geometry
3D geometry
sphere
tetrahedron
prism

What percentage of the space taken up by closest-packed spheres is [u]not[/u] in the spheres themselves?

[quote="falconwing64"]What exactly do you mean by "the space taken up by closest-packed spheres"?  Do you mean three put together in a tetrahedron-like manner?  Are they all in a prism together?  Please clarify.[/quote]

I meant in space, if you have a number of spheres (probably more than three) packed together so that the space between them is minimized, then to find the percentage of space that is between the spheres. The number can be infinite, but you can take a unit of space that would represent the situation, and thus can find that percentage.

I think i know the way in which spheres can be packed with minimal space.. we'll see what you come up with.

(I don't think the radius of the spheres matters for this problem, so you can call it whatever you want... $1, \ r, \ \text{circle thingy}$.. anything  :P )

[quote="cincodemayo5590"][quote="bpms"][quote="falconwing64"]What exactly do you mean by "the space taken up by closest-packed spheres"?  Do you mean three put together in a tetrahedron-like manner?  Are they all in a prism together?  Please clarify.[/quote]
Closest packed spheres, means the least amount of air possible.
[hide="I have seen this before, but I am unsure of the solution"]27%[/hide][/quote]

I got a number a bit smaller than that. Can you post your solution?[/quote]
I have no clue about the solution, just that it is ridiculously complicated and that this was in AoPS vol. 2 in the big picture (unless I don't remember correctly) in one of the chapters.
I think it should be moved. I also think it involves calc, but maybe not.

[quote="bpms"]
I have no clue about the solution, just that it is ridiculously complicated and that this was in AoPS vol. 2 in the big picture (unless I don't remember correctly) in one of the chapters.
I think it should be moved. I also think it involves calc, but maybe not.[/quote]

Oh.. the solution that I [i]thought[/i] was correct is just some space geometry and volume formulas. Maybe the arrangement of the spheres that i have isnt the closest-packed, or the bulk of the solution in AoPS might be proving that the arrangement is the closest-packed possible.

Do you have the book, or is there a link that you know of to this specific problem/solution from the book? I'm interested in what it says.. i'll try searching for it also.

[b]Mods[/b]: Sorry if this is in the wrong place! Please move it if you deem it necessary.  :)

[quote="lotrgreengrapes7926"]Could you please post your solution? :D
The correct answer is...[hide] $\approx25.95\%$ [/hide][/quote]
I remember it saying
[hide]73% is used. Maybe, it is 74%.[/hide]
It said that dudes spent hundreds of years trying to find a better way than the way they are [hide]stacked in grocery stores[/hide] but then some other dudes eventually proved [hide]what grocers knew all along. It is indeed best that way[/hide]


function
geometry
calculus
integration
algebra unsolved
algebra

Suppose that $f,g:\mathbb{R} \rightarrow \mathbb{R}$ are two bijective and continuous functions s.t.$\forall x\in \mathbb{R} , f(g^{-1}(x))+g(f^{-1}(x))=2x$. Prove that if there exist $x_0\in \mathbb {R}$ s.t. $f(x_0)=g(x_0)$ then $f=g$

[quote="harazi"]Ehsan, one small request: some of the problems you posted really do not belong to this section. Please, from now on post them in analysis forum. As for this problem, it is a particular case of the following problem that we have discussed several times on the forum: if $f:R->R$ is monotonous and $ f(x)+f^{-1}(x)=2x$ then $ f(x)=x+a$ for a certain $a$. You see the connection?[/quote]

 :noo:  I thought this problem should be in olympiad section becuase I found it in the olympiad book  :(

This one still works in this area, but the problems with integrals and other stuff should be put in the college playground. Indeed, they are taken from olympiads in Romania, but this doesn't mean they should be put in the olympiad forum. In college playground we also discuss appropriate problems of this type.


Find the real solutions of the following system :
$ ax\plus{}by\equal{}(x\minus{}y)^{2}$
$ by\plus{}cz\equal{}(y\minus{}z)^{2}$
$ cz\plus{}ax\equal{}(z\minus{}x)^{2}$
where $ a,b,c$ are positive reals.

[hide="Solution"]By adding any two equations and subtracting the other, we arrive at
\[ a\equal{}\frac{(x\minus{}y)(x\minus{}z)}x>0\\
b\equal{}\frac{(y\minus{}x)(y\minus{}z)}y>0\\
c\equal{}\frac{(z\minus{}x)(z\minus{}y)}z>0.\]
[i]Case 1: $ (x\minus{}y)(y\minus{}z)(z\minus{}x)\ne0$.[/i] Without loss of generality, assume that $ x>y>z$; then $ x\minus{}y,x\minus{}z,y\minus{}z>0$ so we conclude that $ x>z>0>y$, which is a contradiction.
[i]Case 2: $ (x\minus{}y)(y\minus{}z)(z\minus{}x)\equal{}0$.[/i] Without loss of generality, assume that $ x\minus{}y\equal{}0\implies a,b\equal{}0\not>0$, which is a contradiction.[/hide]

Sorry for not explaining more; I was rushed.

[list][*]$ x \minus{} y,x \minus{} z > 0$ so for $ a$ to be positive, we must have $ x$ positive.
[*]$ y \minus{} x < 0$ and $ y \minus{} z > 0$, so for $ b$ to be positive, we must have $ y$ negative.
[*]$ z \minus{} x,z \minus{} y < 0$ so to for $ c$ to be be positive, we must have $ z$ positive.[/list]
By assumption, $ x > z$, and because $ z$ is positive while $ y$ is negative, $ x > z > 0 > y$.


geometry integration trigonometry calculus function parameterization inequalities

Find the area of the part of cyclinder $x^2+z^2=a^2$ which is contained in the cyclinder $x^2+y^2=b^2$

The trouble to me is that I get a complicated integral...
Anyone who really compute this problem himself will be appreciated..

[quote="liyi"]The trouble to me is that I get a complicated integral...
[/quote]
Alas, you can do nothing about that. If, for instance, $b>a$, the area equals $2a\int_0^{2\pi}\sqrt {b^2-a^2\sin^2\theta}\,d\theta$, which can be easily expressed in terms of the [url=http://mathworld.wolfram.com/EllipticIntegraloftheSecondKind.html]complete elliptic integral of second kind[/url] but is not an elementary function of $a$ and $b$.

is it just me or does that sound a lot like aliens?

"elliptic integral of the third kind"

it makes me want to hide under my bed.

[quote="fedja"]
Alas, you can do nothing about that. If, for instance, $b>a$, the area equals $2a\int_0^{2\pi}\sqrt {b^2-a^2\sin^2\theta}\,d\theta$, which can be easily expressed in terms of the [url=http://mathworld.wolfram.com/EllipticIntegraloftheSecondKind.html]complete elliptic integral of second kind[/url] but is not an elementary function of $a$ and $b$.[/quote]
How do you get the integral?

[quote="liyi"]
How do you get the integral?[/quote]
Consider the standard parametrization of the cylinder $x^2+z^2=a^2$, which is $x=a\sin\theta$, $z=a\cos\theta$, $y=y$. Note that the part of this cylinder contained in the cylinder $x^2+y^2=b^2$ is described by the inequality $|y|\leqslant\sqrt{b^2-a^2\sin^2\theta}$. The rest is straightforward.

For $a>b$, I get $S=8\iint_{0\leq r\leq b\atop 0\leq \theta\leq \frac{\pi}{2}} \frac{ar}{\sqrt{a^2-r^2\cos^2\theta}}drd\theta$

$\iint_{0\leq r\leq b\atop 0\leq \theta\leq \frac{\pi}{2}} \frac{ar}{\sqrt{a^2-r^2\cos^2\theta}}drd\theta = a\int_0^\frac{\pi}{2} d\theta\int_0^b \frac{r}{\sqrt{a^2-r^2\cos^2\theta}}dr = a\int_0^\frac{\pi}{2} \frac{a-\sqrt{a^2-b^2\cos^2\theta}}{\cos^2\theta}d\theta$

Mathematica shows that
$\int_0^\frac{\pi}{2} \frac{a-\sqrt{a^2-b^2\cos^2\theta}}{\cos^2\theta}d\theta = aE(\frac{b^2}{a^2})+\frac{b^2-a^2}{a}K(\frac{b^2}{a^2})=\int_0^\frac{\pi}{2} \sqrt{a^2-b^2\sin^2 t}+\frac{b^2-a^2}{\sqrt{a^2-b^2\sin^2 t}}dt$

Who can tell me how to reduce the previous one into the latter one?


Let be two triangles ABC & A'B'C', whit BAC + B'A'C'=180, ABC=A'B'C'. Show that aa'=bb'+cc', where a,b,c,a',b',c' are thesides of the triangles.

[hide="Proof"]$\alpha'=\pi-\alpha$
$\beta'=\beta$
$\gamma=\pi-\alpha-\beta$
$\gamma'=\alpha-\beta$

$a=2R\sin\alpha, b=2R\sin\beta, c=2R\sin(\alpha+\beta)$
$a'=2R'\sin\alpha, b'=2R'\sin\beta, c'=2R'\sin(\alpha-\beta)$

Therefore

\begin{eqnarray*}aa'=bb'+cc' &\iff& 4RR'\sin^{2}\alpha=4RR'\sin^{2}\beta+4RR'\sin(\alpha+\beta)\sin(\alpha-\beta)\\ &\iff& \sin^{2}\alpha=\sin^{2}\beta+\sin^{2}\alpha\cos^{2}\beta-\cos^{2}\alpha\sin^{2}\beta\\ &\iff& \sin^{2}\alpha=\sin^{2}\alpha\cos^{2}\beta+\sin^{2}\alpha\sin^{2}\beta\\ &\iff& \sin^{2}\alpha=\sin^{2}\alpha\end{eqnarray*}[/hide]

I think the following proof is worth to be posted.

Place the triangles like in the drawing attached, with $O\equiv A\equiv A'$ and $\angle CBO=\angle C'B'O$ and draw $BC"\parallel B'C'$. Let $BC"=a", OC"=b"$. Clearly $\triangle BOC"\sim\triangle B'OC'$, hence $\frac {c'}c=\frac {a'}{a"}=\frac {b'}{b"}=k\ (\ 1\ )$.
From angle bisector theorem and Stewart (for instance) we get $BO^2=BC\cdot BC"-CO\cdot C"O\ (\ *\ )$.
Multiply each side of $(*)$ by $k$ and get the required relation, using $(1)$ convenientl

Best regards.
sunken rock


Write a program that gets as input a phrase and gives as output the number of each character used in the given phrase.

You can use a 26x1 vector to keep track of how many times letters a..z show up. If your program is not case-sensitive, you can use "tolower" function from <ctype.h> to help out. :)

Oh, and input using getchar(). ;)

OK , good idea, and what would the program be if we used pointers? (it's requested in the exercise, as a matter of fact I didn't add it cause I thought that was the only way :blush: )
That's difficult for me...

Well, it is a C program after all, so vectors can't be used (right?). Lets keep the algorithm, and just make ourselves an array

int no[26];

and of course, we have our string (old-type, this is C after all)

char phrase[100];

Now, you can always use pointers to move through these two arrays (simulation of indexes).


pigeonhole principle
induction
combinatorics unsolved
combinatorics

There are representants from $n$ different countries sit around a circular table ($n\geq2$), in such way that if two representants are from the same country, then, their neighbors to the right are not from the same country. Find, for every $n$, the maximal number of people that can be sit around the table.

Hmm, it's easy to show by Pigeonhole that $n^2+1$ doesn't work... I don't know how to easily explain a construction for $n^2$ though.

Indeed, the maximal number is : $n^2$.

Like explained by [b]paladin8[/b], we know that our maximal number is less or equal than $n^2$.

Now, let us denote the $n$ countries by $C_1, C_2, \ldots, C_n$.
Also, for all $k$, we denote by $n_k$ the number of representants coming from $C_k$.
So $N = \sum_{k=1}^n n_k$ is the total number of representants.
And, $r_k$ denotes a representant of $C_k$.

Ok, so, via induction on $n$, we can make the desired construction for $n^2$.
For $n=2$, it's trivial to get a good construction.
Now, let's take a fixed $n \ge 2$. Let us assume that $(a_1, a_2, \ldots, a_{n^2})$ is a good construction for $n$ countries.
Then, for all $k$, we have $n_k \le n$. Moreover, since $\sum_{k=1}^n n_k = n^2$, it means that : $n_k = n$, $\forall k$.
Since the construction is good, we have that each pair $(r_1, r_1), (r_2, r_2), \ldots, (r_n, r_n)$ appears exactly once in the construction.
We then introduce the representants of the country $C_{n+1}$ in the following way :
the couple $(r_1, r_1)$ is replaced with $(r_1, r_{n+1}, r_{n+1}, r_1, r_1)$.
For $k \ge 2$, each couple $(r_k, r_k)$ is replaced with $(r_i, r_{n+1}, r_i, r_i)$.
It is easy to check that our new construction is good, and, there are $n^2 + 3 + 2(n-1) = (n+1)^2$ representants, which proves the result via induction, and provides the desired result!  :D


geometry
circumcircle
parallelogram
geometry unsolved

Let $ O$ be the circumcenter of acute triangle $ ABC$. Let $ k$ be the circle with center $ O'$ that is tangent to $ O$ at $ A$ and tangent to side $ BC$ at $ D$. Circle $ k$ meets $ AB,AC$ again at $ E,F$ respectively. The lines $ OO',EO'$ meet $ k$ again at $ A',G$. Lines $ BO,A'G$ cut at $ H$.

Prove that $ DF^2\equal{}AF\cdot GH$.

[img]http://i227.photobucket.com/albums/dd72/TPFC/radio.png[/img]
We have $ V_A^k: (O')\rightarrow (O),E\rightarrow B,F\rightarrow C$ then $ EF//BC,EO'//BO$.
But $ AE//A'G$ (two triangle $ AO'E,A'O'G$ are similar) then $ BEGH$ is paralellogram, it mean $ BE \equal{} GH$.
Now two triangle $ ADF,DBE$ are similar because $ \widehat{DAF} \equal{} \widehat{DAE} \equal{} \widehat{BDE},\widehat{ADF} \equal{} \widehat{AEF} \equal{} \widehat{DBE}$
Then $ AF.GH\equal{}AF.BE \equal{} DF.DE \equal{} DF^2$($ DE \equal{} DF$ because $ O'D$ perpendicular with $ BC$ and $ EF$)


linear algebra
matrix
linear algebra unsolved

Prove that for any square complex invertible matrix M and any positive integer n, there is a solution to the equation X^n=M.

You may want to consider the Jordan canonical form of $ M$.

Or write $ M\equal{}D\plus{}N$ where $ D$ is a diagonalizable invertible matrix, $ N$ a nilpotent matrix s.t. $ DN\equal{}ND$. $ M\equal{}D(I\plus{}D^{\minus{}1}N)$. $ X\equal{}D^{1/n}(I\plus{}D^{\minus{}1}N)^{1/n}$. For a value of $ D^{1/n}$ it's easy. For a value of $ (I\plus{}D^{\minus{}1}N)^{1/n}$ use Taylor series (there is only a finite number of non zero terms).


calculus
integration
algebra
polynomial
Rational Root Theorem
144

Solve the  equation: $\sqrt{x+11}+x=\sqrt{x+3}$

Unfortunately this involves the solution of a quartic equation, but I was able to see one root [hide="easiest root"] $-2$ [/hide].
This means you only need to solve a cubic, which is a lot easier, but is still really really painful (at least without a calculator).

[quote="FMako"]easiest root:  $-2$ .
[/quote]
Yes! Now, how solve this equation? :)

Come we play ! I note $: \ x\ .s.s.\ y\Longleftrightarrow x=y=0\ \vee\ xy>0.$

$x\ge -3,\ x=\sqrt{x+3}-\sqrt{x+11}\ .s.s.\ (x+3)-(x+11)=-8<0$

$\Longrightarrow x\in (-3,0),\ x=\frac{-8}{\sqrt{x+3}+\sqrt{x+11}}$

$\Longrightarrow \sqrt{x+3}+\sqrt{x+11}=\frac{-8}{x},\ \sqrt{x+3}-\sqrt{x+11}=x$

$\Longrightarrow 2\sqrt{x+3}=x-\frac 8x\Longrightarrow 4(x+3)x^2=x^4-16x^2+64$

$\Longrightarrow 4(x+2)x^2=x^4-20x^2+64\Longrightarrow 4(x+2)x^2=(x^2-16)(x^2-4)$

$\Longrightarrow (x+2)(x^3-6x^2-16x+32)=0$

$\Longrightarrow x_1=-2;\ f(x)\equiv x^3-6x^2-16x+32=0,\ f(-3)<0<f(-2)$

$\Longrightarrow x_1=-2,\ \underline {\ldots}\ x_2\in (-2,98,-2,97),\ x_{3,4}\not\in R$

$\Longrightarrow x_1=-2,\ x_2=-2,97\ldots\ldots,\ x_{3,4}\not\in R$ (with the two exact digits).

[u]Remark.[/u] The points $\underline {\ldots}$ mean the study of the sign for f' and the study of the monotony of f (the Rolle's sequence).

A little, unpardonable and "catastrophic" mistake ! I admit that you are better, Argady. $\underline {\overline {\left| 1-0\right| }}$ for you. And had so much care !. It is trespass. It was a absence of mind. I can not forgive me !.

[quote="FMako"]I would say a cubic is still hard to factorize.[/quote]
Factor Theorem, Rational Root Theorem, or anything else will allow you to do such things.

[quote="arqady"][quote="dondigo"]
$(x+2)(x^3-6x^2-16x+32)=0$
The rest is for you :)[/quote]
The rest is all exercise! :)[/quote]

Okay. I like exercises. :D

By RRT, $x=\frac{32}{k}$ for some integral $k \implies x \in \{ \pm 32, \pm 16, \pm 8, \pm 4, \pm 2, \pm 1\}$. ... uh... Unfortunately, none of these work so we conclude that there are no rational roots.

Back to the drawing board! :D

[quote="chess64"]
Back to the drawing board! :D[/quote]
O.k. I understand you. If $x\in\mathbb Q$ then $x^3-6x^2-16x+32\neq0.$
Maybe existing real root of equation $x^3-6x^2-16x+32=0$ and that root
is the root of $\sqrt{x+11}+x=\sqrt{x+3}$?  :D  :P  :)

[quote="silouan"]
But we have that $2\sqrt{x+3}\geq 1$ for $x\geq-11$ so f is strictly monotonous so it has only one root which is $x=-2$

Is it true Arqady?[/quote]
I think it's wrong! :P

There is only one root $x=-2$.

The all calculations are made before. Me remains to show what was missed.

From $\sqrt{x+3}-\sqrt{x+11}=x$ we have immediate restriction: $-3<x<0$

Now by squaring and moving from one side of equation to another we have $14-2x-x^{2}=2\sqrt{x^{2}+14x+33}$
Hence $x^{2}+2x-14\le 0$ yielding together with above restriction $1-\sqrt{15}\le x <0$

Squaring again gives the equation which was found by [b]dondigo[/b].
We have to prove that the equation $f(x)=x^{3}-6x^{2}-16x+32=0$ has no root in $[1-\sqrt{15}, 0]$

We have $f'(x)=3x^{2}-12x-16$ with roots $2\pm \sqrt{9+\frac{1}{3}}$
The least root is greater than $1-\sqrt{15}$ and the greatest is positive. Hence $min f(x)$ lies at ends of $[1-\sqrt{15}, 0]$ but both values are positive: $f(0)=32, \ \ f(1-\sqrt{15})=-34+10\sqrt{15}>0$


AMC
AIME
geometry
analytic geometry
graphing lines
slope

does anyone have a solution how to do this problem

I just kind of drew a very large diagram and guessed where the line would go through and it worked  :)

[hide="How a friend did it"]
If two circles of equal radius are tangent, then a line passing through the point of intersection divides the circles into two regions of equal area.

Connect the point of intersection of the two leftmost circles tangent to the x-axis to the point of intersection of the center circle and the one directly above it.  Then there are two circles on either side of the line that are not cut by the line; the other four can be paired off such that the area on either side of the line cancels out. This line happens to have slope 3 and equation $y-\frac12=3(x-1)$, so $2y+5=6x$ and $a^2+b^2+c^2=4+25+36=\boxed{065}$[/hide]

Hehe... I just assumed that the circles were squares and worked it out. :o :D

This reminded me of AIME 1984, #6, but I still didn't get it. :oops:


OK, as simple as it looks, I haven't been able to get this problem after an hour of thought:

An object of mass m1, moving with a linear speed v in a laboratory system, collides with an object of mass m2 which is at rest in the laboratory.  After the collision, it is observed that $ (1\minus{}\alpha^2)$ of the kinetic energy in the CM system was lost in the collision.  What was the percentage loss of energy in the laboratory system?

I've gotten as far as $ v_{cm}' \equal{} \alpha v_{cm}$.

Help would be appreciated.

The velocity of the center of mass doesn't change because there are no external forces. You can start by finding the velocity of the center of mass in the lab frame, which is $ \frac{m_1}{m_1\plus{}m_2} v$. Then you can find the velocities of each object in the CM frame, which happen to be $ \frac{m_2}{m_1\plus{}m_2} v$ and $ \frac{\minus{} m_1}{m_1\plus{}m_2} v$ for objects 1 and 2, respectively. After the collision, these velocities become $ \frac{\minus{} \alpha m_2}{m_1\plus{}m_2} v$ and $ \frac{\alpha m_1}{m_1\plus{}m_2} v$. Now you just have to convert back to the lab frame and calculate the energy loss.

Thanks a lot for your reply!  Wow, I was an idiot for thinking that $ v_{cm}$ would change.


probability
Putnam
expected value
algebra proposed
algebra

You have some sort of machine that generates random numbers between 0 and 1 (all with equal probability), one after another.  You keep a running total of the sum of all the numbers that have been put out so far.

What is the expected number of numbers that have been generated when the sum first exceeds 1?

$e$.

This is an old Putnam problem.  The solution here is quite nice:

[url]http://www.kalva.demon.co.uk/putnam/psoln/psol583.html[/url]


1. [url]http://www.mathlinks.ro/viewtopic.php?p=1299095#1299095[/url]
2. [url]http://www.mathlinks.ro/viewtopic.php?p=1299099#1299099[/url]
3. [url]http://www.mathlinks.ro/viewtopic.php?p=1299114#1299114[/url]
4. [url]http://www.mathlinks.ro/viewtopic.php?p=1299105#1299105[/url]

Thanks for posting them, looking at the number of problems you post I guess sometimes you post the individual ones and sometimes the team ones. Do you have possibly individual and team ones that we can post both of them?


[hide="#3"]
I believe if Fluffy is not dangerous, we have a paradox, so I guess we have to say Fluffy is dangerous. With the wording of the question, I'm not exactly sure what to say.[/hide]

[quote="superwilson"]#1 1792 can be writeen as the sum of the cubes of two positive integers(1792=a^3+b^3, a<b) in two ways. Computer the two possible ordered pairs (a.b)

#2 Brenda is collecting money from her relatives.  She starts with no money, and each relative she goes to gives her either a quarter or a dime, with an equal chance of each.  Compute the probability that at some point Brenda will have exactly 1.00 dollar.

#Harry is a seeker, or Hermione is smart. If Harry is a seeker, then Fluffy is dangerous.  If Hermione is smart, then Voldermort is evil.  If Fluffy is not dangerous, then Voldermort is not evil.  Based on the statements, is Fluffy dangerous or not? Or is it impossible to tell?[/quote]
[hide="Number_1"]I think you actually mean $1729$, which can be expressed as $1729=10^{3}+9^{3}=12^{3}+1^{3}$, so the two ordered pairs are $\framebox{(9,10)}$ and $\framebox{(1,12)}$.[/hide]

[quote="superwilson"]
#Harry is a seeker, or Hermione is smart. If Harry is a seeker, then Fluffy is dangerous.  If Hermione is smart, then Voldermort is evil.  If Fluffy is not dangerous, then Voldermort is not evil.  Based on the statements, is Fluffy dangerous or not? Or is it impossible to tell?[/quote]

[hide="solution"]
Let $A$ denote "Harry is a Seeker", $B$ = "Hermione is Smart", $C$ = "Fluffy is Dangerous", $D$ = "Voldermort is evil"

We know that the following statements are true:

1. $A$ or $B$
2. $(A \Longrightarrow C) \equiv (\lnot C \Longrightarrow \lnot A)$
3. $(B \Longrightarrow D) \equiv (\lnot D \Longrightarrow \lnot B)$
4. $(\lnot C \Longrightarrow \lnot D) \equiv (D \Longrightarrow C)$

(using the fact that for any statements $X$ and $Y$, $X \Longrightarrow Y$ is exactly the same as the statement $\lnot Y \Longrightarrow \lnot X$)

Now if $C$ is false, $A$ and $B$ must be false by (2) and (3).  But this contradicts (1) since either $A$ or $B$ must be true.  Therefore $C$ cannot be false.

If $A$ is true and $B$ is false, $C$ and $D$ could be true, and since $C$ cannot be false, $C$ must be true, and Fluffy is indeed dangerous.
[/hide]

[hide="#3"]The contrapositive of "If Fluffy in not dangerous, then Voldermort is not evil" is "If Voldermort is evil, then Fluffy is dangerous". Then just follow the implications starting from either "Harry is a seeker" or "Hermione is smart" - in both cases they end up with "Fluffy is dangerous".[/hide]

[hide="2"]
There are 2 ways to get to $\$$1: (4 quarters, 0 dimes), (2 quarters, 5 dimes), (10 dimes).

There's a $1/16$ chance that you get 4 quarters

[b]if you don't get 4 quarters[/b], there's a $\frac{1}{2^{10}-2^{6}}$ chance that you get 10 dimes.

[b]if you don't get 4 quarters or 10 dimes[/b], there's a $\frac{{7\choose2}}{2^{7}-2^{3}-1}$ chance that you get 2 quarters and 5 dimes.

If you add those together, you should get the correct answer...  I think

$\frac{21}{2^{7}-2^{3}-1}+\frac{1}{2^{10}-2^{6}}+\frac{1}{16}= \frac{3917}{16320}\approx 24\%$

Notice that this problem is the same as the following random walk problem:  If you're standing at the origin, and you can move North or East, and you move one unit at a time, what is the probability that you pass through (0,4),(5,2),or(10,0)?  Here the $x$ axis is the number of dimes you've collected, and the $y$ asix is the number of quarters you have collected.
[/hide]

[quote="neelnal"][quote="superwilson"]#1 1792 can be writeen as the sum of the cubes of two positive integers(1792=a^3+b^3, a<b) in two ways. Computer the two possible ordered pairs (a.b)
[/quote]

Is this not the problem Ramanujan did while he was terribly ill?

Hardy thought it was an ominous number, but Ramanujan said it was a curious number.[/quote]
Not really, The famous number was $1729$ which is not the same thing as $1792$ , in fact it is $63$ more than the first number.
And yes BTW, $1792=12^{3}+4^{3}$ ! (and there is only [b]one[/b] way)

(Also, as  Hardy noticed, 1729 was sort of unlucky being product of 7,17 and 19 .. 1792 OTOH has nice factors $7$ and $2^{8}= 256$  ...not to mention its product of a holy number (64) and a perfect number (28) :) )
calculuslover800: You may like to check your work.[hide]. answer should come out to be around (233/1024).
[hide]or ($1/2^{4}+1/2^{10}+21/2^{7}$) [/hide]
[/hide]
(Of course, this assumes that  Brenda has 10 or more relatives...:) )


trigonometry
quadratics
algebra
polynomial
Gauss
superior algebra
superior algebra unsolved

There is a unique field K with Q included in K included in Q(e^(2/5)PIi) and [K:Q]=2
Find b in K such that K=Q(b).

$ \cos\left(\frac{2\pi}{5}\right)\equal{}\frac{\sqrt{5}\minus{}1}{4}.$ Or put another way, if $ \omega\equal{}e^{2\pi i/5},$ then $ \omega\plus{}\overline{\omega}\equal{}\omega\plus{}\omega^4\equal{}\frac{\sqrt{5}\minus{}1}{2}.$

A field extension of dimenson $ 2$ can only be given by solutions of a single quadratic equation. The extension which contains the number given above must be the extension of order $ 2.$ Hence, $ K\equal{}\mathbb{Q}\left(\frac{\sqrt{5}\minus{}1}{2}\right),$ or more simply, $ K\equal{}\mathbb{Q}(\sqrt{5}).$

However, to get all of $ \mathbb{Q}(e^{2\pi i/5}),$ we must include all of the roots of the quartic polynomial $ x^4\plus{}x^3\plus{}x^2\plus{}x\plus{}1.$ This means that this field must also contain, for instance, $ \sin\left(\frac{2\pi}5\right)\equal{}\sqrt{\frac{5\plus{}\sqrt{5}}8}.$

This nested square root lies in an extension of dimension $ 2$ of $ \mathbb{Q}(\sqrt{5}).$

$ x^4\plus{}x^3\plus{}x^2\plus{}x\plus{}1$ factors over $ \mathbb{Q}(\sqrt{5})$ as

$ \left(x^2\minus{}\left(\frac{\sqrt{5}\minus{}1}2\right)x\plus{}1\right)
\left(x^2\plus{}\left(\frac{\sqrt{5}\plus{}1}2\right)x\plus{}1\right).$

Claim: Let $ \omega\equal{}e^{2\pi i/17}.$ Then there exist fields $ K_1,K_2,$ and $ K_3$ such that $ \mathbb{Q}\subset K_1\subset K_2\subset K_3\subset\mathbb{Q}(\omega)$ and $ [K_1: \mathbb{Q}]\equal{}[K_2: K_1]\equal{}[K_3: K_2]\equal{}[\mathbb{Q}(\omega): K_3]\equal{}2.$

Notes:

A number which lies in any such tower of quadratic extensions of $ \mathbb{Q}$ is a number whose position on the complex plane can be constructed with compass and straightedge.

The claim is (essentially) due to Gauss, and did a lot for making Gauss known in the mathematical community at a young age.

The original problem on this thread, using $ e^{2\pi i/5},$ shows that the regular pentagon can be constructed with compass and straightedge. This particular fact was known to the school of Pythagoras by 500 B.C.


geometry
linear algebra
matrix
algebra
polynomial
parameterization
geometry proposed

I don't understand. It would be nice if you tell me what are those symbols and what do you mean by summation with no summation index. The identity is not true in general if you take for example $a=0, b=1, c=3, p=4$. Then you get $0=16\cdot (-1)\cdot (-2)\cdot 3=96$ (by assuming that the summation is over a countable set). For the beginning mathematicians, it is a common error to forget to say what do the letters stands for.

I think It is only problem of algebra if you use maple with command [>factor  :D  :D

[quote="puuhikki"][quote="shobber"]Can the side of a triangle be $0$?[/quote]
No, but why you are talking about triangles?[/quote]
Because this is the Geometry section.

[quote="shobber"]
Because this is the Geometry section.[/quote]

That does not prove anything. In mathematics you have always say what do you mean. For example, if I post a question is $AB=BA$ to the algebra forum, what is the correct answer? Yes because it is true if $A,B\in\mathbb{R}$? No because the matrix product is not commutative?

It's just an identity, hence an algebra problem. Maybe it would fit in the Geometry section if it had some geometric meaning (which it probably has), but 'til then it is in the wrong section.

[quote="puuhikki"]The identity is not true in general if you take for example $a=0, b=1, c=3, p=4$. Then you get $0=16\cdot (-1)\cdot (-2)\cdot 3=96$ (by assuming that the summation is over a countable set).[/quote]

He did say that $a+b+c = 2p$ ($0+1+3 \neq 2 \cdot 4$). It ought to be true over all (commutative, of course) fields.

Regarding shobber's remark: If an identity is true for all $a,b,c$ sides of a triangle, then it's obviously true for all reals (it follows from the fact that only the null polynomial has an infinity of roots). So if puuhiki found a counter-example in the reals, it surely wouldn't holds for sides of a triangle. In this case it seems that he made a little mistake, and the identity is actually true.

So it's something like this: \[\sum_\textrm{cyc}a(a-b+c)(a+b-c)(b^{2}-c^{2}) = (a+b+c)^{2}(a-b)(b-c)(c-a) .\] I can get a $(a+b+c)(a-b)(b-c)(c-a)$ factor simply by letting (one at a time) $a=b$, $b=c$, $c=a$ and $a+b+c=0$. But I don't know how prove that $a+b+c$ is a double root, except by opening the parentheses.

[quote="perfect_radio"]
Regarding shobber's remark: If an identity is true for all $a,b,c$ sides of a triangle, then it's obviously true for all reals (it follows from the fact that only the null polynomial has an infinity of roots). So if puuhiki found a counter-example in the reals, it surely wouldn't holds for sides of a triangle. In this case it seems that he made a little mistake, and the identity is actually true.

So it's something like this: \[\sum_\textrm{cyc}a(a-b+c)(a+b-c)(b^{2}-c^{2}) = (a+b+c)^{2}(a-b)(b-c)(c-a) .\] I can get a $(a+b+c)(a-b)(b-c)(c-a)$ factor simply by letting (one at a time) $a=b$, $b=c$, $c=a$ and $a+b+c=0$. But I don't know how prove that $a+b+c$ is a double root, except by opening the parentheses.[/quote]

You are right. I make a mistake in my calculations  But still it is important in mathematics to define the parameters you are using!

To prove $a+b+c$ is a double root is simple. First prove the case $a+b+c=1$. Then you can replace $a$ by $pa$, $b$ by $pb$, and $c$ by $pc$ and calculate what is the exponent of $p$ in both sides of the equation.

[quote="Virgil Nicula"] \[\boxed{\ a+b+c=2p\Longrightarrow \sum a(p-b)(p-c)(b^{2}-c^{2})=p^{2}\cdot (a-b)(b-c)(c-a)\ }\ .\] [/quote][color=darkblue][b][u]Sorry[/u][/b], correctly is : $a+b+c=2p\Longrightarrow \sum a(p-b)(p-c)(b^{2}-c^{2})=\boxed-p^{2}(a-b)(b-c)(c-a)\ .$

An equivalent enunciation is : $\boxed{\ a\ne b\ ,\ b\ne c\ ,\ c\ne a\ ,\ a+b+c=2p\Longrightarrow \sum \frac{a(b+c)(p-b)(p-c)}{(a-b)(a-c)}=p^{2}\ }\ .$

This very simple identity is usually sometimes for proving geometrical inequalities. Indeed, I went to the wrong ... address ![/color]


function
calculus
derivative
real analysis
real analysis unsolved

As already known that,

if $ g(x)$ is a differentiable function then \[ \delta\left[g(x)\right] \equal{} \sum_{i\equal{}1}^{n}\dfrac{\delta(x\minus{}x_i)}{|g'(x_i)|}\] where $ g'(x)$ is the derivative of $ g(x)$ and $ x_i$'s are roots of the function $ g(x)\equal{}0$ and also $ g'(x)\neq 0$. But if in case $ g'(x)\equal{}0$ what happens? For example $ g(x)\equal{}c$ a constant function. Does that mean that we cannot have transformation?

Regards,
bhupala

Oh Ok.

Thanks.

One More doubt can we have something like $ \frac{1}{\delta(x)}$ as distribution?

No.

(1) Multiplication and division don't work for general distributions. You can multiply by a nice function, but that's about it.
(2) You can't divide by zero, which $ \delta$ is almost everywhere.


I think this was asked before, but can we use calculus for the purposes of USAMTS, and if so, if it could recieve a full score.

Thanks.

Tip: There's a search bar. ;)

Yes and yes.

Do any of the problems [i]require[/i] calculus?

I'm pretty sure that USAMTS problems will never require calculus.

You [i]can[/i] use Calculus, but no problems require it.
Just like programming.

Yes, you can use calculus. However, all problems can be solved using pre-calculus methods.


Well I have been hearing a rumour that mr vornicu has left the aops for good? Is that true?

Can some admins throw light on this?

I can't say I blame VV: recently he ws havign too much touble in aops - handling spam and all.

But overall he was a nice guy. The sporty sort. I liked him for that.

Yes. He is around 22 right? So he must have passed out of college this year. But he can't forget this for petty reasons like those.

It must be something deeper ...

Maybe he has found a girl? Maybe she does not like maths? (Just making a random guess, no offense).

[quote="Peter VDD"]Dunno... he's just not been visiting for a while, I guess he's busy with finding a place to work and such?[/quote]

Maybe so, but are the circumstances that extraordinary?  I mean, has he ever refrained from posting THAT long :| ? :o

Maths was his first love, right?

First love is very strong. In fact a man can ditch his first love only in favour of a second love, if the latter is a domineering girl.

Keep away from domineering girls is my motto!

But on a more serious note, can some of the admins help?

Somewhere I've read that

[quote="orl"].. yet as long as Valentin enjoys his break.  :lol:[/quote]

so he is taking a break, is he? Is this his regular habit after each IMO?

Yes, he's in his aftermath sleep. After that he does an autumn sleep, then a winter sleep, followed by a spring sleep. And then he goes to the next IMO.

Stancioiu sorin tells me he already has a girl, so that possbility is left out.

What does he do in the IMOs by the way? Is he in the jury?

[quote="bubka"]What does he do in the IMOs by the way? Is he in the jury?[/quote]

Initially he was an IMO participant. Later on he was a observer from the Romanian team. Afterwards he continued to work as [url=http://imo2006.dmfa.si/sub_Coordinators.html]an IMO coordinator as in 2006.[/url]

[quote="bubka"]How do you become an IMO coordinator? I'll become one when I grow up.[/quote]

First of all you need to apply. Then you should satisfy some requirements.

- Fluent English, especially when it comes to math
- Possibly another language
- Experience with IMO problems, and how to solve them
- Experience with grading challenging high-school problems, e.g. your national MOs, TSTs, internationally regional contests as the APMO etc.
- Time when the IMO takes place
- There must be an actual need for you, most of the problem coordinators are from the host country as you can easily figure out
- Be friendly to people but also strict to team leaders who try to go for more credit than their solutions actually deserve...

Not sure whether I presented an exclusive list of credentials. But just have a try and apply.  :)

maybe another administrator can tell us.

Okay my duty was to set things right for fredbel6. He told me about this problem, and asked me whether I could help. I said "I'll take care of that", and he said, "How?", and I said "Just Watch", and then I asked him to check gff, and he said "Oh no! You haven't - ", alarmed, and I said "Yes I have - that should sort things out." And it has.

Once again I have done it my way, the bubka way, and succeeded!

Yes it is true, someone else told me about it, I told Bubka, and then he made this topic in order to sort it out.
It is true I initially said :"Oh no, you haven't" because I was afraid it would lead to violations of Valentin Vornicu's privacy.
As I said, he is said to be fine and that settles it for me.

I don't get it. Were you worried about his health or something? Like he fell off a cliff and broke a few limbs here and there? Huh?

bubka I think you should remove a s from your 4th post...
Anyway he told me he will be busy these days
Does anyone know when he will come back?

We will see... let him enjoy his big break now. ;)

[quote="fredbel6"][quote="DPatrick"]Valentin is on vacation.

He might choose to address the other rumors and innuendo in this thread when he gets back.  Then again, he might decide that his private life is really none of your business.[/quote]

The latter is probably what I do.  From what DPatrick says, nothing unexpected seems to be going on : he is fine.
Thank you, for me that settles it, he needs not address the rumors or innuendo for me.[/quote]

I must say my private life is yet to reach a stage that I won't share here. Maybe because I am too big hearted to be secretive (or maybe becasue I am yet to find a girl - who knows?)

Just to let you know... he's back and is posting again. :yoda:  :wow:
I think he already saw this post, because his signature requests no questions about his absence :D

He is back! :coool:  :coolspeak:  :yoda:

[quote="bubka"]Now we can sleep at night in peace again![/quote]

Well, come on! who's he? God? I think you're going a little far

I think you're all just spamming up the topic.  :roll:  maybe better delete it.

[quote="Gyan"]Okay guys, let the  the water-well come to the thirsty man, and enlighten everyone if you have to find such things out in future the  proper AoPs way to do is:
 to type in Google:

[quote][url=http://www.google.com/search?hl=en&q=Where+is+Valentine+Vornicu++&btnG=Google+Search]Where is Valentin Vornicu[/url][/quote]

Go ahead  and try it , Or just click the link above,  and hit "I am feeling Lucky" button or  see the top link.

And you will get  this
[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=73512&start=20]web page[/url]

And then if you do a locate on "Valentin" you will get[ ..Just read it

[b]NOT[/b] this part:
[quote]You think it is right, but you don't have to cover yourself. Have you thought about the fact that its unpleasant? So what if one man looks with lust at your sister? Maybe your sister wants to be looked upon! Many girls want to be beautiful and show off! Anyway it should be their right to choose wheter they like or not to wear small skirt or pants or something to cover their faces, as men do. You cannot claim "equality" when clearly there is none.[/quote]

But the Key sequence just below it:
[quote][b]Back after a (long) vacation  :)  Don't ask any questions about it please   :D [/b]

[/quote]
That's the proper way to do any search.
Hope that helps  :lol:[/quote]
um... isn't that his new sig? that he got after coming back?

Just so you know, the google search thing doesn't work.


Find all the injective functions $f:N\rightarrow N$ satisfying :
$f(f(n)) \leq \frac{n+f(n)}{2}$ For all positive integer $n$

http://www.mathlinks.ro/Forum/viewtopic.php?t=5449

  darij

Now, I know why it was sounding familiar to me  ;)

Pierre.


A number can be written in the form (m+5)(n+14), where m and n are two-digit numbers created by using each of the digits in the set {2,3,6,9} exactly once. What is the smallest such number?

A science test has 10 questions worth 5 points each, 7 questions worth 6 points each, and 4 questions worth 2 points each. None of these questions will be given partial credit. How many scores between 0 and 100 are impossible to score?

Enjoy!

[hide]As with the area of the rectangle in relation to the sides you'd rather have one large number and one small number rather than even out the numbers, so 26 (the least possible number available) should be paired up with 5 and 39 should be paired up with 14 so 31*53=1643.

1 and 3 is obviously not attainable. From there on all odds are attainable until 53 and all evens are gone until 50 using only the combination of 5 and 2, 6 and 2. With the combination of all 3 (with 9 5s) all odds until 95 are attainable. For the evens everyone is attainable. So 97 and 99 are unattainable (larger than 95 and is not even). Thus the answer is 4 (1, 3, 97, and 99)[/hide]

Tare, for the first one, your answer may be correct but your reasoning definitely isn't.. if this was an 'explain your working' one, you can't just assume the two numbers have to be as close together as possible or anything like that.. in fact, you haven't even done that, by your reasoning, shouldn't 29 and 36 be better?

Mmm? I was saying the [b]opposite[/b] of what you just suggested...[quote="I"]you'd rather have one large number and one small number rather than even out the numbers[/quote]

Besides I guess you could reason through inversive proportions or something like that but since this is a MATHCOUNTS question (and I'm guessing it's not masters) there's no need for detailed explanation but if you'd like to go for it then by all means go for it por favor...didn't that sound cool? :lol:

um...yes i guess that sounds cool. however, it might (for the simple purpose of discussion) be better to tighten up that reasoning somewhat. for example, what makes you think that adding on the 5 and 14 (or whatever the numbers are) won't change you reasoning? i think you're right, but good explanations are ... well ... good.

Oh yeah. So what about 23 and 96 then :) or 69 if you assume that the first digit is always smaller than the second..

Mmm, I agree.

Anyone want to give a better explanation? Up for grabs is this one, I'm too busy to come up with the solution for that sticky Sticky situation :lol:

I think Tare was closer than you guys give him credit for.

Tare, give another shot at the explanation, doing the following:

First, talk about which digits are the tens digits and which digits are the ones digits and why.

Then, point out that the sum of the numbers will be the same no matter what (prove this).  Take the 14 and the 5 into account for this, as well.

Then, argue that given two positive numbers with a constant sum, the product is smaller when they are farther apart.  Then use this to get the final answer.


Prove that the "Fibonomial Coefficients" defined as:

[tex]\displaystyle {n\choose k}_F = \frac {\prod^n_{i=1}F_i}{\left(\prod^k_{i=1}F_i\right)\left( \prod^{n-k}_{i=1}F_i\right)}[/tex] for [tex]1\le k\le n[/tex]

are all integers, where [tex]\{F_n\}^{\infty}_{n=1}[/tex] is the well known Fibonacci sequence.

We have the beautiful Pascal-like relation :

[hide]C(n,k)_F = C(n-1,k)_F * F(k-1) + C(n-1,k-1) * F(n-k+1)

Found by inspection and proved by induction. [/hide]

Fibonacci numbers never cease to amaze us

Nice! I didn't know the formula you found. Your formula made the problem so simple.

Here is a more general version of the problem:

Prove that the Generalized Binomial Coefficients defined as:

[tex]\displaystyle {n\choose k}_C = \frac {\prod^n_{i=1}C_i}{\left(\prod^k_{i=1}C_i\right)\left( \prod^{n-k}_{i=1}C_i\right)}[/tex] for [tex]1\le k\le n[/tex]

are all integers, where [tex]\{C_n\}^{\infty}_{n=1}[/tex] is a sequence of positive integers such that [tex]\gcd(C_m,C_n)=C_{\gcd(m,n)}[/tex].


[quote="arqady"]Let $ m_a,$ $ m_b$ and $ m_c$ are medians of triangle with sides lengths $ a,$ $ b$ and $ c.$ Prove that
\[ m_a \plus{} m_b \plus{} m_c\geq\frac {3}{2}\sqrt {2(ab \plus{} ac \plus{} bc) \minus{} a^2 \minus{} b^2 \minus{} c^2}
\]
[/quote]
$ \leftrightarrow m_am_b\plus{}m_bm_c\plus{}m_cm_a \ge \frac{9}{4}(ab\plus{}bc\plus{}ca)\minus{}\frac{3}{2}(a^2\plus{}b^2\plus{}c^2)$
it is easy from :
$ m_am_c \ge \frac{9}{4}ac\minus{}2m_b^2$

$ m_cm_b \ge \frac{9}{4}cb\minus{}2m_a^2$

$ m_bm_a \ge \frac{9}{4}ba\minus{}2m_c^2$

$ m_a^2\plus{}m_b^2\plus{}m_c^2\equal{}\frac{3}{4}(a^2\plus{}b^2\plus{}c^2)$.

Sergic Primazon, your proof is true and very nice!

[quote=xzlbq][quote=arqady]Let $ m_a,$ $ m_b$ and $ m_c$ are medians of triangle with sides lengths $ a,$ $ b$ and $ c.$ Prove that
\[ m_a \plus{} m_b \plus{} m_c\geq\frac {3}{2}\sqrt {2(ab \plus{} ac \plus{} bc) \minus{} a^2 \minus{} b^2 \minus{} c^2}
\][/quote]

<=>
\[{\it m_a}+{\it m_b}+{\it m_c}\geq 3\,\sqrt {r \left( 4\,R+r \right) }\][/quote]

strgnger is:

\[{\frac {s-a}{{\it w_a}}}+{\frac {s-b}{{\it w_b}}}+{\frac {s-c}{{\it w_c}}
}\geq 3\,{\frac {\sqrt {r \left( 4\,R+r \right) }}{s}}\]

More stronger is:

In triangle,prove that

\[{\frac {s-a}{{\it w_a}}}+{\frac {s-b}{{\it w_b}}}+{\frac {s-c}{{\it w_c}}
}\geq 3\,\sqrt {\frac{r}{s} \left( 4\,{\frac {R}{s}}+\frac{1}{9}\,\sqrt {3} \right) }\]


I dunno I'm usually the one who isn't excited.
our team:
me: yawns
other person: shrugs
other other person: goes to sleep
this one rely nerdy kid who used to b smart but now is really stupid who MESSED UP ON FIVE TARGETS AND MADE US GET 3RD TEAM!!!!!!!: wow i'm really excited!!! let's do it!!

so no

[quote="Zmastr"]I was just wondering... I know that some of you out there may not have the most enthusiastic team, and may not think they need to practice.  Are there ways I can rejuvenate my team's spirit?[/quote]
If your teammates don't care that much, leave them behind.
Team doesn't really matter that much. Unless you think that there's a chance you'll get first, then don't worry about them. Tell them that if they don't practice, you'll practice on your own.
I mean, seriously, if they don't want to practice, forget about them. They're just gonna end up bringing you down and eating up your time trying to persuade them to practice, when it doesn't really matter to you as an individual.
Just my outlook.

Yeah...
A person in my group was getting lazy and said she didn't want to do math anymore.  :o So I just forgot about it. If a person doesn't care about math, he/she's not gonna succeed anyways, and besides, there's only one team per state for MATHCOUNTS. Might as well shoot for individual (if you haven't already done state).

[quote="tz"]I dunno I'm usually the one who isn't excited.
our team:....
this one rely nerdy kid who used to b smart but now is really stupid who MESSED UP ON FIVE TARGETS AND MADE US GET 3RD TEAM!!!!!!!: wow i'm really excited!!! let's do it!!...[/quote]

He's enthusiastic about math. Maybe he isn't the best, but he has the drive to do well at it. And calling him nerdy doesn't help. Be glad you (1) have a team (2) have someone enthusiastic on your team (3) your team is good enough to get 3rd. Cm'n now, how many teams did your team beat to get 3rd?

I think that "wow I'm really excited" comment was sarcasm from his part, meaning, that after analyzing the "nerd" on his team, he sees no real reason to be excited and enthusiasitic.

If your teammates don't care that much, leave them behind.
Team doesn't really matter that much. Unless you think that there's a chance you'll get first, then don't worry about them. Tell them that if they don't practice, you'll practice on your own.
I mean, seriously, if they don't want to practice, forget about them. They're just gonna end up bringing you down and eating up your time trying to persuade them to practice, when it doesn't really matter to you as an individual.
Just my outlook.

^^^^

Not mocking Klebian, just repeating what he said in agreeance.

You can also tell their parents to force them to work. :P Works for my team. :lol:

On my team, everyone is pumped up except one kid, so my mom(the coach) takes his absolute value.  If that doesn't work (which for some reason it does) then his Mom makes him do it :rotfl:

I have a slight correction for jhollenbeck...i practice with his team and that one kid who isnt excited for mathcounts, (even though he made state) falls asleep when our coach offers Candy and Pizza AND  video game parties, he is just being cruel to her because he knows it gets on her nerves.  If he makes nationals he'll cry.

     So for some kids they just dont wanna do it. :starwars:

Yeah, after I got those tests from biffandoc and fanatic, i've been practicing a lot more often.  I guess nats is what to shoot for.

okay i kno i posted a negetive comment.
I'm enthuastic, just not THAT enthuastic.
I'm only enthuastic at the competition.
and our "captain" doesn't even COME to the practice sessons, he only came 4 times for the school round and celebration parties.
and the only enthuastic person on our team is really stupid. He actually RUNS through the hallway to the mathcounts room. But it was because of him we got 3rd team, not 2nd. I know some of u guys think it was sarcasm, but it was true. He doesn't think he is that stupid, he used to be quite smart. And if you ask him he'll deny it. How we found out he messed up was that he went first when we were disscussing the problems, and he didn't know our answers, so he sed the 'real' ones he got.
Anyways you don't need an enthuastic team, just a smart team.
As long as they're smart.

[quote="robinhe in another topic"]Our school doens't "train" students.  My Mathcounts coach just hopes that there are random smart people that will join Mathcounts.[/quote]
That's the exact same thing with ours.

Wow.  I go to a small magnet school, so there's a ton of smart kids that could join.  They just don't hear about it and don't care.  We could win state if more people than me actually cared an ounce about doing well.

[quote="tz"][quote="robinhe in another topic"]Our school doens't "train" students.  My Mathcounts coach just hopes that there are random smart people that will join Mathcounts.[/quote]
That's the exact same thing with ours.[/quote]

Haha same here. And the stupid thing is, we could have a pretty solid team, instead of like one good person and 3 average/decents. The thing is, our coach made no effort to get the good people in our school to join. And when some people had time conflicts, she just said that she didn't want someone on the team that wouldn't be able to come to many practices....

Hey nicely put.
our mathcounts sessons=euchre/poker/knafta/spitball/joke sesson
Our coach got lucky this year, we haven't made top 3 states in 12 years. No clague person has been to nationals in 20 years.

Yeah, our state's in 10 days.   Yesterday our team got together to decide when to practice and everyone was booked for 3 out of 5 days a week.  You'd think MATHCONTS would become a priority when it's so close.   The other day we did a 2000 state sprint round and my teammates got 8, 9, and 9 respectively.  Not to brag, but I got 24.  That's when your team needs to kinda pick up the pace.

That reminds me of solafidefarms' team. No offense.

If kids don't want to practice, let them fail. Even If your team gets last place, you can still make it to nationals by getting in the top 4.
I didn't practice alot for State MC last year. I just came to the meetings and did one practice test. Fourtunately, that was enough to get 1st. Then I practiced for nationals.

[quote="tz"]Hey nicely put.
our mathcounts sessons=euchre/poker/knafta/spitball/joke sesson[/quote]
now how many times have you said that? :huh:

[quote="tz"]
[quote="robinhe in another topic"]Our school doens't "train" students.  My Mathcounts coach just hopes that there are random smart people that will join Mathcounts.[/quote]
That's the exact same thing with ours.[/quote]

Same here, except there are no random smart people, exactly why our team always fails.

Our team's different. I'm probably the only enthusiastic player on the team. I always get the second highest score in our team on practice and previous competitions we've done. We have state this weekend and i bet none of them r going to practice, but that guy'll still get a higher score

Our school has an excellent motivator for coach, and we usually win state (4 of last 5 years).  However, we are not exceptionally gifted, nor do we practice a lot.  Math team is just fum because of the Countdowns every Monday morning, the donuts, and the good times.  Invite mathletes' friends who are open to trying something new.  You'll discover that these students, while not being near competiive, will bring new life to your team.  But that is a coach's decision, and it may not work for some people.


geometry
3D geometry
tetrahedron
sphere
probability
inradius
circumcircle

A tetrahedron with four equilateral triangular faces has a sphere inscribed within it and a sphere circumscribed about it. For each of the four faces, there is a sphere tangent externally to the face at its center and to the circumscribed sphere. A point $ P$ is selected at random inside the circumscribed sphere. The probability that $ P$ lies inside one of the five small spheres is closest to

$ \textbf{(A)}\ 0\qquad
\textbf{(B)}\ 0.1\qquad
\textbf{(C)}\ 0.2\qquad
\textbf{(D)}\ 0.3\qquad
\textbf{(E)}\ 0.4$

[hide="Solution"]
Let $ h$ be the height of the tetrahedron, $ K$ be the area of one face of the tetrahedron, $ r$ be the length of the inradius, and $ R$ be the length of the circumradius, so that $ r \plus{} R \equal{} h$ and $ \frac {R \minus{} r}2$ is the length of the radius of one of the other four spheres.

Then $ V \equal{} \frac13Kh \equal{} 4\left(\frac13Kr\right)\implies r \equal{} \frac h4\implies R \equal{} \frac {3h}4\implies\frac {R \minus{} r}2 \equal{} \frac h4$.

Without loss of generality, let $ h \equal{} 1$ for ease of computation. Then $ P \equal{} \frac {\frac43\pi\left(5\left(\frac14\right)^3\right)}{\frac43\pi\left(\frac34\right)^3} \equal{} \frac5{27} \equal{} 0.185\ldots\implies\boxed{\textbf{(C)}\ 0.2}$.
[/hide]


Find the polynomial $ f(x)$ such that :

(i) $ x^4f\left(\frac{1}{x}\right)\equal{}f(x)$

(ii) $ f(1\minus{}x)\equal{}f(x)$

(iii) $ f(1)\equal{}1$

From (i) we get $ f(x)\equal{}a_0(1\plus{}x^4)\plus{}a_1(x\plus{}x^3)\plus{}a_2x^2$.
From $ (ii),(iii)$ we get $ f(0)\equal{}f(1)\equal{}1\to a_0\equal{}1,a_1\equal{}\minus{}2a_0\equal{}\minus{}2,a_2\equal{}3$.
It give  $ f(1\minus{}x)\equal{}(1\minus{}x)^4\plus{}1\minus{}2(1\minus{}x)(2\minus{}2x\plus{}x^2)\plus{}3(1\minus{}x)^2\equal{}x^4\minus{}2x^3\plus{}3x^2\minus{}2x\plus{}1 \equal{}f(x)$.

That's right. :)

How did you obtain this result?

[quote="Rust"]From (i) we get $ f(x) \equal{} a_0(1 \plus{} x^4) \plus{} a_1(x \plus{} x^3) \plus{} a_2x^2$.
[/quote]

If $ f(x)\equal{}a_0\plus{}a_1x\plus{}..\plus{}a_nx^n$ and $ x^kf(\frac 1x )\equal{}f(x) \forall x\not \equal{}0$, then $ n\equal{}k,a_i\equal{}a_{n\minus{}i}$.


Is there any $ n$ such that $ n!$ has first (leftmost) digit 9?

[quote="Kouichi Nakagawa"]n! has initial digit '9'.
[url=http://www.research.att.com/~njas/sequences/A045528]A045528[/url] - OEIS[/quote]

what does that   mean? could u elaborate more on that please

That link is the (infinite) list of all numbers whose factorial starts with 9.

But then the problem is solved. Just say, "96! = [insert number here], which, as you can see, starts with 9, QED."

(Note: The above is a joke. Even though it is [b]possible[/b], technically, to do out the calculation of 96!, it is very infeasible :P )


If 6 people take a different colored card and put it into a hat. What is the probability that exactly one person draws their own card? I already figured out that I need to use binomial probability for this but I completely forgot how to use the formula. Maybe you could explain how the binomial probability formula was achieved so I might understand it $ \dbinom{n}{r} \cdot p^{r} \cdot q^{n\minus{}r}$. Or just show how to do the problem.

Unless I am wrong:
(n)
(r)
    is obviously the number of configurations, the probability of getting exactly r successes in         n trials.

'p' is the probability of a success while
'q' is the probability of a failure.

These two numbers should be compliments.

Assuming what I have is correct:
n=6
r=1
p=1/6
q=5/6

P and Q are derived from the simple idea:
If there are six different cards, what's the probability I choose the one blue one?

Then plugging these numbers into the equation:

(6)
(1)=6

p^r=(1/6)^1
q^n-r= (5/6)^(6-1)=(5/6)^5

So then multiply these all together:

(6)(1/6)(3125/7776)

So the probability that only one person draws their own card is:

3125/7776

I really hope that my interpretation of the formula was correct, and that I helped you!

It matters quite a lot whether the cards are drawn with or without replacement.

Then you're asking about permutations by numbers of fixed points, for which the key word to look up is "derangement."


function
number theory unsolved
number theory

$P(n)$ denotes the number of the partition of  as the sum of positive integers (disregarding the order of the parts).Prove that:
$P(11m+6)$ divisible by 11

See [url=http://www.math.wisc.edu/~ono/reprints/018.pdf]B. Gordon and K. Ono "Divisibility of Certain Partition Functions By Powers of Primes"[/url].


function
inequalities
algebra solved
algebra

Find all functions f : N--->N such that for all positive integers n the following relation holds : [tex] f(f(n)) \leq \frac{n + f(n)}{2}[/tex]

You forgot to say that the function is injective. We state $f(n)=n$.
Well, fix an n, and now notice that all $f^k(n)$ are between $0$ and max $(n,f(n)$. That means we must have $f^i(n)=f^j(n)$, for some $i>j\in \mathbb{N}$ and from injectivity $f^{i-j}(n)=n$. If$i-j=1$, we are done. Elsewere, $f^{i-j}(n)\leq max(n,f(n))$, so $n\leq max(n,f(n))$, so if $n\geq f(n)$ we are done. Else take $f^{i-j+1}(n)=f(n)$ and do the same trick. You get $f(n)=n$, for all $n\in \mathbb{N}$.
I got 7 on this problem in the contest, so I can assure you the function must be injective or you get nothing.

oh. Another thing. You must check that the function satisfy the inequality or you wouldn't have get 7 in the contest.


integration
logarithms
trigonometry
calculus
limit
real analysis
real analysis unsolved

More generally, let $ a$ and $ b$ be natural numbers. Then

$ \sum_{k=1}^{\infty}\frac{1}{k(ak+b)}=\frac{1}{b}\left(\psi\left(\frac{b}{a}+1\right)+\gamma\right)$.

We have

$ \frac{1}{k(ak+b)}=\frac{1}{b}\left(\frac{1}{k}-\frac{1}{k+b/a}\right)$.

Now use the following expansion:

$ \psi(z+1)=-\gamma-\sum_{n=1}^{\infty}\left(\frac{1}{n+z}-\frac{1}{n}\right)$, $ z\neq-1,-2,\cdots$

When $ z=\frac{b}{a}$ the desired result follows.

Misan, I didn't understand how you concluded that $ S =-\int^{1}_{0}x^{4}\ln(1-x^{4})\, dx$. Could you explain?
Also, as you have done the sum is infinite.

[quote="Carcul"]$ S = \sum_{n = 1}^{\infty}\frac{1}{n(4n+5)}= \frac{1}{5}\sum_{n = 1}^{\infty}\left(\frac{1}{n}-\frac{4}{4n+5}\right) = \frac{1}{5}S'$.

Now define $ f(x) = \sum_{n = 1}^{\infty}\left(\frac{x^{n}}{n}-\frac{4x^{4n+5}}{4n+5}\right)$. Note that $ f(1) = S'$ and $ f(0) = 0$. Let's now differentiate:

$ f'(x) = \sum_{n = 1}^{\infty}(x^{n-1}-4x^{4n+4}) = \frac{1}{1-x}-\frac{4x^{8}}{1-x^{4}}= \frac{1+x+x^{2}+x^{3}-4x^{8}}{1-x^{4}}$

$ = 4x^{4}+4+\frac{x^{3}+x^{2}+x-3}{1-x^{4}}= 4x^{4}+4-\frac{x^{2}+2x+3}{(1+x)(1+x^{2})}$

$ = 4x^{4}+4-\frac{1}{1+x}-\frac{2}{1+x^{2}}$.

Hence: $ \int^{1}_{0}f'(x)\,dx = f(1)-f(0) = S' = \left[\frac{4x^{5}}{5}+4x-\ln(1+x)-2 \arctan x \right]^{1}_{0}$
$ \Rightarrow S' = \frac{24}{5}-\ln 2-\frac{\pi}{2}$

and finally $ S = \frac{1}{5}\left(\frac{24}{5}-\ln 2-\frac{\pi}{2}\right)$.[/quote]

Hmm, the series does not converge uniformly on $ [0,1]$, can you differentiate it?
Edit: The answer $ S = \frac{1}{5}\left(\frac{24}{5}-\ln 2-\frac{\pi}{2}\right)$ ~ 0.5 is not correct. Yesterday I tried this problem in similar way as Carcul did and got this answer, but correct answer should be ~0.23. Didilica's answer with $ \psi$ is good :)

Very nice solution Misan.

[quote="Pou"]Hmm, the series does not converge uniformly on [0,1], can you differentiate it?[/quote]

Could you explain further? Thanks

[quote="pou"]
Hmm, the series does not converge uniformly on $ [0,1]$, can you differentiate it?[/quote]
Yes. All power series can be differentiated term-by-term [b]even[/b] those that dont converge uniformly.


Prove that,If a number is writen just by $ 2$'s and $ 6$'s it can be writen in the form $ 4k\plus{}2$ ,$ k$ is natural.

[hide="Solution"]It's two digit end is 22,26,62 or 66, which are all 2 mod 4, so the conclusion follows.[/hide]

20 and 60 are both divisible by four, so only the last digit actually matters. 2 and 6 are both equivalent to 2(mod 4) so it can be written in the form 4k+2.

[hide="Solution"]Only the last two digits matter:

$ 22\equal{}4k_1\plus{}2$

$ 26\equal{}4k_2\plus{}2$

$ 62\equal{}4k_3\plus{}2$

$ 66\equal{}4k_4\plus{}2$[/hide]

[hide="Proof that only the last two digits matter"]
Any base 10 integer $ \overline{a_n\cdots a_1a_0}$ can be expanded as $ \sum_{k \equal{} 0}^{n}10^{k}\cdot a_k$. However, for $ k\geq 2$, this is $ 100\cdot 10^{k \minus{} 2}\cdot a_k\equiv 0\bmod 4$, so all of those terms are irrelevant $ \mod 4$.

For this reason, we only consider $ k \equal{} 0$ and $ k \equal{} 1$, which gives $ a_0 \plus{} 10a_1 \equal{} \overline{a_1a_0}$, which are the last two digits of the number.
[/hide]


limit
logarithms
calculus
calculus computations

It converges to 0. Let $f(x)=\left( 1+\frac{1}{x}\right)^{x}$ and use the properties of it.

Yes, use mean value theorem for $f(x)$ on interval $[n,n+1]$.

[quote="pardesi"]also u can use defiion of $e$[/quote]
That remark was not helpful. Sure, $\left(1+\frac1n\right)^{n}$ and $\left(1+\frac1{n+1}\right)^{n+1}$ both tend to $e,$ but what does that tell you about their difference?

My first instinct is to reduce everything to Taylor polynomials. Doing that, I noticed that

$\left(1+\frac1{n+1}\right)^{n+1}-\left(1+\frac1n\right)^{n}= O(n^{-2})$

That means that multiplying by $\sqrt{n}$ is not the interesting question - the limit is zero. Here's the interesting question:

What is $\lim_{n\to\infty}n^{2}\left[\left(1+\frac1{n+1}\right)^{n+1}-\left(1+\frac1n\right)^{n}\right]\ ?$

[quote="didilica"]More generally:

If $\lambda >0$ then

$\lim_{x\rightarrow \infty}x^{2}\left(\left(1+\frac{\lambda}{x+1}\right)^{x+1}-\left(1+\frac{\lambda}{x}\right)^{x}\right)=\frac{\lambda^{2}e^{\lambda}}{2}$,
hence Kent's limits is $\frac{e}{2}$.

I hope there are no mistakes.[/quote]

I didn't evaluated a general case, but I had $\frac{e}{2}$ as the value of $\lim_{n\to\infty}n^{2}\left[\left(1+\frac1{n+1}\right)^{n+1}-\left(1+\frac1n\right)^{n}\right]$, too.

dr kent can you please explain how you got this? $\left(1+\frac1{n+1}\right)^{n+1}-\left(1+\frac1n\right)^{n}= O(n^{-2})$

sorry i'm not much used to big oh and little oh :blush: . i went through the thing that you posted though. any more practice problems on problems using big oh and the like,are appreciated

sorry i almost didn't realise there was a $\sqrt{n}$ over there :o

 :(

Take logarithms: $\ln\left(1+\frac1n\right)^{n}=n\ln\left(1+\frac1n\right)$

$=n\left(\frac1n-\frac1{2n^{2}}+\frac1{3n^{3}}+O(n^{-4})\right)$

$=1-\frac1{2n}+\frac1{3n^{2}}+O(n^{-3})$

Now put that into the exponential:

$e^{1}e^{-\frac1{2n}+\frac1{3n^{2}}+O(n^{-3})}$

$=e\left[1+\left(-\frac1{2n}+\frac1{3n^{2}}+O(n^{-3})\right)+\frac1{8n^{2}}+O(n^{-3})\right]$

$=e\left(1-\frac1{2n}+\frac{11}{24n^{2}}+O(n^{-3})\right)$

By the same method,

$\left(1+\frac1{n+1}\right)^{n+1}= e\left(1-\frac1{2(n+1}+\frac{11}{24(n+1)^{2}}+O(n^{-3})\right)$

$=e\left(1-\frac1{2n}+\left(\frac{11}{24}+\frac12\right) \frac1{n^{2}}+O(n^{-3})\right)$

Subtracting, I get

$\left(1+\frac1{n+1}\right)^{n+1}-\left(1+\frac1{n}\right)^{n}=\frac{e}{2n^{2}}+O(n^{-3})$

When I made the earlier post, I was carrying one fewer term all through this, so I didn't know the details of the $n^{-2}$ term; this time I carried through that detail, confirming what sos440 said. Only slight modifications in the argument would be needed to reproduce didilica's result.

@dr kent-i think there are few typos. for example on the 3rd line, shouldnt it be $\frac{1}{3n^{2}}$, you have an extra minus sign there. but anyway i got a general idea of how to work with Big Oh, so thank you very much for spending time on details.

Yeah, it was pretty buggy. I think it's cleaned up now.


Do we have or not the following

For every real numbers a,b with 0<a<b<2 the system

sinx+siny=a
cosx+cosy=b

has a pair (x,y) of solutions, with 0<x< pi/2, 0<y<pi/2 .
Does something change if a>b?

Hmmm... Let a = 1.6, b = 1.4, so cosx + sinx + cosy + sin y = 3
cosx+sinx \leq (2(cos2x+sin2x))1/2 = 21/2 by arithmetic mean - quadratic mean. The same for y. So 3 = cosx + sinx + cosy + sin y = \leq 2\sqrt 2 < 3


All right, we all know that we didn't get a chance to do this awesome meet at FTW. Lets schedule it again, if anyone has any interest. Next week, when are you all free?

Woah next week?  It'll take some time to round up enough people from IL.

Man, I haven't done much FTW. Too busy with other classes

The forum is in a coma right now... no posts


Let $k>1$ be an odd integer. For every positive integer n, let $f(n)$ be the greatest positive integer for which $2^{f(n)}$ divides $k^n-1$. Find $f(n)$ in terms of $k$ and $n$.

Looking $\mod 2^s$ for any $s$ shows us that the divisibility by $2^s$ doesn't depend on odd factors of $n$. Thus let $n=2^s$ be a power of two now.
For $s \geq 1$ we have $k^{2^s} - 1 = (k^{2^{s-1}} - 1)(k^{2^{s-1}} + 1)$.
Looking $\mod 4$ gives that $k^{2^{s-1}} + 1$ is divisible by $2$ exactly once as long as $s>1$, thus $f(2^s)=f(2^{s-1})+1$ then.
So we need just to find out $f(2^0)=f(1)$ and $f(2^1)=f(2)$ which we get from $k$: $f(1)$ is the number of times $k-1$ is divisible by $2$ and $f(2)$ the number of times $k^2-1=(k+1)(k-1)$ is divisible by $2$.
Thus as result: $f(n)=v_2(n)+v_2(k-1) + v_2(k+1) - 1$ for $v_2(n) >0$ and $f(n)=v_2(k-1)$ oherwise (see the formulary, post #3, for the definition of $v_2$).

Approach by [url=http://www.mathlinks.ro/profile.php?mode=viewprofile&u=29721]Erken:[/url]

Since $ k$ is odd,therefore numbers $ k \minus{} 1$ and $ k \plus{} 1$ are both representable in the form $ 2^{s_1}\cdot r_1$ and $ 2^{s_2}\cdot r_2$,where $ s_1,s_2\geq 1$ and $ r_1,r_2$ are some odd numbers.
Also let $ n \equal{} 2^{s}\cdot r$.
Note that
\[ k^{2^{s}\cdot r} \minus{} 1 \equal{} (k^{2^{s}} \minus{} 1)(k^{2^{s}(r \minus{} 1)} \plus{} \dots \plus{} 1)
\]
Clearly,number $ (k^{2^{s}(r \minus{} 1)} \plus{} \dots \plus{} 1)$ is odd,therefore, it is enough to find $ f(n)$ such that $ 2^{f(n)}\parallel{}k^{2^{s}} \minus{} 1$.
Consider two cases:
1.$ n$ is odd.Then the answer is obviously $ \boxed{f(n) \equal{} s_1}$.

2.$ n$ is even,i.e $ s\geq 1$.It follows that
\[ k^{2^{s}} \minus{} 1 \equal{} (k \minus{} 1)(k \plus{} 1)(k^2 \plus{} 1)\dots(k^{2^{s \minus{} 1}} \plus{} 1)
\]
each number of the form $ k^{2^{i}} \plus{} 1$,where $ 1\leq i\leq s \minus{} 1$ is divisible by $ 2$,but is not divisible by $ 4$,therefore

$ (k \minus{} 1)(k \plus{} 1)(k^2 \plus{} 1)\dots(k^{2^{s \minus{} 1}} \plus{} 1) \equal{} 2^{s_1 \plus{} s_2 \plus{} s \minus{} 1}\cdot c$,where $ c$ is some odd integer.

Therefore the answer is $ \boxed{f(n) \equal{} s_1 \plus{} s_2 \plus{} s \minus{} 1}$.


I'm trying to learn Quantum Mechanics.  I'm reading Griffiths.  Right now I'm trying to understand spin.

Early on I came to understand that (according to QM) every particle has a "wave function" associated with it.  This "wave function" is a function of x,y,z, and t (space and time coordinates) that has certain properties.  Okay, fine.  I'm cool with that.

The "state" of a particle at a time t0, as I understand it, is just that particle's wave function, with the time t0 held fixed.  So the "state" of a particle is a certain function in Hilbert space.  Fine.

Now I get to spin.  And they seem to start contradicting themselves and telling me other things.  Suddenly they are saying that the "state" of a certain electron is a certain VECTOR that has only two components.  How could this be??  Please excuse me, but this blatantly contradicts what I had previously learned the "state" of a particle was.

And now Griffiths (p. 210, 2nd Edition) is trying to tell me "The *complete* state of the electron includes not only its position wave function, but also a spinor, describing the orientation of its spin:  Psi(r)*Chi(s)  "  [I'm sorry, I'm such a noob that I'm not using LaTeX.]

Here Chi(s), as I understand it, is a VECTOR that has two components.

I get the feeling that Griffiths is not telling me the full truth about what a wave function is.  I think he tried to dumb it down for me earlier, and now is afraid to admit that a wave function is in fact some more complicated object.  But Griffiths can't protect me by shielding me from the truth.  If I'm going to understand QM, I must know the full truth about what a wave function is, no matter how complicated that may be.  In his attempt at simplicity, he has simply contradicted himself and confused me.

I'm sure that I'm terribly confused.  Please clarify anything if you can.

By the way, I don't mean to criticize Griffiths, except for this one point I've found confusing.  This is my favorite QM book, and Griffiths is an AMAZING author.  He is my favorite physics textbook author of all, even including Feynman.

I can't answer your question, but try http://www.physicsforums.com
There are a bunch of physicists/grad students/etc. who are more than qualified to answer these kinds of things

Thanks!  That's great.  I didn't know about that forum.


Let $ \overline{MN}$ be a diameter of a circle with diameter $ 1$. Let $ A$ and $ B$ be points on one of the semicircular arcs determined by $ \overline{MN}$ such that $ A$ is the midpoint of the semicircle and $ MB\equal{}\frac35$. Point $ C$ lies on the other semicircular arc. Let $ d$ be the length of the line segment whose endpoints are the intersections of diameter $ \overline{MN}$ with the chords $ \overline{AC}$ and $ \overline{BC}$. The largest possible value of $ d$ can be written in the form $ r\minus{}s\sqrt{t}$, where $ r$, $ s$, and $ t$ are positive integers and $ t$ is not divisible by the square of any prime. Find $ r\plus{}s\plus{}t$.

I kinda guessed on this one, and am surprised that I got the right answer. This was my train of thought:
[hide]
Draw the diagram. I could see that the length is about less than 0.1...maybe 0.07.

Realize that $ 4\sqrt {3}$ is very close to $ 7$, but less than $ 7$.

Assume $ \sqrt {3} \equal{} 1.732$, and compute $ 4\sqrt {3}$, which turns out to be about $ 6.928$.
The answer $ 7 \minus{} 4\sqrt {3}$ approximates itself to be around $ 0.072$.
[/hide]

[hide="Solution"]
[asy]unitsize(7cm);
defaultpen(linewidth(.8pt)+fontsize(10pt));

pair O=(0,0), M=(-.5,0), A=(0,.5), Np=(.5,0);
pair B=intersectionpoint(Circle(M,.6), Circle(O,.5)), C=.5*dir(225);
pair P=extension(B,C,M,Np), Q=extension(A,C,M,Np);

draw(Circle(O,.5));
draw(M--Np);
draw(O--A--C--cycle);
draw(O--B--C);

label("$M$",M,W);
label("$N$",Np,E);
label("$A$",A,N);
label("$B$",B,N);
label("$C$",C,SW);
label("$P$",P,NW);
label("$Q$",Q,1.6*dir(30));
label("$O$",O,SE);[/asy]Let $ ACO = CAO = t$, which varies. Then $ MOC = \frac {\pi}{2} - 2t$. Since $ BCA$ always subtends the same arc, it is constant; let it be $ k$. Angle chasing yields that $ PQC = \frac {\pi}{2} - t$ and $ QPC = \frac {\pi}{2} + k + \theta$. By Law of Sines on $ \triangle COP$ and $ \triangle CPQ$,
\[ \frac {CP}{\cos(2t)} = \frac {\frac12}{\cos(k + t)} \\
\frac {PQ}{\sin(k)} = \frac {CP}{\cos(t)} \\
\]

\begin{eqnarray*} PQ & = & \frac {\cos(2t)\sin(k)}{2\cos(t)\cos(k + t)} \\
& = & \sin(k)\frac {\cos^2(t) - \sin^2(t)}{2\cos^2(t)\cos(k) - 2\sin(t)\sin(k)\cos(t)} \\
& = & \sin(k)\frac {1 - \tan^2(t)}{2\cos(k) - 2\sin(k)\tan(t)} \\
& = & \frac {1 - \tan^2(t)}{2\cot(k) - 2\tan(t)} \end{eqnarray*}
Substitute $ u = \tan(t)$; we must now calculate $ \cot(k)$. Let the midpoints of $ MB$ and $ BA$ be $ X$ and $ Y$, respectively. Note that $ OX\perp MB$ and $ OY\perp AB$. By Pythagoras, $ OX = \frac25$, so the are of $ \triangle OBM$ is $ \frac {3}{25}$ and the altitude from $ B$ to $ \overline{OM}$ has length $ \frac {12}{25}$. Let its foot be $ F$; again, by Pythagoras, $ FO = \frac {7}{50}$. This leaves $ \sin\left(\frac {\pi}{2} - 2k\right) = \cos(2k) = \frac {24}{25}$. By the tangent half-angle formula,
\[ \cot(k) = \sqrt {\frac {1 + \cos(2k)}{1 - \cos(2k)}} = \sqrt {\frac {49}{1}} = 7
\]
We now seek to maximize
\[ PQ = \frac12\frac {u^2 - 1}{u - 7} = \frac12\left(u + 7 + \frac {48}{u - 7}\right) = 7 - \frac12\left(7 - u + \frac {48}{7 - u}\right).
\]
By AM-GM, $ 7 - u + \frac {48}{7 - u}\ge2\sqrt {48}$, so $ PQ\le7 - \sqrt {48}$ or $ PQ\le 7 - 4\sqrt3$. Equality is achieved when $ 7 - u = 4\sqrt3$ or $ t = \tan^{ - 1}(7 - 4\sqrt3)$. This is a valid value for $ t$, where $ C$ is close to diametrically opposite $ A$, so we're done. Our answer is $ PQ = 7 - 4\sqrt3$ for $ r + s + t = \boxed{014}$.[/hide]

Wow....
Obviously I couldn't follow anything you wrote, but you forgot one thing  :P. You forgot to add $ r \plus{} s \plus{} t$ ;)

Very nice solution above, but I think you mean $ MOC \equal{} \frac{\pi}{2}\minus{}2t$, not $ MCO$. The other solution I saw [hide="went along other lines:"]

Let $ x\equal{} \frac{CM}{CN}$. Use the side-sine-side triangle formula to get $ [AMC]: [ANC] \equal{} \frac{3x}{4}$ and $ [BMC]: [BNC]\equal{}x$. Then use base-height to get
\[ \frac{MP}{1\minus{}MP} \equal{} \frac{3x}{4} \Rightarrow MP\equal{}\frac{3x}{3x\plus{}4} \qquad \frac{1\minus{}NQ}{NQ}\equal{}x \Rightarrow NQ\equal{}\frac{1}{x\plus{}1},
\]
which means $ PQ\equal{}1\minus{}\frac{3x}{3x\plus{}4}\minus{}\frac{1}{x\plus{}1}\equal{}\frac{x}{3x^2\plus{}7x\plus{}4} \equal{} \frac{1}{3x\plus{}7\plus{}\frac{4}{x}}$, which can be maximized by various methods. The non-calc way would be to set the denominator equal to $ m$, multiply by $ x$ to get $ 3x^2\plus{}(7\minus{}m)x\plus{}4\equal{}0$, and set the discriminant to zero.
[/hide]

It feels like less computation than the solution above (maybe because ratios feel easier to compute? iunno), but also a lot less intuitive (in my opinion).  :)

Let $ a \equal{} MP$ and $ b \equal{} QN$. After you get $ \frac a{1 \minus{} a} \equal{} \frac34x$ and $ \frac {1 \minus{} b}b \equal{} x$, substitute to get $ \frac {4a}{1 \minus{} a} \equal{} \frac {3 \minus{} 3b}b\implies4ab \equal{} 3 \minus{} 3a \minus{} 3b \plus{} 3ab\implies(a \plus{} 3)(b \plus{} 3) \equal{} 12$
$ \implies(a \plus{} 3) \plus{} (b \plus{} 3)\ge2\sqrt {12}\implies1 \minus{} a \minus{} b\le7 \minus{} 4\sqrt3\implies7 \plus{} 4 \plus{} 3 \equal{} \boxed{14}$ (AM-GM; equality holds when $ a \equal{} b \equal{} 2\sqrt3 \minus{} 3$).

[quote="worthawholebean"]
Let $ ACO = CAO = t$, which varies. Then $ MOC = \frac {\pi}{2} - 2t$. Since $ BCA$ always subtends the same arc, it is constant; let it be $ k$. Angle chasing yields that $ PQC = \frac {\pi}{2} - t$ and $ QPC = \frac {\pi}{2} + k + \theta$. By Law of Sines on $ \triangle COP$ and $ \triangle CPQ$,
\[ \frac {CP}{\cos(2t)} = \frac {\frac12}{\cos(k + t)} \\
\]
[/quote]
It should be $QPC=\pi/2+t-k$. Then by law of sines on triangle COP it should be

\[ \frac {CP}{\cos(2t)} = \frac {\frac12}{\cos(k - t)} \\
\]

Anybody know any simpler way? the solution is extremely long and even uses calculus (!) so any simpler solution would be appreciated.

Problem can be found here: http://www.artofproblemsolving.com/Wiki/index.php/2009_AIME_II_Problems/Problem_15

http://www.artofproblemsolving.com/Forum/viewtopic.php?p=1455383&sid=1e289e443b6689b175a665f21ace1045#p1455383

There are more elementary solutions here, but this is one of the hardest AIME problems (in my opinion after doing the recent ones), and there aren't really "simple" solutions. This of course depends on your definition of simple

I converted a common solution so that it won't require trig.
[hide="Lemma"]
Given a cyclic quadrilateral $ABCD$, $\frac{[ABC]}{[ADC]}=\frac{AB\cdot BC}{AD\cdot DC}$.
[/hide]
[hide="Proof"]
Note that $\angle ABC+\angle ADC=180$. Let $M,N$ be on $AB,BC$ respectively such that $BM=AD$ and $BN=CD$. Construct point $P$ such that $ADC\cong MBP$. $\angle MBP+\angle MBN=\angle ABC+\angle ADC=180$, so $P,B,N$ are collinear. Thus the height from $M$ to $BN$ equals the height from $M$ to $BP$, and $PB=BN$, so $[ADC]=[MBP]=[MBN]$. By same base, $\frac{[ABC]}{[ABN]}=\frac{BC}{BN}$. By same base, $\frac{[ABN]}{[MBN]}=\frac{AB}{BM}$. Multiplying these with substitution gives us $\frac{[ABC]}{[ADC]}=\frac{AB\cdot BC}{AD\cdot DC}$, as desired.
[b]Remark:[/b] We did not explicitly use the fact that $ABCD$ is cyclic. We can reword the lemma as follows:
[b]Lemma:[/b] Given triangles $ABC$ and $DEF$ with $\angle ABC+\angle DEF=180$, we have $\frac{[ABC]}{[DEF]}=\frac{AB\cdot BC}{DE\cdot EF}$.
and we can prove it with the following simpler lemma.
[b]Lemma:[/b] Given triangles $ABC$ and $DEF$ with $\angle ABC=\angle DEF$, we have $\frac{[ABC]}{[DEF]}=\frac{AB\cdot BC}{DE\cdot EF}$.
[/hide]
[hide="Solution"]
Let $BC$ and $AC$ intersect $MN$ at points $P$ and $Q$ respectively, and call $\frac{CM}{CN}=x$. Then
\[\frac{MP}{NP}=\frac{[BMC]}{[BNC]}=\frac{BM\cdot CM}{BN\cdot CN}=\frac{3x}{4}\]
$NP=1-MP$, so $MP=\frac{3x}{3x+4}$. Similarly,
\[\frac{MQ}{NQ}=\frac{[AMC]}{[ANC]}=\frac{AM\cdot CM}{AN\cdot CN}=x,\]
giving $MQ=\frac{x}{x+1}$. Because $MQ-MP=d$, $\frac{x}{x+1}-\frac{3x}{3x+4}=d$, or equivalently, $3dx^2 +(7d-1)x+4d = 0$. The discriminant, $d^2 - 14d + 1\ge0$, and $0\le d < 1$, so the maximum value of $d$ is $7 - 4\sqrt{3}$.
[/hide]

[quote="v_Enhance"][url=http://www.artofproblemsolving.com/Forum/viewtopic.php?f=133&t=506457]Do not repost old problems[/url]. Merging[/quote]
Sorry. I tried to search for it but ended up with no results. It is either the search engine is not powerful enough, or I am simply doing it wrong...

Look through the contests section. It is easy to find and nagivate (at the top of your screen, just keep reading right from private messages and you'll find it). You can access all the problem threads from there. (Also this was said in the first two posts of the thread v_enhance linked too, so reading would be helpful next time)

[hide=projective]Let $D = AC \cap MN$ and $E = BC \cap MN$. By ptolemy on $MBAN$, $BA = \frac{\sqrt{2}}{10}$, so
$$(M,D;E,N) \stackrel{C}{=} (M,A;B,N) = \frac{BA}{BM} \cdot \frac{NM}{NA} = \frac{1}{3}.$$
Then if we let $ME = x$ and $ED = y$, we have
$$\frac{1}{3} = (M,D;E,N) = \frac{ED}{EM} \cdot \frac{NM}{ND} = \frac{y}{x(1-x-y)} \implies y = \frac{-x^2 + x}{3+x}.$$
Take the derivative and stuff to find that this is maximized at $x = -3 \pm 2 \sqrt{3}$, so take $x = -3 + 2 \sqrt{3}$ so $y = 7 - 4 \sqrt{3}$ and the answer is $7+4+3 = \boxed{014}$.[/hide]

Posting official solution because the others might still be confused/not know calculus. (Mods please delete my post if the official pamphlet is copyrighted by MAA.)

sorry for bump, but in the @above solution, what makes the ratio $\frac{MP}{NP}=\frac{BMC}{BNC}$ valid?

Edit:nevermind, i got it[hide]for reference, because $MP$and $PN$ are bases of the triangles $BMP$,$MPC$and$NCP$,$PBN$ richard really explains the intuition well in the corresponding math jam

[hide=Solution]
Let $\angle BMN = a$. Let the intersection of BC and MN be T and the intersection of AC and MN be S. By Law of Sines, we can get that $MT = \frac{3sin(a)}{3sin(a) + 4cos(a)}$, and we can quickly angle chase to get that $\angle CAN = 90-a$, and by Law of Sines again we can get that $SN = \frac{cos(a)}{sin(a)+cos(a)}$. Now, we wish to maximize $1-\frac{3sin(a)}{3sin(a) + 4cos(a)}-\frac{cos(a)}{sin(a)+cos(a)}$. This is the same as maximizing $\frac{4cos(a)}{3sin(a) + 4cos(a)} + \frac{sin(a)}{sin{a} + cos{a}} -1$. Let $y = \frac{4cos(a)}{3sin(a) + 4cos(a)}$. We can get that $cos(a) = \frac{3ysin(a)}{4-4y}$. We can simplify the expression we want to maximize with this substitution and get that we must maximize $y+ \frac{4-4y}{4-y}-1$. This is trivial to do, and we get that the max value in the range we want is $7-4\sqrt{3}$, so our answer is $14$.
Hope there was no typos!(There may be lol)

[/hide]

Wait yay I got a super geometric solution :D Diagram is not to scale because lazy

Let all points be what they look like in the diagram, where $Y'$ and $C'$ are reflections of $Y$ and $C$ over $AO$. Simple angle chase gives that $ABXY'$ is cyclic. We have
$$XY=2TO-TX+TY'\leq 2TO-2\sqrt{TX \cdot TY'} = 2TO-2\sqrt{TA \cdot TB} = 2TO-2\sqrt{TO^2-\frac{1}{4}}.$$
and now we just need to solve for $TO$. By the ratio lemma claim in many of the above solutions, $\frac{TM}{TN}=\frac{4}{3}$, implying $TM=4$, $TN=3$, and $TO=\frac{3+4}{2}=\frac{7}{2}$, as desired.

Let $AC$ and $BC$ intersect $MN$ at $D$ and $E$. Note that
\[\frac{4}{3}=(M,N;A,B)\stackrel{C}{=}(M,N;D,E)=\frac{DM}{DN}\div \frac{EM}{EN}\]
so if $DM=4x\cdot DN$ and $EM=3x\cdot EN$ we find that
\[d=DM-EM=\frac{4x}{4x+1}-\frac{3x}{3x+1}=\frac{x}{12x^2+7x+1}\]
and by taking the derivative (this is actually a very clean one) we find that
\[x=\sqrt{\frac{1}{12}}=\frac{\sqrt{3}}{6}\implies d=7-4\sqrt{3}\implies \boxed{014}\]
and we're done. $\blacksquare$

[quote=OlympusHero]Progress: Firstly, we will compute $\angle DCE$. We can do this by first finding the measure of arc MB, which we will do by using LoC on $\triangle MOB$. If $\cos \angle MOB = p$, then $\frac{1}{4}+\frac{1}{4}-2 \cdot \frac{1}{2} \cdot \frac{1}{2} \cdot p = \frac{9}{25}$, which gives $p = \frac{7}{25}$. Then we have $\cos\left(90-\cos^{-1}\left(\frac{7}{25}\right)\right) = \frac{24}{25}$, and using the half angle formula we get $\cos \angle DCE = \frac{5\sqrt2}{14}$.

How can I get $DC$ in terms of $EC$ (or vice versa) in order to use LoC on $\triangle DEC$?[/quote]

Still curious about this??


If $ p \ge q \ge 5$ and $ p$ and $ q$ are both primes, prove that $ 24 \mid (p^{2}\minus{}q^{2})$.

[hide="solution"] For all primes $ p$ greater than or equal to $ 5$, $ p^2\equiv{1}\pmod{3}$ and $ p^2\equiv{1}\pmod{8}$, so, $ p^2\equiv{1}{\pmod{24}}$. This means $ p^2\minus{}q^2\equiv{1\minus{}1}\equiv{0}\pmod{24}$, so we are done. [/hide]


[color=olive]Use mathematical induction to show that (n to the power 5) - n is divisible by 5 for every positive integer n.[/color]

Sorry again for the expression. Thanks guys

If p prime $p|(n^{p}-n)$, because or $p|n$ or $(p,n)=1\Longrightarrow n^{p-1}=1(mod \ p)$.


Thanos18 and Hxtung play a game for the sake of money. They select a natural number n and determine the biggest odd divisor k. If k can be represented as k = 4z+1 (z \in N) than Thanos18 pays Hxtung one Euro and if not vice versa. After that they select the successor of n and make perform the same steps etc. At a certain point they stop the game.
Prove that Hxtung will win !

mr: decomposition of factors
hr: analysis of the situation, restructuring (arranging according powers of two), systematic attempts

Still nobody, after two years and one day? :)

the names sound interesting.... im not good at game theory, but ill give it a try.

What is the "successor of n?" Is it just another number? Or is it k?


induction
inequalities proposed
inequalities


Can someone else try the TeXer and see if it's working? It stopped working for me yesterday, I really need it for my lab report.  Thanks!

Seems like it is down indeed. I'm moving this to the support forum.


$x\ge 0,\ y\ge 0\Longrightarrow (x+y)^3\le 4(x^3+y^3)$. Generally, for $n\in N^*$ we have: $x\ge 0,\ y\ge 0$ $\Longrightarrow$ $(x+y)^n\le 2^{n-1}(x^n+y^n)$.

[quote="conceive"]Prove the following statement. If $x^3+y^3=2$ for some x,y reals, then $x+y \le 2$.[/quote]This inequality should not be here. We can use the following
\[ \frac{x^3+y^3}2\geq\Bigl(\frac{x+y}2\Bigl)^3. \]
Note that $(x+y)^3=x^3+y^3+3xy(x+y)\leq x^3+y^3+3\bigl(\frac{x+y}2\bigl)^2(x+y)$. The desired inequality follows.

Hey, I didn't tell you that x,y are positive.. Virgil Nicula supposed it. Power mean is not necessary true for negative or zero values.

In the text of problem I posted is written that x,y are reals. Not only positive.


$p$ and $q$ represent numbers, and $p*q$ means $\frac{p+q}{2}$.  what is the value of $3*(6*8)$?

There are three solutions in this thread, and only one is needed. Please don't post a solution unless you have a different way of doing it or you think the previous solution is wrong.


For every $1\leq a,b,c,d\leq 2$, prove that \[\frac{4}{3}\leq \frac{a}{b+cd}+\frac{b}{c+da}+\frac{c}{d+ab}+\frac{d}{a+bc}\leq 2.\]

$1 \ge (a-1)(b-1)$
so $\frac{a}{b+cd}\ge \frac{a}{b+c+d}$
then by Cochy-Shvarc-Bunjakovski
$(a+b+c+d)(\frac{1}{a+b+c}+\frac{1}{b+c+d}+\frac{1}{c+d+a}+\frac{1}{d+a+b}) \ge \frac{16}{3}=4+\frac{4}{3}$

notice that the expression is convex in all the variables so for the maximum value we can check the endpoints :)

btw, I'm 99% sure this was posted before :wink:

and I'm 99% sure that I saw this in one St.Peterburg olympiad, it was there not longer then 5 years ago.

Me too for 2nd ineq , but first ineq I can't  :(
[quote="Albanian Eagle"]notice that the expression is convex in all the variables so for the maximum value we can check the endpoints :)

[/quote]
Thanks both friends.


Identify A-F in the following sequence... :maybe:

Ok....I'll post  the structures of A, B, C the next time I'm online! (dont expect too much! I'm not too sure  :blush: )

Yes! source is an Indian book!  :D  :)

[quote="pardesi"][quote="Carcul"]Ok, then thell us what are their structures. I am very curious. By the way, what is the source of all these problems? An indian book?[/quote]
me posting after a long time....
carcul u must have got used to indian "reactions and chemicals" by now!!! :D  :P[/quote]

You still read the chem forum?  :o

According to me

[hide]A is 2-methyl cyclohexan1,3-dione
B is but3-ene-1-yne
C is but-3-ene-2-one.  :maybe: [/hide]

Two moles of acetylene are coupled with cupric chloride to yield vinylacetylene (this is a well known reaction). Next, what is the product from the reaction of an alkyne with a solution of $ \ce{HgSO4}$ in diluted sulphuric acid?

That is just simple markownikoff addition of water

Yes, that's right. In this case the product should be MVK, which is a very reactive molecule.

[quote="Da Vinci"]

You still read the chem forum?  :o

[/quote]
chem forum roks da reminds me of ma skool days i enjoyed studying oc ....i miss that time  :rotfl:  :rotfl:  :blush:  :(  :(

[quote="hell_ever"]What is reaction C---->B ? What does CuCl2 do?[/quote]
hey hell check out mtg's oc book it has lots of such coupling reaction


inequalities
trigonometry
inequalities unsolved

Let $ABC$ be a triangle, find the smallest possible constant $\lambda$ such that $(\lambda+\cos^2 A)(\lambda+\cos^2B)(\lambda+\cos^2C)$ attains its minimum and determine this minimum.

For $\lambda=1$, the problem is still hard enough.

[quote="zhaobin"]I don't think I can understand you :([/quote]
What do you mean, Zhaobin?

I mean, find the minimum value of
\[ (1+\cos^2A)(1+\cos^2B)(1+\cos^2C).  \]


inequalities
logarithms
function
inequalities proposed

Suppose $ a_{1},  ... , a_{n}$ are positive reals, with $ n \geq 2$, such that $ a_{1} a_{2} ...  a_{n} \equal{} 1$.
Prove that: $ \frac{1}{a_{1} \plus{} 1} \plus{} ... \plus{} \frac{1}{a_{n}\plus{}1} \leq \frac{a_{1} \plus{} a_{2} \plus{} ... \plus{} a_{n} \plus{} n}{4}$.

[quote="epitomy01"]Suppose $ a_{1}, ... , a_{n}$ are positive reals, with $ n \geq 2$, such that $ a_{1} a_{2} ... a_{n} \equal{} 1$.
Prove that: $ \frac {1}{a_{1} \plus{} 1} \plus{} ... \plus{} \frac {1}{a_{n} \plus{} 1} \leq \frac {a_{1} \plus{} a_{2} \plus{} ... \plus{} a_{n} \plus{} n}{4}$.[/quote]
$ \sum_{i\equal{}1}^n\left(\frac{a_i\plus{}1}{4}\minus{}\frac{1}{a_i\plus{}1}\right)\geq0\Leftrightarrow\sum_{i\equal{}1}^n\left(\frac{a_i\plus{}1}{4}\minus{}\frac{1}{a_i\plus{}1}\minus{}\frac{\ln{a_i}}{2}\right)\geq0.$
Let $ f(x)\equal{}\frac{x\plus{}1}{4}\minus{}\frac{1}{x\plus{}1}\minus{}\frac{\ln{x}}{2}.$ Then $ f'(x)\equal{}\frac{(x\minus{}1)(x^2\plus{}x\plus{}2)}{4x(x\plus{}1)^2}.$
Thus,  we obtain: $ f(x)\geq f(1)\equal{}0.$
Id est, our inequality is proved.

[quote="pleurestique"]you only proved that $ f$ is increasing on $ [1,\infty)$, but the $ a_i$ need not be in that interval. am i missing something?[/quote]
Yes, I think, you are missing. $ f$ is a decreasing function on $ (0,1].$ Hence, $ x_{min} \equal{} 1.$ Thus, $ f(x)\geq f(1) \equal{} 0$ for all $ x>0.$ :wink:


No calculators are allowed:

Simplify $\sqrt{\log_{2}3 \times \log_{2}12 \times \log_{2}48 \times \log_{2}192+16}-\log_{2}12 \times \log_{2}48+10.$

[hide="bad progress"]Converting everything to $\log_{2}3$ gives:

$4 sqrt{4(\log_{2}3)^{4}+1}-8 (\log_{2}3)^{2}+10$

Now what?[/hide]


integration
logarithms
calculus
calculus computations

the antiderivative of  (-4x)/(x^2  +  3)  dx

i got -2+c is that right at all, if not can comebody help me

You can use the formula $ \int \frac {f'(x)}{f(x)}\ dx \equal{} \ln |f(x)| \plus{} C.$

$ \frac{d}{dx}\left(\minus{}2\ln (x^2\plus{}3)\plus{}C\right)\equal{}\minus{}\frac{4x}{x^2\plus{}3}$, so your answer is correct.


Let $ P(x)$ be a polynomial of degree $ n, n\in N$, such that $ P(k)\equal{}2^k$ for each $ k\equal{}1, 2,..., n\plus{}1$. Determine $ P(n\plus{}2)$.

Obviosly $ P(x)\equal{}2\sum_{i\equal{}0}^n C_{x\minus{}1}^i$, therefore $ P(n\plus{}2)\equal{}2\sum_{i\equal{}0}^nC_{n\plus{}1}^i\equal{}2^{n\plus{}2}\minus{}2.$


I think this one is a bit difficult imo, feel free to move it if you deem necessary.

prove

[tex]\displaystyle{\Big(1+\frac{1}{2}+\frac{1}{3}+\ldots +\frac{1}{2004}\Big) > 8}[/tex]

Does this have a clever solution?  Cause I 'spect that just integrating 1/x to find the lower bound will do it, but that's boring.

Unfortunately the lower bound is in fact too low...

[hide]That's not a problem though. We can just take enough terms by hand to make it work out and then do the rest with an integral.[/hide]


I've noticed that no one is putting internal links in any of the question statements.  Is there a reason for this?  It seems to me that reasonably frequently a question use a word that won't actually appear in the solution to that question -- in those cases, that word doesn't have an internal link associated with it.  Is this bad?

After thinking about it for a little bit, I think it would be a good idea to have links in the problem.  Let's make that the trend, unless anyone comes up with a reasonable objection.

[quote="MCrawford"]I thought about that a while ago and for some reason my asthetic inclination was to encourage the use of the interesting, internally-linkable words in the solutions.  I'm not sure why, but I felt like it wouldn't look as nice to have internal links in the problems.  But I think I'm changing my mind.  I do think it might be nice to keep the large problem files (like the page with ALL 2006 AMC 12B problems) free of internal links, but to include them on the specific pages with the solutions.  Am I just being weird in that desire?  I still can't put my finger on exactly why I care at all, which is unusual for me.[/quote]

I have exactly those same sentiments.  I suppose there is something about maintaining the exact, original wording of the problem and even so much as linking a word in it seems to be going  against this.  Additionally, it seems a little odd to be encouraging people to go to other places in the wiki while they are practicing problems.  The tests they will end up taking wont be open book, so they shouldn't practice with an open book.

Your suggestion of putting the links on the individual problem pages only is a great solution, in my opinion.  I support doing things that way.

[quote="joml88"]Your suggestion of putting the links on the individual problem pages only is a great solution, in my opinion.  I support doing things that way.[/quote]

I agree with the aesthetic notes, and with this.

The reason it struck me was that I was reading some question which referenced a whole bunch of things (the integers, a function, something else), which then appeared only symbolically in the solution ($\mathbb{Z}, f(x)$, etc.) and so didn't contain any of the nice linkyness that wikis are supposed to be good for.

So, when I stumble across individual question pages in the future, I'll wikify them.

On that note, the process of using the wiki to find, type up, or write solutions to contest problems is starting to take on a level of complication that needs documentation.  If somebody builds a "Contest Problems Standards" page, I'll add to it and also link it to the main page of the wiki (or through a page linked from the main page deealing with contest problems).

One thought: I've been continuing to not link up problems where there is no solution.  Basically, if we can put all the links into the solution, I think that's good, but that requires that we wait to see what the solution actually looks like.

I think either way is fine.  I don't think there's any harm in going ahead and linking the problem statement where there is no solution yet.  At the very least, it helps people know what terms and concepts they need to look up if they have confusion with, or difficulty with the problem.


Find the remainder when $ 8\cdot10^{18}\plus{}1^{18}$ is divided by 9.

It is obviously going to be an $ 8$ and then a whole bunch of $ 0's$, and then a $ 1$.  If a number is divisible by $ 9$, the sum of its digits is also divisible by $ 9$.  $ 8\plus{}1\equal{}9$, so our answer is $ \boxed{0}$