors $(1,1,-b,a)^{T}$, $(1,1,0,0)^{T}$


limit
integration
real analysis
real analysis unsolved

does anybody know where can i find solutions to the romanian national contests?
for example the solution for this problem:

prove that for every $n \in \mathbb{N}$ there is a uniuque solution to the equation
$\sum_{k=1}^n {\frac{1}{\sqrt{nx+k}}} =\sqrt{n}$.

let that solution be $x_n$.
find the limit $\lim_n{x_n}$.

the first part is easy, but i don't know how to handle the second part.
i'm familiar with series so it doesn't have to be an elementary solution.
thanks

Suppose you are already prove that x(n) is convergent to a real b

$\frac{1}{n}\sum_{k=1}^{n}\frac{1}{\sqrt{x(n)+ k/n}}= 1$

when n tends to +oo you have a Riemann sum

$\int_{0}^{1}\frac{1}{\sqrt{b+x}}dx=1$

$\sqrt{b+1}-\sqrt{b}=1/2$ just solve this equation you get the value of b

[quote="Matija"] ...to the romanian national contests?
[/quote]

Could someone post the problem of romanian national contest olympiad 1994, 1993, 1992, 1991, 1990, 1989, 1988, 1987, 1986, 1985

thank you

at first i thought i just have to refresh my memory on some theorems, but now i stil don't see why this is true.
could you please explain?


Find all function $\large f: R^{*} to R^{*}$ satisfying:
$\large f(x+f(y)) = \frac{y}{xy+1}$

Letting y = 0, we see that f(x) = 0 for all values of x; thus, the zero function, f(x) = 0, is one function satisfying the given conditions. Furthermore, letting x = 0, we see that
$f(f(y)) = y$ for all values of y, so the identity map $f: x \rightarrow x$ is another suitable function.

Undoubtedly, there exist numerous other functions that I have not mentioned; I will attempt to determine them later.

man i do not know what do you mean.you can easily see that f(x)=0 for all x does not satisfy the given condition.neither does f(x)=x for all x....and you can not say that x=0 and y=0

i lost almost 2 hours with this easy problem :( the only solution is f(x)=1/x for all real x xnot equal with 0.i will not post my proof because it has 4 cases and it is a bit long.just the main ideas:in the initial relation we put y->x+f(y)
we get that f((x^2y+x+y)/(xy+1))=(x+f(y))/(x^2+xf(y)+1).in here we put x=1 and get that f((2x+1)/(x+1))=(f(x)+1)/(f(x)+2).now we must find x such that (2x+1)/(x+1)=x we solve this equation and get that a=(1+sqrt5)/2 or a=(1-sqrt5)/2.and then we get that f(a)=(-1+sqrt5)/2 or f(a)=(-1-sqrt5)/2.obviously we have four cases.in 2 of them we get contradiction after computing the function and in the other 2 we get that f(x)=1/x.that's all :P

Given $y=1$we have $f(x)=\frac{1}{x-k+1} (k=f(1))$
otherwise,$f(x+f(y))=\frac{y}{xy+1}\Rightarrow k=1\Rightarrow f(x)=\frac{1}{x}$

conclusion:never listen to music when you are doing math.it makes you stupid.proof:my post above


parameterization
calculus
derivative
algebra unsolved
algebra

Help me solve this system:
$ \frac{1}{x^2}<4$
$ x^4\plus{}4x\plus{}m^2\minus{}m\plus{}4>0$
I have no way to solve it.

hello, should we solve the system for $ x$? And what is $ m$ a parameter?
Sonnhard.

[quote="SherlockBond"]Help me solve this system:
$ \frac {1}{x^2} < 4$
$ x^4 \plus{} 4x \plus{} m^2 \minus{} m \plus{} 4 > 0$
I have no way to solve it.[/quote]

Let $ f(x) \equal{} x^4 \plus{} 4x \plus{} m^2 \minus{} m \plus{} 4$. Then $ f'(x) \equal{} 4x^3 \plus{} 4$ and so $ f(x)$ is decreasing over $ ( \minus{} \infty, \minus{} 1)$, minimum at $ \minus{} 1$ and increasing over $ ( \minus{} 1, \plus{} \infty)$

So $ f(x)\geq f( \minus{} 1)$ $ \forall x$ with $ f( \minus{} 1) \equal{} m^2 \minus{} m \plus{} 1 > 0$ $ \forall m$

So $ x^4 \plus{} 4x \plus{} m^2 \minus{} m \plus{} 4 > 0$ $ \forall m,x$ and it remains only the first condition which is $ \frac {1}{x^2} < 4$ $ \iff$ $ x^2 > \frac 14$ and so :

$ x < \minus{} \frac 12$ or $ x > \frac 12$ and any $ m$

Notice that you could also avoid derivatives writing $ x^4 \plus{} 4x \plus{} m^2 \minus{} m \plus{} 4$ $ \equal{}(x^2\minus{}1)^2 \plus{}2 (x\plus{}1)^2 \plus{} (m\minus{}\frac 12)^2 \plus{} \frac 34 > 0$ $ \forall x,m$


lets have a massive quoting party!!

i'll type a small message(such as this) and the next person to post will quote me and add just a little bit.etc.

dude the last time this happened *cough*indiana forum*cough* it froze up everybody's computers and was so spammy that it resurrected even the most inactive of forum mods :roll:


for $a,b,c,d\in\mathbb{R^+}$ prove:

$a^4b+b^4c+c^4d+d^4a\geq abcd(a+b+c+d)$

[quote="towersfreak2006"]Using the scalar product notation,
\[ \left[ \begin{array}{cccc} a^2b & b^2c & c^2d & d^2a\\ a & b & c & d\\ a & b & c & d\\ \end{array} \right]\ge\left[\begin{array}{cccc} a^2b & b^2c & c^2d & d^2a\\ c & d & a & b\\ d & a & b & c\\ \end{array}\right]  \]
I actually don't know the proof behind Engel's scalar product magic though...[/quote]

I don't understand what you are doing here... The sequences $\left(a^2b;\;b^2c;\;c^2d;\;d^2a\right)$ and (a; b; c; d) are not equally sorted...

Anyway, the problem was solved at http://www.mathlinks.ro/Forum/viewtopic.php?t=45218 .

  darij

[quote]$\sum(23a^4b+7b^4c+11c^4d+10d^4a)\ge 51\sum a^2bcd$
[/quote]

that is what i had in mind;
AM-GM

[quote="towersfreak2006"]
I actually don't know the proof behind Engel's scalar product magic though...[/quote]

The proof is right before he introduces the notation - it is simply an application of Rearrangement, which he proves nicely  :)

Then I would like to have this application of rearrangement explained... (unless it is the same as in the other thread, but that would require some more words...)

  darij

$a^3+c^2d+cd^2\ge 3\sqrt [3] {a^3\cdot c^2d\cdot cd^2}=3acd\Longrightarrow a^4b+abc^2d+abcd^2\ge 3a^2bcd\Longrightarrow$
$\sum_{cyc} \left(a^4b+abc^2d+abcd^2\right)\ge 3\sum_{cyc} a^2bcd\Longrightarrow \sum_{cyc} a^4b\ge \sum_{cyc} a^2bcd\ .$


geometry
3D geometry
inequalities
vector
trigonometry
complex numbers

$ 1$ Let $ z_1,z_2,z_3\in\mathbb{C}$ such as $ |z_1| \equal{} |z_2| \equal{} |z_3| \equal{} 1$ and also $ z_1 \plus{} z_2 \plus{} z_3 \equal{} 0$ proove that

$ i)|z_1 \minus{} z_2| \equal{} |z_2 \minus{} z_3| \equal{} |z_3 \minus{} z_1| \equal{} \sqrt3$
$ ii)Re(\overline{z_1}z_2) >\equal{} \minus{} 1$
$ iii)M(z_1)$  $ M(z_2)$ $ M(z_3)$ belongs to $ C: x^2 \plus{} y^2 \equal{} 1$

$ 2$ Let $ z_1,z_2,z_3\in\mathbb{C}$ $ |z_1| \equal{} |z_2| \equal{} |z_3| \equal{} 1$ and also $ z_1 \plus{} z_2 \plus{} z_3 \equal{} 1$ proove that $ z_1 \equal{} 1$ or $ z_2 \equal{} 1$ or $ z_3 \equal{} 1$

$ 3$ Let  Let $ z_1,z_2,z_3\in\mathbb{C}$ $ |z_1| \equal{} |z_2| \equal{} |z_3| \equal{} 1$ and also $ Re(\overline{z_1}z_2) \plus{} Re(\overline{z_2}z_3) \plus{} Re(\overline{z_3}z_1) \equal{} \minus{} 1$ proove that $ z_1 \equal{} \minus{} z_2$ or $ z_2 \equal{} \minus{} z_3$ or $ z_3 \equal{} \minus{} z_1$

$ 4$ Let $ z_1,z_2,z_3\in\mathbb{C}$ such as $ |z_1| \equal{} |z_2| \equal{} |z_3| \equal{} 1$ and $ z_1 \plus{} z_2 \plus{} z_3 \equal{} 3$ and $ z_1z_2z_3 \equal{} 1$ proove that $ z_1 \equal{} z_2 \equal{} z_3$

$ 5$ Let $ z_1,z_2,z_3\in\mathbb{C}$ such as $ |z_1| \equal{} |z_2| \equal{} |z_3| \equal{} 3$ proove that $ (z_1 \plus{} z_2 \plus{} z_3)(\frac {1}{z_1} \plus{} \frac {1}{z_2} \plus{} \frac {1}{z_3}) <\equal{} 9$

$ 6$ Let $ z\in\mathbb{C}$ such as $ |z \minus{} a|^2 \plus{} |z \minus{} a|^2 \equal{} 10|a|^2$ $ a\in\mathbb{R \plus{} }$

$ i$Proove that $ M(z)$ belongs to a circle $ C1$
$ ii$Let $ z_1,z_2,z_3\in$ $ C1$
Proove that $ w \equal{} \frac {z_1 \plus{} z_2}{z_3} \plus{} \frac {z_2 \plus{} z_3}{z_1} \plus{} \frac {z_3 \plus{} z_1}{z_2}$ is real

$ 7$ Let $ z_1,z_2,z_3\in\mathbb{C}$  proove that $ e^{|z_1 \plus{} z_2 \plus{} z_3|} \plus{} |z_1 \plus{} z_2 \plus{} z_3| <\equal{} e^{|z_1| \plus{} |z_2| \plus{} |z_3|} \plus{} |z_1| \plus{} |z_2| \plus{} |z_3|$

1.[hide]$ z_{1}, z_{2} and z_{3}$ are the complex cube roots of unity.
$ z_{1}\equal{}1, z_{2}\equal{}w^{p}, z_{3}\equal{}w^{2p}$, where p is not a multiple of 3.
$ z_{2}\equal{}\frac{\minus{}1\plus{}\sqrt{3}i}{2}  and  z_{3}\equal{}\frac{\minus{}1\minus{}\sqrt{3}i}{2}$[/hide]

4.[hide]here, $ z_{1}\equal{}1, z_{2}\equal{}w^{p}, z_{3}\equal{}w^{2p}$, where p [b]is a multiple of 3[/b]
so, $ z_{1}\equal{} z_{2} \equal{} z_{3}\equal{}1$[/hide]

5.and 7.[hide]$ (z_{1}\plus{}z_{2}\plus{}z_{3})(\frac{1}{z_{1}}\plus{}\frac{1}{z_{2}}\plus{}\frac{1}{z_{3}})
\equal{}(z_{1}\plus{}z_{2}\plus{}z_{3})(\frac{\overline{z_{1}}}{|z_{1}|^{2}}\plus{}\frac{\overline{z_{2}}}{|z_{2}|^{2}}\plus{}\frac{\overline{z_{3}}}{|z_{3}|^{2}})$
which reduces to
$ \frac{|z_{1}\plus{}z_{2}\plus{}z_{3}|^{2}}{9}$
now for any  complex numbers, we have the inequality,
$ |z_{1}|\plus{}|z_{2}|\plus{}|z_{3}|\geq|z_{1}\plus{}z_{2}\plus{}z_{3}|$
and we're done.[/hide]

question 1) hell_ever, it doesn't follow that z1,z2,z3 are cube roots of unity.... e.g. $ z_1 \equal{} e^{i*\Pi/3},z_2 \equal{} e^{3*i*\Pi/3},z_3 \equal{} e^{5*i*\Pi/3}$

q1i) $ z_1 \equal{} cos(\alpha) \plus{} isin(\alpha)$
$ z_2 \equal{} cos(\beta) \plus{} isin(\beta)$
$ z_3 \equal{} \minus{} (cos(\alpha) \plus{} cos(\beta)) \minus{} i(sin(\alpha) \plus{} sin(\beta))$
as $ |z_1| \equal{} 1$
$ 1 \equal{} cos^2(\alpha) \plus{} cos^2(\beta) \plus{} 2cos(\alpha)cos(\beta) \plus{} sin^2(\alpha) \plus{} sin^2(\beta) \plus{} 2sin(\alpha)sin(\beta)$
$ \minus{} 1 \equal{} 2(cos(\alpha)cos(\beta) \plus{} sin(\alpha)sin(\beta))$
$ \frac { \minus{} 1}{2} \equal{} cos(\alpha \minus{} \beta)$
$ \alpha \minus{} \beta \equal{} \frac {2\pi}{3}$ (there are other solutions that must also be considered but I don't have much time... :( )
so using this method we can see that these complex numbers form an equilateral triangle centred at the origin and the result follows...

Considered as vectors we can say that $ z_1$, $ z_2$ and $ z_3$ are the sides of an equilateral triangle(with side $ 1$) (just put the tail of one in the head of another... since they sum zero it would be a close poligonal, with trhee equal sides...)

i) Thus each of the $ |z_1 \minus{} z_2|$ is the lenght of the base of a triangle with two sides with lengh $ 1$ and the angle between them is $ 120$.
From cosine's law:

 $ |z_1 \minus{} z_2|^2 \equal{} 1^2 \plus{} 1^2 \minus{} 2*1*1\cos120 \equal{} 2 \plus{} 1 \equal{} 3$

ii) Let be $ z_1 \equal{} \cos\alpha \plus{} i\sin\alpha$ and $ z_2 \equal{} \cos\beta \plus{} i\sin\beta$, with $ 0\leq\alpha<\beta<2\pi$
$ \overline{z_1}z_2 \equal{} \frac {|z_1|^2z_2}{z_1} \equal{} \frac {z_2}{z_1} \equal{} \frac {\cos\alpha \plus{} i\sin\alpha}{\cos\beta \plus{} i\sin\beta} \equal{} \frac {(\cos\beta \minus{} i\sin\beta)(\cos\alpha \plus{} i\sin\alpha)}{\cos^2\beta \plus{} \sin^2\beta}$
$ \equal{} (\cos\beta\cos\alpha \plus{} \sin\beta\sin\alpha) \plus{} i(\sin\beta\cos\alpha \minus{} \sin\alpha\cos\beta))$
$ \equal{} \cos(\beta \minus{} \alpha) \plus{} i\sin(\beta \minus{} \alpha)$
Just note that the possible values for $ \beta \minus{} \alpha$ are $ 60, \minus{} 60, \minus{} 120, 120, 240,\minus{}240$ so $ Re(\overline{z_1}z_2)\ge \minus{} \frac {1}{2}$
I might be wrong but I think it is never less than $ \minus{} \frac {1}{2}$

iii) What does $ M(z_1)$ means :?:

 :)

2) As $ z_1\plus{}z_2\plus{}z_3\minus{}1\equal{}0$ we have a cuadrilateral with all side's lenghts $ 1$ using triangle inequallity we can prove it is a convex cuadrilateral, from here it follows that it is a parallelogram. It is not hard to see that each of the opposite sides are $ z$ and $ \minus{}z$ so we that one side is the vector $ 1$

4) let's name $ 0 \leq Arg(z_i)\equal{}\theta_i < 2\pi$
We see that $ \cos\theta_1\plus{}\cos\theta_2\plus{}\cos\theta_3 \leq 3$ with equality if and only if $ \cos\theta_i\equal{}1$, so it follows(?) that $ z_1\equal{}z_2\equal{}z_3$

 :)


trigonometry
calculus
calculus computations

We're just starting this topic, and I have some problems isolating r.

Convert to polar form:

a) (x^2+y^2)^1/2 - x = 5

b) x^4+2x^2y^2+y^4=2xy

I know this is about substituting with rcos(t)=x, rsin(t)=y, r^2=X^2+y^2, but like I said I'm having trouble isolating r. Can anyone show me what steps to use in solving these, and some other basic information to keep in mind besides the above. I have the answers for them, so I know I did something wrong, but that's irrelevant, I'm more interested in how to go about solving them.

First of all, you can't always isolate $ r$ or $ \theta$. It's true, however, that most equations in this section of textbook are rigged so that they become much simpler in the polar form, and can often be solved for $ r$.

In (a) you get $ r \minus{} r\cos t \equal{} 5$ - is it hard to isolate $ r$ here?
In (b) it helps to recognize the left side as $ (x^2 \plus{} y^2)^2$. In general, keep an eye open for $ x^2 \plus{} y^2$ - sometimes it is disguised.

Yeah, I know that you can't always isolate, but 95% of the cases we get you can, so that's why I phrased it like that. As for the isolating, I meant to ask getting it in simpler form(s), as I seem to usually stop halfway through.


Problem 1: Given $ a,b,c \geq 0$.Prove that
$ \frac{a^3\plus{}b^3\plus{}c^3}3 \geq abc\plus{}\frac3{4}|(a\minus{}b)(b\minus{}c)(c\minus{}a)|$

Problem 2:Given $ a,b,c \geq 0$ and $ a\plus{}b\plus{}c\equal{}3$.Prove that
$ \sum_{cyc} \sqrt{\frac a{3b^2\plus{}1}} \geq \frac3{2}$

[quote="dduclam"]Problem 1: Given $ a,b,c \geq 0$.Prove that
$ \frac {a^3 \plus{} b^3 \plus{} c^3}3 \geq abc \plus{} \frac3{4}|(a \minus{} b)(b \minus{} c)(c \minus{} a)|$
[/quote]
See the stronger problem here:
http://www.artofproblemsolving.com/Forum/viewtopic.php?t=101916

[b]And [/b]
[quote="dduclam"]
Problem 2:Given $ a,b,c \geq 0$ and $ a \plus{} b \plus{} c \equal{} 3$.Prove that
$ \sum_{cyc} \sqrt {\frac a{3b^2 \plus{} 1}} \geq \frac3{2}$[/quote]
[b]?[/b]


Determine whether it's true or not : there exists infinitely many primes p in which p=n^(2)+1 form (n is integer)

sorry for latex.

These are known as Landau primes.
The fifth Hardy-Littlewood conjecture says there are an infinite number but it is unresolved.


If A, B, C, D, E, and F are all real numbers and A/B = C/D = E/F = 3/5, calculate the   value of

4E+3A-2C
__________
4F+3B-2D

Let $ 3x=A$, $ 3y=C$, $ 3z=E$, and $ 5x=B$, $ 5y=D$, $ 5z=F$. Now, we just substitute this into the fraction:
\begin{eqnarray*} \frac{4E+3A-2C}{4F+3B-2D} &=& \frac{4\cdot 3z+3 \cdot 3x - 2 \cdot 3y}{4 \cdot 5z + 3 \cdot 5x - 2 \cdot 5y} \\ \\
&=& \frac{12z+9x-6y}{20z+15x-10y} \\ \\
&=& \frac{3(4z+3x-2y)}{5(4z+3x-2y)} \\ \\
&=& \boxed{\frac{3}{5}} \end{eqnarray*}

Note that the second step could have been skipped, but that just shows how it factors. Also, this method can be used to show that if $ \frac{a}{b}=\frac{c}{d}$, then $ \frac{a+b}{c+d}=\frac{a}{b}$.

jUst Saying power

You can just assume $ A$=$ C$=$ E$=$ 3$

AND $ B$=$ D$=$ F$=$ 5$

So you can just comute what the given fraction equals

[quote="basketball9"]You can just assume $ A$=$ C$=$ E$=$ 3$ AND $ B$=$ D$=$ F$=$ 5$[/quote]
Technically, you can't assume that (for a proof or complete solution), but it still gives the right answer and always works.


[hide]This time, the robber needs to run for 9 hours (starts out with one mile, gains a mile every hour, needs 10 miles total) and the hound is still running at 10 mph, so it runs a total of 9*10=90 miles.[/hide]

Well, the robber is running for 9 hours, so the hound also runs for nine hours, at 10 mph, so he runs for 90 miles. :D

A+MATH wrote:A robber starts one mile ahead of a cop. The robber runs one mile per hour faster than the cop. The robber must run until he is 10 miles away from the cop to get to safety. A bloodhound runs between them back and forth at 10 miles per hour. (It's like the old one only the robber gets away.)

How many miles will the dog run?

[hide]I'll set it up as an equation.

I'll say that the robber runs at 11 miles per hour, and the cop will run at 10.

x = how long it takes for the robber to get away

11x + 1 = 10x + 10

x + 1 = 10

x = 9

So it takes 9 hours for the robber to get away.  Since the bloodhound is running the same speed as the cop (in my example), all we have to do it multiply 9 by 10, to get 90 miles.[/hide]

The robber runs for 9 hours total, and the bloodhound is running at 10mph, so the answer would be 9*10, or 90.


integration
linear algebra
linear algebra theorems

It's kind of dumb question..
But, I just wanted to make sure.
Are Parseval's equality and Parseval's theorem same thing? (In terms of Fourier series)
i.e. do both mean $ \frac{1}{L}\int_c^{c\plus{}2L}|f(x)|^{2}dx \equal{} \frac{a_0^2}{2}\plus{}\sum_{n\equal{}1}^{\infty}[|a_n|^{2}\plus{}|b_n|^{2}]$

An equally dumb answer: the difference is that Parceval identity is the formula itself and the Parceval theorem is the statement with proof that it holds.


Let $ x_{ij}$  belong to the positive integers.  An epression is:

$ \sqrt [x_{31}]{ \frac {x_{11}}{x_{21}}} \plus{} \sqrt[x_{32}]{\frac {x_{12}}{x_{22}}} \plus{} ... \plus{} \sqrt [x_{3n}]{ \frac{x_{1n}}{x_{2n}}}$

where, because they're more difficult to read, I'm clarifying that the indices are
$ x_{13}, x_{32}, ..., x_{3n}$, respectively.

The number of expressions is:

A) countably infinite

B) not countably infinite

The number of finite words on a countable language is countable.


[color=darkred]Let $ n\ge 3$ be a natural number and let $ x_k\ ,\ k\in\overline {1,n}$ be $ n$ positive numbers for which denote

$ 2s \equal{} \sum_{k \equal{} 1}^nx_k$ . Prove that  $ s\ > \ \max_{1\le k\le n}\ x_k\ \Longrightarrow\ \prod_{k \equal{} 1}^n\frac {(n \minus{} 3)s \plus{} x_k}{s \minus{} x_k}\ \ge\ (n \minus{} 1)^n$ .[/color]

[color=darkblue][b]Particular cases[/b] ([u]standard notations[/u]).

$ \odot\ (n \equal{} 3)$ . In a triangle $ ABC$ exists the inequality $ \frac {abc}{(p \minus{} a)(p \minus{} b)(p \minus{} c)}\ \ge\ 8$ .

$ \odot\ (n \equal{} 4)$ . In a convex quadrilateral  $ ABCD$ exists the inequality $ \prod\ \frac {p \plus{} a}{p \minus{} a}\ \ge\ 81$ .[/color]

Hint:
1)Substitution $ s\minus{}x_k\equal{}y_k$
and 2) AM-GM


vector
analytic geometry
graphing lines
slope
AMC 10

I'd imagine the lengths of the tangents are quite simple. Let $ r$ be the radius of the circle, and $ l$ be the distance from the point to the center of the circle. The length of the tangent is $ \sqrt{l^2\minus{}r^2}$, after applying the Pythagorean theorem.

Finding the equatino for such a line would be slightly more difficult. One possible approach; let $ (x, y)$ be a point on the circle. Let $ (x_c, y_c)$ be the center of the circle, and let $ (x_p, y_p)$ be the point in question. Therefore, the vectors $ (x_c \minus{} x)\vec{i}  \plus{} (y_c \minus{} y)\vec{j}$ and $ (x_p \minus{} x)\vec{i}  \plus{} (y_p \minus{} y)\vec{j}$ must be orthogonal. Then, you can find the points on the circle, and use point-slope form to get the equations.

[quote="Ihatepie"]Is there any quick way to find the equations of a tangent from a point to a circle and/or the lengths of the tangents? I created a technique, but it's very long and I'm sure better ones exist. Also, if you have the time, can you post a practice problem or two?

Thanks in advance.[/quote]

This may or may not be the type of problem you are looking for, but here it is:

[url=http://www.artofproblemsolving.com/Wiki/index.php/2009_AMC_10B_Problems/Problem_16]2009 AMC 10 Problem[/url]


[quote="hien"]Find all function $f: \mathbb{R}\to\mathbb{R}$ satifying condition:
$f(f(x)-y)=f(x)-f(y)+f(x)f(y)-xy$  $\forall x,y\in\mathbb{R}$[/quote]

[hide="A beginning"]
Setting $y = 0$ gives $f(f(x)) = f(x)-f(0)+f(x)f(0)$.

Then setting $y = f(x)$ gives
$f(0) = f(x)-f(f(x))+f(x)f(f(x))-xf(x) = f(x)-(f(x)-f(0)+f(x)f(0))+f(x)(f(x)-f(0)+f(x)f(0))-xf(x) = f(0)-f(x)f(0)+f(x)^{2}-f(x)f(0)+f(x)^{2}f(0)-xf(x)$

so $0 = f(x^{2})(1+f(0))-f(x)(2f(0)+x)$

If $1+f(0) = 0$ then for all $x$ $f(x)(2f(0)+x) = 0$, but putting $x = 0$ gives $-1 \cdot-2 = 0$, a contradiction.

Then for each $x$, $f(x) = 0$ or $f(x) = \frac{2f(0)+x}{1+f(0)}$.

Now, we need to figure out what possible values $f(0)$ can take and on what sets we can have $f(x) = 0$.[/hide]

Put $f(0)=a$. Substitute $y=0$, we have: $f(f(x))=(1+a)f(x)-a$
Again Substituting $y=f(x)\Rightarrow a=f(x)-f(f(x))+f(x)f(f(x))-xf(x)$ (*).
Alternativele, put $x=0,y=f(0)$ we get:
$a^{2}(a-1)=0\Rightarrow a=0$ or $a=1$
Put $a=0 \Rightarrow f(f(x)=f(x)\forall x\in\mathbb{R}$, combine with (*) we have $f(x)=x$ (satified)
Put $a=1\Rightarrow f(f(x))=2f(x)-1$ combine with (*) we have $f(x)=\frac{1}{2}x+1$. Retry with condition of problem then it's not satified.
Rezult: $f(x)=x$ is only needed function!


Find all pairs of consecutive primes $(p,q)$  such that
$p^{2}-pq-q-2$ is also a prime number

Here's a clue that may get you started. If p and q are consecutive primes such that p = q + 2, then p^2 - pq - q - 2 is always a prime. In fact, the expression is equal to q.
p^2 - pq - p - 2
= (q + 2)^2 - (q+2)q - (q+2) - 2
= q^2 + 4q + 4 - q^2 + 2q - q - 2 - 2
= q (when simplified)

[quote="Kab^2"]
p^2 - pq - p - 2
=
[/quote]

A problem...this is $p^{2}-pq-q-2$...
Moreover, what if $p,q$ are not twin primes?

Pierre.

Oh, my bad, but it still won't matter. The end result will be q + 2 and not q as I earlier stated. But q + 2 = p, so it is still a prime.


Prove that for any real p>1 there exist a real number M st
1+1/(2^p)+1/(3^p)+1/(4^p)+....+1/(n^p)+....<M.

This should help: http://en.wikipedia.org/wiki/Harmonic_series_%28mathematics%29#P-series


What are base numbers and where can I learn about them?  Is it an algebra topic, arithmetic, or what?

I'm not sure what it would be classified as, but those would be numbers in different bases, e.g. $13_8$. The way it works is that each place in a number is the number there multiplied by the corresponding power of 10. It's a little complicated to explain, but easy enough one you understand it.

For example, $13_8=8+3=11_{10}$, $327_8=3\cdot 8^2+2\cdot 8+7=215_{10}$, $43_6=4\cdot 6+3=27_{10}$. Base 10 is the standard base that we use. ;) Just remember that any number in one of the places is always less than the base.

Base numbers are normally grouped into number theory.  They are covered thoroughly in the Introductory Number Theory course.  They are covered more briefly in the MATHCOUNTS Problem Series.

Mr. Rusczyk has a really delightful article in his blog about number bases, available [url=http://www.artofproblemsolving.com/Forum/weblog_entry.php?e=1409]here[/url].
I hope that helps!


There is a problem like this:
How to divide a number n into the sum of k integers, such that the product of the k numbers is maximum?
The answer is to choose as many '3' as possible, and the rest choose '2' such that there's no '1'.

What about if the restriction extend from integers to real numbers?
That is:
How to divide a number n into the sum of k real numbers, such that the product of the k numbers is maximum?

The answer should be different.
For example, when n=15, 2.5x2.5x2.5x2.5x2.5x2.5>3x3x3x3x3
But I can't figure out the pattern :(

For $ k\equal{}1$ nothing has to be shown. For $ k\equal{}2$ we have to maximise $ x(n\minus{}x)\equal{}\frac{1}{4}[n^2\minus{}(2x\minus{}n)^2]\leq\frac{n^2}{4}$ and equality occurs, iff $ 2x\equal{}n$. For $ k>2$ the maximum does not exist, e.g., choose $ x_1\equal{}x_2\equal{}\minus{}m,x_3\equal{}\ldots \equal{}x_k\equal{}\frac{n\plus{}2m}{k\minus{}2}$ for some $ m\in\mathbb{N}$, then $ \prod_{i\equal{}1}^kx_i\equal{}m^2\left(\frac{n\plus{}2m}{k\minus{}2}\right)^{k\minus{}2}\rightarrow\infty$ for $ m\rightarrow\infty$.

If you meant a partition by non-negative real numbers, you apply Lagrange multiplicators.

By AM-GM, you want all numbers to be equal. But later in your post, you seem to vary $ k$. Then you also need to find the optimal $ k$, so you want to maximise $ \left(\frac{x}{k}\right)^k$. It' optimal to get $ \frac xk$ close to $ e$:
Substitute $ y\equal{} \frac xk$, so you want $ y^{\frac 1y}$ to be maximal.

Actually, Lagrange multipliers and AM-GM would be too involving for the non-negative solution. For $ k=1$ nothing needs to be shown, so let $ k>1$ and $ x_1,\ldots,x_k$ be a choice, where the maximum is attained. Clearly, we can assume $ \forall i: \,x_i>0$. If $ x_i<x_j$ holds, then we set $ y_s=x_s$ for $ s\not=i,j$ and $ y_i=y_j=\frac{x_i+x_j}{2}$. Then $ \sum_{i=1}^ky_i=\sum_{i=1}^kx_i=n$ and

${{{ \prod_{i=1}^ky_i=\prod_{i=1}^kx_i}\cdot\frac{(x_i+x_j)^2}{4x_ix_j}=\prod_{i=1}^kx_i}\cdot\frac{(x_i-x_j)^2+4x_ix_j}{4x_ix_j}>\prod_{i=1}^kx_i}$, a contradiction. So the maximum is attained for $ x_1=\ldots =x_k=\frac{n}{k}$

AM-GM is involving¿
You did nothing else than proving Am-GM, so instead of using the so involved theorem, you recited some proof (to be exact, a bit but not much less)

Well, in order to be really exact, pls observe the following: the problem post is equivalent to AM-GM, so its post and its particular description at the outset implies that the (true) knowledge of AM-GM was not available, hence, citing it might be considered pointless by some.

So, the problem solution should be geared towards the problem description and the likely knowledge available. Admittedly I did not stick to this rule as well  :mad: , when I cited the Lagrange multipliers (and this explains the motivation of my 2nd post)


[quote="Frozen"]What is the planar graph of a polyhedra?[/quote]

It's planar map of polyhedron.

It's such grapf: its vertices correspond to the vertices of polyhedron and its edges correspond to te edges of polyhedron.
It is planar :)


modular arithmetic
number theory
prime factorization

$ n\in \mathbb{N}$ , if $ \frac{n(n\plus{}1)(2n\plus{}1)}{6}$ is a multiple of $ 2008$, find the smallest possible value of $ n$.

:juggle:

[hide]Prime factorize $ 2008$ to discover that it is $ 2^3*251$.  Notice that $ 2n \plus{} 1$ is always odd, and either $ n$ or $ n \plus{} 1$ is even.  From the prime factorization given and the denominator, the even number must be divisible by at least $ 2^4 \equal{} 16$.  One of the terms must be divisible by $ 251$.  Now we have some casework.

Case 1: $ n \equal{} 251k$
Then, if $ k < 16$, $ 251k \plus{} 1$ is divisible by $ 16$.  $ 251 \equiv \minus{} 5 \pmod {16} \rightarrow \minus{} 5k \plus{} 1 \equiv 0 \pmod {16} \rightarrow \minus{} 5k \equiv \minus{} 1 \pmod {16}$, so $ k \equal{} \frac {16x \plus{} 1}{5}$.  $ x \equal{} 4, k \equal{} 13$ is the smallest solution.

Case 2: $ n \plus{} 1 \equal{} 251k$
If $ k < 16$, then $ n \equal{} 251k \minus{} 1 \equiv 0 \pmod {16} \rightarrow k \equal{} \frac {16x \plus{} 1}{251}$.  The first solution of this occurs when $ x \equal{} 47, k \equal{} 3$.

Case 3:  $ 2n \plus{} 1 \equal{} 251k$
Then $ n\equal{} \frac{251k\minus{}1}{2}$.  If $ n \equiv 0 \pmod {16}$, then $ 251k\minus{}1 \equiv 0 \pmod {32}$.  $ k\equal{}\frac{32x\plus{}1}{251}$, whose first solution comes when $ x\equal{}149, k\equal{}19$.  If $ n\plus{}1 \equiv 0 \pmod {16}$, then $ 251k\plus{}1 \equiv 0 \pmod {32} \rightarrow k\equal{}\frac{32x\minus{}1}{251}$.  The first solution occurs when $ x\equal{}102, k\equal{}13$.

Thus, when $ k \equal{} 3$, we have the $ 752*753*1505$, so $ n \equal{} 752$.[/hide]


Prove that for every integer $n > 0$ there exists an integer $k > 0$ such that $2^nk$ can be written in decimal notation using only the digits $1$ and $2$.

A nice problem.

[hide]
Suppose that for some n there exists k such that 2^n * k has only 1's and 2's in it's decimal expansion.  Also assume that it's decimal expansion has n digits.  We know that 2^n * k = 0 or 2^n mod 2^(n+1).  Also, 10^n = 0 mod 2^n, and thus we can find that 10^n = 2^n mod 2^(n+1).
Suppose 2^n * k = 2^n mod 2^(n+1).  Then 10^n + 2^n * k :equiv: 2*2^n :equiv: 0 mod 2^(n+1).  Thus 10^n + 2^n * k is a n+1 digit number divisible by 2^(n+1) made up of only 1's and 2's.
Now suppose 2^n * k :equiv: 0 mod 2^(n+1).  Then 2*10^n :equiv: 0 mod 2^(n+1).  So 2*10^n + 2^n * k :equiv: 0 mod 2^(n+1).  2*10^n + 2^n * k is an n+1 digit number divisible by 2^(n+1) made up of only 1's and 2's.
However, 2^1 * 1 is a 1-digit number with only 1's and 2's in it's decimal expansion.  So by induction, we are done.

Note how this gives us an algorithm to generate such numbers:

1.  Start with an n-digit number 2^n * k.
2.  Find a = 2^n * k mod 2^(n+1).
3.  If a=0, then 2 folllowed by 2^n * k is the n+1-digit number we are looking for.  If a=2^n, then 1 followed by 2^n * k is our n+1-digit number.

Example:

12 is a 2-digit number divisible by 2^2.
12 mod 2^3 = 4 = 2^2, so 112 is divisible by 2^3.
[/hide]

exactly it was like my solution ;)
and we can generalize the problem Prove that for every integer $n>0$ there exists an integer $k>0$ such that $2^nk$ can be written in decimal notation using only the digits $1$ and $2$,and it has just $n$ digits. :D


group theory
abstract algebra
superior algebra
superior algebra unsolved

Let $G$ be a group,and $H$ is a subgroup of $G$,and $G' \subset C(H)$,prove $H' \subset C(G)$

$G',H'$ be the commutator groups of $G,H$,respectively,
$C(H),C(G)$ is the center.$C(H)=\{g \in G| gh=hg,\forall h \in H\}$

[quote="Soarer"]I don't quite understand. Isn't $H' \subset G'$ and $C(H) \subset C(G)$?[/quote]
No,$C(G) \subset C(H)$... :D

[quote="zhaobin"][quote="Soarer"]I don't quite understand. Isn't $ H' \subset G'$ and $ C(H) \subset C(G)$?[/quote]
No,$ C(G) \subset C(H)$... :D[/quote]

Still not right. $ Z(G)\cap H\subseteq Z(H)$ is correct, but $ Z(G)\subseteq Z(H)$ is no true in general. At least if we're talking about the center. But the starter defines $ C(X)$ as the [i]centralisator[/i] of $ X$ in $ G$ (which is very different from the center). $ C_G(G)\equal{}Z(G)\subseteq C_G(H)$ is indeed correct.

[quote="zhaobin"]Let $ G$ be a group,and $ H$ is a subgroup of $ G$,and $ G' \subset C(H)$,prove $ H' \subset C(G)$

$ G',H'$ be the commutator groups of $ G,H$,respectively,
$ C(H),C(G)$ is the center.$ C(H) \equal{} \{g \in G| gh \equal{} hg,\forall h \in H\}$[/quote]

Here are two ways of phrasing a common proof:

If [G,G,H] = 1, then [H,G,H]=[G,H,H]≤[G,G,H]=1, so by the three subgroups lemma, [H,H,G]=1.

If h,k in H, g in G, then [h,g] in G' centralizes h and k, so [ h, k ] = [ h[h,g], k ] and similarly [h,k] centralizes h and k so [ h, k ] = [ h[h,g], k[k,g] ] = [ h^g, k^g ] = [ h, k ]^g, and g centralizes [h,k].


Gretchen has eight socks, two of each color: magenta, cyan, black, and white. She randomly draws four socks. What is the probability that she has exactly one pair of socks with the same color?

An alternate solution:

Because there are 4 socks chosen and exactly one pair of socks, there are two socks that do not match. There are $ (8$ choose $ 2)\minus{}4\equal{}28\minus{}4\equal{}24$ ways to pick two socks that do not match, because every group of 2 socks selected (other than the four matching pairs of socks) are mismatched. This number must be multiplied by 2, because there are 2 remaining colors to pick a matching pair from, so there are $ 48$ ways to pick 4 socks with exactly one matching pair. Since there are four socks being chosen and 8 to choose from, there are $ (8$ choose $ 4)\equal{}70$ possible outcomes. Therefore, the probability of picking exactly one matching pair of socks is $ \frac{48}{70}\equal{}\boxed{\frac{24}{35}}$

math154's complementary counting- balls and urns solution falters since the probability of different possiblities are not equal; for example, it is more likely for Gretchen to obtain a magenta and a black than to obtain two whites.

More concise, different wording of mathluver2.718's solution:

There are $ \binom{8}{4} \equal{} 70$ total possibilities.

For our desired possibilities:

$ 4$ ways to choose the pair of sock's color

$ \binom{3}{2} \equal{} 3$ ways to then choose the colors of the single socks

$ 2 \times 2 \equal{} 4$ ways to choose which single socks to use

Final probability is

$ \frac{4 \times 3 \times 4}{70}$

$ \boxed{24/35}$

I can only find 12 valid combinations (m = magenta sock, c = cyan sock, b = black sock, w = white sock):

mmcb
mmcw
mmbw

ccmb
ccbw
ccmw

bbmc
bbmw
bbcw

wwcb
wwmc
wwmb

and that would be 12/70 = 6/35.

What am I missing here? Sorry it is probably really simple I just cant seem to get it right now.

EDIT: ahh nevermind, there 2 socks of each color so lets say wwcb, theres 4 ways to choose which cyan and black socks to use, should have thought a little more before posting ;p

(4* 6*(2*1)*(6*4))/(8*7*6*5)

The 4 is the 4 different colors to choose from.
The 6 is the 6 different positions to choose from.
The 2*1 and 6*4 are the socks
and (8*7*6*5) is the total number of ways

= 24/35

Why wont this work? The first sock is selected with probability $8/8$. the next sock (finish the pair) has probability $1/7$.   Next is $6/6$, and the last sock is $4/5$. Then the number of permutations is $12$, so the answer is $48/35$.

I dont know why this isn't working.

@above how can a probability for this thing be over 1?

Why can't each pair of socks be the same?(Left sock magenta=Right sock magenta)

[hide=Super easy symmetry ]
First lets find the number of possible outcomes which is 8 choose 4
To find the number of successes lets  do casework based upon which color we are choosing to be a pair
Case 1:Magenta
There are 6 ways to choose the first non pair sock and 4 ways to choose the second non pair sock(you can't choose the other color on the first pick) since order does not matter we divide by 2 getting 6*4/2 ways to do this case.

Now notice that each of the 4 cases are symmetric so we just multiply 12 by 4 to get our # of successes which is 48
so our desired probability is 48/70=24/35
[/hide]

[hide=Solution]We can also do complementary counting.

There are only two possible cases where this cannot happen:

[b]Case I.[/b] All the socks have different color.

Since all the socks have different color, one sock from each color must have been chosen. This gives $2^4$ ways, so the probability is
$$\frac{2^4}{\binom84}=\frac8{35}$$

[b]Case II.[/b] There are two pairs of socks with the same color.

This means, there are only two colors present, and both of the socks from those colors are chosen. There are $\binom42$ ways to choose the colors, so the probability is
$$\frac{\binom42}{\binom84}=\frac3{35}$$

This gives us our answer:
$$1-\left(\frac8{35}+\frac3{35}\right)=\boxed{\frac{24}{35}}$$

[hide= different logical solution]
we have $4$ options for which color to choose the pair of socks
then we have 6 choices for the next sock: imagine there are these socks and from our 4 choices we pick white:
[s]WW[/s] GG BB RR
there are $6$ choices for the next sock and then for the next one there are $4$ choices, not 5 because we can't have two of the same color. However, order doesn't matter!!!!! so we have to divide by two:
$\frac{6 \cdot 4}{2} = 12$
combining that with the 4 different choices for the color of the sock we have $4 \cdot 12 = 48$
finally there are $8$ choose $4$ $=70$ ways to pick the socks so $48/70 = \boxed{24/35}$
[/hide]

[quote=math154]This means that $ m\plus{}c\plus{}b\plus{}w\equal{}4$, where $ 0\leq m,c,b,w\leq2$. The total number of cases for this is $ \dbinom{4\plus{}4\minus{}1}{4\minus{}1}\minus{}4\minus{}4\cdot3\equal{}19$ (subtract the case in which one of $ m,c,b,w$ is $ 4$, and the case in which one of $ m,c,b,w$ is $ 3$ and another is $ 1$). The number of cases in which there is exactly one pair is there the variables $ m,c,b,w$ represent some permutation of $ 2,1,1,0$, so there are $ \frac{4!}{2!}\equal{}12$ cases with exactly one pair. Thus, our final answer is $ \boxed{\frac{12}{19}}$.[/quote]

You made a mistake

[quote=V_V]i think people corrected it after...
plus it was in 2008 lol :P[/quote]

Yeah but he was last active yesterday so he might still see the message

this is my new favorite problem on Alcumus

because it gave me "counting master"

[quote=Tong.Qiu]The problem is way too easy for Level 25 +125 XP. Idk if anyone solved the problem as I did, but if not:
[hide=Alternate solution]
There are a total of $\binom{8}{4} = 70$ combinations of socks. There are 4 ways to choose the color of sock that will be chosen twice, and there are $\binom{6}{2} = 15$ to choose the rest of the socks, so there are $4\cdot15=60$ ways to choose the socks, but we also want to exclude the ways that she chooses 2 pairs of socks, so we subtract 4*3=12 from 60 to get 48.
Therefore, the answer is $\frac{48}{70} = \boxed{\frac{24}{35}}$

[/hide][/quote]

bruh i totally agree this is wayyyy too easy for lvl 25 125 xp except i misread at first and thought it said "at least one pair of socks with the same color" ooooooops

[hide=Solution]Assume all socks are distinguishable. 8C4 ways total. 4 ways to choose the matching color, 3 ways to choose 2 colors, 2*2 ways to choose socks from those colors. Answer is 4*3*2*2/70=24/35.[/hide]

[quote=Tong.Qiu]The problem is way too easy for Level 25 +125 XP. Idk if anyone solved the problem as I did, but if not:
[hide=Alternate solution]
There are a total of $\binom{8}{4} = 70$ combinations of socks. There are 4 ways to choose the color of sock that will be chosen twice, and there are $\binom{6}{2} = 15$ to choose the rest of the socks, so there are $4\cdot15=60$ ways to choose the socks, but we also want to exclude the ways that she chooses 2 pairs of socks, so we subtract 4*3=12 from 60 to get 48.
Therefore, the answer is $\frac{48}{70} = \boxed{\frac{24}{35}}$

[/hide][/quote]

I don't get it, why do we subtract 4*3? Wouldn't we just subtract 4c2=6 because it doesn't matter what order the pairs are chosen? (For example, we could choose the magenta sock pairs then the cyan sock pairs, or the cyan sock pairs then the magenta, but it would get counted as the same case.)

Oh nvm, I understand, it's because the order matters when we do $4*15.$

Were we supposed to assume the socks of each color were distinct from each other? :huh:


AMC USA(J)MO USAMO probability algebra polynomial quadratics

heeey everyone. okok so i have a bunch of these proof questions, and they're basically USAMO-related questions. can anyone help me with any of these? thanks~ ;)

1. Solve in real numbers the system of equations.
    (3x+y)(x+3y)((xy)^(1/2))=14
    (x+y)(x^2+14xy+y^2)=36
2. Find the probability that in a group of n people there are two with the same birthday.  Ignore leap years.
3. Show that all real roots of the polynomial P(x)=x^5-10x+35 are negative.
4. Prove that for any nonnegative integer n, the number 5^5^(n+1)+5^5^n+1 is not prime.
5. If a, b, c are positive numbers, prove that
   9*a^2*b^2*c^2 <= ((a^2)b+(b^2)a+(c^2)a)((a(b^2)+b(c^2)+c(a^2)).

Thanks again.

[hide="#1"]
$ (3x \plus{} y)(x \plus{} 3y)\sqrt {xy} \equal{} 14$

$ (x \plus{} y)(x^2 \plus{} 14xy \plus{} y^2) \equal{} 36$

Let $ x \plus{} y \equal{} p$ and $ xy \equal{} q^2$ ($ xy \ge 0$ since the first equation has a $ \sqrt {xy}$ term)

Then we have $ (3p^2 \plus{} 4q^2)q \equal{} 14$ and $ p(p^2 \plus{} 12q^2) \equal{} 36$

Adding 2 of the first to the second, we get $ (p \plus{} 2q)^3 \equal{} 64$ <--> $ p \plus{} 2q \equal{} 4$
Taking the second, and subtracting from it 2 of the first, we get $ (p \minus{} 2q)^3 \equal{} 8$ <--> $ p \minus{} 2q \equal{} 2$

Therefore, we get $ p \equal{} 3$ and $ q \equal{} \frac {1}{2}$. Plugging in to the first and the second equations, we see that this pair of p and q works.

Therefore, we must have $ x \plus{} y \equal{} 3$ and $ xy \equal{} 1/4$, so x,y are the 2 roots of $ t^2 \minus{} 3t \plus{} \frac {1}{4}$ which has roots by the quadratic equation $ \frac {3\pm\sqrt {8}}{2}$.

Hence, the 2 ordered pairs of $ (x,y)$ are $ (\frac {3 \plus{} \sqrt {8}}{2},\frac {3 \minus{} \sqrt {8}}{2})$ and $ (\frac {3 \minus{} \sqrt {8}}{2},\frac {3 \plus{} \sqrt {8}}{2})$[/hide]

[hide="#5"]
By AM-GM,
$ a^2b \plus{} b^2c \plus{} c^2a\ge 3\sqrt [3]{a^2b*b^2c*c^2a} \equal{} 3abc$
$ ab^2 \plus{} bc^2 \plus{} ca^2\ge 3\sqrt [3]{ab^2*bc^2*ca^2} \equal{} 3abc$
Inequality follows from multiplying these 2 inequalities together.
[/hide]

[hide="2"]
The probability that no two have the same birthday is $ \frac{365 \cdot 364 \cdots (365 \minus{} n \plus{} 1)}{365 \cdot 365 \cdots 365}$. Thus, the answer is $ 1 \minus{} \frac{365!}{(365 \minus{} n)!365^n}$.
[/hide]

[hide="3"]
If $ x \le 3.5$ then obviously $ P(x) > \minus{}10x \plus{} 35 \ge 0$; if $ x > 3.5$ then $ P(x) > x^5 \minus{} 10x > 0$. So no positive roots.
[/hide]

[hide="4"]
Let $ P(x) \equal{} x^5 \plus{} x \plus{} 1$ (here we have $ x \equal{} 5^{5^n}$).

Let $ \omega \equal{} e^{2i\pi / 3}$; since $ P(\omega) \equal{} P(\omega^2) \equal{} 0$, it follows that $ (x\minus{}\omega)(x\minus{}\omega^2) \equal{} x^2 \plus{} x \plus{} 1 | x^5 \plus{} x \plus{} 1$. Hence $ \left(5^{5^n}\right)^2 \plus{} \left(5^{5^n}\right) \plus{} 1 | \left(5^{5^n}\right)^5 \plus{} \left(5^{5^n}\right) \plus{} 1$.

[/hide]


A square pyramid has a base with area of 100 sq. cm. The slant height of the pyramid is 13 cm. A plane which is parallel to the base of the pyramid cuts off the top of the pyramid exactly half way between the apex of the pyramid and the base of the pyramid creating two pieces, a new pyramid and a frustum. How many cubic centimeters are in the volume of the frustum?

[hide="Solution, I think!"]We can see that the height of the smaller pyramid created is going to be 6, and the side lengths of the base are 5.  So the volume of the top is 1/3*6*25= 50, and the volume of the whole thing is 1/3*12*100= 400.  Thus 400-50, which is 350 sq cm.[/hide]

[hide]I got 350 in my head while doing this and with the same method.   :coolspeak: [/hide]

[hide]Or that the small pyramid is half the large pyramid in every way and since it is 3D, it is 1/2^3=1/8 the volume of the large pyramid.[/hide]

Actually, I looked up some information about frustums.  Here it is.


Find the number of lattice points in side the circle $ x^{2} \plus{} y ^{2} \equal{} a^{2}$.

Is there a single expression as an answer?? I'm getting something like this. If this is right, then I'll work on it and give you the answer.

$ \sum_{r\equal{}1}^n [\sqrt{a^2 \minus{} r^2}] \plus{} f(a)$

where $ [.]$ denotes the greatest integer function.

[quote="Euclidean Geometer"]Wats $ f$ macha?[/quote]

I'll give you the answer. Its actually something like 2(a-1) and then you have to multiply four times. Will edit this with the answer next time I come online.

But my question is, do you have a single answer or an expression (closed form) for the answer or this is the answer.

BTW Its from 0-> n-1 not n.


Posts in Games and Fun Factory don't count towards your post count.

And the objective of this forum is not to vote on things, I don't get why you care about it so much...

Dude... POSTS HERE DON'T COUNT.

Go solve a math problem  :rotfl:

If you haven't noticed, your post count is still 0.

[quote="Fantabulate"]i think that it's not fiar to have to post25 times to vote on messages.

what!!!!!!??? i can't do consecuive posting!!!!??? :cursing:

At this rate it will take more than four hours!!!???

 :spam:  :spam:[/quote]

Okay,

1.) Post count does not really matter.
2.) Like the number of people above me, posts in G&FF do not count.
3.) You can answer MULTIPLE topics within 10 minutes, you can't just post in the same thread within 10 minutes.

Best laugh I've had for a while.

Why do you want to vote so badly, anyway?


Prove that there exist infinitely pair of natural numbers $ (m,n)$ such that $ 2m\plus{}6n\minus{}1$ and $ 26m\plus{}90n\minus{}1$ are both perfect square.

[quote="dyd"]Prove that there exist infinitely pair of natural numbers $ (m,n)$ such that $ 2m \plus{} 6n \minus{} 1$ and $ 26m \plus{} 90n \minus{} 1$ are both perfect square.[/quote]
I has found that there exist $ (m,n)\in N$ such that both of them  is p th power of a number?
I will post solution if no one reply it.

@TTsphn: Do you mean you got infinitely many pair of such $ (m,n)$? I'd love to see your solution, maybe you got another approaching. :)

consider system:
$ 2m \plus{} 6n \minus{} 1 \equal{} a^p$
$ 26m \plus{} 90n \minus{} 1 \equal{} b^p$
$ D \equal{} 24$
$ D_{m} \equal{} 90(a^p \plus{} 1) \minus{} 6(b^p \plus{} 1) \equal{} 90a^p \minus{} 6b^p \plus{} 84$
$ D_n \equal{} 2(b^p \plus{} 1) \minus{} 26(a^p \plus{} 1) \equal{} 2b^p \minus{} 26a^p \minus{} 24$
So
$ m \equal{} \frac {D_{m}}{D} \equal{} \frac {15a^p \minus{} b^p \plus{} 14}{4}$
$ n \equal{} \frac {D_{n}}{D} \equal{} \frac {b^p \minus{} 13a^p \minus{} 12}{12}$
Now chose : $ (a,b)\in N$ to
$ 4|15a^p \minus{} b^p \plus{} 14\Leftrightarrow 4|a^p \plus{} b^p \plus{} 2$
$ 12|b^p \minus{} 13a^p \minus{} 12\Leftrightarrow 12|b^p\minus{}a^p$ Chose $ a\equiv b \equiv 1(\mod 12)$then
Then $ (m,n)\in Z$


:ninja: i figured this problem in the book out, but I don't get part of the solution.

"1/3^-2" I figured that I could use the reciprocal and figure out the answer "9," but the book figured it out like:
1/3^-2 = 1/1/3^2 (1 over 1 over 3 squared) = 1/1/9 (1 over 1-ninth) = 9.
how does 1 over one-ninth equal nine? I'm on the verge of figuring out  :blush: ...but please help! :lol:

Remember:

1/x is the reciprocal of x

x=(1/9)

1/(1/9)=9

....because the reciprocal of 1/9 is 9/1, or 9.

or by division

1=numerator

1/9=denominator

if a number is divided by a fraction then the fraction must be its reciprocal and multiply to the number

1/(1/9)= 1*(9/1)

Or just count: how many $ \frac{1}{9}$ make up a $ 1$. The name obviously says $ 9$.

as you know 1/1/9 multiply by nine and yep your answer is 9. If you found this hard then graphing is much more difficult but fun after you ace your exam.

Another way to view it would be to multiply the fraction $ \frac{1}{\frac{1}{9}}$ by $ 1$ in the form of $ \frac{9}{9}$ to clear the demominator. We are left with $ \frac{9}{1}$ which is $ 9$.


real analysis
calculus
integration
function
probability
parameterization
probability and stats

How to sum two iid normal distribution by using convolution?
I tried but cannot handle the complex integration in the computation.

Also, how to sum two iid negative binnomial distribution by using convolution?
(e.g. X~NegBin(r1,p) & Y~NegBin(r2,p)
I also tried but I cannot continue when it is needed to prove
[url]http://img195.imageshack.us/img195/3463/200911191025.jpg[/url]

In both cases, using a transform method (either characteristic function or moment generating function) is far, far easier then trying to do the convolution directly. I strongly recommend using such methods.

Here's a second idea for the negative binomial. We have $ X$ is distributed according to the negative binomial with probability $ p$ and count parameter $ r.$ And $ Y,$ independent of $ X$ is so distributed with the same $ p$ and with count parameter $ s.$ Go back to the roots: $ X$ counts how many terms it takes to have $ r$ successes, and $ Y$ counts how many terms it takes to have $ s$ successes. That actually means that $ X$ is itself the sum of $ r$ independent geometric distributions and $ Y$ is the sum of $ s$ independent geometric distributions. Add then together and we get something that is the sum of $ r\plus{}s$ geometric distributions, which is then a negative binomial with count parameter $ r\plus{}s.$

OK, you're really stubborn and insist on doing analysis instead of probability - what then?

$ P(X\equal{}x)\equal{}\binom{x\minus{}1}{r\minus{}1}p^r(1\minus{}p)^{x\minus{}r}$ for $ x\ge r$.

$ P(Y\equal{}y)\equal{}\binom{y\minus{}1}{s\minus{}1}p^s(1\minus{}p)^{y\minus{}s}$ for $ x\ge s$.

The convolution is:

$ P(X\plus{}Y\equal{}z)\equal{}\sum_{x\equal{}r}^{z\minus{}s}\binom{x\minus{}1}{r\minus{}1}\binom{z\minus{}x\minus{}1}{s\minus{}1}p^z(1\minus{}p)^{z\minus{}(r\plus{}s)}$

Note that the powers of $ p$ and $ (1\minus{}p)$ are constant in that sum. All we have to do is prove a binomial coefficient identity:

$ \sum_{x\equal{}r}^{z\minus{}s}\binom{x\minus{}1}{r\minus{}1}\binom{z\minus{}x\minus{}1}{s\minus{}1}\equal{}\binom{z\minus{}1}{r\plus{}s\minus{}1}.$

I'll let you think about that identity, but with the provision that this now has something to do with the hypergeometric probability distribution.

I know using characteristic function will make Q1 easier but I haven't learn characteristic function formally so I should use the only learnt method -- convolution.
...
Actually, I have no any idea to prove $ \sum_{x\equal{}r}^{z\minus{}s}\binom{x\minus{}1}{r\minus{}1}\binom{z\minus{}x\minus{}1}{s\minus{}1}\equal{}\binom{z\minus{}1}{r\plus{}s\minus{}1}.$


There's no motivation in using a geometric series since the terms are so small and few, so it's just $ 7^0\plus{}7^1\plus{}7^2\equal{}1\plus{}7\plus{}49\equal{}50\plus{}7\equal{}\boxed{57}$.


geometry
ratio
topology
inequalities
logarithms
advanced fields
advanced fields open

I wonder if we can find fractal objects with some kind of non-zero measures

For example in dimension 1, the lengths of two straight lines are comparable. Are there any two fractal objects with the same fractional dimension and their "fractal measure" (i.e. length or area) being comparable? :?:

Yeah, there should be.

See [url]http://mathworld.wolfram.com/HausdorffMeasure.html[/url].

The easiest way to construct a fractal so that you know its dimension is if it's self-similar. Then, it's just the log of the ratio for the size of the similarity to the number of points. So, the Cantor set has Hausdorff dimension $\frac{\log(2)}{\log(3)}$ since it is self-similar with the next layer being 2 copies of itself and 1/3 size.

So you can easily make two fractals of the same dimension by making them self-similar in the same pattern, but just starting with different shapes. For example, consider the Cantor set constructed from $[0, 1]$ and the Cantor set constructed from $[0, 2]$.

Unfortunately it is not as easy to determine the actual measure of a fractal set. I'm pretty sure the Cantor set constructed from $[0, 1]$ has measure 1, though. So it would be pretty hard to take two arbitrary fractals and compare their measure. It seems reasonable to conjecture that if you have two sets that are constructed in the same way (so that their dimension is the same), but ratio of measures of the parent sets is $r$, then the ratio of the measures of the fractals should still be $r$, but fractals are sometimes sort of nonintuitive, so I would hesitate to assert that without proof.

Suppose that $(X,d_{X})$ and $(Y,d_{Y})$ are metric spaces and $f\colon X\to Y$ is a [b]bi-Lipschitz[/b] map, which means that there exist constants $C_{1}$ and $C_{2}$ such that \[C_{1}d_{X}(a,b)\le d_{Y}(f(a),f(b))\le C_{2}d_{X}(a,b)\] for all $a,b\in X$. Then for any $s>0$ and any set $A\subset X$ we have \[C_{1}^{s}\mathcal H^{s}(A)\le \mathcal H^{s}(f(A))\le C_{2}^{s}\mathcal H^{s}(A),\] where $\mathcal H^{s}$ denotes the $s$-dimensional Hausdorff measure of a set. (Warning: this "measure" is in general an [b]outer measure[/b] - it need not be additive.)
This inequality is a fairly easy exercise in applying the definition of $\mathcal H^{s}$. But it implies, for example, that the $\frac{\log 2}{\log 3}$-dimensional measure of the middle-third Cantor set built on $[0,2]$ is $2^{\frac{\log 2}{\log 3}}$ times the measure of the the middle-third Cantor set built on $[0,1]$.

So, could we have boundaries of fractals

Boundary of a volumes are surfaces
Boundaries of surfaces are cureves

How about boundaries of fractals??


[quote="Sakis"]Εστω $ ABC$ τριγωνο με $ AD,BE,CF$ και $ H,O$ ορθοκεντρο και περικεντρο αντιστοιχως. Εστω $ L$ σημειο τομης των $ FE,BC$ και $ M$ των $ AB,ED$.
ΝΔΟ: $ HO$ καθετη στην $ LM$.[/quote]

Me $ AD,BE,CF$ ti???

Οταν εχει το σχημα και την λυση μπροστα του, μερικες φορες κανεις ξεχναει να αναφερει μερικες βασικες λεπτομεριες (θελω να πιστευω) :blush: .

$ AD,BE,CF$: υψη του τριγωνου

Υποθέτω ότι $ AD,BE,CF$ είναι ύψη....
Τότε τα $ FBCE$ και $ EABD$ είναι εγγράψιμα.
Άρα από γνωστό θεώρημα ξέρουμε ότι η πολική του $ M$ θα έιναι η $ CF$ και αντίστοιχα η πολική του $ L$ θα είναι η $ AD$.Άρα το $ H$ είναι το σημείο τομής των δύο πολικών,αρα η $ ML$ αίναι η πολική του $ H$,αρα $ OH\bot ML$. :D  :lol:

[b]H ΛΥΣΗ ΕΙΝΑΙ ΑΦΙΕΡΩΜΕΝΗ ΣΤΟ ΡΑΠΑΝΟ ΚΑΙ ΤΟΝ Κ.ΚΩΣΤΑ[/b]

Ειδική περίπτωση θεωρήματος Brocard... To περίκεντρο είναι το ορθόκεντρο του τριγώνου που σχηματίζεται από τους πόλους.

Δημητρη δεν συμφωνω με τροπο λυσης διοτι νομιζω πως το θεωρημα La Hire εφαρμοζεται στον ιδιο κυκλο.
Εδω οι κυκλοι (FBCE) και (EABD) ειναι διαφορετικοι.
Το δε Ο δεν ειναι περικεντρο κανενος απο τους δυο κυκλους.

Δημήτρη, αν είμαι ο σωστός [b][size=100]ΚΩΣΤΑΣ[/size][/b], σ' ευχαριστώ πολύ για την αφιέρωση αλλά ο Σάκης έχει δίκιο. Αν και είναι σωστό το σκεπτικό, υπάρχει αδυναμία στην απόδειξη, η οποία όμως ξεπερνιέται.

[hide]Το κλειδί για τη λύση αυτής της όμορφης άσκησης, είναι η γνωστή ιδιότητα σε κάθε τρίγωνο, σύμφωνα με την οποία τα συμμετρικά του ορθοκέντρου ως προς τις πλευρές του, ανήκουν στον περιγεγραμμένο κύκλο.

Έτσι, εάν $ D',\ E',\ F'$ είναι τα σημεία τομής του περιγεγραμμένου κύκλου $ (O)$ του δοσμένου τριγώνου $ \bigtriangleup ABC,$ από τις ευθείες $ AD,\ BE,\ CF$ αντιστοίχως, έχουμε $ HD \equal{} DD'$ $ ,(1)$ και $ HE \equal{} EE'$ $ ,(2)$ και $ HF \equal{} FF'$ $ ,(3)$

Από $ (1),\ (2),\ (3)$ $ \Longrightarrow$ $ D'E'\parallel DE$ $ ,(4)$ και $ E'F'\parallel EF$ $ ,(5)$

Από $ (4),\ (5)$ $ \Longrightarrow$ $ L'M'\parallel LM$ $ ,(6)$ όπου $ L'\equiv BC\cap E'F'$ και $ M'\equiv AB\cap D'E'$.

$ \bullet$ Επειδή τώρα οι πολικές ευθείες των σημείων $ L',\ M',$ ως προς τον κύκλο $ (O),$ περνάνε από το σημείο $ H,$ συμπεραίνουμε ότι τα σημεία αυτά, ανήκουν στην πολική ευθεία του $ H$ ως προς τον ίδιο κύκλο.

Έτσι, έχουμε ότι $ OH\perp L'M'$ $ ,(7)$

Από $ (6),\ (7)$ $ \Longrightarrow$ $ OH\perp LM$ και η πρόταση έσει αποδειχθεί.[/hide]
Κώστας Βήττας.

Εξαιρετικη οπως παντα η λυση σας κυριε Βηττα.
Θα δωσω μια λυση σε αυτην την ασκηση βασισμενη σε ριζικους αξονες
και την αφιερωνω στον κ. Στεργιου που ξερω οτι του αρεσει το συγκεκριμενο κεφαλαιο.

Οριζω $ c$ τον περιγγεγραμμενο κυκλο του $ ABC$ και $ k$ τον κυκλο Euler.
Επειδη $ LB.LC \equal{} LF.LE$ το $ L$ ανηκει στον ριζικο αξονα των $ c, (EFBC)$ και στον ριζικο αξονα των $ k,(EFBC)$.
Αρα το $ L$ ειναι σημειο του ριζικου αξονα των $ c,k$.
Με εντελως παρομοιο τροπο αποδεικνυεται οτι και το $ M$ ειναι σημειο του ιδιου ριζικου αξονα.
Επειδη ομως το κεντρο του Euler ειναι σημειο της $ HO$, η $ HO$ ειναι η διακεντρος των $ c,k$.
Αρα $ HO$ καθετος στην $ LM$.

Αν ορισω και το σημειο $ K$ σημειο τομης των $ AC,FD$ αφου θα ανηκει στον ιδιο ριζικο αξονα τα $ K,M,L$ θα ειναι συνευθειακα
πραγμα που ειναι και ευθεια εφαρμογη του θεωρηματος Desarques.

Σάκη πολύ όμορφη απόδειξη, τη ζήλεψα.   :)

Ως μία μικρή μόνο σημείωση, να πούμε ότι από τη σχέση $ LB\cdot LC \equal{} LF\cdot LE$ που μας δίνει το εγγράψιμο τετράπλευρο $ BCEF$, συμπεραίνεται άμεσα ότι το σημείο $ L$ ανήκει στον ριζικό άξονα των κύκλων $ (c)$ και $ (k)$.

Κώστας Βήττας.

[quote="Sakis"]Εξαιρετικη οπως παντα η λυση σας κυριε Βηττα.
Θα δωσω μια λυση σε αυτην την ασκηση βασισμενη σε ριζικους αξονες
και την αφιερωνω στον κ. Στεργιου που ξερω οτι του αρεσει το συγκεκριμενο κεφαλαιο.

......................................................................................................
.[/quote]

Σάκη, πολύ ωραία η λύση αλλά και η άσκηση.Ευφυής επίσης και όμορφη είναι και η λυση του Κώστα.
 Θα τις συμπεριλάβω σε επόμενο τεύχος(ελπίζω στο Νο 2 , με τις μετρικές σχέσεις) με τα ονόματά σας βέβαια !
 Μήπως έχεις πηγή για την άσκηση ; Είναι μήπως θέμα κάποιου διαγωνισμού ;

Μπάμπης


trigonometry
geometry
circumcircle
conics
geometric transformation
reflection
angle bisector

A colleague of mine offered me the following problem:

It is given circle k(O, r) and a segment AB. Find (the locus of) all points M from k, such that the line OM is angle bisector of the angle AMB.

I think the problem is not easy at all. I tried trygonometry and homothethy. The only thing I proved is the locus is not an arc (using trigonometry). He said the problem is with practical application. He invented the problem when he programmed software for road mirrors. Thank you for the time spent.

Just an idea...

Consider the perpendicular bisector of $ AB$.  For any point $ P$ on this line, the second intersection of $ PO$ with the circumcircle of $ ABP$ is such a point $ M$.

Note that this gives all points $ M$ such that $ MO$ is either the internal angle bisector or the external angle bisector.

[quote="borislav_mirchev"]It is given circle $ k(O, r)$ and a segment $ AB$. Find (the locus of) all points $ M \in k$,
such that the line $ OM$ is angle bisector of the angle $ AMB$.[/quote]
If you mean you want to construct the points by ruler and compass, see [url=http://www.mathlinks.ro/Forum/viewtopic.php?t=143900]here[/url]. The problems are equivalent, we cannot construct the points and we stop here.

I tried to find the locus of [u]all[/u] points $ M$ (not only on the circle). But I found only parametric equations. Here are my results:

Let $ A(-b,0), \ B(b,0)$ and $ O\left(\frac{b^{2}}{a}, \frac{b^{2}}{\lambda a}\right)$

For any $ p\in \mathbb R$ we define the points $ P(p,0)$ and $ T(0,t)$ where $ t=\lambda(p-a)$

If $ M(x,y)$ then

$ x = \frac{p}{t^{2}+p^{2}}\cdot(t^{2}+b^{2})$

$ y = \frac{t}{t^{2}+p^{2}}\cdot(p^{2}-b^{2})$

How can we eliminate the variable $ p$ without going through a complicated 3rd degree equation  :?:

Here are the equations of the lines where the point $ M$ lies:

$ xp-yt=b^{2}$
$ xt+yp=pt$

Thank you! I had lots of difficulties with this problem by using "classical" geometry. Do you know something about the inventor of the problem? And/or or where it appears for the first time?

About your question:
If I'm understanding you - you mean the system: - try to divide first equation by p, second equation by t, and sum resulting equations.

No, I don't know the inventor. I saw this problem only here, in mathlinks. Also [url=http://www.mathlinks.ro/Forum/viewtopic.php?t=130695]this[/url] problem has the same dificulties - I think it is equivalent too

[quote="borislav_mirchev"]If I'm understanding you - you mean the system: - try to divide first equation by p, second equation by t, and sum resulting equations.[/quote]
Then you'll get again the 2nd linear equation. Actually I started from the linear system and from this I found

$ x = \frac{p}{t^{2}+p^{2}}\cdot(t^{2}+b^{2}) \ \ (1)$

$ y = \frac{t}{t^{2}+p^{2}}\cdot(p^{2}-b^{2}) \ \ (2)$

I wrote the linear equations, because, if someone wants to try, he can get directly $ y$ in terms of $ x$

Now, if we want to eliminate the variable $ p$ in $ (1),(2)$, we set $ t=\lambda(p-a)$ at $ (1)$

Then, we have a cubic equation of $ p$ which contains $ x$. We have to solve it for $ p$. Say that the result is $ p=f(x)$

Then we put the obtained $ p=f(x)$ in one of the linear equations

I tried various ways, but always I end in a really complicated cubic equation, which requires work to be solved

P.S.
This is an interesting problem! The locus is a curve which is definetely not a conic


inequalities
calculus
derivative
function
inequalities proposed

[quote="romano"]Let $ a,b,c >0 $ . Prove that :
 $ \sqrt{\frac{a^3}{a^3+(b+c)^3}} + \sqrt{\frac{b^3}{b^3+(a+c)^3}} + \sqrt{\frac{c^3}{c^3+(b+a)^3}} \geq 1 $[/quote]

I'm a Korean who immigrated to Canada 3 years ago. I've never seen this problem, but I solved it in a few seconds.
Let $f(x)$ be $\sqrt{\frac{x^3}{x^3+(a+b+c-x)^3}}$. Then the problem becomes $f(a)+f(b)+f(c)\geq 1 $.
By Jensen inequality, we get,
\[f(a)+f(b)+f(c)\geq 3f(\frac{a+b+c}3)=1\]
So we are done

[quote="fleeting_guest"]That only proves the equivalence of the inequality with the statement "$f(x)$ is convex".
That is certainly an improvement over the original statement, but some other argument,
such as differentiation, is needed in order to prove convexity.[/quote]
I thought the differentiation would turn out to be simple but it doesn't seem to be simple. Maybe I'll look for another solution... :(
______________________________________________________
EDITED: But I'm quite sure that this is a convex function.

[quote="siuhochung"]fleeting guest, i think you reversed the sign in the last step.

Anyway, my solution is to use the holder's inequality
let the given l.h.s. be A, we are going to prove $A \ge 1$.
By holder's inequality, $A^2 (a^3(b+c)^3+b^3(c+a)^3+c^3(a+b)^3) \ge (a^2+b^2+c^2)^3$
Now we have to prove $(a^2+b^2+c^2)^3 \ge a^3(b+c)^3+b^3(c+a)^3+c^3(a+b)^3$
(All sums are symmetric)
After expanding, we have to prove $3 \sum a^4b^2 + \sum a^2b^2c^2 \ge 3 \sum a^3b^2c + \sum a^3b^3$
By amgm we have $a^4b^2+a^4c^2+2a^2b^2c^2+2a^3b^3 \ge 6a^3b^2c^2$
Adding up similar inequalities we have
$\sum a^4b^2 + \sum a^2b^2c^2 + \sum a^3b^3 \ge 3 \sum a^3b^2c$
By muirhead, we also have $2 \sum a^4b^2 \ge 2 \sum a^3b^3$
Adding up these two inequalities and done.[/quote]

  Maybe you mean  $ A^2 (a^3+(b+c)^3+b^3+(c+a)^3+c^3+(a+b)^3) \geq (a^2+b^2+c^2)^3$

@ Lightrhee: Your solution is, alas, wrong: The function $f\left(x\right)=\sqrt{\frac{x^3}{x^3+(a+b+c-x)^3}}$ is not convex on the whole interval [0; a + b + c].

@ Siuhochung: Very nice solution, but note that the following passage:

[quote="siuhochung"]By holder's inequality, $A^2 (a^3(b+c)^3+b^3(c+a)^3+c^3(a+b)^3) \ge (a^2+b^2+c^2)^3$
Now we have to prove $(a^2+b^2+c^2)^3 \ge a^3(b+c)^3+b^3(c+a)^3+c^3(a+b)^3$[/quote]

should actually read:

By Hölder's inequality,

$A^2\left(a^3\left(a^3+\left(b+c\right)^3\right)+b^3\left(b^3+\left(c+a\right)^3\right)+c^3\left(c^3+\left(a+b\right)^3\right)\right)\geq \left(a^2+b^2+c^2\right)^3$.

Now we have to prove

$\left(a^2+b^2+c^2\right)^3\geq a^3\left(a^3+\left(b+c\right)^3\right)+b^3\left(b^3+\left(c+a\right)^3\right)+c^3\left(c^3+\left(a+b\right)^3\right)$.

All the rest is actually correct.

Anyway, the inequality has already been discussed at http://www.mathlinks.ro/Forum/viewtopic.php?t=28280 and http://www.mathlinks.ro/Forum/viewtopic.php?t=28294 . Please be warned that you might get a shock when seeing the almost 1-line solutions (I mean the ones by Hardsoul and by Vasc). At least, I did...

  Darij

[quote="fleeting_guest"]It may reduce to convexity of a simpler function$g(x) = \frac {1}{\sqrt{1+x^3}} = h(x^3)$, evaluated at points $(b+c)/a$.
This function is convex because $h(x)$ is ($x^a$ is convex when $a < 0$).
The sum of the points is easily seen to be at least 6, with equality when $a=b=c$, so...[/quote]

I think it's a bit futile.  We inevitably get a factor of $(5x^3-4)$ in the numerator, so we don't have strict concavity.  We could use some sort of smoothing argument, I guess, but none immediately springs to mind.  The Holder solution seems to be the most natural.

Here is my soln(i hope it's correct!):Because the given inequal is homognizous,we can assume that:[tex]a+b+c=1[/tex] and it becomes:
[tex]\sum \sqrt{\frac{a^3}{3a^2-3a+1}}\geq 1[/tex]
I guess that: [tex]f(x)=\sqrt{\frac{x^3}{3x^2-3x+1}}[/tex] is convex where [tex]x<1[/tex] and applying Jensen's ineq and we are done!

[quote="keira_khtn"]I guess that: [tex]f(x)=\sqrt{\frac{x^3}{3x^2-3x+1}}[/tex] is convex where [tex]x<1[/tex] and applying Jensen's ineq and we are done![/quote]

I thought I had said clearly enough that it isn't convex... (E. g., we have $f^{\prime\prime}\left(0.9\right)\approx -0.556...<0$.)

  darij

[quote="keira_khtn"]Oh!May this way is wrong.Here is my solution:It's not hard to see:
[tex]\sqrt{\frac{a^3}{a^3+(b+c)^3}}\geq \frac{a^2}{a^2+b^2+c^2}[/tex]
Suming up ,and we are done![/quote]

But its not easy for me!!!So please show me that how did you obtain it? :?:


calculus
integration
MIT
college
vector
geometry
geometric transformation

I have the idea that the cross product, since it is defined just for $ R^3$ is not really meaningful, I know there are some calculus things as Stokes thm, surface integrals, that are described through it. Nevertheless I still feel is not a powerful theoretical thing such as the dot product is. My thread is maybe not very well explained, but if someone has some remarks I would appreciate them.

This was born in some sense because the calculus class I'm taking ($ 18.022$ for those who know MIT numbers) is not theoretical at all, and constrained to $ 2$ and $ 3$ dimensions (more suitable for engineers) that has made me thing about the real usefulness of cross product. I've begun to "dislike" calculus for that.

Thanks,

That's why I dropped 18.022 :)  There is some [url=http://en.wikipedia.org/wiki/Cross_product#Cross_product_as_an_exterior_product]underlying theory[/url] behind the cross product, but nothing that I understand particularly well.  There are also certain ways to generalize it.

The interpretation that appeals most to me as a number theorist is the relationship between the cross product and multiplication in the [url=http://en.wikipedia.org/wiki/Quaternion]quaternions[/url].  In other words, the existence of the cross product in dimension $ 3$ is intimately related to the existence of a [url=http://en.wikipedia.org/wiki/Clifford_algebra]Clifford algebra[/url] of dimension $ 4$.  There is also a cross product in dimension $ 7$ induced by the multiplication in the octonions, a Clifford algebra of dimension $ 8$.  And in [i]every other dimension[/i] all we've got is the wedge product.  Curious, isn't it?

The right idea is that the $ \mathbb{R}^3$ that the result of the cross-product lives in isn't the same $ \mathbb{R}^3$ that its arguments came from.

What do two vectors in $ \mathbb{R}^3$ determine? They determine a plane, a sense of rotation in that plane, and a numerical quantity equal to the area of the parallelogram spanned by those two vectors. For the sake of convenience, we represent this information as a vector, but the direction of that vector isn't particularly meaningful - it's just a marker for a plane (being the normal vector to that plane) and for a sense of rotation in that plane. That's that way to look at something like torque. The vector we write for torque point in a direction that nothing is going in - but it's just a marker for a plane.

For $ n>3,$ we need more information than that contained in one vector to indicate a plane and a sense of rotation. The 3-dimensional coincidence that allows us to write the cross product that way is that $ \binom31\equal{}\binom32.$

Thanks Quiaochu :) for your answer, I regret of taking that class. Sometime ago when I found wedge product and differential forms, i reinvindicated with this part of math. Now I was just wandering.... finals are close. I prefer Algebra.


14. In triangle $ABC$, it is given that angles $B$ and $C$ are congruent. Points $P$ and $Q$ lie on $\overline{AC}$ and $\overline{AB}$, respectively, so that $AP = PQ = QB = BC$. Angle $ACB$ is $r$ times as large as angle $APQ$, where $r$ is a positive real number. Find the greatest integer that does not exceed $1000r$

The AIME (American Invitational Mathematics Exam) has all of it's answers as non-negative integers less than 1000. Also this should be in the Highschool Basics forum.

@castigioni: It looks as though the number we are looking for needs to be less than 1000 because the angle APQ will have to be greater than 90 in order for the triangle APQ to be isosceles, as the question implies. angle PAQ and angle AQP have to be equal for sides AP and AQ to be equal, which means that angle APQ is greater than 90, which means that r will thus be less than 1000.

@ 1=2, You mean the The intermediate forumn right?


Here's a problem I had trouble on:

One day I found a receipt from my grandfather's store.  The biill was faded so that I couldn't read the price of the gadgets at all, and i could read only 3 of the 5 digits in the total. The bill showed the following:

                     72 gadgets at $\$$ _. _ _  each =$\$$_74.5_.

How many dollars did one gadget cost?  Express your answer as a decimal to two decimal places.

p.s. I didn't check to see if this problem had been posted.  Sorry if I posted it again.

[hide]
To be divisible by $72$ the number has to be divisible by $8$ and $9$. To be divisible by $8$ the last 3 digits have to be divisible by $8$, and the only digit that works is $6$. To be divisible by $9$ the digits have to sum to $9$. You have $7$, $4$, $5$, and $6$, which sum to $22$. The first digit has to be $5$. So you know that the $72$ gadgets cost $\${574.56}$ so each one costs $\frac{\$574.56}{72} = \${7.98}$.
[/hide]

[hide]
The last 3 digits have to be divisible by 8.
The only 3 digits that work are 456:
7+4+5+6=22
27-22=5
57456/72=798
$\boxed{7.98}$[/hide]


This was on one of last year's nciml (nassau county math league) tests, I figured it was at the level of harder amc problems

The diameter of a circle is AB. Point C is the midpoint of one of the semicircles.  Point D is on arc AC and lines segments AD and DB are drawn. Point E is on DB with CE perpendicular to DB⊥. If AD has length 5 and DE has length 2, find length of BE.

Um -- you need to at least describe the diagram here.  "The semicircles" -- does that mean the semicircles with endpoints at A and B?  Or are there other semicircles?

I didn't change the wording of the original problem, but i hope the diagram clears things up

[hide] draw DC, get CDB=45 by inscribed angles. so CE=2. now draw CB. again by inscribed angles, CBA=45. let x=DBA, so 45-x=CBD. tanx=5/(BE+2) , tan(45-x)=2/BE. tan(45-x)=[1-tanx]/[1+tanx], so (BE-3)/(BE+7)=2/BE <-> 2BE+14=BE^2 -3BE,

BE^2 -5BE-14=0, and BE=7[/hide]

hooray for trig

good job

I liked the official solution to this problem, it started out as:
This little known theorem is called the Broken Chord Theorem. In general, AD + DE = BE so BE = 7. There are several proofs (none very easy).

interesting...never heard of it. whats the proof of it?

that's probably why its "little known." though it is very intriguing and makes this problem so much easier... gotta keep that one handy :D

yep, lol. i never hear any of these "interesting results"

since mAD <= mBD, choose F s.t. AD  :cong: FB.  C is the midpoint of arc AB, so AC :cong: BC, and mAngle DAC = mAngle CBD, since both are inscribed angles intercepting the arc DC.  Then Triangle ADC :cong: BFC by side-angle-side, which tells us that DC :cong: FC, and triangle DCF is isosceles.  CE is the altitude of an isosceles triangle so it bisects DF, and DE :cong: FE, and we add to see that mAD + mDE = mFE + m FB.  But, mFE + mFB = mEB, so mAD + mDE = mEB

another way to think of it is that E is the midpoint of the broken chord ADB.


geometry
ratio
search
function
area of a triangle
Heron\u0027s formula
algebra

Point $P$ is inside $\triangle ABC$. Line segments $APD$, $BPE$, and $CPF$ are drawn with $D$ on $BC$, $E$ on $AC$, and $F$ on $AB$ (see the figure at right). Given that $AP=6$, $BP=9$, $PD=6$, $PE=3$, and $CF=20$, find the area of $\triangle ABC$.

[asy]
size(200);
pair A=origin, B=(7,0), C=(3.2,15), D=midpoint(B--C), F=(3,0), P=intersectionpoint(C--F, A--D), ex=B+40*dir(B--P), E=intersectionpoint(B--ex, A--C);
draw(A--B--C--A--D^^C--F^^B--E);
pair point=P;
label("$A$", A, dir(point--A));
label("$B$", B, dir(point--B));
label("$C$", C, dir(point--C));
label("$D$", D, dir(point--D));
label("$E$", E, dir(point--E));
label("$F$", F, dir(point--F));
label("$P$", P, dir(0));[/asy]

massssssssssssssss points is being yelled at me from this problem

I'll try to make a solution later

edit> Alright, applying mass points, we set a weight of $1$ to $AE$, which is the weight on $C$ and thus the weight on $PF$. Similarly, we get a weight of $3$ for $CP$, so they are in ratio of $3: 1$ and thus $CP=15$ and $PF=5$. From here, you know that $CE=2AE$ and $BF=2AF$ by Mass Points/Menelaus. I'm not sure excatly how to go from here.

there was some good discussion here, and later on in the topic, i put a pretty understandable solution

[url]http://www.artofproblemsolving.com/Forum/viewtopic.php?t=57750[/url]

[hide="Solution"]Because we're given three concurrent cevians and their lengths, it seems very tempting to apply mass points. :yup: We immediately see that $w_E=3$, $w_B=1$, and $w_A=w_D=2$. Now, we recall that the masses on the three sides of the triangle must be balanced out, so $w_C=1$ and $w_E=3$. Thus, $CP=15$ and $PF=5$.

Recalling that $w_C=w_B=1$, we see that $OC=OB$ and $OP$ is a median to $BC$ in $\triangle BCP$. Applying Stewart's theorem, $BC^2+12^2=2(15^2+9^2)$, and $BC=6\sqrt{13}$. Now notice that $2[BCP]=[ABC]$, because both triangle share the same base and the $h_{\triangle ABC}=2h_{\triangle BCP}$. Applying Heron's formula on triangle $BCP$ with sides $15$, $9$, and $6\sqrt3$, $[BCP]=54$ and $[ABC]=\boxed{108}$. [/hide]

[quote="jelly119"][quote="4everwise"][hide="Solution"]Because we're given three concurrent cevians and their lengths, it seems very tempting to apply mass points. :yup: We immediately see that $w_E=3$, $w_B=1$, and $w_A=w_D=2$. Now, we recall that the masses on the three sides of the triangle must be balanced out, so $w_C=1$ and $w_E=3$. Thus, $CP=15$ and $PF=5$.

Recalling that $w_C=w_B=1$, we see that $OC=OB$ and $OP$ is a median to $BC$ in $\triangle BCP$. Applying Stewart's theorem, $BC^2+12^2=2(15^2+9^2)$, and $BC=6\sqrt{3}$. Now notice that $2[BCP]=[ABC]$, because both triangle share the same base and the $h_{\triangle ABC}=2h_{\triangle BCP}$. Applying Heron's formula on triangle $BCP$ with sides $15$, $9$, and $6\sqrt3$, $[BCP]=54$ and $[ABC]=\boxed{108}$. [/hide][/quote]

I thought $BC$ should be $6\sqrt{13}$. ;)[/quote]Fixed. :P

[quote="pkerichang"]what is mass points??  Although I think that's what I call "area ratio" chasing...[/quote]

Can anyone explain him??????

[quote="shobber"]I don't think that it is necessary to learn mass point after you master Area Ratios and Menelaus and Ceva.[/quote]

Well, but he just want to know, so I think that somebody have to help him.....

yes, I would also benefit from learning what Mass points is :)

[quote="K81o7"]Essentially, mass points applies laws of torque to geometry to find lengths...I probably can't explain sufficiently here, but there are some good topics which explain it...[/quote]

Sorry to tell anybody, but this is the 1000th time a MathLinker asked for mass point, and all people say: "it is not necessary,..., use search function,", etc, but can anyone post a LINK???????????????

[quote="4everwise"][hide="Solution"]Because we're given three concurrent cevians and their lengths, it seems very tempting to apply mass points. :yup: We immediately see that $ w_E \equal{} 3$, $ w_B \equal{} 1$, and $ w_A \equal{} w_D \equal{} 2$. Now, we recall that the masses on the three sides of the triangle must be balanced out, so $ w_C \equal{} 1$ and $ w_E \equal{} 3$. Thus, $ CP \equal{} 15$ and $ PF \equal{} 5$.

Recalling that $ w_C \equal{} w_B \equal{} 1$, we see that $ OC \equal{} OB$ and $ OP$ is a median to $ BC$ in $ \triangle BCP$. Applying Stewart's theorem, $ BC^2 \plus{} 12^2 \equal{} 2(15^2 \plus{} 9^2)$, and $ BC \equal{} 6\sqrt {13}$. Now notice that $ 2[BCP] \equal{} [ABC]$, because both triangle share the same base and the $ h_{\triangle ABC} \equal{} 2h_{\triangle BCP}$. Applying Heron's formula on triangle $ BCP$ with sides $ 15$, $ 9$, and $ 6\sqrt3$, $ [BCP] \equal{} 54$ and $ [ABC] \equal{} \boxed{108}$. [/hide][/quote]

What's O agian? Did you draw a circle on this graph?

Assign a mass of $3$ to point $E$, a mass of $1$ to $B$, and a mass of $2$ to points $A$ and $D$. It follows that $P$ has a mass of $4$. Hence $C$ must have a mass of $1$ and a mass of $3$ on $F$, since $A$ and $B$ have masses of $1$, so $D$ is the midpoint of $\overline{BC}$. Hence we have $\frac{CP}{PF}=3\implies CP=15, PF=5$.

Applying Stewarts Theorem on cevian $\overline{BC}$ in $\triangle BCP$ yields $BC^2+12^2=2(15^2+9^2)\implies BC=6\sqrt{13}$. Now, observe that $2[BCP]=[ABC]$, since they share the same base in $\overline{BC}$. It follows that the heights share the same ratios, so $2PD=2AD$. We have $\triangle BCP$ has side lengths $15, 9$ with $BC=6\sqrt{13}$. Using Heron's formula and difference of squares yields $[BPC]=54\implies [ABC]=2[BPC]=2\times 54=\boxed{108}$

EDIT: oops just realized its same as 4everwise :blush:

[color=darkred][b][u]Lemma[/u].[/b] For the point $P$ which belongs to the inside of $\triangle ABC$ denote $\left\{\begin{array}{c}
D\in AP\cap BC\\\\
E\in BP\cap CA\\\\
F\in CP\cap AB\end{array}\right|\implies$ $\boxed{\ \prod\frac {PA}{PD}=2+\sum\frac {PA}{PD}\ }\ (*)$[/color]

[b][u]Proof[/u].[/b] $(\exists ) \{x,y,z\}\subset \mathbb R^*_+$ so that $\left\{\begin{array}{c}
\frac {DB}{DC}=\frac zy\\\\
\frac {EC}{EA}=\frac xz\\\\
\frac {FA}{FB}=\frac yx\end{array}\right|$ . Apply the [u][b]van Aubel[/b]'s relations[/u] $\left\{\begin{array}{c}
\frac {PA}{PD}=\frac {EA}{EC}+\frac {FA}{FB}=\frac {y+z}{x}\\\\
\frac {PB}{PE}=\frac {FB}{FA}+\frac {DB}{DC}=\frac {z+x}{y}\\\\
\frac {PC}{PF}=\frac {DC}{DB}+\frac {EC}{EA}=\frac {x+y}{z}\end{array}\right|$ .

 Show easily that the relation $(*)$ becomes a simple identity $\prod (y+z)=2xyz+\sum yz(y+z)$ .

[color=darkred][b][u]PP[/u].[/b] For the point $P$ which belongs to the inside of $\triangle ABC$ denote $\left\{\begin{array}{c}
D\in AP\cap BC\\\\
E\in BP\cap CA\\\\
F\in CP\cap AB\end{array}\right|$ .

Suppose that $\left\{\begin{array}{c}
PA=6\ ;\ PD=6\\\\
PB=9\ ;\ PE=3\end{array}\right|$ and $CF=20$ . Find the area $S=[ABC]$ .[/color]

[b][u]Proof[/u].[/b] Denote $\frac {PC}{PF}=r$ and apply the relation $(*)\ :\ \frac 66\cdot$ $ \frac 93\cdot r=2+\frac 66+\frac 93+r\implies r=3\implies \frac {PC}{PF}=3$ . Observe that

$\frac {PB}{PE}=\frac {PC}{PF}=3$ , i.e. $D$ is the midpoint of $[BC]$ and $CF=20\implies \boxed{\ PC=15\ \ \wedge\ \ PF=5\ }$ . If $L$ is  the midpoint of $[PB]$ ,

then $\left\{\begin{array}{c}
PD=6\\\\
DL=\frac {15}{2}\\\\
LP=\frac 92\end{array}\right|$ .  Since $\frac {6}{4}=\frac {\frac {15}{2}}{5}=\frac {\frac 92}{3}\ ,\  PDL$ is $P$-right angled $\implies [PDL]=\frac 12\cdot PD\cdot PL$ , i.e.  $\boxed{\ [PDL]=\frac {27}{2}\ }$

Since $\frac {[PDL]}{[ACB]}=\frac {[PDL]}{[PDB]}\cdot\frac {[PDB]}{[PCB]}\cdot\frac {[PCB]}{[ACB]}=$ $\frac {PL}{PB}\cdot\frac {DB}{CB}\cdot\frac {PD}{AD}=\frac 18\implies$ $S=8\cdot [PDL]=8\cdot \frac {27}{2}\implies$ $\boxed{\ S=108\ }$ .

[u]See[/u] [b]PP16[/b] (with its [u]generalization[/u]) from [url=http://www.artofproblemsolving.com/blog/66866][color=darkred][b]here[/b][/color][/url].

This problem is simplified by area lemma a good explanation is in the book 106 geometry problems.

Here's a surprisingly easier solution I discovered when doing this problem (the solution is now on the wiki too):

EDIT: Oops I was being stupid :P lucky coincidence...

[hide="Solution"]

Note that $[BPD]=[APB]=[APC]=[CPD]=\frac{1}{4}[ABC]$ since $AP=PD=6$. Using the sine area formula with regard to angles $APB, BPD,$ we realize both these are equal hence right. So the answer is $4*\frac{1}{2}*6*9=108$.

Surprised nobody has posted a Menelaus solution yet.

Using Menelaus's Theorem on $\triangle BPD$, we see that $\frac{BC}{CD}\cdot\frac{12}{6}\cdot\frac{3}{12}=1$. This implies $BC=2\cdot CD$ which implies $D$ bisects $BC$. \\Now, using Menelaus's Theorem on $\triangle CPD$, we have $\frac{2}{1}\cdot\frac{12}{6}\cdot\frac{FP}{20}=1$, which implies $FP=5$ and $PC=15$. \\Now let $BD=DC=x$. Using Stewart's Theorem on $\triangle BPC$ with cevian $PD$, we have $81x+225x=72x+2x^3$ which implies $x^2=117$. However, this also implies $\angle BPD=90^\circ$ by the Pythagorean Theorem. \\ Now, let's carefully compute areas. Note that $[APB]=\frac{6\cdot 9}{2}=27, [APE]=\frac{6\cdot 3}{2}=9, [BPD]=[PDC]=\frac{6\cdot 9}{2}=27$. All that remains is $PEC$. By area ratios, we find that $\frac{[PEC]}{9}=\frac{[PEC]+54}{36}$. This yields $[PEC]=18$. Now, since $[ABC]=[APB]+[APE]+[BPD]+[PDC]+[PEC]$, we find that $[ABC]=27+9+27+27+18=108$ as desired.

Interesting problem.

[hide=Solution]Let $PF=x$. Applying Gergonne's Theorem, we get $\frac{6}{12}+\frac{3}{12}+\frac{x}{20} = 1 \implies x = 5$. Menelaus's Theorem on $\triangle BPD$ gives $BC=2BD$, so $CD=DB$. Now, let $CD=DB=y$. From Stewart's Theorem, we solve to get $y=3\sqrt{13}$, and $BC=6\sqrt{13}$. Applying Heron's Formula, we get $[CPB]=54 \implies [ABC]=\boxed{108}$.[/hide]

Midpoint of $EB$ and $CF$ (+ heron bash) kills the problem.

[quote=OlympusHero]Interesting problem.

[hide=Solution]Let $PF=x$. Applying Gergonne's Theorem, we get $\frac{6}{12}+\frac{3}{12}+\frac{x}{20} = 1 \implies x = 5$. Furthermore, we have $[ADP]=\frac{1}{4}[ABC]$, $[CAP]=\frac{1}{4}[ABC]$ and $[CPB]=\frac{1}{2}[ABC]$. Therefore, $[CPB]=\frac{1}{2}[ABC]$ and we have $CD=DB$. Now, let $CD=DB=y$. From Stewart's Theorem, we solve to get $y=3\sqrt{13}$, and $BC=6\sqrt{13}$. Applying Heron's Formula, we get $[CPB]=54 \implies [ABC]=\boxed{108}$.[/hide][/quote]

How do you know $CD=DB$. I mean, how does $[CPB] = \frac{1}{2}[ABC]$ imply $CD=DB$?

I have no idea what I was talking about back then. I will edit what I said.

Masspoints and midpoint formula kill this problem.

We can immediately let the mass points of $B$ be $1,$ $E$ be $3,$ $D$ be $2,$ and $A$ be $2.$ Hence, it follows that the mass points of $C$ is $1$ and $F$ is $3.$ Since $CF=20$ and $CP \colon PF = 3 \colon 1,$ $CP=15$ and $PF=5.$ Note that since the masses of $B$ and $C$ are both $1,$ $D$ is the midpoint of $\overline{CB}.$ By Stewart's Theorem on $\triangle BPC,$ $$2BD^3 + 2\cdot 6^2 BD = 15^2BD + 9^2BD \implies BD = 3\sqrt{13}.$$ Thus, $BC = 6\sqrt{13}.$ Because $[ACP] = [DCP]$ and $[BAP] = [BDP],$ $[BPC]$ is half of $[ABC].$ By Heron's formula on $\triangle BCP$ with sides $15$, $9$, and $6\sqrt{13},$ $[BPC] = 54.$ Therefore, $[ABC] = \boxed{108}.$


Now, could someone tell me what Ceva's theorem is?  You might remember Hilbert saying a bit of it last Saturday in the math class.

If AD, BE, and CF are cevians of ABC, they are concurrent iff (AF/BF)*(BD/CD)*(CE/AE)=1.

Not quite - they don't have to be medians, you just need D,E,F on sides BC, CA, AB respectively.. medians are always concurrent ;)

sorry i'm a tad bit sleepy. as long as you're here, do you want to do menelaus and signed distances?

A random note:  none of the three things that have been mentioned here (Menelaus, Ceva, or signed distances) will ever be necessary for Mathcounts.  They are, however, pretty darn cool.

I think I used Ceva for mathcounts once...just a tad bit faster than the official soln though. Can't remember what problem, sorry.


group theory
abstract algebra
floor function
superior algebra
superior algebra unsolved

show every p sylow subgroup of Sp is generated by p cycle,p prime and there exist (p-2)! such subgroups.

There's a huge difference between posted before and solved...

The p-sylows of $S_{n}$ can be constructed in the following way:

Let $n=\sum_{i=0}^{m}{c_{i}\cdot p^{i}}$ with $c_{i}\in\{0,...,p-1\}$. Then define $W_{0}: =\{0\}$ and $W_{n+1}: =W_{n}\wr\mathbb{Z}_{p}$ (a wreath-product of the two groups with $\mathbb{Z}_{p}$ operating on itself by multiplication or addition if you want).
The direct product $\prod_{i=0}^{m}{W_{i}^{c_{i}}}$ is isomorphic to the p-sylowgroups of $S_{n}$.

Proof: Maybe tomorrow...

I've no time now to finish this (have to prepare some stuff), but  started typing 3 hours ago, and what Gockel posted seems more or less to be something similar.

Let $n=pq+r, 0 \leq r < p$ and take the subgroup generated by the following:
Write any integer $s$ in $p$-adical base, thus $s=...+s_{2}p^{2}+s_{1}p+s_{0}$.
After having written numbers that way, we define $s \oplus t$ by simply adding the digits in base $p$, but $\mod p$ (represented by the smallest nonnegative integers of that residue class).
For those $m$ with $p^{m}\leq n$ define the subgroup $T_{m}$ generated by all elements $\sigma_{x}$ with $x=(..., x_{m+2}, x_{m+1}, x_{m})$ a sequence of $x_{i}\in \{0,1,2,...,p-1\}$)  of the following type: $\sigma_{x}(s)=s \oplus p^{m-1}$ if $s_{i}=x_{i}$ for $i \geq m$.
In other words: we map exactly those $s$ with the given leading digits from $x$ to $s \oplus p^{m-1}$ (so only add $1$ to the $(m-1)$-th $p$-adical digits, seen $\mod p$).

Each $T_{m}$ is ismorphic to $C_{p}^{k_{m}}$ then, with $k_{m}$ being the biggest integer with $k_{m}p^{m}\leq n$, thus $k_{m}=\left\lfloor \frac{n}{p^{m}}\right\rfloor$.

Now let $H$ be the subgroup generated by all the $T_{M}$. Using that $ab$ with $a \in T_{m_{1}}, b \in T_{m_{2}}$ can also be written as $b^\prime a^\prime$ with $a^\prime \in T_{m_{1}}, b^\prime \in T_{m_{2}}$, we can write every element of $H$ as product $a_{1}a_{2}a_{3}...$ and $a_{i}\in T_{i}$.
Conversly, this representation is unique, thus we get that $|H| = |T_{1}| \cdot |T_{2}| \cdot |T_{3}| \cdot ... = p^{\left\lfloor \frac{n}{p^{1}}\right\rfloor+\left\lfloor \frac{n}{p^{2}}\right\rfloor+\left\lfloor \frac{n}{p^{3}}\right\rfloor+... }$, giving that $H$ is a $p$-sylow.

Now using that all other sylows are conjugated, I will leave the counting itself to some ambitionated combinatorian ;)