MATHCOUNTS
number theory
least common multiple

I came across with this problem: Can any one give the clue?

A group of 25 people were discussing a large positive integer. "It can be divided by 1" said first person. "It can be divided by 2" said the second person. "and by 3" by third person. "And by 4" added fourth person. This continued until everyone had made such comment. If exactly two persons were incorrect and those two persons said consecutive numbers, what was the least positive integer they were discussing?

[hide] think that the correct answer to this problem is 5354228880.

If you look at the problem by writing out all 25 numbers, you see that you can eliminate some of the earlier terms. For example, if you make 2 one of the wrong numbers, the 4 and 8 etc are still factors of that number. However, an integer that is not a multiple of 2 cannot be a multiple of 4 or 8 or 16 etc.

Therefore, the number that are false must not have any of their multiples in the 25 numbers. Thus, the biggest numbers you can take out are 25 and 24. The number that I have put up is the LCM of 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17, 18, 19, 20, 21, 22, 23.
[/hide]
Am I right?

Also, it took a lot of time to find the LCM of 23 numbers. Is there a way to find which numbers I can eliminate before searching for the LCM?  :(

If a number is divisible by both 12 and 16, it is divisible by 24 anyway.  Therefore, you must remove some other numbers that the number isn't divisible by from factors of other numbers, such as primes like 23 and 19.

I'll post the answer once I'm done figuring out the LCM...

[quote="kyyuanmathcount"]I think it would be best to take out 23 and 22 (or 24 i dont think it matters) because you need to take out a large prime (it isnt possible to take out two primes, there are no consecutive primes except 3 and 2).[/quote]

You couldn't take out 22, as if it were divisible by 2 & 11, then it would be divisible by 22.  I listed the numbers, and I found 16 and 17 to be the only ones you could take out...

If you find the LCM of the remaining numbers, I think you'd get:
$2^3\cdot3^2\cdot5^2\cdot7\cdot11\cdot13\cdot17\cdot19\cdot23$

I remember this problem from somewhere...  Was this from the MC handbook last year?

Yes, that makes the most sense anyway. Think about it... since you never have two consecutive primes after $2,3$, which the number must obviously be divisible by, you must have at least one composite number. Logically, the given composite number that cannot divide must have more of a certain factor than every other number given. So, the first choice I would go with is $2^4=16$. Then, since $15$ obviously must divide the number since both $3$ and $5$ do, we use $17$, and that is the answer, since neither must necessarily divide the number.

Oh, yeah, sorry, didnt think of that.  sweet problem

This was a national team round question from some year, and yes, you take out 16 and 17


Hola Gente,

Despues de una larga ausencia, pues nuevamente me encuentro por aqui. Que sorpresa la mia al ingresar y darme cuenta que ya no era moderador de Spanish Communities y que ahora en mi lugar habian dos personas, incluido Pascual! Jeje! Talvez como moderador ahora Pascual entienda por que un par de veces edite sus post!

En fin, aun cuando no creo tener ahora el tiempo para volver a hacer tantos post como solia hacer antes, pues voy a ver si puedo volver a ser un miembro activo del foro, tanto en Español como en Ingles. En especial por que me parece que el foro en Español esta un poco mas activo que antes.

Por cierto Pascual, felicitaciones (me imagino que ya te han de haber felicitado mil veces antes que yo) me dijeron que sacaste oro en la Ibero. La verdad mi investigacion y otras obligaciones me han dejado tan poco tiempo disponible que no he chequeado, sin embargo, quien me lo dijo, pues lo considere una fuente confiable!

Ciao,

Gracias David, si es correcto.

Pues esperamos siga contribuyendo como lo hacia antes.

Con suerte no tendre que editar nada, solo ponerle latex a las cosas y ya. jaja.

Bueno, nuevamente, felicitaciones.

Pues como dije, en realidad esta medio dificil que vuelva pronto a participar tanto, aun cuando me parece genial que ahora haya mas actividad en los foros en Español. Por cierto, cuando den luz verde a la discusion de la Putnam (Dr. Merryfield dio un veto a la discusion de la misma hasta que el mismo de los enunciados) me gustaria comentar tambien aqui uno de los problemas... que o me volvi idiota y lo malentendi por completo... o es el problema mas facil que ha salido en una Putnam en los 66 anios que lleva de hacerse!

En fin, alli hare el comentario cuando sea apropiado!

Ciao,

Saludos David...

me alegra q este devuelta por la comunidad.... como veras ..
la comunidad esta creciendo.....
y realmente requerimos de personas que aporten a la comunidad como usted lo ha hecho...

gracias David por todas las iniciativas q ha tenido ...
y se q recibiremos de usted ... sus mejores iniciativas e ideas para seguir mejorando esta comunidad..
...... bueno claro cuando cuente con el tiempo suficiente...
y suerte con sus proyectos y tareas ...

Carlos Bravo :)
Lima -PERU


topology
vector
geometry
3D geometry
sphere
linear algebra
matrix

I'm following Lee's proof (Smooth Manifolds), example 1.24 (pg. 22).  $ V$ is an $ n$ dimensional real vector space, $ P$ is a $ k$ dimensional subspace, with $ V \equal{} P \oplus Q$, and $ Q'$ is any $ (n\minus{}k)$-dim subspace.  Lee states that

$ \{A \in L(P, Q) : \Gamma(A) \cap Q' \equal{} 0\}$

Is an open set in $ \mathbb{R}^{k(n \minus{} k)}$, where $ \Gamma(A)$ is the graph of $ A$.  However, he offers no reason, and I can't quite see why this is.

Just wanted to give this a little bump.  I'm starting to think that it's not as trivial as Lee seems to think.

This thread wouldn't be more appropriate in the linear algebra section, would it?

(Picking up my copy of Lee to glance at this)

The graph of any such $ A$ is a subspace of $ P\oplus Q\equal{}V$. It intersects the subspace $ Q'$ nontrivially if and only if it intersects $ Q'$ on the unit sphere $ S$ (in some convenient norm). The intersection of $ \Gamma(A)$ with $ S$ is a compact set; if it doesn't met $ Q'$, there's some nonzero distance $ \epsilon$ between them. Now, if $ \|A\minus{}A'\|<\epsilon$ in the operator norm, $ |Ax\minus{}A'x|<\epsilon$, and $ |(x,Ax)\minus{}(x,A'x)|<\epsilon$. This means that $ (x,A'x)$ can't be in $ Q'$, and we have constructed an open ball around $ A$ in the desired set. Repeating everywhere, the set is open.

Alternatively, we could show that the complement is closed. Choose a basis $ v_1,v_2,\cdots,v_k$ for $ P$ and a basis $ w_{k\plus{}1},w_{k\plus{}2},\cdots,w_n$ for $ Q'$. $ \Gamma(A)$ intersects $ Q'$ nontrivially if and only if the $ n$ vectors $ v_1\plus{}Av_1,\cdots,v_k\plus{}Av_k,w_{k\plus{}1},\cdots,w_n$ are linearly dependent. Put them in an $ n\times n$ matrix $ M(A)$; they're linearly dependent if and only if the determinant is zero. The entries and the determinant of $ M(A)$ are continuous functions of $ A$, so the set $ \{A: \det M(A)\equal{}0\}$ is closed.


geometry
geometric transformation
reflection
inequalities
induction
rectangle
combinatorics proposed

Let $ m,n$ be two natural nonzero numbers and sets $ A \equal{} \{ 1,2,...,n\}, B \equal{} \{1,2,...,m\}$. We say that subset $ S$ of Cartesian product $ A \times B$ has property $ (j)$ if $ (a \minus{} x)(b \minus{} y)\le 0$ for each pairs $ (a,b),(x,y) \in S$. Prove that every set $ S$ with propery $ (j)$ has at most $ m \plus{} n \minus{} 1$ elements.

[color=#FF0000]The statement was edited, in order to reflect the actual problem asked. The sign of the inequality was inadvertently reversed into $ (a \minus{} x)(b \minus{} y)\ge 0$, and that accounts for the following two posts.[/color]

We'll prove it by induction on $ n\plus{}m$.  For $ n\equal{}m\equal{}1$ the statement is true.
Let $ S$ be a set satisfying the condition and $ (a,b)\in S$ be any point other than $ (1,1)$ or $ (n,m)$.  (If there are no such points, then $ S$ has at most 2 elements and so the statement is true for this $ S$).

$ (a\minus{}x)(b\minus{}y)$ means that $ (x,y)\in S$ must be either in the rectangle $ (1,\ldots,a)\times(1,\ldots,b)$ or in the rectangle $ (a,\ldots,n)\times(b,\ldots,m)$.
These rectangles are of sizes $ a \times b$, $ (n\minus{}a\plus{}1) \times (m\minus{}b\plus{}1)$.
Since $ a\plus{}b<n\plus{}m$ and $ n\minus{}a\plus{}1\plus{}m\minus{}b\plus{}1<n\plus{}m$, the statement is true for both of these rectangles, so that there are at most $ a\plus{}b\minus{}1$ elements from $ S$ in the first one, and at most $ n\minus{}a\plus{}1\plus{}m\minus{}b\plus{}1\minus{}1$ elements from $ S$ in the second one.
Hence, the total number of elements in $ S$ does not exceed $ (a\plus{}b\minus{}1)\plus{}(n\minus{}a\plus{}1\plus{}m\minus{}b\plus{}1\minus{}1)\minus{}1\equal{}n\plus{}m\minus{}1$. (note that we substracted 1 because $ (a,b)$ was so far counted twice as it's in the intersection).

Wojtek

Let $ S_k \equal{} \{(a, b) \in A \times B \mid a \plus{} b \equal{} k\}$.  Then $ S_2 \cup S_3 \cup \cdots \cup S_{m \plus{} n} \equal{} A \times B$, and $ S$ can contain at most one element from each $ S_i$.

In fact the property was misquoted; it was its exact reversal $ (a - x)(b - y)\le 0$ for any pairs $ (a,b),(x,y) \in S$.
Funnily enough, the answer $m+n-1$ is the same! :)

Here is a nice geometrical interpretation:

Let $S=\{(a_i,b_i)|a_i\in A,b_i\in B\}$ be a set that has propriety $(j)$ and let $|S|=k$.Suppose that $a_1\le a_2\le...\le a_k$.From the hypothesis we know that $b_1\ge b_2\ge...\ge b_k$.

Consider in the geometrical plane the set of points $\mathcal{P}=\{(a,b)|a\in A,b\in B\}$.By a [i]lattice path[/i] we will refer to a path between two points $(a,b),(c,d)\in\mathcal{P},a\le c,b\ge d$ such that we move only to the right and\or down along the lattice grid starting from point $(a,b)$ and finishing at point $(c,d)$.It's pretty obvious that any [i]lattice path[/i] from $(a,b)$ to $(c,d)$ has constant length(more precisely $c-a+b-d$).

From $a_1\le a_2\le...\le a_k$ and $b_1\ge b_2\ge...\ge b_k$ we know that there is a [i]lattice path[/i] $\mathcal{L}$ from $(a_1,b_1)$ to $(a_k,b_k)$ that passes through $(a_2,b_2),...,(a_{k-1},b_{k-1})$ and so $|\mathcal{L}|\ge k-1$($|\mathcal{L}|$ is the length of $\mathcal{L}$).But it's obvious that the longest [i]lattice path[/i] between two points from $\mathcal{P}$ is the [i]lattice path[/i] from $(1,m)$ to $(n,1)$,which has length $n-1+m-1=m+n-2$,so $$m+n-2\ge |\mathcal{L}|\ge k-1$$ meaning that $k\le m+n-1$ or $|S|\le m+n-1$.This ends our proof.


A ninja has a certain amount of swords, all of them different, at home. When she wakes up, she realizes that she is late from work, and so she picks eight swords at random from her pile of swords. (She's an octopus? I don't know.) If the probability that she leaves a certain sword at home is 1/3, how many swords does she have?

[hide]The probability of leaving a certain sword home is
$\frac{n-1}{n}\cdot \frac{n-2}{n-1}\cdots \frac{n-7}{n-6}\cdot \frac{n-8}{n-7}=\frac{1}{3}$, and everything cancels except $\frac{n-8}{n}=\frac{1}{3}$. Solving, we get $n=12$.[/hide]

wouldn't it be easier to say that if she left 1/3 of her swords, she took 2/3, so 12?

Seriously, Treething, you need to think of coming up with a mock MathCounts test.  ;)

I kinda like the skater question better ;)


Sorry These were the first forums I found and I thought maybe since you all seem high in math to be able to help me. And people post on these regulary unlike the other empty math forums.

Im having a really hard time Factoring by grouping...

Can someone explain to me how it works I understand Foil somewhat of how the order goes so no need to clarify on that part.
2t^2+3t-2  (for those of you not following the way my teacher taught as to start was to do this )
2 * -2 = -4
4 * -1 = -4
4 + -1 = 3 I don't even understand this first step. I kno there muliplying the outnumber and then finding out what multiplies to get those numbers combined and what ADDS up to get the Middle number , well not that complicated. Now my teacher said I write it like this (2t^2-t) + (4+-2) next line is the 3rd step t(2t-1) + 2(2t-1) . Now that the parenthese are the same theres one more step but I don't kno if I did it right 2t^2-1 + 4t-1. See thats totally wrong I have no idea what im doing here PLEASE any help is greatly appreciated for I have a test on it tommorow.

If that problem was too difficult or strange can you help me with this one
x^2+5x - 6 . (factor the expression)

Thank you

[quote="D@nny"]
2t^2+3t-2  (for those of you not following the way my teacher taught as to start was to do this )
2 * -2 = -4
4 * -1 = -4
4 + -1 = 3 I don't even understand this first step. I kno there muliplying the outnumber and then finding out what multiplies to get those numbers combined and what ADDS up to get the Middle number , well not that complicated. Now my teacher said I write it like this (2t^2-t) + (4+-2) next line is the 3rd step t(2t-1) + 2(2t-1) . Now that the parenthese are the same theres one more step but I don't kno if I did it right 2t^2-1 + 4t-1. See thats totally wrong I have no idea what im doing here PLEASE any help is greatly appreciated for I have a test on it tommorow.[/quote]

I don't really understand what the point of the first two multiplications are.  Basically, what you're trying to do is to find $a, b$ such that

$(2t + a)(t + b) = 2t^2 + 3t - 2$

If you expand this, it becomes

$(2t^2 + (a + 2b) t + ab = 2t^2 + 3t - 2$

So

$a + 2b = 3$
$ab = -2$

At this point it's mostly a matter of guess-and-check, and you find that $a = -1, b = 2$.  So

$2t^2 + 3t - 2 = (2t - 1)(t + 2)$

If you wanted to factor this by grouping, you could notice that

\begin{eqnarray*} 2t^2 + 3t - 2 &=& 2t^2 + 4t - t - 2 \\ &=& 2t(t + 2) - (t + 2) \\ &=& (2t - 1)(t + 2) \end{eqnarray*}

Or, the way you seem to have grouped it,

\begin{eqnarray*} 2t^2 + 3t - 2 &=& 2t^2 - t + 4t - 2 \\ &=& t(2t - 1) + 2(2t - 1) \\ &=& (t + 2)(2t - 1) \end{eqnarray*}

If you've factored enough expressions, it will come naturally. For example, $2t^2+3t-2$. This is how I would do it: First, look at the factors of $-2$ (the constant term). This could be $-1,2$ or $-2,1$. Now look at the factors of $2$ (the leading coefficient), which would be $1,2$. When you multiply one of the first pair's numbers with the second pair's, then add them, it should equal $3$. The combination that works is $2\times2+1\times-1=3$. So the factored form is $(2t-1)(t+2)$. :D


Here are some probability problems from AoPS on  probability:

1.  There are 11 students in a Chemistry class, including the Barnes triplets:  Anna, Bob and Katy.  The teacher calls all 11 students in random order one at a time to her desk to tell each student his or her final grade.  What is the probability the Bob is the first triplet that she calls to her desk?

2.  The same 11 students from the previous problem make up a History Class, again with the the Barnes triplets:  Anna, Bob and Katy.  As before, the teacher calls all 11 students in random order one at a time to his desk to tell each student his or her final grade.  However,the first time he calls a student to his desk, Bob is in the bathroom and therefore can't be called up, but he's back in time for the rest of the students to be called.
a.  What is the probability that Bob is the first of the triplets to be called to the desk?
b.  What is the probability that Anna is the first of the three called to the desk?

3.  A fair 6-sided die is tossed 8 times.  The sequence of 8 results is recorded to form an 8-digit number.  For example. if 3, 5, 4, 2, 1, 1, 6, 5 are tossed, then we have the 8-digit number 25, 421,165.  What is the probability that the number formed is a multiple of 8?[hide="Hint"]remember divisibility rules[/hide]

[hide]Q1. 1/11

Q2a. 8/100
Q2b. 1/10

Q3. [quote]3. A fair 6-sided die is tossed 8 times. The sequence of 8 results is recorded to form an 8-digit number. For example. if 3, 5, 4, 2, 1, 1, 6, 5 are tossed, then we have the 8-digit number 25, 421,165.[/quote] Shouldn't it be [u][b]3[/b][/u]5,421,165?

11/108?[/hide]


Decode the following quote by the famous science fiction author Orson Scott Card.

"ODC  LAWB  HC  ZFUWI  CF  ADUEB  OWHBPI, CAW  FBMX  CXTW  FY    ZEDIW  CAEC  UECCWKI  HI  YHBEM  ZEDIW, CAW  TDKTFIW.  LAEC  E   TWKIFB  AEN  HB  UHBN. FBZW  XFD  DBNWKICEBN   LAEC  TWFTMW   KWEMMX  LEBC, XFD  ZEB'C  AECW   CAWU  EBXUFKW. XFD  ZEB YWEK CAWU, ODC  XFD  ZEB'C  AECW  CAWU,  OWZEDIW  XFD  ZEB  EMLEXI   YHBN  CAW  IEUW  NWIHKWI  HB  XFDK  FLB  AWEKC."

[hide="hint 1"]This is a simple substitution cipher.[/hide]

[hide="hint 2"]I=S[/hide]

"but when it comes to human beings, the only type of cause that matters is final cause, the purpose. what a person had in mind. once you understand what people really want, you can't hate them anymore. you can fear them, but you can't hate them, because you can always find the same desires in your own heart"


Let

\[ \omega = \frac{-1+i\sqrt{3}}{2}. \]
The non-zero complex numbers $a$, $b$ and $c$ satisfy

\[ \frac{a}{b} = \frac{b}{c} = \frac{c}{a}. \]
What is the value of

\[ \frac{a+b-c}{a-b+c}? \]
(A) 1
(B) $\pm \omega$
(C) 1, $\omega$, or $\omega^2$
(D) 1, $-\omega$, or $-\omega^2$

Solution in spoiler:

[hide]First, note that  and  are two of the cubic roots of unity.

Since a/b = b/c = c/a, (a/b)^2 = a/b * b/c = a/c = 1/(c/a) = 1/(a/b)

So (a/b)^3 = 1.  Similarly, (b/c)^3 = 1, so (c/b)^3 = 1, so c/b is one of the three roots of unity (1, \omega^2\omega\omega^2$.  Thus the answer is (C).[/hide]


1) In the triangular grid (Attachment), all nearest neighbors are one unit apart.  Express as a common fraction the probability that two distinct, randomly-selected points are a rational distance apart.

[hide]ANS: 2/3[/hide]

2) Two points determine on line and three noncollinear points determine three lines.  How many lines are determined by 18 points, no there of which are collinear?

[hide]ANS: 153[/hide]

[hide="2"]Wecan choose two point to make a line in $ \binom{18}{2}\equal{} 9\cdot{17}\equal{} 153$ ways.[/hide]

[hide="#1 solution"]counting!! 31 [/hide]

the answer

EDIT: atleast, its my answer

It says express as a common fraction. And probability.


If $ \triangle ABC$ is inscribed in a semicircle whose diameter is $ AB$, then $ AC \plus{} BC$ must be

$ \textbf{(A)}\ \text{equal to }AB \qquad\textbf{(B)}\ \text{equal to }AB\sqrt {2} \qquad\textbf{(C)}\ \geq AB\sqrt {2}$
$ \textbf{(D)}\ \leq AB\sqrt {2} \qquad\textbf{(E)}\ AB^2$

[hide="Click for solution"]
The inscribed triangle with maximum perimeter is the isosceles right triangle, in this case $ AC\plus{}BC\plus{}AB\sqrt{2}$.  Thus, in general, $ AC\plus{}BC \ge AB\sqrt{2} \implies \boxed{\textbf{(D)}}$.
[/hide]


linear algebra
matrix
linear algebra unsolved

Let $ C$ be an $ n\times n$ matrix such that whenever $ C=AB$ then $ C=BA$. What is $ C$?

suppose that $ P$ is invertible. Then $ CP=(CPP^{-1})P=P(CPP^{-1})=PC$, hence $ C$ commutes with every invertible matrix..

The converse is not quite true: $ C=0$ commutes with anything, but does not satisfy the above assumption. However, this is the only exception.


Find all values of the real m, for which there exists a triangle with sides a,b,c and angles A,B,C, such that a/b=m=<A/<B.

[hide]
$a = 2R\sin A$

$b = 2R\sin B$

$\frac{A}{B}= \frac{a}{b}\Rightarrow  \frac{A}{B}= \frac{\sin A}{\sin B}\Rightarrow \frac{\sin A}{A}= \frac{\sin B}{B}$

The only solution is $m=1\Rightarrow a=b\Rightarrow \triangle ABC$ isosceles

[color=blue][b]Indeed:[/b][/color]

$f(x) = \frac{\sin x}{x}\Rightarrow  f'(x) = \frac{x\cos x-\sin x}{x^{2}}$

We will show that $f(x)$ is strictly decreasing in $(0,\pi)$

$g(x) = x\cos x-\sin x$

$g(0) = 0$

$g'(x) = \cos x-x\sin x-\cos x =-x\sin x<0,\ \forall x\in (0,\pi)$

$\Rightarrow g(x)<0, \ \forall x\in (0,\pi)$

$\Rightarrow  f$ strictly decreasing in $(0,\pi)$
[/hide]


Will MIT admission host a pre-HMMT 'get-together' this year?

They did last year:

http://www.mitadmissions.org/topics/misc/miscellaneous/small_gettogether_friday.shtml

I'm not sure... I saw Matt (McGann) in the halls yesterday, but I forgot to ask about this.
If there is something, you'll see it on the mit blogs or you could just email him. That's your best bet.

I was hoping to do another small gathering this year, but it looks like I'll be tied up in Selection Committee all weekend, unfortunately.

Oh, and from another thread:

[quote]I didn't mean like, once you win you join instantly, I meant that like if you win HMMT, then when you apply they let you in for sure. I heard it from a coach of mine...[/quote]

As joml88 said, this isn't true.  I don't know how these crazy rumors get started.

Have an awesome time on campus!


Too well known. We have \[ \dfrac{1}{a_j} - 1 = \dfrac{a_1 + \cdots + a_{j - 1} + a_{j + 1} + \cdots + a_n}{a_j} \geq (n - 1) \cdot \dfrac{\sqrt[n - 1]{a_1\cdots a_{j - 1}a_{j + 1}\cdots a_n}}{a_j}.  \]

Multiplying these inequalities for $j = 1,2,\cdots,n$ gives that \[ \prod \left(\dfrac{1}{a_j} - 1\right) \geq (n - 1)^n.  \]


[quote="Math pro"]x,y,z>0  and x+y+z=3
prove that

  x/(xy+1) + y/(yz+1) + z/(zx+1) >=3/2[/quote]
See here:
http://www.artofproblemsolving.com/Forum/viewtopic.php?t=21689


geometry
transformation
algebra
Translation
IMO Shortlist

We call a set $ S$ on the real line $ \mathbb{R}$ [i]superinvariant[/i] if for any stretching $ A$ of the set by the transformation taking $ x$ to $ A(x) \equal{} x_0 \plus{} a(x \minus{} x_0), a > 0$ there exists a translation $ B,$ $ B(x) \equal{} x\plus{}b,$ such that the images of $ S$ under $ A$ and $ B$ agree; i.e., for any $ x \in S$ there is a $ y \in S$ such that $ A(x) \equal{} B(y)$ and for any $ t \in S$ there is a $ u \in S$ such that $ B(t) \equal{} A(u).$ Determine all [i]superinvariant[/i] sets.

it would seem that it is sufficient for $ S$ to be be closed under applications of $ A^{\minus{}1}B$ and $ B^{\minus{}1}A$. Note $ A^{\minus{}1}B \equal{} (B^{\minus{}1}A)^{\minus{}1}$.

$ A^{\minus{}1}B \equal{} \frac{(x\plus{}b)\minus{}x_0}{a}\plus{}x_0$
$ B^{\minus{}1}A \equal{} x_0\plus{}a(x\minus{}b\minus{}x_0)$

not sure if this is enough to characterize the sets, or if something more explicit is wanted, it certainly doesn't 'feel' complete.

That's pretty much a restatement of the problem.

Is $ x_0$ fixed?

the problem reads as though $ A$ (and its variables $ a,x_0$) is fixed, and we need to find sets $ S$ for which there exists a $ B$ that satisfies the necessary closures. but it seems to me that its easier (and equivalent) to determine a collection of set $ S$'s for a fixed $ B$, correct me if my logic is flawed.

what i should have said in my previous post was that for a fixed $ A$ and $ B$, then for any subset $ X \subset \mathbb{R}$ and take $ S$ to be the closure of $ X$ under the applications of $ A^{\minus{}1}B$ and $ B^{\minus{}1}A$.

but the problem seems to suggest that there is a nice 'presentation' for such sets $ S$.

We claim that the superinvariant sets are $\varnothing$, $\mathbb{R}$, $\{p\}$, $\mathbb{R}\backslash\{p\}$, $(p,\infty)$, $[p,\infty)$, $(-\infty,p)$, $(-\infty,p]$. Clearly all these are superinvariant. Conversely, suppose $S$ is a superinvariant set.

[b]Case A:[/b] $S$ is periodic, i.e., there exists a [i]period[/i] $t > 0$ such that $S+t = S$ (here $S+t$ means $\left\{s+t : s\in S\right\}$).
If $t$ is a period, then scaling $S$ by a factor of $a$ shows that $at$ is also a period of $S$. Hence, either $S$ is empty, or else we can pick $s\in S$ and then deduce that $s-at, s + at\in S$ for all $a > 0$, and by varying $a$ we see that $S = \mathbb{R}$.

[b]Case B:[/b] $S$ is not periodic.
We note that, in this case, for a given stretching $A$, we have a unique choice of the constant $b$ in the translation $B$. Now, let $a_1,a_2$ be positive reals not equal to $1$. We know there exist $b_1,b_2\in\mathbb{R}$ such that $S$ is invariant under the transformations $A_1(x) = a_1x-b_1$ and $A_2(x) = a_2x-b_2$. If we first apply $A_1$ and then $A_2$, we get the transformation $x\mapsto a_1a_2x - (a_2b_1 + b_2)$. If we first apply $A_2$ and then $A_1$, we get the transformation $x\mapsto a_1a_2x - (a_1b_2 + b_1)$. Both of these transformations are stretching by the same factor, and $S$ is invariant under both transformations, so by the afore-mentioned uniqueness we see that $a_1b_2 + b_1 = a_2b_1 + b_2$, i.e., $b_1/b_2 = (a_1 - 1)/(a_2 - 1)$. By varying $a_1$ and $a_2$ we can find a constant $p$ such that, for any $a > 0$, $S$ is invariant under $A(x) = ax - p(a-1)$, i.e., $A(x) = p+a(x-p)$. In other words, any stretching about $p$ preserves $S$. From this we deduce that:
(i) if $S$ contains any point $q > p$, then $S$ contains the entire interval $(p,\infty)$.
(ii) if $S$ contains any point $q <p$, then $S$ contains the entire interval $(-\infty,p)$.
Combining these observations we may conclude that $S$ is a union of some of the following sets: $(-\infty,p)$, $\{p\}$, $(p,\infty)$, which gives us the answers.


In a triangle ABC ,K,L,M are points on the sides BC,CA,AB respectively.

AK,BL andCM are concurrent at the point P.

Prove PK/AP + PL/BP +PM/CP >= 3/2

Denote the area of triangle $ABC$ by $S$, area of triangle $PBC$ by $S_1$, area of triangle $APC$ by $S_2$ and area of triangle $ABP$ by $S_3$

we have
$PK/AP = S_1/(S-S_1), PL/BP = S_2/(S-S_2), PM/CP  = S_3/(S-S_3)$

change variable: $S-S_1=x,S-S_2=y,S-S_3=z$
then $x+y+z=3S-(S_1+S_2+S_3)=2S$, $S=(x+y+z)/2$.

the inequality becomes
\[
\frac{\frac{y+z-x}{2}}{x} + \frac{\frac{x+z-y}{2}}{y} + \frac{\frac{x+y-z}{2}}{z} \geq \frac{3}{2}
\]
or
\[
\frac{y+z}{x} - 1 + \frac{x+z}{y} - 1 + \frac{x+y}{z} - 1 \geq 3
\]

\[
\frac yx + \frac zx + \frac xy + \frac zy +\frac xz +\frac yz\geq 6
\]
which is easy to prove by AM-GM ineq.

[quote="liyi"]\[\frac yx + \frac zx + \frac xy + \frac zy +\frac xz +\frac yz\geq 6\]which is easy to prove by AM-GM ineq.[/quote]

You're kidding right?  :lol:  \[\frac ab + \frac ba \geq 2\] ;)

[quote="tetrahedr0n"]Whats the problem with what liyi wrote?[/quote]

Nothing, Peter knows that it's a well known inequality.


Let $ a,b,c$ be positive number such that $ a^2\plus{}b^2\plus{}c^2\equal{}3$. Prove that
$ 1 \plus{} \frac {1}{3}abc \geq \frac {4}{9}(ab \plus{} bc \plus{} ca)$  :)

The inequality can be written as
\[ 9\plus{}3abc\ge 4(ab\plus{}bc\plus{}ca).\]
Now it is well known that
\[ a^2\plus{}b^2\plus{}c^2\plus{}3(abc)^\frac 23\ge 2(ab\plus{}bc\plus{}ca)\]
and can be easily proved from Schur's inequality and AM-GM. Hence, letting $ t\equal{}(abc)^\frac 13$, we'll be done if we can show that
\[ 9\plus{}3t^3\ge 6\plus{}6t^2\Leftrightarrow (t\minus{}1)(t^2\minus{}t\minus{}1)\ge 0,\]
which is true since $ t\le 1$ and $ t^2\minus{}t\minus{}1\le \minus{}t<0$.

[quote="Gibbenergy"]Let $ a,b,c$ be positive number such that $ a^2 \plus{} b^2 \plus{} c^2 \equal{} 3$. Prove that
$ 1 \plus{} \frac {1}{3}abc \geq \frac {4}{9}(ab \plus{} bc \plus{} ca)$  :)[/quote]
My proof, so long :oops:
$ \begin{array}{l}
 \frac{{a^2  \plus{} b^2  \plus{} c^2 }}{3} \plus{} \frac{1}{3}abc \ge \frac{4}{9}(ab \plus{} bc \plus{} ac) \\
  \\
  \Leftrightarrow \frac{3}{2}((a \minus{} b)^2  \plus{} (b \minus{} c)^2  \plus{} (a \minus{} c)^2 ) \plus{} 3abc \minus{} ab \minus{} bc \minus{} ac \ge 0 \\
  \\
  \Leftrightarrow \frac{3}{2}((a \minus{} b)^2  \plus{} (b \minus{} c)^2  \plus{} (a \minus{} c)^2 ) \plus{} \sum {ab\frac{{c^2  \minus{} \frac{{a^2  \plus{} b^2  \plus{} c^2 }}{3}}}{{c \plus{} \sqrt {\frac{{a^2  \plus{} b^2  \plus{} c^2 }}{3}} }} \ge 0}  \\
  \\
  \Leftrightarrow \sum {(\frac{3}{2} \minus{} \frac{{b(a \plus{} c)(a \plus{} c \plus{} \sqrt {\frac{{a^2  \plus{} b^2  \plus{} c^2 }}{3}} )}}{{3(c \plus{} \sqrt {\frac{{a^2  \plus{} b^2  \plus{} c^2 }}{3}} )(a \plus{} \sqrt {\frac{{a^2  \plus{} b^2  \plus{} c^2 }}{3}} )}}} )(a \minus{} c)^2  \ge 0 \\
 \end{array}$
And it's easy to prove $ \frac{3}{2} \minus{} \frac{{b(a \plus{} c)(a \plus{} c \plus{} \sqrt {\frac{{a^2  \plus{} b^2  \plus{} c^2 }}{3}} )}}{{3(c \plus{} \sqrt {\frac{{a^2  \plus{} b^2  \plus{} c^2 }}{3}} )(a \plus{} \sqrt {\frac{{a^2  \plus{} b^2  \plus{} c^2 }}{3}} )}} \ge 0$%  :wink: Is that right?


On agreement, we will coordinate our tournament at 7:30 EST on Thursaday, February 21.

The teams that will be playing each other are: OhioIsBallerz vs AZNPride.

Plesae use the other post to decide rules.

Thanks

Just to let you know, I just sent a PM to everyone participating in the Feb. 2008 Tournament.  You should have gotten one too!
:)

First, THE RULES ARE WRONG.  We ARE using verizonwires's rules: 1 normal with everyone, and 9 countdowns.  Failure to do so will result in an automatic win for the opposing team.  Ties settled by ONE countdown.  Thanks now to minicon, all of that got screwed up.
Second,  [size=200][b][u]ADD MY CREDIT TO THE IMAGE[/u][/b][/size].
Thirdly, all tournament information should be posted on this forum, not on another, or on another message board.  If the information is spread all over the place, nothing will ever get done.  If you decide to do this differently, you will be hosting your own tournament seperate from mine.

Fourth, minicon: stop it.  Don't send PMs with false information.

I like minicon's rules much better, who put you in charge?

Wasn't I the one who started the tournament? :lol:

I put myself in charge, and no one else, and honestly, now is too late to change rules.  Like I said, this isn't going to happen if we don't post all the information on this forum.  I can't see what happens when forms are submitted and I don't have the time to check the other message board.  If you guys want to post elsewhere, that's fine with me so long as you update here first!

I'm sorry, but if you decide not to play by the rules, Aznpride will get an automatic advance to Round 2.

Edit: To answer your question about the rules, here there is no settling a tie and no countdowns so that is why we've decided to use verizonwires's rules.

Who decided that anyone got to run this tournament? As of now, if anirudh wants to proclaim himself dictator, fine, but I'm not participating without any type of consensus.

[quote="MustItAlwaysBeMe"]#H34N1: This is what happens; listen to others.  Thanks Catalyst, I hope everyone else feels the same way.

And for anyone who cares, these are the rules that we have agreed by:

[quote="verizonwires"]each round has 3 matches:

(first to 2 points advances to the next round)

the first is ranking normal where the you add scores and see the winner. (winner gets 1 point)*

the second is countdown with each player going head to head. (winner gets 1 point)

the third is an option if a winner is not decided. The third is a response normal game. in which a winner will be decided. (winner get 1 point) *

* If there is a tie, it will be broken by a one-one countdown with the best player on the team.[/quote][/quote]

We'll play with those rules

[quote="MustItAlwaysBeMe"][quote="perfect628"]I don't remember agreeing to that. In fact, I remember objecting.[/quote]
Sorry, then, I thought it was unanymous.  Unfortunately, it is fixed for the Feb. Tournament, but start a poll for March, and see what people think.[/quote]

I already started a poll for march

i'll be the self proclaimed dictator! since i'm not in anyteam thats participating...

=(

Having the two coordinators running the tournament constantly disagreeing over what's happening, two confusing websites, and bad use of the English language (i.e. "we will coordinate our tournament at 7:30 EST on Thursaday, February 21" I have no idea what this means, does this mean the first match is that time or we are picking times during that time?) is slowing this tournament down and confusing people.

I think this has already been settled ;-)

Quite frankly, I'm not happy with perfect at all! He decided to join a tourny run by MustItAlwaysBeMe, and now he's making demands. perfect, make ur own tourny if u want but quit harassing others.
If this continues to happen, I'm hoping that MustItAlwaysBeMe kicks perfect out of the tourny for good. I'm not trying to get my team to win, but I just want fair game and i really like verizonwires' rules.
If there is ne one not agreeing to wat i am saying, i suggest u speak up now!

returnofkiladi - I don't see what point you have there. Anyone has the right to petition for a change of rules; the response to this petition is up to MustIt and others. He still has the right to speak what he wants to if he believes it is going to speed up and make the tournament flow better.

returnofkiladi, I believe I made two comments, one of which was a clarfication of the other. I'm curious as to the "behavior" of which you speak.

You know... err... they didn't have to make the tournament. We can just make our own rules and our own website... H34N1 can do it for us!! =).  I think the two teams should agree on their own rules, but whatever... Yo carl, when we playing

flaming is bad..

I liked minicon's rules for its simplicity, but I also liked verizonwires' rules for its accuracy/thoroughness.

If we're going to manage the tournaments on a separate site (namely #H34N1's), we should post all stuff related to the tournaments ON THAT SITE and nothing there on AoPS. I thought we would do most of the management on the forum here so I didn't even bookmark #H34N1's site, and now I can't even find the link anymore.

So first off, we should decide the rules so everyone can compete in the same way. I was under the impression we were going by these:
--2 Normal Games, 20 probs, 45 secs, ranking.
--Tie broken by countdown best vs. best

comment ONLY IF you have thoughts on the rules, and in a format similar to the above. Flaming can be done on AIM. Thanks.

The site is http://mersenne.zapto.org/FTW/.

The rules aren't very clear, so we need to come to consensus

[quote="undefined117"]So as of now, we're going by the rules I stated two comments ago.

and #H34N1, it says access denied on your page. dunno why. I can't login either.[/quote]

I think you have to register.

yeah that's why i'm wondering. I did register. meh i'll register again..


It is known (see also http://www.mathlinks.ro/Forum/viewtopic.php?t=311341) that if $ a_k\ge 0,$ then
$ \sum a_k$ converges if and only if $ \sum a_k/S_k$ converges ($ S_k \equal{} \sum^k a_n$) and provided that
$ S_k\ne0.$

Give an example of a convergent series $ \sum a_k$ such that $ \sum a_k/S_k$ diverges ($ S_k\ne0$)

$ 1\minus{}\frac{1}{2}\minus{}\frac{1}{4}\minus{}\frac{1}{8}\minus{}\frac{1}{16}\minus{}\dots$
Here $ a_k/S_k\equal{}\minus{}1$ for all $ k\ge 2$.

nice and simple. The example I had in mind was much less simple: $ a_k\equal{}(\minus{})^k/ln(k)$ whose proof requires "more than one line"

[quote="mlok"]$ 1 \minus{} \frac {1}{2} \minus{} \frac {1}{4} \minus{} \frac {1}{8} \minus{} \frac {1}{16} \minus{} \dots$
Here $ a_k/S_k \equal{} \minus{} 1$ for all $ k\ge 2$.[/quote]


This is probably easy...but i can't get it. :wallbash_red:

The polynomial p(x) has a remainder 3 when divided by x-1 and remainder 5 when divided by x-3. What is the remainder when divided by (x-1)(x-3)?

good luck!!!!!  :ninja:

$P(x)=Q(x)(x-1)(x-3)+(ax+b)$

Knowing that $P(1)=3$ and $P(3)=5$ and substituting $x=1$ and $x=3$, we get

$3=a+b$
$5=3a+b$

The solution of the system is $a=1, b=2$. Hence the desired remainder is $R(x)=x+2$

Reainder has to be of  first degree and thus suppose the  remainder is $ax+b$

[hide="hint"]Do you get $a+b=3$ and $3a+b=5$ ?
[hide="because"] write the polynomial as $(some \ thing) (x-1)(x-3)+ax+b$ what are the remainders if divided by $x-1$ ? by $x-3$ ? [hide](same as putting x=1 (or 3 ) in the expression) [/hide][/hide][/hide]
[hide="answer should be"] $x+2$[/hide]

Hint:

Use Division Algorithm, and make sure you know how to substitute values for x into polynomials.


function
logarithms
Support
real analysis
real analysis solved

Let $f$ be a $C^{\infty}$ class real function such that for all $x\in (\frac{1}{2},2)$ we have $f(x)>0$ and $f$ is 0 everywhere else. Define $g(x)=\sum_{j\in Z}{f(\frac{x}{2^j})}$ for positive $x$ and $h(x)=\frac{f(|x|)}{g(|x|)}$ if $x$ is different from 0 and 0 if $x=0$. Prove that $h$ is also $C^{\infty}$.

If we ignore terms that are zero, then for each fixed $x>0$ there is a neighborhood of $x$ on which the sum that defines $g$ is a finite sum. Hence $g$ is infinitely differentiable on $(0,\infty).$ At each $x$ there is at least one nozero (hence positive) term in the sum, so $g(x)>0$ for $x>0.$

$g$ sastifies the scaling property that $g(x)=g(2x).$ That is, $g(\ln t)$ is a periodic function of $t.$ It then follows that there exist constants $0<c<C$ such that for all $x>0,$ $c\le g(x)\le C.$

Note that this all shows that $g$ is discontinuous at $0.$ It is smooth in neighborhoods that do not include zero.

The support of $h$ is the set $\left[-2,-\frac12\right]\cup \left[\frac12,2\right].$ On its support, this is the quotient of two $C^{\infty}$ functions, with the denominator bounded away from zero. Hence $h$ is $C^{\infty}$ on its support and hence everywhere.

harazi didn't ask it, but there's an obvious point to this:

$\sum_{j=-\infty}^{\infty}h\left(\frac x{2^j}\right)=1$ for all $x\ne0.$ This collection of functions acts as a smooth partition of unity with some nice scaling properties. I can think of several possible uses for such an object in Fourier analysis (or Paley-Weiner theory, or pseudodifferential operators).


calculus
derivative
inequalities
inequalities proposed

Let $a, b, c, d > 0$
Prove that:
$a^2 + b^2 + c^2 + d^2 + 2\sqrt{abcd} \geq ab + bc + ca + da + db + dc$

Well, I've solved it using theorem about sum of partial derivatives twice, and my proof is not-so-long, but I would like to see something without calculus.

I may be wrong, but isn't this Turkevici's inequality? I'm pretty sure it's been discussed here, if you want to search. Incidentally, what technique (sum of partials) are you talking about?

Ah, you are right, I haven't thought about that before :).
And about that theorem I don't know if it has any name, but I was thinking about that:

Let $x_1, x_2, ..., x_n$ be non-negative reals.
We want to prove $f(x_1, x_2, ..., x_n) \geq 0$
If the following two conditions are satisfied:
1.If $x_k$ is 0 for $k=1,2,..,n$ then $f(x_1, x_2, ..., x_{k-1}, 0, x_{k+1}, ..., x_n) \geq 0$ for all non-negative reals $x_1, x_2, ..., x_n$
2.If $\sum f_{x_i}'(x_1, x_2, ..., x_n) \geq 0$
then it implies that $f(x_1, x_2, ..., x_n) \geq 0$ for all non-negative reals $x_1, x_2, ..., x_n$.

I think this problem can be solved by Karamata 's ineq  ;)

[quote="siuhochung"]I think the "theorem" you mentioned in posted the the theorems and formulaes section already?[/quote]

Indeed, here's a link

http://www.mathlinks.ro/Forum/viewtopic.php?t=20851


it is known that the order of stability of alkene is directly propotional to the number of alkyl groups attached to them ...but using this fact now how oes one prove the relative reactivities of each towards addition?
well the statement is that obvious that more substituted gives higher degreee carbocation  and better is the charge distributed...
But the relative reactivities depend on the change in potential energy of molecules with repect to the molecules out of which they are formed.so how can one be so sure that the stability determine s relative reactivity??

The relative reactivity of alkenes also depend on the nature of the electrophile, because if the rate determining step of the addition is the first (carbocation formation) then the more substituted the double bond more favourable the addition will be, and so steric effects are not important; however, if the rate determining step is the second (for example, in hydroboration and oxumercuriation), then steric effects are more important and the less substituted the double bond the better.


quadratics
number theory
Diophantine equation

Does anyone know how to solve a generalized form of the Markov diophantine equation?

(x_1)^2 + (x_2)^2 + ... + (x_n)^2 = n*x_1*x_2*...*x_n

Okay, the thoughts that I had are definitely not very helpful before.  Sorry.

For the standard Markov equation -- how do you prove that all solutions are generated by the chain?

Hmm, actually, I found the answer (sorta). For

(x_1)^2 + (x_2)^2 + ... + (x_n)^2 = k*x_1*x_2*...*x_n

if there is a solution (x_1, ..., x_n) then (x_1, ..., x_(n-1), k*x_1*x_2*...*x_(n-1) - x_n) is also a solution. For k=n, we can find the trivial solution (1, ..., 1) and work from there.

Yeah, it would. I don't know how to prove that though, nor how someone would come up with that formula. I think Mathworld mentioned something about it being quadratic in any one of the variables and so we could generate the solutions by noting that, but I didn't really get how.


Let $a_1,a_2,\ldots a_n$ be real numbers such that their sum is equal to zero. Find the value of \[ \sum_{j=1}^{n} \frac{1}{a_j (a_j +a _{j+1}) (a_j + a_{j+1} + a_{j+2}) \ldots (a_j + \ldots a_{j+n-2})}.  \]
where the subscripts are taken modulo $n$ assuming none of the denominators is zero.

Please help me with this!!!! I havent figured out anything !!!!

[hide=" Not spam! "]( Please this is NOT Spam!! - just a bit frustrated with the problem!) [/hide]


[color=red][i]Για δείτε και αυτό !

 Ήθελα από καιρό να το βάλω για λύση, αλλά με τη δουλειά στη γεωμετρία όλο και το ανέβαλα ![/i][/color]

 Τα $ P_a,P_b,P_c$  είναι τα συμμετρικά ενός τυχαίου εσωτερικού  σημείου $ P$ ενός τριγώνου ως προς τα μέσα των

πλευρών  του τριγώνου $ ABC$.Να αποδειχθεί ότι οι ευθείες

 $ AP_a, BP_b,CP_c$ συντρέχουν !

 Μπάμπης

[hide]Παλιό γνωστό θέμα, νομίζω καλύτερα για τις σχολικές ασκήσεις.

Από $ BP_{c}\parallel \equal{} AP\parallel \equal{} CP_{b}$ έχουμε ότι οι $ BP_{b},\ CP_{c}$ αλληλοδιχοτομούνται κ.ο.κ.[/hide]
Κώστας Βήττας.

Εάν θεωρήσουμε τα μέσα Κ,L,Μ των ΑΒ , ΒC , CΑ αντίστοιχα έχουμε ότι LM//ΑΒ, και LM//PaPb, άρα LM//AB διότι τα L και K είναι και μέσα των PPa και PPb αντίστοιχα. Ομοίως PbPc//BC και PcPa//CA. Οπότε το PaPbPc είναι ομοιόθετο του ABC ως προς κάποιο κέντρο συμμετρίας R στο οποίο οι PbB, PcC και PaA συντρέχουν.


Explain this apparent contradiction:

For some pair of matrices $ A$ and $ B$ we can have $ AB \neq BA$, yet in this calculation we have reversed the terms $ A_{ik} B_{kj}$:

$ (AB)_{ij} \equal{} \sum_{k\equal{}1}^{n} A_{ik}B_{kj} \equal{} \sum_{k\equal{}1}^{n} B_{kj} A_{ik}$

Where's the contradiction?  The last expression you obtained is different than the definition of $ (BA)_{ij}$.


search
inequalities unsolved
inequalities

There are already two posts on this problem. Just search the Olympiad section of Intermediat forum.

sorry, ok

is there any way i can delete the topic?


prove that (a+b)^n < or equal to 2^(n-1) *(a^n+b^n) for positive integer n.

The title should say something about the problem.
Or do you think that "prove" is such a rare word in mathematics that your thread will be the only one where something is proved¿
Or do you think that you re the only one that want's help¿ (refering to your other threads mit titles like "help me...").

Please change your titles (there's an Edit-Button) and use better ones in future (at the moment, all your threads here have no good title). For example, "(a+b)^n <= 2^(n-1) *(a^n+b^n)" could be the title of this one.

Oh my god. All these problems are the same as posted in [url=http://www.mathlinks.ro/Forum/viewtopic.php?t=111466]this topic[/url]. The original poster didn't post anything after that and now we get 9 topics of this completely new member itazzirun, spamming up the forum with stupid titles. I'm beginning to wonder why those problems are that urgent  :wink:


geometry
inradius
circumcircle
trigonometry
geometry unsolved

Let R and r denote the circumradius and inradius of a triangle ABC, respectively.  Let O denote the circumcenter of the triangle.  Let the feet of the perpendiculars to the sides AB, AC, BC, be N, M, and L.  Prove that OL+OM+ON=R+r.

Please help.

Thanks

Solution:
Let $ h_{1},h_{2},h_{3}$ be the length of perpindiculars from $ O$ to $ AB,AC,BC$ respectively.And $ \angle BAC\equal{}\angle BON\equal{}\angle AON\equal{}\alpha\Rightarrow$
$ \cos{\alpha}\equal{}\frac{h_{1}}{R}$$ \Rightarrow h_{1}\plus{}h_{2}\plus{}h_{3}\equal{}R(\cos{\alpha}\plus{}\cos{\beta}\plus{}\cos{\gamma})$,where $ \angle ABC\equal{}\beta,\angle ACB\equal{}\gamma$,then we must only prove that
$ R(\cos{\alpha}\plus{}\cos{\beta}\plus{}\cos{\gamma}\minus{}1)\equal{}r$,which is equivalent to $ (\cos{\alpha}\plus{}\cos{\beta}\plus{}\cos{\gamma})\equal{}\frac{R\plus{}r}{R}$,which is obviously true.If you want i could post the proof to this.


Consider the equation:
$x^{9}+3x^{8}+3x^{7}+2x^{6}+2x^{5}+x^{4}=x$

Let $a$ be the sum of the real solutions to the equation
Let $b$ be the product of the real, non-zero solutions to the equation
Let $c$ be the sum of the non-real solutions to the equation
Let $d$ be the product of the non-real solutions to the equation

Find the largest proper divisor of $|1000a+100b+10c+d|$

[hide="Big hint"]There's a reason the $x$ hasn't been factored out and is on the other side of the equation even though the $x=0$ solution doesn't affect the answer.
[hide="Massive hint"]$2x^{6}=x^{6}+x^{6}$[/hide][/hide]

Oh. OH. You are a terrible person  :yup:

[hide="Solution"]
$(x^{9}+3x^{8}+3x^{7}+x^{6})+(x^{6}+2x^{5}+x^{4})=x$
$(x^{3}+x^{2})^{3}+(x^{3}+x^{2})^{2}=x$

Thus, $x^{3}+x^{2}=x$ is a root equation.
We may divide out $x$ to achieve $x^{2}+x-1=0$, which has two real solutions.

Factoring out the $x$ and the root quadratic, the resulting sixth degree has all complex roots. This can be proven easily by re-writing it in the form $(x+a_{1})^{6}+c_{2}(x+a_{2})^{4}+c_{3}(x+a_{3})^{2}+c_{4}=0$

Thus, the only real solutions are those to the quadratic, and $a=-1$ and $b=-1$.

We get that $c=-3-a=-2$ and $d=-1/b=1$

$|-1000-100-20+1|=1119$, and the largest proper divisor is $\boxed{373}$
[/hide]

[quote="1=2"][hide="partial soluiton"]From Vieta's, a+c=-3. Since x=0 is a solution, we can divide both sides by x and get

$x^{8}+3x^{7}+3x^{6}+2x^{5}+2x^{4}+x^{3}-1=0$

The product of the roots of that is 1, so bd=1[/hide][/quote]

And how does this help you solve the problem? I'm pretty sure the whole point of the problem is that you don't use Vieta's (well, not right off the bat anyway).

Haha, yeah, I know, really.

Amy, I'm not that terrible  :( am I? :stink:

Btw, that was exactly my solution.


linear algebra
matrix
algebra
polynomial

Let's  say I'm given the matrix A and its determinant:

$ A\equal{}
\begin{bmatrix}
t\minus{}1&\minus{}1&0\\
0&t\minus{}1&\minus{}1\\
\minus{}1&3&t\minus{}4\\
\end{bmatrix}
, |A|\equal{}t^3\minus{}6t^2\plus{}12t\minus{}8$

and I'm asked to show that the two are equal. Where have I gone wrong in my workings?

$ \begin{bmatrix}
t\minus{}1&\minus{}1&0\\
0&t\minus{}1&\minus{}1\\
\minus{}1&3&t\minus{}4\\
\end{bmatrix}$

$ (t\minus{}1)R_3\plus{}R_1\rightarrow R_3$

$ \begin{bmatrix}
t\minus{}1&\minus{}1&0\\
0&t\minus{}1&\minus{}1\\
0&3t\minus{}4&(t\minus{}4)(t\minus{}1)\\
\end{bmatrix}$

Thus the determinant should equal $ (t\minus{}1)\left|\begin{array}{cc}t\minus{}1 & \minus{}1 \\3t\minus{}4 & (t\minus{}4)(t\minus{}1)\end{array}\right|$

Further evaluating it we have $ (t\minus{}1)((t\minus{}1)^2(t\minus{}4)\plus{}(3t\minus{}4))\equal{}(t\minus{}1)^3(t\minus{}4) \plus{} (t\minus{}1)(3t\minus{}4)$

Which is obviously wrong as it is a polynomial of degree 4, and the answer given is of degree 3, so where did I go wrong?

The action $ (t\minus{}1)R_3\plus{}R_1\rightarrow{R_3}$ multiplies your det by $ t\minus{}1$. Thus you must divide your final result by $ t\minus{}1$.

[quote="loup blanc"]The action $ (t - 1)R_3 + R_1\rightarrow{R_3}$ multiplies your det by $ t - 1$. Thus you must divide your final result by $ t - 1$.[/quote]

Okay, so speaking in general terms, if I have a matrix A

$ A = \begin{bmatrix}a_1 & a_2 & a_3 \\
b_1 & b_2 & b_3 \\
c_1 & c_2 &{ c_3} \\
\end{bmatrix}$

and I use the following row manipulations to form a matrix B

${ \frac {a_1}{b_1}}R_2 - R_1 \rightarrow R_2$

$ \frac {a_1}{c_1}R_3 - R_1 \rightarrow R_3$

would the determinant of A in terms of B be

$ |A| =\frac{|B|}{\frac{a_1}{b_1}+\frac{a_1}{c_1}}= \frac {b_1c_1|B|}{a_1(b_1 + c_1)}$

I would want to say
don't do never that follows : $ aR_1\plus{}bR_2\plus{}cR_3\rightarrow{R}_1$
but do the following:
$ R_1\plus{}bR_2\plus{}cR_3\rightarrow{R}_1$
Read your book !

[quote="loup blanc"]Read your book ![/quote]

My book only has  unnecessarily hard techniques to find determinants, whereas if my last post is correct (can somebody please confirm?), I will have the easy technique I have been looking for.

$ \det(A)\equal{} \det \begin{bmatrix}t\minus{}1&\minus{}1&0\\ 0&t\minus{}1&\minus{}1\\ \minus{}1&3&t\minus{}4\\ \end{bmatrix} \equal{} \det(tI \minus{} B) \equal{}$ characteristic polynomial of B, with

$ B \equal{} \begin{bmatrix} 1& 1&0\\ 0&1&1\\ 1&\minus{}3&4\\ \end{bmatrix}$.


The next decimal term in the sequence:

$ 0,\sim  0.5, \sim  0.\overline{6},\sim 0.75, \ldots$

Well you have to find the pattern first and the pattern i found was:
0/1=0, 1/2=0.5 2/3=0.666666etc 3/4=0.75 so the next one is 4/5=0.8
The answer is 0.8


let ABC be triangle .

draw aline wich intersects AB and AC in points P and Q,such that;  BP=PQ=QC

very easy.i wondered that nobody solved it. :o

what is the queston?

i mean. certainly if P,Q were mid points of an equilateral triangle it'd work.

ure right.but what u said is a very special form of it :|

So, I guess you want to see some ruler and compass construction?

yes Arne.i saw two solution for it ,that werent nice. :)

and im waiting for a good solution ;)


Eight lamps, each with an on-off switch, are arranged in a circle. A lamplighter can flip (in other words, change the state of ) the switches, but he is not allowed to flip just one at a time.
When he switches lamps on or off, he has to flip three in a row, for instance, he can switch 1, 7, and 8, but not 1, 7, and 6. Prove that no matter what combination of switches is on and off, he can always find some way to turn all of the lights on.

That's weird, I did finish, but my computer must have deleted some of it because it hates me. I finished it now.

[hide]WLOG, let light 8 be off

Flip lights:

1,2,3
2,3,4
4,5,6
5,6,7
7,8,1

Now light 8 will be turned on and every other light will stay the same. You can do this to all the off lights to turn them on. [/hide]


quadratics trigonometry calculus geometry algebra polynomial function

Just wondering, but how many of you took algebra 1 in middle school?  :D

Uhh I'm not from the states and don't really know what Algebra 1 encompasses but if my memory serves me correctly in grade 6 we could solve linear equations and in grade 7 we could solve quadratics (through factoring and formula) and we did some basic trig etc. etc. Then I thought it would be a good idea to jump into grade 9 math in grade 8... and everything went downhill from there  :P.

I am doing algebra 2 in middle school right now.(grade 10 in canada) I am getting a credit for the course to(I already did grade 9/algebra1)

i guess i should've specified. In the US, middle school is 6th-8th grade, although it depends in many schools..:)

Took algebra I first in 6th grade with a helpful elementary school math teacher (my middle school was 7th-8th grade), then the school forced me through it a second time in 7th grade.

Took Alg 1 in 7th grade...currently taking Alg 2 this year (as 8th) and plannign to take Geo next year (as 9th)....

for fun I am reviewing pre-calc and taking a beginning calculus course

I took it in 6th, which was middle school for me

Im not exactly in Middle School yet, but I am planning to take it.

[quote="tjhance"]I took Algebra I in 7th grade. Most advanced people were only taking Pre-Algebra. I don't even want to know what the regular kids were doing.[/quote]

haha yeah some people in 8th grade are taking "Algebraic Thinking" which is like what GT people learn in 5th grade

some people take algebraic expressions at my school after algebra 1

[quote="karthik_k"]er..... Could someone give me a rough idea of what Algebra 1 consists of ? Since I'm not from the US, I don't quite know what you mean by "Algebra 1".[/quote]

A lot of Algebra I is focused on solving system's of equations and matrices. You do little geometry and angle stuff but at the end of the year you will get into polynomials and factoring.

well........ in that case..... most of us do some bit of solving equations and things in middle school... Matrices though, somehow seems to remain as high school domain even though basic operations on matrices seem easy enough for middle school... polynomials and factoring are generally covered by 8th grade in India..

Lol im being taught pre algebra by my big bro

i took it in the eighth grade.  the weird thing is that in sixth grade, by about April, the teacher would say, "today's linear equations!"  and a large chunk of the class would go, "GASP! I can't do those!"

the next year, in seventh grade, In about February, the teacher would say, "today's linear equations!"  and a large chunk of the class would go, "GASP! I can't do those!"

in the eighth grade, in about November, the teacher would say, "today's linear equations!"  and a large chunk of the class would go, "GASP! I can't do those!"

and this is the advanced class.

this year, in ninth grade, i am in Algebra III.  This is a class mainly for seniors and advanced juniors. (i doubled up.)
in about september, the teacher would say, "today's linear equations!"  and a large chunk of the class would go, "GASP! I can't do those!"

THESE ARE SENIORS!

THEY HAVE THEIR OWN JOBS!

THEY DON'T KNOW HOW TO GRAPH A LINE!

AAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAHHHHHHHH!!!!

[quote="7h3.D3m0n.117"]haha youd be amazed at the number of people who forget basic things taught to them 2 years ago[/quote]

Lol my friend already forgot his basic multiplication facts  :P

Took Algebra I in 7th, Geometry over the Summer, and Advanced Algebra now in 8th...  It seems like i'm in the only school that teaches geometry in between Algebra and Advanced Algebra...

[quote="star99"]most schools teach alg1, geo, and then alg2[/quote]

yeah but then there are the idiots who modify the curriculm to alg1, alg2, geo

I took Algebra in 6th grade, switched schools, took Geometry in 7th, algebra II in 8th ...  now I'm a senior in Analysis.


inequalities
search
geometric inequality

[color=purple] Sorry for this post, but this is a my open problem, about a Chinese book or magazine ( I don't know :( ) "[b]Geometric Inequality in Chinese[/b]"in which there are a lot of nice problems about geometric inequalities.
 I have searched desperately for this book ( magazine ) in Viet Nam but it is not to be found anywhere ( I don't know why that happens  ).
  On Internet I can search it but I can't read Chinese so  I would like to contact anybody who can help me to find such very nice book ( magazine ) about geometric inequalities.
 Thank you very much ![/color]

[b][i]Geometric Inequality [/i][/b]written by Bottema, that may be helpful

[color=purple] Thank you tobeno_1  :)  but about the book of Bottema I have too.
   Could you please tell me how to get the book "[b]Geometric Inequality in Chinese[/b]" ? Thank you very much ![/color]


let a, b, c be a primitive pythagorean triple, with b even
find all solutions for equation a2 + b4 = c8
if you analize, it is not too hard :)  :)  :)

You probably mean $ a^2\plus{}b^4 \equal{} c^8$, I guess... :wink:

Pierre.

yes, you are right.
i don´t know how to post exponents. can you say me it?


I have been a new member. I need your help. I wish my English is better. Can you help me?

I'm not Vietnamese, but you may find this helpful:
[url]http://www.artofproblemsolving.com/Forum/index.php?f=256[/url]

Try watching You-Tube videos in English. You'll pick up grammar and usage that way, while being entertained.

Good luck!

[quote="haphuong345"]Thank you very much. I hope that I can speak and write English well soon and I can communicate with people around the world.[/quote]
I am a Vietnamese,just a sophomore at HUT,you know?I am glad to know a Vietnamese here,espically a girl If I don't misunderstand.
Can I make friend with you?You can exchange with me by privating message.

Well, since the subject of this thread is the English language:
[quote="nkht-tk14"]privating message.[/quote]
"Private message" is an adjective-noun phrase.  Often, nouns in English can be converted into verbs -- in this case, "to message" is a verb meaning "to send a message."  Then the "-ing" ending is how we form the gerund form of a verb ("messaging" is the act associated with the verb "to message," just as "swimming" is the act associated with the verb "to swim"), so the phrase I quoted should be "private messaging."  It happens in this particular case that the noun "private message" would be even better (the thing you may exchange is the private message; "private messaging" is the act of exchanging, and so is not itself exchanged).


function
linear algebra
matrix
graph theory
algebra
rational function
combinatorics proposed

Here goes the problem:
 We have to arrange three kinds of objects in a row, in such a way that no two objects of the same kind be together. If we have D_j objects of the kind j (j=A,B,C), how many distinct arrangements do we have?
 First of all note that the numbers D_j must satisfy certain kind of relationship to become possible making the arrangements with the specified constraint above remarked. If D_1<=D_2<=D3 is an ordering of such numbers, I think it suffices that D_1 + D_2 >= D_3 - 1 for the existence of such arrangements.
 I would thank you for any clue on the solving of such an enumeration problem.

I do not think there is a particularly "nice" answer.  Let $ f(i, j, k)$ denote the number of ways to do this if there are $ i$ objects of type $ A$ and so forth.  Then the generating function $ F(a, b, c) \equal{} \sum_{i, j, k \ge 0} f(i, j, k) a^i b^j c^k$ is something like half the sum of the coefficients of the matrix

$ \left( \mathbf{I} \minus{} \left[ \begin{array}{ccc} 0 & b & c \\ a & 0 & c \\ a & b & 0 \end{array} \right] \right)^{\minus{}1}.$

This is due to an interpretation of the question as counting the number of walks on the complete graph on $ 3$ vertices visiting each vertex a specified number of times.  In particular, it follows from some general results that $ F(a, b, c)$ is a rational function.  But although it is quite easy to write down the coefficients of a one-variable rational function, the situation in two or three variables is much more complicated and I don't think you'll get a significantly better answer than this.


Ex. 4-4 Given that x is directly proportional to y and to z and is inversely proportional to w, and that x=4 when
(w,y,z)=(6,8,5), what is x when (w,y,z)=(4,10,9)?

Solution: Because x is inversely proportional to w, when all other variables are constant, xw is constant. Similarly, when the other two variables are constant, each of x/y and x/z is constant. We can combine all these by saying xw/yz is constant. .......and blah blah blah

I don't get the solution. I understand how he got xw, x/y, and x/z, but I don't get the rest of the stuff like  other variables being constant and how he got  xw/yz from combining those. Could someone please explain this to me?

When a variable, say $ x$ is directly proportional to another variable, say $ y$, it means that for some constant $ k$, $ x \equal{} ky$. When a variable, $ x$, is inversely proportional to a variable, $ w$, it means that for some constant, let's say $ k$ again, that $ x \equal{} \dfrac{k}{w}$. Now given your problem, $ x \equal{} \dfrac{kyz}{w}$. Plugging in our variables for the first set we have $ 4 \equal{} \dfrac{k8*5}{6}$ so $ k \equal{} \dfrac{3}{5}$. Using this variable again with the second set, we have $ x \equal{} \dfrac{\dfrac{3}{5}*10*9}{4} \equal{} \dfrac{27}{2}$

this doesn't go in intermidiate, it goes in HSB. mods?


I just finished and turned it in (I don't want to get in the jam tomorrow...), when will we be allowed to discuss them(I'm not patient....)???

If I'm not mistaken, the round is still going on till tomorrow night (I sure hope so, as I haven't submitted yet...) and the jam isn't till the 30th.  I believe you can discuss them on Tuesday, but they'll make it known.

The math jam is on wednesday. You can discuss them after that.

Please do not discuss the problems until Wednesday, Nov 23.

Nov. 23???? yeah!!!!!I thought it won't be allowed until 30 (after the math jam), thats good :lol:  :lol:


geometry Euler Gauss algorithm quadratics calculus probability

Which best represents the way you like mathematics? (And don't say stuff like "All math eventually gets applied... for example number theory... crytography...", that's besides the point, how do you [i]view[/i] math? (You can tell I'm for pure...(YAY! nested parentheses ( :P ))))

PURE LIKE SO FRICKIN PURE IT'S CRAZY

I suspect you'd get more applied on AoPS though, overall.

[quote="archimedes1"]Pure I guess for me, but I really like combinatorics as well though...[/quote]

 :huh: Combinatorics isn't pure?

Applied.

I enjoy pure math but I cannot see it as a profession. I feel like if I were to be doing pure math all my life, I'd be selfishly exploring my own thoughts without doing anything for others. I find it more satisfying to use my mathematical abilities to help others rather than to explore my own thoughts. I suppose you can still call my desire selfish, but I feel that it is a better that the alternative.

I suppose that the areas that I want to learn more about are medical technologies such as diagnostic tools and the underlying physics, and such.

[quote="MysticTerminator"]that's not applied[/quote]

For example, how many ways can the UN select a 5 member group (assume out of 200 total countries) to lead an international health program?

[quote="Altheman"]Applied.

I enjoy pure math but I cannot see it as a profession. I feel like if I were to be doing pure math all my life, I'd be selfishly exploring my own thoughts without doing anything for others. I find it more satisfying to use my mathematical abilities to help others rather than to explore my own thoughts. I suppose you can still call my desire selfish, but I feel that it is a better that the alternative.[/quote]

Now you have me feeling like such a selfish person  :( ... come to think of it, none of my interests are of much help to the world at all... (math, music, linguistics,...)

[quote="Cronus"]I never thought that anyone in this site will ever porefer applied over pure!
I prefer pure.[/quote]

That's exactly what I thought, until a couple weeks ago.

[quote="MysticTerminator"][quote="7h3.D3m0n.117"]Combinatorics can be applied, since you can prove many identities using the algebraic method, or a counting method (hence, applied).

I generally prefer pure, since a lot of the applications my past teachers have given me was tedious to the point where it turned me off of doing anything applicable.[/quote]

that's not applied[/quote]
Oh.  Am I mixing up discrete mathematics and applied mathematics?

[quote="MysticTerminator"]yah. discrete math is still (:() a part of math[/quote]

You don't like discrete math?

HAH HELL NO

I think it is by far my least favorite area of maths

*sigh* No one seems to like discrete math very much... it's mostly pure, I think. And IMO it's like half of all mathematics.

Yes; that is what I was trying to say, about the "statistics and computing".

Statistics is not math??? I don't see how.

And computing has a lot of mathematical aspects to it.

Granted, I felt the same way as most of you about "applied math" until I read about some of the great mathematicians of the past and met some of the great "applied mathematicians" of today like Ingrid Daubechies, Andrew Odlyzko, or Donald Knuth. A lot of their work is decidedly applied, but their knowledge and use of what you would call "pure" mathematics is more "pure" than the work of some "pure" mathematicians. And the problems they work on are certainly what I would call beautiful.

Knowledge of statistics, computing, and even physics are also very important parts of modern attacks on the Riemann Hypothesis, as well as many other of the Millennium Prize Problems.

All I'm saying is, as a young mathematicians, don't close yourself off to areas of mathematics because you have a limited perception of what "applied" mathematics is and what "applied" mathematicians do. Math is just math. You should attack any question that you feel is interesting and/or useful regardless of what category it falls into.

[quote]Statistics is not math??? I don't see how.

And computing has a lot of mathematical aspects to it[/quote]
In pure math there's no real need for computations or statistics.Just the developing of theorems andstuffs...That's more fun than statistics and statistics is scary.So ...

[quote="gauss202"]There is no clear distinction between pure and applied mathematics. It's a very grey line. I guess I generally think of applied mathematics though as the development and study of mathematical algorithms - but even that isn't really a broad enough description. For example, is the theory of fourier transforms pure mathematics or applied mathematics? I can't see how it can be put exclusively into either category.[/quote]

I don't think this is a very good example.  The theory of fourier transforms is pure mathematics, regardless of whatever applications it may have.  This is analogous to asking whether the theory of differential calculus is pure or applied mathematics.

I tend to think of applied mathematics as mostly falling into one of two categories.  One of them is the algorithmic/computation side, which is very clearly applied.  The second is mathematics aimed at understanding some model rooted in physics or some non-mathematical discipline.  Of course, this manages to encompass just about all of PDE as well as a good portion of probability, so it's a bit of a lacking definition, but what aspects of the model you focus on determine whether it is pure and applied.  'quantitative' aspects like looking at what function spaces you can expect a solution to lie in given initial data and estimates in various norms are more 'pure', numerical solutions and monte carlo techniques are more 'applied', and other things lie in a sort of nebulous grey area that isn't really worth naming.

well, I think probability theory is certainly mathematics. I think statistics in and of itself is usually not considered to be a separate discipline, however. I agree with blah^3 regarding Fourier theory; in fact, I would even think of PDEs and most mathematical physics as pure math since yeah, it may happen to have some applications, but it stands on its own as a coherent mathematical theory (whereas I would say all these algorithmic stuff do not -- the only point of them really is so that they can be implemented by some computers).

[quote="Cronus"]In pure math there's no real need for computations or statistics[/quote]

I don't think that's true.

I agree that fourier theory is pure mathematics, but it is also just as significant an aspect of applied mathematics (fast fourier transform, spectroscopy, solutions to PDE's, etc.). My point being, that many areas of mathematics can't be put into this artificial distinction between pure and applied. To me, all math is JUST math - some of it (either pure and applied) is pretty, and some of it is not.

Note, btw, jmerry's advice in this thread: http://www.artofproblemsolving.com/Forum/viewtopic.php?p=1169866#1169866

could you give an example of needing computation or statistics in pure math, then?

and it makes sense to say fast fourier transform is applied math. obviously for algorithmical purposes. basic fourier theory is pure math. solutions to PDE, in my view as in my last post, are pure math. spectroscopy in terms of FTIR and such is just using math; I don't think there's any further theory that's being developed there.

One example is work on the Riemann-Hypothesis that conjectures that the zeros of the Zeta function share some statistical properties with the eigenvalues of random matrices. See the [url=http://en.wikipedia.org/wiki/Hilbert%E2%80%93P%C3%B3lya_conjecture#1970s_and_random_matrices]Hilbert-Polya conjecture[/url]. I know that there are several mathematicians at the Institute for Advanced Studies and at Princeton who are very interested in lines of attack line this using ideas from statistics and quantum mechanics.

You should also read about [url=http://en.wikipedia.org/wiki/Andrew_Odlyzko]Andrew Odlyzko[/url], one of our generations finest mathematicians, who helped develop the [url=http://en.wikipedia.org/wiki/Odlyzko-Sch%C3%B6nhage_algorithm]Odlyzko-Schönhage algorithm[/url] for computing the zeros of the Zeta function. He has also used computers and computations to establish (and refute) a lot of important conjectures in number theory.

the reason that the fourier transform is applicable to problems in applied math is precisely because of all of the properties one proves in a typical course in harmonic analysis.  for example, it's useful for PDE because it transforms differential operators to multiplication by polynomials.  as noted before, i think that finding representations for solutions to PDE (explicit or otherwise) is not really 'applied math' anyways.

of course there's not a lot of value in nitpicking excessively here but i think that one should be careful in labelling something as 'applied' just because it has applications.

A self-contradictory devil's response: Well if "applied" math isn't defined based on it being math that is "applicable", then why even use the term "applied"? You could argue that all "applied" math is based on some "pure" mathematical principles - so why should we even make any distinctions between the two?

well, all math is applicable, as in fact the OP noted, I think, so using the former definition clearly wouldn't lead us anywhere. and obviously, applied math is based on math. that's why it's got the "math" at the end there. I may be wrong, but I think we're making the distinction based on the [i]intended[/i] use of the mathematical theory being developed -- if it has further uses in math in and of itself, or if the only possible use is in things outside of math, such as computing.

I think the division of math into pure math and applied math is very unnatural. Most areas of math have some applications (there are very few exceptions like set theory). Moreover, the development of math throughout the history was very influenced by applications in physics, engineering and other disciplines.  On the other hand, I am not aware of any mainstream area of math whose only intended use is outside of math. (In particular, certainly the research in the theory of computation is only partially motivated by applications.)

I think it doesn't matter if the math is pure or applied, it matters if it is nice. I like the pure math and the beauty involved in it, but I also like the physics and the mathematics that appears in it, the amazing reasonings involved in the proof of a physical equation. Math is math, and this is what matters.

What is the difference? I mean, what are some examples of fields that are respectively pure or applied?


Me and my friends at IMO had a drinking competition and I am looking for winner. The question for you is: how many bottles (0,5 litre ones) did you drink in first and second day? (and just informative question: how many times you visited the toilet ;)).

One guy from Slovakia drunk 3 and 3 (1st and 2nd day).
One Czech guy (CZE1) drunk 5 and 6 or something like that...
And me... 3 and 8 ;) (the second day 5 times to the toilet ;))

Is there anyone better than us? ;)

When I was in IMO, in the rather strict UK, only one of us six was over eighteen, getting a coke could be hard at times.

I hope nobody 'beat you', the true winner is the one who walks away from such a game before the bottles have been opened.

I think that I don't understand you or you don't understand me :D
so I'll repeat myself: we were drinking clear water: the water we were given while solving the problems. And we count only those bottles you drunk in 4,5 hours of solving problems. Hope I am clearer :) or not?

:blush:  Yes, that was quite a misunderstanding.
I think we got two bottles of water to begin with, and I never asked for more, and usually I am thirsty man.  I did not need 'comfort stops' as some US'ers say.
Maybe I wasn't so thirsty because of the Glasgow climate : very very moist.  Which most people hate but very good for me  :mad:

Yes, I get quite upset when people glorify excessive alcochol consumption, sorry.

It's okey

btw: I was not thirsty, I just wanted to beat everyone ;) and I wrote problems 4 and 5 in first hour and couldn't solve 6, so I was trying to beat everyone in drinking ;) It was quite funny.

*could tell some story about NoName and drinking water but doesn't*

I think I beated all of you now 7 and 8 bottles for the first and second day.
But as far as I remember there was also CAN 1 that was drinking a lot.

Btw ( 4 times to the toilette each day :wink: )

[quote="son of Skanderbeg"]
But as far as I remember there was also CAN 1 that was drinking a lot.
[/quote]
Go Canada!!! :rotfl:

The real drinking competition was held every night after the second day of contest and until the end of IMO near the entrance to DIC =) Big hello goes to eveyrbody who was there! =)

I drank only 2 bottles each contest day. But a few times during the week, I tried to drink a 0,5 l bottle as fast as I can. The only time I had a timer was at the banquet after the closing ceremony - final result 17.34 seconds (of course, the time isn't really that accurate). This could be kind of  drinking contest too, though I don't know if anyone else pariticipated... oops:

Well our IRL 4 beat all records something like 5 bottles, not including the coke he drunk, so you can declare him the winner :lol:


AMC USA(J)MO USAMO trigonometry SFFT geometry trapezoid

I will post my solution.

Unfortunatley, I realized I did some rather unnescessary complex number stuff on #5 instead of just using sin x = sin (180-x)...

actually, #5 is very similar to an old USAMO #1. I think it was in 1996?

1,2,4,5 were fine. I almost have a complete solution for 3. So 24 probably.

13 pages for over-proving stuff + rushing on the night before...(all solutions are similar to the official ones, so I'm satisfied)

Yay, my solutions!

I'm pretty satisfied with everything except 5 :blush:

But I'm definitely going to get more than ... what I got last time :blush:

Is my solution to number 3 rigorous? (The degree-part.)

Haha, you think you wrote a lot... I had 24 pages!  I suffer from a severe case of rigor mortis, leaving absolutely nothing unproven (if I can help it).

For number 2, why the heck is everyone assuming that the diagram looks like the one in the official solution?  That's not rigorous!  I spent two pages solving number 2 while assuming that diagram, and like 7 pages proving that that diagram was the only one possible.  Anyone want to take a glance at my number 2 solution and tell me why all that work was unnecessary?

[quote="solafidefarms"]So ... post your solutions/ comment on others here? :)

[I messed up #4. But my totally made-up solution on #3 turned out to get the right answer??! :D][/quote]

I made the same mistake as yours on #4 before I checked my solutions on the deadline date. Then I started number-theory bruteforcing it...

PS: your #1 is hilarious :rotfl: ...

Dang, I should've bashed out the trig in #5... it actually wasn't too bad, reading the other solutions above. I think everything else is right, though.

ok, everyone has to read my #5. Its pathetic, but i got the right answer :P.

#3, i read wrong, as you can probably tell.

#4 was stupid mistakes.

My attachment is too big (7.8 MB), so I'll just explain what I did.
1. Same as everyone else.
2. Same as pretty much everyone else.
3. This was my weird solution (in particular, I found that $ p(x)\equal{}\sqrt{x\plus{}\frac{1}{4}}\minus{}\frac{1}{2}$ also works.)
[hide="solution"]
We have $ \frac{1}{a_0}\plus{}\frac{1}{a_1}\plus{}\cdots\plus{}\frac{1}{a_{n\minus{}1}}\plus{}\frac{1}{p(a_n)}\equal{}1$
and
$ \frac{1}{a_0}\plus{}\frac{1}{a_1}\plus{}\cdots\plus{}\frac{1}{a_{n\minus{}1}}\plus{}\frac{1}{a_n}\plus{}\frac{1}{p(a_{n\plus{}1})}\equal{}1$
Setting these equal, we have
$ \frac{1}{p(a_n)}\equal{}\frac{1}{a_n}\plus{}\frac{1}{p(a_{n\plus{}1})}\implies a_np(a_{n\plus{}1})\minus{}p(a_n)p(a_{n\plus{}1})\minus{}p(a_n)a_n\equal{}0$
Time for SFFT!
$ (a_n\minus{}p(a_n))(p(a_{n\plus{}1})\minus{}p(a_n))\equal{}p(a_n)^2$ (1)
Since after we get $ a_0\equal{}2$ and $ p(a_1)\equal{}2$, we see that $ p(a_1)$ is prime, so we only have to consider $ 1$, $ p(a_n)$, and $ p(a_n)^2$ for every subsequent n.
Therefore, from (1), we have
$ a_n\minus{}p(a_n)\equal{}1 \implies p(a_n)\equal{}a_n\minus{}1$
$ a_n\minus{}p(a_n)\equal{}p(a_n) \implies p(a_n)\equal{}\frac{a_n}{2}$
$ a_n\minus{}p(a_n)\equal{}p(a_n)^2 \implies p(a_n)^2\plus{}p(a_n)\equal{}a_n\implies (p(a_n)\plus{}\frac{1}{2})^2\equal{}a_n\plus{}\frac{1}{4} \implies p(a_n)\equal{}\sqrt{a_n\plus{}\frac{1}{4}}\minus{}\frac{1}{2}$
Then we throw out the last solution because it isn't a polynomial.
[/hide]
4. Pretty much same as official solution.
5. I got the answer by looking at small cases and finding a pattern, but I didn't submit it, because I thought it was kind of pointless wasting the grader's time by giving that reasoning.

Solutions, practically the same as official ones.

Hmm I'm a little bit surprised it wasn't necessary to show $ C_1C_2C_3C_4$ was an isosceles trapezoid, since the fifth circle was a constraining condition (then again, its fairly simple to do so).

[asy]pen f = fontsize(10);pair C1 = MP("1",(0,0),NE,f), C2 = (2,0), C4 = MP("3",4*expi(100*pi/180),SE,f), C3=IP(CR(C2,4),CR(C4,6));D(CR(C1,1));D(CR(C2,1));D(CR(C3,3));D(CR(C4,3));D(D(C1)--D(C2)--D(C3)--D(C4)--cycle,linewidth(0.7));[/asy]

yeah, you can show its an isoceles trapezoid pretty easily becuase OO1O4 is congruent to OO2O3 and we get a lot of angle equalities...

andersonw: I don't think that's a polynomial...

I can tell that you didn't open my spoiler tag, or completely read through the solution.

Edit: Although, I should have been more clear in my post.

question for Erin Schram, on #5, i wrote $ \frac{1004^2}{\binom{2008}{3}}$ as my answer which is equivalent to the correct answer, but will i get points taken off for not simplifying?

[quote="cognos599"]question for Erin Schram, on #5, i wrote $ \frac {1004^2}{\binom{2008}{3}}$ as my answer which is equivalent to the correct answer, but will i get points taken off for not simplifying?[/quote]
I don't know. Unlike MCTP's answer, your answer can obviously be simplified. But when we graders create the rubric for Problem 5, we might decide that simplifying isn't important enough to cost a point. We mostly care about the solution having a solid line of reasoning. In some problems, good procedure in calculation is the key to getting the correct answer, so we grade on good procedure.

Erin Schram
USAMTS grader

crap i didn't prove it was isoceles because I thought all those equal angles kind of made it obvious... :(

[quote="Yongyi781"][quote="tenniskidperson3"]
5. Used a TI-83 plus program to find a pattern, hopefully a 4.[/quote]

Probably a 3, since you had the wrong answer as well :([/quote]

wait, i had the right answer, but my logic was complete bs(i also found a pattern). How many points would i get?

i did the same thing. i think you will get either a 2 or 3. maybe a 1.

Hm, what if you did all this directed angle stuff and got a sum, and you said it evaluates to $ 1506\cot \frac {\pi}{2008}$, and dividing by $ \binom{2007}2$ because there are that many distinct triangles with $ X$ at $ (1, 0)$, and got the answer? (For more information, read my solution.)

am i the only one who used ptolemy's theorem for number 2? I hope I don't get points off for not proving it...

How did you use Ptolemy's Theorem on it? When I first saw the problem, I incorrectly used it because I assumed that the centers of $ C_1$, $ C$, $ C_3$ were collinear. Did you get the right answer? If so, how did you use Ptolemy's?

[quote="Erin J. Schram"][quote="azjps"]Hmm I'm a little bit surprised it wasn't necessary to show $ C_1C_2C_3C_4$ was an isosceles trapezoid, since the fifth circle was a constraining condition (then again, its fairly simple to do so).[/quote]
If I were grading the official solution to Problem 2/2/20, I would take a point off for not explaining why the centers of the four circles form an isosceles trapezoid (unless the other graders overrule me when we create the rubric).

As azjps said, the fifth circle is a constraining condition. The angles from the center of that circle to the centers of each set of adjacent circles of radius one and radius three must be congruent. Thus, the quadrilateral that surrounds that fifth circle breaks down into a triangle, an isosceles triangle, the mirror images of the first triangle, and another isosceles triangle, which makes an isosceles trapezoid.[/quote]I agree with this comment, and we have added this to the "official" solutions posted on http://www.usamts.org.

[quote="grn_trtle"][quote]I agree with this comment, and we have added this to the "official" solutions posted on http://www.usamts.org.[/quote]

Does this mean that if we mentioned that the two lines were parallel and that the other two sides were congruent, forming an isosceles trapezoid, but didn't actually prove that the two lines are parallel, we will lose points?[/quote]That's not enough to show that you have an isosceles trapezoid -- you might have a parallelogram, for example.

I attempted to use Ptolemy's at first as well; since the trapezoid is isosceles, it is cyclic as well. You can then use Ptolemy to find the diagonal, and then I suppose you could trig bash. However, I just used the same solution as the official solution.

So... would I lose a point for this?
I didn't assume it was an isosceles trapezoid, but I did assume that C_1C_2 and C_3C_4 were parallel, and the center of the small circle was collinear with the intersection point of circles 1 and 2, and the intersection point of circles 3 and 4.

Edit: Hmmm... that pretty much implies that it is a isosceles trapezoid, so I probably will lose a point. I still want to make sure, though.

Uh... for #2 would $ \frac{\minus{}6\plus{}3\sqrt{6}}{2}$ be accepted?

Umm... I see no reason for them not to...


function
integration
real analysis
real analysis unsolved

Are there any general method to differentiate this type of function?

$ g(x)\equal{}\int_a^x f(x,y)dy$

Thanks!

Thanks to the multivariable chain rule, the problem splits into two: $ g'(x_0)\equal{}\frac{\partial}{\partial x}\int_a^x f(x_0,y)dy\plus{}\frac{\partial}{\partial x}\int_a^{x_0} f(x,y)dy$.
If the function $ f$ is nice enough, this is equal to $ f(x_0,x_0)\plus{}\int_a^{x_0} f_x(x_0,y)dy$. The second term is the one that needs most work.

[quote="balto"]Are there any general method to differentiate this type of function?

$ g(x) \equal{} \int_a^x f(x,y)dy$

Thanks![/quote]

balto, how about this\[ g\left( x \right) \equal{} \int_a^{\alpha \left( x \right)} {f\left( {x,y} \right)dy}
\]


[hide]Use the property $\log_ab+\log_ac=\log_abc$ to get:
$\log(\log_2{3}\times \log_3{4}\times ...\times\log_{1023}{1024})$

Then use the property that $\log_ab=\frac{\log a}{\log b}$

The terms will cancel out and you get $\log(\log_21024)=\log10=\boxed{\boxed{1}}$[/hide]

Two people said that $\log_a b=\frac{\log a}{\log b}$...I thought it was reversed...

[quote="lkryptonicl"][quote="VenDiddy"]You use the fact that $\log{a}+\log{b}=\log{(ab)}$ and $\log_{b}{a}=\log{b}/\log{a}$.
\begin{eqnarray*} \log(\log_2{3})+\log(\log_3{4})+\log(\log_4{5})+...+\log(\log_{1023}{1024}) &=& \log{\left( \frac{\log{3}}{\log{2}} \times \frac{\log{4}}{\log{3}} \times \frac{\log{5}}{\log{4}} \times\cdots\times \frac{\log{1023}}{\log{1024}} \right)} \\ &=& \log{\left(\frac{\log{1023}}{\log{2}}\right)} \\ &=& \log{\left(\log_{2}{1023}\right)} \\ &=& \log{10} = \boxed{1} \end{eqnarray*}[/quote]

$log_{2}1023$ doesnt equal 10 so your answer isnt one..[/quote]

Obviously a typo.

[hide]
I think this is like a chain.

You can simplify it to
$\log_2{1024}=\boxed{10}$[/hide]

[quote="mathgeniuse^ln(x)"][hide]Using the chain rule for logarithms I get 10.
$10$[/hide][/quote]
Chain rule is for multiplication, not addition.

But logarthmic addition is multiplication as long as the logarthmic terms are in the same base.

\[ \mbox{Log Law: }log_{b}(a)+log_{b}(c) = log_{b}(ac) \]
\[ \mbox{Log Law: }log_{b}(a) = \frac{ln(a)}{ln(b)} \ NOT\ \frac{ln(b)}{ln(a)} \]

[quote="leonariso"][quote="VenDiddy"]You use the fact that $\log{a}+\log{b}=\log{(ab)}$ and $\log_{b}{a}=\log{b}/\log{a}$.
\begin{eqnarray*} \log(\log_2{3})+\log(\log_3{4})+\log(\log_4{5})+...+\log(\log_{1023}{1024}) &=& \log{\left( \frac{\log{3}}{\log{2}} \times \frac{\log{4}}{\log{3}} \times \frac{\log{5}}{\log{4}} \times\cdots\times \frac{\log{1023}}{\log{1024}} \right)} \\ &=& \log{\left(\frac{\log{1023}}{\log{2}}\right)} \\ &=& \log{\left(\log_{2}{1023}\right)} \\ &=& \log{10} = \boxed{1} \end{eqnarray*}[/quote]

it should be
\begin{eqnarray*} \log(\log_2{3})+\log(\log_3{4})+\log(\log_4{5})+...+\log(\log_{1023}{1024}) &=& \log{\left( \frac{\log{3}}{\log{2}} \times \frac{\log{4}}{\log{3}} \times \frac{\log{5}}{\log{4}} \times\cdots\times \frac{\log{1023}}{\log{1022}} \times \frac{\log{1024}}{\log{1023}} \right)} \\ &=& \log{\left(\frac{\log{1024}}{\log{2}}\right)} \\ &=& \log{\left(\log_{2}{1024}\right)} \\ &=& \log{10} = \boxed{1} \end{eqnarray*}[/quote]

Whoops, I fixed it now.  :ninja:


Prove the followings
\[ \sum \frac{a}{\sqrt{a\plus{}b}} \ge \frac{\sum \sqrt{a}}{\sqrt{2}}
\]
\[ \sum\sqrt{\frac{a^3}{a^2\plus{}ab\plus{}b^2}} \ge \frac{\sum \sqrt{a}}{\sqrt{3}}
\]
for any $ a,b,c \ge 0.$

We'd like to see a solution without using rearrangement inequality. :)

[quote="can_hang2007"]Prove the followings

\[ \sum \frac {a}{\sqrt {a \plus{} b}} \ge \frac {\sum \sqrt {a}}{\sqrt {2}}
\]

We'd like to see a solution without using rearrangement inequality. :)[/quote]

I think that one was old, we can square it.,then use s.o.s :wink:

\[ \sum\sqrt {\frac {a^3}{a^2 \plus{} ab \plus{} b^2}} \ge \frac {\sum \sqrt {a}}{\sqrt {3}}
\]
for any $ a,b,c \ge 0.$

is weaker than:
\[ \sum\frac {a^3}{2a^2 \minus{} ab \plus{} 2b^2} \ge \frac {\sum {a}}{3}
\]
for any $ a,b,c \ge 0.$
Which is famous.  :lol:

But my proof of above inequality is ugly. :(

[quote="can_hang2007"]

We'd like to see a solution without using rearrangement inequality. :)[/quote]

But how to use rearrangement inequality?

The first reduces to showing that
\[ 2(a^2\plus{}b^2\plus{}c^2)^3\geq (a\plus{}b\plus{}c)^2(a^4\plus{}b^4\plus{}c^4\plus{}a^2b^2\plus{}c^2a^2\plus{}b^2c^2).\]

[quote="hitek"]The first reduces to showing that
\[ 2(a^2 \plus{} b^2 \plus{} c^2)^3\geq (a \plus{} b \plus{} c)^2(a^4 \plus{} b^4 \plus{} c^4 \plus{} a^2b^2 \plus{} c^2a^2 \plus{} b^2c^2).
\]
[/quote]
I am sorry but this inequality is not true.

[quote="shfdfzhjj"][quote="can_hang2007"][quote="can_hang2007"]Prove the followings
\[ \sum \frac {a}{\sqrt {a \plus{} b}} \ge \frac {\sum \sqrt {a}}{\sqrt {2}}
\]

\[ \sum\sqrt {\frac {a^3}{a^2 \plus{} ab \plus{} b^2}} \ge \frac {\sum \sqrt {a}}{\sqrt {3}}
\]
for any $ a,b,c \ge 0.$

We'd like to see a solution without using rearrangement inequality. :)[/quote]
[hide="hint"]The first one follows from
\[ \sum \frac {a^2(a \plus{} b)}{3a^2 \plus{} 2ab \plus{} 3b^2} \ge \frac {1}{4}\sum a \quad \forall a,b,c \ge 0
\]
The second one follows from
\[ \sum \frac {a^3}{2a^2 \minus{} ab \plus{} 2b^2} \ge \frac {1}{3}\sum a \quad \forall a,b,c \ge 0
\]
Two of them are good exercises for SOS. :)
[/hide][/quote]

But how to use s.o.s?[/quote]
My solution for the first :)
For the second, we can prove it similar. :)


Twenty five boys and twenty five girls sut around a table. Prove that it is always possible to find a person both of whose neighbors are girls

The boys are determining $25$ intervals around the circle. The girls have to be disposed into these intervals (not necessary in each of them, and two or more girls can be in the same interval).

- If more than $12$ intervals are used by the girls, then it is easy to see that at least two of the intervals have a common 'end-boy', which is the desired person.

- at most $12$ intervals are used, then from the pigeon-hole principle, one of these intervals contains at least $3$ girls (without any boy between them). Then one of the girls in the middle is the desired person.

Pierre.

Consider $50$ seats around the circle. Divide them into $2$ groups where adjacent seats are in different groups. There is a group that contains at least $13$ girls. There must be $2$ girls in this group who sit next to each other. "Next to" means next to by considering only seats in their group.. In fact, they are seperate by a seat. That's seat is a required person.


geometry
analytic geometry
perpendicular bisector
geometry unsolved

Let $ ABC$ be an acute triangle and $ O$ be its circumcenter.The tangents at $ A$ and $ B$ to $ (O)$ intersect the tangent at $ C$ at $ D$ and $ E$ respectively.$ AE$ meets $ BC$ at $ P$ and $ BD$ meets $ AC$ at $ R$.Let $ S,Q$ be the midpoints of $ BR,AP$ respectively.Prove that $ \angle ABQ\equal{}\angle BAS$.

[b]Equivalent formulation:[/b] Let $ M,N$ be the feet of the B-symmedian and C-symmedian of $ \triangle ABC.$ $ D,E$ are the midpoints of $CN$ and $ BM.$ Then $ \angle DBC \equal{} \angle ECB.$

Using barycentric coordinates with respect to $\triangle ABC,$ we get

$ M(a^2: 0: c^2) \ , \ N(a^2: b^2: 0) \ , \ D(a^2: b^2: a^2 \plus{} b^2) \ , \ E(a^2: a^2 \plus{} c^2: c^2)$

$ BD \equiv za^2 \minus{} x(a^2 \plus{} b^2) \equal{} 0$ and $ CE \equiv a^2y \minus{} x(a^2 \plus{} c^2) \equal{} 0$ meet at the point

$ P \ ( a^2: a^2 \plus{} c^2: a^2 \plus{} b^2)$

Clearly, coordinates of $P$ satisfy the equation of the perpendicular bisector $\ell_a$ of $ BC,$ namely

$\ell_a \equiv (b^2 \minus{} c^2)x \plus{} a^2(y \minus{} z) \equal{} 0$

Thus, $ \triangle PBC$ is P-isosceles $\Longrightarrow$ $ \angle DBC \equal{} \angle ECB,$ as desired.


LaTeX Asymptote geometry parameterization function AMC AIME

I have the Chinese version of several of the AMC tests. Should I try to translate the problem, if the question isn't already in the wiki? If so, the solution as well?

It would be very much appreciated if you translated the problems. It's probably not a good idea to translate the solutions directly, though - try to paraphrase them or solve the problems yourself.

So far this is what I have done.

http://www.artofproblemsolving.com/Wiki/index.php/2002_AMC_10A_Problems

Any suggestions for improvements? I've never used LaTeX or made any heavy edits to the wiki before, so I'm not sure if I'm doing it correctly.

I have English copies (pdf) of the 2002 AMC 10A problems and solutions, but I don't have time to edit the wiki. Do you want a copy of them?

That looks great. Feel free to add asymptote diagrams for some of the geometry problems if you know how, like I did for problem 25.

Thanks everyone for your help.

If you have any AMC questions that aren't on the wiki, send them to me and I'll put them on the wiki. I have too much spare time, and I'm wasting too much of it.

...

I have more than enough problems to put on the wiki for now. Much more than enough...

http://www.artofproblemsolving.com/Wiki/index.php/2004_AMC_10B_Problems

is what I did. I haven't added some questions yet, because I don't know how to add pictures?

Edit:

Just added another test. Except for the questions that need pictures.

http://www.artofproblemsolving.com/Wiki/index.php/2005_AMC_10B_Problems

How do you do those lines on top of something? Like in a length, AB, I want to put the line on top of it. I need it for problem 23...

[quote="1=2"]you're gonna regret that offer, chenhsi.  :diablo:[/quote]

Dude, just to note that not only 1 does not equal 2 but also I also disrespect your view. People offer their efforts to make ML/AoPS a more useful community. And that is the result they get: Selfish people like you.

http://www.artofproblemsolving.com/Wiki/index.php/AMC_12_Problems_and_Solutions/Tasks

has a list of ideas on the bottom.

What is the view on those ideas so far?

Hopefully 1=2 is talking about his own experiences.  :wink:

Links to forum topics are available via resources, which are in turn [now] linked through the boxes at the bottom of each problem page. There's also some other parameter which was supposed to link it, probably
[code] {{ .. box | .. | t=####### }} [/code]
to link to thread ###### of the url, but I didn't implement it yet (parser function issues?).

PDFs are already available through resources, and are directly linked from [[Template:AIME Problems]], and answer keys are useful and, for example, all of the AIMEs have them (example format: [[1984 AIME Answer Key]]). Speaking of which, there are only maybe three AIMEs that are missing on the wiki now; see [[User:Azjps/sandbox]].

And $ \overline{AB}, \stackrel{\frown}{AB}$.
[code]$\overline{AB}, \stackrel{\frown}{AB}$.[/code]

There's also [[Template:ProbAMC]] for inputting AMC questions, but the syntax is a tad complex, and plus I'm not sure that the "answers" section works.

[quote="azjps"]Hopefully 1=2 is talking about his own experiences.  :wink:[/quote]

well yeah.

There are many many AMC problems to put onto the Wiki, I was just pointing out that he would get many many many many problems to upload.

Might I point you to the [url=http://www.artofproblemsolving.com/Forum/resources.php?c=182&cid=43]Contests[/url] section? Pretty much all the AMC 10 Problems have been posted there and the diagrams are already in asymptote so it's pretty much copying and pasting.

This is kind of reviving, but, here is my AMC inventory:

[hide="AMC"]
[hide="AJHSME/AMC8"]
1985 - Problems, Solutions
1986 - Problems, Solutions
1987 - Problems, Solutions
1988 - Problems, Solutions
1989 - Problems, Solutions
1990 - Problems, Solutions
1991 - Problems, Solutions
1992 - Problems, Solutions
1993 - Problems, Solutions
1994 - Problems, Solutions
1995 - Problems, Solutions
1996 - Problems, Solutions
1997 - Problems, Solutions
1998 - Problems, Solutions
1999 - Problems, Solutions
2000 - Problems, Solutions
2001 - Problems, Solutions
2002 - Problems, Solutions
2003 - Problems, Solutions
2004 - Problems, Solutions
2005 - Problems, Solutions
2006 - Problems, Solutions [hide="Note"][i]I have two versions for 2006, one with the official cover, and one with the solutions at the bottom of the page, with a hint[/i][/hide]
2007 - Problems, Solutions
2008 - Problems, Answers
(that is everything 1985-2008)
[/hide]
[hide="AMC10"]
2000 - Problems, Solutions
2001 - Problems, Solutions
[hide="A"]
2002 - Problems, Solutions
2003 - Problems, Solutions
2004 - Problems, Solutions
2005 - Problems, Solutions
2006 - Problems, Solutions
2007 - Problems, Solutions
2008 - Problems, Solutions
2009 - Problems, Solutions
[/hide]
[hide="B"]
2002 - Problems, Solutions
2003 - Problems, Solutions
2004 - Problems, Solutions
2006 - Problems, Solutions
2005 - Problems, Solutions
2007 - Problems, Solutions
2008 - Problems, Solutions
[/hide]
[/hide]
[hide="AHSME/AMC12"]
1980 - Problems, Solutions
1981 - Problems, Solutions
1982 - Problems, Solutions
1983 - Problems, Solutions
1984 - Problems, Solutions
1985 - Problems, Solutions
1986 - Problems, Solutions
1987 - Problems, Solutions
1988 - Problems, Solutions
1989 - Problems, Solutions
2000 - Problems, Solutions
2001 - Problems, Solutions
[hide="A"]
2002 - Problems, Solutions
2003 - Problems, Solutions
2004 - Problems, Solutions
2005 - Problems, Solutions
2006 - Problems, Solutions
2007 - Problems, Solutions
2008 - Problems, Solutions
[/hide]
[hide="B"]
2002 - Problems, Solutions
2003 - Problems, Solutions
2004 - Problems, Solutions
2005 - Problems, Solutions
2007 - Problems, Solutions
2008 - Problems, Solutions
2009 - Problems, Answers
[/hide]
[/hide]
[hide="AIME"]
1993 - Problems, Solutions
2001 - Problems, Solutions
2002 - Problems, Solutions
2003 - Problems, Solutions
2004 - Problems, Solutions
2005 - Problems, Solutions
2006 - Problems, Solutions
2007 - Problems, Solutions
2008 - Problems, Solutions
2009 - Problems, Solutions
[/hide]
[hide="USAMO"]
1989 - Problems, Solutions
1990 - Problems, Solutions
1991 - Problems, Solutions
1992 - Problems, Solutions
1993 - Problems, Solutions
1994 - Problems, Solutions
1995 - Problems, Solutions
1996 - Problems, Solutions
1997 - Problems, Solutions
1998 - Problems, Solutions
1999 - Problems, Solutions
2000 - Problems, Solutions
2001 - Problems, Solutions
2002 - Problems, Solutions
2003 - Problems, Solutions
[/hide]
[/hide]

And I wanted to see if anyone is willing to type it in the Wiki. I don't have time myself.


Let $A$ and $B$ be arbitrary finite sets and let $f: A\longrightarrow B$ and $g: B\longrightarrow A$
be functions such that $g$ is not onto. Prove that there is a subset $S$ of $A$ such that
$\frac{A}{S}=g(\frac{B}{f(S)})$.

What do you mean by $\frac A{S}$ and $\frac B{f(S)}$?

An "onto" function is a surjective function, I suppose (I'm not sure we use the same terminology).

Edit. Well, I think $\frac A{S}$ is $A\backslash S$, i.e. set difference.


Hello Everyone,

I am posting a MATHCOUNTS Problem of the Week in this forum. This will hopefully be a challenge to you all. What better way to get smarter than rackin' your brains?! Well, here it is:

What is the median of the first 2006 positive integers? Express your answer as a decimal to the nearest tenth.

The two middle numbers are 1003 and 1004, so the median of the first 2006 positive numbers is 1003.5


$ f(x) \equal{} \begin{cases} \frac {q}{q \plus{} 1}\mbox{ if }x \equal{} \frac {p}{q}\mbox{ is a rational number} \\
0\mbox{ if }x\mbox{ is irrational or }0 \end{cases}$
It is assumed here that $ \gcd(p,q) \equal{} 1$ and $ q > 0$ so that the function is well-defined and nonnegative.

It has to be shown that there is no such closed interval on which $ f(x)$ has maximum although it is bounded.

[hide="Solution"]
WLOG, consider the sequence $ S = \{ a_i | a_i = \frac{1}{i} \}$. We consider this sequence in particular because multiplying any of the numbers by some constant integer, we arrive at a different number with the same output. That is, $ f \left(\frac{1}{i} \right) = f \left(\frac{k}{i} \right), \ k \in \mathbb{N}$ and thus all numbers in the domain that output a nonzero number may be reached.

This sequence is monotonically decreasing, whereas the sequence given by $ R = \{ f(a_i) | a_i \in S \}$ is monotonically increasing because the forward difference $ f(a_{i+1}) - f(a_i) = \frac{i+1}{i+2} - \frac{i}{i+1} \geq 0$.

Now, we take the limit as $ i$ approaches infinity. That is, $ \lim_{i \rightarrow \infty}{f(a_i)} = \lim_{i \rightarrow \infty}{\frac{i}{i+1} = 1}$.
However, we notice that $ f(a_i)$ can never reach $ 1$. Thus, $ R$ is open and bounded and so no maximal element occurs in it.

Therefore, for any closed interval $ I$ of $ f(x)$, there occurs no maximal element, though the range $ R$ is bounded.
[/hide]