[hide="easy"] there are 1024 contestants in a tennis tournament single elimination. how many games must be played to determine the winner?[/hide]

[hide="same thing with twist"] still 1024 contestants. what is the minimum games needed to play to decide the winner. [/hide]

[hide="small change"]what if there were 209570 contestants.(same way as the 1st one)[/hide]

[hide="1 and 3"]You just subtract one from each of the original contestents because all but one of them needs to lose.
1:$ 1023$
3:$ 209569$[/hide]

couldnt you solve this the same way you solved 1 and 3

[quote="hunter34"]Yeah whoops there should only be 1 $ 2^{0}$ so it is 1024-1.[/quote]

i knowq but its double elimination

[quote="Prime factorization"][quote]
$ \ 2^{9}+2^{8}+2^{7}+2^{6}+2^{5}+2^{4}+2^{3}+2^{2}+2^{1}+2^{0}+2^{0}\&\equiv\&2^{10}$[/quote]

I'm pretty sure you could just say

$ \sum_{k=0}^{9}2^{k}= 2^{(9+1)}-1$[/quote]

Because summation isn't MATHCOUNTS level :P

Other wise I would have.

i don't understand how the answer to part one can be
[hide]1023[/hide]
i mean, it asks for the number of games that must be played to determinet the winner, not the number of contestants that must be eliminated?..

In each game, one contestant must be eliminated.

@abacadaea

That is the number of contestants-2. What the problem is asking is the number of games.

[quote="erdogankerem123"]no, that means the same thing, only using a summation sign to phrase it.[/quote] no its not the same thing. im realy confused can someone just post a solution

OK guys, I think all of these questions should be really easy.
[hide]As for the confusion about #1 and #2, II would just solve the problems by counting the nubmer of problems in round 1, 2, 3,....., and 10.  Obviously, the total number of games would be 512+256+128+64+32+16+8+4+2+1=1023.   :D [/hide]
@abacadea: Where's the "twist."  I see the slightly reworded but technically same question as number 1 used for number two.

In a single elimination tournament with n people (n doesnt have to be a power of 2) the number of games played is always n-1 because like some of you hvae already alluded too, each game there must be 1 person eliminated and we want n-1 pepole eliminated to have 1 champion.

[quote="abacadaea"]GEEZ! what is so hard about this!!!! the first one is single elimination. i know how to do that. the second one is DOUBLE elimination....now everyones confused. :?:[/quote]

well maybe you shouldve mentioned that when you typed the question in the first place.  Although you did say that in a post in the middle of the thread, usually poeple only read the first post and like the last post because they want to either respond to the last post or look at the first post which is the question. and if you didnt include that in your question, thats why everyone is getting it wrong.  you jsut have in your origianl question, same thing with a twist and the twist is "minimum" not double elimination in which you intended, because you never mentioned doulbe elimination.

oh woops i didnt realize that. sorry. :blush:


How many integers can be represented as a difference of two
distinct members of the set {1, 2, 3}?

[quote="isabella2296"]There are 3 numbers, so we have $ 3 \times 2 \equal{} 6$.[/quote]
But you're overcounting some differences such as: $(2,1)$ and $(3,2)$ which both have a difference of $1$.

We can see that there are $\boxed{2}$ unique differences which are: $1$, and $2$ which can be made by: $(2,1)$ and $(3,1)$ respectively.

But FTW said that the correct answer was $4$ which I believe to be incorrect because I see no way how $4$ differences can be made with those three numbers.

FTW is correct. JSGandora, you missed 1 - 2 = -1 and 1 - 3 = -2.

First we can remember $a-b$ = $-(b-a)$
so for each positive difference $d$, there is another difference that is it's opposite.
Knowing this, we can just count the number of positive differences, then multiply that by 2.
so there are obviously just 2 positive differences: $1, 2$
so $2\cdot2=4$
4 is the answer.


geometry
trigonometry
inequalities
cyclic quadrilateral
geometry unsolved

Let $ A_1A_2A_3A_4A_5A_6A_7, B_1B_2B_3B_4B_5B_6B_7, C_1C_2C_3C_4C_5C_6C_7$ be regular heptagons with areas $ S_A, S_B$ and $ S_C$, respectively. Let $ A_1A_2 \equal{} B_1B_3 \equal{} C_1C_4$. Prove that

$ \frac{1}{2}  < \frac{S_B \plus{} S_C}{S_A} < 2 \minus{} \sqrt{2}$

$ a_2 = A_1A_2 < a_3 = A_1A_3 < a_4 = A_1A_4$
$ b_2 = B_1B_2 < b_3 = B_1B_3 < b_4 = B_1B_4$
$ c_2 = C_1C_2 < c_3 = C_1C_3 < c_4 = C_1C_4$
$ b_3 = a_2\ \Longrightarrow\ b_2 = \frac {b_2b_3}{b_3} = \frac {a_2^2}{a_3},\ \ \ c_4 = a_2\ \Longrightarrow\ c_2 = \frac {c_2c_4}{c_4} = \frac {a_2^2}{a_4}$

$ q(7) = \frac {S_B + S_C}{S_A} = \frac {b_2^2 + c_2^2}{a_2^2} = \frac {a_2^2}{a_3^2} + \frac {a_2^2}{a_4^2}$
for regular heptagons, similarly $ q(6)$ for regular hexagons. Sides/areas of hexagons $ (A_i)_1^6 \longrightarrow (B_i)_1^6 \longrightarrow (C_i)_1^6$ shrink slower then for heptagons $ (A_i)_1^7 \longrightarrow (B_i)_1^7 \longrightarrow (C_i)_1^7$ $ \Longrightarrow$
$ q(7) < q(6) = \frac {1}{3} + \frac {1}{4} = 2 - \frac {17}{12} = 2 - \sqrt {2} \sqrt {\frac {289}{288}} < 2 - \sqrt {2}$

$ a_2a_3 + a_2a_4 = a_3a_4$ by Ptolemy theorem for the cyclic quadrilateral $ A_1A_2A_4A_7 \Longrightarrow$ $ \frac {a_2}{a_3} + \frac {a_2}{a_4} = 1 \Longrightarrow$ $ q(7) = 1 - \frac {2a_2^2}{a_3a_4}.$ By the sine theorem for the $ \triangle A_1A_3A_4$ with $ a_2 = A_3A_4$ and easily calulated angles $ \angle A_1A_3A_4 = \frac {4\pi}{7},\ \angle A_3A_4A_1 = \frac {2\pi}{7},\ \angle A_4A_1A_3 = \frac {\pi}{7},$
$ q(7) = 1 - \frac {2\sin^2 \frac {\pi}{7}}{\sin \frac {2\pi}{7} \sin \frac {4\pi}{7}} = 1 - \frac {1 - \cos \frac {2\pi}{7}}{2\left(1 - \cos^2 \frac {2\pi}{7}\right) \cos \frac {2\pi}{7}} = 1 - \frac {1}{2(x_0 + 1)x_0}$
where $ x_0 = \cos \frac {2\pi}{7}$ is the only positive root of $ Q(x) = 8x^3 + 4x^2 - 4x - 1 = 0$ (the other roots are $ \cos \frac {4\pi}{7},\ \cos \frac {6\pi}{7},$ both negative). Q(x) is obtained similarly as in [url=http://www.mathlinks.ro/viewtopic.php?t=173185]Regular heptagon[/url], by substitution ${ \frac {(2k + 1)\pi}{7}\longrightarrow \frac {2l\pi}{7}} = \pi - \frac {(2k + 1)\pi}{7},\ x \longrightarrow - x,$ where $ k = 0, 1, 2$ and $ l = 3 - k = 1, 2, 3.$ We have to show $ (x_0 + 1)x_0 > 1.$

[color=blue][b](1)[/b][/color] $ P(x) = (x + 1)x - 1 = 0$ has roots $ x_{12} = - G,\ 1/G$ where G is the golden ratio. Using $ G^2 = G + 1$ repeatedly,
$ Q(x_2) = Q(1/G) = \frac {3 - 2G}{G^3} < 0 \Longrightarrow$ $ 1/G = x_2 < x_0 = \cos \frac {2\pi}{7} \Longrightarrow$ $ P(x_0) > 0,\ (x_0 + 1)x_0 > 1.$

[color=blue][b](2)[/b][/color] $ Q(x) = 4(2x - 1)(x + 1)x - 1,$ $ Q(5/8) = 1/64 > 0 \Longrightarrow x_0 < 5/8,\ 4(2x_0 - 1) < 1 \Longrightarrow$ $ (x_0 + 1)x_0 = \frac {1}{4(2x_0 - 1)} > 1.$

As a result, $ q(7) = 1 - \frac {1}{2(x_0 + 1)x_0} > \frac {1}{2}.$ $ 1/G < x_0 < 5/8$ allows to improve the right inequality to $ \frac{1}{2} < q(7) < \frac{1}{2} + \frac{1}{130}.$


Find number of letter of $ 3^{100}$[size=150] ?[/size]

$ \log3^{100} \plus{}1\equal{}$ number of digits of that number it equals to $ 100* \log3 \plus{}1$ and $ \log3\equal{}0.47...$ so the answer is 48.

[quote="queen"]$ \log3^{100} \plus{} 1 \equal{}$ number of digits of that number it equals to $ 100* \log3 \plus{} 1$ and $ \log3 \equal{} 0.47...$ so the answer is 48.[/quote]

Just a minor correction: it's $ 1\plus{}\lfloor\log3^{100}\rfloor\equal{}1\plus{}\lfloor100\log_3\rfloor$.


I am trying to run Imagemagick on a .gif file, specifically to make it 1/4 as large.  However, when I run the program I create to perform the task, it simply loops apparently infinitely C:\[path]\[filename]>[.bat program instructions] (see screenshot).  I have no clue what it is doing, although it is clear that it is not what it is supposed to do.  The original .bat program that I was using to convert it was
[code]convert tesseract.gif -resize 25% tesseract2.gif
pause[/code]
Do you have any idea of what is going on, or alternatively, is there another way to resize .gif files that does not lose the animation?

You haven't called your .bat program convert.bat have you? Because that would cause a loop as convert.bat would then run convert.bat which would run convert.bat .... whereas it should call convert.exe. So either rename the batch file and/or use .exe and .bat extensions when running the programs.

That was the problem.

I am working with a 625kb .gif file, and when I try to resize it or separate it into the component images, it seems to lose some of the information.  Is there any way to work around this?  Also, is there some way to delete every other component image?

[quote="roadnottaken"]I am working with a 625kb .gif file, and when I try to resize it or separate it into the component images, it seems to lose some of the information.  Is there any way to work around this?[/quote]Try coalesce - see [url=http://wiki.flux-cms.org/display/BLOG/Resizing+animated+GIFs+with+ImageMagick]Resizing animated GIFs with ImageMagick[/url]
[quote="roadnottaken"]Also, is there some way to delete every other component image?[/quote][url=http://www.fz-juelich.de/vislab/software/www/FAQ.html#C32]How do I extract a single image from a multi-image file?[/url]


In a $\triangle{ABC}$, $I$ is its incenter, $B$ its orthocenter and $O$ its circumcenter. If $\angle{A}-\angle{C}=35^\circ$, find $\angle{IOB}$.

Hint: $ABC$ is a right triangle. Just finish it with some angle-chasing.

Does anyone have a full solution? I need it... thanks anyway!

Actually the answer is $15.6^\circ$, hence...

Best regards,
sunken rock


The following people are looking at one animal enclosure.  Each makes a statement, and each is correct regarding his/her
statement.
-------------------------------------------------------------

Chris:  I see a panther.

Dana:  I see a cougar.

Nora:  I see a catamount.

Pat:  I see a puma.

Ray:  I see a mountain lion.

Carrie:  I see a mountain cat.

Sam:  I see a painter.

-------------------------------------------------------------
What COULD be the fewest number of animals that these people
are looking at in the one animal enclosure?

[hide="Solution"] Two.  The cougar and the painter.  (Presumably, cougars don't paint.) [/hide]

Incorrect.

A good go at it.

But look up all meanings of the word you discarded in your hidden solution, such as in the free online dictionaries, for example.

[hide] Is the answer 3? The panther, the cougar, the catamount, the puma, and the mountain lion are alike and as one, the mountain cat is a smaller version of a leopard, and so is not one of the above groups, and a painter, being a human? [/hide]

[hide="Well, of course then..."] [i]pain·ter 3   (pān'tər)  Pronunciation Key
n.   Chiefly Upper Southern U.S.
See mountain lion.

[Alteration of panther.] [/i]

Quite curious.  Anyway, since this is a riddle, the obvious answer is $\boxed{1}$.  "Mountain cat," turboturtle, is an acceptable name for a mountain lion. [/hide]

The poster who had an alternate definition of "painter" with his follow-up post is correct with his number of animals.


Euler
integration
calculus
function
calculus computations

Let ${ a_n = \sum_{i = 1}^{n}{\sqrt {i}} - \frac {2}{3}n\sqrt {n}} (n\in \mathbb{N} ^ + )$
Prove that:
$ a_n \to + \infty$ when $ n \to + \infty$

Euler-Maklarena Formula (I'm not sure about Maklarena, Maybe in english it's not correct  :maybe: ) :
$ \sum_{0}^{n}f(t) \minus{} \int_{0}^{n}f(t)dt \equal{} \frac {f(0) \plus{} f(n)}{2} \plus{} \int_{0}^{n}f'(t)(\{t\} \minus{} \frac {1}{2})dt$
hence we see that :
$ \sum_{1}^{n}\sqrt {i} \minus{} \frac {3}{2}n\sqrt {n} \equal{} const \plus{} \frac {\sqrt {n}}{2} \plus{} \frac {1}{2} \int_{1}^{n} \frac {1}{\sqrt {x}}(\{x\} \minus{} \frac {1}{2})\approx B \plus{} C\int_{1}^{n}\frac {\{x\}dx}{\sqrt {x}}$
I use that $ \frac {\sqrt {n}}{2} \equal{} A \plus{} \frac {1}{4}\int_{1}^{n}\frac {1}{\sqrt {x}}dx$
and $ A,B > 0$ and $ C$ is some constant.
Last integral $ \int_{1}^{n}\frac {\{x\}dx}{\sqrt {x}}$ diverge.
Or I said that : Euler Maklarena formula can be rewrited as
$ \sum_{0}^{n}f(t) \minus{} \int_{0}^{n}f(t)dt \equal{} f(0) \plus{} \int_{0}^{n}f'(t)\{t\}dt$

His name is Colin Maclaurin  :lol:
I 'll show more simple proof, but using the same idea.
Because $ (\sqrt{x})''<0$, then $ \sqrt{x}$ - concave function, so $ \sqrt{x-y}+\sqrt{x+y}\le 2\sqrt{x}$.Then:
\[ \int_{n-1}^{n} \sqrt{x}\, dx = \int_{n-1}^{n-\frac{1}{2}} \sqrt{x}\,dx+\int_{n-\frac{1}{2}}^{n} \sqrt{x}\,dx=
\int_{0}^{\frac{1}{2}}\sqrt{n-\frac{1}{2}-t}\,dt+\int_{0}^{\frac{1}{2}}\sqrt{n-\frac{1}{2}+t}\,dt =
\int_{0}^{\frac{1}{2}}\left(\sqrt{n-\frac{1}{2}-t}+\sqrt{n-\frac{1}{2}+t}\right)dt \le 2\int_{0}^{\frac{1}{2}}\sqrt{n-\frac{1}{2}}=\\
=\sqrt{n-\frac{1}{2}}\]
So,
\[ \sqrt{n}-\int_{n-1}^{n} \sqrt{x}\, dx \ge \sqrt{n} - \sqrt{n-\frac{1}{2}} \ge \frac{n-(n-\frac{1}{2})}{\sqrt{n} + \sqrt{n-\frac{1}{2}}}=
\frac{1}{2\left(\sqrt{n} + \sqrt{n-\frac{1}{2}}\right)} \ge\\
\ge \frac{1}{4\sqrt{n}}\]
and finally
\[{ a_n = \sum_{k = 1}^{n}{\sqrt {k}} - \frac {2}{3}n\sqrt {n}}=\sum_{k = 1}^{n}{\sqrt {k}}-\int_{0}^{n} \sqrt{x}\, dx=
\sum_{k = 1}^{n}\left(\sqrt {k}-\int_{k-1}^{k} \sqrt{x}\, dx\right) \ge\\
\ge \sum_{k = 1}^{n}\frac{1}{4\sqrt{k}} \ge \frac{\sqrt{k}}{4}\to \infty\]

[quote="greenvert"]His name is Colin Maclaurin  :lol:
[/quote]
 :blush:
.
Here is generalization :
prove:
if $ \alpha \in (0,1]$ then,
$ \sum_{1\leq k \leq n}k^{\alpha} \equal{} \frac {n^{\alpha \plus{} 1}}{\alpha \plus{} 1} \plus{} \frac {n^{\alpha}}{2} \plus{} C_{\alpha} \plus{} o(1)$
Check:
if $ \alpha\equal{}\frac{1}{2}$ then $ C_{\frac{1}{2}}$ is irrational  :lol:.


If h=a+3b^2+5c^3, and if a is multiplied by 64, b is multiplied by 8 and c is multiplied by 4, how will the value of h be changed.

i put in numbers for a, b, and c but couldn't figure out how the value of h is changed...any ideas?
pls and thanks =]

[hide]Basically, what it's saying is to plug in $64a$ for $a$, $8b$ for $b$, and $4c$ for $c$, and see how it affects the value.

Like this:

$h_{1}= a+3b^{2}+5c^{2}$
$h_{2}= (64a)+3(8b)^{2}+5(4c)^{3}$
$h_{2}= 64a+64*3b^{2}+64*5c^{3}= 64(a+3b^{2}+5c^{2}) = 64h_{1}$.

So that will multiply the value of $h$ by 64.[/hide]

(I would say that this falls more under basics than intermediate.)

Double post in [url]http://www.mathlinks.ro/Forum/viewtopic.php?t=132140[/url]
I'm going to delete my post there since someone already answered here and the other one should be locked.


As figure shown, $ O_1,O_2$and $ O_3$ three circles touch pairwise to each other. $ \overline {AB}$ is the common tangent to $ O_1,O_2$.
$ \widehat{AB}$ is a semi-circle,and touches each circle at different point. If circle $ O_1$ and circle $ O_2$ both have the radii with length 1 unit, then how about the radius of circle $ O_3$ ?

[color=red]It is edited[/color]

Do you mean the radius of circle $ O_{3}$?  You told us that $ O_{1}$ has radius 1 and then you asked us to find the radius of that circle.

Let $ AB \equal{} 2R$
$ r$ = the radii of the small circle

Look at my attachment,
$ R \equal{} OC \equal{} OD$
$ OC \equal{} OD$
$ 1 \plus{} \sqrt{2} \equal{} r \plus{} 1 \plus{} \sqrt{(r\plus{}1)^2 \minus{} 1^2}$
$ \sqrt{2} \equal{} r \plus{} \sqrt{r^2 \plus{} 2r}$
$ 2 \equal{} r^2 \plus{} r^2 \plus{} 2r \plus{} 2r\sqrt{r^2 \plus{} 2r}$
$ 1 \equal{} r^2 \plus{} r \plus{} r \sqrt{r^2 \plus{} 2r}$

$ r$ can be found by solving the equation

NB : maybe there is a simpler solution.
CMIIW

I did it the same way, except you want to subtract r before squaring.
$ (\sqrt{2}\minus{}r)^2\equal{}r^2\plus{}2r$
$ r\equal{}\frac{1}{\sqrt{2}\plus{}1}\equal{}\sqrt{2}\minus{}1$

You could also easily show that the two triangles are congruent by RHS. Then, through corresponding sides,

$ r \plus{} 1 \equal{} \sqrt{2}$
$ r \equal{} \sqrt{2} \minus{} 1$

[quote="kurt.math"]You could also easily show that the two triangles are congruent by RHS. Then, through corresponding sides,

$ r \plus{} 1 \equal{} \sqrt {2}$
$ r \equal{} \sqrt {2} \minus{} 1$[/quote]

Well, so elegant your works!!
 :lol:  :D
 :thumbup:


There are five chairs on stage, arranged in a straight line parallel to the front of the stage. Big Kahuna, Major Domo, Em See, Lou Tenant and Captain Marv El will occupy the five chairs, but Major Domo cannot sit next to Big Kahuna. How many seating arrangements are possible?

We count the number of ways that they can be seated with the given condition. If one of them sits on the far left end, then there are $ 3$ possible places for the other person. If one of them sits in the second seat from the left, there are $ 2$ possible places for the other. Finally, if one sits in the middle, then the other has $ 1$ possible place it may go. There are $ 2$ of them and there are $ 3!$ ways to arrange the other people.

$ (6)(2)(3!)\equal{}\boxed{72}$

Alternatively, we could have used complementary counting.

Multiplication sign can be done in two ways:

[code]6\cdot3[/code]

Will net $6\cdot3$

[code]6\times3[/code]

Will net $6\times3$

[quote=HYP135peppers]We can accomplish this question with some complementary counting.

There are 120 ways ($5!$) to arrange the five people. There are 4 ways to seat Major Domo and Big Kahuna together, and the others can be seated $3!$ ways, which makes $4*3*2$ ways to seat Major Domo and Big Kahuna together.

In conclusion, there are $120-24=72$ ways to arrange the five people with the restriction stated in the question.

P.S. Can someone tell me how to show the multiplication sign with $LaTeX$?[/quote]

120-24 = 72 lol, you forgot to multiply your 4*3*2 by 2 because there are two ways that major domo and big kahuna can be seated together. 120-48 = 72


vector
linear algebra
matrix
quadratics
algebra
polynomial
function

Hello There !

Given f(x) where x is a vector [ x1, x2, x3...xn]  from here,  it is easy to calculate the gradient of f(x) and the Hessian of f(x).

Now supose that I perform a change of variables such that x=Sy, where S is an nxn inversible matrix.

How can I express the gradient of f(x) and the Hessian of f(x) in terms of the y variable ?

Let $g(y)=f(Sy)$. The [url=http://www.math.hmc.edu/calculus/tutorials/multichainrule/]multivariable chain rule[/url] gives $\nabla g(y)=(\nabla f(Sy)) S$, where the gradients are understood as row vectors. Matrix $S$ multiplies a row vector from the right and produces another row vector. Let's check if this makes sense:
$\frac{\partial g}{\partial y_{i}}=\sum_{k=1}^{n}\frac{\partial f}{\partial x_{k}}\frac{\partial x_{k}}{\partial y_{i}}=\sum_{k=1}^{n}\frac{\partial f}{\partial x_{k}}s_{ki}$. OK.

Now differentiate with respect to $y_{j}$, using the chain rule again:
$\frac{\partial^{2}g}{\partial y_{j}\partial y_{i}}=\sum_{k=1}^{n}\sum_{l=1}^{n}\frac{\partial^{2}f}{\partial x_{l}\partial x_{k}}s_{lj}s_{ki}$. There ought to be a cleaner way to write this. :maybe:

Thanks for the answer. Lets think in the following way:

For convention lets call grad( f(x)) the gradient of f(x)  and F(x) the Hessian of f(x) .

if x=Sy then by the multivariable rule: grad(f(Sy)) =  transpose[grad(f(x))].S
Well, for the same reason : F(x) = grad(grad(f(Sy))= S.F(x).S as the transpose [F(x)]= F(x) .

Is it correct ? if so , writting SF(x)s would be the cleaner way to write the expression mlok gave in the thread above?

If it is correct, why it must be in this sequence SF(x)S ? Why it can't be for instance F(x).S.S ?

Your formula is almost correct, but it should say $S^{T}F(x)S$. The transpose moves $S^{T}$ to the left of $F(x)$. [url=http://www.met.rdg.ac.uk/~ross/Documents/Chain.html]See also here[/url].

Consider the special case when $f(x)$ is a quadratic polynomial $f(x)=x^{T}A x$ (with $x$ a column vector). The Hessian of $f$ is simply $2A$. Also, $f(Sy)=y^{T}S^{T}A S y$ , with Hessian $2S^{T}AS$. This special case is in fact not so special, since a smooth function $f$ can be replaced by the second-order terms of its Taylor expansion, as far as the Hessian is concerned.


The four lines x =0, y=0, 5x+3y=30, and 5x-6y=0 form two triangles in the plane. Compute the area of the smaller triangle.

Suppose that x and y are positive real numbers such that 1/x 2 + 1/y 2 =8 and xy=3. Compute x+y

For the second:

$\frac {1}{x^2}+\frac {1}{y^2} = \left(\frac {1}{x}+\frac{1}{y}\right)^2-\frac {2}{xy}$

Can you finish it off from here?

Okay then, let me try to solve this problem:

$\displaystyle \frac {1}{x^2}+\frac {1}{y^2}$

$xy=3$

If we plug $x=\displaystyle \frac {3}{y}$ into the first equation, we obtain:

$\displaystyle \frac {y^4+9}{9y^2}= 8$

However, if we solve for y, I get a really weird answer from my TI-89, so this is where I am stuck...

[quote]The four lines x =0, y=0, 5x+3y=30, and 5x-6y=0 form two triangles in the plane.
Compute the area of the smaller triangle.[/quote][hide]

The line $5x = 3y = 30$ has intercepts $A(6,0),\; B(0,10).$
The line $5x - 6y = 0$ intersects AB at $C(4,\;10/3).$

Triangle OAB has area: $\frac{1}{2}(6)(10) = 30$

Triangle OAC has the same base but one-third the altitude of triangle OAB.
Therefore: area OAC = $\frac{1}{3}\times30 = 10$[/hide]

[quote]Suppose that x and y are positive real numbers such that $\frac{1}{x^2} + \frac{1}{y^2} = 8$ and $xy = 3.$
Compute $x+y.$[/quote]
I thought I found a truly clever solution
then I see that it's just a spinoff of blahblahblah's idea . . . *sigh*[hide]
We have: $\frac{1}{x^2} + \frac{1}{y^2} = 8\;\;\Rightarrow\;\;\frac{x^2 + y^2}{x^2y^2} = 8\;\;\Rightarrow\;\;x^2 + y^2 = 8x^2y^2$

$.\qquad x^2 + y^2 \;= \;8(xy)^2 \;= \;8(3^2) \;= \;72$

Add $2xy$ to both sides: $x^2 + 2xy + y^2 \;= \;72 + 2xy$

And we have: $(x + y)^2 \;= \;72 + 2(3)  \;= \;78$

Therefore: $x + y \;= \;\sqrt{78}$[/hide]

[quote="splintercell990"]$\displaystyle \frac {y^4+9}{9y^2}= 8$

However, if we solve for y, I get a really weird answer from my TI-89, so this is where I am stuck...[/quote]

Continuing along this path, y^4+9=72y^2, and y^4-72y^2+9=0. Your calculator should be able to handle the last equation.


Suggestion? Put all of these into one topic.


Traian Băsescu has designed the course of a cart in an asumement park (as seen in [url=http://www.mathlinks.ro/Forum/album.php?mode=attach&id=6570]this picture[/url]). The last part is an almost-circle of radius $R = 10 \, \textrm m$. We know that the maximum acceleration a person in the cart can withstand is $a_\textrm{max}= 80 \, \textrm m \slash \textrm s^{2}$ and that loaded cart has mass $M = 300 \, \textrm{kg}$. Determine:
a. the maximum possible height of the inclined plane;
b. the reaction force of the loop upon the cart when it passes through $B$, if $h = 35 \, \textrm m$;
c. the speed of the cart when it passes through $C$, if $h = 35 \, \textrm m$.

PS: Friction neglected shall be.
PPS: I'm interested in a solution for a.

a) $mgh = \frac{1}{2}mv^{2}$ by conservation of energy, thus $v^{2}= 2gh$.

Now, the centripital acceleration at the bottom is $\frac{v^{2}}{R}= \frac{2gh}{R}$.

Thus, $\frac{2gh}{R}= a_{max}$

So, $h = \frac{a_{max}R}{2g}$.

According to the official solution, we actually have that $h_\textrm{max}= \frac{R \left( a_\textrm{max}-g \right)}{2g}$. Why is that?

The upward force accelerating the cart is provided by the normal force due the cart's weight, as well as the centripetal force. So actually:

$a_{max}= \frac{2 g h_{max}}{R}+g$

$h_{max}= \left(\frac{a_{max}-g}{2g}\right) R$


Show that 3^2008 + 4^2009 can be written as product of two positive integers each of which is larger than 2009^182.

hasn't this been posted like 17 times already? just use the Sophie Germain identity.


function
Euler
number theory proposed
number theory

Find all numbers [i]m[/i] such that Euler`s function of [i]m[/i] divides [i]m[/i].

[quote="marko avila"]all the numbers of the form: 2a3b where a and b are exponents and b>=1.

the problem is straightforward from the formula for eulers function using the prime factorisation of n. the rest is easy. :D

this was a pre-tst in my state in mexico. :P[/quote]

are you sure? I think you are wrong. Take n=120 as an example. $\phi (n)$ = (5-1)(3-1)(2-1)=8. 8|120.

[quote="ZetaX"]$\phi (120) = (5-1)(3-1)4(2-1) = 32$ and that doesn't divide $120$ ![/quote]

oops, sorry for being idiotic..

i think the solution [code]2^a2^b[/code] without the condition of b>=1 coz even 2 powers are solutions

yes you are right the exponent a is the one thats >=1 because powers of three are not solutions :D


If students must take 2 exams on a day and there are 5 possible periods in which to take the exams, what is the possibility that a student with a randomly selected schedule has both finals during consecutive periods?

25%
33 1/3 %
40%
44 4/9%
50%

[hide]
Consider both the exams as one unit to find the no. of such favourable cases  :)
[/hide]

This question is actually from the TMSCA (Texas Math and Science Assc.) 2007 state math test.
Question #22. :|

I'm an idiot when it comes to probability, but I would say that
[hide]your numerator is the number of ways which you could have consecutive periods.  I see 8 different ways: 1,2; 2,1; 2,3; 3,2; 3,4; 4,3; 4,5; 5,4.  Then your denominator would be the number of possible arrangements for the 5 peiods taken two at a time with order taken into account (since we care about time) so I would use $ _5$$ P_2$ which equals $ \frac{5!}{(5\minus{}2)!}$ and then this leaves you with $ \frac{8}{20}$ which is equivalent to $ 40$%[/hide]
But I am a dumbace whenever it comes to probability so I would not be the least bit surprised if I did this easy problem absolutely incorrectly.


white wine @ room temp 70F is chilled in ice (32F). It takes 15 mins for wine to chill to 60F, how long will it take for the wine to reach 56F

Let $T(t)$ be the temperature at time $t$. Then we know (as the subject states) that

$\frac{dT}{dt}=-k(T-T_{f})$ so the solution will end up (I'll show more on solving if you want) being $T(t)=T_{f}+(T_{0}-T_{f})e^{-kt}$. Need to now find $k$. $T(15)=32+(70-32)e^{-k(15)}=60\Rightarrow k=0.0204$, so our equation is $T(t)=32+38e^{-0.0204t}$, now plug in $56=32+38e^{-0.0204t}\Rightarrow t=22.5$ minutes.

already solved it. got same answer. :D thnx, appreciate it!


Το πρώτο με συντεταγμένες γινόταν πολύ εύκολο (αρκεί να θυμόταν κανείς απ' έξω τις συντεταγμένες βασικών σημείων και να μην τις ψάχνει εκείνη την ώρα γιατί είναι χάσιμο χρόνου). Γεωμετρικά ήταν σίγουρα δύσκολο.

Nick οι πολικές να πούμε... :D

Για το δεύτερο φαντάστηκα ότι θα λυνόταν με βασικές ανισότητες αλλά απ' ότι φαίνεται δε φτάνει αυτό (αλλιώς δε θα ήταν θέμα Βαλκανιάδας).

Το τρίτο και το τέταρτο και να ασχοληθώ δε θα προσφέρω και πολλά στον κόσμο των μαθηματικών οπότε λέμε ένα Q.E.Not.D. :D

Παρακαλώ όποιος μάθει μόλις βγουν τα αποτελέσματα πώς πήγαν τα παιδιά να δημοσιεύσει τα 6 ευελπιστούμε μετάλλια.

Καλή επιτυχία. :)

Τα θέματα ήταν πολύ δύσκολα . Να δούμε που θα πάνε οι βάσεις των μεταλλίων. Αυτό θα εξαρτηθεί και από το πώς έγραψαν συνολικά τα παιδιά. Δεν αποκλείεται να δοθούν μετάλλια και με λιγότερο από 10(με άριστα το 40). Θα περιμένουμε με αγωνία.

   [color=red][b]Καλά αποτελέσματα παιδιά και καλή επιστροφή  ![/b][/color]

 Μπάμπης

Αλέξανδρε ,

 καλά τα νέα σου  :lol: . Να μας κρατάς ενήμερους , για ό,τι  νέο  μαθαίνεις  από τα Σκόπια !

 [u]Μπάμπης[/u]

Apo oti tha kserete ta apotelesmata einai sto:  http://www.smm.org.mk/index.php?option=com_content&view=article&id=123&Itemid=146

Sinxaritiria se ola ta paidia! Polla metalia girizoun!!!...

(kai epidi exo mia idiaiteri protimisi sti geometria apla na sinxarw ton Nick Rapanos gia tin lisi pou ekane ston prwto kai tin thaumasan oloi. O Silouan einai kai official balkanionikis!CONGRATULATIONS!!!)


Let x, y, n be positive integers, with n > 1.  How many solutions are there to the equation x^n - y^n = 2^100


simplify  xsquare rooted^6

[color=green][size=75]Please make your titles less "spammish".  Thanks.
-robinhe[/size][/color]

do you mean

$\sqrt[6]{x}$????

or

$(\sqrt{x})^{6}$?

to solve $(\sqrt{x})^{6}$

[hide]
basically that means $\sqrt{x}\cdot \sqrt{x}\cdot \sqrt{x}\cdot \sqrt{x}\cdot \sqrt{x}\cdot \sqrt{x}$....or just $x \cdot x \cdot x$..which is $x^{3}$.[/hide]

[quote="Ihatepie"]simplify  xsquare rooted^6[/quote]

[hide]$\left(\sqrt{x}\right)^{6}=\sqrt{x}\times\sqrt{x}\times\sqrt{x}\times\sqrt{x}\times\sqrt{x}\times\sqrt{x}=x^{3}$, because $\sqrt{x}\times\sqrt{x}=\sqrt{x^{2}}=x$[/hide]

Alternate method to simplify $\sqrt{x}^{6}$:
[hide]
\[\sqrt{x}^{6}=x^{\frac{6}{2}}=x^{3}\]
[/hide]

Note that $x^\left(\frac{1}{2}\right) = \sqrt{x}$.
Similarly, $x^\left({\frac{1}{n}\right) = \sqrt[n]{x}$, where $n$ is known as the [i]index[/i].

In the standard [i]square[/i] root, the index is $2$.

Therefore, $(\sqrt{x})^{6}=\left(x^\left(\frac{1}{2}\right)\right)^{6}$. We can then multiply the two powers.

$x^{\left(\frac{6}{2}\right)}= x^{3}.$

Excuse my $\text{\LaTeX}$ :rotfl:


A dart is thrown at the square target shown. Assuming the dart hits the target at a random location, what is the probability that it will be in the shaded region? Express your answer as a common fraction.
[asy]size(150);
void drawSquare(real side, pair center = (0,0), Label l = ""){
 pair A = (center.x - 0.5*side,center.y - 0.5*side);
 draw(A -- (A.x + side, A.y) -- (A.x + side, A.y + side) -- (A.x, A.y + side)--cycle);
 label(l,(A.x + side, A.y + side) -- (A.x, A.y + side),S);
}
drawSquare(2, "$2''$");
drawSquare(3, "$3''$");
drawSquare(4, "$4''$");[/asy]

To the AoPS admins, this figure isn't shaded.

Well. the answer is 5/16, so we can assume that the area between the 3-side and 2-side is the shaded thing.
So.
The shaded is 9-4 = 5
the total is 4^2 = 16

probability = 5/16

Admins: game 305412
i answered 0 first
and got wrong
other person memmed this
got right
i ended up losing the cd
4-3
i should've won
4-2
i lost 26 rating
give me my rating points back please
thank you

first of all, are you complaining about the question or the other person memming?

i don't really see how one would complain about the question in that sense and memming is a natural part of FTW, you can't stop it

second, admins can't give rating points
you'll have to simply win games to get your rating back up

i'm complaining about horrible problem with incorrect answer. it's not my opponent's fault.

The problem is glitched, and I submitted the correct answer, $ 0$
It thinks that the right answer is like $ \frac{5}{16}$ or something.
This is seriously unfair
I thought you're supposed to try to answer questions correctly
Not remember wrong answers.

oh!

i see what you mean

well, that's an issue with the original submitting of problems into the FTW database
you could ask admins to change it, but that takes a while and they have far more important stuff to do (ex. FTW2)

mostly likely, they won't get to it in the near future unless they're REALLY bored

thanks, but i've forgiven this problem. It's ok dart and unshaded target, i don't care anymore.
I'm basically forgiving everything i used to be angry at(including people that muted me like 10 times in a row)


vector
real analysis
real analysis unsolved

Prove that $R^n$ cannot be written as a countable union of proper subspaces.
$R^n$ is considered to be a vector space over $R$

Look at [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?highlight=partition&t=25908]this[/url] thread. The proof can easily be extended to show that $\mathbb R^n$ is not a countable union of more than one non-empty closed subsets. Since all subspacesof $\mathbb R^n$ are closed, that's that.

Ok but I thought my following idea will yield better results.
The completements of the subspaces are dense.Then why not use Baire category theorem

[quote="the game"]Ok but I thought my following idea will yield better results.
[/quote]

It does, but that's a pretty strong generalization :).

Hmm Could you plz post that proof because though my idea is correct my proof is wrong  :(

Post your proof and we'll see what's wrong with it :).

Hmm I will if if only i write something which will really prove the fact that complements of subspaces are denses.Actually thats where I got stuck

So, basicly, all you need now is to prove that the complement of a proper subspace of $\mathbb R^n$ is dense, or, equivalently, that a proper subspace of $\mathbb R^n$ has empty interior.

Assume a subspace contains a ball $B(x,r)$ of $\mathbb R^n$. Then it must also contain $B(x,r)-x=\{y-x|y\in B(x,r)\}$, which is equal to $B(0,r)$. This means that our subspace contains all the vectors of norm less than $r$ of $\mathbb R^n$, so it must be the entire $\mathbb R^n$.


graph theory
number theory unsolved
number theory

A set $ P$ consists of $ 2005$ distinct prime numbers. Let $ A$ be the set of all possible products of $ 1002$ elements of $ P$, and $ B$ be the set of all products of $ 1003$ elements of $ P$. Find a one-to-one correspondance $ f$ from $ A$ to $ B$ with the property that $ a$ divides $ f(a)$ for all $ a \in A$.

In the language of graph theory, we're looking for a matching in the graph whose vertices are subsets of size $ n$ and $ n\plus{}1$ in a set $ S$ of size $ 2n\plus{}1$, and whose edges correspond to set inclusion (so edges connect a set $ A$ of size $ n$ with a set $ B$ of size $ n\plus{}1$ iff $ A \subset B$). Since $ \binom{2n\plus{}1}{n} \equal{} \binom{2n\plus{}1}{n\plus{}1}$, the number of vertices of each kind is the same, and a matching is at least feasible. In what follows, an "$ n$-set" is a set of size $ n$.

To prove that a matching exists, we use the standard "marriage theorem" criterion: every set of $ k$ vertices is connected to at least $ k$ others. If this fails for some set, then a matching is clearly impossible; Hall's theorem guarantees that this necessary condition is also sufficient.

To check the condition, let $ X$ be a set of $ n$-element subsets of $ S$, with $ |X|\equal{}k$. Let $ Y$ be the collection of all $ (n\plus{}1)$-sets which contain at least one member of $ X$, and let $ Y\equal{}l$. Let us count how many pairs $ (A,B)$ are there with $ A \in X$ and $ B \in Y$ such that $ A \subset B$.

On the one hand, for every $ A \in X$ there are obviously $ n\plus{}1$ ways of extending it to subset $ B$ of size $ n\plus{}1$, since there are $ n\plus{}1$ elements of $ S$ outside of $ A$. Thus the number of pairs is $ k(n\plus{}1)$.

On the other hand, for every $ B \in Y$ there are $ n\plus{}1$ ways of removing an element from $ B$ to obtain an element of $ X$, so the number of such pairs is $ l(n\plus{}1)$. So in fact, $ l\equal{}k$, and by Hall's theorem there is a matching as required.

Incidentally, the graph we just analyzed is conjectured to be Hamiltonian, but this is still an open problem (sometimes called "the mid-level conjecture", "[url=http://arxiv.org/abs/math/0608485]the middle level problem[/url]" or "Erdős' revolving doors conjecture".)


Let $ n$ be a positive integer. Determine if there exist positive integers $ a_1,a_2,...,a_n,b_1,b_2,...,b_n$ so that $ \sum a_i \equal{}\sum b_i$ and $ n\minus{}1>\sum\frac{a_i\minus{}b_i}{a_i\plus{}b_i}>n\minus{}1\minus{}\frac{1}{1998}$

[hide="Hint"]
If $ a_1, a_2, \cdots, a_n, b_1, b_2, \cdots, b_n$ are $ 2n$ distinct positive integers s.t. $ a_1\plus{}a_2\plus{}\cdots\plus{}a_n\equal{}b_1\plus{}b_2\plus{}\cdots\plus{}b_n$, then
\[ n\minus{}1>\sum_{i\equal{}1}^n \frac{a_i\minus{}b_i}{a_i\plus{}b_i}
\]
[/hide]


What is the maximum value of $ x$ for which there exist real numbers $ x$ and $ y$ which satisfy the inequalities $ \minus{}10\le x\plus{}y\le4$ and $ x^2\plus{}y^2\minus{}36(x\plus{}y)\le2xy$?

In the second inequality, move the 2xy term over to the RHS, so that it will become (x-y)^2  -36(x+y) is less than or equal to 0. If (x+y) is negative than (x-y) will be imaginary, so x and y will not be real solutions. So (x+y) has to be positive. Checking (x+y) values from 0 to 4, it is clear that the maximum value of x occurs when (x+y) = 4.

x+y=4
so x-y = + or - 12

let x-y=12

solving the system, x= 8, y=-4


Median is dividing the triangle in two parts of equal area.
If we draw the triangle $ ABM_b$ we easily draw the triangle $ ABC$. We know the angle, side and area of the triangle $ ABM_b$. So we know the angle, side and altitude ($ AH$). Draw a segment $ KL$ such that $ |KL|\equal{}|BM_b|$. Then draw two lines at the distance $ AH$ from the line $ KL$. After that draw two circle arcs, with the grade mesaure equal to the angle $ BAC$. Then take one of the intersection  points ($ M$). Triangle $ KLM$ is what we need :)
[geogebra]9a1cd843982d9396dbf21f450431743cb210f9ca[/geogebra]


Let $ X=\left\{1,2,...,n^{2}\right\}$. How many permutations $ \pi$ of $ X$ have the property that there does not exist a strictly monotonically increasing sequence $ \left(a_{1},a_{2},...,a_{n+1}\right)$ of $ n+1$ distinct elements of $ X$ satisfying $ \pi\left(a_{1}\right)<\pi\left(a_{2}\right)<...<\pi\left(a_{n+1}\right)$ or $ \pi\left(a_{1}\right)>\pi\left(a_{2}\right)>...>\pi\left(a_{n+1}\right)$ ?

There is an answer in the OEIS, [url]http://www.research.att.com/~njas/sequences/A079402[/url], but no proof (although a proof is claimed).  Also, please try to name your posts in a more useful way.


Two positive integers a and b are relativley prime if the only common divisor of a and b is 1. How many positive integers less than 2004 are relativley prime to 2004?

This is a typical inclusion-exclusion problem :

$2004 = 2^2 * 3 * 167$ so count numbers divisible by $2$, add those divisible by $3$, those divisible by $167$, substract those divisible by $2*3$, $2*167$,... add those divisible by $2*3*167$.
This counts numbers not relatively prime with 2004

[hide]I find 664 numbers relatively prime with 2004[/hide]

Here is my solution, but I am not sure if it is correct or not.

[b]Solution:[/b]

[hide]To find the number of relatively prime integers, find those which are not - that is, all those which share a prime divisor with $2004$; then subtract this from $2004$.

Now the prime factors of $2004$ are $2, 3$ and $167$. If we take all multiples of $2$, and all multiples of $3$, we will have included $6$ twice. And if we also take all multiples of $167$, we will have included $1002$ three times! Therefore, to count all numbers which are multiples of $2, 3$ or $167$, we must include some sets of multiples and exclude others.

We can calculate the number of relatively prime factors, by letting $A , B , C$ be, respectively, the sets of multiples of $2, 3, 167$ which are less than or equal to $2004$. Then the set of integers which share with 2004 a common factor 2, 3 or 167, is $A \cup B \cup C$, which would be $n(A \cup B \cup C)=1002+668+12-334-4-6+2=1340$, giving $2004-1340=664$ integers relatively prime to $2004$.[/hide]

Okay, then I am correct!

Thanks for the help!

Another way to think is to count "not divisibly by 2,3,167) ..
  (1/2) of the numbers  are divisible by 2 ..  so  (1/2) are not divisible by  2..
Similarly  ... 2/3 of the numbers are not divisible by 3
and 166/167 of the numbers are not divisible by 167

so required answere = $(1/2)(2/3)(166/167)(2004)$

euler's totient function is basically this problem...


I am wondering if the message boards are deleted after the classes end. That would be a shame, seeing how it takes a while to type up good LaTeX solutions.

They are put in an archive that you will be able to see.

Nobody can post in them two weeks after the class ends. I have created ways for people in the algebra 1 class to keep in touch, but nobody seems interested.  :(


$ABCD$ is a convex quadrilateral with $AB$ parallel to $CD$. The diagonals $AC$ and $BD$ meet at $E$. $X$ is the midpoint of the line joining the orthocenters of $BEC$ and $AED$. Show that $X$ lies on the perpendicular to $AB$ through $E$.

[color=red][Moderator edit: The problem is from http://www.kalva.demon.co.uk/aus-pol/apsol/asol974.html .][/color]

Let the orthocenters of $ABE, BCE, CDE, DEA$ be $P, Q, R, S$ respectively. $P$ and $Q$ both lie on the perpendicular from $B$ to $AC$, so $PQ$ is perpendicular to $AC$. Similarly, $R$ and $S$ both lie on the perpendicular from $D$ to $AC$, so $RS$ is also perpendicular to $AC$ and hence parallel to $PQ$. Similarly, $QR$ and $PS$ are both perpendicular to $BD$ and hence parallel to each other. So $PQRS$ is a parallelogram. Hence the midpoint of $QS$ is also the midpoint of $PR$. But $P$ and $R$ both lie on the perpendicular from $E$ to $AB$. Hence the midpoint of $QS$ lies on the perpendicular from $E$ to $AB$, as required.


Si alguien la tiene la del 2º nivel podría ponerla porfavor, que la he buscado pero no la encuentro; se los agradecería  :blush:


$ f(x)\equal{}acos3x\plus{}bcox2x\plus{}ccosx\plus{}sinx$.
Prove that $ f(x)\equal{}0$ has solution in $ [0,\pi]$ for all $ a,b,c \in R$

[quote="sunlight"]$ f(x) \equal{} acos3x \plus{} bcox2x \plus{} ccosx \plus{} sinx$.
Prove that $ f(x) \equal{} 0$ has solution in $ [0,\pi]$ for all $ a,b,c \in R$[/quote]

Unfortunately, that's wrong :

For $ a\equal{}0$, $ b\equal{}\frac{1}{2}$ and $ c\equal{}0$, we get :

$ f(x)\equal{}\frac{\cos(2x)}{2}\plus{}\sin(x)\equal{}\minus{}(\sin(x))^2\plus{}\sin(x)\plus{}\frac{1}{2}$

For $ x\in[0,\pi]$, we have $ \sin(x)\in[0,1]$ and the polynomial $ \minus{}x^2\plus{}x\plus{}\frac{1}{2}$ is always $ \geq\frac{1}{2}$ when $ x\in[0,1]$, so $ f(x)\equal{}0$ has no solution for $ x\in[0,\pi]$


If $F_{n}$ is the nth term of the fibonacci's sequence, find all pairs $(m,n)$ of positive integers such $F_{n}\cdot F_{m}=mn$

$n-1\le F_{n}$ and $F_{n}>2n$ if n>7. Therefore solution is only $(m,n)=!,1) \ or \ (1,5) \ or \ (5,1)$.

there are another solutions to this equation.
We have $F_{n}/n=m/F_{m}$
We know that $F_{k}>k$ for all $k>5$
If $F_{n}>n$ we have that $m>F_{m}$ so we know that $m<5$ so the quotient involving the m terms can just be:
1,2,3/2,4/3.
Lets study each case:
$F_{n}/n=1$ just possible for n=5 (1,5), (5,1)
$F_{n}/n=2$ We have that if $F_{n}>2n$ for n>8 so we only have to look for some answer until$F_{8}$ and there is none!
$F_{n}/n=3/2$ We have until $F_{6}$ to look, and again there is no solution.
$F_{n}/n=4/3$ We now have $F_{6}$ as answer, so we have the solutions:

$(5,1);(1,5);(4,6);(6,4);(1,1);(5,5)$


Triangle $ ABC$ with bisector $ AD,BE,CF$ ($ D,E,F$ on $ BC,CA,AB$ respectively). Call $ D',E',F'$ are symetric point of $ D,E,F$ over $ EF,FD,DE$ respectively. Prove that $ AD',BE',CF'$ are concurrent!

Look at PDatK40SP's message for the generalization: http://www.mathlinks.ro/viewtopic.php?p=1112049#1112049


I'm trying to use the interactive asymptote to learn, but I can't import the olympiad package. I downloaded the package, saved it to /usr/share/asymptote (this is the correct place - don't worry), but when I type in import olympiad; I get the following error messages:

/usr/share/asymptote/olympiad.asy: 159.48: no matching variable 'currentpicture.xunitsize'
/usr/share/asymptote/olympiad.asy: 159.97: no matching variable 'currentpicture.xunitsize'
/usr/share/asymptote/olympiad.asy: 160.49: no matching variable 'currentpicture.yunitsize'
/usr/share/asymptote/olympiad.asy: 160.98: no matching variable 'currentpicture.yunitsize'
/usr/share/asymptote/olympiad.asy: 173.48: no matching variable 'currentpicture.xunitsize'
/usr/share/asymptote/olympiad.asy: 173.96: no matching variable 'currentpicture.xunitsize'
/usr/share/asymptote/olympiad.asy: 174.48: no matching variable 'currentpicture.yunitsize'
/usr/share/asymptote/olympiad.asy: 174.96: no matching variable 'currentpicture.yunitsize'
/usr/share/asymptote/olympiad.asy: 212.19: no matching variable 'currentpicture.xunitsize'
/usr/share/asymptote/olympiad.asy: 228.19: no matching variable 'currentpicture.xunitsize'
/usr/share/asymptote/olympiad.asy: 262.19: no matching variable 'currentpicture.xunitsize'

Can someone please help me fix this? It should be a rather easy fix - it's all the same error message.

Edit: I fixed it. all of them should've been xsize/ysize rather than x/yunitsize. I don't know if this is a problem with all linux distributions, or just mine. You may want to note that this is a possible thing that needs to be fixed though.

Edit2: Actually I don't know if they should be x/y unitsize, because they probably mean something different. Can anyone confirm/deny that I did the right thing?

Edit3: I changed back to xunitsize/yunitsize, and of course it's broken again. I don't want xsize/ysize because that's the maximum sizes, but I don't see any equivalent. Anyone know of a fix?

Can you give us a minimal example of a file you are trying to compile?
I'd be surprised if it's a Linux problem - the only difference between the operating systems is likely to be in finding files and Asymptote seems to have found olympiad.

This is how I can recreate it: $is a bash line, > is asymptote
$ asy
> import olympiad;

at which point I get the error messages.

I think xunitsize & yunitsize are defined in plain_picture.asy which is provided as part of the Asymptote distribution. Can you check the file exists in /usr/share/asymptote?

it does, but xunitsize and yunitsize aren't in the file.

I wonder if that's because you have an older copy of Asymptote. The latest Linux version 1.28 does have xunitsize in plain_pictures.asy. Here is my Windows version to compare with and use, though it may be an idea to check you have the latest version.

Okay, I downloaded it from the official site and it worked. The version in the ubuntu/debian repositories was out of date. Thanks.


pigeonhole principle
number theory
prime numbers
combinatorics proposed
combinatorics

Take any 26 distinct numbers from {1, 2, ... , 100}. Show that there must be a non-empty subset of the $ 26$ whose product is a square.

[hide]

I think that the upper limit for such subset is 37.[/hide]

[quote="tmrfea"]Take any 26 distinct numbers from {1, 2, ... , 100}. Show that there must be a non-empty subset of the $ 26$ whose product is a square.[/quote]

EDIT: I got the complete solution.

We have $ 2^{26}\minus{}1$ non-empty subsets.
Also, there are $ 25$ prime numbers less than $ 100$.
Every product in the subset can be written as "$ x^{2}\cdot k$", where $ k$ is the combination of prime factors less that $ 100$ in which every factor occurs $ 0$ or $ 1$ times.
There are $ 2^{25}$ such combinations.
So, by the Dirichlet's principle (box principle, pidgeonhole principle) there are two subsets $ A$ and $ B$ that have the same combination.
Let their products are $ a$ and $ b$, respectively.
Now we have three cases:
1) $ B\subset A$. We take the difference $ A\setminus B$ and it's good, because $ \frac{a}{b}$ is a square.
2) $ A\cap B \equal{}\emptyset$. We take the union $ A\cup B$ and it's good, because $ ab$ is a square.
3) $ A\cap B \equal{} R$. In this case, we take the sets $ M \equal{} A\setminus R$ and $ N \equal{} B\setminus R$. They have the same "combination" so their union $ M\cup N$ is good (like in case 2).


How do you get to take the test (I'm talking about the FIRST one you take, which I believe consists of multiple part and free response)? Is it by nomination from your teacher (and if so, how?) or you just send email that you want to take it? Also, do you have to pay for it (or your school)?

Thanks.  :)

I [i]think[/i] you have to be recommended by your physics teacher (I too found out after the deadline, about mid-Jan).  It's not a real recommendation, to my understanding, just some form the teacher fills out.  Check out [url=http://www.aapt.org/Contests/olympiad.cfm]www.aapt.org/Contests/olympiad.cfm[/url], the official website.  There's a link somewhere in there to the form you have to send in as well, as well as one to previous qualifying exams.   There's also a physicsbowl competition, which you may want to look into.

Actually, just ask your teacher to recommend you for the competition, and I'm sure he will be happy to sign you up. You can even offer to pay the 20 dollar fee if you want.

[quote="saturnlife"] You can even offer to pay the 20 dollar fee if you want.[/quote]
 :o Hey now, come on.  Our school paid for the AMC tests, this 20 fee should be paid by them right?  I think it can reasonably be argued that a student (me:P) doing well would just be giving my school free advertising.

So who do I contact with the information?

Also, can I still compete if I have little amount of knowledge? I"m starting physics next year and wondered if I should do it (money shouldn't be a problem but it's olympiad..). My goal is to just qualify from the first round. That's all I ask.

[quote]For additional information about the U.S. Physics Team, contact the AAPT Programs and Conferences Department at (301) 209-3344 or email aapt-prog@aapt.org.[/quote]
If you take physics C next year, you can at least make the semis... as long as you don't take the preliminary test the morning after the super bowl. :roll:


geometry
cyclic quadrilateral
geometry unsolved

[quote="en one"]ABCD :SQUARE  ,  FIND X ?

[url=http://forogeometras.com/index.php?PHPSESSID=9df718e60edd83d000cb2e88080eba24&topic=326.msg1120#new][img]http://img529.imageshack.us/img529/1228/cuadradomu8.jpg[/img][/url][/quote]
it would be better if you post the problem completely,not just the picture...

Let us to present a ''constructive'' solution.

Through vertex $ D,$ of the given square $ ABCD,$ we draw an arbitrary line which intersects its diagonal $ AC$ at one point, so be it $ E.$

We denote as $ E',$ the intersection point of $ AB,$ from the line through $ E$ and perpendicular to $ DE.$

Because of $ DE\perp EE'$ and $ DB\perp AC,$ we have $ \angle EDB \equal{} \angle AEE'$ $ ,(1)$

From cyclic quadrilateral $ ADEE'$ $ \Longrightarrow$ $ \angle AEE' \equal{} \angle ADE'$ $ ,(2)$

From $ (1),$ $ (2)$ $ \Longrightarrow$ $ \angle ADE' \equal{} \angle EDB$ $ ,(3)$

From $ (3)$ $ \Longrightarrow$ $ \angle EDB \plus{} \angle BDE' \equal{} \angle ADE' \plus{} \angle BDE' \equal{} 45^{o}$ $ \Longrightarrow$ $ \angle EDF \equal{} 45^{o}$ $ ,(4)$

Similarly we have that $ \angle FDF' \equal{} 45^{o}$ $ ,(5)$ where $ F\equiv AC\cap DE'$ and $ F',$ the intersection point of $ BC,$ from the line through $ F$ and perpendicular to $ DE'.$

So, from $ (4),$ $ (5)$ we conclude that the points $ D,$ $ E,$ $ F',$ are collinear.

$ \bullet$ We denote the point $ P\equiv EE'\cap FF'$ and it is easy to show that $ \angle EPF' \equal{} 45^{o},$ from the cyclic quadrilateral $ DEPF.$

Hence, for any line through the point $ D,$ we can have the configuration as the problem states, only with $ \angle x \equal{} \angle EDF \equal{} 45^{o}$ and the solution is completed.

Kostas Vittas.

vittasko interesting solution constructive.

another direct solution  :maybe:

Dear Mathlinkers,
an article concerning "Geometric miniatures on a square" with original proofs has been put on my website
http://perso.orange.fr/jl.ayme    vol. 7 Miniatures géométriques sur un carré  p. 38
Sincerely
Jean-Louis


geometry
perimeter
geometric transformation
homothety
geometry unsolved

Squares $ABB_{1}A_{2}$ and $BCC_{1}B_{2}$ are externally drawn on the hypotenuse $AB$ and on the leg $BC$ of a right triangle $ABC$ . Show that the lines $CA_{2}$ and $AB_{2}$ meet on the perimeter of a square with the vertices on the perimeter of triangle $ABC .$

Let $A_{2}E \bot AC$ We have $\triangle A_{2}EA=\triangle ACB$ so $A_{2}E/EC=A_{2}E/(EA+AC)=AC/(AC+BC)$
Otherwise $FC/CC_{1}=FC/BC=AC/(AC+BB_{2})=AC/(AC+BC)$
So $A_{2}E/EC=FC/CC_{1}$ Thus triangle $A_{2}EC$ and $FCC_{1}$ are similar
So that $\angle A_{2}CE=\angle FC_{1}C$. Thus $A_{2}C//FC_{1}$
Let $AD$ be the bisector of angle $ACB$
$AD/DB=AC/CB=AC/CC_{1}=AM/MF$ so $DM//BC$ $(1)$
$AD/DB=AC/CB=AC/BB_{2}=CF/FB$ so $DF//AC$ $(2)$
Form (2) $CF/CB=AD/AB=DF/BB_{2}$ so $CF=DF$ $(3)$
From (1) (2) (3) and $\angle ACB=90$ we have $CNDF$ is a square $q.e.d$

I labelled randomly.
Let $ b\equal{}CB$ and $ a\equal{}AC$. First, note by homothety that NM, BC, and AE are concurrent.
Now, we prove that AE ML CH are concurrent, by clarifying O to be the intersection of ONLY AE and CH, and declaring L as the foot of the perpenicular from O. By find the length LC, we can then prove that this is the same as the side length of the square.

First: OL/AL = DE/DA = $ \frac{b}{a\plus{}b}$.
Next: OL/LC = JA/(AC+JH) = $ \frac{a}{a\plus{}b}$
Thus AL/LC = a/b = AC/BC = AL/LN $ \implies \boxed{LC\equal{}LN}$
QED


geometry
circumcircle
power of a point
radical axis
geometry unsolved

Let $ ABCD$ be a cycilc quadrilateral with circumcenter $ O$ and with diogonal intersection point $ X$.Let $ l_A,l_B,l_C,l_D$ be an angle bisectors of $ \angle A,\angle B,\angle C,\angle D$ respectively.Denote $ K\in l_A\cap l_B$,$ L\in l_B\cap l_C$,$ M\in l_C\cap l_D$,$ N\in l_D\cap l_A$.Let $ O_l$ be a circumcenter of the quadrilateral(not necessary convex))$ KLMN$.Prove that $ O_l,O,X$ are collinear.

Denote $ E \in AB \cap CD, F \in AD \cap BC$. We have $ E \in NL, F \in KM$
We have:
\begin{eqnarray*} 2 (\overrightarrow{KA},\overrightarrow{KF}) & \equiv & (\overrightarrow{KA}, \overrightarrow{KF}) + (\overrightarrow{KA}, \overrightarrow{AD}) + (\overrightarrow{AD}, \overrightarrow{KF}) \\ & \equiv & (\overrightarrow{KA},\overrightarrow{KF}) + (\overrightarrow{AB}, \overrightarrow{KA}) + (\overrightarrow{KF}, \overrightarrow{CB}) \\ & \equiv &  (\overrightarrow{AB}, \overrightarrow{CB}) \\ & \equiv & (\overrightarrow{DA}, \overrightarrow{DC}) \\ & \equiv & 2(\overrightarrow{DA},\overrightarrow{DM})  \ \text{ (mod } 2 \pi \text{ ) } \end{eqnarray*}
$ \Rightarrow (KA,KF) \equiv (DA,DM) \ \text{ (mod }  \pi \text{ ) } \\ \Rightarrow (KA,KM) \equiv (KA,KF) \equiv (DA,DM) \ \text{ (mod }  \pi \text{ ) }$
It means that $ D,A,K,M$ are concyclic
So $ \overline{FK}. \overline{FM} = \overline{FA}.\overline{FD}$, then $ F$ lies on the radical axis of $ (ABCD)$ and $ (KLMN)$
Similarly with $ E$, and we have $ EF$ is the radical axis of $ (ABCD)$ and $ (KLMN)$
Thus, $ EF \perp OO_l$
We've known that $ O$ is the orthocenter of $ \Delta XEF$, so $ EF \perp OX$
And we have the desired result  :)


Let $ a, b, c$ non-negative real numbers. Prove or disprove that

$ \sqrt{\frac{a\plus{}b}{c}} \plus{} \sqrt{\frac{b\plus{}c}{a}} \plus{} \sqrt{\frac{c\plus{}a}{b}} \ge 2(a\plus{}b\plus{}c)\sqrt[4]{\frac{2}{\sqrt[3]{abc}(a\plus{}b)(b\plus{}c)(c\plus{}a)}}$.

_________
 :?:  :?:

[quote="arqady"][quote="sandumarin"]If $ a,b,c>0 $   prove that
$ \sum_{cyc}{\sqrt{\frac{b+c}{a}}}\ge 2(a+b+c)\sqrt[4]{\frac{3}{a^2(b+c)^2+b^2(c+a)^2+c^2(a+b)^2}}\ge 3\sqrt{2} $
_____
Marin Sandu[/quote]
I think your inequality is much weaker.  :wink:[/quote]

 :oops:  :wink:

[quote=sandumarin]Thanks Arqady and Ligouras !!
If $ a,b,c>0 $ then:
$ \sum_{cyc}\sqrt{\frac{b+c}{a}}\ge\sqrt{2}\sqrt[4]{\frac{3(a+b+c)^3}{abc}} $
____
Marin Sandu[/quote]
See here:
http://www.artofproblemsolving.com/community/c6h1099056


Given an angle cut in a sheet of paper, find a procedure to divide the angle in 3 equal parts only folding the paper.

You can generalize it to 5,6,7,..... parts.

Strange. I thought that folding is just a symmetry  :?

By folding you can solve the "impossible" trisection problem.

About 15 years ago I read a beautiful book about the use of origami technique to solve "impossible" problems but unfortunately I don't remember neither the title of the book nor the proofs. :(

This article may interest you (though I haven't read it myself):
[url=http://journals.cms.math.ca/cgi-bin/vault/public/view/CRUXv23n2/body/PDF/page81-88.pdf?file=page81-88]Folding the Regular Heptagon[/url]

In his wonderful book GEOMETRIC CONSTRUCTIONS (Springer 1998), George E.Martin gives two different ways to trisect an angle by paperfolding : one is Lemma 10.12 (Dayoub-Lott 1977), the other one is Exercise 10.14 (Hisashi Abe 1980).
Kind regards. Jean-Pierre


group theory
abstract algebra
analytic geometry
superior algebra
superior algebra unsolved

Hi

I couldn't solve these problems and I want your help.

[b]Question 1[/b]

Let $ G$ be a finite group with exactly one Sylow $ p$-subgroup. Prove that every subgroup and every factor group of $ G$, too, has exactly one Sylow $ p$-subgroup.

[b]Question 2[/b]

Let $ G$ be a finite group and $ H$ be a subgroup of $ G$. Let $ P_1$ be a Sylow $ p$-subgroup of $ H$. Show that there is a Sylow $ p$-subgroup $ P$ of $ G$ such that $ P \cap H \equal{} P_1$

[b]Question 3[/b]

Let $ G$ act on $ X$ and $ H$ act on $ Y$. Prove that the direct product $ G \times H$ acts on the cartesian product $ X \times Y$.

[b]Question 4[/b]

Assume $ G$ acts on $ X$, and let $ K$ be the kernel of the permutation representation of this action. Suppose that $ H$ is a normal subgroup of $ G$ and subgroup of $ K$ ($ H \triangleleft G$ and $ H<K$) Show that $ G/H$ acts on $ X$ when we put $ x(Hg) \equal{} xg$ for all $ x \in X$, $ Hg \in G/H$. What is the kernel of the permutation representation?

[quote="ZetaX"]1:
To have only one Sylow subgroup $ P \leq G$ is equivalent to that subgroup being normal. Note that $ P\cap H$ and $ P/N$ are Sylow subgroups of $ H$ and $ G/N$, respectively; you need to show that they are normal.

2:
Ever $ p$-group is contained in a $ p$-Sylow subgroup.

3:
Let the first coordinate act on the first coordinate, and likewise for the second ones.

4:
That's just checking the defintion of action (in fact, well definedness is the crucial point); the kernel will be $ K/H$.[/quote]

sorry but I need more explanation.

Are you experienced with double cosets? Because (2) can be solved easily using double cosets. Can you guess what two subgroups are good to use for double cosets?

If you don't know much about them, check out http://www.artofproblemsolving.com/Forum/weblog_entry.php?t=237985.


integration
trigonometry
calculus
LaTeX
logarithms
calculus computations

How to find the reduction formula for $tan^{n}x$

what ive done so far
[hide]$In$=$tan^{n}x$
$\int \tan^{n-2}x\tan^{2}x\,dx$
$\int \tan^{n-2}x(\sec^{2}x-1)\,dx$

$\int \tan^{n-2}x\sec^{2}x\,dx$ $-$$\int \tan^{n-2}x\,dx$
 [/hide]

How can i proceed?

-----------------------------------------------------------------------------

$\int \frac{sinx}{2+cosx}\,dx$ $<----$ requires knowledge of inverse fns?

thanks

[quote="Kent Merryfield"]Back up to the step

$\int \tan^{n-2}x\sec^{2}x\,dx-\int\tan^{n-2}x\,dx$

Erase everything you wrote below that; it doesn't go anywhere, and looking at it would only confuse you.

Now, use this substitution: $u=\tan x.$ Then $du=\sec^{2}x\,dx$ and we can rewrite that first integral as

$\int u^{n-2}\,du.$

Now go forward from there.

---

jmerry and I were trying to tell you some things about $\text{\LaTeX},$ including using \tan and \sec insted of tan and sec, and about using \, to create small amounts of space before the dx. Please look at other people's code - you can see it by holding your mouse over the symbols.[/quote]

kk ill learn the latex thing/thanks 4 explanation

You can't use the recursion forever if you want to compute something. When you get to $I_{1}$, the integral is just $\int \tan x\ dx$, and we have
\[\int_{0}^{a}\tan x\ dx=\ln |\sec a|\]


MCrawford wrote:If , find  in terms of .

[hide]

(2^3)^x = 8^x = k

8^{x+1} = 8k

[/hide]

I remember there was a question analogous to this in the 2003 National MATHCOUNTS Countdown Round.  Both participants answered incorrectly.  (I'm pretty sure this is true; my memory irks me.)

[hide]If 2^(3x) = k

then k is equivalent to 8^x because 8 is 2 cubed.

Therefore 8^(x+1) is k mulitiplied by 8 because the power is increased by 1. For example... 8^2 (64) is 8 times greater than 8^1 (8) because you have to multiply by 8.

So therefore... the answer is 8k

8k. There's your answer.[/hide]

[quote=avn]imagine bumping a 16 year old thread[/quote]
but i posted a legitimate solution, which is allowesd

[quote=asp211]According to the basic forum rules...
[quote=jellymoop]Avoid bumping old posts.[/quote][/quote]
but as long as you post a legitimate solution it is allowed (its better to bump an old thread than to create a new thread).

I got [hide]$8k$[/hide] (why are we bumping a thread from 2004?)

[quote=avn][quote=shake9991][hide] 2^3x can be rewritten as: (2^3)^x... which simplifies to 8^x. Because 8^(x+1) is equal to 8*8^x, and 8^x=k:

8^(x+1)=8*8^x=8k[/hide][/quote]

[quote=intelligence_20]solution - we split up $2^{3x}$ into $(2^3)^x$ which works due to basic exponent properties, resulting that $8^x = k$. Now we multiply both sides by 8 to get our final answer $8k$[/quote]

please explain to me the difference between these 2 solutions.[/quote]

It’s possible that they used the same method :/

avn is saying that intelligence posted the same solution.

aaaaaayyyyyyyyy thanks for the 16 year old bump @intelligence_20

[quote name="Einstein394" url="/community/p21622624"]
Technically, it is greater than 16 years
[/quote]

ayyyyy but thanks for the 34 days, 15 hours, 46 minutes bump if that's good enough for you


calculus
integration
trigonometry
logarithms
calculus computations

Let $ a,\ b$ be postive constant numbers.

Evaluate $ \int_0^{\frac {\pi}{2}} \frac {a\cos x \minus{} b\sin x}{b\sin x \plus{} a\cos x}\ dx$.

hello, here you must use the substitution $ z\equal{}\tan\left(\frac{x}{2}\right)$.
Sonnhard.

Ohh! Your mean! It took me five minutes to realize!!

Let $ u\equal{}b\sin(x)\plus{}a\cos(x)$

$ \int\frac{a\cos(x)\minus{}b\sin(x)}{b\sin(x)\plus{}a\cos(x)}\,dx\equal{}\ln|b\sin(x)\plus{}a\cos(x)|$

hello, but $ u^'=b\cos(x)-a\sin(x)$ and this is different from $ a\cos(x)-b\sin(x)$.
Sonnhard.

Oops  :blush: ...sorry my bad..its a little late over here  :D

Well then I went through the sub $ x\equal{}2\arctan(t)$ and got

a really long answer, I will post it tomorrow.

Goodnight

hello, your result ist correct.
Sonnhard.

Yes, your answer is right. But my solution is not same as yours at all.

[quote="hsiljak"][b]Lemma[/b]. $ \int \frac {\sin (x \minus{} t)}{\sin (x \plus{} t)}dx \equal{} x \cos 2t \minus{} \log(\sin (x \plus{} t))\sin 2t \plus{} C$.

Proof of the lemma:

$ \int \frac {\sin (x \minus{} t)}{\sin (x \plus{} t)}dx \equal{} \int \frac {\sin (x \plus{} t \minus{} 2t)}{\sin (x \plus{} t)}dx \equal{}$
$ \equal{} \int \frac {\sin (x \plus{} t)\cos 2t \minus{} \cos (x \plus{} t)\sin 2t}{\sin (x \plus{} t)}dx \equal{}$
$ \equal{} \int \cos 2tdx \minus{} \sin 2t \int \cot (x \plus{} t) dx \equal{}$
$ \equal{} x \cos 2t \minus{} \log(\sin (x \plus{} t))\sin 2t \plus{} C$

Now, dividing both numerator and denominator in our integral with $ \sqrt {a^2 \plus{} b^2}$ and denoting $ \frac {a}{\sqrt {a^2 \plus{} b^2}} \equal{} \sin t$, $ \frac {b}{\sqrt {a^2 \plus{} b^2}} \equal{} \cos t$, we have the integral from the lemma.

So, the indefinite integral becomes (after simplyfying)

$ \frac {x(a^2 \minus{} b^2) \plus{} 2ab\log(a\cos x \plus{} b\sin x)}{a^2 \plus{} b^2}$

and after plugging in the bounds

$ \frac {\frac {\pi}{2}\left(a^2 \minus{} b^2\right) \plus{} 2ab\log\left(\frac {b}{a}\right)}{a^2 \plus{} b^2}$[/quote]

I believe that you have should said $ \int \frac {\cos (x \minus{} t)}{\cos (x \plus{} t)}\ dt$.

Please make a comfirmation.

[quote="Timestopper_STG"]kunny,can you show your solution? thank you :)[/quote]

Here is a hint.

Consider the following definite integrals.

Let $ I\equal{}\int_0^{\frac{\pi}{2}} \frac{\cos x}{b\sin x\plus{}a\cos x}\ dx,\ J\equal{}\int_0^{\frac{\pi}{2}} \frac{\sin x}{b\sin x\plus{}a\cos x}\ dx$.

The $ \cos$ integral has the solution:

$ \int \frac {\cos (x \minus{} t)}{\cos (x \plus{} t)}dx \equal{} \minus{} x \cos 2t \plus{} \log(\cos (x \plus{} t))\sin 2t \plus{} C$

That could have been used, but for transforming our integral in this form, we'd have to take

$ \frac {a}{\sqrt {a^2 \plus{} b^2}} \equal{} \cos t$.

Correct me if I'm wrong (of course, there could be some silly mistake I made in the process, but I'm pretty sure that it's ok).

For $ \cos t \equal{} \frac {a}{\sqrt {a^2 \plus{} b^2}},\ \sin t \equal{} \frac {b}{\sqrt {a^2 \plus{} b^2}}$, let $ r \equal{} \sqrt {a^2 \plus{} b^2}>0$, we have

$ a\cos x \minus{} b\sin x \equal{} r\left(\frac {a}{r}\cos x \minus{} \frac {b}{r}\sin x\right) \equal{} r\cos (x \plus{} t)$,

$ b\sin x \plus{} a\cos x \equal{} r\left(\frac {b}{r}\cos x \plus{} \frac {a}{r}\cos x\right) \equal{} r\cos (x \minus{} t)$.

Sorry Double Posting.

[quote="pankajsinha"]Quite elementary this is:

Let  acosx+bsinx = P(acosx+bsinx) + Q(-asinx+bcosx).

By equating coefficients of cosx and sinx, P cand Q can be easily solved for . Consequently, the integral becomes

Px + Q[b] [i]ln[/i]|[/b]acosx+bsinx[b]|[/b] +c

On plugging the limits we get

P(pi)/2 +Q [i]ln[/i](b/a)

P=(a^2 -b^2)/(a^2 +b^2)   and    Q = 2ab/(a^2-b^2)[/quote]

Yes, that's the background of the problem, although I set $ a\cos x \minus{} b\sin x \equal{} P(b\sin x \plus{} a\cos x) \plus{} Q(b\sin x \plus{} a\cos x)'$.

One Comment:

Why $ a\cos \theta \plus{} b\cos \theta \equal{} 0\Longleftrightarrow a \equal{} b \equal{} 0$ for all real $ \theta$?

[quote="kunny"] [color=darkred]Let $ a,\ b$ be postive constant numbers. Evaluate $ \int_0^{\frac {\pi}{2}} \frac {a\cos x \minus{} b\sin x}{b\sin x \plus{} a\cos x}\ \mathrm {dx}$ .[/color] [/quote]

[color=darkblue][b][u]Proof.[/u][/b] Suppose w.l.o.g. $ a\ne 0\ \ \vee\ \ b\ne 0$ , i.e. $ a^2 \plus{} b^2\ne 0$ . Denote $ f(x) \equal{} b\sin x \plus{} a\cos x$ . I look for the constant $ A$ , $ B$ so that for any $ x\in\mathbb R$ , $ a\cos x \minus{} b\sin x \equal{} Af(x) \plus{} Bf'(x)$ , i.e. $ a\cos x \minus{} b\sin x \equal{}$ $ A(b\sin x \plus{} a\cos x) \plus{} B(b\cos x \minus{} a\sin x)$ . Obtain easily that $ A \equal{} \frac {a^2 \minus{} b^2}{a^2 \plus{} b^2}$ and $ B \equal{} \frac {2ab}{a^2 \plus{} b^2}$ . Thus, $ \frac {a\cos x \minus{} b\sin x}{b\sin x \plus{} a\cos x}\ \mathrm {dx} \equal{}$ $ \int\frac {Af(x) \plus{} Bf'(x)}{f(x)}\ \mathrm {dx} \equal{}$ $ Ax \plus{} B\ln|f(x)| \plus{} \mathrm C$ , i.e. $ \int\frac {a\cos x \minus{} b\sin x}{b\sin x \plus{} a\cos x}\ \mathrm {dx} \equal{}$ $ \frac {a^2 \minus{} b^2}{a^2 \plus{} b^2}\cdot x \plus{} \frac {2ab}{a^2 \plus{} b^2}\cdot\ln|b\sin x \plus{} a\cos x| \plus{} \mathrm C$ . In conclusion, $ \int_0^{\frac {\pi}{2}}\frac {a\cos x \minus{} b\sin x}{b\sin x \plus{} a\cos x}\ \mathrm {dx} \equal{}$ $ \frac {a^2 \minus{} b^2}{a^2 \plus{} b^2}\cdot\frac {\pi}{2} \plus{} \frac {2ab}{a^2 \plus{} b^2}\cdot\ln\left|\frac ab\right|$ .[/color]

Thank you very much, Virgil Nicula.  :)

That's I wanted to describe.

[quote="kunny"] [color=darkred]Let $ a,\ b$ be postive constant numbers. Evaluate $ \int_0^{\frac {\pi}{2}} \frac {a\cos x - b\sin x}{b\sin x + a\cos x}\ \mathrm {dx}$ .[/color] [/quote]

[color=darkblue][b]Method I ([u]Pankajsinha[/u]).[/b] Suppose w.l.o.g. $ a\ne 0\ \ \vee\ \ b\ne 0$ , i.e. $ a^2 + b^2\ne 0$ . Denote $ f(x) = b\sin x + a\cos x$ . I look for the constant $ A$ , $ B$ so that for any $ x\in\mathbb R$ , $ a\cos x - b\sin x = Af(x) + Bf'(x)$ , i.e. $ a\cos x - b\sin x =$ $ A(b\sin x + a\cos x) + B(b\cos x - a\sin x)$ . Obtain easily that $ A = \frac {a^2 - b^2}{a^2 + b^2}$ and $ B = \frac {2ab}{a^2 + b^2}$ . Thus, $ \frac {a\cos x - b\sin x}{b\sin x + a\cos x}\ \mathrm {dx} =$ $ \int\frac {Af(x) + Bf'(x)}{f(x)}\ \mathrm {dx} =$ $ Ax + B\ln|f(x)| + \mathrm C$ , i.e. $ \int\frac {a\cos x - b\sin x}{b\sin x + a\cos x}\ \mathrm {dx} =$ $ \frac {a^2 - b^2}{a^2 + b^2}\cdot x + \frac {2ab}{a^2 + b^2}\cdot\ln|b\sin x + a\cos x| + \mathrm C$ . In conclusion, $ \int_0^{\frac {\pi}{2}}\frac {a\cos x - b\sin x}{b\sin x + a\cos x}\ \mathrm {dx} =$ $ \frac {a^2 - b^2}{a^2 + b^2}\cdot\frac {\pi}{2} + \frac {2ab}{a^2 + b^2}\cdot\ln|\frac ab|$ .

[b]Method II ([u]Kunny[/u]).[/b]  Denote $ I = \int_0^{\frac {\pi}{2}}\frac {\cos x}{b\sin x + a\cos x}\ \mathrm {dx}$ and $ J = \int_0^{\frac {\pi}{2}}\frac {\sin x}{b\sin x + a\cos x}\ \mathrm {dx}$ . Observe that $ \{\begin{array}{c} a\cdot I + b\cdot J = \frac {\pi}{2} \\
 \\
b\cdot I - a\cdot J = \ln|\frac ba|\end{array}$ . Obtain $ I = \frac {a}{a^2 + b^2}\cdot \frac {\pi}{2} + \frac {b}{a^2 + b^2}\cdot\ln|\frac ba|$ and $ J = \frac {b}{a^2 + b^2}\cdot \frac {\pi}{2} - \frac {a}{a^2 + b^2}\cdot\ln|\frac ba|$ . In conclusion, $ \int_0^{\frac {\pi}{2}} \frac {a\cos x - b\sin x}{b\sin x + a\cos x}\ \mathrm {dx} =$ $ a\cdot I - b\cdot J$ , i.e. $ \int_0^{\frac {\pi}{2}}\frac {a\cos x - b\sin x}{b\sin x + a\cos x}\ \mathrm {dx} =$ $ \frac {a^2 - b^2}{a^2 + b^2}\cdot\frac {\pi}{2} + \frac {2ab}{a^2 + b^2}\cdot\ln|\frac ab|$ .

[b][u]A similar exercises:[/u][/b] $ \int\frac {3e^x + 5\sin x + 10\cos x}{e^x + 4\sin x + 3\cos x}\ \mathrm {dx}$ ; $ \int\frac {\sin x}{e^x - \sin x - \cos x}\ \mathrm {dx}$ .   [/color]


If for some $ n\in \mathbb{N}$ such that $ f(n)\equal{}n^4\minus{}360n^2\plus{}400$ represents

for prime number , find the sum of these primes.

This is AMC12, please use the resources, don't repost.


modular arithmetic
superior algebra
superior algebra unsolved

Let $\sigma$ be the m-cycle $(1 2 \hdots m)$. Show that $\sigma^{i}$ is also an m-cycle if and only if $(i,m)=1$.

I managed to prove this using group actions acting on the elements. But that's insane! I don't see the obvious aproach for one or the other obscure reason. Can someone refresh my mind.

$\sigma^{i}$ is an $m$-cycle iff the cycle in $\sigma^{i}$ containing $1$ is of the length $m$.
The cycle containing $1$ is $\alpha=(1 \;[i+1]\; [2i+1]\;\cdots \;[(a-1)i+1])$, where $[bi+1]$ is the residue of $bi+1$ modulo $m$.
If $\sigma^{i}$ is not an $m$-cycle then there is $1<a<m$ so that
\[ai+1\equiv 1 \mod m\]
Hence $m\mid ai$. If $(m,i)=1$ then $m\mid a$ (Not Possible since $1<a<m$). Hence $(m,i)=1$.

Conversely if $(m,i)=k>1$, then $\frac{m}{k},\frac{i}{k}\in\mathbb{Z}$. Then
\[\frac{m}{k}i\equiv o\pmod{m}\]
Then $(1 \,[i+1]\,\cdots\,[( \frac{m}{k}-1)i+1])$ is a cycle of length less than $m$. So $\sigma^{i}$ is not an $m$-cycle.


$ p,m$ and $ n$ are the positive integers so that $ p$ is prime number and $ p\equal{}m^{2}\plus{}n^{2}$. Find all $ p$, for which there exists such pair of positive integers $ (m,n)$ that $ m^{3}\plus{}n^{3}\minus{}4$ is divisible by $ p$.
(Very nice problem. I have two different ways to solve it. I will post it later :wink: )

I think, it posted before.
Let $ s\equal{}m\plus{}n$, then $ s^2\equal{}2mn\plus{}p,s^3\plus{}8$ $\equal{}s(2mn\plus{}p)\plus{}8$ $\equal{}sp\plus{}2(smn\plus{}4)$ $\equal{}3sp\minus{}2(sp\minus{}smn\minus{}4)$ $\equal{}3sp\minus{}2(m^3\plus{}n^3\minus{}4)$.
Therefore $ p|(s\plus{}2)(s^2\minus{}2s\plus{}4)$. For $ p\equal{}2,5$ $ p|(s\plus{}2)$. If $ p>5$, then $ p\ge 13$ and may be only $ p|s^2\minus{}2s\plus{}4$ or $ p|mn\minus{}m\minus{}n\plus{}2$. If $ p>13$ $ |mn\minus{}m\minus{}n\plus{}2|<m^2\plus{}n^2\equal{}p$. for p=13 work $ m\equal{}\minus{}3,n\equal{}\minus{}2$.

Hey dude, i have a solution to this.

 $m^{2}+n^{2}|m^{3}+n^{3}-4$ $\rightarrow m^2+n^2|(m^3+n^3-4)^2=m^6+n^6-8(m^3+n^3-4)+2(m^3n^3-8)$ $\Rightarrow m^2+n^2|2(mn-2)(m^2n^2+2mn+4)$. We have $m^2+n^2=2\Rightarrow m=n=1$ or the expression equals $0$, when $mn-2=0$. The solutions are $(1;1), (1;2), (2; 1)$. i can't get solutions from $(m^2n^2+2mn+4)$

[quote="maths_lover5"]Hey dude, i have a solution to this. ... I can't get solutions from $(m^2n^2+2mn+4)$.[/quote]
These two statements are contradictory.

And poor [b]Rust[/b], finding a nice solution $13 = (-2)^2 + (-3)^2 \mid  (-2)^2 + (-3)^2 - 4 = -39$; all in vain, since twice the problem setters asked for $m,n$ to be positive  :P

My Solution:

$\ m^2+n^2\mid m^3+n^3-4$

$\Rightarrow p\mid (m+n)(p-mn)-4$

$\Rightarrow p\mid p(m+n)-mn(m+n)-4$

$\Rightarrow p\mid mn(m+n)+4$

$\Rightarrow p\mid 3mn(m+n)+12$

and since $p\mid m^3+n^3-4$, we get:

$p\mid m^3+n^3+3mn(m+n)+12-4$

$\Rightarrow p\mid (m+n)^3+2^3$

$\Rightarrow p\mid (m+n+2)((m+n)^2-2(m+n)+4)$

and since $p$ is prime,we have two cases:

[color=#f00]Case 1:[/color]$p\mid m+n+2$

Since $p\mid m+n+2$,we have:
$m^2+n^2\leq m+n+2$
$\Rightarrow m(m-1)+n(n-1)\leq 2$
and since $m,n\in N$,we have $(m,n)=(1,1)$ or $(m,n)=(1,2)$

[color=#f00]Case 2:[/color]$p\mid (m+n)^2-2(m+n)+4$

$p\mid (m+n)^2-2(m+n)+4$
$\Rightarrow p\mid m^2+n^2+2mn-2(m+n)+4$
$\Rightarrow p\mid 2mn-2(m+n)+4$
$\Rightarrow m^2+n^2\leq 2mn-2(m+n)+4$
$\Rightarrow (m-n)^2+2(m+n)\leq4$
note that $(m-n)^2+2(m+n)\geq 0$
and in this case the only solution is: $(m,n)=(1,1)$

Finally,we have two solutions: $(m,n)=(1,1);(1,2)$.

P.S:only such prime numbers are $2$ and $5$  :)


limit
geometry
incenter
analytic geometry
geometry proposed

Given triangle $ABC$ inscribed $(O)$, centroid $G$, lines $GA,GB,GC$ intersect $(O)$ at $A_{1},B_{1},C_{1}$ resp, $G_{1}$ is centroid $\triangle A_{1}B_{1}C_{1}$ lines $G_{1}A_{1},G_{1}B_{1},G_{1}C_{1}$ intersect $(O)$ at $A_{2},B_{2},C_{2}$ resp similarly ... we have $\triangle A_{n}B_{n}C_{n}$ with sequence centroid $\{G_{n}\}_{n=1}^{\infty}$ prove that
$\lim_{n\rightarrow \infty}G_{n}=O \Leftrightarrow \lim_{n\rightarrow \infty}\triangle A_{n}B_{n}C_{n}=Equilateral\text{ }triangle$

I saw incenter of triangle has similar property, but a big question, can we show all point of triangle satify this property, if we use fix barycentric coordinate it means you choose $I_{n}(\alpha,\beta,\gamma)$ with triangle $A_{n}B_{n}C_{n}$ ($\alpha,\beta,\gamma$ don't depend triangle) then that property is true for thia case, but other center of triangle can't show as above. Who has any idea?


geometric series
calculus
calculus computations

Hello, I have a quick question

I want to find the sum $ b*(1\plus{}p/100)^{\minus{}k}$ $ p>0$
as k goes from 0 to infinity. I can't figure it out. Could someone guide me through it with a solution please?

thanks in advance

I'm assuming you want $ k$ to be natural. $ \sum_{k\equal{}0}^ {\infty}b(1\plus{}\frac{p}{100})^{\minus{}k}p\equal{}\sum_{k\equal{}0}^ {\infty}bp\frac{1}{(1\plus{}\frac{p}{100})^k}\equal{}bp\sum_{k\equal{}0}^ {\infty}\frac{1}{(1\plus{}\frac{p}{100})^k}$ (if that sum exists, of course). However, what you now want is just a sum of geometric series and we know (and it is easily provable) that is $ \frac{1}{1\minus{}\frac{1}{1\plus{}\frac{p}{100}}}$. So, that means the answer you're looking for is $ b\cdot p\cdot \frac{1\plus{}\frac{p}{100}}{\frac{p}{100}}\equal{}b(100\plus{}p)$, if I didn't calculate something incorrectly. (Of course, we are working on the assumption that $ |1\plus{}\frac{p}{100}|>1$, otherwise we'll have problems with the sum of geometric series. Case $ p\equal{}0$ is trivial, though and in other cases, the sum does not converge.)

Hi [b]mornik[/b] i think that  [b]bigman[/b] want to say:

compute the sum : $ \sum_{k \geq 0} b \left( 1 \plus{} \frac {p}{100} \right)^{ \minus{} k}$ with $ \color{red}{p \succ 0}$

than obviously $ \forall p \succ 0$ we have $ \left( 1 \plus{} \frac {p}{100} \right)^{ \minus{} 1} \prec 1$ than our serie  converge and:

$ \sum_{k \geq 0} b \left( 1 \plus{} \frac {p}{100} \right)^{ \minus{} k} \equal{} b \sum_{k \geq 0} \left( \frac {1}{\left(1 \plus{} \frac {p}{100} \right)}\right)^k \equal{} 100 \frac {b}{p} \plus{} b$ ....

and thanks :)

Ah, I see, $ p$ is not the part of the series, it was ment to mean $ p>0$. Obviously, everything still stays the same, we only have to divide the final solution by $ p$ and get exactly what you got :)


limit
geometry
trigonometry
induction
perimeter
Pythagorean Theorem
calculus

1.) Let $ (x_n)_{n \geq 1}$ be a sequence such that $ x_1 \equal{} 1$, and $ x_{n \plus{} 1} \equal{} x_n \plus{} \sqrt {x_n^2 \plus{} 1}, n \geq 1$. Evaluate $ \lim_{n \to \infty}{\frac {2^n}{x_n}}$.

2.) Evaluate $ \lim_{n \to \infty}{2^n \sqrt {2 \minus{} \sqrt {2 \plus{} \sqrt {2 \plus{} \dots \plus{} \sqrt {2 \plus{} \sqrt {3}}}}}}$, where we have $ n$ radicals.

For (1), numerically, we have $ \lim_{n\to\infty}\frac{2^n}{x_n}\equal{}\frac{\pi}2.$

(It does depend on the initial condition.)

Since the answer is so nice, there's probably a nice explanation involving things like algebraic identities for tangents or secants. I haven't found that nice explanation yet.

What might the relation between 1 and Viete's formula be?  :idea:

As for 2, I don't understand the expression.

About 1), Kent, could you give me more detalies? What exactly numerical method means?

And about the second one... I don't know what you don't understand. The dots area shows that there are mising some radicals $ \plus{} \sqrt {2 \plus{} \sqrt {2\plus{}...}}$. It means that the total number of radicals that are in the expresion is $ n$.

[quote="TheNightTime"]would the second diverge because the radical expression tends to a finite limit?[/quote]

Nope.The radical expression tends to $ 0$.Notice that $ \sqrt{3}\equal{}2\cos \frac{\pi}{6}$, $ \sqrt{2\plus{}\sqrt{3}}\equal{}\sqrt{2\plus{}2\cos \frac{\pi}{6}}\equal{}\sqrt{4\cos ^2 \frac{\pi}{6\cdot 2}}\equal{}2\cos \frac{\pi}{6\cdot 2}$ ..and so on.we can prove using induction that
$ a_n\equal{}\sqrt{2\plus{}\sqrt{2\plus{}\cdots \sqrt{2\plus{}\sqrt{3}}}}\equal{}2 \cos \frac{\pi}{6\cdot 2^{n\minus{}1}}$ ( $ n$ radicals).So $ a_n \to 2$.

Thus we have to compute $ \lim_{n\to \infty} 2^n \sqrt{2\minus{}a_n}$.But $ \sqrt{2\minus{}a_n}\equal{}2 \sin \frac{\pi}{6 \cdot 2^n}$ (here I used $ \cos 2 \alpha \equal{}1\minus{}2 \sin^2 \alpha$ ). So

$ \lim_{n\to \infty} 2^n \sqrt{2\minus{}a_n} \equal{}\lim_{n\to \infty} 2^{n\plus{}1} \sin \frac{\pi}{6 \cdot 2^n} \equal{} \frac{2\pi}{6}\lim_{n\to \infty} \frac{\sin \frac{\pi}{6 \cdot 2^n}}{\frac{\pi}{6 \cdot 2^n}}\equal{}\frac{\pi}{3}$

Hope I didn't make any computation mistakes  :D

[quote="turcas_c"]About 1), Kent, could you give me more detalies? What exactly numerical method means?[/quote]
Just the most simple-minded method: run the recursion on a spreadsheet for 20 or 30 terms, divide by $ 2^n$ (or vice versa) and see if you can recognize the resulting number. The convergence appears to be geometrically fast, or what the numerical analysts call first order. In particular we seem to have $ \left|\frac{x_{n\plus{}1}}{2^{n\plus{}1}}\minus{}\frac2{\pi}\right|\approx\frac14\left|\frac{x_n}{2^n}\minus{}\frac{2}{\pi}\right|.$

One note: with the spreadsheet, I can play around with different initial conditions. With $ x_1\equal{}\sqrt{3},$ we get $ \frac{2^n}{x_n}\to\frac{\pi}3$ and with $ x_1\equal{}\frac{1}{\sqrt{3}},$ we get $ \frac{2^n}{x_n}\to\frac{2\pi}3.$ There's probably a clue in there somewhere.

I think that a connection can be made between (2) and the perimeter of a regular polygon with $ 6\cdot 2^n$ sides inscribed in a circle. I'm still looking for that sort of geometric or trigonometric connection for (1).

The numerical evidence given by Kent suggests that $ \lim_{n \to \infty} \frac{2^n}{x_n} \equal{} 2 \arctan x_1$.  Writing the recurrence in reverse gives

$ x_n \equal{} \frac{x_{n\plus{}1}^2 \minus{} 1}{2x_{n\plus{}1}}$

and now it is clear what to do from here; if $ x_n \equal{} \cot \theta_n$ then $ \theta_n \equal{} 2 \theta_{n\plus{}1}$, hence

$ \lim_{n \to \infty} \frac{2^n}{x_n} \equal{} \lim_{n \to \infty} 2^n \tan \frac{\theta_1}{2^{n\minus{}1}} \equal{} 2 \theta_1$

as desired.

The geometric explanation of (2):

Let $ x_n$ be the length of one side of a regular polygon with $ 6\cdot 2^n$ sides inscribed in a unit circle. From what we know of the regular hexagon, we have that $ x_0\equal{}1.$

Draw that polygon in the circle, and then bisect one of the sides. The bisector intersects the side at a distance of $ \sqrt{1\minus{}\left(\frac{x_n}2\right)^2}$ from the origin. Look outside of that original side to see two right triangles with legs equal to $ \frac{x_n}2$ and $ 1\minus{}\sqrt{1\minus{}\left(\frac{x_n}2\right)^2}$ and hypotenuse $ x_{n\plus{}1}.$ By the Pythagorean theorem,
\[ x_{n\plus{}1}^2\equal{}\left(\frac{x_n}2\right)^2\plus{}\left[1\minus{}\sqrt{1\minus{}\left(\frac{x_n}2\right)^2}\right]^2\equal{}2\minus{}2\sqrt{1\minus{}\left(\frac{x_n}2\right)^2}.\]
Hence, with a little more algebraic simplification.
\[ x_{n\plus{}1}\equal{}\sqrt{2\minus{}\sqrt{4\minus{}x_n^2}}.\]
Running the first few steps of this recursion, we get that

$ x_0\equal{}1$
$ x_1\equal{}\sqrt{2\minus{}\sqrt{3}}$
$ x_2\equal{}\sqrt{2\minus{}\sqrt{2\plus{}\sqrt{3}}}$
$ x_3\equal{}\sqrt{2\minus{}\sqrt{2\plus{}\sqrt{2\plus{}\sqrt{3}}}}$

We can show that this is (apart from the power of $ 2$ in the front) the very sequence we are asked to consider.

Since $ 6\cdot 2^nx_n$ is the perimeter of the $ n$-gon, we clearly have that $ \lim_{n\to\infty}6\cdot2^nx_n\equal{}2\pi,$ meaning that $ 2^nx_n$ tends to $ \frac{\pi}3.$

This sequence and process has been known since antiquity and was behind most ancient attempts to approximate $ \pi$ - including, I think, Archimedes's result that $ \textstyle 3\frac{10}{71}<\pi<3\frac17.$ Note that computing it requires only addition, subtraction, and the extraction of square roots - lots of square roots.

Let $ y_n\equal{}3\cdot 2^nx_n,$ so that $ y_n$ converges to $ \pi.$ Here are the values I get for $ y_n$ for $ n\equal{}0,1,\dots,7:$

$ 3.00000000$
$ 3.10582854$
$ 3.13262861$
$ 3.13935020$
$ 3.14103195$
$ 3.14145247$
$ 3.14155761$
$ 3.14158389$

If we then let $ z_n\equal{}\frac{4y_n\minus{}y_{n\minus{}1}}3$ (the first step in a Richardson extrapolation), we get that $ z_7\equal{}3.14159265356,$ which approximates $ \pi$ to within 10 decimal places. (I can't do too much better than that without doing something to avoid the bad roundoff error caused by the subtractions in the recursion for $ x_n.$)

What if we start with a square instead of a hexagon? Then we have $ x_0\equal{}\sqrt{2}$ but the recursion is the same: $ x_{n\plus{}1}\equal{}\sqrt{2\minus{}\sqrt{4\minus{}x_n}}.$ That gives us

$ x_0\equal{}\sqrt{2}$
$ x_1\equal{}\sqrt{2\minus{}\sqrt{2}}$
$ x_2\equal{}\sqrt{2\minus{}\sqrt{2\plus{}\sqrt{2}}}$
$ x_3\equal{}\sqrt{2\minus{}\sqrt{2\plus{}\sqrt{2\plus{}\sqrt{2}}}}$

and so on.

This time, it is $ \lim_{n\to\infty}2^{n\plus{}1}x_n\equal{}\pi.$


The theorem states;

Let $ (X_n: n\in\mathbb{N})$ a sequence of i.i.d. r.v. Then the sequence $ (\frac {1}{n}\sum_{k \equal{} 1}^nX_k: n\in\mathbb{N})$ converges a.s. iif $ \mathbb{E}[X_1]$ is finite. In that case
\[ (a.s.) \lim_{n\rightarrow \infty}\frac {1}{n}\sum_{k \equal{} 1}^nX_k \equal{} \mathbb{E}[X_1].
\]
I have a question regarding the proof of this theorem, I will write a part of the proof first;
Let's assume that $ (\frac {1}{n}\sum_{k \equal{} 1}^nX_k: n\in\mathbb{N})$ converges a.s. From $ \frac {X_n}{n} \equal{} \frac {1}{n}\sum_{k \equal{} 1}^nX_k \minus{} \frac {n \minus{} 1}{n}(\frac {1}{n \minus{} 1}\sum_{k \equal{} 1}^{n \minus{} 1}X_k)$, we have that $ \frac {X_n}{n}\rightarrow 0$ a.s. $ n\rightarrow \infty$. We then have that $ \mathbb{P}(\overline{\lim_n}\{|\frac {X_n}{n}|\geq 1\}) \equal{} 0$, which is equivalent with $ \mathbb{P}(\overline{\lim_n}\{|X_n|\geq n\}) \equal{} 0$. Since $ \{|X_n|\geq n\}$ are independent events $ n \equal{} 1,2,\dots$, for Borel 0-1 (???) law ( also know as Borel-Cantelli lemma) ), we have that $ \sum_{n \equal{} 1}^{\infty}\mathbb{P}(|X_n|\geq n) < \infty$....

I will stop with the proof. Now, my question is; how in the world can we can conclude that $ \sum_{n \equal{} 1}^{\infty}\mathbb{P}(|X_n|\geq n) < \infty$, and it states that that conclusion is from Borel 0-1 law! What I know about Borel 0-1 law is that, if we have a sequence of events $ (A_n: n\in\mathbb{N})$,  if $ \sum_{n \equal{} 1}^{\infty}\mathbb{P}(A_n) < \infty$ then $ \mathbb{P}(\overline{\lim_n}A_n) \equal{} 0$ ( in case we add assumption that $ A_n$ are independent and $ \sum_{n \equal{} 1}^{\infty}\mathbb{P}(A_n) \equal{} \infty$ then $ \mathbb{P}(\overline{\lim_n}A_n) \equal{} 1$ ). Now I really can't see how we can conclude the reverse statement, nor I have ever seen the proof for such statement and I can't say that it's obvious. Even my book doesn't give the proof for this statement, but what bothers me, is that the statement is frequently used in important theorems such as this one and in the Kolmogorov's theorem on three sequences. I hope someone can clarify this. Thanks in advance

Ok, I remembered this, we have $ \mathbb{P}(\overline{\lim_n}\{|X_n|\geq n\}) \equal{} \mathbb{E}[K_{\overline{\lim_n}\{|X_n|\geq n\}}]$ $ \equal{} \mathbb{E}[\limsup_n K_{\{|X_n|\geq n\}}]\geq \limsup_n\mathbb{E}[K_{\{|X_n|\geq n\}}] \equal{} \limsup_n \mathbb{P}(\{|X_n|\geq n\})$, but cause $ X_n$ are equally distributed we would have $ \mathbb{P}(\{|X_n|\geq n\}) \equal{} \mathbb{P}(|X_1|\geq n)$, but this would lead me to $ 0 \equal{} \mathbb{P}(\overline{\lim_n}\{|X_n|\geq n\})\geq \limsup_n \mathbb{P}(|X_1|\geq n)$, so this would imply that $ \mathbb{P}(|X_1|\geq 1) \equal{} 0$, but this make's even less sense, cause it would imply that $ \mathbb{P}_{X_1}$ "lives" only on $ ( \minus{} 1,1)$.
Please help me before I go mad with this  :wallbash_red:

Let us assume to the contrary that the sum diverges.  Since the events $ |X_i| > i$ are independent , it follows from Borel-Cantelli that almost surely infinitely many of the $ X_i$ are at least $ i$ in absolute value.

On the other hand, we know that
\[ |\frac {1}{n \plus{} 1} \sum_{i \equal{} 1}^{n \plus{} 1} X_i \minus{} \frac {1}{n} \sum_{i \equal{} 1}^n X_i| \equal{} |\frac {X_{n \plus{} 1}}{n \plus{} 1} \minus{} \frac {\sum_{i \equal{} 1}^n X_i}{n^2 \plus{} n}| \geq |\frac {X_{n \plus{} 1}}{n \plus{} 1}| \minus{} \frac {1}{n \plus{} 1} |\frac {\sum_{i \equal{} 1}^n X_i}{n}|. \]

Rearranging, we have
\[ |\frac {X_{n \plus{} 1}}{n \plus{} 1}| \leq |\frac {1}{n \plus{} 1} \sum_{i \equal{} 1}^{n \plus{} 1} X_i \minus{} \frac {1}{n} \sum_{i \equal{} 1}^n X_i| \plus{} \frac {1}{n \plus{} 1} |\frac {\sum_{i \equal{} 1}^n X_i}{n}|.
\]
If the average of the $ X_k$ is converging a.s., then the right hand side is tending to $ 0$ a.s., so the left hand side must also do so, and in particular must a.s. be less than $ 1$ only finitely often.


modular arithmetic
algebra unsolved
algebra

Consider sums of the form $\sum^{n}_{k=1}e_{k}k^{3}$ with $e_{k}\in \{-1,+1\}$. Is there such a sum with value $0$ for
a) $n = 2000$;
b) $n = 2001$.

Prove that it is only possible for $n \geq 12$ and $n = 0,-1 \pmod{4}$.

[i]Bundeswettbewerb 2000, Second Stage, problem 4[/i]

[i]Edited by Myth[/i]

I assume it is : e_k  {-1,+1}.

If I'm not wrong, it's possible to see that :

. k^3 + e1 (k+1)^3 + e2 (k+2)^3 + e3 (k+3)^3
is even for every combination   :D of ei  { -1,+1 }

. (4*k+1)^3,  (4*k+1)^3 + e(4*k+2)^3 (e = +-1) are odd

So the sum is odd for n=1 or 2 [4] as thus can't be 0.

[quote="belenos"]... combiniaison ... [/quote]
please don't be upset with me, but "combination" is the right word :)

Hi belenos,

could you write please more explicitely which part of the question you answer. Please also give some more arguments concerning your ideas.

I think belenos proved that no such sum can be 0 for n=1 or 2 (mod 4):

Let's assume n=4k+1. S(our sum) = S_1+S_2, where S_1= \sum{k=1->4k-3; 4|k-1}[e_k*k3+e_(k+1)*(k+1)3+e_(k+2)*(k+2)3+e_(k+3)*(k+3)3] and S_2=e_(4k+1)*(4k+1)3

It's easy to see that S_1 is even because every sum e_k*k3+e_(k+1)*(k+1)3+e_(k+2)*(k+2)3+e_(k+3)*(k+3)3 with 4|k-1 is even (2 of the terms are even and 2 are odd). It's also obvious that S_2 is odd because 4k+1 is odd. This gives us an odd S.

We do the same for n=4k+2. S_1 will be written in the same way, so it'll be even, and S_2 will be e_(4k+1)*(4k+1)3+e_(4k+2)*(4k+2)3, so it will be odd, S=S_1+S_2 being odd.

These 2 cases show that in order to have such a sum equal to 0 we must have n=0 or 3(mod 4).

Hi Orlando,

>> write please more explicitely
>> "combination" is the right word

Oh my god, so it's just like at school  :D

"Blnos ! Once again, check your spelling, write COMPLETE proofs, otherwise  \downto  \varepsilon^3 "

Thank you Grobber for your explicit proof.

please don't be upset with me :) I was just poiting out your French origins :D:D

I had a question about this which I forgot to ask:

Are we just supposed to prove that n must be >=12 and 0 or 3(mod 4) in oprder for a solution to exist, or are supposed to prove that for ANY n>=12 and 0 or 3(mod 4) a solution exists?

Prove the following equivalence:

\sum^(n)_(k=1) e_k * k^3 = 0 with e_k \in {-1,+1} <=> n = 0,-1 (mod 4) for n \geq 12.

Using the identity $n^{3}-(n+1)^{3}-(n+2)^{3}+(n+3)^{3}=12n+18$ we can show all $n=16k$ satisfy the condition. I guess the aim to invent  as many such identities as possible. It is not very integesting.
Other ingenius ideas?


geometry
parallelogram
search
angle bisector
geometry unsolved

In  a  quadrilateral $ ABCD$  where $ AD\equal{}BC$, $ AC$  and $ BD$ neets at $ O$ , and the angle  bisector of the angles $ DAB$ and $ CBA$  meets  at $ I$.  Prove that the midpoints  of $ AB,CD,$ and $ OI$  are colinear.

$ S \in AD \cap BC$, we have $ SI$ is the bisector of $ \angle ASB$
Let $ M,N$ be the midpoint of $ AB,CD$. Construc the parallelogram $ AMED, BMFC$. We have $ DECF$ is a prallelogram easily, so $ N$ is the midpoint of $ EF$. Note that $ ME\equal{}MF$, then $ MN$ is the bisector of $ \angle EMF$. So $ MN \parallel SI$. In addition, $ MN$ passes through the midpoint of $ OS$, so $ MN$ is the middle line of $ \triangle{OSI}$, and it passes through the midpoint of $ OI$, as desired

[quote="Hong Quy"]In  a  quadrilateral $ ABCD$  where $ AD \equal{} BC$, $ AC$  and $ BD$ neets at $ O$ , and the angle  bisector of the angles $ DAB$ and $ CBA$  meets  at $ I$.  Prove that the midpoints  of $ AB,CD,$ and $ OI$  are colinear.[/quote]

 This is problem

 http://www.mathlinks.ro/viewtopic.php?search_id=2018813849&t=167953

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 Babis

¿Why is the intersection $ MN$ & $ SO$ the midpoint of $ SO$?