Let $n \geq 2$ be an integer. Let $S$ be a subset of $\{1,2,\dots,n\}$ such that $S$ neither contains two elements one of which divides the other, nor contains two elements which are coprime. What is the maximal possible number of elements of such a set $S$?

[quote="dzeta"]It seems like this year they decided not to kill us with the last probem. :D
The aplication $f:S \rightarrow \{[n/2]+1,...,n\}$ which maps $x \in S$ into $2^{k}x \in (n/2,n]$ is a one-to-one mapping (since there are no $x,y \in S$ s.t. $x|y$). Also, since $(x,y)>1$ and $(x,y)|(f(x),f(y))$, it follows that $f(S)$ cannot contain two consecutive numbers. It follows easily that $|S| \leq [(n+2)/4]$. This can be achieved by $S=\{k$ even: $k > n/2\}$.[/quote]

It seems to great :!:

How can we prove that l$ S$l $ \leq [(n \plus{} 2)/4]$ :?:

[quote]What didn't you understand in dzeta's post?[/quote]

This conclusion about number of elements in $ S$?

Can anyone give a proof that does not depend on dzeta's solution?

BTW, does anyone have the actual competition solutions? (Specifically to this problem?)

Also, nice solution dzeta!

*To address Albanian eagle's question, I think dzeta showed that f had a bijective mapping between its elements and the even integers greater than $\lfloor\frac{n}{2}\rfloor + 1$. Hence, it suffices to maximize the latter set (which S was mapped to via the bijection), and that's just $\mid S\mid \leq \frac{n+2}{4}$.


I think u do [code] sqrt{2} [/code] maybe?

This should be moved to the [url=http://www.artofproblemsolving.com/Forum/index.php?f=123]LaTeX Forum[/url].

The code is [code]$\sqrt{x}$[/code] which shows up as $ \sqrt{x}$.

[quote="RoFlLoLcOpT"]This should be moved to the [url=http://www.artofproblemsolving.com/Forum/index.php?f=123]LaTeX Forum[/url].

The code is [code]$\sqrt{x}$[/code] which shows up as $ \sqrt {x}$.[/quote]
Just a note: The brackets arent necessary if you need just one number or variable under the sqrt. For more than one number or variable you need a bracket.

dude you should just click on an image and it probably tells you the latex code

[quote="maybach"]I can't seem to figure out LaTex for square roots.  For example, when I try to write "square root of two" in LaTex, it turns out as $ sqrt2$.  What do I need to write to get the LaTex right?[/quote]

$ \sqrt [3] {27}$

 It is \sqrt [3] {27}

oh, you just forgot to put a backward slash thingy.
you do [code]$sqrt4$[/code]
so this becomes $ sqrt4$
it's supposed to be [code]$\sqrt4$[/code]
which comes out $ \sqrt4$
and if you want to add radicals, you just do [code]$\sqrt[7]4$[/code]
which comes out $ \sqrt[7]4$

Yes!  It worked.  I can now write $ \sqrt[3]{y}$.  Thanks Mathfreak77


function
integration
real analysis
real analysis unsolved

find the continious function have the following property

$ f(x)\equal{}(1\plus{}e^x)(2008\plus{}\int^{x}_{0}\frac{f(x\minus{}t)}{1\plus{}e^{x\minus{}t}}dt)$

Differentiate both sides of equation and you'll get
$ f' \equal{} f \plus{} e^x \frac{f}{1\plus{}e^x}$
hence $ f \equal{} e^{x\plus{}C} (1\plus{}e^x)$
put it into equation and solve it for C.
The final answer is $ f(x) \equal{} \boxed{2008 \, e^x (1\plus{}e^x)}$.


Circle A is inscribed in a semicircle so that the diameter of circle A is the radius fo the semicircle. Circle B is externally tangent to circle A, and circle B is tangent to both the arc and teh diameter of the semicircle. If the length of the radius of the circle B is $ 3\sqrt{3}$, find the area inside the semicircle but outside of the 2 circles.

[hide="Hint"]Setting up an equation, we can see:
\[ {\left(2r \minus{} 3\sqrt {3}\right)}^2 \plus{} {\left(r \minus{} 3\sqrt {3}\right)}^2 \equal{} {\left(r \plus{} 3\sqrt {3}\right)}^2
\]Then:
\[ r\equal{}3\sqrt{3}\plus{}\frac{9}{2}\](We throw out the negative root.)
[/hide]

[hide="Solution"]Setting up an equation, we can see:
\[ {\left(2r \minus{} 3\sqrt {3}\right)}^2 \plus{} {\left(r \minus{} 3\sqrt {3}\right)}^2 \equal{} {\left(r \plus{} 3\sqrt {3}\right)}^2
\]Then:
\[ r \equal{} 3\sqrt {3} \plus{} \frac {9}{2}
\]
(We discard the extraneous negative root.)
Then, the region requested is:
\[ \frac{1}{2} \cdot \pi{\left(6\sqrt{3}\plus{}9\right)}^2\minus{}\pi{\left(3\sqrt{3}\plus{}\frac{9}{2}\right)}^2\minus{}\pi{\left(3\sqrt{3}\right)}^2 \\
\equal{}\pi\left(\frac{189\plus{}108\sqrt{3}}{2}\minus{}\left(27\plus{}27\sqrt{3}\plus{}\frac{81}{4}\right)\minus{}27\right) \\
\equal{}\pi\left( \frac{81\plus{}162\sqrt{3}}{2} \plus{} \frac{81}{4}\right) \\
\equal{}\boxed{\pi\left(\frac{243\plus{}324\sqrt{3}}{4} \right)}\]It feels like this answer is too complicated? Did I do something wrong?[/hide]

The answer should be one of the following:

A. 64$ \pi$
B. 81$ \pi$
C. 100$ \pi$
D. 132$ \pi$
E. 216$ \pi$

[quote="hohoho11"]Circle A is inscribed in a semicircle so that the diameter of circle A is the radius fo the semicircle. Circle B is externally tangent to circle A, and circle B is tangent to both the arc and teh diameter of the semicircle. If the length of the radius of the circle B is $ 3\sqrt {3}$, find the area inside the semicircle but outside of the 2 circles.[/quote]
Ummm, it's not written correctly, because if circle A is inscribed in the semicircle, then it is tangent to the arc of the semicircle and the diameter, however, if circle b is externally tangent to circle A, it cannot be also tangent to the arc and the diameter of the semicircle...
Unless I'm interpreting this wrong somehow.

[quote="123456789"][hide="Hint"]Setting up an equation, we can see:
\[ {\left(2r \minus{} 3\sqrt {3}\right)}^2 \plus{} {\left(r \minus{} 3\sqrt {3}\right)}^2 \equal{} {\left(r \plus{} 3\sqrt {3}\right)}^2
\]
Then:
\[ r \equal{} 3\sqrt {3} \plus{} \frac {9}{2}
\]
(We throw out the negative root.)
[/hide][/quote]
Where did u get the $ (2r\minus{}3\sqrt{3})$? You're implying that the external tangent is equivalent to the semi-circle's diameter minus the radius of circle B, which is not.

@zhangpeijin:

That's what I thought @ first, but its completely possible if B is inside the semicircle as well.  :)

No, but then they would be [b]internally[/b] tangent, but not [b]externally[/b] tangent.

no, I see how it can be done.  Just because circle B hits the arc doesn't mean it hits the center of the arc

To clarify,
[asy]size(200);
real r = 12 * 3^.5;
pair O=(0,0),A=(-r,0),B=(r,0),C=(0,r/2),D=(-(6*12*3)^.5,r/4),F=r*expi(160.528*pi/180);
picture semi = new picture;
draw(semi,Circle(O,r));
clip(semi,A--B--B+(0,50)--A+(0,50)--cycle);
add(semi);
draw(A--B); dot(O); dot(C); dot(D);
draw(Circle(C,r/2));
draw(Circle(D,r/4));
draw(O--C--D);
draw(O--F); draw((-(6*12*3)^.5,0)--D--(0,r/4));
label("\(A\)",C,NE);[/asy]

[hide="Solution"]
The correct equation should be, by the Pythagorean Theorem twice,
\[ (2r \minus{} 3\sqrt {3})^2 \minus{} (3\sqrt {3})^2 \equal{} (r \plus{} 3\sqrt {3})^2 \minus{} (r \minus{} 3\sqrt {3})^2 \Longrightarrow r \equal{} 6\sqrt {3}
\]
The area is $ \pi\left(\frac {1}{2}(12\sqrt {3})^2 \minus{} (6\sqrt {3})^2 \minus{} (3\sqrt {3})^2\right) \equal{} 81\pi$.
[/hide]

Yeah, thats what i meant. I still don't really post diagrams, so I should have pointed out that b is only externally tangent to a, not anything else.

hey, how do you make such good diagrams!!! the only thing that i know how to use is paint...and well it doesnt turn out so well using that!

Eh, I've never learned asymptote...
I just draw a picture on MS paint or something.

Its not hard, its basically like other programing languages (it was written in C I believe), except with different functions.

Look at the .pdf on their website - it has a lot of basic and advanced information.


trigonometry
limit
integration
calculus
calculus computations

question i posed to some high school students (15 - 17 yr olds) yesterday ... time limit is [b]30[/b] minutes.... (i am too generous ! :D )

find the sum
\[{ \sum_{k=1}^ \infty \;\{ \dfrac{1} {e^2 + (2k\cdot \pi + 3)^ 2 }} \;+\; \dfrac {1}{e^2 + (2k\cdot \pi - 3)^ 2 }\}
\]

Let $ P(z) \equal{} e^2 \plus{} (2 z \pi \minus{} 3)^2$ . Note that your original sum is equal to $ \sum \frac {1}{P(k)}$ where $ k$ is summed over [u]non-zero[/u] integers. Now by the residue theorem this last expression is equal to $ \minus{} \pi \sum {Res} {\frac {1}{P(z)} \frac {\cos \pi z}{\sin \pi z}} \minus{} \frac {1}{P(0)}$ where the sum is over the two roots of $ P(z) \equal{} 0$. Computing this last (three terms sum) gives the answer (I am too lazy to do it, the constants looks ugly).

My Method
[hide]
Use $ \ce{sigma}$ from 1 to n and apply opp. of limit of a sum.
Then take limit n--->$ \ce{infinity}$
[/hide]

is there any general formula for $ \sum_{k\equal{}1}^{\infty}\frac{1}{x^2\plus{}ax\plus{}b}$?

high school students don't learn complex variables in this country......  :maybe:

[quote="0714446459923"]is there any general formula for $ \sum_{k \equal{} 1}^{\infty}\frac {1}{x^2 \plus{} ax \plus{} b}$?[/quote]
I know there is a general form for $ \frac{1}{k^2\plus{}a^2}$ for $ |a|<1$. But Im not sure about this.

Question:

Does $ \sum_{k\equal{}0}^{\infty}$ mean $ \lim_{n \to \infty} \sum_{k\equal{}0}^{n}$ ?

Similar question for $ \int_{0}^{\infty}$

[quote="FOURRIER"]Question:

Does $ \sum_{k \equal{} 0}^{\infty}$ mean $ \lim_{n \to \infty} \sum_{k \equal{} 0}^{n}$ ?

Similar question for $ \int_{0}^{\infty}$[/quote]
Yes.


calculus
derivative
function
algorithm
real analysis
real analysis solved

My question concerns excercise 3.18 in Rudin's Principle of mathematical analysis:

In it one is asked to define a recursive series by setting

x[n+1]= ((p-1)/p)*x[n]+ (a/p)*x[n]^(-p+1) ,

where p>2 and a>0,

and to describe the behaviour of the resulting sequences {x[n]}.

I have managed to show that x[n] converges to the pth root of a as n tends to infinity.

My question is: What can be said about the rate of convergence? I am not able to find any rate for p larger or equal to 3.

All help is needed and welcome.

-Laura S

Based on numerical experimentation, convergence is clearly second-order.

The iteration is of the form $x_{n+1}=f(x_n),$ with $f(x)=\frac{p-1}px+\frac apx^{1-p}.$ Assume the fixed point (there is clearly exactly one positive fixed point for any $p\neq0$) is $y.$ We should have second order convergence if $f'(y)=0$ and $f''(y)\neq0$

Calculating the appropriate values, we have $\frac yp=\frac apy^{1-p},$ so $y^p=a$ and $f'(y)=\frac{p-1}p+\frac {a(1-p)}py^{-p}=\frac{p-1}p+\frac{1-p}p=0.$
Finally, $f''(y)=-\frac{a(1-p)}{y^{p+1}}=\frac{1-p}{\sqrt[p]{a}}\neq0.$
This all applies for any $p\neq0,1$ (For $p=1$ convergence is instant). Use Taylor's theorem and the fact that $f$ has plenty of derivatives to finish the result.

If I understood correctly, in the above you showed that indeed the iteration converges to the pth root of a.

If so, I'm sorry for not being clear enough: What I need is a proof of the rate of convergence. Or does what you showed imply second order convergence?

-Laura

[quote]Or does what you showed imply second order convergence?
[/quote]
Yes, it does. Reviewing, we have the following:
$f(y)=y.$
$f'(y)=0.$
$f''(y)\ne 0.$ In what follows, let $b=\frac{f''(y)}2.$

Then by Taylor's Theorem, $f(y+h)=y + bh^2 + o(h^2).$

Let $x_n=y+e_n$. Then $x_{n+1}=y +be_n^2+o(e_n^2).$ Thus, $e_{n+1}=be_n^2+o(e_n^2).$

Hence, $\frac{e_{n+1}}{e_n^2}\to b.$ That's second order convergence.

Incidentally, the iteration can be algebraically rearranged to

$x_{n+1}=x_n-\frac{x_n^p-a}{px_n^{p-1}}.$

This form makes it clear that this is nothing more or less than Newton's method applied to the equation $x^p-a=0.$

As long as we are looking for a simple zero of a function (which we are here), Newton's method is always second order or better.

The $p=2$ case is an ancient (Babylonian?) algorithm for finding square roots.


topology
invariant
advanced fields
advanced fields unsolved

let $ k$ be a field and $ \mathbb{A}^n : \equal{} Spec ~ k[x_1 , ... , x_n]$. I want to show that $ \mathbb{A}^1 \times \mathbb{A}^1$ is not homeomorphic to $ \mathbb{A}^2$. any ideas?

uhm..
what are closed sets in $ \mathbb{A}^1$? so what are closed sets in $ \mathbb{A}^1\times\mathbb{A}^1$?

[deleted]

I misunderstood the question (see -oo-'s response below). Terribly sorry.

@grobber: :?:  what should be the spectrum of a topological space like $ \mathbb{A}^1 \times \mathbb{A}^1$? your proof seems to show that $ Spec k[X,Y]$ is not homeomorphic to $ Spec (k[X] \times k[Y])$, but that's not the question. $ \mathbb{A}^1 \times \mathbb{A}^1$ and $ \mathbb{A}^2$ are both connected (besides, also irreducible with exactly one generic point).
@ma_go: http://en.wikipedia.org/wiki/Spectrum_of_a_ring
@all: I first thougt that a homeomorphism must map the generic point to the generic point, and that this makes the map unique since the points are dense (and hence equal to the canonical continuous map $ \mathbb{A}^2 \to \mathbb{A}^1 \times \mathbb{A}^1$, which is not injective). but this argument is nonsense in this non-hausdorff situation :(.

Sorry for that.

I hope I understand correctly now: you want to show that the product topology on $ \mathbb A^1\times\mathbb A^1$ (where each $ \mathbb A^1$ has the Zarsiki topology) does not coincide with the Zariski topology on $ \mathbb A^2$? This seems to work then:

The closed irreducible subsets of dimension $ 1$ of $ \mathbb A^1\times\mathbb A^1$ are precisely the "lines", i.e. the sets of the form $ \{p\}\times\mathbb A^1$ or $ \mathbb A^1\times\{p\}$ for some (non-generic) point $ p$ in one of the $ \mathbb A^1$'s. It follows that no three distinct such sets can have a point in common. Clearly, this is a property that a homeomorphism would preserve. In $ \mathbb A^2$, however, we can easily find three such sets: the sets of prime ideals which contain $ x, y$, and $ x\plus{}y$ respectively will do.

The supremum of the lengths of all chains of closed irreducible subsets, but that's not crucial. Forget about dimension, if you prefer, and replace "closed irreducible subsets of dimension $ 1$" with "closed, irreducible, non-empty subsets which are neither the whole set nor singletons". In $ \mathbb A^1\times\mathbb A^1$ these are the lines.

ok. I think I understand your classification of these sets in $ \mathbb{A}^1 \times \mathbb{A}^1$ if we assume that $ k$ is algebraically closed. but it wasn't  :maybe:

edit2: after a couple of hours, I still have no idea why the sets look like that ..

What is the problem about $ k$ not being algebraically closed? The underlying topological space of $ \mathbb A^1$ consists of a set of cardinality $ |k^\mathrm{alg}|$ carrying the cofinite topology, and the generic point. owk

how does this help me? I have just not understood grobbers proof, k beling algebraically closed or not.

I still do not understand what you meant by

[quote="-oo-"]I think I understand ... if we assume that $ k$ is algebraically closed[/quote]

Every nontrivial closed subset $ Z$ of $ \mathbb A^1\times\mathbb A^1$ is contained in $ F_1\times\mathbb A^1\cup\mathbb A^1\times F_2$ for some nontrivial closed (hence finite) subsets $ F_1,F_2\subset\mathbb A^1$. If $ Z$ is irreducible, it must be contained in one of these lines. owk

I think I've got it. thank you grobber and owk :).

another invariant: in $ \mathbb{A}^1 \times \mathbb{A}^1$, two closures of non-generic non-closed points interesect in almost one point, but not in $ \mathbb{A}^2$.


search
number theory
advanced fields
advanced fields theorems

I don't really understand how mathematicians today use the word "technical." Most of the time, I see things like "this theorem can be generalized in such-and-such way, but we only prove a simpler case because it is just a technical matter to make the generalization." The idea here seems to be "the generalization requires nontrivial effort but doesn't provide any new insights." On the other hand, I remember that Tao or someone wrote about a recent groundbreaking paper in number theory and referred to it as a "technical tour-de-force" (in what seemed to be a very complimentary fashion). So in this case, does it mean that the authors overcame huge "technical" obstacles to prove a relatively simple result? Or does it mean that the technical details that they worked out actually resulted in new and useful techniques?

What does "technical" really mean? Is it good or bad?  :o

If you search Mathscinet ("Review text" contains "technical"), you'll find many more examples of how the word is used. To me, "technical" describes not a result (which could be very important or not so), but its proof. It's a proof full of details the meaning of which is not clear to the reader. It's entirely possible, though, that another reader will not find the same proof all that technical. When I see a proof that begins with
[quote]pick $\Phi_{3,1}$ so that its Fourier transform is equal to $1$ on the set $[-13/10, 8/(5\sqrt{2})]$ and supported in $[-13/100-1/100,8/(5\sqrt{2})+1/100]$[/quote]
followed by six pages of computations, I would call that technical. [Source: Lemma 5 in the paper "The disc as a bilinear multiplier" by Grafakos and Li]. But since it eventually leads to a nice result referred to in the title, one can view that $8/(5\sqrt{2})+1/100$ as a necessary evil.


Μου φάνηκε τόσο όμορφη αυτή η κατασκευή τριγώνου που της αξίζει ένα νούφαρο. Για να μην την χαλάσω αφήνω το παραμύθι να μιλήσει μόνο του στο attachment

Αν δεν αποκρυπτογραφήαατε το παραμύθι σας προτείνει την κατασκευή τριγώνου σπό τα $ a , bc\equal{}k^2 , B\minus{}C\equal{}\phi$ Η λύση σε λίγες μέρες για να φανερωθεί το νούφαρο!!


Hi everyone [currently at the Chemistry Olympiad Scholar School].

I don't quite understand how to apply MO Theory to Polyatomic atoms [note I have limited resources]. So can someone explain? Thanks.

Well, firstly find the basic shape of the molecule(from vsepr or other theories)
Calc the no of electrons used in sigma bonds and lone pairs and by subtracting this from the total determine no of electrons which can participate in pi bonding.
Count the no of AOs which can take part in pi bonding. Combine these to give the same no of mo which are delocalized over all atoms. Decide whether mos are bonding are anti bonding and feed the  appropriate no of pi electrons in to  the mos (2 per mo) the orbitals with lowest energy are filled first. The no of pi bonds can be determined from mos which have been filled

My problem is with how to determine whihc orbitals overlap. How do we apply SDLC? How do we determine whether the MO's are degenerate or not? How do we draw the MOs?


I doubt it.  That would be cruel, and it can mess up participants.  After all, what's worse than trying to prove something that's false?

I suspect all problems on the current round are true...

[quote="ZzZzZzZzZzZz"]Are there ever any trick questions on the USAMTS that ask to prove something that doesn't exist or is untrue?  Like a trick question to see if people can catch the error?[/quote]
Imagine what a pain that would be to grade.

Over a decade ago, the USAMTS was mailed out with an impossible problem due to a typographical error. (That was back when the contest was distributed by mail.) We sent out a correction a few days later when the error was spotted. Nowadays we independently check the problems for errors before they are posted.

The purpose of the USAMTS is to challenge your mathematical abilities. Trick questions do not serve that purpose.

Erin Schram
USAMTS grader


algebra
polynomial
function
algebra unsolved

[quote="Akashnil"]$ a_1,a_2,\cdots a_{2009}$ are not constants, they are variables and the polynomial is supposed to be [url=http://mathworld.wolfram.com/MultivariatePolynomial.html]multi-variable polynomial[/url] whose variables are $ a_1,a_2,\cdots a_{2009}$. So $ P$ is not a constant function in your example.[/quote]
Is it so?? :maybe:
 i think the polynomial was of the form $ P_i(x_1,x_2,\cdots,x_n)$and $ P_i(a_1, a_2, .. , a_{2009})$were its value when$ x_i \equal{} a_i$
please check that
[quote="ll931110"]...

with[b] all real numbers[/b] $ a_1, a_2, .., a_{2009}$?[/quote]
if you are correct,

most probably he meant" [i][b]for[/b][/i] all real numbers $ a_1, a_2, .., a_{2009}$?" :maybe:
But putting $ a_i \equal{} 0$except for $ a_1$ we have $ a_1 (P_1(a_1,0,0,...) \plus{} 2009P_2(a_1,0,0,...) \plus{} ... \plus{} 2P_{2009}(a_1,0,0,...)) \equal{} 1$ which is not possible since $ \frac {1}{a_1}$ is not polynomial. :wink:

At first, I said: 'Just for fun'. So you don't need to be complicated like that.

P/S: sorry, my English is very poor  :(


Let a,b,c be positive reals with ab+bc+ca=1.  Show the following (all sums are cyclic)

\[ \sum \frac{1+bc}{\sqrt{1+a^2}} \leq 2\sqrt{3} \]
\[ (a^2+b^2+c^2+3a^2b^2c^2+6) \sum \frac{1}{(1+a^2)^3} \geq 9 \]
\[ \sum \sqrt{1+bc} \leq \sqrt{3} \sqrt{a^2+b^2+c^2+3} \]

On second thought, some of these might be trivial... oh well

[quote="ThAzN1"]
On second thought, some of these might be trivial... oh well[/quote]

Oh yeah, as the third one..

$\sum \sqrt{1+bc} \leq \sqrt{3(1+bc+1+ca+1+ab)}=\sqrt{3(3+1)}=2\sqrt{3}$

while

$\sqrt{3} \sqrt{a^2+b^2+c^2+3} \geq \sqrt{3} \sqrt{(ab+bc+ca)+3}=\sqrt{3} \sqrt{4}=2\sqrt{3}$

Done :D


Given that $ 3^8\cdot5^2 \equal{} a^b$, where both $ a$ and $ b$ are positive integers, find the smallest possible value for $ a \plus{} b$.

$ \textbf{(A)}\ 25 \qquad \textbf{(B)}\ 34 \qquad \textbf{(C)}\ 351 \qquad \textbf{(D)}\ 407 \qquad \textbf{(E)}\ 900$
[hide]
In this problem, a+b will be smallest when they are closest together.

Since there is a $ 5^2$, b must be 2. Therefore, $ a \equal{} 405$.

$ 405 \plus{} 2 \equal{} 407, \boxed{(D)}$[/hide]

$ 3^8\cdot5^2 \equal{} (3^4\cdot5) (3^4\cdot5) \equal{} 9^4\cdot25 \equal{} 405^2$

$ 405 \plus{} 2 \equal{} 407$ $ \longrightarrow \boxed{D}$

Thats the best I can do you guys. Probably a better answer.

[hide]
From the fundamental theorem of arithmetic $ a\equal{}3^{e_1}5^{e_2}$ so the only possible sollutions are
$ (3^85^2)^1$ and $ (3^45)^2$ so the answer is $ 3^45\plus{}2\equal{}407\longrightarrow \boxed{D}$[/hide]

[hide="Solution"]We have $ 3^85^2\equal{}3^6(3\cdot5)^2\equal{}(3^3(3\cdot5))^2\equal{}(27\cdot15)^2\equal{}405^2$
So $ a\plus{}b\equal{}405\plus{}2\equal{}407\implies\boxed{\text{D}}$.[/hide]


If $ p$ is a positive integer, then $ \frac {3p \plus{} 25}{2p \minus{} 5}$ can be a positive integer, if and only if $ p$ is:

$ \textbf{(A)}\ \text{at least }3 \qquad\textbf{(B)}\ \text{at least }3\text{ and no more than }35 \qquad\textbf{(C)}\ \text{no more than }35$
$ \textbf{(D)}\ \text{equal to }35 \qquad\textbf{(E)}\ \text{equal to }3\text{ or }35$

$ \frac{3p\plus{}25}{2p\minus{}5} \equal{} 1 \plus{} \frac{p\plus{}30}{2p\minus{}5}$

$ 2p\minus{}5 | p\plus{}30$ implies $ 2p\minus{}5 \le p\plus{}30$ so $ p \le 35$

$ \frac{3p\plus{}25}{2p\minus{}5} > 0$ implies $ p > 2.5$ so $ p \ge 3$

now we can test numbers in the interval... i guess
better yet, set $ \frac{p\plus{}30}{2p\minus{}5}\equal{}k$, $ k$ is between $ 1$ and $ 33$
$ 2pk\minus{}5k \equal{} p\plus{}30$ so $ (2k\minus{}1)p \equal{} 30\plus{}5k$ ...

$ p\equal{}5$ works, so does $ p\equal{}9$ so I'll go with $ B$.


floor function
ceiling function
inequalities

If $ a,b,c$ are the sidelenghts of a triangle and $ x\equal{}\frac{(a^2\plus{}b^2\plus{}c^2)(a\plus{}b\plus{}c)}{a^3\plus{}b^3\plus{}c^3}$ , find $ [x]$ .

You can't, because $ a\equal{}b\equal{}c\equal{}1$ gives $ [x]\equal{}3$ but $ a\equal{}3$ $ b\equal{}4$ $ c\equal{}5$ gives $ 25/9$ whose floor is 2.

I think... $ [x]$ might mean ceiling...

in the future use [code]\lfloor text \rfloor[/code] or [code]\lceil text \rceil[/code] to make it clearer.

Solution
[hide]
$ \dfrac{(a^2 \plus{} b^2 \plus{} c^2)(a \plus{} b \plus{} c)}{a^3 \plus{} b^3 \plus{} c^3} \minus{} 2 \equal{} \dfrac{(b \plus{} c \minus{} a)(c \plus{} a \minus{} b)(a \plus{} b \minus{} c) \plus{} 2abc}{a^3 \plus{} b^3 \plus{} c^3} > 0$

and

$ 3 \minus{} \dfrac{(a^2 \plus{} b^2 \plus{} c^2)(a \plus{} b \plus{} c)}{a^3 \plus{} b^3 \plus{} c^3} \equal{} \dfrac{(a^3 \plus{} b^3 \plus{} c^3 \minus{} a^2b \minus{} b^2c \minus{} c^2a) \plus{} (a^3 \plus{} b^3 \plus{} c^3 \minus{} ab^2 \minus{} bc^2 \minus{} ca^2)}{a^3 \plus{} b^3 \plus{} c^3}\ge 0$

The first inequality is true for $ a$, $ b$ and $ c$ being triangles sides. The second inequality is true because of rearrengment inequality.
Both of them prove that the answer is $ 2$.
[/hide]

:)

[quote="rofler"]You can't, because $ a \equal{} b \equal{} c \equal{} 1$ gives $ [x] \equal{} 3$ but $ a \equal{} 3$ $ b \equal{} 4$ $ c \equal{} 5$ gives $ 25/9$ whose floor is 2.[/quote]

Well this was the idea , rofler . I didn't say $ [x]$ is unique . This was the solution $ [x]\equal{}3$ , if the triangle is equilateral , and $ [x]\equal{}2$ in the others case .

oooh you are right I forgot the to see the equality case...


A,B,C are three positive real numbers such that ABC=1

Prove that A/(A+1)(B+1)  +B(/B+1)(C+1)  +C/(C+1)(A+1)  >=3/4

[hide]

The inequality becomes

$4(a+b+c+ab+bc+ca)\geq 3(abc+a+b+c+ab+bc+ca+2)$

$a+b+c+ab+bc+ca\geq 6$, that is true for AM-GM, using $abc=1$[/hide]


geometry rectangle perimeter conics parabola calculus derivative

Please can anybody help me with this problem?

The area of a rectangle with base of length x and perimeter 100 is A(x)=x(50-x). Find the maximum possible area of this rectangle by finding the point on the parabola A= x(50-x) at which the tangent line is horizontal.
A=xy

If you know derivatives, then use the following fact: the slope of the tangent line at $x$ is $A'(x)=50-2x$. Equating this to $0$ gives $x=\dots$.

If you don't know derivatives... the vertex of a parabola is halfway between two points that lie on its intersection with a horizontal line (such as $y=0$). So, the tangent line is horizontal in the middle of the segment $[0,50]$.


I think everythink can =a^2\(b+c) +b^2\(a+c) + c ^2(a+b) :lol:

Do not double post!  Continue your discussion here: [url]http://www.artofproblemsolving.com/Forum/viewtopic.php?p=660068[/url]


It Says:

Find an expression for the fugacity coefficient when the attractive interaction is dominant in a Van der Waals gas, and the pressure is low enough to make the approximation
\[ \frac {{4ap}} {{(RT)^2 }} < < 1
\]
Evaluate the fugacity for ammonia at 10.00atm and 298.15 K.

[b]Hint:[/b] Assume b=0 and neglect 4ap as compared to
\[ (RT)^2
\]
in the expression of
\[ V_m
\]
from the Van der Walls equation.

Sorry for the mess I don't know how to properly align the equations with the text. But the "Hint" is one sentence and so is the problem.

Thank You

I don't know if Im on the right track but here is what i got so far:

\[ \ln \varphi  \equal{} \int\limits_0^p {{{Z \minus{} 1} \over p}} dp\]
And
\[ f \equal{} \varphi p\]
And you know that:
\[ 4ap <  < (RT)^2\]
The Van der Waals equation states that (ignoring b of course):
\[ p \equal{} {{RT} \over {V_m }} \minus{} {a \over {V_m^2 }}\]
Using algebra you get:
\[ V_m  \equal{} {{\sqrt {R^2 T^2  \minus{} 4ap}  \plus{} RT} \over {2p}}\]
or
\[ V_m  \equal{}  \minus{} {{\sqrt {R^2 T^2  \minus{} 4ap}  \minus{} RT} \over {2p}}\]
neglecting 4ap and using first (positive) equation for Vm you get:
\[ V_m  \equal{} {{\sqrt {R^2 T^2 }  \plus{} RT} \over {2p}}\]
which reduces to:
\[ V_m  \equal{} {{RT} \over p}\]

This is as far as i get get, it does not make sense since Vm=RT/p is the molar volume of a perfect gas and Im dealing with a real gas.
I just need help in solving for Z, this will let me plug it into the very top equation and get the fugacity coefficient.
\[ Z \equal{} {{pV_m } \over {RT}}\]

I finally managed to answer this question. If you what me to post the answer then I will do so, just leave a reply and I will post it when I visit this site again.

Please post your answer, I think we are in the same class with the same professor


1. A set of seven integers has a median of 73, a mode of 79, and a mean of 75. What is the least possible difference between the maximum and minimum value in the set?
[hide="Answer"]8[/hide]
2. If $ f(x)\equal{}\frac{4x\plus{}1}{3}$, what is the value of $ (f^{\minus{}1}(1))^{\minus{}1}$?
[hide="Answer"][b]2[/b], but I get $ \frac53$..[/hide]

[quote="Ihatepie"][quote="Bomboyage"]1. A set of seven integers has a median of 73, a mode of 79, and a mean of 75. What is the least possible difference between the maximum and minimum value in the set?
[hide="Answer"]8[/hide]
2. If $ f(x) \equal{} \frac {4x \plus{} 1}{3}$, what is the value of $ (f^{ \minus{} 1}(1))^{ \minus{} 1}$?
[hide="Answer"][b]2[/b], but I get $ \frac53$..[/hide][/quote]

71,71,73,73,79,79,79[/quote]
Can you express your thought process? How can you do this so fast?

[quote="Ihatepie"]Well first, try to fit as many 79s as you possibly can since you want the lowest number to be as high as possible.
so you have _,_,_,73,79,79,79

so the remaining three have to add up to 215.a third of that is 72 1/3
so 71,71,73 (or 71,72,72)[/quote]

If you fit in too many 79's, the lowest number is [i]lower[/i]. :roll:

The reason 72 and 73 do not work as the lowest number is that they rely on exactly 2 79's in their set, and that would clearly violate 79 being the mode.


Let be $ a,b,c>0$ such that $ abc\equal{}\frac{4}{9}$ . Show that :

$ \frac{a^4}{ab(a\plus{}b)}\plus{}\frac{b^4}{bc(b\plus{}c)}\plus{}\frac{c^4}{ca(c\plus{}a)}\ge \frac{3}{(a\plus{}b\plus{}c)^2}$

AM-GM:
$ LHS \geq 3\sqrt [3]{\frac {a^2b^2c^2}{(a \plus{} b)(b \plus{} c)(c \plus{} a)}}$
So we have to prove that:
$ (a \plus{} b \plus{} c)^6\geq \frac {81}{16}(a \plus{} b)(b \plus{} c)(c \plus{} a) \rightarrow[(a \plus{} b) \plus{} (b \plus{} c) \plus{} (c \plus{} a)]^6 \geq 324(a \plus{} b)(b \plus{} c)(c \plus{} a)$

We set $ a \plus{} b \equal{} x,b \plus{} c \equal{} y,c \plus{} a \equal{} z$
So we have to prove that:

$ (x \plus{} y \plus{} z)^6 \geq 324xyz$
But $ (x \plus{} y \plus{} z)^6 \geq 3^6x^2y^2z^2$
So we have that
$ xyz \geq \frac {4}{9}$
Which is:
$ (a \plus{} b)(b \plus{} c)(c \plus{} a)\geq abc$
Which is true....
Please check it.....

I think your inequality has no equality case. A slight refinement can be made with Holder's inequality resulting in
\[ \sum{\dfrac{a^3}{b(a\plus{}b)}}\ge\dfrac{6}{(a\plus{}b\plus{}c)^2}\]

With equality only when $ a\equal{}b\equal{}c$.

:)


real analysis
complex analysis
complex analysis solved

let $(a_n)$ be a sequence of complex number and $S_n=a_1+a_2+..+a_n$.
Suppose that $limsup |a_n|^{1/n}=L$. Prove that $limsup |S_n|^{1/n}=L$

I think its not correct: shouldn't it be $\limsup |S_n|^{1/n}=\max(1,L)$ ¿
Counterexample: $a_n=\left(\frac{1}{2}\right)^n$

Proof: look at the the power series (and let $a_n$ start with $a_0$)
$A(z)=\sum_{n=0}^{\infty} a_n z^n$ and $B(z)=\frac{1}{1-z}=\sum_{n=0}^{\infty} z^n$
Then we have $C(z)=A(z) \cdot B(z) = \sum_{n=0}^{\infty} S_n z^n$
After a theorem of Cauchy we have that the convergence radius of $A(z)$ is $\frac{1}{L}$, that of $B(z)$ is $1$ and so we get for that of $C(z)$ the value $\min\left(1,\frac{1}{L}\right)$ and so $\limsup |S_n|^{1/n}=\max(1,L)$

(The theorem says: the convergence radius of $\sum_{n=0}^{\infty} b_n z^n$ is exactly $\frac{1}{\limsup |b_n|^{1/n}}$ were $0$ and $\infty$ are allowed to occur)

Edit: it's necessary that (when converging) $A(1) \neq 0$

[Moderator edit: just tweaking the TeX code a little: \limsup and \max instead of limsup and max, \left( and \right) instead of ( and ).]

[quote="ZetaX"]
Then we have $C(z)=A(z) \cdot B(z) = \sum_{n=0}^{\infty} S_n z^n$
After a theorem of Cauchy we have that the convergence radius of $A(z)$ is $\frac{1}{L}$, that of $B(z)$ is $1$ and so we get for that of $C(z)$ the value $min(1,\frac{1}{L})$ and so $limsup |S_n|^{1/n}=max(1,L)$[/quote]
oh sorry it was as you said. Nevertheless I think you've just prove that the radius of C(z) >= min(1,\frac{1}{L}). You've probably used a theorem that I don't know, isn't it. In any case it was the way I found to solve the pb  :)

Yes alekk, you are right.

New version:
Since $A(1) \neq 0$ (as assumed) $C(z)$ has a pole at 1 ( from $B(z)$ ).
Changing $A(z)$ and $C(z)$ and seting $B(z)=1-z$ in the original proof gives that the radius of convergency of $A(z)$ is greater or equal to that of $C(z)$.  Both statements together then give the rest of the proof.

Now it's also easy to see that in the case $A(1)=0$ the radius of the circles of convergency of $A(z)$ and of $C(z)$ are equal.

Edit: I changed this post, because it was wrong.


analytic geometry
geometry
parallelogram

The coordinates of three vertices of a parallelogram are (-3, 1), (2, 5), and (4, 1). Find the sum of the coordinates of the fourth vertex which is in the third quadrant.

I think i may have done this wrong.
[hide]$-4$[/hide]

[quote="Ubemaya"][hide]
(2, 5) is 2 to the left and 4 up from the point (4, 1). The point 2 to the left and 4 up from (-3, 1) is (-5, 5). The sum is 0.[/hide][/quote]
Third quadrant...
That tells you that the answer is negative...
What quadrant is it counted as if it's on an axis?

If it's on an axis it's not in any quadrant.

[hide]From $(2,5)$we go down $4$ and left $5$ to get $(-3,1)$. So from $(4,1)$ we go down $4$ and left $5$ to get $(-3,-1)$. The sum is $-3-1=-4$.[/hide]


search
linear algebra
linear algebra unsolved

$ A,B \in M_n (K)$. Prove that:

1.  $ {det (adj A)} \equal{} (det A)^{n\minus{}1}$
2. $ {adj (adj A)} \equal{} (det A)^{n\minus{}2}A$
3. $ {adj(AB)} \equal{} {adj(B)adj(A)}$
4. $ {adj(A^{T})} \equal{} {(adj A)}^{T}$

Most of these have been posted. I haven't yet found the most productive thing to search for.

[color=red][Two links :
http://www.mathlinks.ro/Forum/viewtopic.php?t=153199 (I used "Kent Merryfield" for the author field and "adj" as a keyword)
http://www.mathlinks.ro/Forum/viewtopic.php?t=171741 (I used "darij grinberg" for the author field and "adj" as a keyword)
-mathmanman][/color]

it s done with 1. and 2.
But wat bout statement 3 and 4?^^
anyway thank you alot

no one interested in statement 3 and 4?? :(


geometry
floor function
trigonometry
function
probability
complex numbers
angle bisector

Hi,

This is the last round of the COP, namely Championship Round. There are 15 questions and you'll be given a week to finish and send me the answers. The answers may include radicals, $\pi$, or any other mathematical symbols because your answer must be [b]exact[/b]. Use of calculators or online site (but not online mentor or help from other people) are allowed. Any approximation not mentioned on the problem will counted as wrong.

For instance: If the answer was 2.57, making it to 3 or 2 will give you no credit.

Your score will only be based on the number of correct answer. You're free to guess any.

[b]Answers must be send to me by 4/1/05.[/b]

[hide="Problems"]

1. Let $z_1$ and $z_2$ denote two complex numbers such that their product is 53 and sum is 14, find the value of $z_1^4 + z_2^4$.

2. On $\Delta {COP}$, $\angle {O} = 45^\circ$ and $\angle {C} = 60^\circ$. The angle bisector of $\angle {C}$ meets at $A$ on $\overline {OP}$ with $\overline {CA} = 20$. From $A$, we mark point $B$ that $B$ is 21 units from $A$ and $AB \perp AP$. From $O$, we mark point $D$ that $D$ is 10 units from $O$ and $\overline {OD} \perp \overline {OA}$. Let area of 6-sided polygon $CODABP$ as $\beta$. Find $\lfloor \beta \rfloor$.

3. Let $p$ be "principle number" such that $p_n = p_1 \cdot p_{n-1}$. Given that $p_1 = p_2 = 4$, how many digits are in $p_{2005}$?

4. James has a number $rstu$. Changing the order of the digit, we can find 23 other numbers with digits $r,s,t$ and $u$. The sum of those 23 numbers is 59871. Find the maximum sum that can be resulted between $rstu$ and a number with digit $r,s,t$ and $u$. (For example, if $rstu = 1234$, then the maximum sum is $1234 + 4321 = 5555$.)

5. In $\Delta {CAR}$, $\tan {\angle {ACR}} = \frac {333}{652}$ and altitude from $\angle C$ meet at $\overline {AR}$ at $B$. If $\overline {AB} = 5$ and $\overline {BR} = 13$, then what is $\lfloor \sqrt {[CAR]} \rfloor$?

Such that $[x]$ denotes the area of the triangle.

6. There are 12 points distributed around the circle so that all $\binom {12}{2}$ pairs of points are connected by segments, no three of them being concurrent. Denote the number of intersection of points inside the circle as $\psi$. What is $\psi$ in base 100?

7. Find the sum of all values such that $r(x) = 0$ where $0 \leq x \leq 2\pi$.

$r(x) = \frac {\csc^4 x - 2 \csc^2 x + 1}{\csc x (\csc x - \sin x) + \frac {\sin x - \cos x}{\sin x} + \cot x}$

8. Given that $c_1,c_2,c_3, ... c_n$ such that $c_1+c_2+c_3+c_4...+c_n = 17$. For each $z > 0$, let $s(z)$ be the minimum sum of the following expression:

$\sqrt {c_n^2 + d_n^2} + \sqrt {c_{n-1}^2 + d_{n-1}^2} + .... \sqrt {c_1^2 + d_1^2}$.

such that $d_n + d_{n-1} + d_{n-2} + ... d_1 = z$. Suppose that both $s(z)$ and $z$ are both positive integers. What does $z$ equal in this case?

9. The society of Free-Math-Solving-Society is looking for 6 members. Let's say that there are 10 people who are going for the candidate. How many ways can this possible such that Larry and Jane serve together and Laura and John don't serve with each other?

10. Find the exact value of:

$1 - \frac {1}{10} - \frac {1}{18} - \frac {1}{28} - \frac {1}{40} - \frac {1}{54} - \frac {1}{70} ...$

11. David are playing "three-coin-or-one-head" game. In "three-coin-or-one-head" game, each person flips three coins simulataneously and the process keeps going until at least one head is shown. What is the probabibility that David wins by other two coins also coming up with heads on this last flip?

12. Find all the values of $y$ such that $y = 1 - y + y^2 - y^3 + y^4 - y^5 + ...$.

13. Amanda has 8-digit phone number such that first four digit of her phone number is 1 seven, 1 eight, and 2 nines. The fifth digit is either 2,3, or 5. There is no information available for her sixth digit number but her seventh digit is 9 minus whatever the six digit is. The eighth digit is any of odd digits. If the total possible number of her phone number is $\varphi$, what is $\sqrt {\varphi}$?

14. Let's say there are three guys with different functions. Abe has function $\cos^2 x$. Bob has function $\cos^2 {2x}$. Cyn has function $\cos^2 {3x}$. Given that period of $x$ is $[0, \pi)$, find all values of $x$ such that when Abe's function, Bob's function, and Cyn's function is added, it equals the smallest positive integer. List them.

15. *Let $S$ be the set of all values assumed by $\left\{\frac{p}{q}\right\}$ for $q$ a positive integer not exceeding 2005. The sum of the elements of $S$ can be expressed as $\frac{m}{n}$, where $m$ and $n$ are relatively prime positive integers. Determine the remainder obtained when $m+n$ is divided by 1000. ($\{x\}$ denotes the fractional part of $x$, given by $x - \lfloor x \rfloor$ where $\lfloor x \rfloor$ is the floor of $x$.)*

*Thanks to TMildorf for contributing this problem.

[/hide]

Are you going to post the solutions to the proof round?

EDIT: You're right. I changed it. Thanks.

Please do not look at this thread until the champion from the proof is drawn.

I have to grade them so please be patient.

Just note that you can look at this thread now. :lol:

Good luck to our three contestant! :)

A number.

I do not much about that problem because it's not made by myself. I just know it's pretty *tough*. He said it's not too bad but yup.

P can be any number... It really doesn't matter.

Wow, I completely forgot about the deadline, what time today are they due?
Also, could you explain what #11 is saying?
Thanks.

#11:

I have to admit that the although this is nice question, it's poorly written. So, I'll try to sound it better.

The probability that a person would win the game by having three heads is what the question is asking. [b]BUT[/b], you have to know that this probability may go on forever, and nobody would win the game until like 12134th flips (just random number). So, find this infinite probability and call it $\frac {j}{k}$.

I know I kind of gave this away (maybe or maybe not) but this question DOES need some rephrasing and I'm making it fair. :D


Al can build a toy house in $ A$ hours, and Bob can build the same model toy house in $ B$ hours. Al works alone until a toy house is one-third finished, and Bob then helps him to finish the toy house. How many hours were required to build this toy house? Give the answer as a single factored fraction in terms of $ A$ and $ B$.

[hide]In 1 hour, Al accomplishes $ \frac{1}{A}$ of the task and Bob accomplishes $ \frac{1}{B}$ of the task.  Working together, they accomplish $ \frac{1}{A}\plus{}\frac{1}{B}$ of the task in an hour, so it takes $ \frac{1}{\frac{1}{A}\plus{}\frac{1}{B}}$ hours to complete it.  However, they only work together for 2/3 of the time.  For the other 1/3, Al works $ \frac{A}{3}$ hours.  So the total time is $ \frac{A}{3}\plus{}\frac{2}{3(\frac{1}{A}\plus{}\frac{1}{B})}$.[/hide]

If you take lingomaniac88's answer and factor and combine over one denominator, you get:
[hide="Answer"]$ \frac {a(a \plus{} 3b)}{3(a \plus{} b)}$[/hide]


advanced fields
advanced fields unsolved

Hello everyone. Can you help me with this?

I am looking for examples of elliptic curves over rational numbers of the form Y^2 = X^3 + B, B integer,  with rank > 0.
I did a brief research and I suspect that the curves of that form can't have rank>1.
Is that true?
If it is, how can we prove this?

Thank you in advance!

The curve $ y^2\equal{}x^3\minus{}11$ has a rank equal to $ 2$.


find all natural number have 5 digits  such that :
n divisible by 2 ; n+1 divisible by 3;n+2 divisible by 4;n+3divisible by 5.

You are a long tme on this forum now, but your titles are as dumb as at the beginning!
1. the words "prove that" don't even occure or have anything to do with the problem.
2. this title has no or even less meaning to the reader.
3. this is a combinatorics problem¿! (I suggest moving it to Intermediate)


The area of square $ ABCD$ is 100 square centimeters, and $ AE \equal{} 2$ cm. What is the area of square $ EFGH$, in square centimeters?

[asy]pair a,b,c,d,e,f,g,h;
a=(0,17);
b=(17,17);
c=(17,0);
d=(0,0);
e=(5,17);
f=(17,12);
g=(12,0);
h=(0,5);
label("$A$",a,N);
label("$B$",b,N);
label("$C$",c,S);
label("$D$",d,S);
label("$E$",e,N);
label("$F$",f,E);
label("$G$",g,S);
label("$H$",h,W);
draw(a--b--c--d--cycle);
draw(e--f--g--h--cycle);[/asy]

By similarity \[ \overline{AB}\equal{}\overline{BF}\equal{}\overline{CG}\equal{}\overline{DH}\equal{}2\] Because the area of the square is $ 100$ units, the sides are each 10 units. Hence, sides longer than $ \overline{AB}$'s length are of length 8.

Through Pythagorean Theorem, we get all sides of square $ EFGH$ to be $ \sqrt{68}$.

Hence the area is \[ \sqrt{68}^2\equal{}\boxed{68}\]

Or we can have the area of ABCD = the area of 4 triangle and 1 square inside it
Since $ AE \equal{} 2$, $ EB \equal{} 10 \minus{} 2\equal{} 8$
Thus, we have:
$ 100 \equal{} 4*(\frac{1}{2} * 2 *8) \plus{} x$
$ 100 \equal{} 32 \plus{}x$
$ x \equal{} 68$


if in a triangle ABC (with sides a,b,c and angles A,B,C)
    cos a = cos b. cos c  + sin b. sin c. cos A
    cos b = cos a. cos c  + sin a. sin c. cos B
    cos c = cos b. cos a  + sin b. sin a. cos C
THEN PROVE THAT----
       |a/(sin a)|   = |b/(sin b)| =  |c/(sin c)|

wait i don't understand...the statement u are telling us to prove is exactly the law of sines...


For triangle $ABC$, $P$ and $Q$ satisfy $\angle BPA + \angle AQC=90^{\circ}$. It is provided that the vertices of the triangle $BAP$ and $ACQ$ are ordered counterclockwise(or clockwise). Let the intersection of the circumcircles of the two triangles be $N$ ($A \neq N$, however if $A$ is the only intersection $A=N$), and the midpoint of segment $BC$ be $M$. Show that the length of $MN$ does not depend on $P$ and $Q$.

[quote="lightrhee"]This is the second problem of [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=422487]the second round of 19th Korean Mathematics Olympiad[/url].

2.
For triangle ABC, P and Q satisfy $\angle BPA + \angle AQC=90^{\circ}$. It is provided that the vertices of the triangle BAP and ACQ are ordered counterclockwise(or clockwise). Let the intersection of the circumcircles of the two triangles be N($A \neq N$, however if A is the only intersection $A=N$), and the midpoint of segment BC be M. Show that the length of MN does not depend on P and Q.[/quote]

Under the problem's situation we have $\angle (BP {,\;} PA) + \angle(AQ {,\;} QC) = 90^\circ \Rightarrow$
$\angle(BN {,\;} NC) = \angle(BN {,\;} AN) + \angle(AN {,\;} NC)= \angle(BP {,\;} PA) +\angle(AQ {,\;} QC) = 90^\circ \Rightarrow$
in any case $\angle BNC = 90^\circ \Rightarrow$ as $MN$ is the median of a triangle with one right angle ( could you say the name of such triangle , please ? :blush: ) so $MN = \frac{1}{2} BC =$ const.


If lim(a_n)=a , lim(b_n)=b and a<b then proof that a_n<b_n starting from some n.
Specially, if lim(a_n)=a<b then  starting from some n a_n<b

P.S. in lim n goes to infinity

I need help with this homework. Please, help me  :(

And your question is? :maybe:

Besides, limit problems go in High School Basics/Intermediate Problems.

I accidentally skipped one word when I was typing. Sorry.

@Moderators: Please move this to the appropriate forum.


Convergence
calculus
derivative
algebra
polynomial
limit
real analysis

Just to check that I understood the problem right: the terms of each series are the partial sums of the previous series, correct?

Then the answer is "Yes, of course!". Indeed, we need the sum of each series to be $0$, so, replacing $x_{0}+x_{1}+\dots+x_{m}$ by $-\sum_{k>m}x_{k}$, we see that the second series can also be written as $-\sum_{k\ge 0}kx_{k}$ (provided that we can justify the regrouping of the terms) and so on. Thus, it is enough to find $x_{n}$ decaying faster than any power of $n$ such that $\sum_{k\ge 0}k^{p}x_{k}=0$ for all $p$. This is a very classical problem (so classical that I am surprised they published it as a problem in AMM :?). One way to look at it is to consider the function $f(t)=\sum_{k\ge 0}x_{k}e^{ikt}$, which reduces the question to that about the existence of an infinitely smooth up to the boundary analytic in the unit disk function all whose derivatives vanish at $1$. $f(z)=e^{-(1-z)^{-1/2}}$ would be a classical example of such thing.

I want to raise a possible objection to this solution.  I'm a little confused by the notation here, but if we define $ f(z) \equal{} \sum_{k\geq 0} x_k z^k$, I agree that we must have all left-sided derivatives of $ f$ vanish at $ 1$.

What I don't agree with is that this condition is sufficient.  I think that we are implicitly moving a limit inside a sum somewhere, which is not valid in general.

fedja seems to suggest that the coefficients of $ e^{\minus{}1/(1\minus{}z)^{\minus{}1/2}}$ solve the problem, but it actually does not appear that $ SSA$ converges in this case: http://www12.wolframalpha.com/input/?i=Series[e^(-1.%2F(1-x^2))%2F(1-x)^2%2C+{x%2C0%2C200}]

(sorry, couldn't get the URL to highlight correctly)

OK, let's do it in more detail. Suppose that $ \sum_{k\ge 0} k^p x_k\equal{}0$ for all integer $ p\ge 0$ and $ x_k$ decay faster than any power. Let $ S_k\equal{}\sum_{j\equal{}0}^k x_j\equal{}\minus{}\sum_{k\plus{}1}^\infty x_j$.

[b]Claim 1[/b] Since $ x_k$ decay faster than any power, so do $ S_k$.

[b]Claim 2[/b] $ \sum_{j>k}k^p|x_j|<\plus{}\infty$ for all $ p>0$.

[b]Claim 3[/b] For every $ p$, we have $ \sum_{k\ge 0}{k^p S_k}\equal{}\sum_{j>k}k^p x_j\equal{}\sum_{j\ge 0}\Bigl(\sum_{k\equal{}0}^{j\minus{}1}k^p \Bigr)x_j$.

[b]Claim 4[/b] The sum in parentheses is a polynomial in $ j$ of degree $ p\plus{}1$.

If all claims are valid, we conclude that $ \sum_{k\ge 0}k^p S_k\equal{}0$ for all $ p$ and we can repeat everything again and again. Which claim seems suspicious?

As to Wolframalpha and $ SSA$, the conditions are terribly unstable. If you just make an error $ \varepsilon$ in $ x_0$ and have all other $ x_j$ correct, in $ SA$, you'll have error $ \varepsilon$ in each term and in $ SSA$ the error will grow linearly, and so on. I am extremely suspicious about computer simulations with such instability issues ;).

Yes, these were the details I was looking for, thanks.  You make a very good point about error building up in that power series.

No, wait, my original objection remains.  This is the claim I am contesting:

[b]Claim:[/b] If $ f(x) \equal{} \sum_{k\geq 0} a_k x^k$ is a power series with radius of convergence $ 1$, and $ \lim_{x\to 1^{\minus{}}} f^{(p)} (x)\equal{}0$ for all nonnegative integers $ p$, then $ \sum_{k\geq 0}k^p a_k \equal{} 0$ for all nonnegative integers $ p$.

It may be that you're using a slightly different condition.

Strangely enough, I posted the power series for a different one than what you mentioned, but I am not seeing what, if any, properties distinguish the two...

Well, I'm looking at it as at a Fourier series of a $ C^\infty$ function on the circle, so I take the derivatives along the circle rather than along $ [0,1]$. The analiticity inside the circle just ensures that all coefficients with negative indices are $ 0$. The condition you mentioned (Abel-Poisson summability) is, of course, insufficient for convergence without extra assumptions but, if we know in advance that the coefficients decay fast, everything is OK. The power $ \frac 12$ is there exactly to ensure that $ f\in C^\infty$ on the circle. If I had written $ e^{ \minus{} 1/(1 \minus{} z)}$ instead, the radial derivatives would all be $ 0$ but the series would diverge.

Thanks for the explanation–this should serve as a good reference to those of us who need work on our Fourier analysis...


1,2,2,3,3,3,4,4,4,4........... Find the 2008th number in the sequence....
Kindly explain the method employed and its variants.....

Usually, looking at small cases helps with sequence problems.

Let's find the last position a number $ n$ occurs.

1 occurs last at position 1
2 occurs last at position 1+2 = 3
3 occurs last at position 1+2+3 = 6
4 occurs last at position 1+2+3+4=10

From here, we can deduce that $ n$ occurs last at position $ 1 \plus{} 2 \plus{} 3 \plus{} \dots \plus{} n \equal{} \frac{n(n\plus{}1)}{2}$

Therefore, we have to find $ n$ such that all other smaller numbers occur last before 2008. If we approximate $ n(n\plus{}1)$ by $ n^2$, we get that $ n \approx 64$. Then it's a simple matter of trial and error to reach $ \boxed{n \equal{} 63}$. If the problem requires proof, you'll need to show that $ \frac{62 \cdot 63}2 < 2008 \le \frac{63 \cdot 64}2$ (after explaining what this has to do with the sequence at hand).

Generalized as [url=http://www.svsu.edu/mmpc/mmpcep2c.pdf]Problem 2 on the MMPC Part 2[/url].


limit
ceiling function
integration
calculus
calculus computations

\[ \mathop {\lim }\limits_{n \to \infty } \frac{1}
{n}\sum\limits_{k \equal{} 1}^n {\left( {\left\lceil {\frac{n}
{k}} \right\rceil  \minus{} \frac{n}
{k}} \right) \equal{} \gamma }\]

It is not hard to show that the limit is equal to $ \int_{0}^{1} \left( \left\lceil \frac{1}{x} \right\rceil \minus{} \frac{1}{x} \right) \, dx$, and the rest calculation is straightforward.


inequalities
function
calculus
derivative
linear algebra
matrix
symmetry

$ a,b,c$ are reals ,$ 0<\equal{}a,b,c<\equal{}1$.then prove that $ a/(1\plus{}bc) \plus{}b/(1\plus{}ac)
\plus{}c/(1\plus{}ab) <\equal{}2$[/b]

[hide="Solution"]$ \displaystyle \frac {a}{1 \plus{} bc} \plus{} \frac {b}{1 \plus{} ca} \plus{} \frac {c}{1 \plus{} ab} \leq \frac {a}{1 \plus{} abc} \plus{} \frac {b}{1 \plus{} abc} \plus{} \frac {c}{1 \plus{} abc} \equal{} \frac {a \plus{} b \plus{} c}{1 \plus{} abc} \leq 2$[/hide]

[hide="Alternatively..."]This function is convex in each of its variables, and therefore must attain its extremum at a boundary point, so it suffices to check that this is true for ordered pairs of 0's and 1's, from which the result is obvious. [/hide]

Zhero, this isn't true -- consider, for example, $ (x \minus{} 1/2)^2$ over $ [0,1]$ -- it achieves a global minimum at $ x \equal{} 1/2$, but it is convex throughout.

The possible conditions that imply that the extrema are on the endpoints are monotone increasing and monotone decreasing.

Ack, I keep making that typo; my apologies. :blush: I meant to say [i]maximum[/i].

Convex functions do attain their [i]maxima[/i] at their boundaries, correct? In which case, this method would be correct?

Yeah, that's true, and that's all that matters here. I guess what you'd have to show then is that the function is everywhere convex. How do you plan on doing that?

If by, "[i]Take the second derivative?[/i]" you mean finding the eigenvalues (in terms of a,b,c) of the Hessian matrix, and showing that they are all positive for all $ a,b,c \in [0,1]$, then yes, that works.

There's really no need to get into the details of that here and now, but if [b]Zhero[/b] (or anyone else) has another approach in mind, it's certainly worth sharing.

What? I don't think you need to show that it is everywhere convex. You can argue along the lines of the following. Suppose b,c are fixed. Then $ f(x,b,c)$ is a function in $ x$ in a single variable which you can easily show to be convex.

Thus $ f(a,b,c)\le \text{max}\, \{ f(0,b,c),f(1,b,c)\}$

Then $ f(0,b,c)\equal{}b\plus{}c\le 2$.

Then maximize $ f(1,b,c)$ holding $ c$ fixed, etc.

[quote="Zhero"]Could you tell why we need to show that it is convex everywhere? Does it not suffice to show that it is convex over $ [0,1]^3$?  :maybe:

My multivariable calc is very shabby; basically, the only convexity test I know (which I [i]believe[/i] is correct, though it may not be; could you verify?) is that a function is convex if it is convex in each of its variables.

In this instance, by symmetry, it suffices to show that this function is convex in $ a$ when the variables lie in $ [0,1]^3$, which is clearly true.

Would that be enough for a proof?[/quote]

If you fix two variables we can show that
$ \frac{a}{1\plus{}a}$ is convex and then the result follows, that the function is convex and the maximum occurs at endpoints.


function
algebra
domain
logarithms
complex analysis
absolute value
complex analysis unsolved

In preparation for discussion of carleson's theorem and what not, there is an exercise where I can't get the details right, so I would like some help.

Let $ \gamma$ be an arc with one endpoint $ e^{it}$ and all other points contained in some Stolz angle $ \Gamma_\sigma(e^{it})$. If $ f$ is bounded and analytic in $ \mathbb{D}$ and $ f(z)$ has the limit $ L$ as $ z$ approaches $ e^{it}$ along $ \gamma$, then $ f$ has the nontangential limit $ L$ at $ e^{it}$.

There are several proofs but I have no idea of what you already know. Can you show what you attempted and we'll discuss if it can be finished along those lines?

I am trying to choose a particular sequence of functions analytc in the Stolz angle which converge pointwise on the arc to $ L$, so that we can apply Vitali and get locally uniform convergence of the sequence, so that we can hopefully well enough compare the limit on nontangential sequences with that of the particular sequence.

Okay, I just tried to formulate my idea for the past 10 minutes. Do you understand what I mean? As for what I know, I basically only know what I learned in a half-semester course in complex analysis 1.5 years ago (and I don't remember everything that well, unfortunately). So not very much advanced. On the convergence of analytic functions we did basic theorems like Montel, Vitali and the other even more basic ones.

The difficulty with that approach is that you cannot get a [b]fixed[/b] arc on which you have convergence to $ L$. What you get instead is that you have a sequence of uniformly bounded functions in the right half-plane (just map the disk to half-plane) such that they are closer and closer to $ L$ on [b]variable[/b] arcs starting from $ 0$ and contained within a fixed angle. Still, not everything is lost. You can look, for instance, at the intersection points of your arcs with circles of rational radii and assume without loss of generality that those intersection points also converge to some fixed points thus giving you an infinite set in the interior of the domain on which the convergence takes place (you'll have to use uniform equicontinuity of your functions inside the plane to justify passing to the limit, of course), which is enough for locally uniform convergence. Not the most elegant way, but definitely a workable one.

[quote="fedja"] You can look, for instance, at the intersection points of your arcs with circles of rational radii and assume without loss of generality that those intersection points also converge to some fixed points thus giving you an infinite set in the interior of the domain on which the convergence takes place[/quote]
I don't understand this. Do you mean that the infinite set consists of the fixed points? What do you mean by fixed point here? Why do the intersection points converge to them? Heh, sorry, I guess I don't understand what you mean :)

OK, let me be more explicit. I'll split the solution into 5 steps. See if you can fill in the details for each step. If not, we'll discuss it.

1) You can replace the disk with the right half-plane $ \mathbb C_\plus{}$ and assume that your boundary point is $ 0$.
2) Consider the family of functions $ f(tz)$ with $ t>0$. It suffices to show that they converge to $ L$ uniformly on compact subsets of $ \mathbb C_\plus{}$ as $ t\to 0\plus{}$.
3) To this end, it suffices to show that every sequence $ t_n\to 0\plus{}$ has a subsequence $ t_m$ such that $ f(t_mz)$ converge to $ L$ on a set $ E\subset C_\plus{}$ having an accumulation point in $ \mathbb C_\plus{}$ (if you don't know this theorem, we'll discuss it but I believe it is one of the theorems you mentioned or something equivalent to it).
4) $ f(t_n z)$ are close to $ L$ on some arcs $ \gamma_n$ emanating from $ 0$, contained in a fixed angle, and crossing the unit circle. Let $ z_{nj}$ be some point on those arcs such that $ |z_{nj}|\equal{}1\minus{}\frac 1j$. We can pass to a subsequence and assume that $ z_{nj}\to z_j$ for each $ j$ as $ n\to\infty$.
5) $ f(t_n z_{nj})$ tend to $ L$. Hence $ f(t_nz_j)$ tend to $ L$.
The end.

Ok, I understand now. Thank you very much!

The problem is that I was trying to model a solution thinking that it was basically the same as if $ \gamma$ was radial. When $ \gamma$ is radial then you don't have the problem of dealing with different $ \gamma_n$'s, although there's not really very much of a difference in the solution.

What can be said about this if we merely have $ f \in H^p$, I would presume for $ p\geq 1$ in order to be able to say anything.

all functions from $H^p$ where $0<p \leq \infty$ has nontangential limits almost everywhere.
The sense of this theorem is that function from $H^p$ is less than nontangential maximal function wich in angle is bounded by Hardy's maximal function wich maps $L^p \to L^p$ and hence is weak type $1-p$. So by standart methods around proving existence of limit a.e. works alse here. In the case $p<1$ just by Blashke product everything will be fine. (because it do not change nontangential limits.)
of course the proof of this theorem include more then 50 pages, but the sense all of this page is this this 5 string.

If you meant an analogue of your original problem, it doesn't hold in $H^p$. You may have the radial limit at a point without having the non-tangential one. Just think of what the Poisson extension of the wave
[asy]
size(100);
D((-10,0)--(10,0),black);
DPA((-10,0)--(-4,0)^^(-4,-4)--(-2,-4)^^(-2,4)--(2,4)^^(2,-4)--(4,-4)^^(4,0)--(10,0),red+linewidth(1));
[/asy]
looks like in the upper half-plane and combine such waves of different sizes with appropriate coefficients to get the logarithm of the absolute value of an $H^p$ function that is bounded on the imaginary semi-axis but not on $y=x/10$, say. Then multiply it by something bounded and slowly decaying as $z\to 0$.


Find all the natural solutions of the equation:

x^2+8*y^2+6xy-3x-6y=3.

Well... I guess I have a solution...

x^2+8*y^2+6xy-3x-6y=3 <=> (x+2y)(x+4y)-3(x+2y)=3 <=>
(x+2y)(x+4y-3)=3

So we have four cases:
1) x+2y=3 and x+4y-3=1, with  the solutions: x=2 and y=1/2
2) x+2y=1 and x+4y-3=3, with the solutions: x=-4 and y=5/2
3) x+2y=-3 and x+4y-3=-1, with the solutions: x=-8 and y=5/2
4) x+2y=-1 and x+4y-3=-3, with the solutions: x=-2 and y=1/2.

so we don't have natural solutions for this diophantine equation.
Hope I am right! :D

Sorry for that.. I guess there should only be two cases to study because cases 3 and 4 can't have natural solutions because the sum of two natural numbers can never be negative so you don't have to solve the two systems...  :D


[quote="abdurashidjon"]Here I will give problem and solution lets discuss it how to guess thjis  way
 [color=blue]Problem[/color]
    $f(x+y)=f(x)(y)$ find all functions $f$ for all real numbers.
[color=blue]Solution[/color]
    Here  the solution starts
Setting $x=y=\frac{t}{2}$ we get the function as follows $f(t)=f^2(\frac{t}{2})$
Now take both sides of functions as $ln$ after setting it becomes as
 $lnf(x+y)=lnf(x)+ln(y)$.. Let $lnf=g$ then our statement becomes $g(x+y)=g(x)+g(y)$ it can be found as $g(x)=cx$ so $lnf(x)=cx$ then $f(x)=e{}^c{}^x$ where $c$ is constant. But seems easy others too I will post them but how to guess is the problem[/quote]
There are many ($2^C$,C - continium) functions satisfyed: $g(x+y)=g(x)+g(y)$. They are constucted by Hamel's basis in R(Q). If exist x, were g(x) is continuosly, then g(x) continiusly in R and g(x)=cx.

Tahnks for reply Rust But my question is that how to guess touse $ln$,$log$,$lg$. And one more problem how can we get to prove and find is the function bounded or not?
 Abdurashid


Let $ a;b;c$ be real numbers satisfying $ a\plus{}b\plus{}c \geq 3$
Prove that $ a^4\plus{}b^4\plus{}c^4\plus{}3ab\plus{}3bc\plus{}3ca \geq 2a^2\plus{}3b^2\plus{}2c^2\plus{}3abc$

[quote="chien than"]Let $ a;b;c$ be real numbers satisfying $ a \plus{} b \plus{} c \geq 3$
Prove that $ a^4 \plus{} b^4 \plus{} c^4 \plus{} 3ab \plus{} 3bc \plus{} 3ca \geq 2a^2 \plus{} 3b^2 \plus{} 2c^2 \plus{} 3abc$[/quote]

[quote="Kouichi Nakagawa"]$ a^4 \plus{} b^4 \plus{} c^4 \plus{} 3ab \plus{} 3bc \plus{} 3ca \minus{} (3a^2 \plus{} 3b^2 \plus{} 3c^2 \plus{} 3abc)$
$ \equal{} (p^4 \minus{} 6q^2 \minus{} 4(p^2 \minus{} 2q)q \plus{} 4pr) \plus{} 3q \minus{} (3(p^2 \minus{} 2q) \plus{} 3r)$
$ (p \equal{} a \plus{} b \plus{} c, \ q \equal{} ab \plus{} bc \plus{} ca, \ r \equal{} abc)$
$ \equal{} p^4 \minus{} 4qp^2 \minus{} 3p^2 \plus{} 4rp \plus{} 2q^2 \plus{} 9q \minus{} 3r$
$ \ge \frac {1}{9} \left(5 p^4 \plus{} 3 p^3 \minus{} 20 q p^2 \minus{} 27 p^2 \minus{} 12 q p \plus{} 18 q^2 \plus{} 81 q\right)$
$ \left(r \ge \frac {1}{9}\left(4pq \minus{} p^3\right) \ \ \text{by Schur's ineq}\right)$
$ \ge \frac {1}{9} \left(\frac {p^4}{3} \minus{} p^3\right) \ \ \left(q\in\left( \minus{} \infty, \frac {p^2}{3}\right] \ \text{by} \ q\ge\frac {p^2}{3}\right)$
$ \equal{} \frac {1}{27} p^3 (p \minus{} 3) \ge 0 \ \ (\because p\ge 3)$[/quote]

I think you didn't notice some condition(and mistake):
   First of all $ a,b,c$ are real numbers. So therefore $ \left(r \ge \frac {1}{9}\left(4pq \minus{} p^3\right) \ \ \text{by Schur's ineq}\right)$  - is not valid.(becouse Schur's ineq. is for nonnegative real numbers!)
  Secndly, $ \ \ \left(q\in\left( \minus{} \infty, \frac {p^2}{3}\right] \ \text{by} \ q\ge\frac {p^2}{3}\right)$ is wrong becouse
$ \frac{p^2}{3} \ge q$
$ \leftrightarrow$
$ a^2\plus{}b^2\plus{}c^2 \ge ab\plus{}ac\plus{}bc$

[quote="SaYaT"]Secndly, $ \ \ \left(q\in\left( \minus{} \infty, \frac {p^2}{3}\right] \ \text{by} \ q\ge\frac {p^2}{3}\right)$ is wrong becouse
$ \frac {p^2}{3} \ge q$
$ \leftrightarrow$
$ a^2 \plus{} b^2 \plus{} c^2 \ge ab \plus{} ac \plus{} bc$[/quote]

Sorry.
I forgot to be nonnegative real numbers about Schur's ineq.

And I was typo, secondly.
Because, $ q\in\left( \minus{} \infty, \frac {p^2}{3}\right]$ is $ q\le\frac {p^2}{3}$.
So, I think that secondly is right essentially.


Dude wth is up with people coming on from other states and asking stupid questions, yet forgetting to include stupid answers?

LOL mewto. missouri is probably the most lively forum and everybody goes to it

everyone should come to virginia now, since it's like the most coolest state.

P.S. Virginia is going to win Nats


calculus
integration
algebra
polynomial
algebra unsolved

given that $ a_0,a_1,....a_n$ are integers and the equation
$ a_0x^n \plus{} a_1x^{n \minus{} 1} \plus{} ..... a_n \equal{} 5$ has 4 distinct integral roots,
find the no of integral roots of the equation
$ a_0x^n \plus{} a_1x^{n \minus{} 1} \plus{} ..... a_n \equal{} 7$

Let $ P(x) \equal{} a_0x^n\plus{}\cdots \plus{}a_n\minus{}5$.
Then, there exist pairwise distinct integers $ a,b,c,d$, and positive integers $ u,v,w,t$, and a polynomial $ Q$ with integer coefficients such that
$ P(x) \equal{} (x\minus{}a)^u(x\minus{}b)^v(x\minus{}c)^w(x\minus{}d)^tQ(x)$ and $ Q(x)$ has no integer roots (otherwise $ P$ would have more than four integer roots).

For any integer $ m$ :
- If $ m \in \{a,b,c,d \}$, we have $ P(m)\equal{}0 \neq 2.$
- If $ m \notin \{a,b,c,d \}$, the integers $ m\minus{}a,m\minus{}b,m\minus{}c,m\minus{}d$ are nonzero and pairwise distincts, thus $ |(m\minus{}a)(m\minus{}b)(m\minus{}c)(m\minus{}d)| \geq 4$. since $ Q(m)$ is a nonzero integer, we deduce that $ |P(m)| \geq 4$. It follows that $ P(m) \neq 2$.

Thus, the equation $ a_0x^n\plus{}\cdots \plus{}a_n \equal{} 7$ has no integer root.

Pierre.


limit
trigonometry
real analysis
real analysis unsolved

hint:
[hide]If $a_{1}=\cos x$,then $a_{2}=\cos \frac{x}{2}$[/hide]
 :P


Find all ordered pairs $a,b,c\in(0,1)$, satisfying  $a^2+b^2+c^2+2abc=1.$

Let a=cosA,b=cosB,c=cosC such that

    A+B+C=180degree,   0<A,B,C<90degree

then a*a+b*b+c*c+2abc=1[/list][/code][/quote][/u][/i][/b]


modular arithmetic
LaTeX
algorithm
number theory
relatively prime
Euclidean algorithm

Sorry for number 2  :blush: I meant the last 2 digits with the question marks so your guest was perfect.
And I'm looking forward to the solution of the last problem.  :lol:

[quote="abacadaea"][quote="Anaxerzia"][quote="dogcat"][list]$ 42^{600}\equiv? (\mod 13)$
$ 39^{39^{39}}= ...??$
$ 105x\equiv 741 (\mod 809)$ $ x\equiv ?$[/list][/quote]

[hide]1) We can write $ 42^{600}\equiv3^{600}\equiv(3^{12})^{50}\pmod{13}$. Since $ 3^{12}\equiv1\pmod{13}$ from Fermat's Little Theorem, $ (3^{12})^{50}\equiv1^{50}\equiv\boxed{1}\pmod{13}$.

2) Here you have not provided a mod to work in, and it is pretty much impossible to calculate this without a computer. However, I'm going to assume you want a tens and units digit:

The problem is equivalent to finding the modulo 100 residue of $ 39^{39^{39}}$, so let $ 39^{39^{39}}={x}$ Note that $ \phi(100) = 40$ so $ (a^{40})^{n}\equiv1\pmod{100}$, where $ a$ is relatively prime and $ n$ is an integer, from Euler's theorem. Also note that $ 39^{39}+1$ is a multiple of 40, since ${ 39^{39}\equiv{-1}^{39}\equiv-1\pmod{40}}$. Thus from Euler's theorem $ 39^{39^{39}+1}\equiv{1}\pmod{100}$. Also note that $ 39^{39^{39}+1}= 39^{39^{39}}\cdot39 = 39x$. So we have $ 39x\equiv1\pmod{100}$. Solving for $ x$, we find that $ x\equiv{59}\pmod{100}$, so the tens and units digit of $ 39^{39^{39}}$ is $ \boxed{59}$. Another (very long) way to do this is to find a pattern in tens and units digits of powers of 39. It should repeat after the 40th term  :wink:.

3) Takes too long to do..  :ninja: [/hide][/quote]

look for a pattern and put #2 in mod 100[/quote]

What? Did you even read my solution?

what? oh whoops. i thought that was for number 3 :blush:

[quote="Anaxerzia"]Another (very long) way to do this is to find a pattern in tens and units digits of powers of 39. It should repeat after the 40th term  :wink:.[/quote]

Actually, it will always repeat after the $ 20^{th}$ term  :wink:

[hide="Proof"] $ \bmod 25$ the last two digits repeat every $ 20$ terms, whereas $ \bmod 4$ the last two digits repeat every $ 2$ terms; hence, $ \bmod 100$ the sequence repeats $ \text{lcm}(20, 2) = 20$ terms by Chinese Remainder Theorem. [/hide]
#3 is tedious to do in $ \text{\LaTeX}$.  Read up on the [url=http://en.wikipedia.org/wiki/Extended_Euclidean_algorithm]Extended Euclidean algorithm[/url].

Sorry for asking, but by what property is $ 42^{600}\equiv3^{600}$? $ 42\equal{}3*2*7$...so...$ 42^{600}\equal{}(3^{600})(2^{600})(7^{600})$...so...basically I don't know how you made that step...

[quote="Icy"]Sorry for asking, but by what property is $ 42^{600}\equiv3^{600}$? $ 42 \equal{} 3*2*7$...so...$ 42^{600}\equal{} (3^{600})(2^{600})(7^{600})$...so...basically I don't know how you made that step...[/quote]

$ 42\equiv 3\bmod 13$, hence $ 42^{k}\equiv 3^{k}\bmod 13$.  (Alternately, $ 42^{600}\minus{}3^{600}\equal{} (42\minus{}3)(...)$)  This is a corollary of an important basic result in modular arithmetic.

[b]Lemma:[/b]  $ a\equiv b\bmod m, c\equiv d\bmod m\implies ac\equiv bd\bmod m$

In fact, $ \equiv$ is an equivalence relation and can be treated like $ \equal{}$.


algebra
difference of squares
special factorizations

Find integer n $n^{4}-4n^{3}+22n^{2}-36n+18$ is perfect square.

${n^{4}-4 n^{3}+22 n^{2}-36 n+18=k^{2}}$

Substituing

${n=\pm \sqrt{s-8}+1}$

Siplifying

${k^{2}+63=s^{2}}$

That the only interger solution is

${(k=\pm 31\land s=32)\lor (k=\pm 9\land s=12)\lor (k=\pm 1\land s=8)}$

Substituing we take the following n's

${n=3}$
${n=\pm 1}$

How did you come up with that substitution??

It does seem a bit contrived.  Here's a little bit easier substitution to see.

$n^{4}-4n^{3}+22n^{2}-36n+18 = (n-1)^{4}+16n^{2}-32n+17 = (n-1)^{4}+16(n-1)^{2}+1$ and from here set $a=(n-1)^{2}$ and then we have $a^{2}+16a+1=k^{2}$ which we have a familiar method for solving.  Dealing only with values of $\ a$ that are perfect squares we do infact get $\ \pm 1, 3$

Thank you, that clears it all up!

From there, $(a+8)^{2}-63 = k^{2}$, and then proceed as you would by factoring the difference of squares and comparing it to the positive factors of 63.

thanks for replying.but I have an another question.
Find integer n $n^{4}+6n^{3}+11n^{2}+3n+31$ is perfect square.
By the same way.
$(n+3 / 2)^{4}-(5/2x^{2}+21/2x-415/16)$
How do I find the solution?

[hide="how you could do the second"]
you can compare the expression to $(n^{2}+3n)^{2}$ and $(n^{2}+3n+9)^{2}$. it's greater than the first (since $2n^{2}+3n+31$ has negative discriminant) and less than the second:
\[(n^{2}+3n+9)^{2}=n^{4}+6n^{3}+27n^{2}+54n+81,\ 16n^{2}-51n+50>0\]
But that leaves [b]eight[/b] cases to check... there's probably a better way.[/hide]


If you've recently received an email (like I did) that looks like this: [url=http://yuethomas.vze.com/images/paypalscam.gif]http://yuethomas.vze.com/images/paypalscam.gif[/url], do [b]NOT[/b] - I repeat, do [b]NOT[/b] fill in your personal info. It's a scam. The link in the email goes to a site that has no affiliation with paypal whatsoever. If you give them your real debit card number, you're going to be scammed SO bad that even your posterity will feel it.

Ah, thanks. I heard about that on TechTV.

Kinda ironic they tell you [b]not [/b]to do what it's asking to do in the right side text...:)

Yes, I noticed that they even said "only sites beginning with http://www.paypal.com are real PayPal sites"

Orisis, would that be posterity or prosperity?

Prosperity makes sense to me. If you fall for this scam, you will become poorer.

Meh it sounds weird...prosperity is a state of matter or condition so it can't feel anything while posterity is a group of people so they can feel things.

I knew I was rushing things. Yeah, it should've been posterity.


group theory
abstract algebra
superior algebra
superior algebra unsolved

find all the $p$-sylow subgroup of $S_n$.

Any $p^{k}$-cycle $\Delta = [a_{1},\dots,a_{p^{k}}]$ of $S_{n}$ (where $p$ is prime and $k$ is the power of $p$ in the decomposition of $n$) is a $p$-Sylow. But $\forall \sigma \in S_{n}, \sigma \Delta \sigma^{-1} = [a_{\sigma(1)},\dots,a_{\sigma(p^{k})}]$, whence the $p$-Sylows of $S_{n}$ (which are all conjugated) are exactly its $p^{k}$-cycles.

Your $p^k$ isn't the highest power of $p$ that divides the order of $S_n$...

So, I read that again, we take $k$ maximal such that $p^{k} | n!$ and the proof is not affected (the fact that $p^{k} \leq n$ being clear by Legendre's formula, or just using the group structure of $S_{n}$). Ok?

Why is $p^k \leq n$¿
Take e.g. $p=2;n=4$, then you get $p^k=2^3 >4=n$.


geometry
circumcircle
trigonometry
power of a point
radical axis
similar triangles
geometry proposed

Assume $A,B,C$ are three collinear points that $B \in [AC]$. Suppose $AA'$ and $BB'$
are to parrallel lines that $A'$, $B'$ and $C$ are not collinear. Suppose $O_1$ is circumcenter of circle passing through $A$, $A'$ and $C$. Also $O_2$ is circumcenter of circle passing through $B$, $B'$ and $C$. If area of $A'CB'$ is equal to area of $O_1CO_2$, then find all possible values for $\angle CAA'$

Let $C'$ be the intersection of the circles $(O_1), (O_2)$ other than the point $C$. The radical axis $CC'$ of these 2 circles is perpendicular to their center line $O_1O_2$ and the center line cuts the segment $CC'$ at its midpoint $D$. Since the lines $AA' \parallel BB'$ are parallel, the angles $\angle CAA' = \angle CBB'$ are equal. Thus the angles spanning the major arcs $CA', CB'$ of the circles $(O_1), (O_2)$ are equal. In particular, the angles $\angle CC'A' = \angle CC'B'$ are equal, which means that the points $A', B', C'$ are collinear. The triangles $\triangle CC'O_1, \triangle CC'O_2$ are both isosceles. Since $O_1O_2 \perp CC'$, the center line $O_1O_2$ bisects the angles $\angle CO_1C', \angle CO_2C'$, i.e.,

$\angle CO_1O_2 = \frac{\angle CO_1C'}{2},\ \ \angle CO_2D = \frac{\angle CO_2C'}{2}$

The angles $\angle CA'C', \angle CB'C'$ span the major arcs $CC'$ of the circles $(O_1), (O_2)$ with the central angles  $\angle CO_1C', \angle CO_2C'$. Hence, we also have

$\angle CA'B' \equiv CA'C' = \frac{\angle CO_1C'}{2},\ \ \angle CB'C' = \frac{\angle CO_2C'}{2}$

Thus the internal angles $\angle CA'B' = \angle CO_1O_2$ and the corresponding external angles $\angle CB'C' = \angle CO_2D$ of the triangles $\triangle A'CB', \triangle O_1CO_2$ are equal, which means that these 2 triangles are always similar, regardless of the angle  $\angle CAA' =  \angle CBB'$.

The similar triangles $\triangle A'CB' \sim \triangle O_1CO_2$ are congruent, iff their corresponding sides $CO_1 = CA'$ are equal. The segment $CO_1$ is the circumradius of the triangle $\triangle CAA'$ and the segment $CA'$ its side against the angle $\angle CAA'$. Using the extended sine theorem for this triangle,

$\sin{\widehat{CAA'}} = \frac{CA'}{2 CO_1}$

Hence, $CO_1 = CA'$ iff $\sin{\widehat{CAA'}} = \frac 1 2$. i.e., iff $\angle CAA' = 30^o$ or $150^o$.


An aircraft,which can fly at a maixmum speed of $30$ km/h in still air,intends to fly in the shortest time possible from point A to point B,with B lying $850$km to the north-west of A.the wind is blowing at a speed of $20$km/h in the direction due north.Find the course that the pilot must steer his aircraft.

 :)

Not to be mean, but can an aircraft flying $30$km/h even lift off??? I'm sure some professional bicyclist can bike at that speed...

[hide]I got $73^\circ$ counterclockwise from the north at $30\text{ km/h}$...[/hide]

[quote="arjunlandes"]Oh yeah, I was wrong. Here's how you do it.
[hide]Think of the direction vectors as sides of a triangle. Then we have the $20$ km/hr vector at an angle of $45$ degrees with the course vector (which must point in the direction of B from A, which is $135$ degrees). Then, the $30$ km/hr vector must be at such an angle as to make the resultant of the $30$ km/hr vector and the $20$ km/hr vector point at $135$ degrees. Draw triangle ABC with AC=$20$, BC=$30$, and angle A=$45$ degrees. Then $\frac{sin 45}{30}=\frac{sin B}{20}$, from which $B=sin^{-1}\frac{\sqrt{2}}{3}$. Then C=$180-45-sin^{-1}\frac{\sqrt{2}}{3}=16.9$ degrees north of west.[/hide][/quote]
couldn't have said it better myself

[quote="shyong"][quote="arjunlandes"]Oh yeah, I was wrong. Here's how you do it.
Think of the direction vectors as sides of a triangle. Then we have the $20$ km/hr vector at an angle of $45$ degrees with the course vector (which must point in the direction of B from A, which is $135$ degrees). Then, the $30$ km/hr vector must be at such an angle as to make the resultant of the $30$ km/hr vector and the $20$ km/hr vector point at $135$ degrees. Draw triangle ABC with AC=$20$, BC=$30$, and angle A=$45$ degrees. Then $\frac{sin 45}{30}=\frac{sin B}{20}$, from which $B=sin^{-1}\frac{\sqrt{2}}{3}$. Then C=$180-45-sin^{-1}\frac{\sqrt{2}}{3}=16.9$ :huh:  :huh:  degrees north of west.[/quote]

My calculator give me $C=180-45-\sin^{-1}\frac{\sqrt{2}}{3}=106.9$ if follow your computation ??[/quote]
He said 16.9 degrees north of west. Yours is from the north, so just subtract 90 to get the north of west degree measure. $106.9-90=16.9$


Hello friends,

One of my friends asked me to prove the following identity
\[ \sum\limits_{k \equal{} 1}^n {{k^2}\left( {\begin{array}{*{20}{c}} n \\
k \\
\end{array} } \right){p^k}{{(1 \minus{} p)}^{n \minus{} k}}} \equal{} np(1 \minus{} p)\]where $ 0 < p < 1$ but I am not able. Could you please give me some hints? Thanks.

Really?
I think $ \sum_{k\equal{}1}^n k^2 \binom{n}{k} p^k (1\minus{}p)^{n\minus{}k} \minus{} \left(\sum_{k\equal{}1}^n k \binom{n}{k} p^k (1\minus{}p)^{n\minus{}k}\right)^2 \equal{} np(1\minus{}p)$

[url=http://mathworld.wolfram.com/BinomialDistribution.html]Binomial Distribution[/url] - MathWorld


this is similar to one of the MOEM problems

Two of Mr. Patrick's classes took the same test. His class of 20 students had an average score of 80. His other class of 30 students had an average of 70. What was the average score for all 50 students?

[hide]20*80+30*70 = 1600+ 2100 = 3700      3700/50 = 74[/hide]

[hide]80 * 20 = 1600

70 * 30 = 2100

the total is 3700

so you divide it into 50

which is...

74

[/hide]

Without even doing any algebra, you should know that the grade will be between 70 and 75 because the average of 70 and 80 is 75 and there are more 70s than 80s, weighting it towards the 70.

Duh....

but we want the answer....

so I did it anyway....

But it's also a good idea to use that just to make sure your answer makes sense, in case you made an error, and got a weird number like 3000.

Two of Mr. Patrick's classes took the same test. His class of 20 students had an average score of 80. His other class of 30 students had an average of 70. What was the average score for all 50 students?

his first class has a total score os 1600
his second class has a total score of 2100
1600+2100=3700
3700/50=74

idk which 1 is correct but i think its either 70 or 74

Look at the answerse before you...

Using common sense, we get $ \text{AVERAGE}\neq70$.

Please don't revive old threads (particularly those which are more than 3 years old).

You're a new member, so I suppose you wouldn't necessarily know, but please keep that in mind from now on.


[hide="Hint"] Use: $\cot x-\tan x=2\cot 2x$ we have:
\[\frac{1}{2^{n}}\cot\frac{a}{2^{n}}-A=2\cot 2a \]
[/hide]

Yes, that's right.
But $A=\frac{1}{2^{n}}\cot\frac{a}{2^{n}}-2\cot2a$


$ \lim_{n\to \infty} \frac{1}{\sqrt n}\Biggl \{1\plus{}\frac{1}{\sqrt2} \plus{}\frac{1}{\sqrt3} \plus{}\cdots \plus{}\frac{1}{\sqrt n} \Biggr\} \cdots\cdots (*)$

[hide="Hint"] For $ n > 1$, the following holds:

$ 2\sqrt {n} \minus{} 2 < \sum_{k \equal{} 1}^{n}\frac {1}{\sqrt {n}} < 2\sqrt {n}$
[/hide]

[hide="Solution"]Multypling both sides with $ \frac {1}{\sqrt n}$ and letting $ n\to \infty$ gives: $ 2\leq L \leq 2$ where L denotes our limit, so: $ L \equal{} 2$.[/hide]


geometry
analytic geometry
graphing lines
slope
trigonometry

Let the points of contact are$(x_1,y_1) and (x_2,y_2)$ thus the slopes are $\frac{y_1}{x_1} and \frac{y_2}{x_2}$

Let us represent the point of contact as (x,y)
By Pythagoreas Theorem,
$(x^2 +y^2) + 2^2 = 3^2 +4^2$
$\rightarrow x^2 +y^2=21$
From th equation of the circle and above equation we get ,
$x+4y=21$
Thus,
$21(x^2 +y^2)=(x+4y)^2$
$21x^2 +21y^2 = x^2 + 16y^2 +8xy$
$20x^2 +5y^2 -8xy=0$
$5(\frac{y}{x})^2 - 8\frac{y}{x} + 20 =0$
From here we will get the two slopes.:idea:

The tangent lines from the origin have the equation y= mx where m has two values.

  Each of these lines intersects the circle in one point

 Find the intersection of y=mx and (x-3)^2 + (y-4)^2 = 4

 By substitution of y=mx into equation for circle and simplification
we get the equation (m^2+1)x^2  -(8m+6)x + 21 = 0

 To have one point of intersection we must have equal roots

 (8m+6)^2 -84(m^2 +1)  = 0

 Solve for m and you get two values of m

Note: I am outlining the solution to save typing and to leave something for the reader to complete.

Alternatively,
[hide]We know the center of the cirlce is $(3,4)$ with radius $2$.  The distance the center of the circle is from the origin is $5$.  Let $O$ be the origin, $O\prime$ be the center of the cirlce, and $T$ and $T\prime$ the tangency points.  Let $\theta =$angle $O\prime$ $O$ forms with x-axis and $\phi =$ angle $T\prime$ $O$ $O\prime =$angle $O\prime$ $O$ $T$. Therefore the two slopes are $\tan(\theta + \phi)$ and $\tan(\theta-\phi)$.By some trig, $\tan(\theta + \phi) = \frac{\tan(\theta)+\tan(\phi)}{1-\tan(\theta)\tan(\phi)} = \frac{\frac{4}{3}+\frac{2}{\sqrt{21}}}{1-\frac{4}{3}\frac{2}{\sqrt{21}}} = \frac{12+2\sqrt{21}}{5}$ The other slope is then $\tan(\theta-\phi) = \frac{\tan(\theta)-\tan(\phi)}{1+\tan(\theta)\tan(\phi)} = \frac{\frac{4}{3}-\frac{2}{\sqrt{21}}}{1+\frac{4}{3}\frac{2}{\sqrt{21}}} = \frac{12-2\sqrt{21}}{5}$.[/hide]

[quote="p.c"]What are the slopes of the tangents from the origin to the circle $(x-3)^2+(y-4)^2=4$[/quote]

I usually am awful at geometry, so correct me if my answer is wrong. Please check over my pitiful attempt.
 [hide]We have a circle centered at $(3,4)$ with radius $2$. The distance from the radius,$C$, to the origin,$O$ is $5$.
The distance from the origin at which this line first intersects the circle at point $A$ is $5-2=3$.
The second intersection occurs at point $B$ with distance $5+2=7$ from the origin.
Let the points of the tangents from the origin to the circle be known as $C$ and $D$.
So, ${OD}^2={OC}^2=(OA)(OB) =3*7=21$ by Power of Point.
So $OD=OC= \sqrt{21}$
Now $RC=RD=2$, so a bit of trig yields arctan(ROC)= 2/5
Now I am stuck. I can see a brute force method, but I don't want to go that route. [/hide]


Day 1 has started! I have sent out the role PM's.

[u]Players[/u]
Extra
kheg_k
Klebian
LadyKn1ght
noneoftheabove
Peter VDD
pianoforte
S2
Sunny
SuperNova
Sly Si
Tiaan

[u]Roles[/u]

4 mafia: as usual, only night communication allowed.

1 angel

1 inspector

1 mathematician: an innocent role with the power to choose 1 night and pick at least 3 players that are still alive to inspect. I will then tell him how many mafia there are in the group.

1 assassin: read his job description in Rules and Sign-up

4 citizens: note that one of the citizens will be the miller.

To make the game more interesting, one of the mafia members and one innocent player, not necessarily a citizen, will have a fake ID. The mafia member knows he has a fake ID, but the innocent player does not. When either of them die, it will only be revealed that this person had a fake ID.

[u]Voting[/u]
Accusation and seconding does not count as a vote. You may not retract a vote, but you may switch your vote ONLY ONCE if it is within 24 hours of your initial vote.

good morning everyone. :) I'm glad to see us all again in a small game, this one should go rather quick.

Hello everyone.

I just hope that this day won't take about 10 days (as it did before). :sleeping:

Goodbye everyone. :bye:

Oh wow that seems small.  Hiya all!

Wow...mafia game 15 is only on 3rd day...I bet this game will end sooner.

Melissa, please forward me a list of the role assignments.

[hide][quote="Klebian"]To start off,
I sent this PM to melissa.

To start the game off, I just came up with a random number.
I am giving you my secret number, $x$.
I will ask somone to guess a number.
The number they guess will be added to $x$, and the sum will be taken (mod 12).
Then, the final sum will be taken, and we will count down the playing list that many spots.
The person there is the person I will accuse.

So I sent my number to Melissa.
Someone, guess a number, lets get this game started.[/quote][/hide]

Eh?

O watever,  x=x+20148

[quote="Sly Si"]Yikes.  Four mafia means we're at a big disadvantage--we lynch two innocent people and we're doomed barring a successful strike by the assassin or doctor.  I would advise the assassin to be very careful in choosing a target.  Not that that's not a good idea anyway, but it could prove especially crucial here.

Also, Klebian, I don't think it's legal for Melissa to interfere with the game by confirming what your number is.  You'll have to rely on us to trust you.[/quote]

Actually that's quite true...
4 mafia is a large number, and in a game of 12, I think that it'd be hard to not lose...
So 4 mafia, 4 citizens, and 4 innocent power roles... hmm...
Interesting to see how this goes eh?

As to my system extra, what I meant was I take the number you say, add x, and take it mod 12, go down the list to that person, and he's killed.
For example, say x is 9 and you say the number 8. That gives 17, which means 5.
Going down the list gives noneoftheabove.
So I'd accuse noneoftheabove.
But yeah...
See the thing is Sly Si, when I suggested something simliar to that in the last game, samath told me to PM the number so I wouldn't be lying...
But, eh... I already PMed Melissa the number...
What to do now then?

[quote="Klebian"]What to do now then?[/quote]

You know, I like the idea if we can't come up with anything to work with.  But....I'd still prefer lynching the most inactive person for day 1.

Most inactive person?
It's day 1!
And so far, there is Supernova, Pianoforte, S2, and Ladykn1ght who haven't posted.

[quote="Klebian"]To start the game off, I just came up with a random number.
I am giving you my secret number, $x$.
I will ask somone to guess a number.
The number they guess will be added to $x$, and the sum will be taken (mod 12).
Then, the final sum will be taken, and we will count down the playing list that many spots.
The person there is the person I will accuse.[/quote]Somehow I don't like the strategy and somehow I don't like people asking for the strategy... ;)

OK, I will tell you what Klebian's number was: 5. You guys can decide whether you want to use it or not :D

Wow.

And how do you expect people to believe you this time?  Honestly, how do we know you aren't just lying again?

i will now explain my thinking

my original plan(when i thought there were masons) was that i would get on the block, be forced to roll claim, role claim mathematician.
then in the night...
i told every1 in the day that the masons should induct me and angel should save men in night while i do job
then melissa said:
Please do note that there are no masons in this game :)

i will now say my plan.
after the day where the masons inducted me,
if pianoforte(who i purposely pretended i was suspicious of) was a mason, i would say that i inspected him along with two masons and found him innocent.

if he wasn t a mason, i would say i inspected one of the people who actually was a mason with two masons, and that one mason was innocent.

What i would actually do:
i would not inspect any1, saving my power when needed, and ensure that mafia would not kill me.

then of course i found there was no mason.
so i tried to make mafia think i used my power by saying there was 1mafia out of klebian and extra.
i would try to convince every1 that klebian was miller and extra was really the angel. if klebian or extra were in trouble of getting lynched, i had the thought of telling every1 i was trying to fake out the mafia

then my parents would no longer let me go on games & fun factory

slysli and tiann are mafias.

My name is Tiaan, not tiann!!
Guys, now I'm sure that I am the miller. This means that only Sly Si is guilty (told ya I had a feeling about him all along!)
Obviously I vote [color=green]nay[/color] on me and all you other innocents should do so quickly, before the 4 mafia vote aye, because then we will lose.

On the block: Tiaan
[color=red]Aye (3):[/color] SuperNova, pianoforte, kheg_k
[color=green]Nay (1):[/color] Tiaan
1 to lynch, 3 to save

If Tiaan is not lynched, I will be lynched.  Therefore, I have no choice but to vote [color=red]AYE[/color].

OK, that's it. Tiaan is lynched and he was a citizen.
Game 16 is over, and mafia wins.

[hide="My original role list"]
Extra - angel
kheg_k - mathematician
Klebian - citizen
LadyKn1ght - citizen
noneoftheabove - mafia with fake ID
Peter VDD - assassin
pianoforte - mafia
S2 - citizen with fake ID
Sunny - mafia
SuperNova - inspector
Sly Si - mafia
Tiaan - miller[/hide]

[hide="What happened during the two nights"]
Night 1: SuperNova inspected Peter VDD and the mafia killed Peter VDD. Extra (angel) saved himself.
Night 2: SuperNova inspected Tiaan and the mafia killed S2. Extra didn't try to save anybody. kheg_k inspected Extra, Tiaan and Sly Si.[/hide][/hide]

W00t evil wins!
heh sorry I'm still in evil mode from the previous game!

Wow. Kheg_k lost the whole game for us with his little "plan"...

Yay!!  We win!

Hmm.  You mighta had a chance if klebian wasn't...sacrificed. lol

Sorry, I was frustrated, no one was talking, kheg had made an idiotic mistake, and it seemed pointless...

Woo!  We win!

kheg definitely helped us out there.


trigonometry
geometry
trig identities
Law of Sines

The whooping crane is Canada's tallest bird. When the angle of elevation of the sun decreases from 30 degrees to 25 degress, the length of a whooping crane's shadow increases by 62 cm. How tall is the whooping crane, to the enarest centimetre?

The book's answer says 150 cm but I got a different answer. Can someone please tell me how to approach this problem?

First, draw the 30-60-90 right triangle, label the hypotenuse $z$, and extend the leg apposite the 60 degree angle by 62 cm; then connect that with the top of the 30-60-90 triangle. Since the bottom angle 25, we can conclude that the top angle is 65. From here use the Law of Sines on the smaller triangle as follows:

\[\frac{\sin 25}{z}=\frac{\sin 5}{62}\]
\[z=\frac{62 \sin 25}{\sin 5}\]
\[z \approx 300.64\]

Since $z$ is the hypotenuse of the 30-60-90 right triangle and the height of the bird is opposite the 30 degree angle, we can conclude that the bird is $\frac{z}{2} \approx \fbox{150}$ cm in height.

Hmm, this isn't all that hard, in particular, but I'm going to move it to Intermediate since it contains Trig.


calculus
integration
trigonometry
calculus computations

prove that if f is continuous, then integral  0 to a f(x)dx = integral 0 to a f(a-x)dx. thx for you help...

thanks for your useful hint. i solved it very easily. Again, thank you very much

You are welcome. :)

Here is a problem from Japanese University entrance exam

1974 Keio University entrance exam/Medical

(1) Prove tHat $ \int_{\minus{}a}^a f(x)\ dx\equal{}\int_0^a \{f(x)\plus{}f(\minus{}x)\}\ dx$.

(2) Evaluate $ \int_{\minus{}\frac{\pi}{4}}^{\frac{\pi}{4}} \frac{\sin ^ 2 x}{1\plus{}e^{\minus{}x}}\ dx$.

hello
$ \int_{\minus{}a}^a f(x)\ dx\equal{}\int_{\minus{}a}^{0}f(x)dx\plus{}\int^{a}_{0} f(x)dx$
$ \int_{\minus{}a}^a f(x)\ dx\equal{}\int_{0}^{a}f(\minus{}x)dx \plus{}\int^{a}_{0} f(x)dx$
  ...substituting $ x\equal{}\minus{}t$ and changing it again
$ \int_{\minus{}a}^a f(x)\ dx\equal{}\int_{0}^{a}(f(x)\plus{}f(\minus{}x))dx$
thank u

hello
[hide="2"]
using the result of 1st.
$ \int_{ \minus{} \frac {\pi}{4}}^{\frac {\pi}{4}} \frac {\sin ^ 2 x}{1 \plus{} e^{ \minus{} x}}\ dx \equal{} \int^{\frac {\pi}{4}}_{0} [ \frac {\sin ^ 2 x}{1 \plus{} e^{ \minus{} x}} \plus{} \frac {\sin^2( \minus{} x)}{1 \plus{} e^{x}} \equal{} \int^{\frac {\pi}{4}}_{0}sin^2x dx \equal{} \int^{\frac {\pi}{4}}_{0}\frac {1 \minus{} cos2x}{2}dx \equal{}\frac{\pi\minus{}2}{8}$
[/hide]
thank u

[quote="kunny"]
(1) Prove tHat $ \int_{ \minus{} a}^a f(x)\ dx \equal{} \int_0^a \{f(x) \plus{} f( \minus{} x)\}\ dx$.

[/quote]

$ I \equal{} \int_{\minus{}a}^a f(x)dx \equal{}\int_{\minus{}a}^a f(a\minus{}a\minus{}x)dx \equal{}\int_{\minus{}a}^a f(\minus{}x)dx
\\
\\
\therefore 2I \equal{} \int_{\minus{}a}^a (f(x) \plus{} f(\minus{}x))dx
\\
\\
Let, g(x) \equal{} (f(x)\plus{}f(\minus{}x)).
Hence, \; g(x)\; is \; even \; function.
\\
\\
2I \equal{} 2\int_{0}^a (f(x) \plus{} f(\minus{}x))dx$


Note that gcd(2004, 999) = 3.  Now let the period of a prime p be the smallest natural n such that p | $a_n$.  For any prime p dividing $a_{999}$, its period must be a factor of 999.  If it is to be a factor of 2004 as well, it must be 3 or 1.  $a_1 = 1, a_2 = 2, a_3 = 5$ hence the gcd is 5.

I don't understand. Why does it divide in period?


[quote="FieryHydra"][quote="Jumbler"]Anybody solve it?  :)[/quote]
Yes, but it's kind of tedious. Hope you solved it on your own.[/quote]

I'm not sure I know how to solve it correctly, so please show your idea at least...


Suppose that a spherical conductor of radius $ R_{2}$ has a net charge $ Q$, and concentric cavity of radius $ R_{1}$. How does the charge distribute itself inside the conductor? Describe the electric field and electric potential within the cavity.

Suppose a non-conducting sphere of radius $ R_{2}$ has a net charge $ Q$ distributed evenly, and concentric cavity of radius $ R_{1}$. Describe the electric field and electric potential within the cavity.

For the spherical conductor: the charge will distribute itself at the outer surface of the conductor. By applying Gauss' law to a sphere with radius $ <R_{1}$ you get that the field inside the cavity is zero, because there is no charge inside; and applying Gauss' law to a sphere with radius $ \in (R_{1}, R_{2})$ where the electric field is zero, you get that there will be no charge on the inner surface. (Actually, that's not quite right, you also have to use the fact that the curl of $ \vec E$ is zero, but never mind...) The potential is constant everywhere within the outer surface and equals $ kQ/R_{2}$.

For the spherical insulator: same as the above, the field inside the sphere of radius $ <R_{1}$ is zero, and the potential is therefore constant within the cavity. It can be found by Gauss' law, but it's not trivial, I think -- maybe I'll have time to calculate it later...


[quote="puuhikki"]We have been given a set of eight distinct non-negative integers $A=\{0,a_{1},a_{2},\cdots,a_{7}\}.$ Let $B=A+A+A.$. It is knows that $\{1,2,\cdots,70\}\subset B.$. Determine $a_{1},a_{2},\cdots,a_{7}.$[/quote]
First of all, what do you mean by "$B=A+A+A$"?
Second, I can already tell that this problem looks too hard for MathCounts, so I may move this problem to an upper-level forum later.  :wink:

[quote="puuhikki"]It would be ok if you move this to the another forum. By $B=A+A+A$ I mean the set $B=\{a+b+c|a,b,c\in A\}$.[/quote]
All right, I'm moving this problem to Pre-Olympiad, which I hope is the best forum for it.  :wink:

[quote="daermon"][hide]are they powers of two?  I'll post my solution if I'm right[/hide]

btw nebula if I"m right I don't think this is really too hard for mathcounts.  I may have misunderstood the problem though  :maybe:[/quote]
No, I don't think that works, unfortunately.  :wink:  You thought of this because of base $2$, right? Well, that actually won't work because the condition $B=\{a+b+c|a,b,c\in A\}$ means that you can only have $3$ nonzero digits in your number, which clearly does not cover all numbers (such as $63$).

:maybe:
doesn't it mean you just choose one from each set?


Express the following as a common fraction:
$ \sqrt[3]{4\div 13.5}$.

$ 4 \div 13.5 \equal{} 4 \div 13\frac{1}{2} \equal{} \frac{8}{27}$. Ah, how handy! It just so happens that $ \sqrt[3]{\frac{8}{27}} \equal{} \frac{\sqrt[3]{8}}{\sqrt[3]{27}} \equal{} \boxed{\frac{2}{3}}$.


Let $ a_i>0,b_i>0$ such that $ a_i b_i \equal{}c_i^2\plus{}d_i^2$ $ (i\equal{}1,2,\ldots , n)$，then prove that:
$ (\sum_{i\equal{}1}^{n}{a_i})(\sum_{i\equal{}1}^{n}{b_i}) \geq (\sum_{i\equal{}1}^{n}{c_i})^2\plus{}(\sum_{i\equal{}1}^{n}{d_i})^2$

[color=darkblue]By Cauchy-Shwartz we have:
$ (\sum a_i)(\sum b_i) \ge (\sum \sqrt {a_ib_i})^2 = \bigg(\sum\sqrt {c_i^2 + d_i^2}\bigg)^2$.
By Minkowsky we have:
${ \bigg(\sum\sqrt {c_i^2 + d_i^2}\bigg)^2\ge \bigg(\sqrt {(\sum c_i)^2 + (\sum d_i)^2}\bigg)^2 = (\sum c_i)^2 + (\sum d_i)^2}$[/color]