factorial
logarithms
vector
group theory
superior algebra
superior algebra unsolved

I'm looking for a bijection:

$\varphi: \mathbb{Z}_{n!}\rightarrow S_{n}$

But it has to be something programmable, i.e. based on a formula of some sorts.  Any ideas?

[quote="mlok"]If $S_{n}$ means the symmetric group on $n$ elements, then $S_{n}$ has $n!$ elements, not $2^{n}$. Are you actually looking for an [url=http://people.howstuffworks.com/lethal-injection.htm]injection[/url]?[/quote]

I do not understand $\mathbb{Z}_{n!}$ is the set $\{ 0 , 1 ,2 ,... ,n!-1\}$ which surly has $n!$ elements.  :blush:

My OP originally had 2^n instead of n! because I wasn't thinking.

Step 1: represent each $k<n!$ as a sum $0+a_{1}\cdot 1!+a_{2}\cdot 2!+\cdots+a_{n-1}\cdot (n-1)!$, where $0\le a_{j}\le j$. This can be done uniquely by the greedy algorithm.
Step 2: Convert this sequence into a permutation by working top down (starting at $a_{j-1}$); at each stage, append the $(a_{j}+1)$st element of $\{1,2,\dots,n\}$ not already included.

This isn't fast for large $n$; it's hard to avoid slow recursion when doing anything with factorials. There's no reason to expect a nice formula.

Of course, why didn't I think of that!  Kinda like base-n!... cool.

It may not be fast, but hopefully I'm working with small enough numbers for it to work in reasonable time.  I don't even think I'll be going above 6! (but I am doing it for 4 independent permutations, making it (6!)^4 overall).
Thanks!

The obvious implementation is $O(n^{2})$ to calculate the permutation that goes with a particular number. The first step is $O(n)$ as long as you store a list of factorials (or other tricks), and the second is $O(n^{2})$, possibly $O(n\log n)$ with the right methods.

You don't hit factorial time unless you try to list everything.

[quote]Step 2: Convert this sequence into a permutation by working top down (starting at $a_{j-1}$); at each stage, append the ($a_{j}+1$)st element of $\{1,2,\dots,n\}$ not already included. [/quote]

I don't believe that step is correct.  I end up with the cycle:

(abcd...n)

Where each letter represents a unique number (since you are removing from the set {1, 2, ..., n} each time).  This of course can never represent all permutations, even the identity is not in it.

Edit:

Oh, you weren't giving me a cycle.  1 gets sent to the element in the 1st position, 2 gets sent to the element in the 2nd, and so on.  Ok, now it makes sense.

Ok, got it working.  I was able to combine step one and two into the same step.  This probably doesn't make much of a difference in the end, as it's just doing twice the work in half the steps.

[code]
#include <iostream>
#include <vector>
#include <math.h>

using namespace std;

int factorial(int n) {
  int result = 1;
  while (n > 1) result *= n--;
  return result;
}

void disp(vector<int> v) {
  if (v.size() == 0) {
    cout << "{}" << endl;
    return;
  }
  cout << "{";
  for (int i = 0; i < (signed)v.size()-1; i++) {
    cout << v[i] << ", ";
  }
  cout << v[v.size()-1] << "}";
}

vector<int> permute(int k, vector<int> v) {
  vector<int> result(v.size()-1, 0);
  int i;
  for (i = v.size()-1; i > 0; i--) {
    int m = factorial(i);
    result[result.size() - i] = v[k / m];
    v.erase(v.begin() + k / m);
    k -= k / m * m; //note the flooring making this not just equal to k
  }
  result.push_back(v[0]); //append the very last element (as there is now only one in v) to the end

  return result;
}

int main() {
  vector<int> v;
  v.push_back(1);
  v.push_back(2);
  v.push_back(3);
  v.push_back(4);

  int i;
  for (i = 0; i < 24; i++) {
    disp(permute(i, v));
    cout << endl;
  }
  return 0;
}
[/code]


Hi.
I don't know if some one can help me, with a web-site where I can find problems for the trig-star contest ( trigonometry ). I already solve all the examples that I found it on line. I will participate on the national trig-star competition.
Send me the info to leonus3@msn.com
Tank you

AIME sounds abuot right:
http://www.artofproblemsolving.com/Forum/resources.php?c=182&cid=45

Here are some problems:

$ \blacksquare$ 1. Prove that $ \cot^2 \frac {\pi}{7} + \cot^2 \frac {2\pi}{7} + \cot^2\frac {3\pi}{7} = 5$

$ \blacksquare$ 2. Prove that if $ \arccos a + \arccos b + \arccos c = \pi$ then $ ab\sqrt {1 - c^2} + bc\sqrt {1 - a^2} + ca\sqrt {1 - b^2} = \sqrt {(1 - a^2)(1 - b^2)(1 - c^2)}$

$ \blacksquare$ 3.Prove the inequality $ \sec^{2n} x + \csc^{2n} x\geq 2^{n + 1}$

$ \blacksquare$ 4. Prove that if $ \beta \geq 1$ then ${ (1 + 2\sin^2\alpha)^{\beta} + (1 + 2\cos^2 \alpha)^\{beta}\geq 2^{\beta + 1}$

Hope this would help...


inequalities
trigonometry
inequalities proposed

[quote="Tiger100"]Prove that in any acute triangle we have the inequality

$ sinA.sinB \plus{} sinA.sinC \plus{} sinC.sinB \le (cosA \plus{} cosB \plus{} cosC)^2$[/quote]

It is true for all $ \triangle$ .

$ \sin A \sin B \plus{} \sin A \sin C \plus{} \sin C \sin B \equal{} \frac{r^2\plus{}p^2\plus{}4rR}{4R^2}$

$ \cos A \plus{} \cos B \plus{} \cos C\equal{}\frac{R\plus{}r}{R}$

So : $ \Leftrightarrow p^2 \le 4R^2\plus{}4Rr\plus{}3r^2$ - which is well known ( $ IH^2 \equal{} 4R^2\plus{}4Rr\plus{}3r^2\minus{}p^2$)


Let $n$ be a positive integer. If the equation $2x+2y+z=n$ has $28$ solutions in positive integers $x$, $y$, and $z$, then $n$ must be either

(A)$14$ or $15$
(B)$15$ or $16$
(C)$16$ or $17$
(D)$17$ or $18$
(E)$18$ or $19$

[hide]
rewrite as $2(x+y)+z = n$.  Once $x+y$ is determined, $z$ will be unique.  Since all variables are positive, $x+y \ge 2$ and can go up to $\left\lfloor \frac{n}{2}\right\rfloor$.  If $x+y = 2$, there's 1 solution.  if $x+y = 3$, there's 2.  So on and so forth, we see that the total solutions are $1+2+3+4+\dots$, a.k.a. triangular numbers.

Since 28 is the 7th triangular number, $x+y$ can at most be $8$, it follows that $n$ is either $16$ or $17$. The answer is $C$.
[/hide]


Let $ a,b,c\geq 0$, no two of which are zero.Prove that:
$ \frac{a(b\plus{}c)}{a^2\plus{}bc}\plus{}\frac{b(c\plus{}a)}{b^2\plus{}ca}\plus{}\frac{c(a\plus{}b)}{c^2\plus{}ab}\geq 2$.

No one can solve?   :(  :?:

[hide="to arqady"]arqady, please post your's or Vasc's solution.  [/hide]

[quote="Inequalities Master"]Let $ a,b,c\geq 0$, no two of which are zero.Prove that:
$ \frac {a(b \plus{} c)}{a^2 \plus{} bc} \plus{} \frac {b(c \plus{} a)}{b^2 \plus{} ca} \plus{} \frac {c(a \plus{} b)}{c^2 \plus{} ab}\geq 2$.[/quote]
I only repeat

$ Ineq \leftrightarrow  2abc(\sum_{sym} a^2(b\plus{}c)) \plus{} \sum_{sym} a^4(b^2\plus{}c^2) \plus{} abc\sum_{sym} b^2(a\plus{}c) \ge\ 4a^2b^2c^2 \plus{} 2\sum_{sym} a^3b^3 \plus{} 2abc \sum_{sym}a^3$

$ \rightarrow  LHS \minus{} RHS \equal{} (a\minus{}b)^2(b\minus{}c)^2(c\minus{}a)^2 \plus{} 2a^2b^2c^2 \plus{} abc\sum_{sym} a^2(b\plus{}c)$

We have done!

[quote="onlylove_math"][quote="Inequalities Master"]Let $ a,b,c\geq 0$, no two of which are zero.Prove that:
$ \frac {a(b \plus{} c)}{a^2 \plus{} bc} \plus{} \frac {b(c \plus{} a)}{b^2 \plus{} ca} \plus{} \frac {c(a \plus{} b)}{c^2 \plus{} ab}\geq 2$.[/quote]
I only repeat

$ Ineq \leftrightarrow 2abc(\sum_{sym} a^2(b \plus{} c)) \plus{} \sum_{sym} a^4(b^2 \plus{} c^2) \plus{} abc\sum_{sym} b^2(a \plus{} c) \ge\ 4a^2b^2c^2 \plus{} 2\sum_{sym} a^3b^3 \plus{} 2abc \sum_{sym}a^3$

$ \rightarrow LHS \minus{} RHS \equal{} (a \minus{} b)^2(b \minus{} c)^2(c \minus{} a)^2 \plus{} 2a^2b^2c^2 \plus{} abc\sum_{sym} a^2(b \plus{} c)$

We have done![/quote]
Nice my friend.Here is my proof
Let$ a\geq b\geq c$
We only prove$ \frac {a(b \plus{} c)}{a^2 \plus{} bc} \plus{} \frac {b(c \plus{} a)}{b^2 \plus{} ca} \geq 2$
It is true because $ \frac {a(b \plus{} c)}{a^2 \plus{} bc} \plus{} \frac {b(c \plus{} a)}{b^2 \plus{} ca} \minus{}2\geq \frac{(a\minus{}b)^2}{(a^2\plus{}c^2)(b^2\plus{}c^2)}$

[quote="manlio"][quote="Allnames"]Nice my friend.Here is my proof
Let$ a\geq b\geq c$
We only prove$ \frac {a(b \plus{} c)}{a^2 \plus{} bc} \plus{} \frac {b(c \plus{} a)}{b^2 \plus{} ca} \geq 2$
It is true because $ \frac {a(b \plus{} c)}{a^2 \plus{} bc} \plus{} \frac {b(c \plus{} a)}{b^2 \plus{} ca} \minus{} 2\geq \frac {(a \minus{} b)^2}{(a^2 \plus{} c^2)(b^2 \plus{} c^2)}$[/quote]

I think it is wrong  :wink:[/quote]
No manlio,I am quite sure my proof is not wrong,you can check it again,
With my proof,i think this problem is not strong (mean there is a stronger problem).Perhasp,i have seen it ,but i can remember sorry :lol:

I made all computations and I obtained

 $ \frac {a(b \plus{} c)}{a^2 \plus{} bc} \plus{} \frac {b(c \plus{} a)}{b^2 \plus{} ca} \minus{} 2 \equal{} \frac {(a\minus{}b)^2(ab\plus{}c^2\minus{}2ca\minus{}2cb)}{(a^2 \plus{} bc)(b^2\plus{}ca)}$ and this expression can be less than 0  :wink:

Am I wrong?

Sorry manilo,i haven't tried that way(and now i have some troubles),and i cant check yours care fully
Here is my full proof

$ \frac {a(b \plus{} c)}{a^2 \plus{} bc} \plus{} \frac {b(c \plus{} a)}{b^2 \plus{} ca} \minus{} 2 \equal{} \frac {b(a \minus{} b) \plus{} c(a \minus{} c)}{b^2 \plus{} c^2} \minus{} \frac {a(b \minus{} a) \plus{} c(b \minus{} c)}{a^2 \plus{} c^2}$
With$ a\geq b\geq c$ $ \frac {b(a \minus{} b) \plus{} c(a \minus{} c)}{b^2 \plus{} ca} \minus{} \frac {a(b \minus{} a) \plus{} c(b \minus{} c)}{a^2 \plus{} bc}\geq \frac {(a \minus{} b)^2(ab \minus{} c^2)}{(a^2 \plus{} bc)(b^2 \plus{} ab)}\geq 0$
Is it true?

[quote="Allnames"]Sorry manilo,i haven't tried that way(and now i have some troubles),and i cant check yours care fully
Here is my full proof

$ \frac {a(b \plus{} c)}{a^2 \plus{} bc} \plus{} \frac {b(c \plus{} a)}{b^2 \plus{} ca} \minus{} 2 \equal{} \frac {b(a \minus{} b) \plus{} c(a \minus{} c)}{b^2 \plus{} c^2} \minus{} \frac {a(b \minus{} a) \plus{} c(b \minus{} c)}{a^2 \plus{} c^2}$
With$ a\geq b\geq c$ $ \frac {b(a \minus{} b) \plus{} c(a \minus{} c)}{b^2 \plus{} c^2} \minus{} \frac {a(b \minus{} a) \plus{} c(b \minus{} c)}{a^2 \plus{} c^2}\geq \frac {(a \minus{} b)^2(ab \minus{} c^2)}{(a^2 \plus{} c^2)(b^2 \plus{} c^2)}\geq 0$
Is it true?[/quote]

Why do you write at denominator $ (a^2 \plus{} c^2)$ and $ (b^2\plus{}c^2)$ ?

I think it is wrong.  :wink:

[quote="Inequalities Master"]Let $ a,b,c\geq 0$, no two of which are zero.Prove that:
$ \frac {a(b \plus{} c)}{a^2 \plus{} bc} \plus{} \frac {b(c \plus{} a)}{b^2 \plus{} ca} \plus{} \frac {c(a \plus{} b)}{c^2 \plus{} ab}\geq 2$.[/quote]

This inequality is true but weak. I use the identity
$ \frac {a(b \plus{} c)}{a^2 \plus{} bc} \plus{} \frac {b(c \plus{} a)}{b^2 \plus{} ca} \plus{} \frac {c(a \plus{} b)}{c^2 \plus{} ab}\minus{} 2
\equal{} \frac{(a\minus{}b)^2(b\minus{}c)^2(c\minus{}a)^2\plus{}abc(a\plus{}b)(b\plus{}c)(c\plus{}a)}{(a^2\plus{}bc)(b^2\plus{}ca)(c^2\plus{}ab)}$

Allnames solution is not correct. There are the case that the sign is reverse
If $ a \geq b \geq c$ Then
$ \frac{a(b\plus{}c)}{a^2\plus{}bc}\plus{}\frac{b(c\plus{}a)}{b^2\plus{}ca}\minus{}2 \equal{} \frac{(a\minus{}b)^2[(a\minus{}c)(b\minus{}2c)\plus{}(\minus{}bc\minus{}c^2)]}{(a^2\plus{}bc)(b^2\plus{}ca)}$

[quote="Inequalities Master"]Let $ a,b,c\geq 0$, no two of which are zero.Prove that:
$ \frac {a(b \plus{} c)}{a^2 \plus{} bc} \plus{} \frac {b(c \plus{} a)}{b^2 \plus{} ca} \plus{} \frac {c(a \plus{} b)}{c^2 \plus{} ab}\geq 2$.[/quote]

As I did this problem, I found another interesting problem

$ \frac{a^7(b\plus{}c)}{a^2\plus{}bc}\plus{}\frac{b^7(a\plus{}c)}{b^2\plus{}ac}\plus{}\frac{c^7(a\plus{}b)}{c^2\plus{}ab} \leq (a^6\plus{}b^6\plus{}c^6)$

Sorry every body,in all time ,i have seen an other problem,(it is not  Inequalities Master'prblem)and i have posted the proof for it,.very sorry
Here is the problem that I have prove   :blush:   :blush:  :blush:
$ \frac {a(b \plus{} c)}{c^2 \plus{} b^2} \plus{} \frac {b(c \plus{} a)}{a^2 \plus{} c^2} \plus{} \frac {c(a \plus{} b)}{a^2 \plus{} b^2}\geq 2$
And it is very easy
Sorry

[quote="Allnames"]

$ \frac {a(b \plus{} c)}{c^2 \plus{} b^2} \plus{} \frac {b(c \plus{} a)}{a^2 \plus{} c^2} \plus{} \frac {c(a \plus{} b)}{a^2 \plus{} b^2}\geq 2$

[/quote]

This inequality is absolutely corrected.  The following variance of it is also interesting:
$ \sum_{cyc} \frac{a^3(b\plus{}c)}{b^2\plus{}c^2} \geq \sum_{cyc}(a^2)$

[quote="Allnames"]
$ \frac {a(b \plus{} c)}{c^2 \plus{} b^2} \plus{} \frac {b(c \plus{} a)}{a^2 \plus{} c^2} \plus{} \frac {c(a \plus{} b)}{a^2 \plus{} b^2}\geq 2$
[/quote]
For positive $ a,$ $ b$ and $ c$ we obtain:
$ \frac {a(b \plus{} c)}{c^2 \plus{} b^2}\geq\frac {2a^2}{a^2 \plus{} b^2 \plus{} c^2}.$ :wink:
I didn't find the link.
[quote="Gibbenergy"] The following variance of it is also interesting:
$ \sum_{cyc} \frac {a^3(b \plus{} c)}{b^2 \plus{} c^2} \geq \sum_{cyc}(a^2)$[/quote]
It's equivalent to
$ \sum_{cyc}\frac {(a \minus{} b)^2(a^2 \plus{} b^2 \plus{} c^2 \plus{} ab)ab}{(a^2 \plus{} c^2)(b^2 \plus{} c^2)}\geq0.$

Sorry every body,i have just fonund a strobger ineq than wsp57.Here is it
$ \frac {a(b \plus{} c)}{a^2 \plus{} bc} \plus{} \frac {b(c \plus{} a)}{b^2 \plus{} ca} \plus{} \frac {c(a \plus{} b)}{c^2 \plus{} ab}\geq 2\plus{}\frac{8a^2b^2c^2}{\sum {a^2\plus{}bc}}$
It is true and quite easy, :lol:

[quote="Allnames"]Sorry every body,i have just fonund a strobger ineq than wsp57.Here is it
$ \frac {a(b \plus{} c)}{a^2 \plus{} bc} \plus{} \frac {b(c \plus{} a)}{b^2 \plus{} ca} \plus{} \frac {c(a \plus{} b)}{c^2 \plus{} ab}\geq 2 \plus{} \frac {8a^2b^2c^2}{\sum {a^2 \plus{} bc}}$
It is true and quite easy, :lol:[/quote]
It is quite easy . Here 's my proof
$ Ineq \leftrightarrow \sum a^2b^2(a^2 \plus{} b^2) \plus{} 3abc[ \sum ab(a \plus{} b) ] \ge\ 12a^2b^2c^2 \plus{} 2\sum a^3b^3 \plus{} 2abc(\sum a^3)$
But we have $ abc[ \sum ab(a \plus{} b) ] \ge\ 6a^2b^2c^2$
Need prove
$ \sum a^2b^2(a^2 \plus{} b^2) \plus{} 2abc[ \sum ab(a \plus{} b) ] \ge\ 6a^2b^2c^2 \plus{} 2\sum a^3b^3 \plus{} 2abc(\sum a^3)$
$ \leftrightarrow (a^2b \minus{} ab^2 \plus{} b^2c \minus{} c^2b \plus{} c^2a \minus{} a^2c)^2 \ge\ 0$
We have done! :wink:

Let $a+b+c=3u$, $ab+ac+bc=3v^2$ and $abc=w^3$.
Hence, it's obvious that our inequality is a linear inequality of $w^3$.
Hence, $uvw$ kills it.

[quote=luofangxiang]$ \frac {a(b \plus{} c)}{c^2 \plus{} b^2} \plus{} \frac {b(c \plus{} a)}{a^2 \plus{} c^2} \plus{} \frac {c(a \plus{} b)}{a^2 \plus{} b^2}\geq 2$

The following inequality is also true
[img]http://s8.sinaimg.cn/middle/0018zOAxgy70BDP9Vj1c7&690[/img][/quote]

you see this:

\[\frac{x(y+z)}{y^2+z^2+yz}\geq\frac{x(3xy+3xz-y^2+2yz-z^2)}{(x+y+z)\sum{yz}}\]

<=>
\[(xy+xz-y^2-z^2)\geq 0\]
give a sum,done.
BQ


Does anybody know any general formula for the sum:
$ 1^k \plus{} 2^k \plus{} ... \plus{} n^k$ where $ n$ and $ k$ are positive integers.
(The formula, if exist,  would be a function of n and k)

http://mathworld.wolfram.com/FaulhabersFormula.html

See also my latest blog entry.


inequalities
combinatorial geometry
absolute value
real analysis
real analysis unsolved

Suppose $\sum_{n\geq 0}{a_n}$ is a  convergent series of complex numbers. Prove that for any $0<r<1$ the series $\sum_{n\geq 0}{a_n z^n}$ converges uniformly on the convex hull of the closed disc of radius $r$ and center $O(0)$ to which we add the point $1$.

I don't understand the statement of the problem clearly.

Convergence on the disk $|z|< 1$ is easy to prove:
Since $\sum a_n$ converges, then $a_n\to 0$ and hence $|a_n|$ is bounded, suppose $|a_n|\leq L$.
Then $\sum |a_n z^n| = \sum |a_n| |z|^n\leq L \sum |z|^n$.

[quote="liyi"]I don't understand the statement of the problem clearly.
[/quote]
You have to show [b]uniform[/b] convergence in a certain region, not just pointwise convergence in the open disk :).

We have proved that $\sum a_n z^n$ absolutely converges on the circle $|z|=r(r<1)$ then we can easily show that it converges uniformly on the disk $|z|\leq r$ by Weierstrass Theorem:
$\sum |a_n z^n| =\sum |a_n| |r^n| \left|\frac{z}{r}\right|^n \leq \sum |a_n r^n|$

What does '...to which we add point $1$' mean?

[quote="liyi"]
What does '...to which we add point $1$' mean?[/quote]
It means that the region is the convex hull of a circle of radius $r<1$ and the point $1$ on the complex plane (so it looks like a circle with a small "hat" on its side). Another way to describe such a region is to define it by the inequality $1-|z|\geqslant c|1-z|$ with some small $c>0$ (it is not exactly the same region but it gives an equivalent version of the problem)

I'll have to look again to be sure, but a picture of this region may be the only figure in all of Antoni Zygmund's classic [i]Trigonometric Series.[/i]

We're talking about the concept of "nontangential convergence."

It's easy to prove that for any $m$ and any $z$ with $|z|<1$ the two series $\tilde f_m(z)=a_mz^m+a_{m+1}z^{m+1}+\ldots$ and $f_m(z)=a_m(z^m-z^{m+1})+(a_m+a_{m+1})(z^{m+1}-z^{m+2})+\ldots$ have the same value (just compare the partial sums). We must thus show that for any $\varepsilon>0$, there is an $m_0$ s.t. $|f_m(z)|<\varepsilon,\ \forall m\ge m_0$ and for all $z$ in our region. From now on $\forall z$ means all $z$ in our region with absolute value less than $1$.

There is an $M>0$ s.t. $|1-z|\le M(1-|z|),\ \forall z$, and there's also an $m_0$ s.t. $|a_m+\ldots+a_{m+p}|\le \frac\varepsilon M,\ \forall m\ge m_0,\ \forall p\ge 0$.

For $m\ge m_0$ we will have $|f_m(z)|=|(1-z)[a_mz^m+(a_m+a_{m+1})z^{m+1}+\ldots]|$, which is $\le |1-z|\cdot\sum_{n\ge m}\frac\varepsilon M|z|^n\le M(1-|z|)\cdot\sum_{n\ge m}\frac\varepsilon M|z|^n\le\varepsilon$.

I hope it's Ok.


ratio
geometry
3D geometry
tetrahedron
octahedron

[quote="M4RI0"]A regular tetrahedron and a regular octahedron have edges of the same length. What is the ratio between their volumes?[/quote]
Just apply the formulas for the volumes and get the ratio of volumes equal to $1: 4\sqrt{3}$. I might have messed up, but I think this problem should be moved to a lower level because their is nothing hard about this problem. :!:

[quote="M4RI0"]It is from the chapter 1: Tricks of the trade 1.1 slicing and dicing; Notes on Euclidean Geometry, Kiran Kedlaya. I think it is supposed to be done without the formulas. So my question is: How :?: Also the ratio is 1:4.[/quote]
Maybe because you could arrange 4 tetrahedrons within an octahedron?

[quote="M4RI0"]It is from the chapter 1: Tricks of the trade 1.1 slicing and dicing; Notes on Euclidean Geometry, Kiran Kedlaya. I think it is supposed to be done without the formulas. So my question is: How :?: Also the ratio is 1:4.[/quote]
I am prbably doing something wrong but I do not get the ratio of 1:4. :(

The tetrahedron has a volume of $\frac{s^{3}\sqrt2}{12}$ and the octahedron has a volume of $\frac{s^{3}\sqrt2}{3}$.


linear algebra
matrix
vector
search
algebra
polynomial

\[ A \equal{} \left( \begin{array}{ccccc} n & n & 0 & ... & 0 \\
0 & n \minus{} 1 & n & ... & 0 \\
... & ... & ... & ... & ... \\
0 & 0 & 0 & 2 & n \\
0 & 0 & 0 & ... & 1 \\
\end{array} \right)
\]
Find eigen values and eigen vectors for $ A$ ( or give me a link to this problem which has certainly been solved already, but I did not know how to search for it  :) ) . Thanks !

It appears to me that this matrix is upper triangular (correct me if it isn't).  Therefore, taking the determinant and hence finding the characteristic polynomial is easy as well.  The characteristic polynomial will be
$ f(t)\equal{}(t\minus{}n)(t\minus{}(n\minus{}1))\cdots (t\minus{}2)(t\minus{}1)$
from which it is easy to see what the eigenvalues are.

[quote="JRav"]It appears to me that this matrix is upper triangular (correct me if it isn't).  Therefore, taking the determinant and hence finding the characteristic polynomial is easy as well.  The characteristic polynomial will be
$ f(t) \equal{} (t \minus{} n)(t \minus{} (n \minus{} 1))\cdots (t \minus{} 2)(t \minus{} 1)$
from which it is easy to see what the eigenvalues are.[/quote]

but what about the eigenvectors? ( foreach eigen value which is $ k$, $ k \equal{} 1,2, .., n$ )

I'll take the $ t\equal{}n$ case:  Just like any other case, subtract $ n$ from each entry on the diagonal and set to 0.  You end up with
$ \minus{}1x_{1}\equal{}0$
$ \minus{}2x_{2}\equal{}0$
:
:
$ (2\minus{}n)x_{n\minus{}1}\equal{}0$
$ (1\minus{}n)x_{n}\equal{}0$

from which it is clear that the zero-vector is an eigenvector.

No, it isn't. Eigenvectors are nonzero by definition.

Those are also the wrong equations. You actually get $ 0x_1\equal{}nx_2$, $ 1x_2\equal{}nx_3$, $ 2x_3\equal{}nx_4$, ..., $ (n\minus{}1)x_n\equal{}0$. The eigenvectors for $ n$ are multiples of $ (1,0,\dots,0)^T$.


Find all integers $\frac{1}{r}$ such that:

$\sum_{n=0}^{\infty}\frac{F_{n}}{r^{n}}$ is an integer, given $F_{0}= 0$ and $F_{1}= 1$.

[hide="i must be being stupid"]
$\sum_{n=0}^{\infty}\frac{F_{n}}{r^{n}}=\sum_{n=0}^{\infty}F_{n}\frac{1}{r^{n}}=\sum_{n=0}^{\infty}integer\cdot integer$

did you mean that $r$ is an integer, not $\frac{1}{r}$?
[/hide]

I'd assume he meant $r$ is an integer, not $r^{-1}$ because that makes the proof trivial.

[hide="So then"]stick it into the closed form for $F_{n}$ and carry it out as the sum of two infinite geometric series. You get that
$\sum = \frac{r}{r^{2}-r-1}\implies \frac{r}{r^{2}-r-1}= k\; : \; k \in \mathbb{Z}$

You also have $|r|>phi$

This gives the solution $r=2 \implies \sum = 2$

[/hide]

Yeah, that's what I had meant.  I actually sort of posted this without thinking.
When solving it, one ends up with the equation $r = \frac{1+s+\sqrt{5s^{2}+2s+1}}{2s}$, where r is the ratio and s is the sum.

Of course, I thought the equation would have more solutions than 2 and 2, but it didn't.  :oops:


A helicopter carrying Dr. Evil takes off with a constant upward acceleration of 5.0m/s^2 . Secret agent Austin Powers jumps on just as the helicopter lifts off the ground. After the two men struggle for 10.0 s, Powers shuts off the engine and steps out of the helicopter. Assume that the helicopter is in free fall after its engine is shut off and ignore effects of air resistance.

What's the maximum height of the helicopter?

Here's what I thought.
a = -5.0 m/s^2
t = 10 sec
s = ?
Velocity (Final) = 0.

s = V(f)t - (at^2)/2 = - (at^2)/2 = -(-5.0 * 100) / 2 = 250 m.

Unfortunately, that's not right. Someone help me please?

[color=blue]It's a simple problem. The equation of the helicopter's velocity is
$\large v=v_0+at=at$
At the moment power shuts off the engine, $\large v=50\frac{m}{s}$
The distance between the helicopter and the ground is $s_1=\large\frac{at^2}{2}=250m$
Then, the equation of the helicopter's velocity is $v'=v-gt$
The helicopter reach the maximum height when its velocity is zero $\large v'=0\Longrightarrow s_2=\frac{v^2}{2g}=125m$
And we have the maximum height of the helicopter is $\large s=s_1+s_2=375m$
Is that the answer?[/color]

Yes! That's the right answer. Much thanks man!


vector
calculus
combinatorics proposed
combinatorics

dogs stand at the vertices 1, 2 and 3 of an equilateral triangle of side 1. Dog on
vertex 1 runs towards dog 2. Simultaneously dog 2 runs towards dog 3 and dog 3 towards
dog 1. They all run with the same speed. Find the distance traveled before they meet.

First observe that anytime those 3 dogs form an equilateral triangle with the same center as in the first one. Transform tha vector of speed into 2 vectors, one to the center, and one from the center to another dog. Let us have that their velocity is $ a$. Than the vector component should be $ \frac {a\sqrt {3}}{3}$ Thus they will meet in $ t \equal{} \frac {s}{V} \equal{} \frac {1}{\frac {a\sqrt {3}}{2}} \equal{} \frac {2}{a\sqrt {3}}$ The distance travelled should be $ a*t \equal{} a\frac {2}{a\sqrt {3}} \equal{} \frac {2}{\sqrt {3}}$
:wink: It looks more like physics to me  :P
(Note: we can proceed like this because we know that when they dogs will meet the triangle becomes a point :wink;)

hey thanks for super fast reply ;)....but i guess that the answer should be greater than
sqrt(3) .
If they all meet at the center of the equilateral triangle in straight line (suppose ;)).. then total distance traveled is 3*(sqrt(3) / 3 ) = sqrt(3). But they travel in a curve, so it should be more than sqrt(3).

polskimisiek:  what you've computed is their [i]displacement[/i], not the distance they've traveled.  The dogs should actually be traveling on curves somewhat like tractrixes.

By the way, why is this in Combinatorics?  This is almost certainly a calculus question.

This was posted in the physics forum, although the question asked for the time, not for the distance traveled: http://www.mathlinks.ro/viewtopic.php?t=247023

However, knowing the time ($ \frac {2}{3v}$) multiply it by the velocity v to get the distance traveled, or $ \frac {2}{3}$.

EDIT: This is the distance for each dog, counting all three we get $ 2$.

Hey, as soon as I read the problem, I thought.. " Is the answer (i.e., total distance traveled) dependent on the time at which each dog starts ?". If not, then let 2 & 3 stay a their respective vertices. Now, 1 travels to 2 ( distance = 1 ), and 3 travels to 1 (which is at vertex 2 now ;) ) (distance = 1 ), so total distance = 2   :roll:
.
I'm now surprised(?) to see the answer is 2 indeed ;) . . . can some one explain the reason .


Newton sums are a method to calculate the sum of the nth powers of the roots of a polynomial. (For example, we might ask what the sum of the squares of the roots of x^2-1 is. Either we can compute it directly by finding the roots, which are 1 and -1, and squaring each, or we can use an algorithm. Suppose instead we ask what the sum of the fourth powers of the roots of x^5-6x^3+5x^2-2 is. Unless this is a very special case and the roots can be found by ingenuity, you will have great difficulty doing this problem unless you use Newton sums.) A decent explanation can be found at [url]http://www.mathpages.com/home/kmath034.htm[/url].

by roots, do u mean real roots or all roots? and do newton sums calculate nth powers of all of the roots, or can we limit the # of roots?

That's all the roots, including the complex roots. Also, for multiple roots, it includes them multiple times. For example, if we want to find the sum of the squares of the roots of x^2+2x+1, we will get 2 and not 1.


So, today I was informed that my school wouldn't be taking the AMC this year because it is "too expensive".

Anyone know anyplace where I can take the test? Preferably northwest suburbs. If not, I may just pay for it myself.

You can take it a university!  There is a university location in naperville..  I actually contacted a university to register because I didn't think my school was giving it out.

Click [url=http://www.unl.edu/amc/b-registration/b1-archive/2008-2009/CU2009/2009-CU-list.shtml]here[/url] and scroll down to IL.

You should probably contact the coordinator in case you have to register (instead of just stopping by that day).

I'm taking the 12 B at a university in Naperville.  Probably the one suma's talking about...

Hey, if you guys go to North Central to take the AMC, you should come visit me at my school. It's only 5 minutes away, walking...

http://maps.google.com/maps?daddr=301-445+Aurora+Ave,+Naperville,+IL+60540+(Naperville+Central+High+School)&geocode=CbFk4J8Mc3ADFdtVfQIdhdq--iHtFlfwWXmKxg&dirflg=&saddr=30+N+Brainard+St,+Naperville,+IL+60540&f=d&dq=Naperville+Central+High+School&sll=41.768411,-88.155515&sspn=0.006295,0.006295&cid=41768411,-88155515,14306380593487091437&ie=UTF8&t=h&z=16

I'm taking the AMC 10A (probably) on Feb. 10, at my own school.

Thanks everyone for the help!

Yea, I've emailed the university site director and it says everything is on site. It's gonna be about an hour drive, but at least I'll get out of school and stuff.
Hopefully it will be worth it. But I'm going to meet some AoPSers, so I'll try to pass down some sagely wisdom or whatever is appropriate.

So...

There were a lot of people at North Central College. I was the one in the blue hoodie who didn't know anyone.


Needing the help of the fellow mathematicians...

Sorry for not posting the problem in the post coz i cant seem to type the problem properly. So i attach the problem....

Let $ n\geq2$ , $ 0 < x_{1} < x_{2} <$...$ < x_{n \minus{} 1} < x_{n}$ , for $ x_{k}, k \equal{} 1,2,...,n$ to be integers.

Define $ S_{n} \equal{} \left(1 \plus{} \frac {1}{x_{1}}\right)\left(1 \plus{} \frac {1}{x_{2}}\right)\ ... \left(1 \plus{} \frac {1}{x_{n}}\right)\ \minus{} 1$
then
(i) Show that the recurrence $ S_{n \minus{} 1} \equal{} \frac {x_{n}}{x_{n \minus{} 1}}S_{n} \minus{} 1$

(ii)If $ S_{n}$ and $ S_{n \minus{} 1}$ are positive integers then $ S_{k}$ is also a positive integer for every $ k \equal{} 1,2,3,...,n$

[quote="r0ger"]Let $ n\geq2$ , $ 0 < x_{1} < x_{2} <$...$ < x_{n \minus{} 1} < x_{n}$ , for $ x_{k}, k \equal{} 1,2,...,n$ to be integers.

Define $ S_{n} \equal{} \left(1 \plus{} \frac {1}{x_{1}}\right)\left(1 \plus{} \frac {1}{x_{2}}\right)\ ... \left(1 \plus{} \frac {1}{x_{n}}\right)\ \minus{} 1$
then
(i) Show that the recurrence $ S_{n \minus{} 1} \equal{} \frac {x_{n}}{x_{n \minus{} 1}}S_{n} \minus{} 1$
[/quote]

Maybe I misunderstand your problem (sorry if it occurs) but this seems to me obviously wrong. According to me, the good recurrence relation would be $ S_{n\minus{}1}\equal{}\frac{x_n}{1\plus{}x_n}S_n\minus{}\frac 1{1\plus{}x_n}$

And, in order to show that your recurrence formula is wrong, just choose $ x_1\equal{}1$ and $ x_2\equal{}2$. Then :
$ S_1\equal{}1$
$ S_2\equal{}2$
And obviously $ \frac {x_{2}}{x_{1}}S_{2} \minus{} 1\equal{}3\neq S_{1}$

You should check the exact problem statement or better explain this current statement.
Thanks.

[quote="pco"]
Maybe I misunderstand your problem (sorry if it occurs) but this seems to me obviously wrong. According to me, the good recurrence relation would be $ S_{n \minus{} 1} \equal{} \frac {x_n}{1 \plus{} x_n}S_n \minus{} \frac 1{1 \plus{} x_n}$

And, in order to show that your recurrence formula is wrong, just choose $ x_1 \equal{} 1$ and $ x_2 \equal{} 2$. Then :
$ S_1 \equal{} 1$
$ S_2 \equal{} 2$
And obviously $ \frac {x_{2}}{x_{1}}S_{2} \minus{} 1 \equal{} 3\neq S_{1}$

You should check the exact problem statement or better explain this current statement.
Thanks.[/quote]

Well, thanks for the information and what you said really make sense...but i check it again thoroughly and that's what my teacher gave me. Any idea? Should i go on with the question or not?

[quote="pco"]

Maybe I misunderstand your problem (sorry if it occurs) but this seems to me obviously wrong. According to me, the good recurrence relation would be $ S_{n - 1} = \frac {x_n}{1 + x_n}S_n - \frac 1{1 + x_n}$

And, in order to show that your recurrence formula is wrong, just choose $ x_1 = 1$ and $ x_2 = 2$. Then :
$ S_1 = 1$
$ S_2 = 2$
And obviously $ \frac {x_{2}}{x_{1}}S_{2} - 1 = 3\neq S_{1}$

You should check the exact problem statement or better explain this current statement.
Thanks.[/quote]

Hey,guess what pco  :lol: ! I discusses the question and it is proved that the question is wrongly written. When you know about the correct recurrence, the method becomes clear.  :idea: So, the correct recurrence is $ S_{n - 1} = \frac {x_n}{1 + x_n}(S_n+1) - 1$ or $ S_{n - 1} = \frac {x_n}{1 + x_n}S_n - \frac 1{1 + x_n}$


abstract algebra
group theory
number theory
relatively prime
superior algebra
superior algebra unsolved

Let G be a group of order 225 which  has a cyclic 5-sylow subgroup
prove that G is abelian.

I don't know if this is the most efficient solution, but it follows a pretty standard path in analyzing groups of small order. Let me know if any of the statements I'm making is not clear to you; I assume you're familiar with Sylow's theorems, and with the fact that a group of order $ p^2$ ($ p$ prime) is abelian.

1. How many $ 5$-Sylow groups are there? Their number is congruent to $ 1$ modulo $ 5$, and divides $ 225$. The only such number is $ 1$, so there's a single $ 5$-Sylow subgroup, and it's normal. Call it $ N$.

2. The group $ G$ acts by conjugation on $ N$. In other words, there's a homomorphism $ \theta : G \to \mbox{Aut}(N)$, defined by $ \theta(g)(n) \equal{} g^{\minus{}1}ng$ for $ g \in G$ and $ n \in N$. What is $ \mbox{Aut}(N)$? This is where we use the fact that $ N$ is cyclic: the automorphisms of a cyclic group of order $ m$ are simply "raise to a power relatively prime to $ m$". In our case $ N$ is cyclic of order $ 25$, So $ \mbox{Aut}(N)$ is a group of order $ \varphi(25) \equal{} 20$.

Now, in the action $ \theta$, the group $ N$ itself acts trivially since it's abelian, namely $ N \leq \mbox{ker}(\theta)$. Therefore $ G / \mbox{ker}(\theta)$ is a group of order dividing $ 9$, and it's isomorphic to a subgroup of $ \mbox{Aut}(N)$ which has order $ 20$. This can only happen if this subgroup is trivial, namely, $ \mbox{ker}(\theta) \equal{} G$. What this means is that the action of $ G$ by conjugation on $ N$ is trivial, or in other words, $ N \leq Z(G)$ where $ Z(G)$ is the center of $ G$.

3. Now let's move on to the $ 3$-Sylow subgroups. How many are there? Their number is congruent to $ 1$ modulo $ 3$, and divides $ 225$. The only possibilities are $ 1$ and $ 25$.

4. First, let's suppose there are $ 25$ $ 3$-Sylow subgroups, and let $ X$ be the set of those subgroups. $ G$ acts on this set by conjugation, and we know that this action is transitive. However, $ N$ acts trivially since it's in the center, so this leaves only $ G/N$ to really permute the elements of $ X$ around. But $ G/N$ has order $ 9$, which is too small to act transitively on a set of order $ 25$.

5. Therefore, there's a single $ 3$-Sylow subgroup $ K$, and it is normal in $ G$. Now the subgroups $ N$, $ K$ are both normal, they have trivial intersection, and they generate $ G$, so $ G \cong N \times K$, a direct product. but $ N$ and $ K$ are both abelian (having orders $ 25$ and $ 9$; in fact, we know that $ N$ is cyclic), so $ G$ is abelian, too.

There are, indeed, non-abelian groups of order $ 225$ (two of them, actually). The $ 5$-Sylow subgroup is still normal in both of them, of course, but it is necessarily not cyclic, so it's the product of two cyclic groups of order $ 5$. The automorphism group of $ C_5 \times C_5$ is $ \mbox{GL}_2(\mathbb{F}_5)$, a group of order $ 480$, which easily accommodates a non-trivial action by $ G$.

Can we omit 4)?  It seems to me that if $ N\leq Z(G)$, it is already clear that $ G$ is the direct product of $ N$ and a Sylow 3-subgroup.

Heh. Of course.  :) I was trying to reason directly to show that a $ 3$-Sylow subgroup is normal, but that's unnecessary since $ N$ and $ K$ commute ($ K$ being any $ 3$-Sylow subgroup).


If $ x$,$ y$, and $ z$ are real numbers such that \[(x \minus{} 3)^2 \plus{} (y \minus{} 4)^2 \plus{} (z \minus{} 5)^2 \equal{} 0,\] then $ x \plus{} y \plus{} z \equal{}$

$ \textbf{(A)}\ \minus{}12\qquad
\textbf{(B)}\ 0\qquad
\textbf{(C)}\ 8\qquad
\textbf{(D)}\ 12\qquad
\textbf{(E)}\ 50$

[hide]

Since $ a^2\ge 0$ for all real numbers $ a$, $ (x\minus{}3)^2$, $ (y\minus{}4)^2$, $ (z\minus{}5)^2\ge 0$. For them to sum up to $ 0$, they must all be equal to $ 0$, so $ x\equal{}3$, $ y\equal{}4$, and $ z\equal{}5$. Hence, $ x\plus{}y\plus{}z\equal{}12\Longrightarrow\mathrm{ (D) \ }$

[/hide]


The function $ f(n)$ is defined on the positive integers and takes non-negative integer values. It satisfies
(1) $ f(mn) \equal{} f(m) \plus{} f(n)$, (2) $ f(n) \equal{} 0$ if the last digit of $ n$ is $ 3$, (3) $ f(10) \equal{} 0$. Find $ f(1985)$.

Since $ f$ nonnegative and $ 0\equal{}f(10)\equal{}f(2)\plus{}f(5),f(2)\equal{}f(5)\equal{}0$. Then $ f(1985)\equal{}f(397)\equal{}\frac{f(397^3)}{3}\equal{}\boxed{0}$.


Here's a proof I found for the AM-GM inequality in the case of three numbers:

Suppose that $x,y,z$ are three real positive numbers and we want to prove that $(x+y+z)^{3}\geq 27xyz$. Set $a=\sqrt{\frac{xz}{y}},b=\sqrt{\frac{xy}{z}},c=\sqrt{\frac{yz}{x}}$. Then we have:

$(ab+bc+ca)^{4}\geq (3abc(a+b+c))^{2}=9a^{2}b^{2}c^{2}(a+b+c)^{2}\geq 27a^{2}b^{2}c^{2}(ab+bc+ca)$

$\Leftrightarrow (ab+bc+ca)^{3}\geq 27(ab)(bc)(ca)$

which is equivalent to what we wanted to prove.

(The inequality used in the above proof is $(u+v+w)^{2}\geq 3(uv+vw+wu)$ which is true because it's equivalent to $(u-v)^{2}+(v-w)^{2}+(w-u)^{2}\geq 0$.)


Princeton college Harvard Yale Stanford Duke

Hi every one,

It's the first post of mine in this forum, so if it's silly I'm sorry. I want to apply some universities in US and Australia. Sat and Toefl are required, but what's the deadline for me to send them to the universities?  I hope your help,

Thank a lot.

It is different for each university. Usually, they will have an application deadline, and all of your various application materials need to be sent by then. You can look on the websites of the universities to which you'd like to apply. If you have questions about specific universities, we will be more likely to have specific answers for you. ;)

I see the rank from usnews

http://www.usnews.com/usnews/edu/college/rankings/brief/t1natudoc_brief.php

1. Princeton University(NJ)

2. Harvard University(MA)

3. Yale University(CT)

4. California Institute of Technology

4. Stanford University(CA)

4. Massachusetts Inst. of Technology

7. University of Pennsylvania

8. Duke University(NC)

9. Dartmouth College(NH)
....

but could anyone tell me that how is the ranking of universities in Maths and Computer?

Thank a lot,

The top schools are so close, I am always doubtful of some kind of exact ranking. The same schools at the top of that list (1, 2, 3, all the 4s) are very good for maths and computer science I think. Well maybe not so much Yale, but the others surely are among the best.

Who can help me about the concrete rank about computer science and maths? I need some detailed information, thank u.

I searched online but I can't find any ranking of mathematics departments for undergraduates. Here are some sources for graduate schools, which isn't that relevant, but it's something for you to look at anyway:

Math:
http://www.usnews.com/usnews/edu/grad/rankings/phdsci/brief/mat_brief.php
http://math.wlu.edu/GradRank95.pdf
http://www.usnews.com/usnews/edu/grad/rankings/phdsci/brief/matsp3_brief.php
http://www.phds.org/rankings/math/ (this one lets you rank according to certain attributes)

Compsci:
http://www.usnews.com/usnews/edu/grad/rankings/phdsci/brief/com_brief.php
http://www.greguide.com/comps.html
http://www.phds.org/rankings/computer-science/ (also lets you rank according to certain attributes)

I recommend Princeton. :D :D ;)


If a circle has equation $ (x-3)^{2}+(y-4)^{2}=16$, then what is the maximum value of $ \frac{(y+2)}{x}$, where $ (x,y)$ is a point on the circle.

Rewrite $ \frac{y+2}{x}$ as $ \frac{y-(-2)}{x-0}$, which means the slope of the line passing through the fixed point $ A(0,-2)$ and the mobile point $ P(x,\ y)$ on the circle.

Distance formula?  You can do it geometrically, which is much simpler.  Draw a diagram.

[quote="vikas"][quote="kunny"]Rewrite $ \frac{y+2}{x}$ as $ \frac{y-(-2)}{x-0}$, which means the slope of the line passing through the fixed point $ A(-2,\ 0)$ and the mobile point $ P(x,\ y)$ on the circle.[/quote]
You mean (0,-2) right?  Then you would use distance formula and find an equation relating x and y and also use the circle equation and solve for the x then y and plug that into your resulting equation y+2/x?[/quote]

Sorry, i made a typo. :oops:

[quote="kunny"]This problem doesn't make sense, because the slope of the line $ AP$ can be greater as much as.[/quote]

$ \frac{{y+2}}{x}$ is the slope of a line through (0, -2) and some point on the circle.  Hence it will be maximum/minimum when the line touches the circle.  Applying the tangency condition for the line $ mx-y-2=0$, we get two values for m.  Larger $ (0.506)$ value represents the maximum and smaller$ (-5.649)$will represent the minimum.

I dont get how to get that answer just geometrically

[quote="bos1234"]why would it be a max. / min. at the point of contact?[/quote]

There will be two tangents from a point external to the circle.
At one point it the slope will be maximum and minimum at the other point.


inequalities
trigonometry
function
inequalities proposed

Prove the following inequality.

\[\sin \frac{\pi}{4}>\sqrt[3]{\sin \frac{\pi}{16}\sin \frac{\pi}{8}\sin \frac{9}{16}\pi}\]

the function $f(x)=ln(sin(x))$ is defined and concave on $(0,\pi)$, so that
$\frac{ln(sin(\frac{\pi}{16}))+ln(sin(\frac{\pi}{8}))+ln(sin(\frac{9\pi}{16}))}{3}<ln(sin(\frac{1}{3}(\frac{\pi}{16}+\frac{\pi}{8}+\frac{9\pi}{16})))=ln(sin(\frac{\pi}{4}))$
The inequality required follows by exponentiation.


Prove that: $ \forall a;b;c\geq 0$:
     $ (\sum a)^2\sqrt {\sum ab}\geq\frac {9\sqrt {3}}{8}\prod(a \plus{} b)$

[quote="karis"]Prove that: $ \forall a;b;c\geq 0$:
     $ (\sum a)^2\sqrt {\sum xy}\geq\frac {9\sqrt {3}}{8}\prod(a \plus{} b)$[/quote]

Assume $ a \plus{} b \plus{} c \equal{} 1$ and denote $ q\equal{}ab\plus{}bc\plus{}ca,r\equal{}abc$ then the inequality is equivalent with

\[ \sqrt q  \ge \frac{{9\sqrt 3 }}{8}\left( {q \minus{} 3r} \right)
\]

In case $ q \le \frac{1}{4}$ we have

\[ \frac{{9\sqrt 3 }}{8}\left( {q \minus{} r} \right) \le \frac{{9\sqrt 3 }}{8}q \le \frac{{9\sqrt 3 }}{{16}}\sqrt q  < \sqrt q
\]

And in case $ \frac{1}{4} \le q \le \frac{1}{3}$ we have

\[ \frac{{9\sqrt 3 }}{8}\left( {q \minus{} r} \right) \le \frac{{9\sqrt 3 }}{8}\left( {q \minus{} \frac{{4q \minus{} 1}}{9}} \right) \equal{} \frac{{\sqrt 3 }}{8}\left( {5q \plus{} 1} \right) \le \sqrt q
\]

let ab+bc+ca=1,remember that equation:(a+b)(b+c)(c+a)=(a+b+c)(ab+bc+ca)-abc
and Use lagrange multiplier,the inequality become very easy.

Please tell about langrange multipliers.


In a isosceles triangle with base b and sides a, draw a height against one of the equal sides. This height slits the side in two, x and y. Give x and y in terms of a and b.

we can call this newly drawn height, h.  so, h^2 + y^2 = b^2 and x^2 + h^2 = a^2.
if we subtract these equations, we get y^2 - x^2 = b^2 - a^2.
we also know that a = x + y.
so, we can solve for y, y = a - x.
simply substitute into our equation: (a - x)^2 - x^2 = b^2 - a^2.
a^2 - 2ax + x^2 - x^2 = b^2 - a^2
2a^2 - b^2 = 2ax
x = a - b^2/2a
and y, its counterpart, is merely y = b^2/2a since x + y = a.


What is the smallest perfect square that ends in $9009$?

my solution is the same, $x^2 \equiv 9 \mod 1000 \implies x=t(500)\pm 3$ so $x^2 = (500t \pm 3)^2 \equiv 9009 \mod 10000$

which implies $\pm t \equiv 3 \mod 10$ now because we want the smallest solution we shall put $t=3$ and $x=500t+3=1503$


Can only school teams register for MOEMS, or can individual students register themselves to MOEMS?

If you wanted to, you could come up with a realistic school name and use that to compete.  But you would have to deal with everyone else to register.   Are you homeschooled or at a school but want to do this by yourself :?:

I go to a real school, but I wanted to do this by myself. Same as AMC 8. I thought of this question to ask because my school didn't want to register for those competitions and I'm really dying to join them.

hmm, tricky situation!  As I said, you could register as a new school [url=http://moems.org/enroll.htm]Here[/url].
Or look at this topic [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=168408]Starting a new MOEMs group at your school.[/url]
Other than that, I don't know what to do.  :roll:

oh and don't forget you would have to give a valid address for the tests to be emailed to and i think there might be a fee for amc8 . I think its like 2 dollars per test

[quote="gauss1181"]Like the AMC 8?[/quote]
Are you trying to do it individually?
I think that depends on what grade and school you are in..because from my experience, the school provides you with AMC 8 without fee up to 8th.

OK Folks, as a genuine math team sponsor/adult, I think I can conclusively answer your questions.
1) MOEMS costs at least 89 for a classroom of up to 35 to participate in one division(6th grade and under, or under 8th grade). Registration closes at the end of September. If you are interested in participating but your school does not, you need to find a teacher or other adult who is willing to run the tests for you. You also need to raise the money to pay for the test. You can ask other participants to pay something for it, or you can pay for it out of your / your parents' pocket.

2) AMC 8 is similar - you must have a sponsor (at school if you go to school), pay for the exam, and have a non student address to send the packet. It costs as much as 53 plus 11 for each packet of 10 exams, which must be paid by Nov 6.

I homeschool my children, and I started a math club for homeschooled students so that they may participate in these exams. All you need are the funds and a willing sponsor. Math knowledge is optional. <wink> :surrender:

Don't revive old topics, especially since this was already answered.


1) Whats the units digit of 123^321?

2) A 110 sided polygon has an X amount of diagnols. Find X

3) If you randomly open a 100 page book. What is the probability that the sum of the two pages that you are looking at is a perfect square?

4) How many even five digit numbers are there that don't contain a 5 as one of their digits?

5) How many solutions are there to this.  a+b+c=25        -100 < a,b,c < 100

 :D

[hide="1"]The units digits of powers of 3 go in a cycle of 3,9,7, and 1. Since $\frac{321}{4}$ leaves a remainder of 1, the units digit is 3.[/hide]

[hide="2"]The number of diagonals in a convex polygon is $\frac{s^2-3s}{2}$ where $s$ is the number of sides. So there are $5885$ diagonals.[/hide]

[hide="3"]Assuming the even numbered pages are on the left page, the only perfect squares that can be seen are 1, 9, 25, 49, 81, 100, 121, and 169. Since the even pages are on the left, pages 1 and 100 will not have an accompanying page, making 51 possible places to flip to. The probability is $\frac{8}{51}$[/hide]

[hide="4"]There can be 5 different digits in the units place, 8 digits in the 10,000s place, and 9 digits in each of the others. There are $8\cdot9^3\cdot5=29160$ numbers.[/hide]

[quote="dimak11214"]For number three I got a different answer. There are 101 diferent places to open the book. In any part you open you have two consecutive numbers. Unless you open the first or last pages( then you would have the number 1 and the cover, or 100 and the cover). Lets say one page is X and the other is X+1. The sum would be 2X+1. So 2X+1=a perfect square. This works for any odd perfect square, 9,25,49,81,121,169(197 is the highest sum) But if we open the first or last pages we get 1 or 100(which are perfect squares) So there are 8 possibilities.
My answer is 8/101 (but I might be wrong) :oops:[/quote]

[hide]I agree that there are 8 possibilities, but I don't think you can open the book to 101 different places. For example, if you can see pages 2 and 3 at the same time, you cannot see pages 3 and 4 at the same time, since they would be on opposite sides of the same sheet of paper. So there are only 49 ways to see 2 pages, and 2 ways to see 1 page (1 and 100). I think...[/hide]

[quote="Hokkage"][quote="catcurio"][hide="4"]There can be 5 different digits in the units place and 9 digits in each of the others. There are $9^4\cdot5=32805$ numbers.[/hide][/quote][hide="I think you made one little mistake"]For the very first digit, it can't be a 0, correct? Or else the number would be a 4-digit number. So there should be $8 \cdot 9^3 \cdot 5$ such numbers.[/hide][/quote]

Yes, you're right. I always do that... :(


You are driving a pickup truck that has a tall, vertical container in the trunk filled with an explosive liquid.  What is the maximum acceleration you can give the truck without tipping the container if it is

a) a cylinder with radius 2m and height 15m
b) a rectangular box with square base of side length 4m and height 15m
c) Two rectangular boxes as described in (b) are placed 1m apart, one behind the other.  They are joined at the top with a rigid bar of negligible mass.

The coefficient of friction between all surfaces is $\mu$.

Is translational movement (ie. sliding across the trunk) allowed?

Actually, I hadn't considered it sliding.  I just thought that it would tip before it slid.  (After all, static friction is greater than kinetic friction.)  But suppose it [i]could[/i] slide; would it?

The condition for non-sliding is easy: $\mu mg \geq ma \Rightarrow a \leq \mu g$. Let's suppose this is fulfilled; otherwise, the following calculation gets messier.

The condition for non-rotating is obtained by equating torques around an arbitrary non-moving point; let this point be the point around which the container rotates (assuming no sliding); let's call this point $O$ and let $C$ be the center of mass of the container. Let $\phi$ be the angle between $OC$ and the vertical. The torque of the inertial force is $mal\cos\phi$, where $l \equiv |OC| = \sqrt{r^{2}+(h/2)^{2}}$. The torque of the gravitational force is $mgl\sin\phi$, which leads to $\tan\phi = a/g$. Now we have that $\tan\phi \in [0, r/(h/2)]$. The first limit means that $a$ is zero and that the container is balancing on one point, maintaining an instable equilibrium. The second limit means that the container is practicaly touching the floor with its base (but it's not touching it because in that case the above equations wouldn't be valid). Thus, we have $\tan\phi \leq 2r/h \Rightarrow a \leq 2rg/h \approx 2.67 \text{m}/\text{s}^{2}$.

The second case is done in the same way, but there are two possibilities: the container balances on one edge or it balances on just one point. The third case is just a minor modification of the second one.

If there is sliding, we'd have to calculate the translational acceleration and enter the non-inertial referential system of the container in order to equate the torques as we did here. There would be an additional force (the inertial force due to the container's acceleration) which would provide a torque, though.


Euler
modular arithmetic
number theory
prime factorization

Oh man,never mind... :oops:  :blush: ,how could i forget the coefficients!!I don't think i can show myself on AOPS again.

what shady missed is that in the binomial thm. there are coefficients. The next to last coefficient is 400, so 400*80 ends in two zeros, only the last term matters.

[quote="ShoeFactory"][quote="chess64"][hide="Euler has a more beautiful method"]
By Euler's some theorem, $a^{\phi(n)}\equiv 1 \pmod {n}$, if $a\perp n$. So in this case, $n=1000$ and $\phi(1000)=400$. So $3^{400}\equiv 1 \pmod {1000}$. This means that the last three digits are $001$ and the last two are $\boxed {01}$.
[/hide]

Or you could also just realize that the last two digits are just the same as the last two digits of $3^{400\pmod {100}} = 3^0 = 1$. Either way, your answer is wrong. :([/quote]

How would you know $\phi(1000)=400$ ?[/quote]Time to review a formula:

If the prime factorization of a positive integer $n$ is given by $n=\prod_{j=1}^Np_j^{a_j}$, with $p_j$ being primes and $a_j>0$ being the exponents for $j=1,\ldots,N$, then

\[ \phi(n)=\prod_{j=1}^Np_j^{a_j-1}(p_j-1)  \]

[quote="shadysaysurspammed"][hide="my beautiful method"]$3^{400}=81^{100}$
$=(80+1)^{100}$
So applying bino theorem....
$80^{100}+80^{99}........80^2+80+1$
Now we can clearly see that addition of all terms till $80^2$ will give something with 2 zeros in the end,hence adding 81 to it,the l;ast 2 digits are ${\boxed{\boxed{\boxed{\boxed{81}}}}}$[/hide][/quote]

Sorry, if the error is already pointed out, but there is an error..
$(1+80)^{100} = 1 + 100 \cdot 80 + \frac{100 \cdot 99} 2 (80)^2 + \frac{100 \cdot 99 \cdot 98 } 6 (80)^3 + \dots$
That way you can see that last 2 digits  are $01$
Actually it is not that hard, if you want to show off to, see the last three digits are  $001$ or last  5 digits are $88001$ :)

One thing you may like to know that $3^{20} \equiv 1 \bmod 100$

(or $3^{100} \equiv 1 \bmod 1000) etc)$

Another "fun" way is to notice that $100 = 4 \cdot 25$ so find the value mod 4 and 25  separately ...
(Euler can make things very easy but other ways may be fun too ... for example ...

Mod 4
$3 \equiv (-1)$  and since $400$ is even $3^{400} \equiv (-1)^{400} \equiv 1$

Mod 25:
Say you notice $3^3 \equiv 27 \equiv 2$
and notice       $2^7 \equiv 128 \equiv 3$
so $(3^3)^7 \equiv 3^{21} \equiv 3$  which tells you
$3^{20} \equiv 1$ which of course means  $3^{400} \equiv 1$

Now if a number is $1 \bmod 4$ and $1 \bmod 25$  it has to be $1 \bmod 100$
Hope that is fun.

[quote="10000th User"][quote="ShoeFactory"]
How would you know $\phi(1000)=400$ ?[/quote]Time to review a formula:

If the prime factorization of a positive integer $n$ is given by $n=\prod_{j=1}^Np_j^{a_j}$, with $p_j$ being primes and $a_j>0$ being the exponents for $j=1,\ldots,N$, then

\[ \phi(n)=\prod_{j=1}^Np_j^{a_j-1}(p_j-1)  \][/quote]

Or simpler, $n: =p_1^{e_1}p_2^{e_2}\cdots p_k^{e_k}$, \[ \phi (n) = n\left( 1-\frac{1}{p_1}\right)\left( 1-\frac{1}{p_2}\right)\cdots\left( 1-\frac{1}{p_k}\right). \]

Yes yes simpler. I have thought of editing my post  :blush:

[quote="me@home"]is it true that you are the 10^5th user?[/quote]To show you the 'proof' of it, click on [u]Members[/u] then select Sort By Joined Date and I'm the 10000th User :D

Here is a link to it

[url=http://www.mathlinks.ro/Forum/memberlist.php?mode=joindate&order=ASC&start=9975]www.mathlinks.ro/Forum/memberlist.php?mode=joindate&order=ASC&start=9975[/url]

Upon checking it, I'm off by one spot because there was a user before me that had a bad name, he was removed and I moved up a spot  :roll:

[quote="me@home"]is it true that you are the 10^5th user?[/quote]
He's the 10^4th user, not the 10^5th, just to clarify... :roll:

sweet i used this on a math competition today, after just reading about it yesterday.

finding the remainder when $4^{87}+6^{87}$ is divided by $25$..

That's really similar to #6 on 1983 AIME... :roll:

[quote="Kalva"]6. What is the remainder on dividing $6^{83}+8^{83}$ by $49$? [/quote]

Yeah Binomial Thm works best in that case. Write \[ (5-1)^{87} + (5+1)^{87} \pmod {5^2} \] and almost every term reduces to 0.

so, has anyone gotten tired of trick problems yet, or are they still fun?

someone should start a topic of the funnest math problems...

You can do that in other problem solving topics i guess.


geometry
incenter
trigonometry
angle bisector
geometry proposed

The incircle of triangle $ABC$ with incenter $I$, is tangent to sides $BC,CA,AB$ at $M,N,P$ respectively. $Q$ is a point on the incircle of triangle $ABC$ such that $NQ$ is perpendicular to $CQ$. The tangent to the incircle in point $Q$ intersects $AC$ and $AB$ at $S,T$ respectively. If $SI$ and $TI$ pass through the point $P$ and $N$ respectively then Find angle $ACB$. :D

Let $w$ be the incircle of $\triangle ABC$ and $z$ the circle with diameter $NC$.
The circles $w,z$ are perpendicular to each other, so the tangent line to $w$ at $Q$ will pass through the center of $z$. Hence the center of $z$ is the point $S \Rightarrow  SC=SN.$

The incircle of $\triangle TAS$ is the circle $w$.
So the line $SI$ is the internal angle bisector of the angle $S$ in $\triangle TAS$. But $SI$ passes through $P$, so $SI\perp AT\Rightarrow  \angle SAT = \angle STA$

Similarly $TI$ is an internal angle bisector of $\triangle TAS$ and $TI \perp AS \Rightarrow  \angle SAT = \angle AST.$

So the triangle $TAS$ is equilateral $\Rightarrow  \angle NSI = 30^{o}\Rightarrow  \frac{IN}{NS}=\frac{\sqrt 3}{3}$

$\tan \frac{\angle ACB}{2}= \frac{IN}{NC}= \frac{IN}{2\cdot NS}= \frac{\sqrt 3}{6}$

So $\angle ACB = 2 arctan\left(\frac{\sqrt 3}{6}\right)$

Beautiful Solution :) , and it was a beautiful problemo :lol:

Sorry, I don't like the "tangled" enunciation of this problem because is obviously that the triangle $TAS$ is equilateral. From these reason, I"ll suggest an equivalent enunciation which change the old problem in one very easy (at the most for the Pre-Olympiad section) :

[color=darkred][b]An equivalent enunciation.[/b] Let $ABC$ be an equilateral triangle with the incircle $w=C(I,r)$ which touches the sides $[BC]$ and $[CA]$ in the points $D$ and $E$ respectively. Denote the point $L\in AC$ for which $LD\perp AB\ .$ Ascertain the value $\phi$ of the angle between the tangents from the point $L$ to the incircle $w\ .$[/color]
[color=darkblue][b]Proof.[/b] Suppose w.l.o.g. $AB=1\ .$ Thus, $r=\frac{\sqrt 3}{6}$, $EL=1$ and $\tan\frac{\phi}{2}=$ $\frac{IE}{LE}=$ $\frac{r}{1}=\frac{\sqrt 3}{6}\ .$ Therefore, $\phi=2\cdot\arctan\frac{\sqrt 3}{6}\ .$[/color]

[color=darkblue][b]Remark.[/b] In the old problem was used the following very simple property :
$\blacktriangleright$ If for the $Q$- right triangle $QNC$ exists a point $S\in NC$ so that $SQ=SN$, then the point $S$ is the middlepoint of the side $[NC]\ .$[/color]


function
geometry
geometric transformation
reflection
complex analysis
complex analysis unsolved

Let $f(z)$ be regular in  $D=\{z: |z|<1 \}$, and continious in $D \bigcup G$, where $G$ is an arc of positive measure of the unit circle such that $f|_G =0$. Prove that $f_{D \bigcup G}=0$.

Since it's zero on $G$, of course it's continuous. Maybe you meant to say that it's continuous on $D\cup G$? In that case, we can apply the Schwartz reflection principle to extend it to a function holomorphic in a neighborhood of a point $z\in G$, and by the maximum modulus principle, this function will have to be identically equal to $0$.

Yes, surely, thank you, Grobber, i meant that $f$ is continious on $D \bigcup G$. I've got it.
Thanks again.


umm I have to derive a formula to calculate for how much cash to charge a buyer who pays through a merchant account with 3.9% + 0.80 usd fee so that i recieve the desired amount for the product. This is not related to my syllabus, but my teacher wanted to make us more math minded, but i can't seem to derive a viable formula.

[u]Example[/u]

I'm selling a product for 10 usd
Merchant account charges 3.9% + 0.80 usd
So i would only receive 8.81 usd
So i have to charge an amount so that I'll get 10 usd

Help me derive a formula, thanks :)

Let $ x$ be the price you will sell the item for.  What you want to happen is:

$ x \minus{} (.039x \plus{} .8) \equal{} 10$.

Now you can solve for $ x$.

How's that?

so to find x i would have to do the following.

x = 10 + (.39x - .60)?

but there is a problem with, i can't equate x because the value of x is required in (.39x + .60), any suggestions?

umm I'm a bit weak when it comes to solving equation. we are thought about around 11th grade here.

Where does $ .39$ come from?  Don't you mean $ .039$?  Where does $ \minus{}.60$ come from?  Isn't that supposed to be $ .8$?

Be very careful about little mistakes like that.

So, you took the equation I wrote and manipulated it, and got $ x \equal{} 10 \plus{} (.039x \plus{} .8)$.  (Correcting mistakes here.)  And as you point out, there's a problem with this: $ x$ is equal to the right hand side, but you can't compute the right hand side without knowing what $ x$ is!

Yes, that is a problem, sort of a wonderful problem.  What do we do about this?  How do we find out what $ x$ is?

This is a wonderful puzzle.  I am almost jealous of you for not knowing the answer, the same way I'm jealous of people who have never read Surely You're Joking, Mr. Feynman, because they now get to discover it for the first time.

I am reluctant to tell you the answer.  Let me just say, there is a way to figure out what $ x$ is!  It is nice, it is simple, it is excellent!  Centuries ago, the first mathematicians asked themselves the question you're asking yourself now, and when they figured out the answer, it was a momentous discovery!

I apologize.  Somebody else will chime in and be more helpful.

yes those were careless mistakes on my part ^^'

[hide="Hint"]
Subtract .039x from both sides...[/hide]

[hide="Solution"]
$ x\equal{}10\plus{}(.039x\plus{}.8)$

$ .961x\equal{}1.8$

$ x\equal{}\frac{1800}{961}$.
[/hide]

Additional stuff:

You did a step you didn't need to. If you had just left it as singularitarian had given to you the first time, it would have been easier. This is also true because when singularitarian corrected your mistakes, he made a mistake of his own. The negative wasn't distributed all the way. The .8 should have been subtracted.

I believe my algebra was correct.  Fishy, your solution starts with the same equation I wrote down, so I'm not sure why you thought I had a mistake.  You have a slight arithmetic error though--where you have written $ 1.8$, it should have been $ 10.8$.


AMC
AIME
AMC 12
USA(J)MO
USAMO
geometry
geometric transformation

index probably not
floor value probably not, either
so sorry :(
but you did a good job on AMC

again since you are a nith garder your index doesnt matter
all you can hope for is that the floor for AIMEII be a 7, then you will probably qualify since you have  high AMC12 score, and you will proabably qualify even in the case of tie-braking

I made some pretty stupid errors too, ending up with an 8 :( . This seemed like a really easy AIME, easiest one I have ever seen, so I'm doubtful that 8 wiill do it. I thought 1-11 and 14 were easy enough to be AMC 12 problems, with 1-3 being easy enough to be AMC 10 problems. I expect the floor to be around a 9, but I wouldn't be surprised if it climbed as high as 11.

Maybe this year, we will have the same situation as last year:
2 different indexes for 2 AIME, and 2 different floors also.

AMC 12B: 132.5

AIME: 9

Index: 222.5

I think I made it to the USAMO, especially with the expansion.

Good Luck to everyone else!  :P

[quote="Phelpedo"][quote="nat mc"]I got a 7 with 6 stupid mistakes. Ugh, I wish I had taken the 1.

Grade:9
AMC 12A: 143
AIME II: 7
Index:213

Will a 213 index make it?[/quote]

By the way, I thought I saw in some other topic that you did take the AIME I (and got a 9)?[/quote]
I took it unofficially because our stupid school corporation had a two hour delay the day of the test, and I got a 10. Meh, I just wanted to go to MOSP since I'm a ninth grader

Alright...first time taking these tests...
Grade: 8
AMC 10A: 138
AIME: 1!  :P

So my chances at USAMO are nonexistent...at least I have 4 more years.

[quote="JGeneson"]Getting an 8 and calling it easy is silly.

I got a 12 (13 answered, 1 error), which means a 259.5 index. I am hoping (for friends) that USAMO index = 216, AIME floor = 7.[/quote]
please vote, everyone, so we can make an accurate  conjecture about the floor

I got an 8, so I will probably qualify for USAMO :)

I solved 1-7, and brute-forced #11, no errors.

Grade: 9
AMC 12 A Score: 122
AIME II Score: 8

It seems to me like the floor will be a 7 or 8, maybe a 9.

lol I brute forced number 11 too. I got 1,2,4,5,10,11,12
3- forgot 81 was a power of 3
6- forgot to do $2-\sqrt3$ minus it
7- did only 3 digit numbers
8- forgot rotations, etc.
9- read $C_3$'s center was at (0,24) instead of (24,0)
13- didn't read [i]odd[/i]

Didn't have time for 14 and 15, I was too busy checking my wrong answers. Though 15 looks pretty simple

I would conjecture floor of 8, index of 215-220.

For reference:

Grade: 10
AMC 12A: 123.0
AIME: 10
Index: 223.0

My uneducated guess right now is that the index will be 210-220, with the floor at either 7 or 8.

I got a 12.  Didn't see the similar triangles on #12, missed three out of the four cases on #13, and couldn't finish #15 after doing quite a bit of work.

Grade: 12
AMC 12A: 144.0
AIME II: 12
USAMO Index: 264.0

Lol I suspect you guys are taking it a tad bit too seriously. :P But then again, there are those blokes who're refreshing the AMC page every 15 seconds...

[quote="white_horse_king88"]You guys should relax a little... this is important, but I believe you can't live without sleeping. So go to bed. :)[/quote]

Actually, it's only 9 here, so.......another 2 hours of anxiousness! :lol:  :rotfl:

I'm sure there's no one at the AMC office anymore. :-) But if it comforts some of you guys to wait, lol...go for it. :-)

Are we allowed to vote if we took the AIME-I but took the II unofficially, but still using bubble sheets and everything?

Well, it's too late. I already did. I got a 13 with 2 stupid mistakes. This time, they actually were stupid, since I didn't check my work; I just used the full 3 hours to solve the problems. Actually, it might have been only 2:45 or so, since I had to leave at a fixed time anyway. Oh well. On #7, I wrote "times 2" and circled it, but forgot to multiply some of my numbers by 2. On #12, I did this monster brute force thing (but didn't use coordinates) and messed up in the arithmetic, not dividing by 2 at one point in my calculations. Otherwise, I would have gotten perfect.

So yeah, the AIME-II definitely was MUCH easier this time around. I got a 10 with 3.5 stupid mistakes on the AIME-I. And yes, I did brute force #11. :)

[quote="nat mc"][quote="SamE"]So yeah, the AIME-II definitely was MUCH easier this time around. I got a 10 with 3.5 stupid mistakes on the AIME-I. And yes, I did brute force #11. :)[/quote]
You can't really say that because you took it unofficially. Anyways according to AMC, the AIME I and AIME II were "statistically identical" or at least close enough to only make one index. AND thomas Jefferson took the AIME II this year instead of AIME I.[/quote]

Of course, the possible reason the amc might have said that the aimes were "statistically identical" was because TJ took the aime II.  Do a significance test and see what you get!  :lol:  ;)

For what it's worth, Percent of students scoring 15, 14, 13, 12, 11 and 10 --

AIME I:
0.02%, 0.06%, 0.05%, 0.13%, 0.24% and 0.37% altogether for 0.86% of the total

AIME II:
0.02%, 0.07%, 0.33%, 0.60%, 1.00% and 1.32% altogether for 3.35% of the total

[quote="h_s_potter2002"][quote="nat mc"][quote="SamE"]So yeah, the AIME-II definitely was MUCH easier this time around. I got a 10 with 3.5 stupid mistakes on the AIME-I. And yes, I did brute force #11. :)[/quote]
You can't really say that because you took it unofficially. Anyways according to AMC, the AIME I and AIME II were "statistically identical" or at least close enough to only make one index. AND thomas Jefferson took the AIME II this year instead of AIME I.[/quote]

No, from what I've heard, statistically, the AIME I was much harder than the AIME II.  Mybe they made the floor score the same because Thomas Jefferson and AAST took the AIME II though?[/quote]
No read Mr Dunbar's post. They were close enough to statistically identical to make only one floor even with Thomas Jefferson taking it

By the way, I was just basing my comment on the AIME-II being easier on my own experience - I got 13 (2 stupid mistakes) on the II and 10 (3.5 stupid mistakes) on the I. You can argue based on stats etc. but if you haven't seen the problems from the other one, take it and see what you would have gotten, then compare.

I have already and I got a 10 on the I (3 stupid mistakes), and a 7 on the II (6 stupid mistakes). I would say that the AIME I was easier than the II because I got a higher score, but since I wasn't under pressure, I had less stupid mistakes

Hmmm... that's interesting. We both made more stupid mistakes on the official one. However, in my case (which I know best), I did completely solve more problems on the II, but I never made any progress on one of the problems and sort of made a reasonable guess on the other (my half a stupid mistake). Also, I checked my work on the I but didn't on the II. I guess then it's just personal opinion as to which test was easier. :huh:

I got a 12 on the II and a 9 on the I.  On AIME I I made one or two real mistakes but on the AIME II I had plenty of time to check my work after solving 12.  I didn't make any mistakes on the II.  The II had many more approachable problems, meaning that.. if you had a decent idea of how to do it, you were going to get it right, barring any mistakes.  I thought the I had alot of traps on it and consequently many people fell into those.
I personally thought the II was easier, but that might be because I just took it after I'd had a little more practice, so maybe I was a little bit.. more skilled or whatever.  I think the two had alot higher scores.


Hi,can anyone sketch the graphs of:

1)$y=2\left(\frac{x}{1-x}\right)^k+\left(\frac{1-2x}{2x}\right)^k$ where $x\in (0,1/2),k=\frac{\ln3}{\ln2}-1$ 2-D graph.
2)$z=\left(\frac{x}{1-x}\right)^k+\left(\frac{y}{1-y}\right)^k+\left(\frac{1-x-y}{x+y}\right)^k$ where $x,y,x+y\in (0,1),k=\frac{\ln3}{\ln2}-1$ 3-D graph

3)$\left(\frac{x}{y+z}\right)^k+\left(\frac{y}{z+x}\right)^k+\left(\frac{z}{x+y}\right)^k=2$ where $x,y,z>0;k=\frac{\ln3}{\ln2}-1$3-D graph.

I have got MATLAB so you can give me which commands to type or u can also send e-mail to hdsoul@gmail.com.
Thanks.

I have Derive and sketching graphs is extremely easy.

You can get Derive (shareware) from [url=http://www.derive.com]www.derive.com[/url]

Hope that it helps you.


function
geometry
geometric transformation
reflection

It's not difficult, but I made an interesting solution. So here is the problem:

[img]http://img201.imageshack.us/img201/7777/asdoy2.gif[/img]
[img]http://www.picvalley.net/u/83/82684_325.GIF[/img]

I liked this problem!  :D
[hide]$f(x) = x^{2}+kx+2$

$g(x) = x^{2}+(2k-1)x+3-k$

Suppose that $f,g$ have some common root $r$.
Then $r$ will is also a root of $h = g-f$

$h(x) = (g-f)(x) = (2k-1-k)x+3-k-2$

$h(x) = (k-1)x+1-k$

$h(x) = (k-1)(x-1)$

$0 = (k-1)(r-1) \ \ (1)$

[b]A)[/b] if $k=1$

Then $f(x)=g(x) = x^{2}+x-2$
So, the two functions have [b][u]two[/u][/b] common roots $1$ and $-2$. So they don't satisfy the conditions of the problem

[b]B)[/b] if $k\not = 1$

From $(1)\Rightarrow r=1\Rightarrow f(1)=0$

But $f(1) = k+3 \Rightarrow k=-3 \Rightarrow$

$f(x) = x^{2}-3x+2 = (x-1)(x-2)$
$g(x) = x^{2}-7x+6 = (x-1)(x-6)$

So for $k=-3$  they have exactly one common root[/hide]

Good work, pontios! Interesting solution, I didn't know it. I told you it is not difficult ;). My decision is a little bit more different. Only a little mistake. When k=1,  $f(x)=g(x) = x^{2}+x+2$ , not $f(x)=g(x) = x^{2}+x-2$. So their roots are common, but they are not real. Of course this mistake doesn't reflect on the answer.
When I have time I will post my solution. :)


Hi, I always think the answer is going to be negative, but none of the choices are below zero. It says that the  $\bigtriangleup H^{\circ}$ is positive, meaning that it is endothermic, right? But the question confuses me.

The value of $\bigtriangleup H^{\circ}$ for the reaction below is $+128.1 kJ$

$CH_3OH_{(l)} \Rightarrow CO_{(g)} + 2H_{2(g)}$

How many $kJ$ of heat are released when $15.5g$ of $CH_3OH_{(l)}$ decomposes as shown in the equation?

$A) 0.48$
$B) 62.0$
$C) 128$
$D) 653$
$E) 326$

I get negative 62, because it does not release heat at all, because it is endothermic. Can somebody help me with this?

Hi!

I am not really sure what is confusing you. I also get 62 kJ of energy released. I am not sure how your negative came about.

My working out:

Moles of Methanol used:
12+1+1+1+16+1=32
15.5/32=0.48

Energy taken in per mole=128.1
therefore, energy taken in by reaction=0.48x128.1=62

Where did your negative come from?

Because the $\bigtriangleup H^{\circ}$ is positive, which means that the reaction is endothermic. If it is endothermic, it wouldn't release heat at all, but gain it. Since it gains heat, and the question asks how much is [i]released[/i], wouldn't that mean it is negative? Or am I just being really stupid lol.

That's correct- no heat is being [i]released[/i], according the the $\Delta H$ given.

You basically answered your own question. The enthalpy shown in the reaction shows that heat [i]is taken in[/i], so, I don't know what you're worried about :)

Blah. It turns out that the teacher informed all the students - while I was retaking a test - that all of the answers are inversed of what they should be  :oops_sign: But thanks for your help  :) I really appreciate it :yoda:


Find the product  :maybe:

[hide]I think intramolecular rn takes place and the product is 2 cjoint cycloprapane rings(sth like that)  :maybe:
but seems very unstable......[/hide]

At first sight I would also vote on spiropentane as the main product.

I would say the joint-cyclopropane ring system would be formed at low temp and at high temp, maybe a cyclohexane with the 4 CH2Br groups at positions 1 and 4??

And what is your explanation for that temperature effect?

The intramolecular reaction is faster and requires low Ea because as soon as the free radicals are formed, a bond is formed bwtween the  carbons.

In the cyclohexane product, there is possibility of the formation of stable product and hence the reaction is slower beacuse at high temp as molecules are far apart, the free radicals formed(on the same molecule) must be stable till such time till aanother free radical is available... :maybe: (this was something along the lines of the function of NBS)

Sodium is a very reactive metal - it will be very unlikely that bromine atoms will survive in the original molecule.


Sa se arate ca pentru orice pereche de permutari $\{\sigma ,\tau\}\subset S_{n}^{2}$

exista lantul de inegalitati : $n\le \sum_{k=1}^{n}\frac{\sigma (k)}{\tau (k)}\le (n+1)\cdot\sum_{k=1}^{n}\frac{1}{k}-n\ .$

Comparăm $\frac{\sigma (i)}{\tau (i)}+\frac{\sigma (j)}{\tau (j)}$ cu $\frac{\sigma (i)}{\tau (j)}+\frac{\sigma (j)}{\tau (i)}$. Diferenţa celor două expresii e $\left( \sigma (i)-\sigma (j) \right) \left( \frac1{\tau (i)}-\frac1{\tau (j)}\right)$. Să presupunem că $\sigma (i)-\sigma (j)$ şi $\tau (i)-\tau (j)$ au acelaşi semn. Dacă schimbăm $\tau (i)$ cu $\tau (j)$, atunci suma creşte. Dacă au semne diferite, suma scade.
Aşadar, dacă vrem s-o minimizăm, trebuie să avem $\sigma = \tau$. Dacă vrem s-o maximizăm, $\sigma (i)+\tau (i) = n+1, \, \forall i \in \overline{1,n}$.

Cam aşa se demonstrează şi inegalitatea rearanjamentelor, aşa că am mai sărit nişte paşi.


I faced this question, and my first thought was that it involves with Pigeon Hole Principle, does it?
I only heard of Pigeon Hole Principle, but don't actually know how to use them.
If it does, how would you use it?
Can you guys post solutions with your answer?

7. Katrina and Abbie start a game with one pile of 40 pennies. They take turns. On each turn, a player must take 1,2,3,4, or 5 pennies from the pile. The player who takes the last penny from the pile of 40 pennies wins the game. If Abbie plays first, what number of pennies must she take from the pile on her first turn in order to guarantee that she can win the game?

[hide="using pigeon hole(probably wrong)"]with the pigeon hole pricipal, we can say there are 6 "holes"(every hole is the number of pennies taken per turn), and 40 "pigeons". there must be one hole with atleast 7 pigeons. i did something wrong[/hide]

I don't know what this has to do with the pigeonhole principle:
[hide]First take 3 pennies; there will be 37 pennies left.
Then, for each move, Katrina will pick a number of pennies between 1 and 5.
For Abbie's move, pick 6 minus the number of pennies Katrina just picked.
So, every two moves, the pile will decrease by six.
Also, notice that, whatever Katrina picks, Abbie will be able to complete six for those two turns.
So at the end, there will only be one ball left for Abbie because 37 mod 6 is 1.[/hide]

[quote="DonkeyKong"][quote="ZhangPeijin"][hide="My Way"]
Well, seeing as.
No matter what the other play picks, you can always pick a number that, when combined with the other number, has a sum of 6.
$ 50 \equal{} 2(mod 6)$
Therefore, she will simply have to pick a 1 in the beginning, and the other play will have to pick the last one.
[/hide][/quote]
Hmm... So you use basically use modulo to figure it out..Right?
And what if the number of pennies are 69, and with the same rule.
Would you do mod 6 on this one, too?[/quote]
Oh god... I thought it was 50 pennies, not 40...
 :mad:
Anyways...
If it were 69, then it would be $ 3(mod 6)$.
So you pick 2 first.

I think we are missing something here while dancing around the solution quite closely :lol:
[hide]
Keeping score as if I was playing

Start with 40, I would pick 4,
36 after my pick.......... you pick 2
34 after your pick..........I pick 4(because 2+4=6)
30 after my pick..........you pick 5
25 after your pick..........I pick 1(because 5+1=6)
24 after my pick and I continue to pick to get a multiple of 6

18 after my pick

12 after my pick

6 after my pick

Now you can't take all 6, but you will leave 5 or less and I will be able to take the rest.  So you want to take 4 because 40 = 4(mod 6).  If you can pick up to n at a time, you want your opponent picking when there is a multiple of n+1 left over.

Hope that helps, and hope it is right :lol:   It is a different game if you want to force your opponent to take the last one though.

[/hide]

Agreed with frost13.  The more challenging variation is to force your opponent to take the last one, as previously stated.  This is a pretty difficult problem to solve.

Edit: [hide="Hint"]We wish to force the other person to choose the last one correct? In order to do this, there must be only one left after your turn.  So there must be at most $ 6$ pennies left during your final turn.[/hide]

I figured it out.
It has to do with mod, instead of pigeon hole principle.

Well, frost13 and mathcrazed got it.

Thanks for all of you for help.

[quote="alkjash"]Wait...
Don't you take three and leave 37 on the first turn?
Then, after the multiples of six, there will be one left that the other person has to pick.[/quote]
No, you tried to make the numbers of pennies to multiples of 6. The reason is whatever the opponent chooses when we have 6 marbles, you are always going to win.

Well, this problem made me think logically...And I finally got it..just today.

[quote="GautamR"]Excuse me, but what IS the pigeon hole principle?

Is it like the chicken nugget theorem? (xy-x-y)[/quote]

I do not know a lot about this theorem, but I think it pertains to Counting section.

I think it's useful when you have question like 'You have red, green, and blue socks in your drawer. How many socks do you have to draw out in order to get all three colors?.'

I am not sure.

I remember when the founder of AoPS came to our school and made us do that game...
First person to win 3 in a row gets a free book.
And everyone who participated got a free T-shirt.
 :lol:

[quote="alanchou"]Try this

[hide]
Start backwards. Let's say it's you and someone else.  You want to have 1-5 on the last part so that you can win.  Therefore, you force your opponent to get to 6.  To make them have 6, you need to have any number from 7-11, to bring their's down to 6.  Then they must have 12.  We just keep going on in this manner and this is the chart.

you     1-5
them   6
you     7-11
them   12
you     13-17
them   18
you     19-23
them   24
you     25-29
them   30
you     31-35
them   36
you     40

So to get from 40 to 36, you take [b]4[/b][/quote]
Umm, you forgot the hide...
And I thought it said you had to make your opponent pick the last one, not you.  :P

[quote="mathcrazed"]Alkjash, we are trying to make ourselves pick the last one, not the other person.  At least in the problem stated by DonkeyKong.  If you take three and leave thirty-seven on the first turn, then you will end up leaving one on the last turn for the OTHER person to take, which is not the desired result for DonkeyKong's problem.[/quote]

Oh oops...  :blush:
My mind went to a similar problem in which you want the other person to pick the last one, hehe.

Pigeonhole Principle is basically like this: If you have $ n\plus{}1$ objects and $ n$ places to put them, then there is at least one place with at least 2 objects. (basically common sense)

Can be generalized: if you have $ kn \plus{} c(c<n)$ objects and $ n$ places to put them, then there is at least one place with at least $ k\plus{}1$ objects.

[quote="mathcrazed"]Your chart's wrong...

You 1-5
Them 6-10
You-7-15.
...[/quote]

no, if you let them have 7, then they choose 1 and you lose.


Given cos(x)=-1/9
with 90o < x < 180o. Find cos(1/2)

could you help me step by step.. much appreciated

Stop posting problems in the wrong forum.  This also looks like homework, and I believe you mean $\cos \frac{x}{2}$.

which forums does this question belong to then?? these are my practice quiz questions just want to know how to solve them.

This is fairly basic trigonometry, and probably belongs here: http://www.artofproblemsolving.com/Forum/index.php?f=149

Anyway, I'm going to assume that the problem is what I think it is, in which case we use the identity

$\cos \frac{x}{2}= \sqrt{ \frac{1+\cos x}{2}}= \sqrt{ \frac{1-\frac{1}{9}}{2}}= \sqrt{ \frac{4}{9}}= \frac{2}{3}$

We pick the positive square root because on this interval $\cos \frac{x}{2}$ is positive.

My attempt to merge this with a previously moved topic failed. I'm moving it to Intermediate; the moderators there can note the redundancy.


abstract algebra
superior algebra
superior algebra unsolved

let $I$ be an infinite set, $A$ an abelian group. consider the abelian groups $A^{(I)}= \bigoplus_{i \in I}A$ and $A^{I}= \prod_{i \in I}A$. under which conditions is $A^{(I)}$ a direct summand of $A^{I}$ ?

for example, this is false if $I = \mathbb{N}$ and $A = \mathbb{Z}$, since it would yield an epimorphism $\mathbb{Z}^{(\mathbb{N})}\cong (\mathbb{Z}^\mathbb{N})^{*}\to \mathbb{Z}^{(\mathbb{N)}}^{*}\cong \mathbb{Z}^\mathbb{N}$. but if $A$ is a field with the same operation, then it is true because of linear algebra.

what about more general conditions? I'm particular interested in the case where $A$ is the multiplicative group of a  field.


function
calculus
derivative
analytic geometry
graphing lines
slope

If $f(x)=2x\sqrt{x+3}$ where $x$ and $f(x)$ are both real, then the range is ${f(x): f(x)\geq k}$. Find the value of $k$.

[quote="jeez123"]hmm, my solution packet has 20 steps in solving the problem which involves using a calculator :(

Could you rephrase the problem a bit for me? what does the $f(x): f(x)\geq k$ means? And what is $f'(x)$? thanks![/quote]

The question (I think) is asking for the smallest real value possible for the function $f(x)=2x \sqrt{x+3}$. $f'(x)$ is the derivative of the function (calculus), and can be easily used to find a minimum or maximum point of a function, since at that point, the slope of the tangent line to the function (the derivative) is zero.

[quote="jeez123"]then, what is the general form (the formula) for derivative? How do you use it to find the maximum?[/quote]

This is all stuff that you will learn in calculus.  If it gives you any comfort, I have no idea what they're talking about when they use derivatives either.


geometry
3D geometry
quadratics
trigonometry
algebra
polynomial
quadratic formula

http://putfile.com/pic.php?pic=12/34721022918.jpg&s=x10

^^any help?

[hide="solution"]you want to solve for $z$, so you should proceed to isolate $z$

$12+\sqrt[5]{z^3+63}=11$
$\sqrt[5]{z^3+63}=-1$
$(\sqrt[5]{z^3+63})^5=(-1)^5$
$z^3+63=-1$
$z^3+63=-1$

$\star$, $z^3=-64=(-4)^3$

$\sqrt[3]{z^3}=\sqrt[3]{-4^3}$
$z=-4$

there are other (complex) values for $z$ that work, but i doubt you are required to find them, but for kicks

using the equation from the $\star$
$z^3+64=0$
$(z+4)(z^2+4z+16)=0$ (sum of cubes)
then use the quadratic formula to solve the quadratic
$z=-4, -2+2\sqrt{3}i, -2-\sqrt{3}i$[/hide]

[hide="also note"]  A more geometric approach for finding the potential cube roots of -64 is through the fact that when the solutions are plotted in the complex plane [x-axis represents real numbers and y-axis represents imaginary numbers], they are equidistant from each other and the origin (0,0).  Immediately, we know that -4 is a solution, which can be rewritten as $-4+0i$, which means that its point in the complex plane is $(-4,0)$. The other [two] solutions can be found by mapping a circle about the origin with a radius 4 and find the points that are $120^o$ away from $(-4,0)$.  We can then either use some trig, or construct some 30-60-90 reference triangles to determe that the other points are $(2, 2\sqrt3), (2, -2\sqrt3)$ which translates as $2\pm2\sqrt3i$

Therefore, the three cube roots of -64 are: $-4, 2\pm2\sqrt3i$[/hide]

Wow, I've never seen that method before... Can you explain?  Why are all of the points located on the circle centered on the origin and including the point we already found?  Also, how do you know that the three solutions are equidistant on the complex plane?

these are what we call roots of unity, and this geometrical approach does not really help us because it takes more work to prove all these things than it takes just to factor it, but you should look at demoivre's theorem, and that basically shows you that they complex numbers are equally spaced out and they have equal distances

When do these roots of unity apply?  In any polynomial with complex roots, or only specific ones?

it is like you have $x^n=1$, and this is a $n$th degree polynomial so it must have n roots, but we only get 1 by using the roots, so what demoivre's theorem is for is that you find that the other roots are evenly spaced around a circle, i would suggest reading something about complex numbers so you can understand how this works better


Let be $ a,b\in \mathbb{R}_\plus{}^*$ such that $ |a|\neq |b|$ . Solve the following equation :

$ \frac{x\minus{}a}{2x\plus{}a\minus{}b}\plus{}\frac{x\minus{}b}{2x\plus{}b\minus{}a}\equal{}\frac{2x\minus{}a\minus{}b}{a\plus{}b}$

hello, your equation is equivalent with $ \frac{8x(\minus{}x\plus{}b)(\minus{}x\plus{}a)}{(2x\plus{}a\minus{}b)(\minus{}2x\minus{}b\plus{}a)(a\plus{}b)}\equal{}0$ from here we get $ x_1\equal{}0$ , $ x_2\equal{}a$ and $ x_3\equal{}b$.
Sonnhard.

Dr Sonnhard could you please show us how you arrived at that equation because the factorization seems rather painful to me... (yes I'm not good at factoring large polynomials... :( )

hello, we have
$ \frac{(x\minus{}a)(2x\plus{}b\minus{}a)\plus{}(x\minus{}b)(2x\plus{}a\minus{}b)}{(2x\plus{}a\minus{}b)(2x\plus{}b\minus{}a)}\equal{}\frac{2x\minus{}a\minus{}b}{a\plus{}b}$
this is equivalent with
$ (x\minus{}a)(2x\plus{}b\minus{}a)(a\plus{}b)\plus{}(x\minus{}b)(2x\plus{}a\minus{}b)(a\plus{}b)\equal{}(2x\minus{}a\minus{}b)(2x\plus{}a\minus{}b)(2x\plus{}b\minus{}a)$ expanding this we have
$ 4\,{x}^{2}a\plus{}4\,{x}^{2}b\minus{}4\,xba\minus{}2\,x{b}^{2}\minus{}2\,x{a}^{2}\minus{}a{b}^{2}\plus{}{a}^{3
}\minus{}{a}^{2}b\plus{}{b}^{3}\equal{}$
$ 8\,{x}^{3}\minus{}4\,{x}^{2}b\minus{}4\,{x}^{2}a\minus{}2\,x{b}^{2}\plus{}4\,xba\minus{}2\,x{a}^{2}\minus{}{a}^
{2}b\plus{}{a}^{3}\plus{}{b}^{3}\minus{}a{b}^{2}$
subtracting the right-hand side from the left-hand side we get
$ 8x^2a\plus{}8x^2b\minus{}8xab\minus{}8x^3\equal{}\minus{}8x(\minus{}x\plus{}b)(\minus{}x\plus{}a)$.
Sonnhard.


That is also my solution  :wink: But what is value of $[a_{1}+a_{2}+...+a_{2005}]$?

Where $[x]$ is integer part of $x$.

[quote="N.T.TUAN"]That is also my solution  :wink: But what is value of [/quote] $[a_{1}+a_{2}+...+a_{2005}]=1$. Use $sinx>x-\frac{x^{3}}{6}$.


inequalities
AMC
USA(J)MO
USAMO
inequalities proposed

[quote="Fermat -Euler"][hide=" my solution"] we have $x^3+y^3\geq xy(x+y)$
then $\frac{1}{x^3+y^3+xyz} \leq \frac{1}{xy(x+y)+xyz} = \frac{z}{xyz(x+y+z)}$
and $\frac{1}{y^3+z^3+xyz} \leq \frac{x}{xyz(x+y+z)}$
$\frac{1}{z^3+x^3+xyz} \leq \frac{y}{xyz(x+y+z)}$
 then $\frac{1}{x^3+y^3+xyz}+\frac{1}{y^3+z^3+xyz}+\frac{1}{z^3+x^3+xyz} \leq \frac{z+y+x}{xyz(x+y+z)} = \frac{1}{xyz}$ [/hide][/quote]
Very nice :clap2:

Hi, this has been posted for at least 10 times already I'm quite sure.

[quote="bomb"]Isn't this from past USAMO

Bomb[/quote]

Yes, it is [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=948]from USAMO 1997[/url]. This is also widely discussed in Vietnam. Here is a solution I know.

We use that fact that $ a^3 \plus{} b^3\geq ab(a \plus{} b)$ which is easily varified. Now we have
\[ \frac1{x^3 \plus{} y^3 \plus{} xyz}\leq\frac1{xy(x \plus{} y) \plus{} xyz} \equal{} \frac1{xy(x \plus{} y \plus{} z)} \equal{} \frac {z}{xyz(x \plus{} y \plus{} z)}.
\]
We can have two other similar inequalities and add them to get the desired inequality
\[ \frac1{x^3 \plus{} y^3 \plus{} xyz} \plus{} \frac1{y^3 \plus{} z^3 \plus{} xyz} \plus{} \frac1{z^3 \plus{} x^3 \plus{} xyz}\leq\frac1{xyz}.
\]


#1- Consider the set $ S\equal{}\{1,2,3,...34\}$. How many ways are there to choose (without regard to order) three numbers from S whose sum is divisible by 3?

#2- Robert has 4 gold coins and 4 silver coins (indistinguishable). Each coin has an engraving on one side, but not on the other. He wants to stack the eight coins on a table into a single stack so that no two adjacent coins are face to face. Find the number of distinguishable arrangements possible of the 8 coins.

[hide="#1"]First of all, note that there are 11 numbers of form $ 3k$, 12 of form $ 3k \plus{} 1$, and 11 of form $ 3k \plus{} 2$.

There are 4 methods that you can choose 3 numbers whose sum is multiple of 3:

1) 3 $ 3k$'s - $ \binom{11}{3}$
2) 3 $ (3k \plus{} 1)$'s - $ \binom{12}{3}$
3) 3 $ (3k \plus{} 2)$'s - $ \binom{11}{3}$
4) Each of $ 3k, 3k \plus{} 1, 3k \plus{} 2 \minus{} 12\cdot11\cdot11$

Summing up, there are 2002 ways that you can pick 3 numbers so that their sum is multiple of 3.[/hide]

@grn_trtle - Edit 2:
Lulz, blanked out

[hide="#2"]Let U denote a coin that's facing upwards, and let D denote a coin that's facing downwards.

There are 9 cases: DDDDDDDD, DDDDDDDB, DDDDDDBB, ..., BBBBBBBB.

You can arrange 4 gold and 4 silver coins in $ \binom{8}{4}\equal{}70$ ways.

Hence, the answer is $ 70\cdot 9\equal{}630$.[/hide]

[hide="Number 1"]Take $ S\bmod 3$ to find $ S\equiv\{1,2,0,1,...,0,1\}$.  We have $ 11$ zeroes, $ 12$ ones, and $ 11$ twos.

If we choose $ 3$ elements that are divisible by 3, we have these cases:

[b]Case 1.[/b] $ 0 \plus{} 0 \plus{} 0$. This, of course, happens in $ \binom{11}{3}$ ways.

[b]Case 2.[/b] $ 1 \plus{} 1 \plus{} 1$. This happens in $ \binom{12}{3}$ ways.

[b]Case 3.[/b] $ 2 \plus{} 2 \plus{} 2$. Happens in $ \binom{11}{3}$ ways.

[b]Case 4.[/b] $ 0 \plus{} 1 \plus{} 2$. $ 11\cdot 12\cdot 11$.

Then our answer is $ \boxed{2002}$.[/hide]

Edit: Oh, beaten. Why do we have to divide by $ 6$ in Case 4?
Edit 2: Heh, apparently we don't. Cool ;)

[hide="Number 2"]
We can have TTTTTTTT, TTTTTTTH, etc, etc, for 9 configurations. $ 9\cdot \binom{8}{4}\equal{}\boxed{630}$.
[/hide]

Edit: Hurrrr. Beaten again.

[quote="modularmarc101"]But it says "so that no two adjacent coins are face to face"[/quote]
Neither solution violates this condition. Note that tail to tail is allowed.


floor function
LaTeX
algebra unsolved
algebra

Let $ a$ be a positive real number and $ n$ a non-negative integer. Determine $ S\minus{}T$, where
$ S\equal{} \sum_{k\equal{}\minus{}2n}^{2n\plus{}1} \frac{(k\minus{}1)^2}{a^{| \lfloor \frac{k}{2} \rfloor |}}$  and  $ T\equal{} \sum_{k\equal{}\minus{}2n}^{2n\plus{}1} \frac{k^2}{a^{| \lfloor \frac{k}{2} \rfloor |}}$

i dont really know latex so i am not gonna post a solution
if u split up the summation into two summations that go in steps of 2
from -2n to 2n and -2n+1 to 2n+1
then the denominator becomes simple to work with
I got 0 for the answer

Subtract the two sums to get $ \sum_{k\equal{}\minus{}2n}^{2n\plus{}1} \frac{\minus{}2k\plus{}1}{a^{| \lfloor \frac{k}{2} \rfloor |}}$. Consider the sum of the terms $ k\equal{}\minus{}2n$ and $ k\equal{}2n\plus{}1$. $ \frac{4n\plus{}1}{a^n} \plus{} \frac{\minus{}4n\minus{}1}{a^n} \equal{} 0$. Pair all the rest terms up in a similar fashion to get $ 0$.


how would a middle school register for the AMC 10/12 if it doesn't have a CEEB code?

The contest manager from the school calls the AMC office, and the office assigns a "fake" CEEB number.  The office creates the new number according to internal rules that we use for coding the school type.

Steve Dunbar
MAA Director, American Mathematics Competitions

Okay, thanks!

But would it also be possible to just take it at a high school that's also taking it, since I'm the only one in my school who has any interest in it.

I did that also.

If there is no interest at your school, I doubt that they would register anyway :wink:


linear algebra
matrix
linear algebra unsolved

Find all linear maps on $M_n(C)$ that preserve the invertibility, that is if $A$ is invertible then $f(A)$ is also invertible. Since this is really difficult, a hint is not bad from the beggining: prove that they preserve the rank and thus we have all of them by Leva's fantastic proof.

Here it is:
First,we will prove that $A$ is invertible iff $f(A)$ is invertible.
 It is easy to see that for a given $A$ of rank $r<n, \exists P\in GL_n(C)$,such that $\forall x, P-xA$ is invertible.
Now let $A$ of rank $r<n$.Let $P$ such that $P-xA$ is invertible $\forall x$.Thus ..$f(P)-xf(A)$ is invertible.If $f(A)$would be invertible then $f(P)-xf(A)$ would not $\forall x\in Sp(f(P)(f(A)^{-1}))$.And we have finished.
Now we will prove that $rank(f(A))\geq rank(A)$.
 It is also easy to see that  for a given $A$ of rank $r<n, \exists P\in GL_n(C)$,such that $P-xA$ is noninvertible for exactly $r$ values of $x$.We deduce from this ,that  for these values $f(P)-xf(A)$ is also noninvertible and their inverses are eigenvalues for the matrix $f(P)^{-1}f(A)$.We get the result.
 This give us also that $f$ is an injective endomorphisme,thus bijective and by applying the previous step to the $f^{-1}$.
we get the result.


group theory
abstract algebra
symmetry
superior algebra
superior algebra unsolved

If $|X|=|Y|$ and $X,Y$ are infinite sets. Is $S_{X}\simeq S_{Y}$

It is much more interesting to ask whether the converse also holds, i.e., whether the sets $X$ and $Y$ are of the same cardinality if and only if $S_{X}$ and $S_{Y}$ are isomorphic as groups ;)

[quote="mlok"]Yes, with an isomorphism easily obtained from a bijection $f\colon X\to Y$.

In general, two sets of the same cardinality can be treated as the same, unless they are equipped with some additional structure or their elements are of interest.[/quote]
Yeah that works. After I posted my question I realized how embarassingly simple it was  :blush:

to fedja: that's more interesting.

answer to fedja's "more interesting" question:

let's notice that any non-trivial normal subroup of $S_{X}$ has cardinality at least $|X|$, since you can take a non-trivial element and "take it around" by conjugation: to be a little more precise, if $\sigma$ sends $x$ to $x'\neq x$, you can conjugate it by $\tau_{y}= (x' y)$ and obtain a different element of $S_{X}$.
now take an isomorphism $\varphi: S_{X}\to S_{Y}$: $S_{X}$ has a normal subgroup of cardinality $|X|$, which is the subgroup of [i]finite[/i] permutations (moving a finite subset of $X$), so does $S_{Y}$, and so $|X|\ge|Y|$.
by symmetry (and cantor-bernstein :p) we have the result.

by the way: is it true that $|S_{X}| = |S_{Y}|$ iff $|X|=|Y|$? in other words, is $|2^{X}| = |2^{Y}|$ iff $|X|=|Y|$?

Nice solution, ma_go! The question of yours was asked by grobber in [url]http://www.artofproblemsolving.com/Forum/viewtopic.php?t=4471[/url]. Gauss202 proposed a solution, which was obviously wrong. Since then, it is one of the "open questions" on this forum, to which I don't know the answer myself.

thanks, fedja ^^

anyway, about the stronger proposition, i've been talking to a friend, and he said our teacher said, during a lesson, that this problem should be actually independent from ZFC.
though generalized continuum hypothesis (GCH) implies it (obviously).

i don't know if it's weaker or whatever, i'll ask someone about it (and let you know something, if someone actually cares :p)