How can I draw diagrams in latex files !!!
What program do I need?
Please tell me step by step as you can and give me an example.  :blush:

[quote="ThinkFlow"]To embed pictures, such as diagrams into $ \text{\LaTeX}$ files, see [url=http://www.artofproblemsolving.com/Wiki/index.php/LaTeX:Pictures]this[/url].

If you specifically use geometric diagrams a lot, try asymptote. Learn about asymptote [url=http://www.artofproblemsolving.com/Wiki/index.php/Asymptote_(Vector_Graphics_Language)]here[/url] and about implementing it in $ \text{\LaTeX}$ [url=http://www.artofproblemsolving.com/Wiki/index.php/Asymptote:_Advanced_Configuration#Using_Asymptote_in_LaTeX]here[/url].[/quote]

Thank you for your help  :lol: .
But I did not understand this point:
[quote]
PDF output configuration

Although gsview can configured as the pdf viewer, we recommend using the default pdf viewer, Acrobat Reader.

Now open TeXnicCenter. Go to the Tools menu and click Customize, then go to the Tools tab. Click to modify your Asymptote tool that you made in the Interactive Mode section. In the Arguments line, simply change latex to pdflatex.[/quote]

Sorry,  but I'm not good in softwares :blush: .

I use Kile, so how do I use Asymptote in it?  :maybe:

mathson, those directions explain how to use GhostScript to output your asymptote images in pdf rather than eps format. The best I can do is to tell you to follow those directions, as I myself am not all too sure about asymptote in LaTeX. :(

1-From where can I download [b]"Olympiad package"[/b] ?
2- How can I download and use this package ([url]http://piprim.tuxfamily.org/asymptote/travaux/modules/geometry_dev.asy.html[/url])

[quote]
.
.
.

real epsgeo=10*sqrt(realEpsilon);/*
Variable used in the approximate calculations.
View index ordered by [name] - [type]
*/

void addMargins(picture pic=currentpicture,
                real lmargin=0, real bmargin=0,
                real rmargin=lmargin, real tmargin=bmargin,
                bool rigid=true, bool allObject=true)
{/*
Add margins to 'pic' with respect to the current bounding box of 'pic'. If 'rigid' is false, margins are added iff an infinite curve will be prolonged on the margin. If 'allObject' is false, fixed-size objects (such as labels and arrowheads) will be ignored.
View index ordered by [name] - [type]
*/
  pair m=allObject ? truepoint(pic,SW) : point(pic,SW);
  pair M=allObject ? truepoint(pic,NE) : point(pic,NE);
  if(rigid) {
    draw(m-inverse(pic.calculateTransform())*(lmargin,bmargin),invisible);
    draw(M+inverse(pic.calculateTransform())*(rmargin,tmargin),invisible);
  } else pic.addBox(m,M,-(lmargin,bmargin),(rmargin,tmargin));
}

real approximate(real t)
{
    real ot=t;
    if(abs(t-ceil(t)) < epsgeo) ot=ceil(t);
    else if(abs(t-floor(t)) < epsgeo) ot=floor(t);
    return ot;
}
.
.
.
[/quote]
3- How can I use import geometry_dev ?? It does not work with me !

< Edited >

Please, teach me. How can I download and use [b]"Olympiad package"[/b]?

[quote="Dojo"]Change the type of the document to .asy instead of .txt[/quote]
and where i but it ?

[quote="undefined117"]See the [url=http://asymptote.sourceforge.net/doc/Search-paths.html#Search-paths]documentation[/url].

For temporary usage, you can put it in the same directory as the file you are editing. For a longer-term installation, you can place the package in C:\Program Files\Asymptote.[/quote]

Thank you, you solved my problem  :oops: .


can anyone show me how can I make a regular pentagon?

Just google, and you'll get a lot of results such as :
http://en.wikipedia.org/wiki/Pentagon

Hi,

You can present the 5-th roots of unity in a trigonometric circle...


logarithms
trigonometry
calculus
calculus computations

[b]1. The problem statement, all variables and given/known data[/b]
Determine whether the series converges and diverges.

$ \sum_{n\equal{}3}^{\infty}\ln \left(\frac{\cosh \frac{\pi}{n}}{\cos \frac{\pi}{n}}\right)$

[b]3. The attempt at a solution[/b]

$ \sum_{n\equal{}3}^{\infty}\ln \left(\frac{1\plus{}\frac{\pi^2}{2n^2}\plus{}O(\frac{1}{n^4})}{1\minus{}\frac{\pi^2}{2n^2}\plus{}O(\frac{1}{n^4})}\right)$

$ \equal{}\sum_{n\equal{}3}^{\infty}\ln \left(\left(1\plus{}\frac{\pi^2}{2n^2}\plus{}O(\frac{1}{n^4})\right)\left(1\plus{}\frac{\pi^2}{2n^2}\plus{}O(\frac{1}{n^4})\right)\right)\equal{}\sum_{n\equal{}3}^{\infty}\ln \left(1\plus{}\frac{\pi^2}{n^2}\plus{}O(\frac{1}{n^4})\right)$

$ \equal{}\sum_{n\equal{}3}^{\infty}\left(\frac{\pi^2}{n^2}\plus{}O(\frac{1}{n^4})\right)$

series converges


That is to say $f(x+y)=f(x)+f(y)$ with $f$ bourned on $[a,b]$. To show that $f$ is continuous.

[quote="grobber"]Here's something cool: unless $f$ is continuous (with the hypothesis $f(x+y)=f(x)+f(y)$, of course), the graph of $f$ is dense in $\mathbb R^2$. Nice, isn't it? :)[/quote]
[b]grobber[/b];can you post your proof of this one?i can't prove :(

I don't know if I'll be able to explain it properly, but geometrically, it's pretty clear.

First of all, prove that for all rational $\alpha,\beta$, and all $a,b$, we have $f(\alpha a+\beta b)=\alpha f(a)+\beta f(b)$. If we assume $\alpha+\beta=1$, we find that for any reals $a,b$, the graph of $f$ is dense on the line formed by the points $(a,f(a)),(b,f(b))$.

Now, the fact that $f$ is not continuous simply means that there are $a,b\in\mathbb R$ (take them different from $0$) s.t. $f(a)=\alpha a,f(b)=\beta b,\ \alpha\ne \beta$. For any disk in the plane we can find $t\in\mathbb Q$ s.t. the line formed by $(a,f(a)),(tb,tf(b))$ cuts the disk (this is a consequence of the fact that the lines with equations $y=\alpha x,y=\beta x$ are different). Since the graph of $f$ is dense on this line, there must be a point of the graph on this line which is also inside the disk. This is what we wanted.


Problem statement:
A rigid body consists of three thin uniform rods, each of mass m and length 2a, held mutually perpendicular at their midpoints. Show that the moment of inertia is the same for any axis passing through the origin.

Attempt at a solution:
I calculated the principal moments of inertia. To show this, do I need to find a coordinate transformation to take one of the principal axes to an arbitrary axis that goes through the origin and then show that the transformation does not change that element on the diagonal of the inertia tensor? Or is there another way to do this?

Let $ \alpha$, $ \beta$, $ \gamma$ be the direction angles of a given axis.

It can be shown that the moment of inertia of a rod about an xis through its centre and at an angle $ \theta$ to it is $ I\equal{}\frac{2}{3}ma^2\sin^2\theta$

Hence, moment of inertia of system about axis $ I\equal{}\frac{2}{3}ma^2 (\sin^2\alpha\plus{}\sin^2\beta\plus{}\sin^2\gamma)\equal{}\frac{4}{3}ma^2$, which is constant

I see. That was very helpful (even though you have an extra factor of 2).

fibbonaci good too see  write the attempt at solution  :)


Unitary quadratic trinomials $ f(x)$ and $ g(x)$ satisfy the following interesting condition: $ f(g(x)) \equal{} 0$ and $ g(f(x)) \equal{} 0$ do not have real roots. Prove that at least one of equations $ f(f(x)) \equal{} 0$ and $ g(g(x)) \equal{} 0$ does not have real roots too.

[i]S. Berlov [/i]

Hello,
[quote="N.T.TUAN"][i]S. Berlov [/i]
Unitary quadratic trinomials $f(x)$ and $g(x)$ satisfy
the following interesting condition: $f(g(x))=0$ and $g(f(x))=0$
do not have real roots. prove that at least one of equations
$f(f(x))=0$ and $g(g(x))=0$ does not have roots too.[/quote]

Assume $f(f(x))=0$ and $g(g(x))=0$ have roots. Then $f(x)$ have roots too, say $x_{1}$ and $x_{2}$ with $x_{1}\leq x_{2}$ and $g(x)$ have roots too, say $y_{1}$ and $y_{2}$ with $y_{1}\leq y_{2}$
We can assume $x_{2}\leq y_{2}$ (else we just have to swap f and g)

Since $f(f(x))=0$ has root $x_{3}$, $f(x_{3})=x_{1}$ or $f(x_{3})=x_{2}$, so $f(x_{3})\leq y_{2}$
But $f(x_{3})\leq y_{2}$ and $f(x) \rightarrow+\infty$  when $x \rightarrow+\infty$  implies it exists $x_{4}$ such that $f(x_{4})=y_{2}$ $\Rightarrow $  $g(f(x_{4}))=g(y_{2})=0$

So, if $f(f(x))=0$ and $g(g(x))=0$ has roots, then either $f(g(x))=0$ or $g(f(x))=0$ has root.

So, if neither $f(g(x))=0$ nor $g(f(x))=0$ has root, either $f(f(x))=0$ or $g(g(x))=0$ don't have roots.
Q.E.D.

--
Patrick

[quote="N.T.TUAN"][i]S. Berlov [/i]
Unitary quadratic trinomials $ f(x)$ and $ g(x)$ satisfy
the following interesting condition: $ f(g(x)) \equal{} 0$ and $ g(f(x)) \equal{} 0$
do not have real roots. prove that at least one of equations
$ f(f(x)) \equal{} 0$ and $ g(g(x)) \equal{} 0$ does not have roots too.[/quote]
Easy problem. Here my solution  :lol: .
If equation $ f(x).g(x) \equal{} 0$ haven't root then equation $ f(f(x) \equal{} 0$ and $ g(g(x)) \equal{} 0$ haven't root.
Let $ x_1,x_2$ be root of equation $ f(x) \equal{} 0$,$ x_3,x_4$ be root of equation $ g(x) \equal{} 0$
Suppose 2 equation $ f(f(x)) \equal{} 0$ and $ g(g(x)) \equal{} 0$ have root.
So $ \triangle_f \minus{} 4x_1\geq 0$; $ \triangle_f \minus{} 4x_2\geq 0$; $ \triangle_g \minus{} 4x_4\geq 0$; $ \triangle_g \minus{} 4x_3\geq 0$
So $ 2\triangle_f \plus{} 2\triangle_g \minus{} 4\sum_{i \equal{} 1}^{4}x_i\geq 0$
From equation $ f(g(x)) \equal{} 0$ haven't root  we have equation $ g(x) \equal{} x_1$ haven't root .So $ \triangle_g \minus{} 4x_1 < 0$
Similary we have  $ \triangle_g \minus{} 4x_2 < 0$;  $ \triangle_f \minus{} 4x_3 < 0$ ; $ \triangle_f \minus{} 4x_4 < 0$
So $ 2\triangle_f \plus{} 2\triangle_g \minus{} 4\sum_{i \equal{} 1}^{4}x_i < 0$ QED.


I just looked at it and the answer popped into my head. :ninja:

but are u sure this is only one answer ?

if x>=4 then 4/Y^2+1 but then y^2=-1(mod 4) but this is impossiblethen if x=3 y^2=5, no
if x=2 y^2=1=>y=1
if x=1=>y=0

[quote="mychrom"]Find x,y natural numbers if:
 x!=y*y+1
[/quote][/quote]

$x!=y^{2}+1$
 The only way I can think of doing this within MOEMS level is by guess and check.

@mychrom: please post such questions at MC or HSB, please


if a sphere with a radius 4 is filled with a liquid until the liquid reaches a height of 4 - 2 :rt2: find the volume the liquid covers.

Hint: [hide]the weird height is essential for solving this problem with reletive ease[/hide]

Not sure about this at all. But thought I'd give it a try. Is it[color=white]:[color=white][/color]
[color=white]94.78oz :^3: ?[/color][/color]

3 comments:
(1) instead of making the color white, use the "Spoiler" tag, which will make it completely invisible.
(2) If you just give a number, no one has any idea how you came to your answer, which is a shame.  So instead of just giving answers, give your solution (an explanation of what you did) along with the answer (what you got at the end).
(3) It's really better to use exact values instead of decimal approximations for numbers, because it gives more information about the answer.

Welcome aboard to Ms. Bonaparte.

JBL (a university student who I think is majoring in math) has just given you good advice about how to post solutions to share your interest in learning math with the other participants here. I see you have just joined the site very recently. If you have any questions about the techniques of posting solutions in spoiler text (which seems to have puzzled quite a few new participants recently) or about anything else, please feel free to ask me about it, or you can ask either of the other two forum moderators here, Chinaboy or Rep123Max. I'm the "token adult" here, so I usually have somewhat more time to answer questions than some of the other forum moderators, who are busy with their school studies.

Enjoy.

[hide]

Since spheres are symmetrical when cut down the middle, you can take the 2*Sqrt(2) that is being taken away and make it so that the radius is 4-Sqrt(2) instead of 4. Then, using V_sphere = 4/3 *  :pi: * r :^3: , the volume is 4/3 * :pi: * (4-Sqrt(2)):^3:

About 72.4213421

[/hide]

I don't really know what I'm doing, so this might be wrong.

[hide]Imagine a sphere is packed with very thin disks. Considering a cylinder with base radius r, the height 2r, and a cone with base radius r, height r.

Take a disk from the cone with base radius x and height  r. The volume is x^2 :pi: h.

Take a disk from the cylinder. The volume of the disk is r^2 :pi: h

Take a disk from the sphere that its distance from the center is x. The radius of the base is  :sqrt: (r :^2: -x :^2: ). Volume is (r :^2: -x :^2: ) :pi: h.  That just equal to the difference of the disks from the cylinder and the cone.

The volume of liquid= the sum of those disks in the sphere

=cylinder with radius 4of base, height 4- 2:sqrt: (2)

- a truncated cone with bottom base radius 4 and height 4-2 :sqrt: (2) and top base radius 2 :sqrt: 2

=(32 :pi: -16 :pi:  :sqrt: (2))-(1/3)(16 :pi: (4)-8 :pi: (2 :sqrt: 2)

=(16 :pi: /3)(8-5 :sqrt: 2)

(The disk idea is from the proof of volume of a sphere)[/hide]


A function $f$ is defined on the positive integers by \[\left\{\begin{array}{rcl}f(1) &=& 1, \\ f(3) &=& 3, \\ f(2n) &=& f(n), \\ f(4n+1) &=& 2f(2n+1)-f(n), \\ f(4n+3) &=& 3f(2n+1)-2f(n), \end{array}\right.\] for all positive integers $n$.  Determine the number of positive integers $n$, less than or equal to 1988, for which $f(n) = n$.

f(n) is always odd. If $ n \equal{} b_{r\plus{}1}b_r...b_2b_1b_0$ in binary and n is odd, so that $ b_{r\plus{}1} \equal{} b_0 \equal{} 1$, then $ f(n) \equal{} b_{r\plus{}1}b_1b_2...b_rb_0$. If n has $ r\plus{}2$ binary digits with r > 0, then there are $ 2[\frac{r\plus{}2}{2}]$ numbers with the central r digits symmetrical, so that f(n) = n (because we can choose the central digit and those lying before it arbitarily, the rest are then determined). Also there is one number with 1 digit (1) and one number with two digits (3) satisfying f(n) = 1. So we find a total of $ 1 \plus{} 1 \plus{} 2 \plus{} 2 \plus{} 4 \plus{} 4 \plus{} 8 \plus{} 8 \plus{} 16 \plus{} 16 \equal{} 62$ numbers in the range 1 to 1023 with f(n) = n. 1988 = 11111000011. So we also have all 32 numbers in the range 1023 to 2047 except for 11111111111 and 11111011111, giving another 30, or 92 in total.

It remains to prove the assertions above. f(n) odd follows by an easy induction. Next we show that if $ 2^m<2n\plus{}1<2^{m\plus{}1}$, then $ f(2n\plus{}1) \equal{} f(n) \plus{} 2^m$. Again we use induction. It is true for m = 1 $ (f(3) \equal{} f(1) \plus{} 2)$. So suppose it is true for 1, 2, ... , m. Take 4n+1 so that $ 2^{m\plus{}1 }< 4n\plus{}1 < 2^{m\plus{}2}$, then $ f(4n\plus{}1) \equal{} 2f(2n\plus{}1) \minus{} f(n) \equal{} 2(f(n) \plus{} 2m) \minus{} f(n) \equal{} f(n) \plus{} 2m\plus{}1 \equal{} f(2n) \plus{} 2m\plus{}1$, so it is true for $ 4n\plus{}1$. Similarly, if 4n+3 satisfies, $ 2m\plus{}1 < 4n\plus{}3 < 2m\plus{}2$, then $ f(4n\plus{}3) \equal{} 3f(2n\plus{}1) \minus{} 2f(n) \equal{} f(2n\plus{}1) \plus{} 2(f(n) \plus{} 2^m) \minus{} 2f(n) \equal{} f(2n\plus{}1) \plus{} 2^{m\plus{}1},$ so it is true for 4n+3 and hence for m+1.

Finally, we prove the formula for $ f(2n\plus{}1)$. Let $ 2n\plus{}1 \equal{} b_{r\plus{}1}b_r...b_2b_1b_0$ with $ b_0 \equal{} b_{r\plus{}1} \equal{} 1$. We use induction on r. So assume it is true for smaller values. Say $ b_1 \equal{} ... \equal{} b_s \equal{} 0$ and $ b_{s\plus{}1} \equal{} 1$ (we may have s = 0, so that we have simply $ b_1 \equal{} 1$). Then $ n \equal{} b_{r\plus{}1} ... b_1$ and $ f(n) \equal{} b_{r\plus{}1}b_{s\plus{}2}b_{s\plus{}3}...b_rb_{s\plus{}1}$ by induction. So $ f(n) \plus{} 2^{r\plus{}1} \equal{} b_{r\plus{}1}0...0b_{r\plus{}1}b_{s\plus{}2}...b_rb_{s\plus{}1}$, where there are s zeros. But we may write this as $ b_{r\plus{}1}b_1...b_sb_{s\plus{}1}...b_rb_{r\plus{}1}$, since $ b_1 \equal{} ... \equal{} b_s \equal{} 0$, and $ b_{s\plus{}1} \equal{} b_{r\plus{}1} \equal{} 1$. But that is the formula for $ f(2n\plus{}1)$, so we have completed the induction

This is actually combinatorics. [url=https://artofproblemsolving.com/community/c6h60400p25132790] Solution [/url]


How do you make a % in TeXnic center?
When I write%, it just clears out pretty much everything behind it.

yeah I just figured it out on my own.  :blush:

There's a list of symbols in the $ \text{\LaTeX}$ tutorial at AoPS.

\$
\#
 add the dollar signs....

$ \$$
$ \#$

[quote="Dojo"]Yeah I looked in the $ \text{\LaTeX}$ guide but I couldn't find it...[/quote]
You are right it isn't there so have no added, it along with others, to [[LaTeX:Symbols#Command_Symbols]]


integration
trigonometry
calculus
calculus computations

[hide="hint"]
Lets compute:
$ \int\frac{dx}{\sqrt{x(2-x)}+1}=\int\frac{dx}{x\sqrt{\frac{2-x}{x}}+1}=$ and now $ t=\frac{2-x}{x}$, so:
$ =2\int\frac{tdt}{t^{3}+t+2}=2\int\frac{tdt}{(t+1)(t^{2}-t+2)}=2\int\left(\frac{A}{t+1}+\frac{Bt+C}{t^{2}-t+2}\right)dt$ and now find $ A$ and $ B$ :) [/hide]

Noticing that $ x(2-x)$ is actually $ 1-(x-1)^{2}$,we use this substitution:$ x-1=\sin t$ and get
$ \int_{-\frac{\pi}{6}}^{\frac{\pi}{6}}{\frac{\cos t}{\cos t+1}}dt$,which is similar to Carcul's form.But it doesn't have to be that complicated.


Obviously we should make a linear combination of the two expressions to get the third. Hence for some $\alpha,\beta$ we have

$\alpha(3a_{n}+4b_{n})+\beta(6a_{n}-b_{n})=2a_{n}-5b_{n}$

Equating the coefficients we get the system

$\begin{cases}3\alpha+6\beta=2\\ 4\alpha-\beta=-5\end{cases}$

which has the solution

$\alpha=-{28\over 27},\beta={23\over 27}$

Therefore the desired limit is $8\alpha+\beta=-{67\over 9}$

Limit = calculus = should not be in this forum.


Hi!. Maybe this was suggested before.. I suggest to make an ebook of Mathlinks... What i mean is, everyone will contribute, it will be powerful book because thousands of mathlinkers will propose and solve problems which will be AWESOME. There will not be specific authors.. Yes, there will be dozens of editors which will scan the things that go to ebook.. I mean we can make different ebooks each dealing with different subjects... I think it is great idea since there are so many users in this website...

What do you think?

I like it.
it would be a great eBook because the problems here make you think, instead of the normal "what is 2+2?" kind of problems you get in school which is almost BRAINLESS MEMORIZATION instead of application of concepts. Not only that, it would spread from middle school to college level math.
go with it! :thumbup:  :D  :bruce:


find all a,b,n such
$a^{2}+ab+b^{2}=n^{2}$

and find all m>n  and k such
$m^{3}-n^{3}=k^{2}$

Dear friend,

Let me consider the first equation $a^2+ab+b^2=n^2$. We rewrite the equation in the form
\[(2a+b)^2+3b^2=(2n)^2\]
This leads to the more general form : $x^2+3y^2=z^2$. The problem is that find all integer solution of the equation.
There are some way to solve this equation. In here we use the FOUR NUMBER THEOREM as following
Theorem. Let $a,b,c,d$ are integers. Then $ab=cd$ if and only if there are four integers $m,n,u,v$ such that $gcd(m,n)=1$ and
\[a=um,\,\,\,b=vn,\,\,\,c=un,\,\,\,d=vm\]
The proof of the theorem is simple and i leave it.
Now turn back our equation, we have $3y\cdot y=(z-x)(z+x)$. Then by Four Number Theorem there exist there are four integers $m,n,u,v$ such that $gcd(m,n)=1$ such that
\[3y=um,\,\,\,y=vn,\,\,\,z-x=un,\,\,\,z+x=vm\]
Note that $um=3vn$ and $gcd(m,n)=1$ then we have $n$ divides by $u$. Just write $u=kn$ and note that $2|k$ and $km$ is divisible by $3$ you can easily get all the roots of the equation.
Apply what we have found, the first equation is followed.
The second equation, i still don't know how to finish it.


$ f(x)\equal{}x^{243}\plus{}x^{81}\plus{}x^{27}\plus{}x^{9}\plus{}x^3\plus{}x\plus{}1$
$ h(x)\equal{}x^2\plus{}1,r(x)\equal{}?$
I make $ r(x)\equal{}Ax\plus{}B,p(x)\equal{}f(x)\minus{}r(x),p(1)\equal{}2,p(\minus{}1)\equal{}2$
$ 2\equal{}7\minus{}A\minus{}B$
$ 2\equal{}\minus{}5\plus{}A\minus{}B$
$ A\equal{}6,B\equal{}\minus{}1$
$ r(x)\equal{}6x\minus{}1$
Is it true ? :maybe:

Well, if you want to use the analogy with my solution of the first problem, then notice that $ 1$ and $ \minus{}1$ in that solution were taken because we could split $ x^2\minus{}1$ to $ x\minus{}1$ and $ x\plus{}1$. In this case, you don't get $ x\minus{}1$ and $ x\plus{}1$ but $ x\minus{}i$ and $ x\plus{}i$. If you put $ i$ and $ \minus{}i$ in your calculation instead, you'll get $ A\equal{}0$, $ B\equal{}1$, hence $ r(x)\equal{}1$, which is true.

Note that $ x^4 \equiv 1 \bmod (x^2 \plus{} 1)$, hence $ f(x) \equiv x^3 \plus{} x \plus{} x^3 \plus{} x \plus{} x^3 \plus{} x \plus{} 1 \equiv 3x^3 \plus{} 3x \plus{} 1 \equiv 3x(x^2 \plus{} 1) \plus{} 1 \equiv 1 \bmod (x^2 \plus{} 1)$.

What we are essentially doing here is constructing $ \mathbb{Z}[i]$ as $ \mathbb{Z}[x]_{x^2 \plus{} 1}$.


give an example of mesurable subset $E$ of $\mathbb{R}$ such that, for any $-\infty < a < b <+\infty$,
$\frac{\lambda(E \cap (a;b))}{b-a}=\frac{1}{2}$

There are none :).

Such a set would have to have positive measure. However, it's external measure is equal to its measure, so for every $\varepsilon>0$ there is a sequence of disjoint open intervals $(I_n)_n$ such that $E\subset\bigcup I_n$ and $\sum\lambda(I_n)-\lambda(E)<\varepsilon\ (*)$. If you choose $\delta>0$ and $\varepsilon$ small enough depending on $\delta$, you'll find that $(*)$ implies the existence of an interval $I_n$ such that $\lambda(E\cap I_n)>(1-\delta)\lambda(I_n)$.

Basically, this says that for every measurable set $E$ of positive measure, there is an interval "filled very well" by $E$.

See [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=65640]this topic[/url]. The limit of that quotient as $b\to a$ must be either $0$ or $1$ for almost every $a,$ so there's no way that the quotient can always be $\frac12.$

Grobber could just as easily have showed you an interval that was "filled very poorly" by $E.$


[hide="hint"]Consider the biggest prime $ p\le n$ and check for divisibility by $ p$[/hide]

[quote="mhdi"]Hello,
prove that n! doesn't equal m^3 if  (n;m) are in N² and 1<n[/quote]
[hide]
Let p be the largest prime less then n (since n>1 we're guaranteed that there exists at least one p). 2p and 3p must be less then n! as well, since p|n! implies that $ p^3|n!$, and those are the smallest multiples of p that are less than n. But, by Bertrand's postulate there exists another prime between p and 2p, so we have a contradiction since there is a prime factor larger than p.[/hide]
We can generalize this method easily to see that $ n!$ can't be a perfect power for $ n>1$.


[hide]
let a=1000000 and suppose that
$\sqrt{a-1}+\sqrt{a+1}<2\sqrt{a}$
square both sides
$2a+2\sqrt{a^{2}-1}<4a$
$\sqrt{a^{2}-1}<a$
square both sides
$a^{2}-1<a^{2}$
it's true that 2000 is larger.[/hide]

but where any of his steps not reversable?

[quote="JBL"]Just for the record: what you wrote doesn't prove anything.  It's a trivial matter to begin with any statement and arrive at a true result, which is what you did.  In order to have a proof, you have to begin with true things and arrive at a conclusion (in this case, arrive at one of two statements concerning the relative sizes of two expressions).[/quote]
I don't think you are right.

[quote="Phelpedo"]JBL is right that it's not logically correct, but we all know what gunes means. He's not typing up a solution for a math journal, he's showing how he got to the answer.[/quote]
This is better :) but, I know I am not a mathematician, is there a logical contradiction at my method? The procedure can be done reversely, too. I can write the solution again in reverse order but I don't think it worths cuz true is true.


Sequence
modular arithmetic
IMO Shortlist
graph theory
combinatorics

For a positive integer $n$ define a sequence of zeros and ones to be [i]balanced[/i] if it contains $n$ zeros and $n$ ones. Two balanced sequences $a$ and $b$ are [i]neighbors[/i] if you can move one of the $2n$ symbols of $a$ to another position to form $b$. For instance, when $n = 4$, the balanced sequences $01101001$ and $00110101$ are neighbors because the third (or fourth) zero in the first sequence can be moved to the first or second position to form the second sequence. Prove that there is a set $S$ of at most $\frac{1}{n+1} \binom{2n}{n}$ balanced sequences such that every balanced sequence is equal to or is a neighbor of at least one sequence in $S$.

Please post your solutions. This is just a solution template to write up your solutions in a nice way and formatted in LaTeX. But maybe your solution is so well written that this is not required finally. For more information and instructions regarding the ISL/ILL problems please look here: [url=http://www.mathlinks.ro/Forum/viewtopic.php?t=15623]introduction for the IMO ShortList/LongList project[/url] and regarding[url=http://www.mathlinks.ro/Forum/viewtopic.php?t=15623]solutions[/url]  :)

[quote="mohammad-mahmoodi"]For each balanced sequence a = (a1; a2; : : : ; a2n) let f(a) be the sum of the
positions of the 1’s in a. For example, f(01101001) = 2+3+5+8 = 18. Partition the
¡2n
n ¢balanced sequences into n+1 classes according to the residue of f (mod n+1),
and let S be a class of minimum size. Then jSj · 1
n+1¡2n
n ¢, and we claim that every
balanced sequence is either a member of S or is a neighbor of at least one member of
S. Let a = (a1; a2; : : : ; a2n) be a given balanced sequence. We consider two cases.
Case (i): a1 = 1. The balanced sequence b = (b1; b2; : : : ; b2n) obtained from a by
moving the leftmost 1 just to the right of the kth 0 satisfies f(b) = f(a)+k. (If am+1
is the kth 0 of a, then in going from a to b, the leftmost 1 is moved up m places and
m¡k 1’s are moved back one place each.) Thus we find n neighbors of a so that the
values of f for a and these neighbors fill an interval of n + 1 consecutive integers. In
particular, one of these n + 1 balanced sequences belongs to S.
Case (ii): a1 = 0. This case is similar. Movement of the initial 0 just to the right
of the kth 1 yields a neighbor b satisfying f(b) = f(a) ¡ k.
Hence every balanced sequence is either equal to or is a neighbor of at least one
member of S.

 :)[/quote]

If you copy a solution from somewhere else, then please
[b](1)[/b] tell your source (in this case: the IMO Shortlist 2001 official solutions),
[b](2)[/b] convert it into LaTeX.

  darij

[quote]If you copy a solution from somewhere else, then please
(1) tell your source (in this case: the IMO Shortlist 2001 official solutions),
(2) convert it into LaTeX.

darij

_________________
lift this prayer to heaven's heights, feed my mind with the most glorious thoughts,
sacrificing all my dreams...
for this i will serve and honour to death... or do i just fear the tyrant's wrath?

[/quote]

HOW TO convert it into LaTeX ?
 :blush:

[quote="mohammad-mahmoodi"]HOW TO convert it into LaTeX ?
 :blush: [/quote]

Learn LaTeX here:
Best: http://www.mathlinks.ro/Forum/viewtopic.php?search_id=1602648475&t=188533
Q & A: http://www.mathlinks.ro/Forum/viewtopic.php?search_id=1602648475&t=12230
Official: http://www.mathlinks.ro/Forum/viewtopic.php?search_id=1602648475&t=5261

My personal tip(This is how I learnt.):

Try to search for posts written in LaTeX in the various brunches of this forum (You will find plenty). Then click on "quote" button on the bottom-right corner of the post and see how it looks(You need to be logged on first). You might modify a bit and see the preview yourself. You needn't really post. Just cancel the post when you are done.

For each balanced sequence $ a \equal{} (a_1, a_2, \dots , a_{2n})$ let $ f(a)$ be the sum of the positions of the $ 1$’s in $ a$. For example, $ f(01101001) \equal{} 2\plus{}3\plus{}5\plus{}8 \equal{} 18$. Partition the $ \binom{2n}{n}$ balanced sequences into $ n\plus{}1$ classes according to the residue of $ f\pmod{n\plus{}1}$, and let $ S$ be a class of minimum size. Then $ (n\plus{}1)|S|\leq \binom{2n}{n}$, and we claim that every balanced sequence is either a member of $ S$ or is a neighbor of at least one member of $ S$. Let $ a \equal{} (a_1, a_2, \dots , a_{2n})$ be a given balanced sequence. We consider two cases.
[b]Case (i)[/b]: $ a_1 \equal{} 1$. The balanced sequence $ b \equal{} (b_1, b_2, \dots , b_{2n})$ obtained from $ a$ by moving the leftmost $ 1$ just to the right of the $ k$-th $ 0$ satisfies $ f(b) \equal{} f(a)\plus{}k$. (If $ a_{m\plus{}1}$ is the $ k$-th $ 0$ of $ a$, then in going from $ a$ to $ b$, the leftmost $ 1$ is moved up $ m$ places and $ m\minus{}k$ $ 1$’s are moved back one place each.) Thus we find $ n$ neighbors of $ a$ so that the values of $ f$ for $ a$ and these neighbors fill an interval of $ n \plus{} 1$ consecutive integers. In particular, one of these $ n \plus{} 1$ balanced sequences belongs to $ S$.
[b]Case (ii)[/b]: $ a_1 \equal{} 0$. This case is similar. Movement of the initial $ 0$ just to the right of the $ k$th $ 1$ yields a neighbor $ b$ satisfying $ f(b) \equal{} f(a) \minus{} k$.
Hence every balanced sequence is either equal to or is a neighbor of at least one
member of $ S$.

[hide=" Solution"]
Incorrect, ignore. [/hide]

Let $a_1,a_2,\cdots a_{2n}$ be the sequence of $0$s and $1$s. For some sequence $s$, let $f(s)=\sum_{k=1}^{2n} ka_k$. We claim that for any $m$, the set
\[
S=\{s : f(s)\equiv m\pmod{n+1}\}
\]
is sufficient  (and since there are $n+1$ such sets and $\binom{2n}{n}$ total sequences, at least one must have appropriate size).

Let $s_k$ be the sequence where $a_1$ is put in position $k$, so $s_k=s$ in particular.

If $a_1=0$, then $f(s_{2n})=f(s_1)+n$. Note that $f(s_1),f(s_2),\cdots f(s_{2n})$ is a sequence of integers increasing by $0$ or $1$ each time, so by discrete IVT there must be $k$ with $f(s_k)\equiv m\pmod{n+1}$.

If $a_1=1$, then $f(s_{2n})=f(s_1)-n$. Note that $f(s_1),f(s_2)\cdots f(s_{2n})$ is a sequence of integers decreasing by $0$ or $1$ each time, so again by discrete IVT there must be $k$ with $f(s_k)\equiv m\pmod{n+1}$.

Yes it is.  It's also a bit of a red herring.  In fact mathocean97 and myself did not solve this problem when we tried it some time ago.

[quote=mihirb]Wait is  that the formula for the Catalan numbers?[/quote]

I think it isn't related to Catalan numbers, is it?

[quote=<pi37>]
$f(s)=\sum_{k=1}^{2n} ka_k$
[/quote]
How [b]could[/b] you think of $f(s)$?

For a balanced sequence $s$, let $f(s)$ denote the sum of the positions of the $1$s. It is easy to see that by removing a $1$ and placing it in at all possible $2n$ locations, we get $f$-values from $a,a+1,\ldots,a+n$, where $a$ is the sum of the positions after removing the $1$. Thus, if we fix $0\le k\le n$, then there is some $s'$ adjacent to $s$ such that $f(s')\equiv k\pmod{n+1}$. This implies that the set \[S_k = \{s : f(s)\equiv k\pmod{n+1}\}\] works. To finish, note that \[|S_0|+\cdots+|S_n|=\binom{2n}{n},\] so we can pick some $k$ such that $|S_k|\le\frac{1}{n+1}\binom{2n}{n}$. This completes the solution.

[b]Remark:[/b]  I was originally spending most of my time trying to modify Catalan sequences in various ways to no avail. I think it's pretty clear after trying this approach for a while that it will inevitably bunch up too many sequences close together, thus missing many sequences. The key for this problem is to step away from that approach and look for simpler things.


Given trapezoid  with parallel sides AB and  CD, assume that there exist points E  on line  BC outside segment , and  F inside segment  AD such that $ \angle DAE \equal{}\angle CBF$ . Denote by I the point of intersection of  CD and EF , and by J  the point of intersection of AB  and EF. Let  K be the midpoint of segment EF , assume it does not lie on line CD. Prove that ABIK is cyclic  if and only if CDJK is cyclic

[quote="diegu"]Given trapezoid  with parallel sides AB and  CD, assume that there exist points E  on line  BC outside segment , and  F inside segment  AD such that $ \angle DAE \equal{} \angle CBF$ . Denote by I the point of intersection of  CD and EF , and by J  the point of intersection of AB  and EF. Let  K be the midpoint of segment EF , assume it does not lie on line CD. Prove that ABIK is cyclic  if and only if CDJK is cyclic[/quote]
First we can prove from $ \angle DAE \equal{} \angle CBF$  that $ AFBE$ is cyclic and $ CBFD$ is cyclic.
So $ JF\cdot JE\equal{}JA\cdot JB$ also $ IF\cdot IE\equal{}IC\cdot ID$
$ ABIK$ is cyclic $ \Leftrightarrow$ $ JF\cdot JE\equal{}JI\cdot JK$
$ \Leftrightarrow$ $ JK\cdot IJ\equal{}(IJ\minus{}IF)\cdot (IE\minus{}IJ)$
$ \Leftrightarrow$ $ JK\cdot IJ\equal{}(IF\plus{}IE)IJ\minus{}IJ^{2}\minus{}IF\cdot IE$
$ \Leftrightarrow$ $ IK\cdot IJ\plus{}IF\cdot IE\equal{}2IK\cdot IJ$
$ \Leftrightarrow$ $ IF\cdot IE\equal{}IK\cdot IJ$
$ \Leftrightarrow$ $ CDJK$ is cyclic
QED :D


function
number theory unsolved
number theory

[quote="t0rajir0u"]Er, sorry, what is $p$?[/quote]
I join to the question.
As to my above post I've hardly understood now what I've meant.
My interpretation was as follows (based on the author's example):

$ab \equiv a+b-1 (mod \varphi (ab) )$

If a and b are odd primes and ab = n then phi(n) = (a - 1)(b - 1)

= ab - a - b + 1.

ab (mod (ab - a - b + 1)) in this case is simply

ab - (ab - a - b + 1) = a + b - 1. So it is true...

I wonder if a general case can be drawn from this for n = abcde...?

[What is p? Sorry, I should have written n = a + b - 1 (mod h)]

I have only caught the typo with $p$.

The function $\varphi$ is multiplicative one, hence in particular for distinct primes $a, b, c, ...$ we have
$\varphi (abc...)=\varphi (a)\cdot \varphi (b)\cdot \varphi (c)... = (a-1)(b-1)(c-1)...$
Would you like to consider the difference between this expression and the product $abc...$?  :maybe:
Hope t0rajir0u should  understand better - I bet he knows english better me.  :)


Use vector components to find the magnitude and direction of the vector needed to balance the two vectors. Let $ 625N$ vector be along the -y axis and let the +x axis be perpendicular to it towards the right

[img]http://i27.tinypic.com/2nbvoyu.jpg[/img]

[hide=I found magnitude but how to find the angle?]
[img]http://i31.tinypic.com/2nuriqf.jpg[/img]

$ \vec{R} \equal{} \vec{A} \plus{} \vec{B}$

$ A_x \equal{} \minus{} 875cos30^o \equal{} \minus{} 757.80$
$ A_y \equal{} \minus{} 875sin30^o \equal{} \minus{} 437.5$

$ B_x \equal{} 0 B_y \equal{} 625$

$ R_x \equal{} \minus{} 757.80 R_y \equal{} 189.5$

$ \vec{R} \equal{} 781N$[\hide]

How cna I find the direction?

From looking at your last diagram, the resultant forms an angle of $ 60^{\circ}$ with the vertical in the SW direction.  Therefore, if we translate the vector to the origin, we can say that its direction is $ 60^{\circ}$ West of South.


probability
modular arithmetic
number theory

There is a $ \frac23$ chance that the first spin is $ 1,3\pmod4$, and a $ \frac13$ chance that the first spin is $ 2\pmod4$.

There is a $ \frac12$ chance that the first spin is $ 1,3\pmod5$, and a $ \frac12$ chance that the first spin is $ 2,4\pmod5$.

We must either have $ (1,1),(1,3),(3,1),(3,3)$, which has probability $ \frac23\cdot\frac12\equal{}\frac13$, or $ (2,2),(2,4)$, which has probability $ \frac13\cdot\frac12\equal{}\frac16$, for a total of $ \frac13\plus{}\frac16\equal{}\boxed{\frac12}$.

Or we could just know each square has an equal probability of being picked and 6/12=1/2.

But you don't know that until you do the initial analysis. :wink:

BUt it works, so isnt it a correct solution?

Does anyone know why we are focussing on even and odd remainders in this solution?

[font=times new roman][hide=Listing Possibilities]
S=Shaded
B=Blank
1,1 S
1,2 B
1,3 S
1,6 S
1,7 B
1,9 B

2,1 B
2,2 S
2,3 B
2,6 B
2,7 S
2,9 S

3,1 S
3,2 B
3,3 S
3,6 S
3,7 B
3,9 B

6,1 B
6,2 S
6,3 B
6,6 B
6,7 S
6,9 S

7,1 S
7,2 B
7,3 S
7,6 S
7,7 B
7,9 B

9,1 S
9,2 B
9,3 S
9,6 S
9,7 B
9,9 B

There are an equal amount of S's and B's, so the probability of getting a shaded square is $\tfrac12.$ $$\definecolor{color}{rgb}{0.7, 0.2, 0.9} \scriptsize{\colorbox{color}{\color{black}{A pifinity solution}}}$$

[font=times new roman][hide=Intuitive Solution] Let's say the first spin was a 1. Then, whatever the second spin turns out to be, we know that we are looking in the "1" column, because $(4\times0)+1=1.$ So to find the square for spins starting with 1, we only need to find the remainder for the second spin, and then use that row with column 1.

If the first spin were 2, then all the landing squares for that series would be in the "2" column.

Now that we know how this works, let's take a look at column 3. It's the same as column 1! So that means that all of the results in the series with the first spin being 3 are the same as the results in the series with the first spin being 1!

It turns out, for any number that is the result of the first spin, the resulting series is either the same as if the first spin were one or two. Also, for both series 1 and series 2, there are an equal amount of shaded square results and unshaded square results. This means that there is an equal amount of shaded and unshaded squares in the totals for all the series. Therefore, the probability is 1/2. $$\definecolor{color}{rgb}{0.2, 0.2, 0.4} \scriptsize{\colorbox{color}{\color{white}{A pifinity solution}}}$$

[quote=Bestestsocks][hide]If this showed up on my AMC 10, I would have guessed 1/2. Since it shoed up on alcumus, I actually proved that it is 1/2.[/hide][/quote]

Source:2007 AMC 10

What's wrong with the following approach?

The probability of getting a remainder of 1, 2 and 3 in the first spin is each 1/3. The probability of getting a remainder of 1 and 2 in the second spin is each 1/3 and getting a remainder of 3 and 4 is each 1/6 in the second spin (as there is only one occurrence of 3 and 4 but 2 occurrences of 1 and 2). Applying the fact that the two spins are independent, I simply combined the probabilities for the shaded square as follows:

1/3 * 1/3 + 1/3 * 1/3 + 1/3 * 1/3 + 1/3 * 1/3 (for the combinations of 1 and 2 from the second spin with the first spin) + 1/3 * 1/6 + 1/3 * 1/6 (for the combinations of 3 and 4 from the second spin with the first spin) resulting in...

4/9 + 2/18 = 5/9.


trigonometry
limit
calculus
calculus computations

$ \sum_{n\equal{}1}^{\infty}\frac{n}{\tan{n}}$ converges?

$ \sum_{n\equal{}1}^{\infty}\frac{1}{n\tan{n}}$ converges?

$ n$ can take on values arbitrarily close to $ k\pi$ for large $ n,$ which means that $ \frac1{\tan n}$ is unbounded above. That clearly shows that the first series diverges, as its terms fail to go to zero. It does not so clearly show that about the second series, as we must ask how large $ \frac1{\tan n}$ can be in relation to $ n,$ but it should still be true that the terms do not tend to zero and the series diverges.

Well Kent, L'Hospitaling the nth term of series (2), where n tends to infinity (before LHR, we take n to the numerator in the form of 1/x), an we get that the term tends to 0.

(BTW, the series do converge, but I haven't proven it yet.)

[quote="hsiljak"]Well Kent, L'Hospitaling the nth term of series (2), where n tends to infinity ...[/quote]
What? That makes no sense at all. L'Hôpital's rule requires a $ \frac{0}{0}$ or $ \frac{\infty}{\infty}$ quotient - it requires those limits of the numerator and denominator. But $ \tan n$ does not tend to $ 0$ and $ \tan n$ does not tend to $ \infty.$ Instead, it behaves quite chaotically. L'Hôpital cannot be applied.

[quote](BTW, the series do converge, but I haven't proven it yet.)[/quote]
I can assure you that the first series diverges, and I'm 99% convinced that the second series also diverges.

Is there any chance the $ \tan n$ was supposed to have been $ \tan\left(\frac1n\right)?$ If so, that would change the nature of the problem (and reduce the chaos.)

I absolutely agree with Kent .
In 2-nd problem, indeed there exist $ n_{k}$ such that $ \frac {1}{n_{k}\tan(n_{k})}$ dont tends to 0 :
since $ \lim n|\sin (n)|$ does not exist . This last problems already was posted here , where we find $ p_{i}$ such that $ p_{i}\sin p_{i} \to C$ where $ p_{i} \approx 0 \mod \pi$ hence $ |\cos p_{i}| \approx 1$ thus $ \frac {\cos(p_{i})}{p_{i}\sin p_{i}} \approx C$ where $ |C| > 0$

[quote="Kent Merryfield"][quote="hsiljak"]Well Kent, L'Hospitaling the nth term of series (2), where n tends to infinity ...[/quote]
What? That makes no sense at all. [/quote]

True... mea culpa... (I was still a bit sleepy... 7:45 AM...)

[quote][quote](BTW, the series do converge, but I haven't proven it yet.)[/quote]
I can assure you that the first series diverges, and I'm 99% convinced that the second series also diverges.[/quote]

I was referring to the second. My opinion was that it converges (but now, I don't think so).


Monday Oct 3

4:30 to 6:30

University of Toronto Schools (located on the corner of Bloor and Spadina)

Room 24

all welcome

What happens in a math circle? What ages are welcomed specifically?

In recent years, younger and/or weaker students have been especially encouraged.  The work is generally "Euclid" (the contest) level, I think.

Euclid (the grade 12 contest?)... I know I'm not ready for that.  :blush:   :D

Needless to say, "Mathematical Circles" ( the book ) is great problem solving supplement for those seeking a taste of "math circles".

[quote="Singular"]d.lam i would show up anyways you will improve alot at mcircles[/quote]

  At a glance "mcircles" looks like "miracle". :rotfl:  :rotfl:  :rotfl:  :rotfl:  :rotfl:

 (no offense)

Hey!  You just proved Math-forum advertising works!

I just got back from the first one, I thought it was great. Unless you're already ready to try and get on the canadian imo team TOMMOROW, you would benefit going. Heck, even if you're that good, there are ample people and problems ( and people with problems ) to run into. One word for all people considering it: go.

shoot... I forgot to check this forum on time. Is it too late for me to come on October 17th? Uh-oh, I think I have Choral Ensemble after school... shoot... umm... what time does it go from? (I think I've already seen the time for it but it could be different?)
Is anyone watching Les Miserables or have watched it before? Did anyone like the musical?

On a side note: Did anyone apply for Northern that was out-of-district? I found out that they didn't accept anyone that was out-of-district; that's strange...

Oh and Alex, you're from Leaside right? My friend used to go to Northlee and a lot of his friends went to Leaside which is why he went to North Toronto (to get away from them; ha,ha,ha).

And how many ppl showed up at the Math Circle?

[quote="d.lam.86"]shoot... I forgot to check this forum on time. Is it too late for me to come on October 17th?

Oh and Alex, you're from Leaside right? [/quote]

1) come any days you want

2)  wow im famous

[quote="d.lam.86"]
And how many ppl showed up at the Math Circle?[/quote]
You forgot to answer my last question.  :| How long will these math circles be? Will they occur throughout the year? Will they ever be on another weekday other than Monday? Oh, I pass by Leaside sometimes when I'm going to school, once I get off the highway into Eglinton. The school has a large field; yeah, that's about all I know about Leaside.  ;)

about 20?
every week monday sameplace
430pm to 630pm

Is there a web page on the U of Toronto site that has this information :?:

Hi I am Henry Bai from Jarvis Collegiate Institute and I think that math is cool


Joan tries to solve a really hard problem once each day.  She has a 1/4 probability of solving it each day.  What is the probability that she will solve it before her sixth try?

The probability that she won't solve it by the fifth try (before the sixth) is $ \left(\frac34\right)^5\equal{}\frac{243}{1024}$, so the probability she does solve it before the sixth try is $ 1\minus{}\frac{243}{1024}\equal{}\boxed{\frac{781}{1024}}$.

[hide]We must find the probability that Joan can solve it at any time before the sixth try, so it is the sum of the probabilities that she will solve it on her first, second, third, fourth, and fifth tries. We could evaluate all those cases, but seeing all that casework, we wonder if it will be easier to find the probability that she fails to solve it before 6 tries, and subtract the result from 1.

 In order for her to fail to solve it before her sixth try, she must fail 5 times. The probability of failure on each try is $1 - \frac{1}{4} = \frac{3}{4}$, so the probability that she fails on each of her first 5 tries is $\left(\frac{3}{4}\right)^5 = \frac{243}{1024}$. Therefore, the probability that she succeeds before her sixth try is \[1-\frac{243}{1024} = \boxed{\frac{781}{1024}}.\][/hide]

[quote=brte4][quote=Possible][quote=brte4]Casework Counting sol: [hide]Find the probabilities of all the ways she can solve it before her sixth try:
First try: $\frac{1}{4}$
Second try: $\frac{3}{4} \cdot \frac{1}{4}= \frac{3}{16}$
Third try: $\frac{3}{4} \cdot \frac{3}{4} \cdot \frac{1}{4}=\frac{9}{64}$
Fourth try: $\frac{3}{4} \cdot \frac{3}{4} \cdot \frac{3}{4} \cdot\frac{1}{4}= \frac{27}{256}$
Fifth try:  $\frac{3}{4} \cdot \frac{3}{4} \cdot \frac{3}{4} \cdot \frac{3}{4} \cdot \frac{1}{4} = \frac{81}{1024}$

Adding these cases up, you get  $\boxed{\frac{781}{1024}}$. [/hide][/quote]

How do you add the cases quickly?[/quote]

no i actually did it on paper  but i was feeling to lazy to do it on here lol[/quote]

Oh, I hate bashy addition


greatest common divisor
number theory proposed
number theory

Solve the equation
\[(a^{2},b^{2})+(a,bc)+(b,ac)+(c,ab)=199.\]
in positive integers.
(Here $(x,y)$ denotes the greatest common divisor of $x$ and $y$.)

[i]Proposed by S. Berlov[/i]

I resolved the problem later, but I do not know the origin of it
Can anyone help me?
  Solve the equation :
           $ ab\plus{}c\equal{} \ (a^{2},b^{2})\plus{}(a,bc)\plus{}(b,ac)\plus{}(c,ab)\equal{}239^2$
in positive integers.

[quote="math10"]I resolved the problem later, but I do not know the origin of it
Can anyone help me?
  Solve the equation :
           $ ab \plus{} c \equal{} \ (a^{2},b^{2}) \plus{} (a,bc) \plus{} (b,ac) \plus{} (c,ab) \equal{} 239^2$
in positive integers.[/quote]

We have $ (a,b,c)|\ (a^{2},b^{2}) \plus{} (a,bc) \plus{} (b,ac) \plus{} (c,ab) \equal{}239^2$

If $ 239|(a,b,c)$, then $ ab\plus{}c \ge (239)(239)\plus{}239>239^2$, contradiction.So $ (a,b,c)\equal{}1$.

So let $ (a,b)\equal{}x,(b,c)\equal{}y,(c,a)\equal{}z$ the equation becomes :

$ ab\plus{}c \equal{} x^2\plus{}zx\plus{}xy\plus{}yz\equal{}239^2$ with $ (zx)(xy)|ab$ and $ (yz)|c$

So $ yz|(ab\plus{}c)\equal{}239^2$. As $ (y,z)\equal{}1$ and that $ 239$ is a prime, we have at least one of $ y,z$, say $ y\equal{}1$

$ \Leftrightarrow ab\plus{}c\equal{}(x\plus{}1)(x\plus{}z)\equal{}239^2 \Rightarrow \begin{cases}x\plus{}1\equal{}239 \\ x\plus{}z\equal{}239 \end{cases} \Rightarrow \begin{cases}x\equal{}238 \\ z\equal{}1 \end{cases}$

As $ x^2|ab$, we have $ x^2\equal{}ab$ or $ ab\plus{}c >2 (238)^2>239$, contradiction.

So $ a,b\equal{}x\equal{}238$, $ c\equal{}239^2\minus{}238^2\equal{}477$ is the solution after checking.

[hide="Hint"]Try to prove that: (a,bc)=(a,b)(b,c)[/hide]

my solutions of problem is to prove $ a \equal{} b$.
If $ a,b$ are different.We have:
       $ ab > (a;b)^2 \plus{} a \plus{} b$.
But:$ (a;b)^2 \plus{} (a;bc) \plus{} (b;ca) \plus{} (c;ab)\leq(a;b)^2 \plus{} (a \plus{} b \plus{} c)$
So $ ab\leq(a;b)^2 \plus{} a \plus{} b$.
$ \Rightarrow a\equal{}b$

[quote="mathmanman"]Solve the equation
\[ (a^{2},b^{2}) \plus{} (a,bc) \plus{} (b,ac) \plus{} (c,ab) \equal{} 199.
\]
in positive integers.
(Here $ (x,y)$ denotes the greatest common divisor of $ x$ and $ y$.)[/quote]
Firste note that $ 199$ is a prime. Assume that $ (a,b,c) \equal{} d > 1$. Then $ d^2 \mid (a^2,b^2)$ og $ d \mid (a,bc), (b,ac), (c,ab)$. Then since $ d >1, d \mid 199$ we see that $ d \equal{} 199$. And hence $ LHS \ge 199^2 \plus{} 3 \cdot 199 > 199$. Contradiction. So $ (a,b,c)\equal{}1$.
We can see that $ (a,bc) \equal{} (a,b) \cdot (a,c)$. Proof: Let $ (b,c) \equal{} d$, and $ b \equal{} db_1, c \equal{} dc_1$ (then $ (b_1,c_1)\equal{}1$). Since $ (a,b,c)\equal{}1$ we see that $ (a,d)\equal{}1$ hence $ (a,bc) \equal{} (a,b_1c_1) \equal{} (a,b_1) \cdot (a,c_1) \equal{} (a,b) \cdot (a,c)$.
It's obvious that $ (a^2,b^2) \equal{} (a,b)^2$. So $ LHS \equal{} (a,b)^2 \plus{} (a,b)\cdot (a,c) \plus{} (a,b) \cdot (b,c) \plus{} (a,c) \cdot (b,c) \equal{}$ $ ( (a,b) \plus{} (a,c) ) \cdot ( (a,b) \plus{} (b,c) ) \equal{} 199$
Since $ 199$ is a prime, this can't happen. (Of course $ (a,b), (a,c), (b,c) \ge 1$) So: No solutions.

nice solution.
thank Mathias_DK very much.


probability
invariant
calculus
real analysis
real analysis unsolved

The code for the safe constists of $4$ sign, the first one-letter from English alphabet(a,b,c,d,...,), and the others three are-numbers from $0$ to $9$.
One is supposed not to know the right code, who is trying to open the safe
a). Find the probability that the safe will be opened after the first trial.
b). Find the probaility that the safe will be opened after $k$ trials.

???

And you do a random pick for the code every time or what?

If you pick always the same, your chance remains invariant after the first pick.

If you try something different each time, your chance is k/26000

If you pick a random one each time, your chance is 1-(25999/26000)^k

What is your question? And most of all what has this to do with calculus? :?


[quote="frodon007"]Hi everyone,

can someone help me for that question, where do I start, I'm doing integration at the moment.

[quote]Write down a definite integral that will give the value of the area under the curve y= x^2 cos(1/5x) between x=1/2π and  x= 5/2π.
(the expression x^2 cos(1/5x) takes no negative values for 1/2π≤x≤5/2π.
You are not asked to evaluate the integral by hand.)[/quote]

Thanks again :lol:[/quote]

$ \int_{1/2n}^{5/2n}x^{2}\cos\left(\frac{1}{5x}\right)\ dx$

The location of the $ n$ and $ x$ is sort of ambiguous in the original post.


The seven dwarfs walk to work each morning in single file. As they go they sing their famous song, "High - low - high - low, it's off to work we go ...". Each day they line up so that no three successive dwarfs are either increasing or decreasing in height. Thus, the line-up must go up-down-up-down ... or down-up-down-up ... , If they all have different heights, for how many days can they go to work like this if they insist on using a different order each day?

What if Snow White always comes along too?

easy
you can first asume that theres 4 short dwarves and 3 tall(you can also asume 4 tall and 3 short it doesnt matter)
the you find how many orders for the 4 short ones
which is 4! or 4x3x2x1=24
now you find out how many for the 3
3!=3x2x1=6
now multiply 24by6 and you get 144
answer=144
but if this is rigged and snow white is dumb and wants to work with them
then you add 1 tall person because snow white is tall so you get
24x24=576 :|  :mad:  :lol:  :)

No: the dwarves all have different heights, so (for example) there are arrangements like short < short > short < tall > tall < tall > tall possible.  This question is just an excessively long-winded way of asking for the number of alternating permutations of length seven; it's not terribly hard to find a recursion (consider the position of the largest entry = tallest dwarf), and the associated sequence is [url=http://www.research.att.com/~njas/sequences/A000111]A000111[/url] in the [url=http://www.research.att.com/~njas/sequences]OEIS[/url].

Could someone give an answer to this question?


trigonometry
induction
real analysis
real analysis solved

If x is a complex number on the unit circle that is not root of unity of any order, does it follow that $\sqrt[n]{|x^n-1|}$ goes to 1 when n goes to infinity?

No, it does not.  ;)  :D
Just contsruct counterexample.

[quote="iura"]This problem seems to be too simple.
If $x=\cos \alpha \cdot 2\pi+ i \sin \alpha \cdot 2 \pi$ then $\alpha$ is irrational,
then $x^n$ is dense in the unit circle so then $|x^n-1|$ is dense in $ [0;2] $.
Hence your assumption is wrong.[/quote]
Ah, ah, ah!
Your reasoning is wrong, iura! Maybe you didn't see $\sqrt[n]{\quad}$.

Iura, I know that the level of difficulty of the problems posted by me tends to 0, but even I realize sometimes obvious things.  ;)

[u][b]The author of this posting is : iura[/b][/u]
____________________________________________________________________

Dear Harazi and Myth, I'm terribly sorry for my extremely stupid post.
Now let's construct the counter-example.
(0) For $\alpha$ sufficiently small and positive, we have $\left|z-1\right| <8\alpha$ for $z=\cos {2\pi\alpha}+i \sin {2\pi\alpha}$
Proof: it is $\sqrt{\left(1-\cos {2\pi\alpha}\right)^2+\sin^2 {2\pi \alpha} = 4\sin^4 {2\pi\frac{\alpha}2}+\sin^2 2\pi\alpha}$. Now using the estimation $\sin x < x$ we get the result.
Next we construct the sequence $n_i$ as follows:
$n_0=2004$, $n_i=i!n_i 4^{n_i}$.
We easily deduce the following observations
(1) $n_i | n_j$ for $i<j$
(2) For any $m$ there is a $k$ s.t. $m | n_k$
(3) $n_{i+k} > 2^{n_i+k+1}n_i$ by an easy induction on $k$
(4) From (3) we deduce $\sum_{i=k+1}^{\infty} \frac 1{n_i} < \frac{1}{2^{n_k} n_k}$
(5) From (4) we get $\sum_{i=k+1}^{\infty} \frac{n_k}{n_i} < \frac{1}{2^{n_k}}$

Now let $u=\sum_{i=0}^{\infty} \frac1{n_i}$.
We prove that $z = \cos 2\pi u + i\sin 2\pi u$ satsfies $\sqrt[n_i]{\left|z^{n_i}-1\right|} < \frac23$
Indeed $z^{n_i} = \cos 2\pi n_i u + i\sin 2\pi n_i u = \cos 2\pi t + i\sin 2\pi t$ where $t = \sum_{j=i+1}^{\infty} \frac{n_i}{n_j}$ and by (5) $t<\frac1{2^{n_i}}$
Now using (0) we deduce $\left|z^{n_i}-1\right| < \frac{8}{2^{n_i}}$
Hence $\sqrt[n]{\left|z^{n_i}-1\right|} < \frac{\sqrt[n_i]{8}}{2} < \frac23$ as desired.
Finally, we must check that $z$ is not a root of unity or that $u$ is irrational.
Suppose the contrary. Then $u=\frac{p}{q}$.
Then from (2) we find an $n_k$ s.t. $q | n_k$ so $un_k$ is an integer so $\left\{un_k\right\} = 0$.
But we've shown a few lines above that $0<\left\{n_k u\right\} < \frac 1{2^{n_i}}$. Contradiction
QED

____________________________________________________________________

[color=red][[b]Note by Darij:[/b] This posting contains a number of typos. I have fixed some of them but I could not correct the rest. Thanks for any help.][/color]

[u][b]The author of this posting is : Myth[/b][/u]
____________________________________________________________________

Oooh, why is it so long? As I remember, I send to Harazi nice short counterexample.


$x=tanA$ where $-\frac{\pi}{2}<A<\frac{\pi}{2}$ and $A\ne \pm \frac{\pi}{6}$

Με την ευκαιρια, τις ευχες μου για καλη χρονια (καλη συνεχεια) με υγεια και γαληνη... :)

sostos socrates....ola ksekinan apo to gergonos oti i morfi ton klasmaton 8imizi tin morgi tis ef3a sinartisi tis efa.....
xrisimopiontas malista ton ton tipo gia tin ef(a+b+c)=(efa+efb+efc-efaefbefc)/(1-efaefb-efbefc-efcefa)
kai me mia mikri diereunisi oti den mpori na ine taytoxrona kai o paranomastis miden ,pernume tlk oti ef(a+b+c)=0
diladi a+b+c=kπ opote 3a+3b+3c=3kπ diladi ef(3a+3b+3c)=0
opote prokipti to zitumeno...
[u][b]Simiosi[/b][/u]
O metasximatismos aytos,ine xaraktiristikos otan dinete i sigekrimeni sin8iki x+y+z=xyz...kai voi8ai para poli toso se isotites (opos edo)oso +se anisotites..To mono pu xriazete ine gnosi i kapion trigonometrikon tipon...

 :D

υπεροχη ασκηση.οταντην ειδα  μου ηρθε στο μυαλο η εφαπτομενη .μια ερωτηση:
προσεξτε τη συμετρια σε συνθηκη και ζητουμενη σχεση---> αθροισμα = γινομενο των ιδιων αριθμων .ποιες αλλες παραστασεις μπορει να παρει το κλασμα στην σχεση που ζητα ο ΡΑΚΑΡ ?

λοιπον μπορει καποιος να βαλει καμια ασκηση παρωμοια με αυτη του Ραλλη?δηλ  με ταυτοτητες . ισως ο κ.Στεργιου που στο βιβλιο της Α 'λυκειου για τις ολυμπιαδες  ειχε τετοιες ωραιες ασκησεις
ευχαριστω εκ των προτερων ......

kosta..8a evaza kai mia akomi (anisotita..malista)pu 8a xrisimopiuses ayto ton metasximatismo..ala distixos to vivlio  sto opio ipirxe..to exo danisi se ena pedi.. :(
an vro kapia ali sxetiki 8a tin postaro...

 :D

βιβλιο αλγεβρας ? η ολυμπιαδων ? αν επιτρεπεται... :D

lol....gt na min epitrepete re si??...
biblio algebras tu mpailaki....(exi orea olimpiaka 8emata proetimasias..)

 :D

edit:i sigekrimeni pu evala itan prota8en 8ema gia imo...made in cuba...


integration trigonometry logarithms calculus

Ascertain $ \int_0^{\frac {n\pi}{4}}\frac {|\sin 2x|}{|\sin x| \plus{} |\cos x|}\ \mathrm {dx}$ , where $ n\in \mathbb N^*$ .

Let $ I_n \equal{} \int_0^{\frac {n\pi}{4}} \frac {|\sin 2x|}{|\sin x| \plus{} |\cos x|}dx,$

$ I_{4(k \plus{} 1)} \minus{} I_{4k} \equal{} \int_{k\pi}^{(k \plus{} 1)\pi} \frac {|\sin 2x|}{|\sin x| \plus{} |\cos x|}\ dx$

$ \equal{} \int_0^{\pi} \left(|\sin x| \plus{} |\cos x| \minus{} \frac {1}{|\sin x| \plus{} |\cos x|}\right) dx$

$ \equal{} 4 \minus{} \int_0^{\pi} \frac {1}{|\sin x| \plus{} |\cos x|}\ dx$

$ \equal{} 4 \minus{} 2\int_0^{\frac {\pi}{2}} \frac {1}{\sin x \plus{} \cos x}\ dx$

$ \equal{} 4 \minus{} 2\sqrt {2}\ln (1 \plus{} \sqrt {2}).$

$ \therefore I_{4k} \equal{} I_0 \plus{} k(4 \minus{} 2\sqrt {2}\ln (1 \plus{} \sqrt {2})),\ I_0 \equal{} 0,$ yielding $ \boxed{I_n \equal{} \frac {1}{4}n(4 \minus{} 2\sqrt {2}\ln (1 \plus{} \sqrt {2}))}$

Calculus in the high school forums?  Really, now  :|

Oh, sorry. I haven't seen the section carefully.

Please move these two posts to Calculus Computation and Tutorials.

kunny


Let's say I wanted to save all of this in my own custom package:

%packages
\usepackage{amsmath}
\usepackage{amssymb}
\usepackage{amsthm}
\usepackage[pdftex]{graphics}
\usepackage{mycommands}

%layout
\pdfpagewidth 8.5in
\pdfpageheight 11in
\topmargin 0in
\headheight 0in
\headsep 0in
\textheight 9in
\textwidth 6.5in
\oddsidemargin 0in
\evensidemargin 0in

does that mean that all I would have to do is use my new package and I will have access to the stored packages? Or would I still have to declare the packages in the main .tex file?

[url=http://www.artofproblemsolving.com/LaTeX/AoPS_L_GuidePack.php#own]Our page on making your own packages should answer that question.[/url]


$ a,b,c >0$
$ a\plus{}b\plus{}c\equal{}1$ find max:
$ \sum \frac{a}{a^3\plus{}a^2\plus{}1}$


probability Putnam function Euler factorial algebra polynomial

I made this problem up from the idea from "The Price is Right".

You have a given range of integers, with one of these numbers the target (this is secret), and there are two players. Each player takes a turn to guess a number in that range, and then will be told whether he is too high or too low. Then its the other players turn. Whats the probability that the first player wins, and what is his/her strategy?

I programmed this on my very slow calculator and when the range got to about 60 numbers wide, the probability got to around 0.625. Due to my theorem about nice numbers, I'm guessing that the limit as the size of the range goes to infinity is 1 - 1/e. Any way to prove this? And I haven't worked out the strategy either, but its probably related to that.

A somewhat related problem is B4 on Putnam 2002. I assume the chosen number must be an integer, right?

Yeah, you pick an integer. Basically this is just the normal I'm thinking of a number between 1 and 100 etc game, except with more than one player. It may get even more interesting with 3 players etc..

I'll have a look at the putnam problem right now..

..

Yeah that does look a little similar.. except in that case you chose both strategies to suit you. In my one, their strategy is to suit them..

[hide]If player one chooses a number, lets call it A, he has a 1/n chance of getting it correct. There is a (A-1)/n chance that the number will be between 1 and A-1 inclusive, and an (n-A)/n chance that it will be in the upper bound (excluding A, as we know that that answer is incorrect)

We know what the bound will be, as the director will announce it.

Player two can then choose a number from either 1 to A or A+1 to n. If the number is in the lower bound, knowing that the first player was incorrect, he has a ((n-1)/n)((A-1)/n)(1/(A-1)) = (n-1)/n^2 chance of getting it right. If it is in the upper bound, he also has a ((n-1)/n)((n-A)/n)(1/(n-A)) = (n-1)/n^2 chance of getting it right. He then has the total chance of 2(n-1)/n^2 chance of getting it right.

.....to be continued.

[/hide]

[hide]I don't want to give away the solution to Putnam B4, but it seems to me that I can solve this problem basically the same way. The first person should give away as little information as possible to the second person. Of course, the second person should give away as little information as possible to the first assuming that the first person is playing with the optimal strategy. Giving away the least amount of information possible means choosing a number such that it only cuts off one direction (i.e. choosing the largest or smallest available). I think that's the best strategy.[/hide] Now I need to find a proof.

Slightly editing my calculator program, I can tell you that

my calculator [hide]**********dis[/hide]agrees

Simon, I would tend to agree with you, except for one thing:  if you deny the other player information, you deny yourself information, as well.  Now, there are clearly instances in which your solution is correct -- for example, n = 3.  What this reminds me slightly of is the prisoner's dillema, in that helping your opponent out could also help you out.  Another reason I disagree with you, by the way, is that your strategy is not the limiting case -- the limiting case would be to pick the same number every time.  It gives the opponent 0 new information, and is also going to lose almost all of the time.  So while I think your solution is good for small cases, I doubt it works asymptotically.

Is that legal? And why is that the limiting strategy?

Sorry, it's the limiting case of your strategy.  Well, not really, but it is if we define your strategy in the narrow terms "give as little information as possible."  It isn't exactly a meaningful limit because your chance of winning on each successive pick doesn't keep going up, which it would in any normal strategy.  Anyhow, even if it was legal, it's a sure way to lose.

TripleM, I think that the answer should be 1/2 as n -> :inf:, just because for very large versions of the game, the chances of winning on the first pick are essentially 0, and then the game is still very large, so it doesn't essentially change it very much.  (On the whole, 2 is a nicer number than e, anyhow.)  In fact, I would venture to guess that with n numbers, the first player's best chances will be ceiling(n/2)/n, although I'm not positive.  (This is from only doing out the first 5 cases by hand, mind you.)

Hmm, except case n = 6 destroys my beautiful theory.  Well, I'll keep trying :)

I think the question might also be interesting if we were to just have the people pick randomly from among the remaining numbers at any given point.  I wonder if that would un-even the chances ...

Or what if one person were picking randomly from the ones in the appropriate range (i.e. not the numbers that have already been deemed impossible) and the other person were playing ``strategically''? Would the probability for the person playing strategically winning tend to one? I'll conjecture that the answer is no, but I would't be surprised if I were wrong either.

Wait, I calculated 1/2*1/3 = 1/3.  It does work for n = 6.  And a beautiful computer recursion tells me...

{1, 1/2, 2/3, 1/2, 3/5, 1/2, 4/7, 1/2, 5/9, 1/2, 6/11, 1/2, 7/13, 1/2, 8/15},

which is definitely what I said before.  Which makes sense to me.

The recursion I used to derive it is this:

f[n_] := Max[1/n + (A - 1)*(1 - f[A - 1])/n + (n - A)*(1 - f[n - A])/n, {A, 1, Floor[n/2]}].

I note that this doesn't actually prescribe any particular strategy -- it looks at all possible picks for the first person and takes the biggest.  So what remains is to describe the strategy that gives those values.  In fact, I don't think it matters terribly much which strategy he follows.  At the very least, there's some flexibility.

What was the program you were running, Stephen?

I do think that random question is an interesting one -- my calculation depended on a relatively simple recursion.  I think I can edit it a little bit to do your problem, though -- it'll just be the average of previous values.  I think it's still going to come out to 1/2, or thereabouts.

Okay, so this is my attempt at Simon's problem.  (One person picking randomly among available possibilities, the other picking with strategy and trying to maximize his win percentage.)

Let's look at the case with the random person picking from x numbers.  At each pick, he has 1/x of getting it right.  He has a 2m/x^2 probability of leaving us with m numbers.  (You can do this by looking at what happens as he picks each value.)  Thus, if p(z) is our probability of winning if we're picking with z numbers left, when the random guy picks from x numbers, our probability of winning overall is Sum(2m*p(m))/x^2, m going from 1 to (x - 1).

So I'm running a recursion on it, setting
p(n) = Max(1/n + ((n - A)/n)*Sum(2m*p(m)), m in [1, n - A - 1])/(n - A)^2 + ((A - 1)/n)*Sum(2m*p(m)), m in [1, A - 2])/(A - 1)^2] where the max is taken for A in [1, Floor(n/2)].

So, what my function gives me is this for the first 15 values:
{1, 1/2, 2/3, 1/2, 7/15, 5/9, 4/7, 11/20, 43/81, 27/50, 71/132, 218/405, 1568/2925, 1856/3465, 8669/16200} :approx: {1., 0.5, 0.666667, 0.5, 0.466667, 0.555556, 0.571429, 0.55, 0.530864, 0.54, 0.537879, 0.538272, 0.536068, 0.535642, 0.535123}.

Two things interest me:  you'll probably lose in the case n = 5.  I wonder if this means there's a problem with my algorithm.  Second, that seems to be approaching a weird number.  Let me keep looking at it.  (Mathematica is taking its time calculating p(50).)

It's not 1/sqrt(e). What a shame. But that was just a random wild guess anyway.

It looks like it's approaching a number pretty darned close to 1 - 1/( :pi: - 1), but I don't think that's it.

That would be really cool.

I think it's 1/2.

I will assume that the recursion we are studying is the (possibly simplified) one that JBL last mentioned.  In that case, here is a proof that P(n) stays bounded away from 1/2 for large n.

I will show that P(n) :ge: (5n + 1)/(9n) for every n :ge: 3.  The case n = 3 is easy to check.  Now fix a particular n :ge: 4.  By induction, we can assume that the bound holds for all integers between 3 and n - 2.  We also know that P(1) = 1 and P(2) = 1/2.  In the sum in the recursion, replace P(1) and P(2) by their specific values, and replace all higher terms by the inductive bound.  The resulting sum is an arithmetic series.  When we sum it and simplify, we end up with P(n) :ge: (5n + 1)/(9n), which is exactly what we are trying to prove.  So the bound holds by induction.

We can get better and better lower bounds by assuming that n :ge: 4, n :ge: 5, etc.

[quote="Ian"][quote]I may be mistaken, but it seems to me that in the case of n = 5 it is better to choose three than 1 or 5 (when one is playing against a random player). Sadly, if this is true then the older and more complicated recursion must be used.[/quote]
Can someone verify/disprove this? I generally dont trust myself with calculation errors.[/quote]

Both cases give the strategied player a 3/5 probability of winning.  If he picks 3, his chance of winning is 1/5 + 4/5*(1/2) = 3/5.  If he picks 1, his probability of winning is 1/5 + 4/5*(3/4*(1/2*2/3 + 1/2*(1/3*1 + 2/3*2))) = 1/5 + 3/5*2/3 = 3/5.  If you want me to explain those expressions, I can, but I'm sure that they are correct.

By the way, there is a way to remove the summation in the recursion.  Namely, for every n :ge: 2, we have
[tex] \displaystyle
P(n) = \frac{n-2}{n} P(n - 1) + \frac{2(n-2)}{n(n-1)} P(n - 2) + \frac{1}{n(n-1)}.
[/tex]

That makes it even easier to prove by induction the lower bounds I mentioned in my last post.

Can you show how you got that?  (I haven't really worked with the new recursion, and you couldn't do that with the old one.)

The recursion expresses P(n) as a sum.  By substituting n - 1 for n, we can express P(n - 1) as a sum also.  The two sums are nearly the same; in fact, by splitting off one term, the sums are exactly the same.  In the recursion for P(n - 1), solve for the sum in terms of P(n - 1).  Then plug that solution back into the original recursion for P(n).  We have thus gotten rid of the sum.  Did that make sense?

Anyway, using that non-summation recursion, I believe I now have a reasonably nice closed form for P(n).  Taking the limit of that closed form shows that the limit of P(n) is (1 + e<sup>-2</sup>) / 2.  I will try to supply more details later.

Okay, I figured it was something like that.  Cool.

which is exactly the limiting value James provided with the inverse fit, interesting.

[quote="Ravi B"]Okay, here is the closed form for P(n) that I alluded to.  If n :ge: 1, then
[tex] \displaystyle
P(n) = \frac{1}{2} + \frac{(-2)^n}{n(n+1)!} + \frac{n+2}{2n} \sum_{k = 0}^{n+1} \frac{(-2)^k}{k!}.
[/tex]

It took me many pages of messing around to get that closed form.  But now that it's done, we can easily verify it by induction using the non-summation recursion.

Therefore, the limit of P(n) is (1 + e<sup>-2</sup>) / 2, which is approximately 0.5676676416.[/quote]

I was able to derive the recursive forumla you mentioned above for P(n), but can you explain how you obtained this closed form?

I can give you some ideas of what I did.  But I probably missed a better way.  Even what I am saying below is a cleaned-up version of what I actually did.

First, I set Q(n) = P(n) - 1/2.  That has the virtue of making the (non-summation) recursion homogeneous.  We get the new recursion
[tex] \displaystyle
Q(n) = \frac{n-2}{n} Q(n-1) + \frac{2(n-2)}{n(n-1)} Q(n - 2).
[/tex]

For some reason (probabilistic?), I wanted the coefficients of the two terms on the right to add up to 1, which they don't.  The hardest part was to realize that setting R(n) = nQ(n) / (n + 2) would do the trick.  The recursion becomes
[tex] \displaystyle
R(n) = \frac{(n-2)(n+1)}{(n+2)(n-1)} R(n-1) + \frac{2n}{(n-1)(n+2)} R(n-2).
[/tex]
Note that the two coefficients now add up to 1.

I can now subtract R(n-1) from both sides to get
[tex] \displaystyle
R(n) - R(n - 1) = \frac{2n}{(n-1)(n+2)} (R(n - 2) - R(n - 1)).
[/tex]

Let S(n) = R(n) - R(n - 1).  Then we get the recursion
[tex] \displaystyle
S(n) = \frac{-2n}{(n-1)(n+2)} S(n - 1).
[/tex]

Finally we have reached an equation that can be solved in terms of factorials.  Then we back solve to express R as a sum.  Then we solve for Q, and finally P.

Does anyone know a better way to solve this recurrence?  Of course, after the fact, we get a pretty simple induction.  But was there an easy way to find the closed form (or to even realize there must be a closed form)?

[tex]\displaystyle\frac{1+e^{-2}}{2}[/tex]

This is obviously our answer for the strategy player playing first. (and it's a nice answer, too.) But is it also our answer for the strategy player playing second?

Yes. If we have a sufficiently large game (i.e. one with probability of the strategy player winning as close as we want to [tex]\frac{1+e^{-2}}{2}[/tex]). If the random player goes first, then it will cut off some portion of the board, but with probability 1, it will still be sufficiently large (at least if we choose a REALLY sufficiently large number at the beginning). Then the strategic player will be first again on a sufficiently large set of numbers, so we're back to the same problem again.

Update: After 16 years, two of us have written a paper on the topic, generalizing the problem to trees.  Here is the paper:
[url]https://arxiv.org/abs/2008.11543[/url].  A couple of intriguing conjectures remain (Conjectures 11 and 17).


Let $F_n$ be the fibonacci numbers ($F_1=F_2 = 1$, $F_{n+2} = F_{n+1} + F_n$).

Prove (or disprove) that $F_n  + 1$ is composite for all $n > 3$.

Well, I do not know whether the problem has a solution or not. That's why I put it in the open section.

$F_n+1$ is composite for $n>3$.

Assume that $F_n+1$ is prime for $n>3$;
We use Gelin-Cesaro identity : $F^{4}_n-1=F_{n-2}F_{n-1}F_{n+1}F_{n+2}$.
Thus $F_n+1$ divide $F_{n-2}F_{n-1}F_{n+1}F_{n+2}$.
But $F_{n-1}$ and $F_{n-2}$ are $<F_n+1$ so $F_n+1$ divide $F_{n+1}$ or $F_{n+2}$.
- if $F_n+1$ divide $F_{n+1}$, $F_n+1$ divide $F_{n-1}-1$ then $F_n<F_{n-1}$ wich is impossible,
- if $F_n+1$ divide $F_{n+2}$, $F_n+1$ divide $F_{n-1}-2$ wich is impossible.


So I know that sprint is 30 points, target is 16.  But at national mc, how much is the masters round worth and how is it scored?  Where can i get old masters rounds to practice?

Yes, the masters round is only for the top four from the written round. The masters round is completely separate from written/countdown. it is scored separately by judges.

what is the masters round for??

just to give out awards? xD

Really not sure, but it sure is fun to watch! I learned a lot! I know the winner gets tons of money, though.

And in masters round, you get to write a proof-style solution and present it in front of judges. It's like giving an oral presentation :D

And the public is invited to watch your presentation too. :D

I never happens because the top 4 people at Nationals are obviously really good and they've probably even seen the problem before. :D

Take this year's for example... I bet at least 2 of them have seen the problem before since the problem was so well-known.

Could somebody post this year's masters round?

[quote="tinytim"]One of the questions was something along the lines of:

Prove that you can measure any weight by using only weights that are powers of three (1,3,9,27,.etc.)[/quote]

No, it was more like you had to find the optimum system, that used the least amount of weights. They didn't tell you to use powers of 3. And you also had to prove that that in fact is the system in which the least amount of weights are needed.

System with 1,3,... is optimum, and system 1,2,4,... is optimum [b]if[/b] you are allowed to put weights of 1,2,4,... on [b]only[/b] one side (I saw that somewhere).

[quote="isabella2296"]Just curious - what happens if you're doing the Masters' Round and you have absolutely no idea how to answer the questions?[/quote]

Well, the public gets to hear you verbally wave your arms. '05 featured, iirc, two such.

[quote="math154"]What are you talking about? You don't participate in the Masters' Round unless you're in the Top 4 after the Countdown Round. Otherwise, look at the rules. As for previous tests, just practice proofs.[/quote]

hmm I have some sort of a hunch that it's never the top 4 in cd who actually do it...

After the National Countdown is finished there is a person who goes around asking the top kids if they want to participate in the Master's Round.  You do not have to accept.  They usually keep asking until they have four kids.  The year before Kevin C. won everything, he was asked but declined since he felt he was not prepared for it.  The next year he and his Dad spent a lot of time going over old Mater Round questions to prepare himself for the next year.  It worked.

If your coach has purchased the competition set packets they usually have the Master's Round in the back.  There is no solution, but the question or the topic is included.

Is there an exact judges' rubric as to how the Masters Round is scored?

I don't think so. I'm pretty sure the Master's Round is basically only run by the judges each year, and seeing as they probably change every year (local math professors, etc.), the grading probably changes as well.

[quote="5849206328x"]I can't remember whether it was top 2 written and top 2 countdown not in the top 2 written, or top 2 countdown and top 2 written not in the top 2 countdown (referring to nats)[/quote]
Hmm. I had thought it was the striaght top 4 in written who made masters, then if they did not want to do it, then the 5th place would do it, and then 6th, 7th, 8th, etc. if anybody declined participation

yeah what mr. boyd said is right...it's usually kids in countdown though

though I think Kevin really should have done it the first year he was asked - I personally think it ruins the point if you then practice for it thousands of times, but that's just my view

i think it's the top 2 in cd round and top 2 in written round who get asked first for masters

I'm in 7th grade and in geometry.... I also got 24/25 on the AMC8 contest!!!!!


Find all function $ f : \mathbb{R} \to \mathbb{R}$ such that :

   $ f( xy \minus{} uv ) \equal{} f(xy) \minus{} f(uv) \ \forall \ x;y;u;v \ \in \ \mathbb{R}$

This is the same as $ f(a\minus{}b) \equal{} f(a) \minus{} f(b)$, which is the Cauchy equation; the only solution $ f : \mathbb{Q} \to \mathbb{R}$ is $ f(x) \equal{} kx$. There are *many* extensions to $ \mathbb{R} \to \mathbb{R}$, but the only continuous one is $ f(x) \equal{} kx$  (also the only one not needing the axiom of choice).

[i]Excuse me , The problem should be edited :

  Find all function $ f : \mathbb{R} \to \mathbb{R}$ such that :

   $ f(xy\minus{}uv) \equal{} f(x)f(y) \minus{} f(u)f(v) \ \forall \ x;y;u;v \in \mathbb{R}$[/i] :blush:

$ x\equal{}y\equal{}u\equal{}v$ $ \implies f(0)\equal{}0$ and $ u\equal{}0 \implies f(xy)\equal{}f(x)f(y)$


modular arithmetic
LaTeX
geometry
3D geometry
analytic geometry
trigonometry
graphing lines

[quote="RC-7th"]On three, should the last item in the sequence T_1+T_2+... which we're supposed to solve be T_11?[/quote]

That's what it looks like.  He probably just put T_11 instead of T_{11}.

[quote="RC-7th"]Does any question without an * not require a solution?[/quote]

Yes.

For #5, the answer is pretty big,so can i leave it in another form, like [tex]1889\choose 4[/tex](which i don't think is the answer)?

[quote="Rep123max"]
10.  For how many rational numbers between 1 and 1 will 20! be the resulting product?
[/quote]

There' no number between 1 and 1, do you mean 1 and 0?

Umm...  I'd be really happy if someone could give a tutorial, after the contest, on number six because I'm sure the answer is really simple....and I missed those questions on Getting Started Conest #5...(The only formula I know is a_1/1-r...hmmm)

in #5, does +ve integers include 0 (though i suppose it doesnt)

I'd like to know : are you going to give the solutions after Friday ? I'm new here and it's not easy for me  :?

Is there a good way to print the questions? I prefer to sit and think about a few of these and hate to keep looking at the screen! :-) I noticed that contest #5 had them in PDF format.

Thanks,
Tom

[quote="Rep123max"]Just a clarification, on #15, a is not neccessarily less than b, its just that a and b are less than c.[/quote]

Are a, b and c necessarily positive integers? You forget to wrote ayything about that. Someone who has never heard about pythagorean triplets should think there are infinitely many triplets..

I`m asking this beacuse I`m not sure do you think on all or just on positive integers?

Also, are factors of 70^2 same thing as divisors of 70^2 or sth else?

What do these things mean: lowest terms, numerator and denominator.

Where can I find translations/explanations of terms like those?

gtf, I printed it as pictures in Word, using Print Screen and ACDSee. It`s ok for me.

Since I want more than a 5 on this one...Rep, can you extend the deadline to monday? :-P  I'm kind of busy this week..

[quote="Rep123max"]I fixed it. I meant to put positive integers.

The factors are the same as the divisors

Lowest terms, means it cannot be reduced anymore (i.e. 1/4 is in lowest terms, 3/9 isnt)

Numerator, top of the fraction

Denominator, bottom of the fraction[/quote]

Thanks.

Hey Rep how often do you post contests like this? I think it's time to start frequenting this place because the intermediate stuff is a bit over my head.

[quote="idioteque"]What is leg? Height? Can leg be a cathete (I don`t know is this the right word for what I think - the other, shorter 2 sides of a right triangle)?[/quote]

Yes, in English "leg" is the term for either of the two shorter sides of a right triangle. Usually "height" of a figure is its altitude, that is, the length of a segment perpendicular to one side reaching to the farthest opposite side (or extension of that side) or vertex.

Welcome to the Getting Started forum on the combined AoPS-MathLinks site.

I was wondering if we could discuss the contest now that it's over... :lol:

Since it is over are you going to post the answers?  Or better yet can people put their own answers now?

I ll do a few of the problems:

1.  Just plug in 10 and we get 2310.

2.  70=2*5*7
    70 :^2: =2 :^2: 5 :^2: 7 :^2:
The power of 2 can be 1 or 2, 5 can be one or two and 7 can be one or two.  This makes 2^3=8 choices.

13.  We are basically trying to find \[ 7^{348}+25^{604} \pmod 8 \]

We know that $ 7\equiv-1\bmod 8 $ and that $ 25\equiv1\pmod 8 $.

We can plug these in to get \[ (-1)^{348}+1^{604}=2 \] So 2 is the answer.

I'll share some of the problems that I got, but could someone PLEASE tell me how to do the ones I haven't shown here? Thanks.

3. If you write T_1 ... T_11 on different lines, one beneath the other, a pattern emerges. We have 11 "1's"  2 "10's" ... which becomes 1(11) + 2(10) + 3(9) ... and all the way till 11(1). Adding, we get 286.

4. Ok, here's the proof, but it might be a little confusing because I don't know how to do the summation symbol in a post. To make things easier, let's use * as "summation of." Let's have the number before the * be the number that goes on top of the summation symbol. (I don't know what it's called.) Let the number after be the expression being evaluated.

Ok, the sum of T_1 + ... + T_2n can be written, by the definition of T_n as
2n*n(n+1)/2
=1/2 ( 2n*(n(n+1)) )
=1/2 ( 2n*(n^2 + n) )
=1/2 (2n*n^2 + 2n*n)
=1/2 ( (2n)(2n+1)(4n+1)/6 + (2n)(2n+1)/2)

ok.... then this simplifies to (4n^3 + 6n^2 +2n)/3. AWD.

7. This can be written as a2 + (9)2 = c2 .
a2 + 81 = c2
81 = (c + a)(c-a)

Since a and c must be integers, (c+a) and (c-a) must be factor pairs (their product is 81). There are two pairs that work. "1 and 81" and "3 and 27". Now we have two systems of equations.

c+a=81
c-a=1

Adding,

2c=82
c=41

Similar work for the second factor pair... then we have c=15.
15+41=56.

14. Ok, since 6! to 100! is divisible by 18 (they all have 3 and 6 as factors, and consequently 18) we only need to find 1! + 2!+...+5! mod 18. A little math... voila! the answer is 9.

Ok I'll $\text{\TeX}$ those sigmas up so ppl will be able to under stand it easier.

While in $\text{\TeX}$ mode you can type \sum to get a sigma.  Then to get the index and the thing above it you do \limits_{index}^{thing above}.  You can look at the $\text{\LaTeX}$ section of this website for more information.

What he typed would look like this in $\text{\LaTeX}$
\begin{eqnarray*}
\sum\limits_{n=1}^{2n}\left(\frac{n(n+1)}{2}\right)&=& \frac{1}{2}\sum\limits_{n=1}^{2n}n(n+1)\\
 &=& \frac{1}{2}\sum\limits_{n=1}^{2n}(n^2+n)\\
 &=& \frac{1}{2}\sum\limits_{n=1}^{2n}n^2+\sum\limits_{n=1}^{2n}n\\
 &=& \frac{1}{2}\left(\frac{n(n+1)(2n+1)}{6}+\frac{n(n+1)}{2}\right)\\
 &=&\frac{2n(n+1)(2n+1)}{3}\\
\end{eqnarray*}

[quote]11. A triangle has vertices P=(-8, 5), Q=(-15, -19) and R=(1, -7). The equation of the bisector of angle P can be written in the form ax+2y+c=0. Find a+c. [/quote]

Okay, I know this might be a little above beginner's level, and probably not creative, but I need to brush up on my precalc skills...

Draw out a coordinate plane, and plot roughly each point.

An angle bisector will divide angle P into 2 equal angles.

To find the measure of angle P, first you must find the measure of all of the sides.  Use the distance formula:

[tex]d=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}[/tex]

So, applying:

[tex]d_{PQ}=\sqrt{(-8+15)^2+(5+19)^2}=\sqrt{7^2+24^2}=\sqrt{625}=25[/tex]

[tex]d_{PR}=\sqrt{(-8-1)^2+(5+7)^2}=\sqrt{81+144}=15[/tex]

[tex]d_{QR}=\sqrt{(-15-1)^2+(-19+7)^2}=\sqrt{400}=20[/tex]

Now, we notice that the side lengths are 15,20,25...an arithmetic progression...hmm...just like 3,4,5.  Yep, multiples of the classic pythagorean triple.  You know what that means...right triangle!  You can verify this if you don't believe me using the pythagorean theorem converse:

[tex]15^2+20^2 = 25^2 [/tex]

Now, this means that one of the angles is a right angle.  It's the one between the two legs PR and QR.

Now we pull out some easy right triangle trig to find the measure of angle P:

[tex]\sin{P}=\frac{opp}{hyp}=\frac{20}{25}=\frac{4}{5}[/tex]

So [tex]P=\arcsin{\frac{4}{5}}[/tex].

The bisector will break P up into two equal angles, each of measure [tex]\frac{1}{2}*\arcsin{\frac{4}{5}}[/tex].

Now, there's a formula for the angle between two lines.  Forgot it...okay, google tells me that it is:

[tex]\tan{\theta}=\frac{m_2-m_1}{1+m_1*m_2}[/tex]

So we are interested in finding the slope of this mysterious angle bisector.  We know that [tex]\theta=\frac{1}{2}*\arcsin{\frac{4}{5}}[/tex], and we can find [tex]m_2[/tex], which is the slope of either PQ or QR, we'll use PQ:

[tex]m=\frac{5+19}{-8+15}=\frac{24}{7}[/tex]

Plugging them in to the formula:

[tex]\tan{\frac{1}{2}*\arcsin{\frac{4}{5}}}=\frac{\frac{24}{7}-m_1}{1+m_1*\frac{24}{7}}[/tex]

Now it's all about solving for [tex]m_1[/tex].  Basic cross multiplying and stuff:

[tex]m_1=\frac{\frac{24}{7}-\tan{(\frac{1}{2}*\arcsin{\frac{4}{5}})}}{1+\frac{24}{7}*\tan{\frac{1}{2}*\arcsin{\frac{4}{5}}}}[/tex]

I was really confused about this, but for some reason [tex]\tan{(\frac{1}{2}*\arcsin{\frac{4}{5}})}=\frac{1}{2}[/tex].  I know that [tex]\tan{\arctan{x}}=x[/tex], but I am not sure how to apply it w/the 1/2...  Anyway pretend I didn't use a TI there... Someone please enlighten me on this one...

So, [tex]m_1=\frac{\frac{24}{7}-\frac{1}{2}}{1+\frac{24}{7}*\frac{1}{2}}=\frac{55}{38}[/tex]

So, it's a line with slope 55/38  passing through the point P(-8,5).

[tex]y=mx+b[/tex]
[tex]y=\frac{55}{38}*x+b[/tex]
[tex]5=\frac{55}{38}*(-8)+b[/tex]
[tex]b=\frac{315}{19}[/tex]
[tex]y=\frac{55}{38}*x+\frac{315}{19}[/tex]

Multiplying out w/a common denominator, multiply everything by 38:

38y=55x+315*2

You want it in the form: ax+2y+c=0

Now to get the 2y term, divide everything by 19:

[tex]2y=\frac{55}{19}*x+\frac{630}{19}[/tex]

Into the form we want:

[tex]\frac{-55}{19}*x+2y+\frac{-630}{19}=0[/tex]

[tex]a+c=\frac{-55}{19}+\frac{-630}{19}=\frac{-685}{19}[/tex]

WOW, def. wrong (non-nice answer)...I know it can be done by this method but I must have made a mistake with my arithmetic...

[quote="JS1527"]I was really confused about this, but for some reason [tex]\tan{(\frac{1}{2}*\arcsin{\frac{4}{5}})}=\frac{1}{2}[/tex].  I know that [tex]\tan{\arctan{x}}=x[/tex], but I am not sure how to apply it w/the 1/2...  Anyway pretend I didn't use a TI there... Someone please enlighten me on this one...[/quote]

Here's how you do it:

We use the identity [tex]\tan \frac{\theta}{2} = \frac{1 - \cos{\theta}}{\sin{\theta}}[/tex]. Since we have [tex]\sin{\theta} = \frac{opp}{hyp} = \frac{4}{5}[/tex], we can find that [tex]\cos{\theta} = \frac{adj}{hyp} = \frac{\sqrt{(hyp)^2 - (opp)^2}}{hyp} = \frac{3}{5}[/tex]. Thus [tex]\tan{\frac{\theta}{2}} = \frac{1 - \frac{3}{5}}{\frac{4}{5}} = \frac{\frac{2}{5}}{\frac{4}{5}} = \frac{1}{2}[/tex].

Wow, thanks JSRosen.  Did I get the problem right by any chance?

[quote="JSRosen3"]Not sure...I was too lazy to do the arithmetic ;)

But in general, a non-nice answer doesn't necessarily mean wrong![/quote]

I sure know abotu that.  :maybe:

Um, I think I found an error in JS1257's post. An arithmatic one in solving for m1.

m1=(24/7-1/2)/(1+24/7*1/2)
=[(24(2)-7)/14]/[1+12/7]
=(41/14)/(19/7)
=41/38.

So,  it's a line with slope 41/38 passing through point P(-8,5).
Let's use the point slope formula.

y-5 = 41/38 (x+8)
38(y-5) = 41(x+8)
38y-190=41x+328
41x-38y+518=0

We want it in the form of ax + 2y + c = 0, so we divide the equation by 19.

41/19x-2y+518/19=0

a+c = 41/19 + 518/19 = 559/19

(which is still not a very pretty number anyway)!


How do you put a latex file on your posts? What type file do you attach?

There is a thread at the top of this forum called "LaTeX Help -> How to write in LaTeX and how to use LaTeX".  Please read it -- your answers are in there.


Let ABC be a triangle with circumradius R.
Consider the three lines touching the incircle, parallel to one of the sides.
This way you obtain three small triangles at the corners of the given triangle.
Let Ra, Rb, Rc be the radiuses of the small triangles.
Show that Ra + Rb + Rc = R.

Taken from a flemish math magazine.

I'm not going to justify all of this: it's either well-known or obvious:

1-Ra/R+1-Rb/R+1-Rc/R = 2r/ha+2r/hb+2r/hc = 2, so (Ra+Rb+Rc)/R=1.


Calculate sum for 20 terms

S=(1+2)/(3*4)-(4+5^2)/(9*11)+(7+8^3)/(15*17)-...

where
          11=4+7;
          17=11+6;
and so on.

Here is my program.It is OK for n<7, but if we calculate the sum for more than 7 terms it doesn't give the right output.How is it possible??

Thanks in advance.

PS:Sorry for LaTex.

It's an overflow problem. Try using "long long" instead of "int" or "long int" (the last two mean the same in current C compilers).


geometry
circumcircle
perpendicular bisector
projective geometry
geometry proposed

Triangle ABC has incircle w centered as S that touches the sides BC,CA and AB at P,Q and R respectively. AB isn't equal AC, the lines QR and BC intersects at point M, the circle that passes through points B and C touches the circle w at point N, circumcircle of triangle MNP intersects with line AP at L (L isn't equal to P). Then prove that S,L and M lie on the same line

Perpendicular bisector of $\overline{BC}$ cuts the circle $ \odot(BNC)$ at $ E$ lying on the same side of $ A$ with respect to $ BC.$ Since $ NE$ and $ NP$ are the external and internal bisector of $ \angle BNC,$ it follows that $ EN$ cuts $ BC$ at the harmonic conjugate of $ P$ with respect to $ BC,$ in other words, $ EN$ goes through $ M$ $ \Longrightarrow$  $ \angle MNP \equal{} 90^{\circ}.$  Then $ \angle MLP \equal{} \angle MNP \equal{} 90^{\circ}.$ Since $ LP$ is the polar of $ M$ with respect to the incircle $ (S),$ then $ PL$ goes through $ A.$

Let the circle that touches $k$ and passes thru $BC$ be $\omega$
The homothety centered at $N$ taking  $k$ to $\omega$.It takes $P$ to the midpoint of arc $BC$ thus $NP$ is the angle bisector of $\angle CNB$.$AP,BQ,CR$ are concurrent thus $(B,C;P,M)$ $\implies$ $\odot MNP$ is the Apollonius circle of $\triangle CNB$$\implies$ $\angle MNP=\frac{\pi}{2}$.All we need to prove now is :
$$AP\perp MS$$
$M$ is on the polar of $A$ wrt $k$ and $P$ wrt $k$ thus $AP$ is the polar of $M$ wrt $k$ so $MS\perp AP$ and we are done.


probability
algorithm
search
probability and stats

Here is the problem that I have:

Let ni be a number that has a value between 1 and nmax.

We want to find this value. For this we use repetitive steps.

Each step consists of verifying all plausible values of ni. There is a 50% chance for each value to be proven unplausible and it will be eliminated. And there is a 50% chance that the value be plausible.

If there are 2 or more plausible values for ni, another step is needed, that will test in a similar manner the remaining plausible values.

The formulas I need:
the probablility of finding the value of ni after just one step.
the probablility of finding the value of ni after X steps.
If the number of steps is limited to Y, what value must Y have in order to have 95% probability of finding ni after Y or less steps?

I am looking forward for your answers. Thank you.

first of all.. after each time running your algorithm.. you might have zero possible values.. then what? is there something to prevent this.. or do we have to consider this case too?

ni is something real, that has a value. I am using this algorithm to find that value.

So, if I understand corectlly your question, it is not possible to have zero values. After the last step of my algorithm I will know the value of ni. Did I answer your question?

Thank you for replying!

You haven't described enough to call it an algorithm. What do you mean by "plausible"; what are you testing?
Among other things, you haven't said anything about how your desired value relates to plausibility. What if its value is declared implausible along with the rest?

This problem comes from a real one. It is not one that I have invented. It would be too much to write if I were to explain the whole situation, so I tried to put it as a problem of probabilistics.

The basic thing is that I try to find the value of ni. It is an integer between 1 and nmax.

At the first step, I try all the values between 1 and nmax. For each value I try, there is a 50% chance that I will find out for sure that the value is not the value that I am looking for. This is what I called unplausible. And there is a 50% chance that that value COULD BE the value that I am looking for. That is what I called plausible (sorry if my English or knoledge of probabilistic terminology betraied me).

For the next steps I will try only the values that have remained plausible. The process I use is similar to the one explained above: I verify the plausibility of each value that has not been eliminated at a previous step.

I will have finished my search when only one plausible value remains and that value will be the value for ni. I know from the beginning that ni has a value. I just do not know what value it has.

I hope I have managed to make myself clear. Sorry if my English is not clear enough and thank you for your interest.

What you've said doesn't make sense.  Is this what you mean?

We have a set of size $ N$, with a special element we are trying to identify.  We have a test for this element with the following properties:
1) If we test the special element, the test correctly tells us that it is the special element. [This was missing from your explanation, and is one of the things jmerry asked about.  I'm guessing that it is what you meant.]
2) If we test something other than the special element, the test correctly tells us that it is not special 50% of the time and incorrectly tells us that it is special 50% of the time.  Moreover, these answers are independent.
In other words, our test has a 50% false positive rate but no false negatives.

Assuming this is correct:
The probability of finding the special value after one step is of course just $ \frac{1}{2^{N \minus{} 1}}$, since you must hit correctly on all $ N \minus{} 1$ of the non-special elements.  For more than one iteration, I vote you use a recursive method: let $ p(n, q)$ be the probability of winning starting with $ n$ numbers in $ q$ steps.  Then $ p(n \plus{} 1, q \plus{} 1) \equal{} \sum_{i \equal{} 0}^n \binom{n}{i} \frac{1}{2^i} \cdot p(n \plus{} 1 \minus{} i, q)$, with initial values $ p(1, q) \equal{} 1$ for all $ q \geq 0$ and $ p(n, 0) \equal{} 0$ for all $ n \geq 2$.  Now see if you can solve this recursion.

JBL, thank you very much for your answer.

I cannot solve this recursion but I can make a small program that would calculate all I need. If you could help me with the solution to the recursion would be great. Anyway, now I am in a totally new situation and your answer really helped me A LOT.

Could I complicate it a little more now? The real situation I am facing is that I have a string of special values. I had noted the special value ni because i is an index which ranges from 1 to L. So, if I know L, n and q, what is the probability of knowing all special values?

Let me put it in a diffrent way so I can be clearer. Lets say you have helped me with calculating the probability of hitting a certain value(let's say 1) after rolling a dice just one time($ \frac{1}{2^{N\minus{}1}}$). You have told me it is 1/6. You have also told me the probability of having hit the value 1 at least one time after having rolled the dice q times (by solving the recursive problem). Now I complicate the problem by multiplying the dices: I do not have just one dice, I have L. I start by rolling all the dices. The ones that have hit 1 will not be rolled again. (in my case this means that I have found my special value). I keep rolling the dices that did not hit 1 until they do. What is the probability that I run out of dices before I do the "dice rolling step" [b]q[/b] times?

JBL, I have a small problem. I have tried to calculate p(2,2). In this case, n=1, q=1.

What I obtain is: $ p(n \plus{} 1, q \plus{} 1) \equal{} p(2,2) \equal{} \sum_{i \equal{} 0}^{1}\binom{1}{i}\frac {1}{2^{i}}\cdot p(2 \minus{} i, 1)$ = 1

This means that the event is sure, and this is clearly false. I believe the correct value is 3/4.


Let A, B, and C represent different digits greater than zero.  Determine the minimum value of the expression:
$ \frac {ABC}{A \plus{} B \plus{} C}$
Note:A is the hundreds digit, B is the tens digit, and C is the ones digit

$ \frac{100 a\plus{}10 b\plus{}c}{a\plus{}b\plus{}c}$ written in mathematical form

just set $ a,b\equal{}0$ and the solution will be $ 1$.

zero solution is impossible, and negative ones too.

for values below 1, the denominator need to be greater than the numerator, but this is impossible.

it said a b and c are greater than 0. $ 0>0$ clearly

don't be mad, my fancy guy.

with the same argument

$ a\equal{}1$ this is logical

$ \frac{100 a\plus{}10 b\plus{}c}{a\plus{}b\plus{}c}\equal{}\frac{9 (b\plus{}11)}{b\plus{}c\plus{}1}\plus{}1$

$ c\equal{}9$ another thing logical

$ \frac{100 a\plus{}10 b\plus{}c}{a\plus{}b\plus{}c}\equal{}10\plus{}\frac{9}{b\plus{}10}$

$ b\equal{}9$ yes, we are donne

Can this question solved by algebraric equations?

considering that theres only one equation and hteres 3 inequlities, i highly doubt it

so the answer is 199/19??

i dont quite get lordykelvin's logic lecture about what would happen if a=1 and b=9.... i get it, kind of, but not quite

[quote="spiders_hate_chestnut"]so the answer is 199/19??[/quote]

That's right. :)

[quote]i dont quite get lordykelvin's logic lecture about what would happen if a=1 and b=9.... i get it, kind of, but not quite[/quote]

screw that
i get it now....