show that  y' = y^2 + x^2,  y(0)=0,   has a solution in |x|<= 1/Sqrt(2).
I have a theorem stating that a solution exists in (-h+0,0+h) where h=min(a,b/M)  M is the maximum value of f(x,y)=x^2+y^2, (in this case) in a closed recangle around (x0,y0) in width 2a ad height 2b
(in this case x0=0 and y0=0).

I was given a hint that  x^2 + y^2 >= 2xy for positive x,y
but that seems to evaluate b/M from the wrong direction for me, i get b/M = b/b^2+a^2  <=  b/2ab = 1/2a

what do i do?

any help is welcome

thanks

If $|x|<\frac1\sqrt{2},$ then

$y^{2}<y'<y^{2}+\frac12.$

You can explictly solve both $y'=y^{2}$ and $y'=y^{2}+\frac12.$

How far will that observation take you?

[quote="The_Seer"]show that  $y' = y^{2}+x^{2}, \, y(0)=0$,   has a solution in $|x|\leq  1/\sqrt{2}$.[/quote]You can do better than that. Clearly $y$ is an odd function of $x$ and $y > 0$ for $x > 0$, so all you need is a positive function $\tilde y$ that satisfies $\tilde y' \ge{\tilde y}^{2}+x^{2}$ on an interval that includes $[0,\, 1/\sqrt{2}]$. So try $\tilde y = \alpha x^{3}$ and choose the right $\alpha$.


geometry
rectangle
combinatorics unsolved
combinatorics

We have a grid of $k^2-k+1$ rows and $k^2-k+1$ columns, where $k=p+1$ and $p$ is prime. For each prime $p$, give a method to put the numbers 0 and 1, one number for each square in the grid, such that on each row there are exactly $k$ 0's, on each column there are exactly $k$ 0's, and there is no rectangle with sides parallel to the sides of the grid with 0s on each four vertices.

This was a competition problem?...

First note that $k^2-k+1=\left(p+1\right)^2-\left(p+1\right)+1=p^2+p+1$, so our grid actually has $p^2+p+1$ rows and $p^2+p+1$ columns. Now we consider the projective plane $\mathbb{P}^2\left(\mathbb{Z}_p\right)$ over $\mathbb{Z}_p$ (the field of residues modulo $p$). This plane has $p^2+p+1$ points and $p^2+p+1$ lines. We can arbitrarily define a bijection between the $p^2+p+1$ rows of our grid and the $p^2+p+1$ points of the projective plane, and a bijection between the $p^2+p+1$ columns of our grid and the $p^2+p+1$ lines of the projective plane. Any square of the grid corresponds to a pair of a row and a column of the grid, and thus to a pair $\left(P;\;\ell\right)$ with $P$ being a point and $\ell$ a line in our projective plane. Let's write a "0" in that square if the point $P$ lies on the line $\ell$, and a "1" otherwise. Then, it is easily seen that the conditions of the problem are satisfied:

- On each row, there are exactly $k = p + 1$ squares with "0"s, since each point in $\mathbb{P}^2\left(\mathbb{Z}_p\right)$ lies on exactly $p + 1$ different lines.
- On each column, there are exactly $k = p + 1$ squares with "0"s, since each line in $\mathbb{P}^2\left(\mathbb{Z}_p\right)$ passes through exactly $p + 1$ different points.
- There is no rectangle with sides parallel to the sides of the grid with "0"s on each of its four vertices; in fact, if such a rectangle would exist, then its two horizontal sides (rows) would correspond to two distinct points $P$ and $Q$, its two vertical sides (columns) would correspond to two distinct lines $\ell$ and $\ell^{\prime}$, and since all four vertices of the rectangle have "0"s written in them, we would have $P\in\ell$, $Q\in\ell$, $P\in\ell^{\prime}$, $Q\in\ell^{\prime}$, so the two distinct lines $\ell$ and $\ell^{\prime}$ would have two points $P$ and $Q$ in common, but this is impossible (two distinct lines intersect at one point only).

Thus we have found the required positioning of "0"s and "1"s on the grid.

  Darij


Given four circles $(O),(O_{1}) ,(O_{2}),(O_{3})$ such that:
- $(O_{1}),(O_{2}),(O_{3})$ touch $(O)$ internaly at the points $A,B,C$ ,respectively.
- $(O_{1})$touch $(O_{2})$ externaly at point $C_{1}$, $(O_{2})$touch $(O_{3})$ externaly at point $A_{1}$, $(O_{3})$touch $(O_{1})$ externaly at point $B_{1}$
Prove that $AA_{1},BB_{1},CC_{1}$ are concurence
I solved this problem with the inversion and the similitude, who can help me with other solution ? Thank you

I'm leaving it to you to find a proof based on http://www.mathlinks.ro/Forum/viewtopic.php?t=67092 . I think your problem has been discussed elsewhere on ML as well...

  darij


LaTeX
algebra
binomial theorem
combinatorics proposed
combinatorics

Find the number of ordered pairs $ (A,B)$ of subsets of $ \{1, 2, ..., n\}$ such that union of $ A$ and $ B$ contains even number of elements.

I'm not sure if my solution is correct but i found
n even ---> 2^(n-1)(2^n +1)
n odd   ---> 2^(n-1)(2^n -1)
I would like to post full solution but I don't know how to use Latex or whatever is the program xD

See [[LaTeX]] or [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=123690]this thread[/url] for the basics of LaTeX.  You don't need to install anything, just surround your properly-formatted commands by a pair of dollar signs.  Also, your answer is correct.

Thanks for the help.
That's my solution:

First we fix one subset of size $ k$, with $ 0<\equal{}k<\equal{}n\minus{}1$ now we will construct the other subset so that the union has an even number of elements. Realize that the intersection of the subsets doesn't really matter so the second subset can have any element that is already in the first. Now for the other $ n\minus{}k\minus{}1$ elements they can be or not in the second subset so multiply by $ 2^{n\minus{}k\minus{}1}$ and by fixing these elements the last one is already determined because he will be in the second subset iff the union of $ A$ and $ B$ has an odd number of elements. So we have:

$ \sum_{k\equal{}0}^{n\minus{}1}\binom{n}{k}2^k2^{n\minus{}k\minus{}1} \equal{} 2^{n\minus{}1}\sum_{k\equal{}0}^{n\minus{}1}\binom{n}{k} \equal{} 2^{n\minus{}1}(2^n\minus{}1)$

But now we just need to see that if $ n$ is even and $ A$ has $ n$ elements then every subset $ B$ works. And if $ n$ is odd none of then can have $ n$ elements, so:

$ n$ odd    ---> $ 2^{n\minus{}1}(2^n\minus{}1)$
$ n$ even  ---> $ 2^{n\minus{}1}(2^n\plus{}1)$

That's nice.  I approached it as follows: we must choose a $ 2k$-element subset of $ \{1, \ldots, n\}$ to be the union of $ A$ and $ B$ for some $ k$.  Then each element of this subset can be in $ A$, in $ B$, or in both.  This gives us $ \sum_{0 \leq 2k \leq n\} \binom{n}{2k}3^{2k}}$ for our desired value.  We can recognize this (via the binomial theorem) as $ \frac {1}{2}\left((1 - 3)^n + (1 + 3)^n\right)$, an equivalent expression to yours.


People, read this article written by a guy from my college about socialism, say what you think about it.

http://www.thetriangle.org/media/paper689/news/2005/07/08/EdOp/Socialism.Consistently.Fails.To.Deliver.Freedom-961352.shtml

Well he's right, but he's only right on the scale of if you wanted to run an entire country on the [i]same[/i] socialist society. Then it would fall apart, as he said. But on a smaller scale, look at how a kibbutz in Israel works. It is basically a [b][i]successful, real[/i][/b] socialist community. A kibbutz just proves that socialism can and does work, but it has to be done right. Small communities where everyone gets along can work. When he talked about his roommate buying the stuff and everything and splitting costs - that doesnt work because they are buying things from a non socialist society where you have to pay for things. In a kibbutz nobody has their own money, and everyone works together towards a collective goal - and it works. Everyone gets as much food as they need and is under no better conditions than anyone else. If the world was all devided up into millions of little kibbutzim - it would work. Also, there could be no 'leader' - to get things started, to get the community started people would do a vote, and one leader would maybe tally the votes or lead it, but afterwards everyone works in peace - you have people volunteer for farm work, some for the kitchen etc. etc. Its a boring life in my opinion, but its peaceful. Also - communist russia partly didnt work because it tried forcing people into it, and there was a leader, and it was done all wrong. And yes, because it believed in humans being innately good it didnt work. But if you have similar people with the same mindset working on a small community scale then it works.  :lol:

[quote="ToBe"]A kibbutz just proves that socialism can and does work, but it has to be done right. [/quote]

I think a kibbutz proves nothing of the sort.  Would a kibbutz work if it were not externally protected my a larger nation?  I can think of no such successful experiment.

[quote="ToBe"]well the kibbutz IS protected by an external government - but if the world ran all on the same small communites then socialism is possible. Its not probable at all and won't happen simply because of the way the world is today - but its possible.[/quote]

Even if everyone joined hands and shared, I think the belief that there won't be people who will feel entitled to reward relative to their effort is a second layer of idealism that seems far from reality.

I know that I wouldn't work as hard as I do if there were no possibility of creating a significantly better life for myself and those around me.

Also -- why is socialism only about wages?  Shouldn't movie stars have to give up some of their overly abundant share of good looks?  There is more to labor than what is monetary and money is nothing more than an exchange of labor.

I showed what you guys posted in thetriangle forum to which I can give you the link, not sure if you will like it or not.

http://www.thetriangle.org/administration/index.cfm

http://www.thetriangle.org/main.cfm?include=forums&message_id=25328&action=postreply

This is the discussion on the forums abotu that article.


Let $ N$ be the set of the whole of positive integers. The mapping from $ N$ to $ N$ is defined as follows: $ p(1)=2,\ p(2)=3,\ p(3)=4,\ p(4)=1,\ \ \ p(n)=n\ (n\geq 5)$,
$ q(1)=3,\ q(2)=4,\ q(3)=2,\ q(4)=1,\ \ \ p(n)=n\ (n\geq 5)$. Answer the following questions.

(1) If you make a mapping $ f: N\rightarrow N$ sucessfully, we have $ f$ such that $ f(f(n))=p(n)+2$. Give an example.

(2) Prove that it is impossible that  $ f(f(n))=q(n)+2$ holds in regardless of any definition for $ f: N\rightarrow N$.

Hi

I'm new here and thought I'd try this little problem here.

I hope my solutions are correct:

$ 1) f(f(1)) \equal{} 4, f(f(2) \equal{} 5, f(f(3)) \equal{} 6, f(f(4)) \equal{} 3, f(f(n)) \equal{} n\plus{}2$ for $ n\geq 5$
$ f(1) \equal{} 5, f(2) \equal{} 1, f(3) \equal{} 9, f(4) \equal{} 7, f(5) \equal{} 4, f(6) \equal{} 11, f(7) \equal{} 3$
$ f(2n) \equal{} 2n\plus{}5, f(2n\plus{}1) \equal{} 2n\minus{}2$ for n$ \geq 4$ This is quite obviously a valid function

$ 2)$
$ f(f(1))\equal{}5, f(f(2))\equal{}6, f(f(3))\equal{}4, f(f(4))\equal{}3, f(f(n))\equal{}n\plus{}2$ for $ n\geq 5$
Let's assume $ f(3)\equal{}a$ and $ f(4)\equal{}b$
$ f(f(3))\equal{}f(a)\equal{}4\to  f(f(a))\equal{}f(4)\equal{}b$
$ a) a\geq 5\to  f(f(a))\equal{}a\plus{}2\equal{}b\to  b\geq  5\to  f(f(b))\equal{}b\plus{}2\equal{}a$ Contradiction!
$ b) a<5$
It's easy to check that this is not possible either.


for $a , b, c \geq 0$

prove that

$(a^3 + b^3 + c^3 + 3abc )^2 \geq 2 (\sum_{cyclic} ab )(\sum_{sym} a^3 b )$

$(a^3 + b^3 + c^3 + 3abc )^2 \geq (\sum_{sym} a^2 b )^2 \leq (\sum_{sym}ab )(\sum_{sym} a^3 b)$

First by Schur's, then by Cauchy Schwartz.

EDIT: Oops it is the wrong way round! I have edited the ineq signs.

please, more detail.. cauchy?

I think you are in wrong way..

using cauchy make opposite inequality

this ineq's gap is much smaller than cauchy ineq.  :D  :D

you are right, i realised just afterwards.


linear algebra
matrix
linear algebra unsolved

let A be a $ 3\times 3$ matrix. trace(A)=8, detA=16, and the sum of each row of A is equal to 4. find all eigenvalues of A

"the sum of each row of A is equal to 4" so 4 is an eignevalue.
To find the other 2, $ x$ and $ y$, you solve $ 4\plus{}x\plus{}y\equal{}8$ and $ 4xy\equal{}16$.

thanks. but can you show me why "the sum of each row of A is equal to 4" so 4 is an eignevalue ?
 :blush:

Take $ X\equal{}(1,1,1)^T$, you have $ AX\equal{}4X$.

If each row (or column) of the matrix A sum up to the same value k, then k is an eigenvalue of A, because then (1,1,1)' is an eigenvector with k as associated eigenvalue. Also, if $ \lambda_1$, $ \lambda_2$, ..., $ \lambda_n$ are the eigenvalues of $ A \in M_{n \times n}$, then

$ tr(A) \equal{} \sum_{k\equal{}1}^n \lambda_k$

and

$ det(A) \equal{} \prod_{k\equal{}1}^n \lambda_k$.


inequalities
logarithms
real analysis
real analysis unsolved

Let $x, y$ be two positive real numbers such that $1\leq y\leq x$.
   Show that \[(x-y)\log\left(1+\frac{1}{x}\right)\log\left(1+\frac{1}{y}\right)\leq\log\frac{x(y+1)}{y(x+1)}.\]

Are you sure the second $log$ is not meant to be $log(1+\frac{1}{y})$ ?

I think the LHS of the inequality needs a minor correction.

Yeah, with $\frac{1}{y}$, the inequality becomes trivial by using $log(1+x) \leq x$ and $log(x) \geq x-\frac{x^{2}}{2}$.

Hi,
I think this is an idea:

The inequality is equivalent to:

$\frac{f(x)-f(y)}{x-y}\geq1$, where $f(x)=\frac{1}{\ln \frac{x+1}{x}}$
Based on the mean value theorem we know that there exists $c\in (y,x)$ such that the difference above is $f'(c)$. Thus we show that
$f'(c)\geq 1$.
A calculation shows that
$f'(x)=\frac{1}{x(x+1)\ln^{2}\frac{x+1}{x}}$
and
based on some computations we get that $f''(x)<0$, thus we get that
$f'(c)>f'(x)>f'(1)=\frac{1}{2\ln^{2}2}>1$.

I hope that there are no computational mistakes, anyway this is the idea.
Didi

Let $f(t)=t-\frac{1}{\log \left(1+\frac{1}{t}\right)}\ (t\geq 1)$

$=\frac{1}{\frac{1}{t}}-\frac{1}{\log \left(1+\frac{1}{t}\right)}\ \left(0<\frac{1}{t}\leq 1\right).$ Thus $f(t)$ always takes negative values and decreasing function.

Now for $1\leq y\leq x,$ we can rewrite the inequality in question as $f(x)\leq f(y)<0,$ which completes the proof.

Using the inequalities I gave above, it is enough to check that:

$\frac{(x-y)}{xy}\leq \frac{(x-y)}{y(x+1)}-\frac{(x-y)^{2}}{2[y(x+1)]^{2}}$, which appears to be true after simplification.


You have $2$ circles, $3$ ellipses and $3$ lines in a plane. What is the maximum number of intersections possible.

go step by step:

2 circles: 2
add a line: 4 more
adda line : 5 more
add a line: 6 more
add ellipse: 12 more
add ellipse: 16 more
add ellipse: 20 more

Max: 55 intersections

correct me if i am wrong

Optimally, each ellipse can intersect a circle four times and a line twice, so the first ellipse should add $2(4)+3(2) = 14$ intersection points, the second $18$, and the last $22$. So the answer should be $61$. Of course, drawing a picture showing the maximum can be attained doesn't sound fun.

Oh oops. Made a dumb mistake. Yup, paladin is right i think.

I get [hide].  but I'm probably not right[/hide]

Nice blog by the way, paladin.

[quote="Franz Joseph"]I think the answer is a bit more than $61$.[/quote]

Maybe $71$; I just added $6$ to neelnal's answer, which had been added incorrectly in the first place.

Anyone bored?

EDIT: I believe I have made $71$ points.

I think we can arrange the objects so that each intersects with each in maximum number of points and no intersection point belongs to more than two objects. Hence

1. The two circles can intersect in at most $2$ points.

2. Every two ellipses can intersect in at most $4$ points, hence we can have ${3\choose 2}\cdot 4=12$ points.

3. Every two lines can intersect in at most one point, hence we can have ${3\choose 2}\cdot 1=3$ points.

4. Every circle and every ellipse can intersect in at most $4$ points, hence we can have $2\cdot 3\cdot 4=24$ points.

5. Every circle and every line can intersect in at most $2$ points, hence we can have $2\cdot 3\cdot 2=12$ points.

6. Every ellipse and every line can intersect in at most $2$ points, hence we can have $3\cdot 3\cdot 2=18$ points.

Therefore the grand total is $71$ points, and for the illustration see paladin8's diagram.


[quote="DaMasterbomb"]Given: Circle $O$ and Circle $Q$ are coplanar, with radii $r$ and $s$, respectively; $r>s$ , $OQ=d$ , $r-s<d<r+s$
Prove: Circle $O$ and Circle $Q$ intersect in two points[/quote]

This is incredibly easy by contradiction - if the circles intersect at one point, r + s = d (or r - s = d, if they're internally tangent), and if they don't intersect at all, r + s < d. (OQ has length d.) If one circle is completely inside another, then d < r - s.

Wow, I completely missed the proof by contradiction. I did a about a page of algebra in order to prove that for three cases of a point A, there would be two points that would fit the equation for the radii of both circles...lolz

[hide]
$r-s<d<r+s$ ($r>0$, $s>0$ and $d>0$), so $r$, $s$ and $d$ form a triangle. Then the two circles obviously intersect at two points.
[/hide]


One of my friends sent me another way of writing 1+1=2. I think it's nice and easy   :D

p.s. it was a pps but i couldn't atach it .:(

It's an extension for a archive. After you extract it you'll see that 1+1=2 in a different form :D

My math teacher is going to be like  :rotfl:

Did you liked it :D? We also know that $\lim_{n\rightarrow\infty}{\frac{1-\cos\frac{\pi}{x_{n}}}{\left(\frac{\pi}{x_{n}}\right)^{2}}=\frac{1}{2}}$. :P

can you write it in latex so i can see it?

[quote="math92"]it won't work for me... :([/quote]

me either

Do you have a program which can extract RAR files? If not, you can get one at [url=http://www.7-zip.org]www.7-zip.org[/url]. After that run the program and it should convert the RAR into several JPGs.

isn't $\displaystyle\lim_{z\to\infty}\displaystyle\left(1+\frac{1}{z}\right)^2$ in the 5th slide supposed to be $\displaystyle\lim_{z\to\infty}\displaystyle\left(1+\frac{1}{z}\right)^z$ instead of to the power of 2? the fourth slide says to the power of z and the fifth to the power of two.. i think it's to the power of z...

but either way, very cool  :play_ball:  :read:  :juggle:

[quote="Klebian"]what's .rar?
I can't open it...[/quote]

It's like zip file to open it you must download [url=http://www.rarlab.com]WinRar[/url] for example!

 :)


geometry
circumcircle
parallelogram
cono sur

Let $H$ be the orthocenter of the triangle $ABC$, and let $M$ be the midpoint of $BC$. $X$ is the point of intersection of $HM$ and the arc $BC$ of the circumcircle which doesn't contain $A$. $Y$ is the second point of intersection of the $BH$ and the circumcircle. Show that $XY=BC$.

[quote=hikaru123][hide="Hint"]
$OM=\frac12AH$
So it's easy to get M is the midpoint of HY
BYCH is a parallelogram
YB is perpendicular to AB
$\widehat{BAY}=90^{o}-\widehat{C}=\widehat{XBC}$
So $%Error. "wide" is a bad command.
{BAC}=\widehat{XBY}$[/hide][/quote]

Instead of $M$ is the midpoint of $HY$ and $BYCH$ is a parallelogram, the correct is
$M$ is the midpoint of $HX$
$BXCH$ is a parallelogram
$XC$ is perpendicular to $AC$.
Thanks


If you're interested in having math chat today, please take poll and post here.

I'm not sure how many people are supportive so this is why I'm doing it.

Thanks.

Where do you have it? But sure, i'm interested.

We are hosting it at 7:30 P.M. Eastern and the room is the usual:

Math Chat

I hope people attend! Based on who comes we might do different problems.

Happy Holidays.

Is the room on AoPS or some chat server?

[quote="sam3.14"]I would really like to participate in some math chats. How do I sign up?[/quote]

Just sign on AIM at the time when they are having the math chat.  Then IM someone involved with the math chat and they will send u an invite.


Let ft is a function which has f', f'', f'''
   Prove that exists a satisfying f(a)f'(a)f''(a)f'''(a)<0

It should be $\le$ (or $\ge$, I can't remember), and this is ptobably the third time this problem is being posted :).

[quote="grobber"]It should be $\le$ (or $\ge$, I can't remember), and this is probably the third time this problem is being posted :).[/quote]
It should be $\ge$.Here is the link:
http://www.mathlinks.ro/Forum/viewtopic.php?t=24592


linear algebra
matrix
probability
linear algebra unsolved

Pick a random $ 2 \times 2$ matrix $ A$ with independent, uniformly distributed entries from $ \{0,1,\dots,N\minus{}1\}$. Let $ p_N$ be the probability that this matrix is invertible modulo $ N$.

a) Show [i]or disprove? [/i] that $ p_N \to 1$ if $ N \to \infty$ and $ N$ is a prime number.

b) Show [i]or disprove? [/i] that $ p_N \equal{} \frac{3}{8}$ if $ N$ is a power of 2.

[i]I have numerical evidence for both statements, but no proofs.[/i]

Only the primes matter. Invertibility mod a power of 2 is exactly the same as invertibility mod 2.

The exact probability that a $ k\times k$ matrix mod $ N$ is invertible:
$ \prod_{p|N}\prod_{j\equal{}0}^{k\minus{}1}\frac{p^k\minus{}p^j}{p^k}\equal{}\prod_{p|N}\prod_{i\equal{}1}^{k}\left(1\minus{}p^{\minus{}i}\right)$


[quote="mizosome"]ok..this site is reely weird. ok
so, I don't how to do number 4 on the Week 3 Plane Geometry class problems. I don't get how to get the radii for the small circle.[/quote]
from jls?
post in the jls mathcounts forum.
and no, this site is cool

[hide="hint_1"]Draw the center of the circle (let's call it P) and the radii to all the tangent points. If we call the tangent point on OA C and the tangent point on OB D, what kind of shape is PCOD?[/hide]
Try the problem using that. If it still doesn't work,
[hide="hint_2"]PCOD is a square with side length r, where r is also the radius of the small circle. Try to prove it. Now, can you find the radius of the big quarter-circle in terms of r?[/hide]

post the problems... so we can do solutions


Potentially useful fact 1: $x^5 + y^5 = (x+y)(x^4 - x^3 y + x^2 y^2 - xy^3 + y^4)$
Potentially useful fact 2: $2004 = 2^2*401$

Sorry not to actually post a solution. I lack the time/intelligence (obviously "rate of completion $\approx$ time x intelligence") to complete the proofs of harder questions here, so I like to just put random stuff to try to help. I'll try to start a proof.

If there are solutions on $N$, we have $\{x,y,z\} \subset \{0,1,2,3,4\}$ because $5^5>2004$.
Now, $0^5 = 0, 1^5 = 1, 2^5 = 32, 3^5 = 243, 4^5 = 1024$. Clearly there are no set of three giving the sum 2004, so at least one of $x,y,z$ must be less than zero.

WLOG assume $x<y<z$ and (by the above) $x<0$.

$x^5 \equiv 0, 1, -1, 7, -7 \bmod 25$
$2004 \equiv 4 \bmod 25$

Edit:  So $x \equiv y \equiv z \equiv 3 \bmod 5$ (then their fifth powers are $\equiv -7 \bmod 25$).  This puts an egregious bound on how close $x, y, z$ can be to each other which almost certainly means no solutions.  :P


In the adjoining figure, circle K has diameter AB; circle L is tangent to circle K and to AB at the center of circle K; and circle M is tangent to circle K, to circle L and to AB. The ratio of the area of circle K to the area of circle M is

(A) 12
(B) 14
(C) 16
(D) 18
(E) not an integer

[hide="Solution"]
First, we can prove that B,M, and K are collinear. MK is a radius of the circle M, so the tangent to circle K is perpendicular to line MK. Similarly, BK is perpendicular to the tangent to circle K at B. However, because circle M is internally tangent to circle K, then the two tangents are the same because there is only one unique tangent to every point on a circle. Therefore, because there is only one line perpendicular to any line in 2D space, the two radii are on the same line and the three points are concurrent.

Now, we can use coordinate geometry to find the radius of the circle. Let the origen be K. We can find three equations representing the tangency of the circle M to circle L, line AK, and circle K. After solving, we discover that the radius of the circle is $\frac{1}{4}$ (we can let radius AK be length 1). Therefore, because the area of circle K is $\pi$ and the area of circle M is $\frac{\pi}{16}$, the ratio we are looking for is 16, so the answer is $\boxed{D}$.[/hide]


isn't anyone interested? or is it too easy for you to think about?

sorry. i forgot to mention that $a,b,c$ are the side lengths of a triangle. :blush:

[quote="nayel"]Let $a\le b\le c$ be the side lengths of a triangle such that $3a>b+c$. Prove that
\[\sum_{cyclic}\frac{a}{3c-a-b}\ge \frac{a+b+c}{\sqrt[3]{abc}}\]
[/quote]

$\sum_{cyclic}\frac{a}{3c-a-b}=\sum_{cyc}\frac{a^{2}}{3ca-a^{2}-ab}\geq \frac{(\sum{a})^{2}}{2\sum{ab}-\sum{a^{2}}}$

so it remains $\frac{(\sum{a})^{2}}{2\sum{ab}-\sum{a^{2}}}\geq \frac{\sum{a}}{\sqrt[3]{abc}}\Longleftrightarrow \sum{a}\cdot \sqrt[3]{abc}\geq 2\sum{ab}-\sum{a^{2}}$

but from schur : $2\sum{ab}-\sum{a^{2}}\leq \frac{9abc}{\sum{a}}$

so it remains: $\sum{a}\cdot \sqrt[3]{abc}\geq \frac{9abc}{\sum{a}}\Longleftrightarrow \sum{a}\geq \sqrt[3]{abc}$, which is just AM-GM

yes, you are right, we only need the condition that mentains the positivenes of our denominator: 3a>b+c

[quote="nayel"]:!:  you didn't use the fact that $a,b,c$ are the side lengths of a triangle so i guess the inequality is true even if $a,b,c$ don't satisfy the condition. and about $3a>b+c$ you only used that $3a-b-c>0$! in my proof i used them to form new conditions.[/quote] :lol: add $3a>b+c$ and $3b>a+c$, and you obtain $a+b>c$. So the conditions imply that $a,b,c$ are sides of a triangle.  :) But not equivalent of course, see (a,b,c)=(5,6,10)  :wink:


Find all pairs of positive integers $(a,b)$ such that $\sqrt{a-1}+\sqrt{b-1}=\sqrt{ab-1}.$

By taking both sides in square we will have:
a+b-2+2(sqrt((a-1)(b-1)))=ab-1, so (a-1)(b-1)=2(sqrt((a-1)(b-1))). Let, x=sqrt(a-1)(b-1).Then, x^2=2x. So,x=0 or x=2.If sqrt(a-1)(b-1)=o, a=1 and b is any positive integer, or b=1 and a is any positive integer. If sqrt(a-1)(b-1)=2, then (a-1)(b-1)=4. So, a=2,b=5 or a=5,b=2 or a=b=3.

in order to aviod many manipulations, set $ x\equal{}a\minus{}1$ and $ y\equal{}b\minus{}1$ substitute these in the given equation and the rest is easy... :wink:


Support
HCSSiM
geometry
geometric transformation
reflection

One of the most frequented e-mail providers,  [url=http://www.gmx.de]GMX[/url], in Germany, let people select either George Bush or John Kerry:

[i]Result[/i]: Bush received 8 % and John Kerry got 92 %.  151539 votes were submitted.

Also have a look at [url=http://217.160.163.211/globalvote2004/]Global Vote[/url]. The winner is John Kerry with an average of 70-80% of the votes. I was surprised by the votes from Middle East though not many people participarted from this region.

What do think about the huge difference of American and non-American behaviour ?

In the words of Robin Williams, presumably quoting Ronald Reagan:

"What would our country be without this great land of ours?"

uhhhh....I don't really seem to understand that quote.  What's it mean?

OR do you mean "be" instead of "by"?

I think the disparity is caused by President Bush's attitute toward the rest of the world.  He has dismissed the United Nations several times in the leadup to the war in Iraq and ridiculed John Kerry's more globally oriented way of approaching things.

[i]Result[/i]: Bush received 8 % and John Kerry got 92 %.  151539 votes were submitted.

wow.  i wonder how bush won then.

Y'know, that's somewhat like DC.  (Kerry got 90% of the vote there)

three sites that kerry auto won

california

DC

new york

if there are more, tell me.

[quote="Dark_Shadow"]Gameworld, that's in Germany.  Last time I checked Germany couldn't vote for American Presidents.[/quote]

Well, it might be true that Germans are not allowed to vore for the American President. But I want to emphasize that American people who wanted to vote outside the USA were blocked from the online election server for some security reasons. In Germany for example people just could vote by means of regular mail. Just the American soldiers in Germany could vote online. They were supposed to vote for Bush in 69% of the submittted votes. I have the impression that in general Americans living outside the country do not support Bush as much as people living in the country currently.

[quote="orl"] I have the impression that in general Americans living outside the country do not support Bush as much as people living in the country currently.[/quote]

Unpatriotic influences (not to say that Kerry is, but Germans generally dont support the war in Iraq)

Where I come from, Bush supporters are seen as some strange subhuman alien tribe. I imagine that's true in $\text{world}\setminus\text{USA}$ as well.

When I looked at the map, and as some one remarked: (Forgot  where but the experts were  commenting on our elections and were suggesting how   we  ought to  solve the whole problem by a simple divison)

USRS   (United States of Red States) Voted for Bush.
USBS   (United States of Blue States) Voted for Kerry.

USBS (NorthEastern/Great Lakes/Pacific West)  and USRS (Midwest and south)  are very clearly defined  except the state where I live in (OHIO), I think the we have to draw  "Berlin wall" right through  Columbus.  :)

What is scarry is  in the land of  some of  our Allies (in War of Terror, NNMA (Non Nato major Allly - Collin Powell's Best Friend's home) Bush's approval rating (less than 10%) was much lower than Osama(65%) .  - According to Galllup Polls...

[quote="cardwacko9999"][quote="gameworld7"]three sites that kerry auto won[/quote]

few more: Hawaii (not very important though)

the Northeast (all the states, dont have to list them all- Massechusets- his own state, Conneticut, etc.)

States that bush auto won:

Texas (his own states)

the Midwest (like above)

Alaska[/quote]

New Hampshire? New Jersey? Pennsylvania?

Colorado?

i think i obviosly stated that 3 states. i didn't know others so i said 3 states.

Someone was talking to me about the election and I guess Bush won most of the states in the centre of the country, and Kerry won the east and west coasts. Is that true?

In my school, Bush had the advantage.  But since I'm in Virginia, I suppose that should be expected.

Yup.

Miller's in California, right? California's a pretty liberal state, so you'd expect a number around that.

I live in Connecticut, and my school's mock elections predictably favored Kerry, about 60% to 40%.

In my school, also in California, Kerry got about 81.6% of the student vote and 63.4% of the staff vote. Naturally. But from what I heard, the more inland parts of the state, places with a lot of farms and stuff, actually favored Bush by a lot--they were just overruled by the highly populated coastal cities. It's an interesting question--why is there this huge difference between city residents and farm owners?

[quote="gameworld7"]i'm just saying...

miller middle school had a vote before the elections.  these were the results: 70% Kerry 30% Bush.

i'm just trying to say that in our school, Kerrry had the initiative.[/quote]

Do you ever have your facts right?

Kerry won Bush 70% to 20% with 10% going for independent.

[quote="MithsApprentice"]your school polls reflect the obvious - CA is a Democratic state...John Kerry won their electorial votes so if Bush got like 60% or something in ur school, that would be an odd school...

vice versa in texas where kerry gets like %20 of the votes in schools...but then of course in texas everyone's got a gun...[/quote]
Actually no...like I said before, the more inland parts of California, like the San Joaquin Valley and stuff, went for Bush on Election Day. A school mock election there would probably have Bush as the winner. And the area of the red regions in California is actually way larger than the area of the blue regions--it's just that we coastal cities have a much higher population density.

Yeah, funny thing, but Kerry didn't get as big a margin as was expected.  (Like Bush's margin in Utah or something, where there are more trees than people, j/k)

[quote="Super Naterul"]I'll tell you why Kerry gets so much from other places in the world:

      IF YOU WERE A TERRORIST WHO WOULD YOU VOTE FOR![/quote]

Oh, okay. So the whole entire world besides for the United States are terrorists. That's a fair and accurate assumption.

[quote="Super Naterul"]I'll tell you why Kerry gets so much from other places in the world:

      IF YOU WERE A TERRORIST WHO WOULD YOU VOTE FOR![/quote]Be carefull, some mods and even admins are not us citizens :P ... and usually they're not terrorists, but they can become for certain users ;)

In other words, not only US is affected by terrorism, but all free-democratic countries. Just this year terrorists struck in Madrid killing a lot of innocent people. And there are many other examples. It's a thing we all have to be afraid of ...  :maybe:

[color=green]Are you guys talking about being born in America or living in America? ;)[/color]

[quote="Super Naterul"]I'll tell you why Kerry gets so much from other places in the world:

      IF YOU WERE A TERRORIST WHO WOULD YOU VOTE FOR![/quote]

It's comments like this one that show the unfortunate American state of mind that elected Bush.

I thought of so many meaner ways of saying that, all of which could only be accepted on a forum other than this one.

While your restraint is appreciated, somehow I don't think some others would appreciate your obvious (as opposed to blatant) implication.  If nothing else, I do ask you to respect the political opinions of others, on behalf of Bush supporters (such as myself) on these forums.  To merely tolerate their opinions, while making snide comments about how idiotic we are and how proud you are of yourself for breaking it to us kindly, is impolite on a site as civil as this one.


The TeXnicCenter website is down, so I can't download it for now. Is it absolutely neccessary?

It seems to be up now -- I just downloaded the new version.

You don't need TeXnic Center; any text editor will work.  The advantage of TeXnic Center is that it has lots of TeX-related stuff preloaded into it.  Other text editors will likely have to be configured to run LaTeX.

I need a little help with TeXnicCenter:

I just got a new laptop and I downloaded the new version of MiKTeX, and it was downloaded and installed successfully.  Then, I tried to download TeXnicCenter, and it downloaded successfully but when I first opened the window, the configuration wizard asked where my executables were located, and I had no idea I even had executables like latex.exe, and tex.exe, let alone where they were.  I tried all sorts of places but it wouldn't work.  Here is some info on my computer:

I am running Windows XP Media Center 2005.  It is Vista-capable, and I do not know whether that has anything to do with it.

Also, the newest version of MiKTeX is 2.5.

My computer is a Inspiron E1705 gotten from Dell.

I installed MiKTeX in the directory C:\texmf.

I think you should also update your LaTeX download instructions, because somehow that wizard isn't acting right and it did not pose that question when I first downloaded it on dad's laptop (a Latitude X1).

HELP PLEASE!

A New topic will be better, if you need help! :wink:

[quote="mgao"]the configuration wizard asked where my executables were located, and I had no idea I even had executables like latex.exe, and tex.exe, let alone where they were.

I installed MiKTeX in the directory C:\texmf.[/quote]
They are probably in C:\texmf\miktex\bin. But all you have to do is use Windows search for latex.exe in C:\texmf and it will tell you where it is.


Hi,I said this inequality in that topic.Because I used a Lemma:
for $a,b,c>0$,then:
\[ a^xb+b^xc+c^xa \le 3  \]
where $x=\frac 3 2$
But I am sorry now I think it is hard to prove...Also I think the lemma is still right.
Arqady,Can you post your nice proof :)

[quote="zhaobin"]Hi,I said this inequality in that topic.Because I used a Lemma:
for $a,b,c>0$,then:
\[ a^xb+b^xc+c^xa \le 3  \]
where $x=\frac 3 2$
But I am sorry now I think it is hard to prove...Also I think the lemma is still right.[/quote]
I agree with you.
There is a way by passing this lemma.  ;)
[quote="zhaobin"]
Arqady,Can you post your nice proof :)[/quote]
I shall post  the proof later.  :)

Thanks,And it is acceptable that you post it later :)

I have another appoach now(But haven't completed)
\[ \sum_{cyc}\frac{a}{1+b^3} \ge \frac 3 2  \]
\[ \iff \sum_{cyc}\frac{\frac{a^2}{b^2}}{a/b^2+ab} \ge \frac 3 2  \]
then use Cauchy-Schwarz:
We should only to prove:
\[ 2(\sum_{cyc} \frac a b)^2 \ge 3(\sum_{cyc} \frac{a}{b^2}+ab+bc+ca)  \]
\[ \iff 2(\sum_{cyc} \frac a b)^2 \ge (a+b+c)(\sum_{cyc} \frac{a}{b^2})+3(ab+bc+ca)  \]
But I haven't prove the last one,So I can't sure it is ture or not :(
Can anybody prove or disprove it?

Unfortunately, I find that the last one is false :(
For a good $x$, the triple $(1-x,1-x,1+2x)$ doesn't satisfy the inequality...

[quote="arqady"]$ a > 0,b > 0,c > 0,a \plus{} b \plus{} c \equal{} 3.$ Prove that
$ \frac {a}{b^3 \plus{} 1} \plus{} \frac {b}{c^3 \plus{} 1} \plus{} \frac {c}{a^3 \plus{} 1}\geq\frac {3}{2}.$[/quote]
I am sorry, zhaobin. I forgot about this inequality.
Here is my proof:
$ \sum_{cyc}\frac {a}{b^3 \plus{} 1}\geq\frac {3}{2}\Leftrightarrow2\sum_{cyc}a(a^3 \plus{} 1)(c^3 \plus{} 1)\geq3\prod_{cyc}(a^3 \plus{} 1).$
But $ 2\sum_{cyc}a(a^3 \plus{} 1)(c^3 \plus{} 1) \equal{} \sum_{cyc}(ac^3(2a^3 \minus{} 3a^2 \plus{} 1) \plus{} 3a^3b^3 \plus{} 2a^4 \plus{} 2a \plus{} a^3b)\geq$
$ \geq\sum_{cyc}(3a^3b^3 \plus{} 2a^4 \plus{} 2 \plus{} a^3b).$
Thus, it remains to prove that:
$ \sum_{cyc}(3a^3b^3 \plus{} 2a^4 \plus{} 2 \plus{} a^3b)\geq3\prod_{cyc}(a^3 \plus{} 1).$
But $ \sum_{cyc}(3a^3b^3 \plus{} 2a^4 \plus{} 2 \plus{} a^3b)\geq3\prod_{cyc}(a^3 \plus{} 1)\Leftrightarrow$
$ \Leftrightarrow\sum_{cyc}(2a^4 \minus{} 3a^3 \plus{} a^3b \plus{} 1 \minus{} a^3b^3c^3)\geq0\Leftrightarrow\sum_{cyc}(a^4 \minus{} a^3c \plus{} 1 \minus{} a^3b^3c^3)\geq0,$ which is obvious.

[quote="xzlbq"][quote="arqady"]$a>0,b>0,c>0,a+b+c=3.$ Prove that
$\frac{a}{b^3+1}+\frac{b}{c^3+1}+\frac{c}{a^3+1}\geq\frac{3}{2}.$[/quote]

$a>0,b>0,c>0,a+b+c=3.$ Prove that
$\frac{a}{b^7+1}+\frac{b}{c^7+1}+\frac{c}{a^7+1}\geq\frac{3}{2}.$

BQ[/quote]

I am sorry,see:
http://www.artofproblemsolving.com/Forum/viewtopic.php?f=52&t=317977

[quote=My_dad][quote="arqady"]$a>0,b>0,c>0,a+b+c=3.$ Prove that
$\frac{a}{b^3+1}+\frac{b}{c^3+1}+\frac{c}{a^3+1}\geq\frac{3}{2}.$[/quote]

Here is a traditional C-S proof.
\[ \left( \sum_{cyc} \frac{x}{y^3+1} \right) \cdot \left( \sum_{cyc} x(y^3+1) \left(z+\frac{x}{4}\right)^2\right) \geqslant \left( \sum_{cyc} x \left(\frac{x}{4}+z\right)\right)^2 \]

Thus it remains to prove
\[ \left(\frac{x+y+z}{3} \right)^2 \cdot \left( \sum_{cyc} x \left(\frac{x}{4}+z\right)\right)^2  \geqslant \frac32 \cdot \left( \sum_{cyc} x(y^3+1) \left(z+\frac{x}{4}\right)^2\right) \]
The rest is obvious. If skeptical, use BW to verify. I am lazy, sorry ![/quote]

Yes,your saing is right!
else doing is this:
By C-S,have
\[\sum{\frac{x}{1+y^3}}\sum{x(z+x)^2(1+y^3)}\geq (\sum{x(x+z)})^2\]

Need to prove

\[{\frac { \left( x \left( z+x \right) +y \left( x+y \right) +z \left( y
+z \right)  \right) ^{2}}{x \left( z+x \right) ^{2} \left( 1+{y}^{3}
 \right) +y \left( x+y \right) ^{2} \left( 1+{z}^{3} \right) +z
 \left( y+z \right) ^{2} \left( 1+{x}^{3} \right) }}\geq \frac{3}{2}\]

This can be proved By SOS,but not nice!


function
real analysis
real analysis unsolved

Let

$f(x) = \sum_{n=1}^\infty \frac{1}{x+3^{n}}$

Then $f(0) = \frac{1}{2}$ and $f$ satisfies the recurrence relation

$f(3^{k}) = \frac{1}{3}\left(f(3^{k-1})+\frac{1}{1+3^{k-1}}\right)$

for integers $k$. Unfortunately I can't see how this will help. :P

Maple and I are both stumped. :(

It seems related to the generating function for the divisor function $d(n) =$ the number of divisors of $n$, but I can't touch that function, either.

will it be a new irrational number like the defitnation of $\pi,e$?


A point $ M$ is inside a convex quadrilateral $ ABCD$. Perpendiculars from $ M$ are dropped onto the sides of $ ABCD$, such that $ MN$ is perpendicular to $ AB$, $ MI$ to $ BC$, $ MH$ to $ CD$, and $ MK$ to $ DA$. Prove that $ 2[NIHK]\leq MA\cdot MC\plus{}MB\cdot MD$, where $ [...]$ denote the area of the polygon.


[quote="chess64"]1) Find $\lfloor kx \rfloor$ for a constant $k$ in terms of $\{ x \} = x - \lfloor x \rfloor$ and $\lfloor x \rfloor$.

2) Find $\lfloor x+y \rfloor$ in terms of $\{ x \}$,$\{ y \}$, $\lfloor x \rfloor$, and $\lfloor y \rfloor$.[/quote]

Im not sure bout this
 :?

2) $\lfloor x+y \rfloor$ = $x+y - \{ x +y\}$
=$\lfloor x \rfloor +\lfloor y \rfloor +\{ x \}+\{ y \}-\{\{x\}+\{y\}\}$

Well, that works I guess, but anyone else? :(

3) Prove that $\lfloor 2x \rfloor + \lfloor 2y \rfloor \ge \lfloor x \rfloor + \lfloor y \rfloor + \lfloor x+y \rfloor$.

i finally got to this in the aops book
here is my solution...
$\lfloor 2x \rfloor + \lfloor 2y \rfloor \ge \lfloor x \rfloor + \lfloor y \rfloor + \lfloor x+y \rfloor$
so x and y can be represented as x = a + b where a is the integer part of the number and b is the fractional part (the curly braces), the same goes for y = c + d
$\lfloor 2(a+b) \rfloor + \lfloor 2(c+d) \rfloor \ge \lfloor (a+b) \rfloor + \lfloor (c+d) \rfloor + \lfloor a+b+c+d \rfloor$
we can take out the integer parts a and b since they do not get affected by the floor function
$\lfloor 2(b) \rfloor + \lfloor 2(d) \rfloor \ge \lfloor (b) \rfloor + \lfloor (d) \rfloor + \lfloor b+d \rfloor$
since $0 \geq b,d>1$
so if $0 \geq b,d>0.5$, then they will be equal since all of the parts will be below 1 so they get floored to 0
if $0.5 \geq b , d>1$, (b or d or both satisfiy the previous) then $2b$ and/or $2d$ will get to one and the other side will be floored to 0 so the left side will be either 1 or 2 so the left is greater
we have shown all cases for $b$ and $d$ ($a$ and $c$ cancel) therefore the statement is correct, the left can either be equal to the right or it can be greater

i am pretty sure that this is rigirous but i am not absolutely sure, i bet it is different from the solutions text, their solutions are always so clever
EDIT: wow, i did almost the exact thing that they did (just in rreverse)...i am getting better at these problems


irrational number
real analysis
real analysis theorems

We start with an interval $ [0,1]$ and we construct sets $ A_n$ for all $ n\in\mathbb{N}$ in the following way:

1. $ A_1\equal{}[0,1]$ and do nothing

2. we divide $ [0,1]$ into $ 4\equal{}2^2$ length-equal parts and delete 2 of them: first and third one (delete without endpoints).
The rest is $ A_2$, so $ A_2\equal{}[0.25,0.5]\cup[0.75,1]$.

3. we divide $ [0,1]$ into $ 9\equal{}3^2$ length-equal parts and delete 3 of them: first, 4-th and 7-th (delete without endpoints).
The rest is $ A_3$.

4. we divide $ [0,1]$ into $ 16\equal{}4^2$ length-equal parts and delete 4 of them: these having numbers 1,5,9,13.
The rest is $ A_4$.

...........................................................

Finally let $ A\equal{}\bigcap_{n}A_n$. What kind of set is $ A$ ???

1) empty ?
2) countable ?
3) of Baire's first category ?
4) of measure 0 ?

I think $ A \equal{} \{1\}$

Clearly, $ 1$ belongs to $ A$.
Also, writing $ \bar{A}$ as the complement of $ A$ in $ [0,1]$:

$ \bar{A} \equal{} \bigcup_{i \equal{} 2}^{\infty}\bar{A_i}$ with $ \bar{A_i} \equal{} \bigcup_{k \equal{} 1}^{i}[\frac {i \minus{} k}{i},\frac {i \minus{} k}{i} \plus{} \frac {1}{i^2}[$

Now, let's consider any rational number $ \frac {x}{y}\in[0,1[$ (i.e. $ x<y$), the question is whether we can choose $ i$ and $ k$ such that $ \frac {i \minus{} k}{i}\le\frac {x}{y} < \frac {i \minus{} k}{i} \plus{} \frac {1}{i^2}$.
$ \frac {i \minus{} k}{i}\le\frac {x}{y}$ is verified for $ i \equal{} \frac {yk}{y \minus{} x}$
By fixing $ k \equal{} y \minus{} x$, we get $ i \equal{} y$, and $ \frac {x}{y}\in\bar{A_i}$
As for irrational numbers, we can always find a rational one as close as we want and just below it, anyway, at a distance smaller than $ \frac {1}{i^2}$, and therefore inside $ \bar{A_i}$, so that $ A \equal{} \{1\}$

Hope it's right...

A real number x is deleted iff there exists n such that  0< {nx} <1/n

Consider some irrational number e: its expansion using simple continued fractions; for each convergent h/k:
 |e-h/k| < 1/k^2;

also the convergents alternate on either side of e:  ie if e<h/k for some convergent then e>h/k for the next;

so we can find h,k with ke>h and ke-h<1/k;  So e does not belong to A

Hence A is non-empty countable;
I havent worked out whether A is finite or infinite

I think there is something wrong in the problem, or I miss something.

To me, either:
$ A_2\equal{}[0.25,0.5[\cup[0.75,1]$ (case 1)
or $ A_2\equal{}\{0\}\cup[0.25,0.5]\cup[0.75,1]$ (case 2)

but not, as is written in the opening post: $ A_2\equal{}[0.25,0.5]\cup[0.75,1]$ (or I don't get the logic of the construction, and basically don't see how $ 0$ get excluded)

The solution I provided earlier consider the case 1.

sorry for nonprecise explanation

here "without endpoints" means: we don't delete the endpoints of these intervals, which stays in our sets $ A_n$.

hence:

$ A_2 \equal{} [0.25,0.5]\cup[0.75,1]$

$ A_3 \equal{} \left[\frac {1}{9},\frac {3}{9}\right]\cup\left[\frac {4}{9},\frac {6}{9}\right]\cup\left[\frac {7}{9},1\right]$

$ A_4 \equal{} \left[\frac {1}{16},\frac {4}{16}\right]\cup\left[\frac {5}{16},\frac {8}{16}\right]\cup\left[\frac {9}{16},\frac {12}{16}\right]\cup\left[\frac {13}{16},1\right]$

etc.

surely $ 1$ will be in $ A$, so $ A$ nonempty.

From the representation you gave, it seems we have

$ A_n \equal{} \bigcup_{i\equal{}0}^{n\minus{}1}\left[\frac{1\plus{}ni}{n^2},\frac{i\plus{}1}{n}\right]\equal{}\bigcup_{i\equal{}0}^{n\minus{}1}\left[1\plus{}\frac{1\minus{}ni}{n^2},1\plus{}\frac{1\minus{}i}{n}\right]$

and therefor Enkidu's solution should be valid.


function
extremal principle
national olympiad

[quote="kikrishn"]HI,
  I am solving "Problem solving strategies " by ARthur Engel.I am able to solve all the chapters , save one i.e. Extremal principle.The approaches seem to me very ingenious and hard to guess.Do you know any of the sites or material where this principle is explained in detail ?
thanks
kishore[/quote]

When I used it, I thought that chapter was very nice. The problem is that solutions using the extremal principle [i]are[/i] ingenious. Look for some function of the variables in your problem to maximize. I don't think you can explain it much more. Seeing it work is perhaps more illustrative like with Sylvester's  problem. Engel's problems are well selected in the sense that they comprise from easy to very hard. I don't know what more to say

Ursula


[quote="hunter34"]This should go to Intermediate Topics.[/quote]
Why?  This could even be in the MATHCOUNTS forum, there was a Nat'l Target Round problems like this.
[hide="Hint"]
$ 1^{1}\equal{} 1$
$ (\minus{}1)^{x}\equal{} 1$ where $ x$ is even
$ n^{0}\equal{} 1$
Can you think of any more?[/hide]

[quote="Bicameral"]Find the sum of the integral solutions of $ (x^{2}\plus{}2x)^{(x^{2}\plus{}7x\minus{}18)}\equal{} 1$

On this one, I wasn't exactly sure what to do. I came out with:

$ x^{2}\plus{}7x\minus{}18 \equal{} (x\plus{}9)(x\minus{}2)$ I was thinking only a number with a zero power could work for this.

I'm really not sure what I did wrong; I put -7 as a answer.[/quote]

Its $ \minus{}8$ you forgot when $ x^{2}\plus{}2x\equal{}\minus{}1$, as long as the exponenet is even, it will still work, when that is equal to -1, $ x\equal{}\minus{}1$, and the exponenet is even, so $ \minus{}9\plus{}2\minus{}1\equal{}\minus{}8$

Oh snap!  :wallbash:

nice explanation, I was wondering how I messed it up.

I think I tried 1, but it didn't work, since $ x^{2}\plus{}2x\minus{}1\equal{}0$ doesn't have an integral solution.

Maybe this will help...

$ (x^{2}\plus{}2x)^{(x^{2}\plus{}7x\minus{}18)}\equal{}1\rightarrow (x\plus{}2)^{(x\plus{}9)(x\minus{}2)}x^{(x\plus{}9)(x\minus{}2)}\equal{}1$

I see $ x\equal{}\minus{}1$, $ x\equal{}\minus{}9$, $ x\equal{}2$, so the sum is $ \minus{}8$


I have no idea wut to do and i jus finished 11th

1. visit colleges over the summer or at least look into their websites, and decide where youre going to apply - this is very important. not too many(10+ is not easy), not too high(you should have at least couple of safeties)
2. start preparing essays - it's a good idea to write something about some general questions like what do you want to study and why, who's your mentor, what did you do last summer and so on. not only youre going to use them, but it's a great practice
3. prepare SAT's - if you have them all, great. if not, look into college websites and see what you're short of. If youre applying ivies or top schools, you should have 700+ SAT IIs as well.

4. instead of saying you dont know wut to do, it would be better if you ask what you [b]CAN[/b] do next time.


trigonometry
logarithms
function
Support
real analysis
real analysis unsolved

Please compute, perhaps using a PC,   $\sin\left({10}^{40}\right) \; \; ,\; \; \; e^{2004}.$

Why do you keep putting the $\text{\TeX}$ tags at the end of your posts? They only lead to trouble.

sure it's simply e^2004 we need to compute? that's got over thousand digits man :?

What does "compute" mean? In this context, these are probably transcendental numbers, and the simplest name for them is the way they've already been presented: as $e^{2004}$ and as $\sin(10^{40})$. I'm taking the most reasonable interpretation as "find a reasonable decimal or scientific notation approximation."

1. $\log_{10}(e^{2004}) = 2004\log_{10}e$
Using a spreadsheet, we find that this is approximately 870.3261417.
Take the fractional part of this and exponentiate it: $10^{.3261417} \approx 2.11905$. This gives us our scientific notation approximation of $e^{2004}$ as $2.11905 \times 10^{870}$. No, Peter, that's only 871 digits to the left of the decimal point, not 1000.

2. We need to find out what $10^40$ is equivalent to mod $2\pi$. That large a number will overflow the "mod" function in Excel, but we can step our way up to it. We find $\mod(100,2\pi) \approx 5.7522\dots$. Now multiply that 5.7522... by 100 and calculate what that is $\mod 2\pi$. We now know that $10^4\approx 3.452\dots\mod(2\pi)$. Keep multiplying by 100 until you get to $10^{40}$. (Spreadsheets are very good with recursive formulas.) We wind up with $10^{40} \approx 1.181285843 \mod(2\pi)$. Then calculate $\sin(1.181285843)\approx 0.925095057$.

This is constrained by the accuracy of the standard version of floating point arithmetic underlying the spreadsheet, which is about 15 digits. We lose three digits in #1, so that answer would be (if I showed you all the digits I have) accurate to about 12 significant figures. In the second one, we're losing a little accuracy in each step, but the answer is still probaby good to something like 8 or 10 digits.

You could do better if you were in a platform that supports higher precision or arbitrary precision arithmetic - something like Maple or Mathematica. But if you were in such a platform, you could do this with a single command.

Kent, that's also the result I get when typing it into my calculator ;)

I thought he wanted a more exact result somehow... :?


calculus
inequalities
inequalities proposed

[quote="Γιώργος"]Let $ x,y,z$ be nonnegative real numbers such that $ x \plus{} y \plus{} z \equal{} 2$. Prove that $ x^{2}y^{2} \plus{} y^{2}z^{2} \plus{} z^{2}x^{2} \plus{} xyz\leq 1$. When does the equality occur?[/quote]
Very sorry  :( ,I got a big mistake in calculus,But silouan,it must be
\[ f(x,y,z) \minus{} f(0,y \plus{} x,z) \equal{} \minus{} (2xyz^2 \minus{} x^2y^2 \minus{} xyz)
\]
Here is my second proof (The third one is same to mai quoc thang's one)
Assume $ z\ge y\ge x$.And let $ f(x,y,z) \equal{} LHS$
we have
\[ A\equal{}f(x,y,z)\minus{}f(x,\frac{z\plus{}y}{2},\frac{z\plus{}y}{2})\equal{}(y\minus{}z)^2(\frac{2x^2\minus{}yz\minus{}\frac{y\plus{}z)^2}{4}\minus{}x}{4}
\]
$ A\le 0$ because
\[ yz\ge x^2
\]
\[ \frac{(y\plus{}z)^2}{4}\ge x^2
\]
So we only need to show that
\[ f(x,1\minus{}\frac{x}{2},1\minus{}\frac{x}{2})\le 1 ,x\le \frac{2}{3}
\]
It is trivial.So we have done.
I hope now my proof is true :(

[quote="Allnames"] Assume $ z\ge y\ge x$.And let $ f(x,y,z) \equal{} LHS$
Now
\[ f(x,y,z) \minus{} f(0,y \plus{} x,z) \equal{} \minus{} (2xyz^2 \minus{} x^2z^2 \minus{} x^2y^2)
\]
[/quote]

I think that $ f(x,y,z) \minus{} f(0,y \plus{} x,z) \equal{}\minus{}( 2xyz^2\minus{}y^2z^2\minus{}xyz)$  .  :wink:

[quote="Γιώργος"]Let $ x,y,z$ be nonnegative real numbers such that $ x \plus{} y \plus{} z \equal{} 2$. Prove that $ x^{2}y^{2} \plus{} y^{2}z^{2} \plus{} z^{2}x^{2} \plus{} xyz\leq 1$ (*). When does the equality occur?[/quote]
$ p\equal{}x\plus{}y\plus{}z\equal{}2 ;$

$ q\equal{}xy\plus{}yz\plus{}zx;$

$ r\equal{}xyz .$

$ (*) \Leftrightarrow  A\equal{}3r\plus{}1\minus{}q^2 \geq 0$

By Schur : $ r \geq max\{0;\frac{8(q\minus{}1)}{9}\}$

* $ q \leq 1$ : $ r\geq 0$ ;

$ A \geq 1\minus{}q^2 \geq 0$

* $ \frac{4}{3}\geq q \geq 1$ : $ r \geq \frac{8(q\minus{}1)}{9}$

$ A \geq \minus{}\frac{1}{3}(q\minus{}\frac{5}{3})(q\minus{}1) \geq 0$

We have done :)

Let's make the inequality we have to prove homogeneous. We have to prove $ (x\plus{}y\plus{}z)^{4}>\equal{}16(x^{2}y^{2}\plus{}y^{2}z^{2}\plus{}z^{2}x^{2})\plus{}8(x^{2}yz\plus{}xy^{2}z\plus{}xyz^{2})$, which is equivalent to $ (x^{4}\plus{}y^{4}\plus{}z^{4})\plus{}4(x^{3}y\plus{}xy^{3}\plus{}...)\plus{}4(x^{2}yz\plus{}zy^{2}z\plus{}xyz^{2})>\equal{}10(x^{2}y^{2}\plus{}y^{2}z^{2}\plus{}z^{2}x^{2})$. By Muirhead inequality, we have $ 4(x^{3}y\plus{}xy^{3}\plus{}...)>\equal{}8(x^{2}y^{2}\plus{}y^{2}z^{2}\plus{}z^{2}x^{2})$. So we are left to prove $ (x^{4}\plus{}y^{4}\plus{}z^{4})\plus{}4(x^{2}yz\plus{}zy^{2}z\plus{}xyz^{2})>\equal{}2(x^{2}y^{2}\plus{}y^{2}z^{2}\plus{}z^{2}x^{2})$. We will prove the stronger inequality: $ (x^{4}\plus{}y^{4}\plus{}z^{4})\plus{}(x^{2}yz\plus{}zy^{2}z\plus{}xyz^{2})>\equal{}2(x^{2}y^{2}\plus{}y^{2}z^{2}\plus{}z^{2}x^{2})$. Assume wlog $ x>\equal{}y>\equal{}z$. Than, the last inequality is equivalent to $ (x\minus{}y)^{2}(x\plus{}y\minus{}z) \plus{} (x\minus{}z)^{2}(x\plus{}z\minus{}y)>\equal{}(y\minus{}z)^{2}(x\minus{}y\minus{}z)$. Obviously, $ (x\minus{}y)^{2}(x\plus{}y\minus{}z)>\equal{}0, (x\minus{}z)^{2}(x\plus{}z\minus{}y)>\equal{}0$. If $ (y\minus{}z)^{2}(x\minus{}y\minus{}z)<0$, the inequality is obviously true. Consider the case $ (y\minus{}z)^{2}(x\minus{}y\minus{}z)>\equal{}0$: Easy to notice that $ (x\minus{}z)^{2}>\equal{}(y\minus{}z)^{2}$ and $ x\plus{}z\minus{}y>\equal{}x\minus{}y\minus{}z$. So, $ (x\minus{}z)^{2}(x\plus{}z\minus{}y)>\equal{}(y\minus{}z)^{2}(x\minus{}y\minus{}z)$, implying the desired inequality. So, we see that equality doesn't hold in case all of the variables $ x,y,z$ are positive (As $ 4(x^{2}yz\plus{}zy^{2}z\plus{}xyz^{2})>\equal{}(x^{2}yz\plus{}zy^{2}z\plus{}xyz^{2})$ where the equality doesn't hold). So we are left to consider the case when one of the variables is $ 0$. As $ z$ is minimal, we have $ z\equal{}0$. Our inequality becomes $ x^{2}y^{2}<\equal{}1\equal{}\frac{(x\plus{}y)^{4}}{16}$, which is true by AM-GM inequality. Equality holds if and only if $ x\equal{}y\equal{}1$. So, equality cases are $ (0,1,1)$ and all possible symmetric permutations.  Lasha Lakirbaia.

just make things a little easier.it's equivalent to
$\sum x^4+4\sum x^3y-10\sum x^2y^2+4\sum x^2yz\ge 0$
or $\sum x^2(x-y)(x-z)+3\sum x^2yz+5\sum x^3y-10\sum x^2y^2\ge 0$
trivialized by Schur and Muirhead.

[quote=Γιώργος]Let $ x,y,z$ be nonnegative real numbers such that $ x \plus{} y \plus{} z \equal{} 2$. Prove that $ x^{2}y^{2} \plus{} y^{2}z^{2} \plus{} z^{2}x^{2} \plus{} xyz\leq 1$. When does the equality occur?[/quote]

My solution.

Let $p=a+b+c=2$, $q=ab+bc+ca$ and $r=abc$. Then our inequality is equivalent to
$$q^2\leq 1+3r$$
Using Schur we get
$$8+9r\geq 8q \Leftrightarrow q\leq 1+\frac{9r}{8} $$
So, it is enough to prove that
$$(1+\frac{9r}{8})^2\leq 1+3r \Leftrightarrow r(81r-48)\leq 0$$
But by AM-GM
$$r\leq \frac{8}{27}<\frac {16}{27}$$
Equality at $x=y=1$, $z=0$ and permutations.

[quote=Γιώργος]Let $ x,y,z$ be nonnegative real numbers such that $ x \plus{} y \plus{} z \equal{} 2$. Prove that $ x^{2}y^{2} \plus{} y^{2}z^{2} \plus{} z^{2}x^{2} \plus{} xyz\leq 1$. When does the equality occur?[/quote]

By computer,we have
$$RHS-LHS=\frac{1}{2}\sum_{cyc}{x(x-1)^2}+\frac{1}{4}\sum_{cyc}{xy(x-y)^2}\ge{0}$$


Hello, I am working through some problems in my textbook, but my teacher made no sense when he was explaining this.  Can anyone help me w/this problem?

[i]The region bounded by the curves y=x and y=x^2 is rotated about the line y=3.  Compute the volume of the resulting solid.[/i]

I keep on getting confused...

First of all, there's a formula. Presumably you don't like the formula, and that's a good thing. I don't like the formula either, and so I don't know what it is.

The way to do this problem is to consider an infinitessimal vertical piece of the graph. When you rotate it around the line, the volume will be the difference of two little round discs. Once you have calculated the volume of one of those, you can integrate from 0 to 1 to get the volume.

I don't know the formula either  :?

Here's what I got.

[tex]\pi*\int_{0}^{1}((3-x^2)^2-(3-x)^2)[/tex]

I had written out all my work but I got disconnected and it was not there when I clicked "back".  Is this correct?

Thanks!


algebra
polynomial
limit
algebra proposed

Find all possible values of $ x_0$ and $ x_1$ such that the sequence defined by:

$ x_{n\plus{}1}\equal{}\frac{x_{n\minus{}1} x_n}{3x_{n\minus{}1}\minus{}2x_n}$ for $ n \ge 1$

contains infinitely many natural numbers.

[hide]$ x_{n\plus{}1}\equal{}\dfrac{x_{n\minus{}1}x_{n}}{3x_{n\minus{}1}\minus{}2x_{n}} \Longleftrightarrow \dfrac{1}{x_{n\plus{}1}}\equal{}\dfrac{3x_{n\minus{}1}\minus{}2x_{n}}{x_{n\minus{}1}x_{n}} \Longleftrightarrow \dfrac{1}{x_{n\plus{}1}}\equal{}\dfrac{3}{x_n}\minus{}\dfrac{2}{x_{n\minus{}1}}$

Let $ y_n \equal{} \dfrac{1}{x_n}$.

(If $ x_n\equal{}0$ for some $ n\in\mathbb{N}_0$, then $ x_{n\plus{}2}\equal{}0$, $ x_{n\plus{}3}\equal{}0$ or is undefined, and $ x_{n\plus{}4}$ is undefined. So we can assume that $ x_n \neq 0$ for all $ n \in \mathbb{N}_0$.)

Then we have: $ y_{n\plus{}1} \equal{} 3y_{n}\minus{}2y_{n\minus{}1}$

The characteristic polynomial on this recursion is $ \lambda^2 \minus{} 3\lambda \plus{} 2$ whose roots are $ \lambda\equal{}2$ and $ \lambda\equal{}1$.

Thus, $ y_n \equal{} A(2)^n\plus{}B(1)^n \equal{} A(2)^n\plus{}B$ for some constants $ A,B \in \mathbb{R}$.

If $ A \neq 0$, then $ \lim_{n \to \infty}y_n \equal{} \pm\infty$. So, there exists an $ M$ such that either $ y_n>1$ for all integers $ n \ge M$, or $ y_n < \minus{}1$ for all integers $ n \ge M$.
Therefore, $ 0<x_n<1$ for all integers $ n \ge M$, or $ \minus{}1 < x_n < 0$ for all integers $ n \ge M$.
Thus, if $ A \neq 0$, then $ x_n$ can only be a natural number for at most $ M$ values of $ n$.
This is a contradiction. Therefore, we must have $ A\equal{}0$.

Then, $ y_n \equal{} B$ for all $ n \in \mathbb{N}_0$, and thus, $ x_n \equal{} \frac{1}{B}$ for all $ n \in \mathbb{N}_0$. This yields at most $ 1$ natural number in the sequence.

Therefore, there are no such values of $ x_0$ and $ x_1$ such that the sequence contains infinitely many natural numbers.[/hide]


algorithm
number theory
greatest common divisor
modular arithmetic
Euclidean algorithm

For how many integers between $ 1$ and $ 2003$ is the improper fraction $ \frac{n^2 \plus{} 4}{n\plus{}5}$ not in simplest form?

[hide="Solution"]
If it is can still be reduced, then $ \gcd(n^2 + 4,n + 5)\neq0.$
By the Euclidean Algorithm,
\begin{align*}\gcd(n^2 + 4,n + 5) & = \gcd\left(n^2 + 4 - (n^2 + 5n),n + 5\right) \\
& = \gcd( - 5n + 4,5n + 25) \\
& = \gcd(29,5n + 25).\end{align*}
Since $ 29$ is prime, the only way for the gcd to not be 1 is when $ 29|5(n + 5).$ We see that $ 29$ must divide $ n + 5,$ so the possible values are
\[ \boxed{n\in\{24,53,82,111,\dots,1967,1996\}.}\]

EDIT: Fixed
[/hide]

[hide]We have $ \frac {n^2 \plus{} 4}{n \plus{} 5} \equal{} n \minus{} 5 \plus{} \frac {29}{n \plus{} 5}.$ Hence

[list]$ \frac {n^2 \plus{} 4}{n \plus{} 5}$ is not in its simplest form[/list]
$ \Leftrightarrow\ \frac {29}{n \plus{} 5}$ is not in its simplest form

$ \Leftrightarrow\ n\equiv \minus{} 5\pmod{29}$

$ \therefore\ n\ \in\ \{24,\,53,\,82,\,111,\,\ldots,\,1967,\,1996\}\quad(69\ \mbox{of them)}$ [/hide]


1.The crew of an 8 member rowing team is to be chosen from 12 men, of which 3 can row on one side only and 2 can row on the other side only. Find the number of ways of arranging the crew with 4 members on each side.

2.The number of positive numbers not more than 10 digits formed using 0,1,2,3 is

3.There are 8 orators A,B,C,D,E,F,G & H. in how any ways can the arrangements be made so that A always comes before B and B always comes before C?

Edit: Removed Soln to #1 because of a big huge whole

[hide="3"]
Isn't it just $5!*{8\choose3}= 56*120 = 6720$, because all you need to do is choose 3 spots out of the 8 spots for A,B, and C, then the other 5 can be ordered in any way.

[/hide]

[quote="fizsle"]3) why not 6! simply????[/quote] Because that implies that $B$ must be [b]directly[/b] after $A$ and [b]directly[/b] before $C$.

[quote="fizsle"]1) the given answer is 60,480... so something seriously missing.....

3) why not 6! simply????- answer given is 8! / 3!... weird...[/quote]

For 1, I just realized I didn't account for the order that they were sitting in  :blush: that's why my answer is so small...  I'll edit that out.

For 3, If we choose 3 out of 8 spots for A,B, and C, then there's only one way A,B and C can be arranged, and 5 ways the rest of the letters can be arranged, so we get

$5!*{8\choose3}= \frac{8!}{3!}= 6720$

why isn't #2 $4^{10}-1$? each digit position can be any of the digits and subtract 0 because it's not positive. or am i missing something here? :maybe:

[quote="fizsle"]yes...it should be (4^10) - 1[/quote]

but i don't see anything wrong with tim1234133's solution.

[quote="tim1234133"][hide="2"]The first digit can be a 1,2,3 (so three choices): after that, each digit can be a 0,1,2,3(four choices).

Number of one-digit numbers: $3$
Number of two-digit numbers: $3\cdot4$
Number of three-digit numbers: $3\cdot4^{2}$
    .
    .
    .
    .
Number of ten-digit numbers:$3\cdot4^{9}$

Total: $3(4+4^{2}+4^{3}+4^{4}\dots+4^{9})$
=$4^{10}-4$

By calculator I get 1048572[/hide][/quote]

That should be $3(1+4+\cdots+4^{9})=3\cdot\frac{4^{10}-1}{4-1}=4^{10}-1$.  You left out the one-digit numbers in your final calculation.


ARML
geometry
3D geometry
calculus
derivative

I found this on the ARML Coaches' letter:

[quote]Calculators will still be allowed this year for the Power and Team rounds, however, they will be prohibited in 2009 on the entire competition. An additional pair of problems will be added to the 2009 Individual Round making ten problems. [/quote]

Yay calculatorlessness! Presumably those individual new 2 problems won't both be harder than the existing problems, so I'll get my best score ever my senoir year (hopefully ;))... What think ya'll?

I guess it means that the questions will get slightly easier

or at least not need calculators. And the 10 questions is an idea

that will have to be tried out before I can make any comments.

To see what effect this will have on the individual competition, look at the trends from the last few years.  Prior to 2005, the tendency was for several students to score 8's, and those students (and only those) would go to tiebreakers.  (1995, 1997, and 2003 being notable exceptions.)  However, since 2005, the difficulty of #8 has increased dramatically, so it's been very difficult to score perfect, and we're seeing dozens of students with 7's going to crazy tiebreakers.  I predict that with the addition of two new problems, they'll have both a tamer weed-out problem like pre-2005 #8's, and a killer problem that ensures that at most one student scores perfect.  The other new problem I'm guessing will be somewhat easier and won't affect the top ranks of individuals much.  At a guess, I'd say we'll see about as many 8's after 2009 as we see 7's now, with three to six 9's, and a perfect 10 every two or three years.  Of course, I'm trying to read the writers' minds here, so take that for what you will.

For the effect on team rankings, here's another statistic: In 2005, there were 9 teams that outscored Lehigh on the individuals, by as much as 11 points.  The team with the highest individual score, Exeter, finished third.  In 2006, North Carolina tied for the top individual score with Georgia, who finished 8th.  Last year, Exeter tied for second on the individual with 5th-place Georgia.  North Carolina, which took 4th overall, beat them out by one point on individuals.  So you can see that lately, the correlation between individual rank and overall rank has been somewhat weak.  If I recall correctly, that correlation used to be stronger.  (If anyone knows enough statistics and has enough free time to actually run numbers on this, that'd be cool!)  Adding two questions to individual will make it worth 55% of a team's total possible score, as opposed to 50%, so in theory the correlation should get stronger.  It should also make the extremely fickle relays somewhat less important to the final result, as teams will have a better chance to build up an insurmountable lead before relays.

And no calculators is great!  I don't think it's that necessary, as ARML has always done a good job writing team round problems that don't require a calculator--and they don't really help at all on the power question, though at least one of the older PQ's wouldn't be possible if calcs were legal.  Nonetheless, I'm happy to see the trend of competitions removing calculators continuing.

By the way, this also means that 2008 is the last year for teams to set scoring records under the current format, since the old records won't be meaningful compared to post-2009 scores.  Since the current scoring system began in 1988, the highest team score is Ontario's 200 in 1991 (if you ever need a self-esteem boost, take the 1991 individual!).  The lowest score by a winning team is Minnesota's 125 in the infamous 1997 competition.  The highest non-winning team scores were 194's in 1991 by Chicago and NYC.  With the new format, I predict we'll see another 200+ score before, say, 2012 or so.

Well, the addition of two more questions will decrease the number of perfect scores.  I would say you want to get a decent number of people into the tiebreaker round.

I don't have a real idea of how tame or insane the "killer" question were pre-2005, since my first ARML year was 2006, and it's been a while since I've seen some ARML individuals.  But if they were as tame as I think they were, I would say make #10 around that level (maybe a drop harder if you want to decrease the number of perfect scores).  Question 9 would become the new question 7 (easy, more time to deal with its difficult companion), and the 7-8 pair would be more feasible.

Sly Si was making some numerical claims and statements. You should take what he is saying also refer to the tables I posted [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=151607]here[/url] as posts #36 and 37 on the second page.

I think that even within the 8 question format, 2006 and 2007 were bad test design. One could do better with the same structure, and look at 2003 and 2005 for examples. What 2006 and 2007 both had were unusually difficult hardest questions coupled with unusually easy second-hardest question. That's what created the tiebreaker logjam at 7. What is needed is a doable #8 coupled with another real challenge at about #4,5, or 6. Maybe that will be easier to accomplish with more questions - but it could be accomplished without more questions.

As Si pointed out, this would move the individual round from half the total points to more than half. I get a little nervous, because one of ARML's strengths is that it is genuinely a team contest. But the real influence of each round isn't just the total possible points, it's also the spread of the points actually awarded. For instance, the team round has usually weighed more than the power round, at least among top teams, because the spread of scores actually awarded is usually wider for the team round than for the power round. The bad structure of the 2007 individual round narrowed the range of scores and in effect made that particular individual round less important.

Finally: I don't trust relays. They're fun and all that, but I'd be in favor of anything that made them less important than the other rounds.

One suggestion: what about expanding the team round? Maybe a 12-question, 25-minute team round? Or even a second team round, a "rapid" team round with a shorter time limit and each question counting less?

Hurrah! I don't like calculators -- they just make it so that the questions can be uglier and have nasty numbers.  No calculators usually means less computation and more math, which is good.

[quote="Kent Merryfield"]What is needed is a doable #8 coupled with another real challenge at about #4,5, or 6.[/quote]
True, but it shouldn't be as hard as #8.

WARNING: The following idea may be crazy and/or stupid.
How about a round that requires mathematical knowledge and physical ability?  Say, you have a bunch of cards with digits on them, bars for forming fractions and radical signs, etc., and you have to run and get them so you can form an answer on something like a Velcro board!  Given that you have sufficient space for team representatives to run and it's safe, of course...

If it's still 2008 and it involves either Rubik's cubes, juggling, or both, I'm all for it.

Well, yeah, in that case Southern California would score about 75 out of 40 for that round . . .

Anyway, I like the idea of 2 more questions.  Personally, I would like to see both of them be on the hard end, but not as hard as #8 the last couple of years.  I think that the results are more accurate if there is a range of scores among the top people, rather than a massive tie.

Elimination of calculators always makes for a better test, even though ARML always did a good job of neutralizing their effect.

[quote="Kent Merryfield"]What 2006 and 2007 both had were unusually difficult hardest questions coupled with unusually easy second-hardest question. That's what created the tiebreaker logjam at 7. What is needed is a doable #8 coupled with another real challenge at about #4,5, or 6. Maybe that will be easier to accomplish with more questions - but it could be accomplished without more questions.[/quote]

I think this is the purpose that the new questions will serve.  What I'd like to see is a test of similar structure to 2006 and 2007, but with one of the new questions being of the difficulty we saw in older #8's (like 2005).  That would serve to break up the tiebreaker logjam, while still establishing a perfect score as a pretty incredible achievement.  I also really like that there's been one problem each of the last three years that's hard enough that people get applauded during the competition just for solving that one problem.  I agree, by the way, that 2005 was excellent test design.  I don't understand how no one aced it, though.

I like the idea of expanding the team round, but I'd say instead of a second "rapid team round", add a second round with [i]harder[/i] questions, with more time.  The team round, IMHO, is pretty speed-based as it is.  And relays are just way too fickle, as I said earlier.  They're fun, and it's also fun to consider the possibility that a team like Chicago or Montgomery that consistently performs well on them could mount a dramatic comeback, but I also really dislike how easy it is for a solid team to blow their lead in the home stretch.

Some other parts of Si's comments may be references to [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=152548]this topic.[/url]

I like the idea of expanding team round.  ARML provides opportunities for math enthusiatic students to work together beyond school or district boundary.  Expanding team round also highlights the importance of working as a team, so that the outcome will not completely dependent on number of super stars on the team.  I believe the intention of "speed-based" (quoted from Sly Si) team round has that considerations too (making it impossible for one or two students to finish all the problems for team in the 20 min).  Speaking of that, has any team taken 2008 ARML Local?  The Local format gives 40 min to team round with same number of problems (10), which could make quite a different kind of experience.

I also share Sly Si's observation of relays "how easy it is for a solid team to blow their lead in the home stretch" - it happend to Washington team last year.

[quote="minghu"]Speaking of that, has any team taken 2008 ARML Local?  The Local format gives 40 min to team round with same number of problems (10), which could make quite a different kind of experience.
[/quote]
Yeah, I was on a team of 5, and me and another person did 9/10 problems for the team. (We got 2 wrong from stupid mistakes, but still...)
The 20 minute version is much more balanced.

i like the running idea/rubiks cube thing, though it is kinda crazy

[quote="minghu"]
I also share Sly Si's observation of relays "how easy it is for a solid team to blow their lead in the home stretch" - it happend to Washington team last year.[/quote]

And I'm pretty sure it happened to Washington the year before that, too ...

[quote="samath"]And I'm pretty sure it happened to Washington the year before that, too ...[/quote]
It's not all that simple. One thing to note about 2006 is that relay scores everywhere ran unusually high that year.

Among western site Division A teams, Washington had both the strongest team round (32, to three other teams at 28) and the strongest individual round (80, to SoCal's 79 and SFBA's 74), but they lost a dozen points to SFBA on the power round (SFBA 39, AL 30, SoCal 29, WA 27). As a result, the standings after three rounds were SFBA 141, WA 139, SoCal 136, AL 130.

(This is not was was reported "live" - this is a revision of history in which the protests and score adjustments had already happened.)

On the first relay, WA was the strongest of the four teams: WA 18, SFBA 16, AL 16, SoCal 14. That pulled them into a tie with SFBA at 157, with SoCal at 150 and AL at 146. But then it changed on the second relay: SFBA 12, AL 8, SoCal 6, WA 4, leading to the final score of SFBA 169, WA 161, SoCal 156, AL 154.

I agree that lately the trend is to have an over-the-top difficult #8. There seems to be a hope amongst the ARML organizers that there will be a very small (but positive) number of perfect scores, with a tiebreaker crowning the winner. Maybe 10 years ago that worked well, but now that close to 2000 students are participating, the only way to do that is to ask at least one question SO difficult that it alone narrows down the field of potential perfect scores to a handful.

Not only do I feel like this is a bad idea from a time-usage standpoint -- you're forcing the 98% of students who have very little chance at getting the answer to sit there for 10 minutes feeling dumb -- but it also [i]increases[/i] the likelihood that a large portion of the correct answers will come from lucky guesses. It's one thing for 5 people to guess the answer to a problem that 60% of the field solved anyway, but quite another if 5 people guess the answer to a problem that only 2 students out of 2000 actually solved.

I advocated moving that ultra-tough problem to Tiebreaker #1. Definitely keep some hard problems in the individuals -- maybe 2 or 3 of the caliber of a typical I-6. Maybe make the last pair in the Individuals contain two I-6-level questions. ARML has to face the facts that with only 8 (soon to be 10) questions and 2000 participants, they're almost always going to have a sizable tie at whatever score is the high for that year. Moving the ultra-tough question to Tiebreaker #1 means that the national-champion-to-be HAS to solve it correctly (unless everyone misses it), and it also gives the audience a chance to solve the problem for fun (instead of agonizing over it hopelessly during the Individuals round).

I think generating's solution is a good one. Last year I hadn't known how tough #8 was supposed to be and was quite surprised to see it. Hopefully the changes will bring the kind of differentiation that's healthy for the competition.

are team + power questions going to be bumped up to 5 points each and relay bumepd up accordingly ?

or are they completely scrapping the idea that individual is 50% of your max score while team + relay + power is the other 50% ?

I have to say, the questions had a much better difficulty curve this year.


The product of two integers is -40.  If the same two integers are added together, the sum is -3.  Find the larger of the two integers?

MY WORK:

I found that -8 + 5 = -3 (the sum).

Also, -8 (5) = -40 (the product).

Negative < positive, right?

My answer is for the larger number is 5.

The book's answer is -8 as the larger number.  How can this be?

Isn't -8 < 5?

Who's right? Wrong?  Why?

Thanks!

Yes, you are right I believe you have found an error in your mathbook.

Hmmmm... is there any difference in "greater than" and "larger than"?  Maybe the math book means by "larger than" which has a greater magnitude?

No, greater than and larger than mean the same thing.

The book may have intended to ask which had the greater absolute value, but Interval is correct.

[quote="Interval"]Larger than = greater than = >

I'm glad to be right this time.[/quote]

 :lol:

also note that proper mathematics uses "greater than" over "larger than"

Notice that you can apply Vieta's formulas in this situation.
[b]Theorem[/b] Vieta's Formula's
Given a polynomial $\sum^{n}_{k=0}a_{k}x^{k}$
with roots $r_{1}, r_{2},...,r_{n}$
$\sum^{n}_{k=1}r_{k}=\frac{-a_{n-1}}{a_{n}}$

[quote="Interval"]bpms,

How do I use this complicated formula in the question given?[/quote]
I'm sorry, I forgot the other part you would need.
$\prod^{n}_{k=1}r_{k}\cdot \prod^{n}_{k=1}-1$
It won't give you a direct answer, but it does simplify stuff :)


blogs
algebra
difference of squares
special factorizations

What is the value of $ \displaystyle\frac{109^2\minus{}100^2}{9}$?

Note: The reason dragon was able to factor those difference of squares is because:

$ a^2 \minus{} b^2 \equal{} (a \plus{} b)(a \minus{} b)$

which is a very useful property. For more information, click the link in my signature and go back a few pages.

Or go to james4l's blog and look for the table of contents


superior algebra
superior algebra unsolved

Consider the lattice of all algebraically closed subfields of the complex field $ \mathbb{C}$ whose transcendency degree (over $ \mathbb{Q}$) is finite. Prove that this lattice is not modular.

[i]L. Babai[/i]

What is the lattice supposed to be¿ "Meet" is probably just intersection, but "joint" is more complicated; is it the algebraic closure of the smallest field containing them¿


If $x,y,z \geq 0$ then the following inequality holds:
$ \sum( \frac{x}{xy+x+1}) \leq 1$.

cheers! :D :D :cool:

After some computation you obtain

(xyz-1)^2 \geq 0


Determine all quadruples $ (a,b,c,d)$ of real numbers satisfying the equation:

$ 256a^3 b^3 c^3 d^3\equal{}(a^6\plus{}b^2\plus{}c^2\plus{}d^2)(a^2\plus{}b^6\plus{}c^2\plus{}d^2)(a^2\plus{}b^2\plus{}c^6\plus{}d^2)(a^2\plus{}b^2\plus{}c^2\plus{}d^6).$

[quote="moldovan"]Determine all quadruples $ (a,b,c,d)$ of real numbers satisfying the equation:

$ 256a^3 b^3 c^3 d^3 \equal{} (a^6 \plus{} b^2 \plus{} c^2 \plus{} d^2)(a^2 \plus{} b^6 \plus{} c^2 \plus{} d^2)(a^2 \plus{} b^2 \plus{} c^6 \plus{} d^2)(a^2 \plus{} b^2 \plus{} c^2 \plus{} d^6).$[/quote]

[b]Case 1:[/b] one of $ a,b,c,d$, say $ a\equal{}0$, then clearly we have $ (a,b,c,d)\equal{}(0,0,0,0)$ as the only solution for this case.

[b]Case 2:[/b] $ a,b,c,d \neq 0$, then by AM-GM we have $ R.H.S. \ge \sqrt[4]{a^6b^2c^2d^2} \cdot \sqrt[4]{a^2b^6c^2d^2} \cdot \sqrt[4]{a^2b^2c^6d^2} \cdot \sqrt[4]{a^2b^2c^2d^6}\equal{} |abcd|^3$

Equality holds $ \Leftrightarrow |a|\equal{}|b|\equal{}|c|\equal{}|d|\equal{}1$ and $ abcd>0$.

So $ (a,b,c,d)\equal{}(0,0,0,0), (1,1,1,1),(\minus{}1,\minus{}1,\minus{}1,\minus{}1),(\minus{}1,\minus{}1,1,1)$ and its premutations are the only solutions for the equation.


inequalities
function
conics
parabola
inequalities unsolved

[i]Find the maximum value of the expression[/i]
$(a - b)^4 + (b -c)^4 + (c - a)^4$,
[i]for any real numbers[/i] $1\le a,b,c\le 2$.

Really nice, but I`m 9 grade and did not understand nothing.

at first,we note that

$(x-y)^4+(y-z)^4+(z-x)^4=2(x^2+y^2+z^2-xy-yz-zx)^2$

(because,expanding both the member, we arrive at the expression $\sum\limits_{sym}x^4-3\sum\limits_{sym}x^3y+2\sum\limits_{sym}x^2y^2$)

so, we have only to find the maximum value of

$f(x,y,z)=x^2+y^2+z^2-xy-yz-zx$

treat it as a function in only x as variable:

$f(x)=x^2-(y+z)x+b^2+c^2-bc$

this function is a convex parabola, so it assume his maximum in one of the extreme points.
the same is for y and z, so

$\max{f(x,y,z)}=\max(f(1,1,1),f(1,1,2),f(1,2,2),f(2,2,2))=1$

so $\max((x-y)^4+(y-z)^4+(z-x)^4)=\max(2(x^2+y^2+z^2-xy-yz-zx)^2)=2$

To Beat.
If $f''(x)\geq0,$ for $x\in[a,b]$ then $f$ is convex on $[a,b].$
Hence,  $f$  reaches the greatest value on the ends of  $[a, b].$ ;)

The original inequality was proposed by Hoang Ngoc Minh, one of my friends.

[quote="Beat"]Is there another way to find the maximum of this function $f(x,y,z)=x^2+y^2+z^2-xy-yz-zx$ if we know that $1\le x,y,z\le 2$?[/quote]

  It isn't hard I think
Denote $x=1+a$ and so on. Because $2\geq x \geq 1$so $2\geq (1+a) \geq 1$ or $1\geq a\geq 0$
$f(x,y,z)=f(a,b,c)$
  $(1-a)(1-b)(1-c)+abc \geq 0$
<--> $1-(a+b+c)+(ab+bc+ca) \geq 0$
<--->$1 \geq (a+b+c)-(ab+bc+ca) \geq (a^2+b^2+c^2 -ab+bc+ca)$
Done :!:

[quote="Beat"]:idea:
$x^2+y^2+z^2-xy-yz-zx= \sqrt[n]{\frac{(x-y)^{2n}+(y-z)^{2n}+(z-x)^{2n}}{2}}$
Is this true for all natural $n$?[/quote]

  I wouldn't think it is correct!
You can try $\sum (a-b)^6$ .


geometry
trigonometry
Pythagorean Theorem

Semicircles are drawn on two sides of square $ ABCD$, as shown.  Point $ E$ is the center of the square, and $ QAP$ is a line segment with $ QA \equal{} 7$ and $ AP \equal{} 23$.  Compute $ AE$.

I believe I have seen a nicer solution to this problem, but I have a less-nice one.

[hide] Let $ \angle{QAD}\equal{}\theta$. Then, $ \angle{PAB}\equal{}90\minus{}\theta$. Construct centers $ Y,Z$ on $ AD$ and $ AB$. From isosceles $ \Delta QAY$, we get $ \cos(90\minus{}\theta)\equal{}\sin{\theta}\equal{}\dfrac{7}{2r}$. From isosceles $ \Delta PAZ$, we get $ \cos{\theta}\equal{}\dfrac{23}{2r}$.

Now, $ \sin^{2}\theta\plus{}\cos^{2}\theta\equal{}1$, so we find that $ 2r^{2}\equal{}289$. We need $ AE\equal{}\sqrt{2}r\equal{}\sqrt{289}\equal{}\boxed{17}$.[/hide]

[quote="13375P34K43V312"]I believe I have seen a nicer solution to this problem, but I have a less-nice one.

[hide] Let $ \angle{QAD} \equal{} \theta$. Then, $ \angle{PAB} \equal{} 90 \minus{} \theta$. Construct centers $ Y,Z$ on $ AD$ and $ AB$. From isosceles $ \Delta QAY$, we get $ \cos(90 \minus{} \theta) \equal{} \sin{\theta} \equal{} \dfrac{7}{2r}$. From isosceles $ \Delta PAZ$, we get $ \cos{\theta} \equal{} \dfrac{23}{2r}$.

Now, $ \sin^{2}\theta \plus{} \cos^{2}\theta \equal{} 1$, so we find that $ 2r^{2} \equal{} 289$. We need $ AE \equal{} \sqrt {2}r \equal{} \sqrt {289} \equal{} \boxed{17}$.[/hide][/quote]

If you mean by nice, you can always use congruent triangles.

[quote="13375P34K43V312"]Where exactly are the congruent triangles? The solution I had in mind invoked inscribed angles.[/quote]

Draw in the other two lines to form two triangles.

They are congruent frmo ASA so we hvae a 7-23-h (side length of square) right triangle,

Then you cna find the hypotenuse, divide that by 2 and multiply that answre by sqrt2

Nice job.

[hide="Quick way"]

I'm not sure if this is meant by "nice way" (I didn't read the all the posts) but doing quick angle chase makes this problem very easy. Obviously, $ \angle QDA \equal{} \angle PAB \equal{} 90^\circ$. Now, here comes the angle chase part: Because $ \angle QAP$ is a straightline, $ \angle QAD \plus{} \angle PAB \equal{} 90^\circ$ and $ \angle QDA \plus{} \angle QAD \equal{} 90^\circ$ so $ \angle PAB \equal{} \angle QDA$. Thus, we can easily see that both triangles are same.  Using Pythagorean Theorem, we see that the side of the square is $ 17 \sqrt {2}$. By dividing by 2 and multiplying by $ \sqrt {2}$, we get length of $ AD$ or 17. [/hide]


Euler search function trigonometry calculus

This is fairly good problem that review what you know about infinite sum.

Given that $S = \frac{1}{1^2} + \frac{1}{2^2} + \frac{1}{3^2} + \frac{1}{4^2} \cdots$, find the exact value of $\frac{1}{2^2} + \frac{1}{4^2} + \frac{1}{6^2} + \frac{1}{8^2} \cdots$ in terms of $S$.

Enjoy. :)

Is this the answer?
[hide]$\frac{S}{4}$[/hide]

[hide]Each term of the first sequence just needs an extra $\frac 14$ to turn into the second sequence so the answer is just $\frac S4$[/hide]

I kinda guessed it would be that.  Also, for those interested, the first sum is equal to $\frac{\pi^2}6$ :)

Guys, explanations... ;) I mean by the good ones so people who don't get this can see how you found the answer.

I have little different approach but basically same.

$\displaystyle \frac{1}{2^2} + \frac{1}{4^2} + \frac{1}{6^2} + \frac{1}{8^2} \cdots = \sum^{\infty}_{n=1} \frac{1}{(2n)^2} = \sum^{\infty}_{n=1} \frac{1}{4n^2} = \frac{1}{4} \sum^{\infty}_{n=1} \frac{1}{n^2} \Rightarrow \frac{1}{4} \cdot S$

QED.

[quote="tarquin"]Woah how'd you get that?[/quote]

I assume you're asking how he got $\frac{\pi^2}{6}$.  Well, I'm pretty sure he didn't just figure it out on the spot without ever seeing that before  :rotfl: . Rather, Euler showed that $\sum_{i=1}^\infty \frac{1}{i^2}=\frac{\pi^2}{6}.$  The proof is actually easy to follow if you know about Taylor series, actually you can still follow it without knowing any Calculus.  I might type up the derivation...if you really want to see it you could probably google it...or if you search it on here you might be able to find it.

Actually, on the first page of this site, http://www.artofproblemsolving.com, it tells you what the sum equals :P

Also, you might be interested to know that $\sum_{i=1}^\infty=\zeta(2).$  The general form of the Zeta function looks like $\zeta(x)=\sum_{i=1}^\infty\frac{1}{i^x}.$  If this has interested you, then you might want to look into the Riemann Hypothesis.  It's an unsolved problem in mathematics but has a lot of literature on it that should be accessible to less experienced people (actually, I haven't read anything on it...but I hope to do so sometime).

[quote="amirhtlusa"][quote="tarquin"]Woah how'd you get that?[/quote]
hmm... that is pretty well known...but anyways, this is the proof:
hehe.
let $P(x)=1-\frac{x^2}{3!}+\frac{x^4}{5!}-\frac{x^6}{7!}.....$
clearly $P(0)=1$. to find the roots of $P(x)$, note that for $x\not=0$:
$P(x)=x[\frac{1-\frac{x^2}{3!}+\frac{x^4}{5!}-\frac{x^6}{7!}.....}{x}]=\frac{x-\frac{x^3}{3!}+\frac{x^5}{5!}-\frac{x^7}{7!}.....}{x}=\frac{\sin{x}}{x}$
so $P(x)=0 \rightarrow \sin{x}=0 \rightarrow x= (+-)k\pi$ for $k=1,2,3,...$ note that $x=0$ is not a solution. because $P(0)=1$
now if you factor $p(x)$ you get:
$P(x)= (1-\frac{x}{\pi})(1-\frac{x}{-\pi})(1-\frac{x}{2\pi})(1-\frac{x}{-2\pi})= [1-\frac{x^2}{\pi^2}][1-\frac{x^2}{4\pi^2}][1-\frac{x^2}{9\pi^2}]+....$
now look what happens when you expand the above equation:
you get
$1-[ (\frac{1}{\pi^2})+(\frac{1}{4\pi^2})+(\frac{1}{9\pi^2})+....]x^2+....$now set this equal to the original function for $P(x)$ you  get
$P(x)=1-\frac{x^2}{3!}+\frac{x^4}{5!}-\frac{x^6}{7!}.....= 1-[ (\frac{1}{\pi^2})+(\frac{1}{4\pi^2})+(\frac{1}{9\pi^2})+....]x^2+....$ now take a look at coefficient for the $x^2$ then set them equal to each other, you get:
$\frac{-1}{3!}=-(\frac{1}{\pi^2}+\frac{1}{4\pi^2}+\frac{1}{9\pi^2}+....)=\frac{-1}{\pi^2}(1+\frac{1}{4}+\frac{1}{9}+\frac{1}{16}+....) $
therefore:
$1+\frac{1}{4}+\frac{1}{9}+\frac{1}{16}+...=\frac{\pi^2}{6}$
so we are done :D[/quote]
is this getting started level? lol seems really complicated at first sight... :huh:

S/4

1/1 coresponds to 1/4, 1/4 coresponds to 1/16, 1/9 coresponds to 1/36, etc. each is 1/4 of its match in S, so therefore the entire thing is S/4.

I didnt learn the sum of the inverse of squares until calculus, so I would not say it is getting started but it is cool.

Well, I think a getting started level person can understand it, after all, that exact sum is discussed in AOPS V.1, in the Big Picture section after the Sequences and Series chapter.  ;)

yeah the proof is very easy to understand, it's not complicated...
the only thing that might not fit the Getting Started level is that knowing:
$\sum_{n=0}^{\infty}\frac{(-1)^nx^{2n+1}}{(2n+1)!}=\sin{x}$

It would belong in getting started, because you don't have to know how to figure out the value of S, you only have to figure it out in terms os S.


geometry
incenter
circumcircle
cyclic quadrilateral
perpendicular bisector
geometry proposed

Given triangle $ABC$ and $A'B'C'$ is orthic triangle $I_{a},I_{b},I_{c}$ are incenter of triangles $AB'C',BC'A',CA'B'$ then circumcenter of triangle $I_{a}I_{b}I_{c}$ are orthocenter of contact triangle of $ABC$.

Let $K$, $L$, $M$ be the points of contact of incircle with $BC$, $CA$, $AB$ respectively. Let $X = A'I_{b}\cap BA$ and $Y = A'I_{c}\cap AC$. Easy angle-chasing reveals that $AXA'Y$ is a cyclic quadrilateral and hence $AXY$ is isosceles, i.e. $AI\bot XY$ ($I$ is the incenter of $ABC$). But triangles $A'C'B$ and $A'CB'$ are similar and $I_{b}$, $I_{c}$ are their respective incenters. Hence $I_{b}I_{c}\parallel XY$ and $I_{b}I_{c}\bot AI$. We thus infer that $KLM$ and $I_{a}I_{b}I_{c}$ are homothetic. A straight-forward computation suggests that $KI_{b}=KI_{c}$ and so the altitude $KK'$ in $KLM$ is the perpendicular bisector of $I_{b}I_{c}$. Repeating the above argument for other sides of $I_{a}I_{b}I_{c}$ we deduce the desired conclusion.


Let $ a,b,c$ be positive real numbers such that $ a\plus{}b\plus{}c\equal{}3.$ Prove that
\[ \sqrt{3a\plus{}\frac{1}{b}}\plus{}\sqrt{3b\plus{}\frac{1}{c}}\plus{}\sqrt{3c\plus{}\frac{1}{a}} \ge 6\]
Equality holds if and only if $ a\equal{}b\equal{}c\equal{}1.$
:)

[quote="arqady"]See here:
http://www.artofproblemsolving.com/Forum/viewtopic.php?t=176214[/quote]
Thanks, arqady.
My solution is the following :)


function
quadratics
LaTeX
analytic geometry

The maximum of a function can be determined by the ordered pair ( -b/2a, -[b^2-4ac]/4a), which we know taken from the quadratic formula.  Can somebody explain why the value of y in the ordered pair is negative? We know that the formula states that it is either + or -. Im confused and please forgive me for not learning latex  :oops: .

first of all, that is not necessarily the maximum of a general function, it is the max or min of a quadratic, also, the y term is not negative if there is a little negative sign there, what if $a$ was negative, etc

If the quadratic is $y(x)=ax^2+bx+c$, then you can just plug in $-\frac{b}{2a}$ for $x$ to get the ordered pair for the extremum: \[ \left(-\frac{b}{2a},\frac{4ac-b^2}{4a}\right)  \] Neither coordinate is necessarily negative, depending on the values of $a,b,c$.


What is the maximum number of possible points of intersection of a circle and a triangle?

$ \textbf{(A) }2\qquad\textbf{(B) }3\qquad\textbf{(C) }4\qquad\textbf{(D) }5\qquad\textbf{(E) }6$

6, just draw three non-concurrent, non-parallel secants that intersect outside the circle, then the intersections are the vertices of the triangle.


inequalities
LaTeX
inequalities unsolved

If $a,b,c$ are real positive numbers, where a+b+c=1, prove inequalities:

[code]$\sqrt{\frac{1-a}{a}} \sqrt{\frac{1-b}{b}}+\sqrt{\frac{1-c}{c}} \sqrt{\frac{1-b}{b}}+\sqrt{\frac{1-c}{c}} \sqrt{\frac{1-a}{a}}\geq \ 6$[/code]

[quote="Fitim"]If $a,b,c$ are real positive numbers, where a+b+c=1, prove inequalities:

$\sqrt{\frac{1-a}{a}}\sqrt{\frac{1-b}{b}}+\sqrt{\frac{1-c}{c}}\sqrt{\frac{1-b}{b}}+\sqrt{\frac{1-c}{c}}\sqrt{\frac{1-a}{a}}\geq \ 6$[/quote]

$LHS = \sqrt{\frac{b+c}{a}}\sqrt{\frac{c+a}{b}}+\sqrt{\frac{a+b}{c}}\sqrt{\frac{c+a}{b}}+\sqrt{\frac{a+b}{c}}\sqrt{\frac{b+c}{a}}= \frac{\sqrt{c(c+b)(c+a)}+\sqrt{a(a+b)(a+c)}+\sqrt{b(b+a)(b+c)}}{\sqrt{abc}}$

$By CS, (a+b)(c+a) \geq (\sqrt{ac}+\sqrt{ab})^{2}= a(\sqrt{b}+\sqrt{c})^{2}$. Simirally, $(b+a)(b+c) \geq b(\sqrt{a}+\sqrt{c})^{2}$ and $(c+a)(c+b) \geq c(\sqrt{a}+\sqrt{b})^{2}$

So, $LHS \geq \frac{a(\sqrt{b}+\sqrt{c})+b(\sqrt{c}+\sqrt{a})+c(\sqrt{a}+\sqrt{b})}{\sqrt{abc}}= \sum_{sym}\frac{\sqrt{a}}{\sqrt{b}}\geq 6$

[quote="toetoe"]Hi,
This is my solution :
$A = \sqrt{\frac{1-a}{a}}\sqrt{\frac{1-b}{b}}+\dots$

$A\geq \frac{1-a}{a}+\frac{1-b}{b}+\frac{1-c}{c}$

[/quote]
Isn't it wrong [Reversed]?
Since $\sqrt{xy}+\sqrt{yz}+\sqrt{zx}\leq x+y+z$

My solutions is good:

$\sqrt{\frac{1-a}{a}}\sqrt{\frac{1-b}{b}}= \sqrt{\frac{b+c}{a}}\sqrt{\frac{a+c}{b}}= \sqrt{\frac{ca+ba+bc+c^{2}}{ab}}\geq \sqrt{\frac{4c sqrt{ab}}{ab}}= 2\sqrt{\frac{c}{sqrt{ab}}}$

Now $2\sqrt{\frac{c}{sqrt ab}}+2\sqrt{\frac{b}{sqrt ac}}+2\sqrt{\frac{a}{sqrt cb}}\geq \ 6$              / 2

$\sqrt{\frac{c}{sqrt ab}}+\sqrt{\frac{b}{sqrt ac}}+\sqrt{\frac{a}{sqrt cb}}/geq / 3\sqrt[3]{\frac{abc}{sqrt{a^{2}b^{2}c^{2}}}}= 3$

Sorry for LATEX!
 I dont knows how use!

Put dollar mark before and after your latex code. ($\$$)

If I am not wrong it is quite easy
[quote="Fitim"]If $a,b,c$ are real positive numbers, where a+b+c=1, prove inequalities:

$\sqrt{\frac{1-a}{a}}\sqrt{\frac{1-b}{b}}+\sqrt{\frac{1-c}{c}}\sqrt{\frac{1-b}{b}}+\sqrt{\frac{1-c}{c}}\sqrt{\frac{1-a}{a}}\geq \ 6$[/quote]

$LHS = \sqrt{\frac{b+c}{a}}\sqrt{\frac{c+a}{b}}+\sqrt{\frac{a+b}{c}}\sqrt{\frac{c+a}{b}}+\sqrt{\frac{a+b}{c}}\sqrt{\frac{b+c}{a}}\geq 3\sqrt[3]{\frac{(a+b)(b+c)(c+a)}{abc}}\geq 6$

yes,you're right.My solution is wrong :( .


[quote="Breezy"]First let me just say that it takes a very gifted person to comprehend chemistry.  This is the 3rd week of school and I am already so confused I don't know which way is out.  Can someone please help me with this?

[code]Write the atomic symbols for isotopes with the following. (Enter the symbol with the first (raised) number in the first box, the second (lower) number in the second box, and the element symbol in the third box.)
(a) a mass number of 24 and 13 neutrons
for the second box I put 11 and it said correct.

(b) a mass number of 70 and 40 neutrons
 Here is put 30 and it was also correct

(c) a hydrogen atom with 1 neutron
 Here I put 1 for the obvious reason.  I don't have a clue what it wants for the other 2 answers in these problems

(d) 16 electrons and 18 neutrons

What will the number of electrons do for you if know nothing else about the subatomic particles?

[/code][/quote]

A hydrogen atom must have 1 proton -- no other element could have exactly 1 proton because the number of protons an atom has determines its atomic number and thus what element it is.

If an atom is electrically neutral, then the number of protons must be equal to the number of electrons.


the last option of 14-15 didnt get added in

Hmm, autocensor didn't get that one, interesting...

Answer: phpBB does not consider "crap" to be a swear, although the s word is.

Anyways AIME is "American Invitational Mathematical Examination" and it comes after the AMC ([url=http://www.unl.edu/amc/index.html]Reference[/url])

Well, that word isn't really a swear...

Anyways, AIME is the American Invitational Math Exam. Its the exam you qualify for by scoring decently on the AMC.

does the censor change the words into (for example) c***?

...

You didn't include an option for those people who score 6-8, which pretty much includes those people who barely miss the USAMO cut.

ya, i agree, thats pretty much where i fall in, but i guess if i didn't make any stupid mistake i could manage 9. So i suppose that would be the 6-8 category  :?:

how bout the people who would usually geta 4, but get a 6-8 without mistakes :?:

Well, they usually get a 3-4, don't they?

i spose, but i was just going by paladin's thinking there.

I got 2 last year :-/ And I'm not sure how I did that well. I know for one of the problems I drew a diagram and [i]measured[/i] the angle. As for the other... Lord only knows how I solved it. ((3!)!)!/(3!)=k*n! where k and n are positive integers and n is as large as possible, find k+n.

Anyway, I really need to do better this year.

I'm with paladin8...the 6-8 category.

I can only wonder how Tare knows so much about the autocensor...

lol

i dont really know where i stand due to never actually timing and stuff on an aime.


$x,y,z>0$ . Prove the inequality :
$\frac{3+\sqrt{3}}{9}\geq\frac{xyz(x+y+z+\sqrt{z^2+y^2+z^2})}{(x^2+y^2+z^2)(xy+yz+zx)}$


floor function
number theory
prime factorization
national olympiad

Please help me with solution of the next problem:

Prove that
[img]http://zaba.ru/images/img813.gif[/img]
is divisible by [b]n![/b] for all positive integers [b]n[/b]

[hide="Hint"]In the prime factorization $ n!\equal{}\prod p_k^{e_k}$, we have $ e_k \equal{} \left\lfloor\frac{n}{p_k}\right\rfloor\plus{}\left\lfloor\frac{n}{p_k^2}\right\rfloor\plus{}\left\lfloor\frac{n}{p_k^3}\right\rfloor\plus{}\ldots \leq \left\lfloor\frac{\frac{n}{p_k}}{1\minus{}\frac{1}{p_k}}\right\rfloor\equal{}\left\lfloor\frac{n}{p_k\minus{}1}\right\rfloor$
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[quote="minsoens"][hide="Hint"]In the prime factorization $ n! \equal{} \prod p_k^{e_k}$, we have $ e_k \equal{} \left\lfloor\frac {n}{p_k}\right\rfloor \plus{} \left\lfloor\frac {n}{p_k^2}\right\rfloor \plus{} \left\lfloor\frac {n}{p_k^3}\right\rfloor \plus{} \ldots \leq \left\lfloor\frac {\frac {n}{p_k}}{1 \minus{} \frac {1}{p_k}}\right\rfloor \equal{} \left\lfloor\frac {n}{p_k \minus{} 1}\right\rfloor$
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Hello! could you explain how can I use your hint to show that
[img]http://zaba.ru/images/img813.gif[/img]
 is divisible by n!, thank you.


Need Help, If you get the answer, show how you got it with an undertandable explanation.

How much water would 1,000 swimming pools hold, relative to Swaziland's daily water consumption of 30 million gallons?  To get a rough estimate, assume the average length of a backyard pool is 15 meters, average width is 4 meters, and average depth is 1 1/2 meters.

Answer: The swimming pools would hold_______ million gallons or _______% of Swaziland's daily consumption.

Give Explanation!!!!!!!!!

The issue with this problem is converting volume into gallons.  Once you find the volume of the pool in meters cubed (easy), you then need to find the conversion, whatever it is.  Then you multiply that by 1000 to get total number of gallons.  Call that number $ G$.  Then we compute $ G/30,000,000$ to get the percentage.

[quote="Nexmus"]where'd u get this problem?[/quote]

I got this problem from a book of mathematical techings by Morris Kline.

This is mostly just computational work, no thinking required.

[quote="shaun ong"]i calculated it. The answer is 90000m^3 of water theyre holding. The percentage = to 90000/30000000 *100. So the answer is 30%[/quote]

Wrong! Wrong! Wrong!

[quote="shaun ong"]lol, then what should the answer be? the data u gave was an approx value, i used the approx value to get and approx answer, so whats wrong?[/quote]
90000/30000000=9/3000=3/1000=0.3%...

Did [b]anyone[/b] read or understand the second post correctly?

$ 90000 \text{ m}^3 \approx 23775485 \text{ gal}$

$ \dfrac{23775485}{30000000} \approx 79 \%$

Not hard.