function
induction
combinatorics proposed
combinatorics

Prove that for each $ n$:
\[ \sum_{k\equal{}1}^n\binom{n\plus{}k\minus{}1}{2k\minus{}1}\equal{}F_{2n}\]

we form generating function of $ \sum_{k \equal{} 1}^n\binom{n \plus{} k \minus{} 1}{2k \minus{} 1}$

$ \sum_{n\geq0}x^n\sum_{k \equal{} 1}^n\binom{n \plus{} k \minus{} 1}{2k \minus{} 1}$=$ \sum_{k\geq0}\sum_{n}\binom{n \plus{} k \minus{} 1}{2k \minus{} 1}x^n$=$ \sum_{k\geq0}x^{1 \minus{} k}\sum_{n}\binom{n \plus{} k \minus{} 1}{2k \minus{} 1}x^{n \plus{} k \minus{} 1}$=$ \sum_{k\geq0}x^{1 \minus{} k}\sum_{r}\binom{r}{2k \minus{} 1}x^r$=$ \sum_{k}x^{1 \minus{} k}\times \frac {x^{2k \minus{} 1}}{(1 \minus{} x)^{2k}}$=$ \sum_{k}(\frac {x}{(1 \minus{} x)^2})^k$=$ \frac {(1 \minus{} x)^2}{1 \minus{} 3x \minus{} x^2}$
but if we open $ {1 \minus{} 3x \minus{} x^2}$ we will see that the root of $ {1 \minus{} 3x \minus{} x^2}$ are the squares of root of $ 1 \minus{} x \minus{} x^2$
as we know that $ f_{n}$=$ \frac {\alpha^n \minus{} \beta^n}{\sqrt5}$ where $ \alpha$ and$ \beta$
are root of $ 1 \minus{} x \minus{} x^2$ after opening $ \frac {1}{1 \minus{} 3x \minus{} x^2}$as generating form we get to the point

I have another approach.  Let $ f_k$ be the number of ways to tile a $ k$-board using one-tiles and two-tiles.  Then $ f_k\equal{}F_{k\plus{}1}$ where $ F_{k\plus{}1}$ is the $ (k\plus{}1)^{th}$ term of the fibonacci sequence.  Specifically, $ F_{2n}\equal{}f_{2n\minus{}1}$.  So let's count the number of ways there are to tile a $ (2n\minus{}1)$ board.  As indicated above, there are $ f_{2n\minus{}1}$ ways.  But let us count these tilings in another way.  Since the length of the board is an odd number we know there is at least one $ 1$-tile.  Furthermore there is always an odd number of $ 1$-tiles.  Say there are $ 2k\minus{}1$ $ 1$-tiles, then there are $ 2n\minus{}1\minus{}(2k\minus{}1)\equal{}2(n\minus{}k)$ tiles left on the board which can be covered with $ n\minus{}k$ $ 2$-tiles.  In all there are $ (2k\minus{}1)\plus{}(n\minus{}k)\equal{}n\plus{}k\minus{}1$ tiles from which we are selecting $ 2k\minus{}1$ of them (the $ 1$-tiles) for placement.  There are
$ \displaystyle \binom{n\plus{}k\minus{}1}{2k\minus{}1}$
ways.  Summing over all possible $ k$, we have
$ \displaystyle \sum_{k\equal{}1}^n \binom{n\plus{}k\minus{}1}{2k\minus{}1}\equal{}f_{2n\minus{}1}\equal{}F_{2n}$
as desired.

it is very easy if we use snake oil method...:D

Note that $\sum_{n+k=m}{\dbinom{n}{k}}$ denotes the $m+1$th fibonacci number.This is easy to see by by induction on $m$ and noting the fact that $F_{m+2}=F_{m+1}+F_{m}=\sum_{n+k=m}{\dbinom{n}{k}}+\sum_{n+k=m-1}{\dbinom{n}{k}}=\sum_{n+k=m}{\dbinom{n}{k}+\dbinom{n}{k-1}}=\sum_{n+k=m}{\dbinom{n+1}{k}}=\sum_{n+k=m+1}{\dbinom{n}{k}}$ where we have used the induction hypothesis assuming the statement to be true for all integers upto $m+1$.Now $\sum_{k=1}^{n}{\dbinom{n+k-1}{2k-1}}=\sum_{k=1}^{n}{\dbinom{n+k-1}{n-k}}=\sum_{j+k=2n-1}{\dbinom{j}{k}}=F_{2n}$ :D

This problem is a good example of showcasing the power of [i]generating functions.[/i]
My solution if basically the same as #$2$, but I am still posting it for completeness.

$\text{Sol}^n$.
 It is sufficient to show that :
$$\sum_{n \ge 0} \sum_{k=1}^{n} \dbinom {n+k-1}{2k-1} X^n = \sum_{n \ge 0} F_{2n} X^n$$

$\bullet$ The Left hand side factors as:
 \begin{align*} \sum_{k=1}^{n} \dbinom {n+k-1}{2k-1} & = \sum_{k \ge 1} \sum_{n \ge 0} \dbinom{n+k-1}{2k-1} X^n\\&
                                                                                = \sum_{k \ge 1} X^{1-k} \sum_{n \ge 0} \dbinom{n+k-1}{2k-1} X^{n+k-1}\\&
                                                                                = \sum_{k \ge 1} X^{1-k} \cdot \dfrac {X^{2k-1}}{(1-X)^{2k}} \\&
                                                                                = \dfrac {X}{1-3X +X^2} \end{align*}

$\bullet$ Now factoring the Right hand side is easy using the [u]Binet's [/u] formula : $F_n = \dfrac{1}{\sqrt{5}}\cdot ( \phi ^{n} - \phi ^{-n}))$.
 \begin{align*}  \sum_{n \ge 0} F_{2n} X^n & = \sum_{n \ge 0} \dfrac{1}{\sqrt{5}} \cdot ( \phi ^{2n} - \phi ^{-2n}) X^n \\&
                                                           = \dfrac {1}{\sqrt{5}} \cdot ( \dfrac {1}{1- \phi ^2X }- \dfrac {1}{1- \phi ^{-2}X})\\&
                                                          =  \dfrac {X}{1 - 3X + X^2}  \blacksquare  \end{align*}

Another similar one:
Prove that
$$\sum_{i=1}^{n}\left(\sum_{k=0}^{2i-2}\binom{2i-k-2}{k}\right)=F_{2n}$$


topology
advanced fields
advanced fields unsolved

I have got a problem in point set topology:

Suppose $ A\subseteq B$ can we prove that the boundary $ \partial A$ of $ A$ is a subset of the boundary $ \partial B$ of $ B$?

Seems it is not true although it is true for closure and interior of sets

What I want is a counter-example  :!:

Consider $ \mathbb{R}$ with the finite complement topology, i.e. $ U$ is open if $ \mathbb{R} \minus{} U$ is finite or all of $ \mathbb{R}$. Take $ A \equal{} \mathbb{N}$ and $ B \equal{} \mathbb{R} \minus{} \{\pi\}$ such that we indeed have that $ A \subseteq B$. I will use the definition $ \partial(X) \equal{} \bar{X} \minus{} X^{\circ}$. Now, the closure of $ \mathbb{N}$ is $ \mathbb{R}$ and the interior is $ \emptyset$ such that $ \partial(A) \equal{} \mathbb{R}$. On the other hand we have that $ \mathbb{R} \minus{} \{\pi\}$ is open and that its closure is $ \mathbb{R}$. Hence $ \partial(B) \equal{} \mathbb{R} \minus{} (\mathbb{R} \minus{} \{\pi\}) \equal{} \{\pi\}$. Clearly we have that $ \partial(A) \nsubseteq \partial(B)$.

A simpler couterexample,
Consier $ \mathbb{R}$ with the natural metric topology,
$ A \equal{} \mathbb{Q}$,$ B \equal{} \mathbb{R}$,
we have:$ \partial A\equal{} \mathbb{R}$,$ \partial B\equal{} \emptyset$


calculus
real analysis
real analysis theorems

Trying to decide between the two for single variable calculus. Does anyone have any experience with one of both of them who can weigh in?

I don't know much about Apostol so I can't tell you anything.  But I have experience with Spivak.  It's a great book, and I think it does a really good job of completely building up calculus from the basics.  There are a lot of problems at the end of each chapter-some of which are easy, many of which are quite difficult.  But you'll definitely learn a lot from that book.

Apostol's book is good. But -for me- is an easy book. Spivak, is very different... More excercises, much more levels of different and interesting excercises.  :!:


Find the length of the common chord between the two circles $ x^2 \plus{} y^2 \plus{} 2kx \plus{} c \equal{} 0$ and $ x^2 \plus{} y^2 \plus{} 2py \minus{} c \equal{} 0$

Ans- $ 2\sqrt {\frac {(k^2 \minus{} c)(p^2 \plus{} c)}{p^2 \plus{} k^2}}$

The line is y = -kx/p
We find point of intersection and get distance.

Now some easier method way out?


superior algebra
superior algebra theorems

Suppose that $G$ is a nonempty set and $*$ is a binary operation defined on $G$. Furthermore, suppose that for all $a$ and $b$ in $G$, the equations $a*x=b$ and $y*a=b$ have solutions.

Prove that $<G,*>$ is a group.

You need to assume associativity. Without that assumption, there are examples like this:
$G=\{a,b,c\}$ with $a=a\cdot a=b\cdot c=c\cdot b$, $b=b\cdot b=a\cdot c=c\cdot a$, and $c=c\cdot c=a\cdot b=b\cdot a$.

With associativity, it's pretty simple.

ok, yes you do need associativity. sorry about that.

can you give me a hint?

Start by solving the equations $ax=a$ and $ya=a$ for some choice of $a$. Show that the solutions are 1-sided identities, and then that they are equal. From here, all you need are inverses.


can someone explain to me what the correct answer to #6 is?
is it 16 or 17?

and how did you do it?

the answer should be 17.
you find the general expression for every number.
n*2^(n-1)+2^n(k-1), n is the number of the row and k is the number of the column.
k<=51-n,
you can get 2k+n=69.
Only n=1, 3, ...33 satisfy.
(1+33)/2=17

[hide="hint"]Also, if you look at 3 numbers in a row,
$ x\minus{}b\hspace{.5in}x\hspace{.5in}x\plus{}b\\
\hspace{.25in}2x\minus{}b\hspace{.5in}2x\plus{}b\\
\hspace{.5in}4x$
you see that a number in a row is 4 times the number 2 above it.[/hide]


Given parabola $f(x)=ax^2+bx+c$ and line $g(x)=-bx$($b \neq 0$). $a,b,c \in R$, $a>b>c$, $a+b+c=0$.
(1) Prove that $f(x)$ and $g(x)$ meet at two different points.
(2) If the two functions meet at $A$ and $B$, the prejections of $A$ and $B$ at x-axis are $A_1$ and $B_1$. Find the range of $|A_1B_1|$.

1
[hide]
The two intersection points are where $f(x)=g(x)\implies ax^2 + bx + c = -bx \implies ax^2 + 2bx + c = 0$.  For there to be two intersection points, the discriminant must be greater than zero, or $4b^2 - 4ac > 0 \implies b^2 > ac$
Now there is simple case checking
Case 1 - $b < 0$
If $b < 0$, c must be less than zero since it is less than b.  $b^2>0$ by the trivial inequality, and the only way the RHS can be greater than 0 is if $a$ is also negative, making $ac$ positive.  However, then all three are negative and can't satisfy $a+b+c=0$.
Case 2 - $b > 0$
Likewise, $a>0$ since it's greater than b.  But $c \not >0$ since then all three are positive and can't sum to 0.  So $c < 0$, and $ac < 0$ again.
Case 3 - a or c = 0
If either is 0, the product is 0, and since $b \not = 0$, $b^2 > 0 = ac$

Thus $b^2 > ac$ is always satisfied and there are always two different intersection points.
[/hide]


geometry
3D geometry
sphere
calculus
function
integration
trigonometry

In the figure, two planes intersect a sphere of radius 12 inches in circles which are tangent at A. Segment AB is a diameter of the sphere, and the planes cut off a 30 degree arc, BC. What is the surface area of the odd-shaped lune? What is the volume of the odd-shaped region?

Treat the the weird shape as the revolution of f(x) around the y axis,

where f(x)  = :sqrt: (144-x :^2:) and 3( :rt2: +  :sqrt:6  )  :ge: x :ge: 0.

I think everyone with some backgrounds in calculus should be able to

evaluate the volume and surface area that way.

No calc! (or at least minimal calc.) This is intermediate! That's like using generating functions to count the number of ways to flip heads with a coin! (ok, so im exaggerating :)) But i assure you that there are easier methods to solve this. (especially the volume part!)

I believe I've seen this done using frustrums, but its late so for now I'll use some calculus.

[hide]First to find the volume of the odd-shaped region we have to find the volume of the top piece.  Let's say the top piece has height H.  From here we can break up the piece into n stacked cylinders with height  :tri: h.  The radius for a given cylinder is  :sqrt: (r :^2: -(k :tri: h ) :^2: ).  So as we let n tend to infinity and  :tri: h tend to zero the sum of the volumes of the cylinders is

 [tex]\int_0^H \pi (r^2-h^2)\; dh[/tex]

thus making the volume of the top piece equal to  :pi: r :^2: H- :pi: /3*H :^3: .  In order to find the volume of the odd shaped region simply subtract the volume of top piece from th volume of the upper hemisphere: V=2/3 :pi: r :^3: - :pi: r :^2: H+ :pi: H :^3: /3.  WE know the radius of the sphere is 12 now we only need to find H.  Since the angle betwee planes is 30 andthe radius of the sphere is 12 we can use the law of cosines or otherwise to find that the diameterof the top circle is 12 :rt3:  making the radius 6 :rt3: .  Using the properties of the 30 , 60, 90 triangle we find that the altitude for triangle AOC to be 6(assuming O is the center of the sphere).  Therefore H=12-6=6.  And so the volume for the odd shaped region is V=2/3 :pi: (12) :^3: - :pi: (12) :^2: (6)+ :pi: (6) :^3: /3=360 :pi: .[/hide]

That's enough for me tonight.  Until I get more sleep I hope this is right.

i don't  know calculus.  but i really should.  so for now:

a noncalc solution for the volume:

[hide]

you realize the area you are searching for is half the volume of the circle minus the portion closed off by the plane going at a 30 degree angle.

so you draw your  AC and your AB, and tilt the sphere so that it looks like a circle

if the center of the sphere is O, now draw AO and OC, and voila, a 120-30-30 triangle

because of this 120 degree angle, the volume of the portion of the triangle enclosed by OA, OC, and minor arc(or whatever you call it) AC would be, maybe 1/9 of the triangle?  this part i'm not so sure about.  i'll figure it out shortly

so now you have the area of that region to be 256:pi:

so now we find the volume of the cone

use the law of cosines to fine the AC to be 12 :rt3:

then drop an altitude from O to AC, and that's just 6

so the volume of the cone is 1/3 (6:rt3:):^2::pi:

subtract this from the 256:pi: to get 220:pi: as the area of the chord (or it's equivalent in 3 dimensions) AC

find the area of the whole sphere, which is 4/3 (:pi:)(12:^3: )=256:pi:x9

take half of this, subtract 220:pi: from it, and that should be the volume of whatever we're looking for.

actually, i wouldn't be surprised if this answer was wrong.  i made some unwarranted assumptions in the solution

[/hide]

i would assume the area could be figured out similarly.  i'd do it if i knew portion of the sphere whos volume we are finding.

Hmm, in the spirit of this thread:

Unwarranted assumption #1: the area is in the shape of a circle.

[hide]Where the center of the sphere is R and the center of circle AC is S

ARC is a 30/120/30 triangle.  RC is a 12 inch radius; by similar triangles, SC, the radius = 6 :sqrt: 3

So, the area is pi*r^2 = 104pi.  Supposedly.

Volume calc: the top part is a spherical cap - it doesn't matter which way it's aligned, so long as the distance from the center , 6 :sqrt: 3, is the same.

Thus, integrate y^2  = -x^2 + 144 between 0 and 6 :sqrt: 3

pi*(-x^2 + 144 )dx

pi*(-1/3x^3 + 144x)dx

pi*(-72 + 864)

=792pi...  I do believe everyone's gotten a different answer so far.. owell..

[/hide]

Will leave in embarressment soon


Good luck for all writing the AHSMC tomorrow.

[quote="soledad42"]It's easier than I expected, especially number 5.[/quote]

Come ooooon... [old man voice] it was pretty hard. xD.

[quote="Mathgod"] Come ooooon... [old man voice] it was pretty hard. xD.[/quote]

I'll say that #1, #2 and #4 are routine, #5 is random but easy, and #3 could be hard if one doesn't see it (not for me though  :D)

So how did everyone do?

[quote="soledad42"]

#3 could be hard if one doesn't see it (not for me though  :D)

[/quote]

Gah.

[quote="Newanda"]I was hoping someone could tell me how to solve this old AHSMC 2005 problem.

ABCD is a quadrilateral inscribed in a circle, with AB=AD. The diagonals intersect at E. F is a point on AC such that the measure of angle BFC is the measure of angle BAD. If the measure of angle BAD is twice that of DFC, determine BE/DE.[/quote]

After some angle-chasing, you'll find that $ \angle CDF\equal{}\dfrac{\pi}{2}$. The rest is trivial. =p


geometry
incenter
inradius
geometry unsolved

Let ABC a triangle with 90º > Aº > Cº, and D a point of AC such that BA = BD, the incircle of ABC meets AB and AC on K and L respectively, J is the incenter of the triangle CBD. Show that the line KL meets AJ in M:midpoint of AJ.

Let $X$ be point on $AK$ such that $KX = KA$ and $Y$ a point on $AL$ such that $LA = LY$. Let $Z$ be a midpoint of $XY$ and$S$ a incenter of $ABC$. By simple angles we have $SBI \sim SKZ$, so $SKB \sim SZI$, so $\angle SZI = 90^\circ$, so$I$ lies on $XY$. Thus by Thales $AM = MI$. :D

Let: $\measuredangle A=2\alpha$, $\measuredangle B=2\beta$, $r$ and $R$ are the inradius of $BCD$ and $ABC$ respectively, $D' \in CD$ such that $D'J=DJ$.

Then: $D'L=r(tg\alpha-ctg\beta)+R.ctg\beta$, $LA=R.ctg\alpha$, $DB=r[tg\alpha+ctg(\alpha-\beta)]$, $AB= R[tg(\alpha+\beta)+ctg\alpha]$

We have to show: $JM=MA$, which is equivalent to: $\boxed{\frac{r}{R}=\frac{tg\alpha-tg\beta}{tg\alpha-tg\alpha tg\beta}}$

And this can be proved considering that: $DB=AB$

Dear Mathlinkers,
an article concerning this problem with an original proofs has been put on my website
http://perso.orange.fr/jl.ayme    vol. 5 (last article)
Use Google translator
Merry Christmas
Sincerely
Jean-Louis

Just to clarify j555's proof above: he renamed the incenter of $\triangle BCD$ as $I$ instead of $J.$


The Happy Valley Kennel has 4 chickens, 2 dogs, and 5 cats.  (Some people in Happy Valley like to keep chickens as pets!)  In how many ways can the 11 animals be placed in a row of 11 cages, such that all of the animals of each type are in adjacent cages?  (Two animals of the same species are considered distinguishable.)

[hide]We will glue the cages together in groups of $ 4,2,$ and $ 5$ such that each species goes in only one group. In the first group, there are $ 4!$ ways to arrange the chickens. In the second group, there are $ 2!$ ways to arrange the dogs. In the last group, there are $ 5!$ ways to arrange the cats. Finally,
\[ 4!\cdot2!\cdot5! \equal{} 24\cdot2\cdot120 \equal{} {5760}\]
But now we must arrange the groups of cages. There are $ 3!$ ways to do this. \[ 5760\cdot3!\equal{}5760\cdot6\equal{}\boxed{34560}\]
$ \Box$[/hide]

Thanks, myyellowducky.

I disagree--someone please confirm.

We have to order the 3 groups of animals, which is 3!
There are 4! was to arrange the chickens, 2! ways to arrange the dogs, and 5! ways to arrange the cats.

$ 3! \cdot 4! \cdot 2! \cdot 5! \equal{} \boxed{34560}$

did anyone notice that it is cool that the product of the factorials of 2,3,4,5 is equal to 34560
both cases(except for the 0) the digits increase by 1

and what do you disagree with

[hide] There are 6 ways to arrange the dogs, cats and chickens. Then they are 120 ways to arrange the car, 24 ways to arrange the chicken and 2 ways to arrange the dog. 120*24*6*2 = 34560 (Notice there are factorials in the problem)[/hide]

I think that this problem should be of easier level than 14. I think this problem should be from Level 5 to Level 9 as it is too easy and straightforward. I've solved Level 10's harder than this.

Some people might think it's hard, some might not. Everyone has strengths and weaknesses. AoPS does a good job of making an estimated level.


[hide]First, draw another line of length four that is perpendicular to the sides of length 8 through the point of tangency of the two circles. Then, we know that the circles split the line into lengths of 2. We can connect the radii of the circle through the point of tangency and form right triangles. We form the equation $\sqrt {r^2 - 4} + r = 4$. Solve, and we get 2.5 as the radius but $\fbox {5}$ as the diameter.[/hide]

I remember doing that problem at some point. I guessed and got it right. 5 was my lucky answer. ;)

I think Tarquin has a much better solution though.

But remember, even if you don't know how to do a problem at a competition, NEVER LEAVE IT BLANK!!!!!! (Unless of course, you're feeling really nice, and want to give the other people an automatic advantage. As all of us know, MATHCOUNTS doesn't have any "blank" answers. I wish they did though.) :)  :)  :)

[quote="Press alt f4"]Amazingly enough for me 13 is my lucky answer and when I guess 13 it happens to be right.
Never say 13 is unlucky!   :lol:[/quote]

13 is also lucky in Penguin! culture

[hide]I connected the two centers.  That itself is the length of the diameter.  I made a right triangle where I knew one of the legs was 4.  Therefore, x(the other leg)+y(diameter)=8 and x^2+16=y^2.  So I did x^2-y^2=-16 which is (x+y)(x-y)=-16.  8(x-y)=-16 so x-y=-2.  We now have two equations, x-y=-2 and x+y=8.  2y=10 so y=5.  The diameter is 5.[/hide]

[quote="kingof21penguins"]my lucky number is 2 but it is almost never right[/quote]

To contrast, it has been statistically proven that when a problem asks for a whole number solution, the answer is most often 2. The chances are still very small, of course.


induction
number theory proposed
number theory

Let $b$ and $c$ be any two positive integers. Define an integer sequence $a_n$, for $n\geq 1$, by $a_1=1$, $a_2=1$, $a_3=b$ and $a_{n+3}=ba_{n+2}a_{n+1}+ca_n$.

Find all positive integers $r$ for which there exists a positive integer $n$ such that the number $a_n$ is divisible by $r$.

I find the following result:
 If $\gcd(b;c)=\prod ^{k}_{i=1}p_i^{t_i},p_i\in\mathbb P,t_i\in\mathbb N$ then all the required numbers are of the form:
    $\prod^{k}_{i=1} p_i^{k_i}.m,\gcd(m;c)=1$.
Would I hope it's right? :(

It's unsolved for months,anybody has the solution?

Whats the problem in been unsolved for months¿ There are lots of problems that weren't solved for years.

But good, here's my solution:

Lemma 1:
Let $d$ be a product of common prime divisors of $b$ and $c$, or equivalently: $d|\gcd(b,c)^k$ for big enough $k$.
Then there is some $N$ such that $d|a_n$ for all $n\geqq$.

Prove:
Lets write $g$ instead of $\gcd(b,c)$.
we have that $g|a_3=b , a_4=b^2+c , a_5=b^4+b^2c+c$. But when $g^l | a_m , a_{m+1} , a_{m+2}$, then $g^{l+1} | a_{m+3} = b a_{m+2} a_{m+1}+ca_m$. thus, by induction on $l$ respectivly $m$, we get that $g^l | a_n$ for all $n \geq 3l$, thus $d|a_n$ for all $n \geq k$ (that $k$ defined in the statement).
q.e.l.

Lemma 2:
Let $d$ be such that $\gcd(d,c)=1$.
Then there are infinetly many $n$ such that $d|a_n$.

Prove:
There are only finitely many tripels $(a_n,a_{n+1},a_{n+2}) \mod d$. This gives that the sequence gets cyclic $\mod d$ after some time with period $p$. Now it's simple to see that we can revert the recursion of the sequence $\mod d$, thus it starts its period at $a_0 \equiv 0 \mod d$, which we get by calculating back from the others. Thus we get that all numbers $a_{pk}$ are divisible by $d$.
q.e.l.

Lemma 3:
Let $d$ be such that it has at least one prime factor $p$ in common with $c$ where $p$ doesn't divide $b$.
Then $d$ divides none of the $a_n$.

Prove:
Looking at the sequence $a_n \mod p$ gives $a_{n+3} \equiv b a_{n+2}a_{n+1}$. Since $p \nmid a_1=a_2=1 , a_3=b$, it doesn't divide any of the $a_n$ by induction. But when $p$ doesn't divide $a_n$, then $d$ doesn't too.
q.e.l.

Conclusion:
The Lemmas give that exactly that numbers $d$ divide some (and infinitely many) of the $a_n$ that have the following property:
every prime divisor of $d$ that divides $c$ divides $b$ too.


Find the period of small oscillations of a water pogo, which is a stick of mass m in the shape of a box (a rectangular parallelepiped.) The stick has a length L, a width w and a height h and is bobbing up and down in water of density ρ. Assume that the water pogo is oriented such that the length L and width w are horizontal at all times. Hint: The buoyant force on an object is given by Fbuoy = ρVg where V is the volume of the medium displaced by the object and ρ is the density of the medium. Assume that at equilibrium, the pogo is floating.

I got the right answer, setting the buoyant force equal to ma. However, I don't understand why mg does not need to be factored in to find the answer. Why is this? Doesn't mg also contribute to ma?

Weirdly enough, I didn't have to use $ ma$ at all, and I got ${ 2\pi \sqrt{\frac{m}{\rho wLg}}}$

Perhaps the $ mg$ is implicitly contained within the concept of the buoyant force, for $ F_{buoy}$ exist only because gravity exerts a downward push against the surrounding water. Therefore, gravity "secretly" contributes to the upward force on the pogo.

$ F \equal{} kx \equal{} \rho Vg \equal{} \rho Lwxg$

$ k \equal{} \rho Lwg$

$ \omega \equal{} \sqrt{\frac{k}{m}} \equal{} \sqrt{\frac{\rho Lwg}{m}}$

$ T \equal{} \frac{2\pi}{\omega} \equal{} 2\pi\sqrt{\frac{m}{\rho Lwg}}$

And I just realized that $ mg$ does appear: $ \rho V$ is $ m$.

Consider a spring with a hanging mass. Note that the spring is already stretched to some equilibrium position due to weight (as opposed to mass). The period of small oscillations around the [i]equilibrium[/i] position is the same, no matter what the weight is -- the only thing that changes is what that equilibrium position already is.

In the same way, here, the weight of the object only affects the equilibrium height (in particular, where weight = bouyant force), and not the period of small oscillations.

In other words, what you're doing here is finding the relationship between [i]additional[/i] bouyant force and [i]displacement from equilibrium[/i], since the equilibrium bouyant force cancels out with the weight.

If you are not already comfortable with this, I suggest you explore it using the equation of motion.


This competition occured on February 1. I was just wondering.  :)

Heh, our school skipped it because of snow... sure sucks


Sixteen people attended a party, and each person brought a gift for everyone else at the party. Altogether, how many gifts were brought to the party?

Each of $ 16$ people bought $ 15$ gifts, so $ 16\cdot15\equal{}8\cdot30\equal{}\boxed{240}$.


Each term of a sequence, after the first term, is inversely proportional to the term preceding it, and the constant of proportionality stays the same. If the first term is 2 and the second term is 5, what is the 12th term?

Inverse proportion can be expressed as $ xy\equal{}k$.
This can be make into $ x\equal{}\frac {k}{y}$.
Let $ x$ be our term, and $ y$ be the PROceeding term.  So we start off with $ 2\equal{}\frac {k}{5}$.  Obviously, we get $ k\equal{}10$.  So then we make $ x\equal{}5$ to find the third term.  $ 5\equal{}\frac {10}{y}$.  We get $ y\equal{}2$.  Now we notice, like modularmarc101, that the pattern is $ 2,5,2,5,2...5$.  Every even term is a $ 5$, so the $ 12th$ term is indeed $ \boxed {5}$


geometry
circumcircle
geometric transformation
reflection
symmetry
incenter
angle bisector

In the acute-angled triangle $ ABC$, $ D$ is the intersection of the altitude passing through $ A$ with $ BC$ and $ I_a$ is the excenter of the triangle with respect to $ A$. $ K$ is a point on the extension of $ AB$ from $ B$, for which $ \angle AKI_a\equal{}90^\circ\plus{}\frac 34\angle C$. $ I_aK$ intersects the extension of $ AD$ at $ L$. Prove that $ DI_a$ bisects the angle $ \angle AI_aB$ iff $ AL\equal{}2R$. ($ R$ is the circumradius of $ ABC$)

$ (O)$ is the circumcircle of the $ \triangle ABC,$ $ (M)$ is the circumcircle of $ B ICI_a,$ $ M \in (O),$ $ M$ is the midpoint of $ II_a.$ Regardles of the problem proposition, $ \angle AI_aB = \frac {\angle C}{2}.$ Reflect the circumcircle in the angle bisector $ AII_a$ into a circle $ (O'),$ $ O' \in AD,$ because $ AD, AO$ are isogonals WRT the angle $ \angle A.$ Also regardles of the problem proposition, let $ AL$ be a diameter of $ (O'),$ $ L \in AD,$ $ AL = 2R.$ By symmetry, $ (O'), (M)$ intersect at $ B' \in AC, C' \in AB,$ the reflections of $ B, C$ in $ AI.$ $ BDLC'$ is cyclic because of the right angles at $ C', D,$ $ \overline{AB} \cdot \overline {AC'} = \overline {AD} \cdot \overline {AL}$ and this is equal to power of $ A$ to $ (M),$ hence $ \overline{AD} \cdot \overline {AL} = AM^2 - MI^2.$ Since $ M \in (O'),$ $ ML \perp AM.$ Then $ I_aL^2 = IL^2 = MI^2 + ML^2 = MI^2 + AL^2 - AM^2 = AL^2 - \overline{AD} \cdot \overline {AL} = \overline{AL} \cdot \overline{DL}.$ It follows that the $ \triangle AI_aL \sim \triangle I_aDL$ with the common angle at $ L$ are similar by SAS and $ \angle DI_aL = \angle I_aAD = \frac {\angle B - \angle C}{2}.$

Let $ I_aL$ cut $ AB$ at $ K.$ $ I_aD$ bisects the angle $ \angle AI_aB\ \Longleftrightarrow$ $ \angle AI_aD = \frac {\angle C}{4}\ \Longleftrightarrow$ $ \angle AI_aK = \angle AI_aL = \angle AI_aD + \angle DI_aL = \frac {\angle B}{2} - \frac {\angle C}{4}\ \Longleftrightarrow$ ${ \angle AKI_a = 180^\circ - \left(\frac {\angle A}{2} + \frac {\angle B}{2} - \frac {\angle C}{4}\right) = 90^\circ + \frac {3}{4}\angle C}.$

Denote the interior angles of $\Delta ABC$ simply as $\angle A, \angle B, \angle C$. Let $O$ be the circumcenter, $I$ the incenter of $ABC$, the incircle touches $BC, CA$ at $E, F$, respectively; $AI$ meets the circumcircle again at $P$, $AO$ meets the circumcircle again at $Q$. $QI$ meets the circumcircle again at $H$. Let the A-excircle with center $I_{a}$ touches $AB, BC, CA$ at $G, R, S$, respectively.
Notice that $\angle QAC = \angle BAD = 90^{\circ} - \angle B$, so $AI$ is also the angle bisector of $\angle DAQ$.
First we are going to show that $\angle DI_{a}A = \angle RI_{a}Q$
It is well known that $H, E, P$ are collinear (Let $PO$ meets the circumcirle again at $T$. Notice that $PB = PI = PC$. As $angle EIP = \angle IAQ$ and $IE/AI = IF/AI = sin(A/2)$ and $IP/AQ = PB/PT = sin(A/2), \Delta IEP\sim \Delta AIQ$, thus $\angle IQA = \angle IPE$, then $H, E, P$ are collinear).
By sinus law in $\Delta AI_{a}C$ we have $AI_{a}/cos(C/2) = AC/sin(B/2) = AQ.sinB/sin(B/2)$ and in $\Delta AIB$: $AI/sin(B/2) = AB/cos(C/2) = AD/sinB.cos(C/2)$.
Thus $AI/AQ = AD/AI_{a},\Delta AI_{a}Q\sim \Delta AID$, this means first $\angle AI_{a}Q = \angle ADI$. Second, if $AD$ meets $PH$ at $U$ then easy to see that $HAIU$ is cyclic and $\angle AUI = 90^{\circ}$, then $UIED$ is a rectangle. Then $\angle AI_{a}Q = \angle ADI = \angle UEI$, it follows that $\angle RI_{a}Q = \angle IPE$ and $AD/AI_{a} = AI/AQ = AU/AP, UP\parallel DI_{a}$. Therefore $\angle DI_{a}A = \angle EPI = \angle RI_{a}Q$.
Now easy to see that $AL = 2R = AQ\iff L$ is the reflection of $Q$ over $AI_{a}$ (then of course $L, P, Q$ collinear) $\iff \angle QI_{a}S = \angle KI_{a}G = 3/4 C\iff I_{a}Q$ is the angle bisector of $\angle RI_{a}C \iff \angle DI_{a}A = C/4\iff DI_{a}$ is the angle bisector of $\angle BI_{a}A (= C/2)$.

It is not hard,if we know this problem:

http://www.artofproblemsolving.com/Forum/viewtopic.php?p=1178972&sid=33b04ae6e251ae0fc66cc926e418dda0#p1178972


$x_{1},x_{2}. . . x_{n}$ is the decimal notation for a number $x$ (so, for example, if $x=135$ then $x_{1}=1, x_{2}=3$ and $x_{3}=5$).

Prove that

$\text{(i)}$ $x$ is a multiple of $3$ if and only if $x_{1}+x_{2}+. . .+x_{n}$ is;
$\text{(ii)}$ $x$ is a multiple of $9$ if and only if $x_{1}+x_{2}+. . .+x_{n}$ is;
$\text{(iii)}$ $x$ is a multiple of $11$ if and only if $x_{1}-x_{2}+x_{3}-. . x_{n}$ (where + and − signs alternate) is.

[hide="i & ii"]

We prove that each number is congruent to its digital sum mod 9: clearly this proves [b]i)[/b] and [b]ii)[/b].

Induction: it holds for 1.
Suppose that it holds for $n=k$. Then if $k$ does not end in a 9 it clearly also holds for $n=k+1$, since both the digital sum and the number have increased by 1.
We prove also that if it holds for $k$, it holds for $10k$. $10k \equiv k \mod 9$, and the digital sum of a number does not change by adding a 0 to the end.
Between these two inductive steps we prove the claim by induction.
[/hide]

[hide]
i) $10 \equiv 1 \pmod{3}$
ii) $10 \equiv 1 \pmod{9}$
iii) $10 \equiv-1 \pmod{11}$.
[/hide]

You can easily not use mods and do this problem...
[hide="solution"]
$xyza...c$ is the number then subtract $x+y+z+a....c$ to find the answer to the 1st two
the last one you have to subtract $x-y+z-a...c$
[/hide]


but yous are better friends!!! :) PLEES answer!!!!!!

[quote="asmallgoat"]Are there any math forum sites other than artofproblemsolving.com :(  cuz im not smart enough for you math people :([/quote]

this is probably the best math forum for people in grade school

Of course you are. Otherwise you wouldn't figure out how to register.

[quote="kstan013"][url=http://www.google.com/search?hl=en&q=math+forums]You should just ask your friend funcia![/url][/quote]
you know i should have expected this to happen when i suggested them to google or wiki
wiki is your friend  :lol:
http://en.wikipedia.org/wiki/Special:Search?search=math+forum&go=Go[/hide]

[quote="asmallgoat"]But Im not smzrt enough 4 it!!! :(  :(  :maybe:[/quote]
well then stay in the classroom math section or something


problem: let $ a;b;c \in R$ . Prove that:
$ (\frac{a\minus{}b}{b\minus{}c})^2\plus{}(\frac{b\minus{}c}{c\minus{}a})^2\plus{}(\frac{c\minus{}a}{a\minus{}b})^2\geq 5$

it is in "bat dang thuc suy luan va kham pha" of Pham Van Thuan; Le Vi but I don't know solve
can you help me?

[quote="litbon"]problem: let $ a;b;c \in R$ . Prove that:
$ (\frac {a \minus{} b}{b \minus{} c})^2 \plus{} (\frac {b \minus{} c}{c \minus{} a})^2 \plus{} (\frac {c \minus{} a}{a \minus{} b})^2\geq 5$
[/quote]
It is equivalent to $ \sum_{cyc}(a^3 \minus{} 2a^2b \minus{} a^2c \plus{} 2abc)^2\geq0.$
See also here:
http://www.artofproblemsolving.com/Forum/viewtopic.php?t=99509


Let E be the point of intersection of the diagonals of convex quadrilateral ABCD, and let P, Q, R, and S be the centers of the circles circumscribing triangles ABE, BCE, CDE and ADE, respectively, prove that PQRS is a parallelogram.

Odd because I know I have posted this exact same probelm before but I cant seem to find my link...

oh well, heres another link though

http://www.artofproblemsolving.com/Forum/viewtopic.php?search_id=1626699351&t=39013


After reading the 9-11 hoax about a article i posted here some times ago about weird 9-11 facts, I clicked familiar hoaxes and found this:

Type up "NYC" in capital letters on MS word and change it to font "wingdings"
A picture of a skull, star of david, and thumbs up sign shows up.

Microsoft claims this was just a coincidence.
Wingdings has over 50 different symbols. Combine that with all the keys on the computer keyboard. Somebody calculated that the chance of this happening was one in some number over a trillion.
Is this really a coincidence?
What do you think?

[quote="Totally Zealous"]Is this really a coincidence?[/quote]

Yes.

Probability calculations of real-life phenomena should be subject to quite a bit of suspicion, since the givens are never explicitly stated.  Any probability refers to the probability of an event happening given some incomplete knowledge of the situation leading up to the event.

What are the givens here?  If you phrase the problem as "What are the chances that NYC would happen to become a superficially meaningful symbolic statement in this particular font, Wingdings," then yes, the probability seems small.  If you phrase the problem as "What are the chances that any small fragment of text is translated to a superficially meaningful symbolic statement in a symbolic font," then coincidences become quite likely, and only seem extraordinary because [b]the negative cases are ignored.[/b]

This is an extremely common phenomenon (I don't know if it has a proper statistical name) that accounts for practically all of what people might call signs of a conspiracy or signs of a paranormal event.


If last month was July, then what month will be $ 22$ months from now?

[b]Last[/b] month was July, so this month is August. There are 12 months in a year, so $ 22 \minus{} 12 \equal{} 2$ months before August is our answer. Thus, our answer is $ \boxed{\text{June}}$.

EDIT: Oops. That's what I meant.  :oops:

[quote="isabella2296"] $ 22 \minus{} 12 \equal{} 2$ [/quote]

hmmm.....

you could just say $ 22\equiv\minus{}2\mod12$ so its 2 months before August


[quote="GameBot"]Given that $ a^2 \plus{} 2ab \plus{} b^2 \equal{} 144$, what is the greatest possible value for $ a \plus{} b$?[/quote]

$ a^2 \plus{} 2ab \plus{} b^2$ factors as $ (a \plus{} b)^2$, so the largest possible value for $ a \plus{} b$ is simply the positive square root of 144, or $ \boxed{12}$.


LaTeX
number theory proposed
number theory

Show that if $x, y, z$ are positive integers, then $(xy + 1)(yz + 1)(zx + 1)$ is a perfect square if and only if $xy + 1, yz + 1, zx + 1$ are all perfect squares.

Masoud Zargar

One of the most-often occuring problems...
Just searching for your first $\LaTeX$ code gave you seven topics with it (now after deleting and merging some are gone; now there are there are only three and this one left):
http://www.mathlinks.ro/Forum/viewtopic.php?t=4542
http://www.mathlinks.ro/Forum/viewtopic.php?t=63419 (link only)
http://www.mathlinks.ro/Forum/viewtopic.php?t=58472 (spanish communities)


Prove that every sequence of real numbers contains a monotone subsequence.

I'm quite sure you skipped a detail (namely, the length of the subsequence).

Otherwise it becomes:
[hide]
There are three cases:
i. All $ a_{i}$ are equal, in which case the sequence is constant
ii. There exists an $ a_{i}$ such that $ a_{i}> a_{i+1}$, in which case the subsequence $ a_{i}, a_{i+1}$ is monotone decreasing.
iii. There exists an $ a_{i}$ such that $ a_{i}< a_{i+1}$, in which case the subsequence $ a_{i}, a_{i+1}$ is monotone increasing.

[i](ii. and iii. are not necessarily mutually exclusive)[/i]
[/hide]


If $x$ and $y$ are each positive integers and if $x>y$, how many distinct ordered pairs $(x,y)$ exist such that $x^{2}-y^{2}=800$?

[hide]
We have (x+y)(x-y) = 800. Factoring 800, we find it is 2^5 * 5^2, so it has 18 factors and 9 pairs of factors so that x is greater than y.

The only pair that doesn't work is x + y = 32, x - y = 25 because 32 and 25 have different parity. All others pairs are both even. So the answer is 9 - 1 = 8.[/hide]

just wondering could you please state all the different sets for the {x,y} please?  I could only find 6 for some reason.

Oops you're right, actually I think there are 7 because 1 and 800 doesn't work either.

is there any equation for solving these kinds of problems?(just interested)


Let $a$, $b$ and $c$ be real numbers such that $a^2+b^2=c^2$, solve the system:
\[ z^2=x^2+y^2 \]
\[ (z+c)^2=(x+a)^2+(y+b)^2 \]
in real numbers $x, y$ and $z$.

[url=http://www.mathlinks.ro/Forum/viewtopic.php?t=52498]the solution is hère[/url]. :)

The second equation is $c^2+2cz+z^2=x^2+2ax+a^2+y^2+2by+b^2$.

After subtracting the given equalities, $2ax+2by=2cz$ and $ax+by=cz$.

Multiplying the two equalities, $(a^2+b^2)(x^2+y^2)=c^2z^2$.

But by the Cauchy-Schwartz Inequality, $(a^2+b^2)(x^2+y^2) \ge (ax+by)^2 = c^2z^2$.

We have equality here, so we must have $y^2 = \frac{b^2}{a^2} \cdot x^2$, and $z^2=x^2+y^2$. These are all the solutions.


calculus
integration
trigonometry
function
real analysis
real analysis unsolved

Can the integral $\int_{0}^{2\pi}\exp(-k\sin{\theta})d\theta$, be solved analytically?

k is a scalar constant

Thanks.

One can find a power series for it. The result will be a Bessel function of $k.$

Step 1: Expand out the power series for the exponential.

Step 2: Determine (in terms of $n$) the integral

$\int_{0}^{2\pi}\sin^{n}\theta\,d\theta$

(Note that this will be zero for all odd $n.$)


algebra
difference of squares
special factorizations

Source: USSR
Prove that the expression
$m^{5}+3m^{4}n-5m^{3}n^{2}-15m^{2}n^{3}+4mn^{4}+12n^{5}$ cannot have the value $33$, regardless of what integers are substituted for $m$ and $n$.

[hide="Hint"]The expression factors as $(m-2n)(m-n)(m+n)(m+2n)(m+3n)$.[/hide]

[hide]
factor by grouping and then difference of squares yields:

$(m+3n)(m^{2}+mn-2n^{2})(m^{2}-mn-2n^{2})$

someone do the case work...
[/hide]

Factoricing the expression we get $(m+3n)(m+2n)(m+n)(m-n)(m-2n)$ but $33$ can't be expressed this way.


algebra
polynomial
function
induction
rational function
algebra open

Well this problem 2 weeks ago came to my mind.
$P(x),Q(x) \in \mathbb{R}[x]$ and we know that for real $r$ we have $p(r) \in \mathbb{Q}$ iff $Q(r) \in \mathbb{Q}$
I want some conditions between $P$ and $Q$.My conjecture is that there exist ratinal $a,b,c$ that $aP(x)+bQ(x)+c=0$

Another thing (very obvious) with the condition $P(r) \in \mathbb{Z}$ iff $Q(r) \in \mathbb{Z}$ there exist $c$ that $P(x)+Q(x)=c$ or $P(x)-Q(x)=c$

I see nobody is intersted in this problem.

Another more general conjecture:
$P(x)\in\mathbb{Q}$ if $Q(x)\in\mathbb{Q}$ $\Rightarrow$ $\exists\,R(x)\in\mathbb{Q}[x]$ s.t. $P(x)=R(Q(x))$.

What do you think, Omid Hatami?

A similar problem was used in a Bulgarian Competition (See Contest from Around the World, some year), but with $deg P(x) = deg Q(x) = 2$

To Myth :Yes this conjecture also came to my mind but I forgot to send it

I think everyone has forgot this problem.

[quote="Myth"]You are wrong  ;)[/quote]

Yep you are wrong, Myth never forgets about a problem  :D

Myth's conjecture is far too beautiful, so spending time on it is a real pleasure. I think I've managed to prove his conjecture if the polynomials have rational coefficients, but I used quite heavy machinery. Was I blind and again this case is very easy pr not? Has anyone a normal solution for it?

I almost have proved the first conjecture (my conjecture) with a simple solution. I only need a theorem $\forall p \in \mathbb{Z}[x]$ there exist $c \in \mathbb{Q}$ that :$p(x)-c$ is irreducible.

Also I really want to see your proof. Could you send it ?

[quote="Omid Hatami"]I only need a theorem $\forall p \in \mathbb{Z}[x]$ there exist $c \in \mathbb{Q}$ that :$p(x)-c$ is irreducible.
[/quote]

Cool! :) I think this has been discussed here. We can even find an integer $c$ s.t. $P(x)+c$ is irreducible.

[quote="Myth"]Another more general conjecture:
$P(x)\in\mathbb{Q}$ if $Q(x)\in\mathbb{Q}$ $\Rightarrow$ $\exists\,R(x)\in\mathbb{Q}[x]$ s.t. $P(x)=R(Q(x))$.[/quote]

[i]Proof.[/i]
We may WLOG assume that leading coefficients of $P$ and $Q$ are positive. There is $M\in \mathbb R$ s.t. $\forall\,x>M,\,x\in\mathbb Q\ \exists y>M,\,y\in\mathbb Q$ that equations $Q(t)-x=0$ and $P(t)-y=0$ have a common root. Note: we consider maximal roots of the equaitions.
It follows that resultant $S(x,y)\equiv S(Q(t)-x,P(t)-y)=0$. This resultant is a polynomial of $x$ and $y$. So we know that for all rational $x\geq M$ the equation $S(x,y)=0$ has a rational root $y$. Let $a_1,...,a_k$ be a Hammel basis of coefficients of $S(x,y)$ over $\mathbb{Q}$. Thus we have $a_1S_1(x,y)+a_2S_2(x,y)+...+a_kS_k(x,y)=0$, where $S_i\in\mathbb{Q}[x,y]$. It means that the equation $S_1(x,y)=0$ also has a rational root $y$ for all fixed rational $x>M$.
Let's prove that then $y$ is a rational function of $x$. Indeed, chose $m\in\mathbb N$ that $T(x,y)=mS_1(x,y)\in \mathbb Z[x,y]$. We can write \[T(x,y)=T_r(x)y^r+T_{r-1}y^{r-1}+...+T_0(x),\] where $T_i\in\mathbb Z[x]$. Perform a substitution $z=T_1(x)y$, then \[z^r+F_{r-1}(x)z^{r-1}+...+F_0(x)=0\] also have a rational root $z$ for all rational $x>M$, $F_i\in\mathbb Z[x]$.
This last equation has a series of real roots, which can be asymptoticaly expressed as $c_1x^{p_1}+c_2x^{p_2}+...+c_lx^{p_l}+o(1)$, where $p_1>p_2>...>p_l\geq 0$ are rational numbers.
Now remember that we consider maximal roots for very large $x$ and $y$. It means that $z$ continuously depends on $x$, so $z(x)=c_1x^{p_1}+c_2x^{p_2}+...+c_lx^{p_l}+o(1)$ :!: for some fixed $p_i,c_i$. Besides of that we know that $z$ is an integer number if $x\in\mathbb Z$, since it is a rational number and satisfies a monic polynomial equation with integer coefficients.
Consider $x=n^{sp}$, where $n,s\in \mathbb N$ and $p$ is a common denominator of $p_1,p_2,...,p_l$.  We obtain that $z(n^{sp})=G(n)+o(1)$, $G$ is a usual polynomial. Suppose $G$ has a nonrational coefficient. Then as we know $\{N\cdot G(n)\}$ is densely distributed on $[0,1]$ for some natural $N$ (we need $N$ in order to kill leading rational coefficients). See [url=http://www.mathlinks.ro/Forum/viewtopic.php?t=18079]Distribution in [0..1][/url] topic for details.

It follows that $z(n^{sp})$ is also densely distributed on $[0,1]$ -- contradiction. Therefore $G$ has only rational coefficients.
Finally, we are to prove that $p_1,..,p_n$ are integer numbers. Suppose the contraty, then consider $\Delta^{j}z(n)$ for an appropriate $j$, and you will see that $\{\Delta^j z(n)\}$ is densely distributed on $[0,1]$.

Thus we proved that $y(x)$ is a rational function, but it means that polynomials $P$ and $Q$ are in a rational dependence with integer coefficients (since $P(t)=y$ rationaly depends on $Q(t)=x$ for infinitely many large $t$), so $P(x)=R(Q(x))$, where $R\in\mathbb Q[x]$.

That's all.

How many mistakes here? :(

P.S. As you see, I have left some details unreavealed for you. I hope you will complete them.
Don't beat me very much.

[Edit: I inserted a phrase "if $x\in \mathbb Z$" into the place, where I say that $z$ should be an integer]

I have just added a link for "Distribution in [0..1]" topic.

[quote="Myth"] How many mistakes here? :( [/quote]
No mistakes. This was a really impressive proof, Myth! And a very nice problem too. The only (very minor) remark is that in the situation when the distance from $Y(x)=c_1 x^{p_1}+\dots c_\ell x^{p_\ell}$ to $\mathbb Z$ tends to $0$ as $x\in\mathbb Z$ tends to infinity, you can give a fully elementary (i.e., avoiding Weyl's criterion) proof of the fact that $c_j$ are rational and $p_j$ are integer similar to that in [url]http://www.artofproblemsolving.com/Forum/viewtopic.php?t=35424[/url].

[quote="fedja"]The only (very minor) remark is that in the situation when the distance from $Y(x)=c_1 x^{p_1}+\dots c_\ell x^{p_\ell}$ to $\mathbb Z$ tends to $0$ as $x\in\mathbb Z$ tends to infinity, you can give a fully elementary (i.e., avoiding Weyl's criterion) proof of the fact that $c_j$ are rational and $p_j$ are integer similar to that in [url]http://www.artofproblemsolving.com/Forum/viewtopic.php?t=35424[/url].[/quote] I did it then I wrote $\Delta^j z(n)$ (the same idea as in the Leva's post; it is quite well known). But I remembered those topic, so when I was writing that part of the proof, I simply gave the link.

Too late. :(    Anyway, great proof, Myth! My solution has the same idea, but I could not extend it to polynomial with real coefficients.  :blush:

I have also  have a solution for the problem. It's not detailed, but I think it's beautiful. I've just seen the Myth's proof. I'll read it.
[i]Lemma 1:[/i]
There exist infinitely many $c \in \mathbb{Q}$ that $q(x)-c$ is irreducible and has a real root.
[b]Proof:[/b] First we explain Hilbert theorem:
[b]Hilbert theorem:[/b] If $p(x,y)\in\mathbb{Q}[x,y]$ and is irreducible in $\mathbb{Q}[x,y]$ then there exist infinitely many $c\in \mathbb{Q}$ that $p(x,c)$ is irreducible in $\mathbb{Q}[x].\ \square $
Now if $deg (q)$ is odd because $q(x)-y$ is irreducible in $\mathbb{Q}[x,y]$, then there exist $c$ that $q(x)-c$ is irreducible and $q(x)-c$ obviously has real root. If $deg(q)$ is even then suppose leading term of $q$ is positive (Without Loss Of Generality). Now consider the polynomial \[A(x,y)=q(x)-y^2-b\] for a sufficiently large $b>0$ (suppose $q(x)-b$ has real root). Now we prove $A(x,y)$ is irreducible. If $A(x,y)=B(x,y)C(x,y)$, then $deg_y(B)\in\{1,2\}$. If $deg_y(B)$ is 2 then calculating the coefficient of $y^2$ in 2 sides we have $C(x,y)$ is constant. So $deg_y(B)=1$. So there exist $f(x),g(x)$ that $B(x,y)=f(x)+y,C(x,y)= g(x)-y$. Calculating the coefficient of $y$ in 2 sides $f(x)-g(x)=0$. So $A(x,y)=f(x)^2-y^2$, so $b=0$ and the contradiction proves that $A(x,y)$ is irreducible.
So $A(x,y)$ is irreducible. So there exist $k\in \mathbb{Q}$ that $q(x)-k^2-b$ is irreducible and has real root and Lemma is proved. $\square$
[i]Lemma 2:[/i]
If $p,q$ have the problem conditions, Then there exist $s,t\in \mathbb{Q}$ that $q(x)-t\ |\ p(x)-s$
[b]Proof:[/b] From the first lemma there exist $t\in\mathbb{Q}$ that $q(x)-t$ is irreducible and has a real root $\zeta$. So $q(x)-t$ is minimal polynomial of $\zeta$ and because $q(\zeta)$ is rational then $p(\zeta)$ is rational. So $\zeta$ is root of $p(x)-p(\zeta)$. Now if $p(\zeta)=s$ then because $q(x)-t$ is minimal polynomial of $\zeta$ so $q(x)-t\ |\ p(x)-s. \ \square$
Now we prove the problem with induction on $deg(p)$. Consider $\displaystyle p_1(x)=\frac{p(x)-s}{q(x)-t} $. Also $p_1(x),q(x)$ have the problem conditions and according to induction hypothetis $(deg (p_1)<deg(p))$ there exist $r_1\in \mathbb{Q}[x]$ that $p_1(x)=r_1(q(x))$. Now define $r(x)=r_1(x)(x-t)+s$. For induction basis if $deg(p)<deg(q)$, because $q(x)-t\ |\ p(x)-s$ then $p(x)-s=0$. Therefore $r(x)=s$ and everything is proved. $\square$

[quote="Omid Hatami"]I have also have a solution for the problem. It's not detailed, but I think it's beautiful.[/quote]
Do I understand right that this solution works only for polynomials with rational coefficients?  :?

Yes, It's only for the polynomials in $\mathbb{Q}[x]$ ?

[quote="Omid Hatami"]I almost have proved the first conjecture (my conjecture) with a simple solution.[/quote]
So, it is not completely correct...

No I completed it, and the solution I posted today is correct for all $p,q\in\mathbb{Q}[x]$

But your conjecture was for $P,Q\in\mathbb R[x]$... That's why I say your statement is not completely correct.

This is what I completely proved :

Suppose $p(x),q(x)\in \mathbb{Q}[x]$. We know if $p(x)\in\mathbb{Q}$ then $q(x)\in\mathbb{Q}$. I proved there exist polynomial $r\in \mathbb{Q}[x]$ that:$p(x)=r(q(x))$

Actually, Omid Hatami's argument works in the general case: the Hilbert irreducibility theorem is true over any number field, and we may work in a number field $K$ containing all coefficients of $P$ and $Q$.

Oh, You made me very glad. :first:  :lol:

Dear vess. I can not generalize my solution to $\mathbb R$. How could you make it? :(


floor function
induction
combinatorics proposed
combinatorics

Denote $S$ as the subset of $\{1,2,3,\dots,1000\}$ with the property that none of the sums of two different elements in $S$ is in $S$. Find the maximum number of elements in $S$.

[hide="Solution"]$S=\{1,\,2,\,4,\,7,\,10,\,13,\,16,\ldots\}$

Except for $2$, $S_{n}=S_{n-1}+3$.

The maximum number of elements in $S$ is $\left\lfloor\frac{1000-1}{3}\right\rfloor+1$, or $\boxed{334}$.[/hide]

That's not a solution, it's an answer lacking justification.  A solution would include a proof that no set with 335 elements is possible.  Unfortunately, that result is false.  In particular, taking all numbers larger than or equal to 500 gives an $S$ with 501 elements.  (Another reason that result is false is that your set actually contains 335 elements, not 334.)

Here's a solution outline:  if 1000 is replaced by $n$, the answer is $\left\lfloor \frac{n}{2}\right\rfloor+1$.  We actually prove something more, by induction:
* If 1000 is replaced by $2n$, the max is $n+1$, and in every maximal set there are a pair of elements which sum to $2n+1$.
* If 1000 is replaced by $2n+1$, the max is $n+1$, and there is some maximal set in which no pair of elements sum to $2n+2$.

Together with the base cases 1, 2, these are not very hard to prove, and they induct to give us the answer above.

I'm sure it's just 501 as JBL's formula would give us:

The set would be $ \{500, 501, 502,\dots,1000\} $

The answer is $501$, which can be achieved with the set $$\{500, 501, 502, \dots 1000\}$$Now, let's prove that this is the maximum. Let $C$ be the greatest element of $S$. Consider the $\left\lfloor \dfrac{C}{2}\right\rfloor$ pairs $$(1, C-1), (2, C-2), \dots \bigg(\left\lfloor \dfrac{C}{2}\right\rfloor, \left\lceil \dfrac{C}{2}\right\rceil\bigg)$$As there are not two elements in $S$ that add up to $C$ (as it belongs to $S$), we can have at most one number of each pair. Note that every number smaller than $C$ is in one of such pairs, and that there can't be an element greater than $C$ in $S$, as $C$ is the greatest. Thus, counting $C$, we can have at most $$\left\lfloor \dfrac{C}{2}\right\rfloor+1\le\left\lfloor \dfrac{1000}{2}\right\rfloor+1=501$$ numbers in $S$. We are done. $\square$

We call a number "special" if it is in $S$.
First notice that the set $S = \{500, 501, ..., 999, 1000\}$ is a valid set, and has $501$ elements. Now we prove that $501$ is the maximal amount of elements that $S$ can have.
Let $d$ denote the smallest special number. It is clear that for any number $n$, if $n$ is special then $n - d$ cannot be special without contradicting the defining property of $S$. The only exception to this rule, however, is if $n = 2d$, in which case the contradiction does not occur.
Now we define:
$I = \{1, 2, ..., 999, 1000\}\backslash\{d, 2d\}$.
Notice that $I$ contains at least $|S| - 2$ special numbers. This occurs in the case where $2d$ is special also. Let $x$ denote the amount of special numbers in $I$. We proceed to show that the function $f(n) = n - d$ is an injective function between special numbers in $I$ and non-special numbers in $I$. This can be done by noticing that in $I$, it is always true that if $n$ is special then $n - d$ cannot be, since $I$ does not include the exception $n = 2d$. Furthermore, we know that $n - d$ is in the set $\{1, ..., 1000\}$ because, by definition, $1000 > n > d$ which implies $1000 > n - d > 0$. Lastly, it is clear that, since all special numbers are distinct, then their images under this function are distinct, and the injection is complete.
This injection tells us that, for every one of the special number in $I$ there is a distinct non-special number in $I$. And it is clear that the intersection of all these pairs is smaller than the set $I$. That is,
$2(|S|-2) \leqslant 2x \leqslant |I| = 1000 - 2$,
from which we obtain
$2|S| - 4 \leqslant 998$
$|S| \leqslant 501$
Which completes our proof. $\square$


After your proof is graded do you get it back?

No, we will not be sending back the papers.  Please keep a copy for yourself.


The accompanying figure shows the correctly worked out addition problem in which each letter represents a different digit from 1 to 9.  What digit does D represent?

83A + DBB = CAC2

[hide]We can immediatly state that C=1, so we are trying to find D in
83A+DBB=1A12.  I don't want to explain more, but the answer is [b]6[/b]. (I think).[/hide]


Ann opened a container of cookies and ate one-fourth of them. Sam then ate one-third of the remaining cookies. If there were 24 cookies left, how many cookies were in the container originally?

You can take the remainder of the portions that got eaten so: $ 3/4*2/3\equal{} 1/2$ so 1/2=24 and to get it back to $ 1$, you do $ 24*2\equal{}1/2*2\equal{} \boxed{48} \equal{}1$


conics ellipse trigonometry integration rotation ratio geometry

given an ellipse defined generally by $\left(\dfrac{x}{a}\right)^2+\left(\dfrac{y}{b}\right)^2 = 1$ or parametrically as $x=a*\sin{\theta}$ and $y=b*\cos{\theta}$

find the circumference of the ellipse formed

The arclength of an ellipse with semiaxes $a$ and $b$ is
$\int_0^{2\pi}\sqrt{a^2\cos^2\theta+b^2\sin^2\theta}\,d\theta= \int_0^{2\pi}\sqrt{a^2+(b^2-a^2)\sin^2\theta}\,d\theta$.

Oh, you wanted an explicit form? Sorry, there's no closed form in terms of elementary functions. We can approximate this accurately with a wide variety of methods, and find series expansions, but that's it.

yes, i thought of that too, and it is much easier to do it with the parametric equations, but too bad there is not an easier way to do it

another thing that i found to be weird is that

$\int_0^{2\pi}\sqrt{b^2+(a^2-b^2)\sin^2\theta}\,d\theta =\int_0^{2\pi}\sqrt{a^2+(b^2-a^2)\sin^2\theta}\,d\theta$

logically this is easy because switching the a and the b would just rotate the ellipse 90 degrees, but i still found that odd

but what would you do if you wanted to say find the arc length using a calculator?

the only way i thought of would be doing $\sqrt{a^2+b^2}\sqrt{2}\pi$, but that approximation is not that accurate (i dont feel like doing the explaination, but it has to do with taking a unit circle with $C=2\pi$ and comparing the ratio of the perimeter of little triangles of that circle, and an ellipse that has been stretched by factors a and b)

To obtain a numerical value of the arc length using a calculator, begin with the integral and numerically integrate; if the eccentricity of the ellipse is not terribly close to one, a relatively small number of function evaluations will produce an accurate result.

Here's a usable approximation.

Assume that $a>b.$ Use the identity $\sin^2t=\frac12(1-\cos2t)$ to write

$b^2+(a^2-b^2)\sin^2t=\dfrac{a^2+b^2}2-\dfrac{a^2-b^2}2\cos2t$

$\sqrt{b^2+(a^2-b^2)\sin^2t}=\sqrt{\dfrac{a^2+b^2}2-\dfrac{a^2-b^2}2\cos2t}$

$=\sqrt{\dfrac{a^2+b^2}2}\left(1-\dfrac{a^2-b^2}{a^2+b^2}\cos2t\right)^\frac12$

Expand the square root by the binomial theorem:

$\left(1-\dfrac{a^2-b^2}{a^2+b^2}\cos2t\right)^\frac12=$

$=1-\frac12\left(\dfrac{a^2-b^2}{a^2+b^2}\right)\cos2t- \frac18\left(\dfrac{a^2-b^2}{a^2+b^2}\right)^2\cos^22t$$-\frac1{16} \left(\dfrac{a^2-b^2}{a^2+b^2}\right)^3 \cos^32t+O\left(\dfrac{a^2-b^2}{a^2+b^2}\right)^4$

Now integrate:

$\int_0^{2\pi}\cos2t\,dt=\int_0^{2\pi}\cos^32t\,dt=0$ as would be the case for any odd power.

$\int_0^{2\pi}\cos^22t\,dt=\pi.$ So we estimate the arclength of the ellipse as

$\sqrt{\dfrac{a^2+b^2}2}\left(2\pi-\dfrac{\pi}8\left(\dfrac{a^2-b^2}{a^2+b^2}\right)^2 + O\left(\dfrac{a^2-b^2}{a^2+b^2}\right)^4\right)$

This estimate should be good for low eccentricities; for higher eccentricities, we can take more terms of the binomial expansion.

[quote="Altheman"]the only way i thought of would be doing $\sqrt{a^2+b^2}\sqrt{2}\pi,$ but that approximation is not that accurate[/quote]
If you'll compare that to what I wrote in post #5, you'll see that it agrees with the first term of the estimate from the binomial expansion. Hence, Altheman's reasoning in constructing his approximation was correct, as far as it went. What I've done in #5 is to indicate how to start from that and improve the approximation by adding more terms.

Using left-endpoint Riemann sums to approximate $\int_0^\pi\sqrt{a^2\sin^2 x+b^2\cos^2 x}\,dx$ and doubling for the full ellipse, I get these values ($a=1,b=2$):
1 point: 12.56637061
2 points: 9.424777961
4 points: 9.679683113
8 points: 9.688412390
16 points: 9.688448219
32 points: 9.688448221

If you've ever had any numerical analysis, you can see that something unusual is happening; the error of this method should normally be proportional to $\frac1n$, which would mean that we halve the error when we double the number of points. The actual results are much better (the $n=16$ value is 18000 times as close as the $n=8$!), which means that the errors are cancelling each other out. Also, we gain nothing by going to the methods which are normally more accurate; the trapezoidal rule gives exactly the same values, and Simpson's rule is less accurate than the left endpoint rule with the same number of function evaluations.

We can show that the error goes to zero faster than any power of $n$, but this behavior is intimately connected to the fact that we're integrating over a full period, and therefore doesn't work for any partial arcs of the ellipse.
This doesn't lend itself well to a formula, but it is very accurate.

With $a=2$ and $b=1$, the ratio $\dfrac{a^2-b^2}{a^2+b^2}$ that governs how rapidly the binomial approximation converges is $0.6,$ which is too large to expect rapid convergence. In that case, the one-term formula (Altheman's formula) gives 9.93459 and the two-tem formula (from what I wrote in #5) gives 9.71106. According to jmerry, that's still in error by 0.0226.

[quote="Kent Merryfield"]
Expand the square root by the binomial theorem:

$\left(1-\frac{a^2-b^2}{a^2+b^2}\cos2t\right)^\frac12=$

$=1-\frac12\left(\frac{a^2-b^2}{a^2+b^2}\right)\cos2t- \frac18\left(\frac{a^2-b^2}{a^2+b^2}\right)^2\cos^22t$$-\frac1{16} \left(\frac{a^2-b^2}{a^2+b^2}\right)^3 \cos^32t+O\left(\frac{a^2-b^2}{a^2+b^2}\right)^4$

[/quote]

i never knew that the binomial theorem extended into rational numbers, but that is interesting

[quote="Altheman"]i never knew that the binomial theorem extended into rational numbers, but that is interesting[/quote]
This was Isaac Newton's first big accomplishment in mathematics, and it became a tool that Newton would wield with great effectiveness for the rest of his career.

When I teach Calculus II, at some point I hand to the class a sheet of paper entitled "Seven Power Series to Memorize." So seven functions have series worth memorizing? Here's the list (all are power series centered at zero):

1. $\frac1{1-x}$ (the geometric series - it always goes first.)

2. $\ln(1+x)$

3. $\arctan x$

(#2 and #3 share the property of being derived from the geometric series by integration)

4. $e^x$

5. $\cos x$

6. $\sin x$

(#4, #5, and #6 are the various aspects of the exponential - $\sinh x$ and $\cosh x$ are part of the mix as well, although less worthy of memorization.)

7. $(1+x)^{\alpha},\,\alpha\in\mathbb{R}.$

#7 is the binomial series, albeit not presented in Newton's notation.

$(1+x)^{\alpha}=1+\alpha x+\frac{\alpha(\alpha-1)}2x^2 + \frac{\alpha(\alpha-1)(\alpha-2)}{3!}x^3+\cdots$

If you use the notation ${\alpha \choose k}$ to mean

${\alpha\choose k}=\frac{\alpha(\alpha-1)(\alpha-2)\cdots(\alpha-k+1)}{k!}$

then we can write:

$(1+x)^{\alpha}=\sum_{k=0}^{\infty}{\alpha\choose k}x^k.$

The radius of convergence of this series is 1 (unless $\alpha$ is a nonnegative integer, in which case the power series simplifies to a polynomial.)

The key to dealing with $(A+B)^{\alpha}$ is to figure out which one of $A$ or $B$ is bigger, and factor that one out. If $A>|B|$, then we write

$(A+B)^{\alpha}=A^{\alpha}\left(1+\frac BA\right)^{\alpha}$

and then apply the binomial series to that. That's exactly what I did above.


I was wondering how you guys study the books. I usually read through the chapter doing the exercises..then do the problems... then go on to the next chapter. The problem with this is that several chapters on, when I go back and try some of the problems, even through I was able to do them right after I read the chapter, I have forgotten how to do them and I have to re-read the chapter... suggestions....

This might fit better in [url=http://www.artofproblemsolving.com/Forum/index.php?f=319]this[/url] forum. :)


Bill has a $ 2\times 2\times 2$, a $ 3\times 3\times 3$, a $ 4\times 4\times 4$, and a $ 5\times 5\times 5$ Rubik's cube. Sticky Fingers sue comes along and steals one all with equal probability, then Pick-pocket Pearl comes and takes either the largest one or the smallest one still there, again with equal probability. Bill comes back, but he only wants to play with his $ 3\times 3\times 3$. What is the probability that he can?

$ \textbf{(A)} 1/8\qquad\qquad\qquad\textbf{(B)}
1/4\qquad\qquad\qquad\textbf{(C)}
3/8\qquad\qquad\textbf{(D)}
1/2\qquad\qquad\textbf{(E)}
5/8$

Split it into the four cases.  When Sue steals the 2 by 2, the probability is one half.  If Sue steals the 3 by 3, it is 0.  If she steals the 4 by 4, the probability is just 1.  If she steals the 5 by 5, the probability is also 1, so we have $ \frac{5}{8}$ which is $ \boxed{E}$


There is given two sets B={x is element of R|log(in base3)(x+2) + log(in base2)(3^x - x)=3^x - 1} and A={x is element of              R|3^x=x+2 where R - set of real numbers. Prove that A[i]C[/i]B and that the set B contains both rational and irrational numbers

can you rewrite in LaTex and solve it
thanks

ACB means that A is subset of B don't forget it!


Does anyone know where I can find past NYSML Problems?  I checked the AoPSWiki, but the [url=http://www.artofproblemsolving.com/Wiki/index.php/NYSML_Problems_and_Solutions]NYSML Page[/url] was blank.  Thanks!

[quote="nkouevda"]Problems from past competitions can be bought at [url=http://www.nysml.org/index.php?n=NYSML.ContestArchive]the official site[/url], which is why you won't find them archived somewhere like in our Wiki.[/quote]

Do you know if there are any free resources?  I can't afford all of the contests they're selling.  :(

[quote="Hyperboloid"]Is it illegal to pirate things?[/quote]
[youtube]04UORA-xB-k[/youtube]
Only watch the first 12 seconds.


[i]On the mean streets of New York City, stockbrokers, lawyers, hobos, and aspiring actors are forced to coexist. They can just fine too, provided no gangs pursue them...[/i]

Welcome to Mafia Game 34: NYC. The 24 of you are the tenants of the two adjacent Apartment Buildings [i]La Muerte[/i] and [i]El Delito[/i]. After the recent killings of several of your members, you are all required to attend day-long meetings held in the [i]El Delito[/i] basement. These are very secret meetings, and one person will be voted to die at each.

The nights, however, have become very different. All the local Gangs have started acting more frequently, as they know they have to defeat the NYCUCP before it's too late. Everyone with the ability to do so is staying out to prevent this, risking their lives even further. But absolutely no one is safe...

The only way you can beat the bad guys is if you can eliminate everyone against your group, the New York City Union of Crime Prevention, or NYCUCP.

Everyone do your jobs to stop the evil of the criminals, and, no matter what,

[b][i]don't stay out all night...[/i][/b]

[b]The Living (24)[/b]
33286
Brut3Forc3
zapi2007
woser456
shaggy75
distracted523
emannes
mz94
Peter
1=2
pacman2812
andersonw
erabel
lingomaniac88
james4l
Temperal
Flame
jackdillon
samath
Berylliumnitride
142857
BOGTRO
iLord
Arcesso

[b]The Dead (0)[/b]

Also, I'm now sending out Role PMs. Once I post the second Post (the rules), you may start playing.

[b][u]The Rules[/u][/b]:[list][*]Votes must be in [b]bold[/b].  If you do not bold your vote, it will not be counted.
[*]Please be attentive and [b]unvote[/b], if necessary, before casting a new vote.  This is not [i]required[/i], but I’d appreciate it.
[*]Executions will require a simple majority of votes.  Once a player has reached the necessary majority, his pleas are useless and any attempts to unvote will be unheeded. You may [b]vote: no lynch[/b] – majority votes of this kind are necessary to end the day without a death.
[*]The game is not to be discussed outside of the thread unless your role specifically states that you may do so. - [b]If your role PM tells you the alignment of other players, you may privately talk to them unless otherwise stated.[/b]
[*]Once your death scene has been posted, you’re dead.  Stop typing.
[*]Don’t edit/delete previously submitted posts
[*]Don’t quote any PMs from me.
[*]If you have a night choice to make, it is due by the posted deadline.  I will not wait for you – if you do not submit a choice to me, tough!  No choice will be made.
[*][b]If you anticipate being unavailable for more than a 48-hour period, please post a notice to that effect in the thread.[/b]  Treat this game as a commitment.  Be considerate – don’t leave us hanging.
[*]Remember, it’s a game.  It’s supposed to played.  It’s supposed to be [i]fun[/i].  Do your part.[/list]

Now, here is just a note:

The following two lines appear in all "town" PMs: if you didn't get them, you are not a part of the majority:

[quote]You are a member of the NYC Union of Crime Prevention.
You win when there is no one left in opposition of the NYCUCP.
[/quote]

That's it. If you don't have a PM, and are on the above list, there's a problem, so tell me.

Go.

[i]24 alive, 13 to lynch

It is now Friday, October 13[/i]

This reminds me of realistic mafia on mafiascum... if it's anything like that, it will be cool game to play.

[quote="perfect628"][i]It is now Friday, October 13[/i][/quote]

That's really disturbing. It really is. :maybe:

NYC?! I can go meet my MathCounts teammates!!! YAY!  :D

Random vote [b]vote: mz94[/b].

At least a game without posting restrictions.

I love the date btw. :) I hope to survive at least until my birthday (October 20).

I guess I'll just [b]Vote: Flame[/b] since I don't really know anyone here.

[b]Vote: Peter[/b]
no one likes Friday 13

and if this setup is similar to realistic mafia (which I mentioned) then there are probably 2 killing groups... perfect also mentions "Gangs", not "Gang"... name of the buildings indicate that too - "Death" and "Crime"

Death and Crime? (I don't know any Spanish)

Well, I guess that two gangs seems viable, especially with perfect using "gangs."

I guess we're going to lose people fast in this game.

[quote="Peter"]I hope to survive at least until my birthday (October 20).[/quote]

Today in real life IS my birthday.

And these groups are Death and Crime.

happy birthday!

hmmm I'm not sure how can you know if groups are called Death and Crime... that is if you are not one of them
[b]Unvote, Vote: 1=2[/b]

I have a spanish translator.

[b]Unvote, vote: Flame[/b] for being stupid

@Flame: I think he thought I was posing a question about Death and Crime.

I'm pretty sure he meant the buildings.

I thought that was pretty obvious, I don't like how you jumped on him so quickly. [b]Vote: Flame[/b]

@1=2: Is your vote serious?

[b]random voting: 33286 [/b]

as his number is smaller than mine :lol:

Cf249 pulled off a good lie with his CFM roleclaim, seeing as it made sense to have 2, which covered him all the way to the end.

That was exactly my reasoning behind that roleclaim, and that's how I would've defended it if pressured to do so.  The only thing that could've gone wrong is if there actually was another CFM and he/she claimed it; then the town would be forced to make basically a 50/50 decision on who was telling the truth.

Surprisingly, the day after the night I was found to be out was the only time I was really under any serious suspicion.

IMO, the town lost because they didn't fully make use of all the resources available to them (psychics, hobos, etc.) that could've given them the edge.

Personally, I would have waited for about 5 minutes until day ended and then lynches someone that wasnt you. Also who suspected sk? I had the impression most thought vig.

[quote="chenhsi"]So did you suspect me, #H34N1, because you were scum, or because you really thought that what I did was scummy?[/quote]
I thought you were red rum and the only way I could win was if I kill a red rum during teh lynching

When I realized that planw as bad, I immediatley saw that I lost.

If this game didn't die the townies would have won by a long shot.

the sad part was that the two hobos were killed very quickly
maybe if we had survived a few more days, town might have won

i've, finally, won!!!!!
Congratulate all my teammate: mz94 (although he is mod-killed), flame (our godfather after mz94 is mod-killed and before he is replaced), lingomaniac68 (although he is replaced), crazychinesemaniac and cf249 (good job finding out the last blood).

Rereading the first 10 pages of the game, I noticed the town got off to an unlucky start (after mz94 was modkilled, that is).  There was quite a bit of suspicion toward Flame (whom I replaced) on Days 1 and 2, but others (33286 and iLord) acted more suspicious, and were lynched instead.  1=2, this is where your comment may come in:  after the initial month-long hiatus, the town forgot their suspicions of Flame.  Also, the town lost two powerful roles (a hobo and a psychic) before Day 2 had begun.

Next time I should read the first few pages when I replace...I only read like 1-8 and like 15-end.

[quote="mewto55555"]Also with all the replacements, people dont as carefully read, if they read at all, the first few pages and are therefore not as suspicious of people who did scummy things.[/quote]That is a major problem indeed. Also for the people alive it's difficult to track mistakes from the precessor, as you can't really hear them about it.

Which really supports my idea of having a list of active/inactive players, and only ones on the active list can join games.

lastpost (this is used to preserve order of locked topic)


geometry
trapezoid
trigonometry
similar triangles

The incircle of a triangle $ABC$ touches the sides $BC$, $CA$, $AB$ at the points $D$, $E$, $F$ respectively. Let the line $AD$ intersect this incircle of triangle $ABC$ at a point $X$ (apart from $D$). Assume that this point $X$ is the midpoint of the segment $AD$, this means, $AX = XD$. Let the line $BX$ meet the incircle of triangle $ABC$ at a point $Y$ (apart from $X$), and let the line $CX$ meet the incircle of triangle $ABC$ at a point $Z$ (apart from $X$). Show that $EY = FZ$.

Thers's a fairly simple computational proof (about half a page or so), but here's a proof using a bit of projective geometry:

ED is the polar of C with respect to the incircle of ABC, so if CX and ED intersect at T then the points X, T, Z, C form a harmonic quadruple, and then so do the lines EX, ET, EZ, EC. These lines are intersected by line AD at X, D, Z' and A resectively, so these four points form a harmonic quadruple as well, so (XD/XA) : (Z'D/Z'A)=-1 (the segments are actually oriented segments), but since XDS/XA=-1, it means that Z'D/Z'A=1, which can only happen if Z' is the infinity point of the line AD, which means that EZ is parallel to AD. In the same way we show that FY is parallel to AD, so EZ and FY are parallel, so the quadrilateral EZYF is an isosceles trapezoid and the conclusion follows.

[[b]Moderator edit.[/b] See also http://www.mathlinks.ro/Forum/viewtopic.php?t=3496 for a stronger problem.]

By many similar triangles,
ZE/EX=CE/CX=CD/CX=ZD/DX
so,ZE/ZD=EX/DX=EX/AX
with <EZD=<EXA,
triangles EZD and EXA are similar
then we have <EAX=<EDZ
arcED -arcEX=arcED-arcZD
arcEX=arcZD
then EZ//DX
so we have FY//DX//EZ
then ZF=EY

Let a = BC, b = CA, c = AB be the sides and s the semiperimeter of the triangle $\triangle ABC.$ Power of the vertex A to the incircle (I) is $AX \cdot AD = AF^2 = (s - a)^2.$ Since $AX = \frac{AD}{2},$ we get $AD^2 = 2(s - a)^2.$ BX is the B-median of the triangle $\triangle ABD,$

$BX^2 = \frac{2BA^2 + 2BD^2 - AD^2}{4} = \frac{2c^2 + 2(s - b)^2 - 2(s - a)^2}{4} =$

$= \frac{c^2 + b^2 - a^2 + (a + b + c)(a - b)}{2} = \frac{c^2 + c(a - b)}{2} = c (s - b)$

Power of the vertex B to the incircle (I) is $BY \cdot BX = BD^2 = (s - b)^2.$ Thus

$\frac{BY}{BX} = \frac{BY \cdot BX}{BX^2} = \frac{s - b}{c} = \frac{BF}{BA}$

It follows that $FY \parallel AD.$ In exactly the same way, we can show that $EZ \parallel AD.$ As a result, EFYZ is a cyclic trapezoid with bases $FY \parallel EZ,$ i.e., isosceles and with equal diagonals, hence $EF = YZ,\ EY = FZ.$

what anice problem. :!:

it is easy by using harmonic.;)

$CE^2=CD^2=CZ.CX$, then it means that $(XZED)=-1$.

$BD^2=BF^2=BY.BX$, then it means that $(XYFD)=-1$.

from the first we find that $E(EDZX)=-1$

and from the second we conclude that $F(XYFD)=-1$

and because $AX=XD$, $ZE$ should be parallel to $AD$.

and again because $AX=XD$, $FY$ should be parallal to $AD$.

and it is concluded that :$EZ,FY$ are parallel.

then they forme an isoceles trapezoid in the incenter.

and it is clear that the diagonals of an isoceles trapezoid are equal to each other.

and it means that: $EY=FZ$

ask my later, each part u didnt figure out. ;)

I tried to do this problem using trigonometry and I failed. We need
to prove that $FY=EZ$ $\Leftrightarrow$ where $\mu = \eta$
$\mu=\angle FXY = \angle YFB$ and $\eta = \angle EXZ = \angle ZEC$.
Let $\varphi = \angle DAC$, $\theta = \angle BAD$, $\xi = \angle FBY$ $\nu = \angle ECZ$ and $s=\frac{a+b+c}{2}$. We have: \[AX AD=AE AE\]
\[AD = (s-a)\sqrt{2}\] From triangle ADC we get
\[\frac{\sin \varphi}{\sin \gamma}= \frac{CD}{AD}\]
\[\sin \varphi= \sqrt{2}\tan\frac{\alpha}{2}\cos^{2}\frac{\gamma}{2}\]
Also we get \[\sin \theta = \sqrt{2}\tan\frac{\alpha}{2}\cos^{2}\frac{\beta}{2}\] In triangle AXB: \[\frac{\sin \theta}{\sin \xi}= \frac{BX}{AX}\] From triangle BFY: \[\frac{\sin \mu}{\sin \xi}= \frac{BY}{FY}\] We have: \[BX BY = BD^{2}\] \[\sin \xi = \sqrt{2}\sin \mu \sin \frac{\beta}{2}\] Also we have \[\sin \nu = \sqrt{2}\sin \eta \sin \frac{\gamma}{2}\] From Ceva' s theorem we have
\[\frac{\sin \varphi}{\sin\theta}\frac{\sin \xi}{\sin \nu}=\frac{\sin(\beta-\xi)}{\sin(\gamma-\nu)}= \frac{CX}{BX}\]
We need to prove that $\sin \mu = \sin \eta$ but I don't know how to
finish my proof.

Let $FD\cap XY=P$ and $YF\cap BC=T$ then using Pascal's theorem on hexagon $DDXYFF$ we obtain $A, P$ and $T$ are collinear, and because $BX$ is median of $\triangle ABD$, then using Ceva's theorem we get ratios that prove that $FY\parallel AD$. Similarly on the other side we would obtain $EZ\parallel AD$, so $EZ\parallel YF$ making $YFEZ$ an iscoceles trapezium. So the diagonals are congruent hence $EY=FZ$.

It is weird this solution did not appear so far.

Let $M$ be the midpoint of $FD$. We know that $XM\parallel AB$ and that $XB$ is the $X-$symmedian in $\triangle XFD$, then
\[
\angle BFY=\angle FXY=\angle MXD=\angle FAD
\]
meaning that $FY\parallel AD$. Analogously $EZ\parallel AD$ and we are done.

why is an error that dont let me post latex? its says that new users cant post images


geometry
3D geometry
LaTeX
Putnam
calculus

On Sarah's suggestion, here I am, trying to revive this dead forum.

Whenever you are reading this post, you had better reply, or you will face dire consequences.

some of you may remember the evil twins' midnight concert..

If you don't contribute in saving this forum, you may also find that we have a lot more things in store..

So.. post something, save the forum!  :D

P.S.-I'm not kidding, guys, this is getting kinda sad!

[quote="jpark"]
some of you may remember the evil twins' midnight concert..

[/quote]

Ask Alex.  He should know firsthand about that. Hehehehe....

[quote="jpark"]On Sarah's suggestion, here I am, trying to revive this dead forum.

Whenever you are reading this post, you had better reply, or you will face dire consequences.
[/quote]

Sorry, I was away for a long time. I'll try to come here more often and randomly post. :D

[quote="lightrhee"]
Sorry, I was away for a long time. I'll try to come here more often and randomly post. :D[/quote]
Whoa David, where were you?!  This site was almost...normal without you!  Well, not quite.  But a lot less random.   :D

Maybe you can save this forum by ... umm .. spreading the word about it to your young padawans?  :P  :roll:

[quote]Maybe you can save this forum by ... umm .. spreading the word about it to your young padawans? Razz Rolling Eyes[/quote]

that would be nice too.. i don't know a lot of "padawans" though.. Just grade 12s and up, mostly.

And also, maybe it'll help if we had a topic in this thread.. How about we come up with a topic.. say... the randomness of david.. or maybe we could talk about rubik's cube, or SCHOOL SUCKS!

Anyways, some people are still not replying and if this keeps up, I will choose my first victim. And beware!! Probably people who are geographically closer to me would have better chance of being chosen! (That means YOU, Elyot Grant!)

[quote="jpark"] How about we come up with a topic.. say... the randomness of david.. or maybe we could talk about rubik's cube, or SCHOOL SUCKS!
[/quote]
How come these three topics are all related to me? :D

By the way, this is my 666th post!! :diablo:

[quote="jpark"]
Anyways, some people are still not replying and if this keeps up, I will choose my first victim. And beware!! Probably people who are geographically closer to me would have better chance of being chosen! (That means YOU, Elyot Grant!)[/quote]

Hmm.  I wonder what would happen is Boris joined the site.  Maybe with the penguins complete, they'll post more!

I forgot how to use Latex!!! :o  I must have been away for a long time!
Can someone quickly show me how to enter fractions?

Hmm, I've been so occupied these days that I no longer visit this place as much as I used to.

Perhaps we'll get more popular once the contest season starts.

Hey... how is everyone doing?  I haven't been here in 3 months... and sadly, nothing much has changed!  People should be posting... come on!  You don't need math contests to talk about math!  And when did you guys ever talk about math?  lol...

Jenn... come to putnam training!

I wish you guys luck on all the math contests this year.  May Massey beat all of you.

It's good to see this forum with little more posts. :) We are doing better!

[quote="jpark"][quote]May Massey beat all of you.[/quote]

Somehow, without you and taotao there, i don't think it's going to be that easy shaun..

[quote]Jenn... come to putnam training! [/quote]

I will show up on the monday after thanksgiving. My family's going to Algonquin park as usual for the long weekend, so I can't be there on the thanksgiving. Anyways, I'm sure the training's not on. but I promise to come to the next training session. And Elyot's going to suffer!![/quote]

There is no putnam training on Thanksgiving monday anyway.

Great to see you're coming!! See yah there :)

Wow...is this what usually goes on in the Canadian forum? :rotfl:

[quote="Singular"][quote="Elyot"]cheese is (like) morphine ...  you can seduce girls with it.
[/quote]
:lol:   excellent tip from elyot:  drug girls then they will be easier to pickup[/quote]

That's just twisted right there... :rotfl:

But you don't need to drug girls if you can make them really really want you through some other means.  Perhaps you can just be a really nice, fun guy.  Girls like that too.  If all else fails, then I guess you can cheese it up.

Guess what!!

I went to the Putnam session, and Elyot and Shaun weren't there!!

The one time I decide to go, you aren't there...  :mad:  :mad:

You two had better be there next week!!

Sorry Jenn!  We had to write our algebra midterm monday night!  Calc was supposed to be next monday but it was postponed so nothing should stop us from coming next week!   :)

Yea we went to our brutal algebra midterm instead.

6 problems... easy, easy, easy, easy, INSANE, easy.

The insane one was:

Prove 2^(1/2) + 3^(1/2) +5^(1/3) is irrational.

Yea... that's not a typo.  Can you say UGLY?!?!?  SOOOO MANY LINES OF ALGEBRA!

It's possible that the timezone setting was not set correctly on her account.

Hmm...it doesn't look quite dead to me...Seen as there are still a handful of people posting every now and then.

Anyway, just pop up and leave a fingerprint because I heard *someone* talking about nobody posting.

BTW, Sarah--have you managed to convince Penguin#3 to come yet? I've been looking forward to the possible drama ever since you first brought this idea up...^^||||

[quote="YY"]Hmm...it doesn't look quite dead to me...Seen as there are still a handful of people posting every now and then.
[/quote]

Well you're a little late!  It was dead until we decided to revive it, hehehe.  See?  Now its not so dead!

And Boris is being stubborn and stupid.  He refuses to join the site!

[quote="lightrhee"][quote="jpark"]lol, for one of my exams, I had to prove the fundamental theorem of algebra!

AND I didn't know much calculus!! :D  :D
You're going to be at Putnam next week though, right?[/quote]
This was posted 7:21am, which means it was 9:21am in her timezone. Don't you go to school on Friday?[/quote]

Hmm.. Is it possible that i happened to be in a computer lab at 9:21am at school, and that I didn't feel like doing my work?

Just a thought there, david..  :lol:

[quote="jpark"]

Hmm.. Is it possible that i happened to be in a computer lab at 9:21am at school, and that I didn't feel like doing my work?

Just a thought there, david..  :lol:[/quote]

 :o

I should do this sometime. :D


Hey guys, I have a few questions:

1. There are 35 desks in a classroom. In how many ways can the teacher configure a seating plan for a class of 30 students?

2. Sixteen people attend a meeting. Each person greets everyone with a single handshake. Find the total number of handshakes that are exchanged.

3. A manufacturer determines that the volume of juice in a juice box is normally distributed with a minimum of 244mL and a maximum of 256mL. What is the best estimate of the standard deviation? Complete it without the use of a calculator.

Thanks

[hide="Problem 1"]
$\binom{35}{30}=324632$
[/hide]
[hide="Problem 2"]
Think of this problem in the context of a regular polygon with 16 sides. A line you draw from one vertex to another is like a handshake. Start with the first point, and you see that a new line can be drawn to 15 other points. Take the next point and you see that a new line can be drawn to 14 other points (you can't draw a line to the first point because this line was already drawn the first time). This pattern goes on. So the number of handshakes is $15+14+\cdots+2+1=\frac{1}{2}(15)(16)=120$
[/hide]
I don't know anything about standard deviation, sorry.

For Problem 1, I assumed that each student isn't distinct. If each student is distinct then you have to multiply it by 30!.

problem 3..maybe?[hide]
well if you take a bell curve, you'll notice that the standard deviations are split up as follows..
$\pm 1$ std = 68%
$\pm 2$ std = 95%
$\pm 3$  std = 99.7%
 (these are estimates)
so..figure 256 - 244 = 12 / 6 = [b]2[/b][/hide]

$1^{st}$ problem:
[hide]There are $\binom{35}{5}$ ways to choose the empty desks, and $30!$ ways to arrange students to sit in those desks. Thus, the answer is $\binom{35}{5}*30!$[/hide]
$2^{nd}$ problem:
[hide]Each person shakes hands with $15$ people, but this overcounts by a factor of $2$. So the answer is $\frac{15*16}{2} = \boxed{120}$[/hide]


quadratics
calculus
derivative
integration
calculus computations

A while ago, I found this book at my local library that taught calculus a very nice way and it showed how to factor/solve quadratics with derivatives. I just can't remember how. Can someone please help me?

[url]http://www.physicsforums.com/showthread.php?t=163508[/url]

It uses integration, not derivatives.


Let a,b,c be positive real numbers. Prove that:

$ \frac {a}{b} \plus{} \frac {b}{c} \plus{}\frac {c}{a} \geq \frac {a\plus{}b}{b\plus{}c} \plus{} \frac {b\plus{}c}{c\plus{}a} \plus{}1$

[quote="manlio"]Let a,b,c be positive real numbers. Prove that:

$ \frac {a}{b} \plus{} \frac {b}{c} \plus{} \frac {c}{a} \geq \frac {a \plus{} b}{b \plus{} c} \plus{} \frac {b \plus{} c}{c \plus{} a} \plus{} 1$[/quote]
<=> $ M(b\minus{}c)^2\plus{}N(a\minus{}b)(a\minus{}c)\ge 0$
In there ($ a\ge bc,$)
$ M\equal{}\frac{1}{bc}$
$ N\equal{}\frac{1}{ab}\minus{}\frac{1}{(a\plus{}c)(b\plus{}c)}\ge 0$
So ineq was proved

I know there is a soution using only $ x^2\plus{}y^2 \geq 2xy$ but I don't remember it :( , so can you find it?

Thank you very much.

[quote="manlio"]I know there is a soution using only $ x^2 \plus{} y^2 \geq 2xy$ but I don't remember it :( , so can you find it?

Thank you very much.[/quote]
Maybe it:
http://www.artofproblemsolving.com/Forum/viewtopic.php?t=107534 ?

[quote="manlio"]Let a,b,c be positive real numbers. Prove that:

$ \frac {a}{b} \plus{} \frac {b}{c} \plus{} \frac {c}{a} \geq \frac {a \plus{} b}{b \plus{} c} \plus{} \frac {b \plus{} c}{c \plus{} a} \plus{} 1$[/quote]
My solution:
We need prove that:
$ \sum \frac {a}{b} \minus{} 3 \geq \frac {a \plus{} b}{b \plus{} c} \plus{} \frac{b \plus{} c}{a \plus{} b} \minus{} 2$
or $ \sum \frac {a \minus{} b}{b} \geq \frac {(a \plus{} b)^2 \plus{} (b \plus{} c)^2 \minus{} 2(b \plus{} c)(a \plus{} b)}{a \plus{} b)(b \plus{} c)}$
<=>$ (a \minus{} b)(\frac {1}{b} \minus{} \frac {1}{c}) \plus{} (a \minus{} c)(\frac {1}{c} \minus{} \frac {1}{a}) \geq \frac {(a \minus{} c)^2}{(a \plus{} b)(b \plus{} c)}$
<=.$ \frac {(a \minus{} c)^2b(a \plus{} b \plus{} c)}{ac(a \plus{} b)(b \plus{} c)} \geq \frac {(a \minus{} b)(b \minus{} c)}{bc}$
Because $ \frac {a \plus{} b \plus{} c}{a} > 1$,we need prove that:
$ \frac {(a \minus{} c)^2b}{c(a \plus{} b)(b \plus{} c)} \geq \frac {(a \minus{} b)(b \minus{} c)}{bc}$
<=>$ (b^2 \minus{} ac)^2 \geq 0$
Done!


[quote="Vasc"]Let $a,b,c$ be positive numbers such that $abc=1$. Prove that:
\[a^{3}+b^{3}+c^{3}+{1 \over a^{3}}+{1 \over b^{3}}+{1 \over c^{3}}+21 \geq 3(a+b+c)\left({1 \over a}+{1 \over b}+{1 \over c}\right). \]
[/quote]
Since the thesis, we write the inequality in equivalent form:
\[\sum_{cyc}{(a-b)^{2}\over ab}\cdot{(a-c)^{2}\over ac}\geq 0. \]

Actually, you wrote the inequality in (a) as
$\sum(\frac a{b}+\frac b{a}-2)(\frac b{c}+\frac c{b}-2) \geq 0$. Nice!

There is still a nicer and shorter solution to this inequality. :P
But the inequality in (b) is much more difficult. :D

[quote="Vasc"]
There is still a nicer and shorter solution to this inequality. :P
[/quote]

Maybe you mean this:
The given inequality can be rewritten as
\[18+\left(\frac{a}{b}+\frac{b}{c}+\frac{c}{a}\right)\left(\frac{b}{a}+\frac{c}{b}+\frac{a}{c}\right) \ge 3\left( 3+\frac{a}{b}+\frac{b}{c}+\frac{c}{a}+\frac{b}{a}+\frac{c}{b}+\frac{a}{c}\right)\]
Note that $18+XY-3(3+X+Y)=(X-3)(Y-3)$. So
\[\left(\frac{a}{b}+\frac{b}{c}+\frac{c}{a}-3\right)\left(\frac{b}{a}+\frac{c}{b}+\frac{a}{c}-3\right) \ge 0\]


quote="gagarin"]$ 2^{\sqrt [12]{sinx}} \plus{} 2^{\sqrt [4]{sinx}} \equal{} 2.2^{\sqrt [6]{sinx}}$[/quote]

OK, if I read this correctly, (I consider $ 2.2^{(sinx)^{\frac{1}{6}}}$ to be equal to $ 2^{1+(sinx)^{\frac{1}{6}}}$), this is the solution.
We shall substitute $ sinx$ with $ t$. So, we need to find all the solutions to
$ 2^{t^\frac{1}{12}}+2^{t^\frac{1}{4}}=2^{1+t^\frac{1}{6}}$. $ t$ needs to be between $ 0$ and $ 1$, including zero and one. We note that by A-G $ 2^{t^\frac{1}{12}}+2^{t^\frac{1}{4}}\geq 2^{1+\frac{t^\frac{1}{4}+t^\frac{1}{12}}{2}}$, which, in turn, again by A-G is greater or equal to $ 2^{1+t^{\frac{1}{4}+\frac{1}{12}}}=2^{1+t^\frac{1}{6}}$. So, the only possible solutions could be when $ t=0$ and $ t=1$ (the equality in A-G only works when all arguments are equal and strictly positive). So, the solutions are $ x=2k\pi$ and $ x=\frac{\pi}{2}+k\pi$, $ k$ being an integer.

I'm writing this in a hurry, so there might be some mistakes. Sorry.

thanhnam2902 has already answered it [url=http://www.mathlinks.ro/viewtopic.php?t=203457]here[/url]:mad:


[hide="Answer"]We can find each coefficient by using the combinations formula ${a\choose b}$, where $a$ is the degree that the expression is raised to, and we are finding the coefficient of the $b^{th}$ term of the sequence. Therefore, in the first term, the coefficient of $x^{18}$ is $\frac{20!}{18!2!}=190$. However, in successive terms, we need to factor in the powers of $x$ that are multiplied to each term. Each successive term will have one fewer term when factored, but we will always be finding the $(a-2)^{nd}$ term, so the sums of the coefficients will be ${20\choose 18}+{19\choose 17}+{18\choose 16}+...+{2\choose 0}=\frac{20\cdot 19}{2}+\frac{19\cdot 18}{2}+...+1$, which, coincidentally :roll:, gives us triangular numbers. Therefore, we have $19+(2\cdot 18)+(3\cdot 17)+...+(19\cdot 1)=(2\cdot 19)+(4\cdot 18)+...+(18\cdot 11)+(10\cdot 10)=1330$.[/hide]


[hide]That means $\frac{x}{100} \times x=4 \implies x^2=400$.

Since you can't have negative percents (at least I think you can't  :D ), $x=20$.

(the biffanddoc move) [hide]${\boxed{\boxed{\boxed{\boxed{\boxed{\boxed{\boxed{\boxed{\boxed{\boxed{\boxed{D}}}}}}}}}}}}$[/hide]
[/hide]

[quote="catcurio"]A positive number $x$ has the property that $x\%$ of $x$ is $4$. What is $x$?

$\text{(A)}\ 2 \qquad\text{(B)}\ 4 \qquad\text{(C)}\ 10 \qquad\text{(D)}\ 20 \qquad\text{(E)}\ 40$[/quote]

[hide]The easiest way in my opinion is guess and check. Always start with C. 10% of 10 is 1. That means it is too small. That also means A and B can't be the answer because they are smaller than C. On the AMC, the answers are always in ascending order or descending. Now check D. 20% of 20 is 4. So, the answer is D.[/hide]

[hide]
Translating that into an equation,
$\frac{x}{100}\times x =4$.
First we can simplify the left
$\frac{x^2}{100}=4$
Multiply by 100
$x^2=400$
Take the square root
$x=20$
So the answer is 20
$\boxed{\boxed{\boxed{\boxed{\boxed{\boxed{\boxed{\boxed{\boxed{\boxed{\boxed{\boxed{\boxed{\boxed{\boxed{D}}}}}}}}}}}}}}}$[/hide]

[quote="Klebian"][quote="robinhe"][hide]${\boxed{\boxed{\boxed{\boxed{\boxed{\boxed{\boxed{\boxed{\boxed{\boxed{\boxed{D}}}}}}}}}}}}$[/hide]
[/hide][/quote]

Owned.[/quote]

lol. :D
So this post isn't spam:
[hide]Write the equation like this: $x/100*x=4$. Multiply both sides by a hundred and you get $x^2$=400, or $x=20$.[/hide]

[hide]D
I used um... guess and check (sad huh) but the numbers were small anyway[/hide]


I'm wondering how coaches use the handbook that is distributed each year.   The book suggests a schedule, which is more or less going through the book in order (WarmUp1, WamUp2, Workout1, etc..)

Do you make copies of the problems and have students work on them individually?
Do you hand out problems, ask students to work on them at home, and then bring their solutions to the next meeting for your review?

Or, do people choose to use old contests or other materials instead?

The schedule given isn't neccessarily the best, I have found. I would aim to have the students complete the handbook in a few weeks. For extremely motivated students, one week. Then, you can always obtain more handbooks. List the answers in the things, and encourage students to ask among themselves about solutions to problems that they had trouble with. If they still can not come up with a solution, you can always help at practices.

As Chapter draws nearer, begin using old tests. Harder handbook problems can be integrated at will, but mostly old tests. AFter chapter, when preparing for state, I would say only old tests. Individual problems at home, team rounds at practice. Of the individual problems, supply the answers, and go over requested solutions at the students' will.

You can get old contests from goodlymaths. TZ's website also has some. Handbooks are great, especially because they are very organized.

The goodlymaths site link I found was broken.

You said TZ had a list of mathcounts problems.
Who is TZ?

Oh Totally Zealous.
EDIT: HIs website is: http://freewebs.com/totallyzealous/

http://goodlymaths.0catch.com/contests/mathcounts/mathcounts_archive.htm is goodlymaths archive

i highly don't recommend following the schedule unless you want to be dismally disappointed at your performance at chapter (unless, of course, you live in Nebraska). Rather, you should aim to do maybe one set a day after school and perhaps two on the weekends.

I've found that the handbook doesn't prepare you for the actual competition. Old comps are much better-if you want to win.

[quote="Treething"]i highly don't recommend following the schedule unless you want to be dismally disappointed at your performance at chapter [b](unless, of course, you live in Nebraska)[/b].[/quote]

What about nebraska?

It seems like he implied that NE is less competitive than others-not that I agree. A state's history can't determine how they'll do in the future.

[quote="goldendomer"]Speaking of goodly maths, when you click on other mathcounts site, you'll see Joshua Frost's Tips and Tricks for MATHCOUNTS. The link is broken. Does anyone know how to get there or what it is?[/quote]

Joshua Frost is my coach  :D  :D  :D

[quote="frost13"]Yeah I am :lol:

Remind me to give that to you Carl!!

I am actually in the process of cleaning that "Tips and Tricks" document up.  When it is all good, I'd be hapyto share again.  Mr. Goodlymaths asked me for it awhile ago when I had my own MC website.  No more site, that's why link is dead.[/quote]

I have it, Mr. Bhat sent it to me last year. At least I think it was the right one.

[quote="mysmartmouth"]Here is the archive of Mr. Frost's site:

http://web.archive.org/web/20040410190412/http://fc.sau16.k12.nh.us/~jfrost/mathcountstipstricks.htm

However, I'm getting a lot of invalid symbols, although I think it's probably just my browser.[/quote]

I'm also getting a lot of invalid symbols. :(

Does anybody has any idea how to fix it? :|

I don't think we'll actually be able to use the archive (it's too messy and...archived), but it gives us a good idea of the material Mr. Frost is working on.

[quote="frost13"]Hello, this is Mr Frost, yes THE Mr Frost:-)  That website isn't active anymore and I am working on cleaning up that document.  When it is ready, I'll post it for all the world!![/quote]

When will it be fixed? It does look good.  :)

[quote="mysmartmouth"]Here is the archive of Mr. Frost's site:

http://web.archive.org/web/20040410190412/http://fc.sau16.k12.nh.us/~jfrost/mathcountstipstricks.htm

However, I'm getting a lot of invalid symbols, although I think it's probably just my browser.[/quote]

This link works for me.
The text is good, but the pictures aren't showing up.


Determine all the functions $f$ such that

$f(x^2) - f(y^2) + 2x + 1 = f(x+y).f(x-y)$

for any real numbers $x$ and $y$.

[quote="JBL"]Tsk, tsk -- you didn't back-substitute to make sure your solutions work![/quote]

Well... okay, the second one doesn't.  :P


$ sin{\theta} \plus{} cos{\theta} \equal{} u$ find $ cotan{\theta}$ in terms of u

What is $ cotan{\theta}$?????

Is it $ tan{\theta}\equal{}\frac{perpendicular}{base}$ or $ cot{\theta}\equal{}\frac{1}{tan{\theta}}$???????

skand: $ \cot \theta \equal{} \frac{adjacent}{opposite} \equal{} \tan(90\minus{}\theta) \equal{} \frac{1}{\tan \theta}$

$ \sin \theta \plus{} \cos \theta \equal{} u$
$ 1 \plus{} \cot \theta \equal{} u \csc \theta$

Now $ \cot^2 \theta \plus{} 1 \equal{} \csc^2 \theta$ so we have
$ 1 \plus{} \cot \theta \equal{} \pm u \sqrt {\cot^2 \theta \plus{} 1}$
$ 1 \plus{} 2\cot \theta \plus{} \cot^2 \theta \equal{} u(\cot^2 \theta \plus{} 1)$
$ (1 \minus{} u)\cot^2 \theta \plus{} 2 \cot \theta \plus{} 1 \minus{} u \equal{} 0$
$ \cot \theta \equal{} \frac { \minus{} 2 \pm \sqrt {4 \minus{} 4(1 \minus{} u)}}{2(1 \minus{} u)} \equal{} \frac { \minus{} 1 \pm \sqrt {u}}{1 \minus{} u}$

I know those but I was not confident whether $ cotan{\theta}$ means same as $ cot{\theta}$

In my country,either $ cotangent{\theta}$ or $ cot{\theta}$ is used

skand: $ \cot$ is the usual way of abbreiviating it here too.  To me, "cotan" just looks like someone tried to guess the abbreviation.  As far as I know, there is nothing else it could mean.


if a,b,c,d nonzero naturals numbers and ad-bc=1 then gcd(a^2+b^2,ac+bd)=1

$ad-bc = 1$ implies that $\gcd(a, b) = 1$. now, if $N | a^2 + b^2$ and $N | ac+bd$, $N$ natural, then $N | a(ac+bd) - c(a^2+b^2) = abd - cb^2 = b(ad-bc) = b$. but then $N | a^2+b^2 - b^2 = a^2$. and this implies $N | \gcd(b^2, a^2) = 1$, so $N = 1$.


solve for the value of x:
16^x - 4^x - 2 = 0

kindly show your solution. and maybe a brief explanation. thanks!

Let $y=4^x$. It is equavalent $y^2-y-2=0\Longrightarrow y=2,x=1/2.$


modular arithmetic
Euler
function
number theory
totient function

What is the remainder when 2008^(2008) is divided by 2009?

Please explain.

[hide="Solution"]
$ 2008^{2008}\equiv (\minus{}1)^{2008}\equiv 1 (mod 2009)$

I used $ n\equiv \minus{}1(mod n\plus{}1)$, $ 1^x\equiv 1(mod m)$ because $ (\minus{}1)^2\equiv 1(mod 2009)$[/hide]

The remainder when 2008 is divided by 2009 is -1, so -1^2008=1

Or find $ a^k\equiv \pm 1(modm)$ so it's easy as $ \minus{}1$ and $ 1$ on any natural degree are $ \minus{}1$ or $ 1$.

[quote="sunehra"]So for for any remainder problem of the form $ a^b \equal{} ? \pmod{n}$ you could just find $ a \equal{} x \pmod{n}$ and then replace $ a$ with $ x$ in the previous problem and repeat until you get a nice number to evaluate?[/quote]

Yep. Also, you may sometimes find it useful to factor an expression. For example, if you're attempting to find the remainder when $ 4^{33}$ is divided by 9, note that 64 has a remainder of 1 when divided by 9. Hence, $ 4^{33}\equal{}(4^3)^{11}\equiv1^{11}\equiv1\pmod{9}$.

what I would recommend is finding patterns (as you know, by the way mods are defined, that there has to be a pattern in less than or equal to n if the a and b are congruent modulo n)
although sometimes, if the problem presents a ridicuously large modulo, which usually isn't the case, theorems (especially the Carmichael theorem, euler totient function, and fermat's little theorem) are also pretty helpful.
Wilson's theorem is very powerful when used, but there aren't a lot of times that you can use it... (although it might occur in contest problems)