inequalities
function
inequalities proposed

$ a, b, c$ are positive real numbers.

Prove that

$ \frac {a^2}{\sqrt [3]{a^6 \plus{} 26b^3c^3 }} \plus{} \frac {b^2}{\sqrt [3]{b^6 \plus{} 26c^3a^3 }} \plus{} \frac {c^2}{\sqrt [3]{c^6 \plus{} 26a^3b^3 } }\ge1$.

This version of inequality can be seen as the similar form as IMO ( I forgot the year) but in the same way, we use the appropriate substitution such as

$ \frac{a^2}{\sqrt[3]{a^6\plus{}26(bc)^3}}\equal{}x ...$ and so on.

Therefore, combined with the given condition, the rest follows smoothly. Nice inequality though! :pilot:

Consider the function $ f(x) \equal{} \frac {1}{\sqrt [3]{x}}$. Clearly, $ f$ is convex on $ (0$, $ \infty)$. Applying Jensen yields
\[ \frac {a^2}{\sqrt [3]{a^6 \plus{} 26b^3c^3}} \plus{} \frac {b^2}{\sqrt [3]{b^{6} \plus{} 26c^{3}a^{3}}} \plus{} \frac {c^{2}}{\sqrt [3]{c^{6} \plus{} 26a^{3}b^{3}}} \ge \sqrt [3]{\frac {(a^2 \plus{} b^2 \plus{} c^2)^4}{\sum(a^8 \plus{} 26a^3b^3c^2)}},\]
which is greater or equal to $ 1$, as $ (a^2 \plus{} b^2 \plus{} c^2) \equal{} \sum a^8 \plus{} f(a$, $ b$, $ c)$, where $ f$ is symmetric in $ a$, $ b$, $ c$. So Muirhead does the work.

[quote="great math"]This version of inequality can be seen as the similar form as IMO ( I forgot the year) but in the same way, we use the appropriate substitution such as

$ \frac {a^2}{\sqrt [3]{a^6 \plus{} 26(bc)^3}} \equal{} x ...$ and so on.

Therefore, combined with the given condition, the rest follows smoothly. Nice inequality though! :pilot:[/quote]

Do you mean 2001, problem 2? I thought so too when I saw this for the first time  :D

By CS, we only need to prove: $ (a\plus{}b\plus{}c)^4 \geq\ a^4\plus{}b^4\plus{}c^4\plus{}26abc(a\plus{}b\plus{}c)$
Obviously trues  :maybe:

It was an easy one, I guess. :(

I'll try to make harder ones.

Maybe you are saying about the problem $ \frac{a}{\sqrt{a^2 \plus{}8bc}}\plus{}\frac{b}{\sqrt{b^2 \plus{}8ca }}\plus{}\frac{c}{\sqrt{c^2 \plus{}8ab}}\ge1$.

I get the idea of this problem too. :blush:

How about this one:

$ \frac{a}{\sqrt{a^2 \plus{}8bc}}\plus{}\frac{b}{\sqrt{b^2 \plus{}8ca }}\plus{}\frac{c}{\sqrt{c^2 \plus{}8ab}}<2$.

[quote="Stephen"]$ \frac {a}{\sqrt {a^2 \plus{} 8bc}} \plus{} \frac {b}{\sqrt {b^2 \plus{} 8ca }} \plus{} \frac {c}{\sqrt {c^2 \plus{} 8ab}} < 2$.[/quote]

If $ a,b,c>0$ and $ k\ge 3$ , then :

$ \frac{a}{\sqrt{a^2\plus{}kbc}}\plus{}\frac{b}{\sqrt{b^2\plus{}kca}}\plus{}\frac{c}{\sqrt{c^2\plus{}kab}}< 2$

[hide="Proof"]Let's observe that $ \sum_{cyc}\frac{a}{\sqrt{a^2\plus{}kbc}}\le \sum_{cyc}\frac{a}{\sqrt{a^2\plus{}3bc}}$

So it'll be enough to prove the inequality for the case $ k\equal{}3$ . Being symmetric and homogeneus assume that $ a\ge b\ge c$ and $ abc\equal{}1$ . We have :

$ \sum_{cyc}\frac{a}{\sqrt{a^2\plus{}3bc}}\equal{}\sum_{cyc}\sqrt{\frac{a^3}{a^3\plus{}3}}$ .

Let be $ x\equal{}a^3 \ ,\ y\equal{}b^3 \ ,\ z\equal{}c^3 \Rightarrow xyz\equal{}1$ and $ x\ge y\ge z$ .

We have that : $ y\plus{}z\le 2x \equal{}\frac{2}{yz}\ , \ yz\le 1$ and $ \sqrt{\frac{x}{x\plus{}3}}< 1$ .

So it'll be enough to prove that : $ \sqrt{\frac{y}{y\plus{}3}}\plus{}\sqrt{\frac{z}{z\plus{}3}}\le 1\Leftrightarrow \frac{y}{y\plus{}3}\plus{}\frac{z}{z\plus{}3}\plus{}2\sqrt{\frac{yz}{(y\plus{}3)(z\plus{}3)}}$ $ \le 1\Leftrightarrow 2\sqrt{\frac{yz}{(y\plus{}3)(z\plus{}3)}}\le \frac{9\minus{}yz}{(x\plus{}3)(z\plus{}3)}\Leftrightarrow 4yz(y\plus{}3)(z\plus{}3)\le (9\minus{}yz)^2$ .

$ 4yz(y\plus{}3)(z\plus{}3)\equal{}4yz(yz\plus{}3(y\plus{}z)\plus{}9)\le 4yz(yz\plus{}\frac{6}{yz}\plus{}9)\equal{}4(yz)^2$ $ \plus{}24\plus{}36yz \le 64$

Also $ yz\le 1\Leftrightarrow 9\minus{}yz\ge 8 \Leftrightarrow (9\minus{}yz)^2\ge 64$ , so it's done .[/hide]

[quote="FelixD"]Consider the function $ f(x) \equal{} \frac {1}{\sqrt [3]{x}}$. Clearly, $ f$ is convex on $ (0$, $ \infty)$. Applying Jensen yields
\[ \frac {a^2}{\sqrt [3]{a^6 \plus{} 26b^3c^3}} \plus{} \frac {b^2}{\sqrt [3]{b^{6} \plus{} 26c^{3}a^{3}}} \plus{} \frac {c^{2}}{\sqrt [3]{c^{6} \plus{} 26a^{3}b^{3}}} \ge \sqrt [3]{\frac {(a^2 \plus{} b^2 \plus{} c^2)^4}{\sum(a^8 \plus{} 26a^3b^2c^2)}},\]
which is greater or equal to $ 1$, as $ (a^2 \plus{} b^2 \plus{} c^2) \equal{} \sum a^8 \plus{} f(a$, $ b$, $ c)$, where $ f$ is symmetric in $ a$, $ b$, $ c$. So Muirhead does the work.[/quote]

I think by applying Jensen's inequality in the RHS  we obtain

$ \sqrt [3]{\frac {(a^2 \plus{} b^2 \plus{} c^2)^4}{\sum(a^8 \plus{} 26a^2b^3c^3)}}$

The have the same result  applying Holder !

For the above inequalilty that Stephen has recommmended, without loss of Generality, we may assume that

$ a\ge b\ge c$.

Clearly, we have

$ \frac{a}{\sqrt{a^2\plus{}8bc}} \le 1$

$ \frac{b}{\sqrt{b^2\plus{}8ac}} \plus{} \frac{c}{\sqrt{c^2\plus{}8ab}} \le \frac{b\plus{}c}{\sqrt{b^2\plus{}8ac}} < \frac{b\plus{}c}{\sqrt{b^2\plus{}c^2\plus{}2bc}}\equal{}1$

We are done!

The nice thing is from this problem, combined with IMO 2001, we have the following interesting proposal

Let a, b,c be positive real integer, prove that

$ \sum_{cyc}\frac{a}{\sqrt{a^2\plus{}8bc}}$ is not an integer.  :coool:

[quote="great math"]
The nice thing is from this problem, combined with IMO 2001, we have the following interesting proposal

Let a, b,c be positive real integer, prove that

$ \sum_{cyc}\frac {a}{\sqrt {a^2 \plus{} 8bc}}$ is not an integer.  :coool:[/quote]

what if $ a\equal{}b\equal{}c$? :D

[quote="earldbest"][quote="great math"]
The nice thing is from this problem, combined with IMO 2001, we have the following interesting proposal

Let a, b,c be positive real integer, prove that

$ \sum_{cyc}\frac {a}{\sqrt {a^2 \plus{} 8bc}}$ is not an integer.  :coool:[/quote]

what if $ a \equal{} b \equal{} c$? :D[/quote]

I think he must edit the problem like this:

Let $ a, b, c$ be positive real numbers, prove that if $ \sum_{cyc}\frac {a}{\sqrt {a^2 \plus{} 8bc}}$ is a interger, then $ a\equal{}b\equal{}c$.

yeah, that's right, my head was spinning but never mind. :oops_sign:

$ \frac{a}{\sqrt{a^{2}\plus{}8bc}}\plus{}\frac{b}{\sqrt{b^{2}\plus{}8ca }}\plus{}\frac{c}{\sqrt{c^{2}\plus{}8ab}}\ge1$ can be proved in a different way.

Let $ \frac{a}{\sqrt{a^{2}\plus{}8bc}}\equal{}x, \frac{b}{\sqrt{b^{2}\plus{}8ca }}\equal{}y, \frac{c}{\sqrt{c^{2}\plus{}8ab}}\equal{}z$.

And let $ x\plus{}y\plus{}z\equal{}p, xy\plus{}yz\plus{}zx\equal{}q, xyz\equal{}r$.

Since $ x^2\equal{}\frac{a^2}{a^2\plus{}8bc}, \frac{x^2}{1\minus{}x^2}\equal{}\frac{a^2}{8bc}$.

So, $ \frac{x^2y^2z^2}{(1\minus{}x^2)(1\minus{}y^2)(1\minus{}z^2)}\equal{}\frac{a^2b^2c^2}{512a^2b^2c^2}\equal{}\frac{1}{512}$.

Which leads to $ (1\minus{}x^2)(1\minus{}y^2)(1\minus{}z^2)\equal{}512x^2y^2z^2$.

So, with a bit of calculation, $ p^2\plus{}2pr\plus{}513r^2\minus{}(q\plus{}1)^2\equal{}0$.

Now, let $ p<1$. So, $ 1\plus{}2r\plus{}513r^2>(q\plus{}1)^2$. Because of AM-GM, $ q\ge3r^\frac{2}{3}$.

So, $ 1\plus{}2r\plus{}513r^2>(3r^\frac{2}{3}\plus{}1)^2$.

Let $ r\equal{}t^3$. So, $ 1\plus{}2t^3\plus{}513t^6\minus{}(3t^2\plus{}1)^2>0$.

If we fatorize this, $ t^2(3t\minus{}1)(171t^3\plus{}57t^2\plus{}16t\plus{}6)>0$.

Since $ t>0$, $ 3t\minus{}1>0$.

But, because of AM-Gm, $ p\ge3r^\frac{1}{3}\equal{}3t>1$. Since we had started with $ p<1$, contradiction.

So, $ p\ge1$, and this completes the proof.


At 7,500 meters an aircraft cabin pressure is about 12 psi (0.41 atmospheres) while outside air pressure is 6 psi.
What net force is acting on a cabin door with a surface area of 20 sq. ft.?
What happens to this force if aircraft ascends to 15,000 meters?
(Use the attach below)

Ok I think I have solved for the pressure on one side of the door and the pressure on the other side of the door.
Do i add the pressures or do youo subtract because both pressure act on each other?

F=pA

F= (6psi) (20ft^2) = 120lbs

F= (12psi) (20ft^2) = 240lbs

The text says "net force", so you should subtract the pressures.


[hide]i tried with 9 and 25 and got the remainder of 1

answer=1[/hide]

[quote="shadysaysurspammed"]What does the division of a square of an odd integer by 8 always leave a remainder equal to ?[/quote]

  [hide="Solution"]

  Lets say our odd number is $2n+1$

$(2n+1)^2=4n^2 + 4n + 1$

  To solve the problem we must re-write the above equation as:

$4(n^2 + n) + 1$

  Now there are two possibilites, either $n$ is odd or $n$ is even.  Now we know that:

  An odd number times an odd number equals and odd number, we also know that and odd number plus and odd number equals and even number.

  An even number times an even number equals an even number, we also know that an even number plus and even number equals a even number.  (Both can be shown algebraicly--but im too lazy to do it right now)

  So in either case we get:

$4(2x) + 1$

or

$8x+1$

  From here it is clear that the square of any odd number leaves a remainder of 1, when divided by 8. [/hide]  ------By the way, what was the ancient greek solution to this problem?  At first I was thinking mods, but when I saw this I thought it could be easier..

Mods are the fastest way to rigorously prove these.
[hide]
odd numbers mod 8 are either 1, 3, 5, or 7.

$1^2 = 1 (mod 8)$
$3^2 = 9 = 1 (mod 8)$
$5^2 = 25 = 1 (mod 8)$
$7^2 = 49 = 1 (mod 8)$

and we're done.
[/hide]

However we can use G-UNIT's solution too, but that would require proving that odd*odd=odd and odd+odd=even. I think I'll prove those so that his proof is complete :D.

[hide]
odd integers are in the form of 2n+1 and even integers are in the form of 2n where n is an integer.

odd*odd:

$(2n+1)(2m+1) = 4nm + 2n + 2m + 1 = 2(2nm + n + m) +1$

Since $2nm + n + m$ is an integer if n and m are integers, we are done.

odd+odd:
$(2n+1) + (2m+1) = 2n+2m+2 = 2(n+m+1)$
Since $n+m+1$ is an integer if n and m are integers, we are done.

and that completes his proof :D
[/hide]

[hide] I tried n being 1 and 3, and I got 1.[/hide]

i guess the best way to solve this is guess and check, proving this is more than the scope of the forum.

Simplification on G-UNIT's proof:

[hide]$4(n^2+n)+1=4(n)(n+1)+1$
Either $n$ or $n+1$ is odd, and the other is even, since $n$ is an integer, so we have some even number multiplied by 4 plus 1, which is $8x+1$.[/hide]

I did the exact same method as G-UNIT,pretty easy...

Factorial,u cant us guesswork in these and seein the difficulty of other questions in this MC forum,this sum was amateurish.

YO guys,whazzup with the mod thing,whta is it?

Mods are actually pretty easy once you get the hang of it. Basically, if you see $a \text{ mod } b$, it means the remainder when $a$ is divided by $b$.

Example:
$5 \text{ mod } 3 = 2$

So now we can write it in different notation:
$5 = 2 (mod\text{ }3)$ which is normally how you'll see it.

There are some properties that I used without explaining:

if $a = b (mod\text{ }c)$
then $ad = bd (mod\text{ }c)$
is the only one you needed to know for my proof.

[quote="236factorial"][hide="basically a repeat, but anyway...."]If n is an even integer, than n+1 is an odd integer, but i want to use 2n+1 because I feel like it  :P

$(2n+1)^2=4n^2+4n+1$.

since n is even, $4n^2$ must be divisible by 8, and so does $4n$.
Thus the remainder is 1. [/hide][/quote]

That proof is not complete. It only shows that integers in the form of $4k+1$ work. You either need to prove that:

$n^2 + n$ is divisible by 2
or
$(2n-1)^2$ works. <-- this is probably easier.

i just used guess nd chek, nd soon got 1.

[hide="My Method"]
$\frac{(e+1)^2}{2^3}$
$\frac{e(e+2)+1}{2^3}$
$\frac{e(e+2)}{2^3}+\frac{1}{2^3}$
Smallest even is $0$ but then $r=1$
so $2$ is smallest even,
$e(e+2)$ is always divisible by $8$
So,
$\boxed{R=1}$
[/hide]

[quote="JesusFreak197"][quote="Ars"]The Ancient greeks already proved that n^2=x*8+1 where n is odd.[/quote]

Umm... your point? He's seeing who else can prove it (although no one has rigorously proven it so far).[/quote]

At least he is giving us some information instead of critising other people.

He's spamming, because his comment is not at all helpful to the question. In fact, he gave the answer away in that post, which he didn't bother to hide.


pigeonhole principle
geometry
3D geometry
sphere
combinatorics unsolved
combinatorics

Given $1000000$ points in space, show that the set of pairwise distances has at least $79$ elements.

Yes, it is very silly. First, consider one point. If there aren't 79 different distances, then all 999999 remaining points must lie on one of 78 spheres. By pigeonhole 12821 must lie on one sphere. Then take a point from that sphere, and the 12820 others must lie on one of 78 circles. Pigeonhole again gives 165 points on one circle. Take one point on that circle, and each of the other 164 must lie on one of 78 lines... but then 3 of them lie on a line which is impossible.


symmetry
function
geometry
geometric transformation
reflection

Here is a problem I post a couple of days ago:

Given $ y \equal{} f(x) \equal{} a^x$ and $ y \equal{} g(x) \equal{} log_ax$. If the graph of $ f$ and $ g$ intersects at three points, find the range of real number $ a$.

[url]http://www.artofproblemsolving.com/Forum/viewtopic.php?t=295081[/url]

I can't help representing it again because the result is simply stunning to me. They truely [b]CAN[/b] intersect at three points. What a surprise! The attached graph is plotted with $ a \equal{} 0.026$.

This example appears in Romanian High-school textbooks, to illustrate precisely this apparent paradox with the symmetry of the function-inverse graphs.

there is no paradox here.   :)

to obtain the graph of the inverse of a function, we must reflect all of its points along the line $ y\equal{}x$.  And if you think about it carefully, if a function has a point in common with the line $ y\equal{}x$ and its inverse exists, then $ f$ and $ f^{\minus{}1}$ must necessarily have a point in common.  The exponential function $ f(x)\equal{}a^x$ is decreasing for $ 0<a<1$ and it will intersect the line $ y\equal{}x$.  Hence for this case $ f$ and $ f^{\minus{}1}$ must intersect.   :)

But here, $ f$ and $ f^{\minus{}1}$ intersect at some points not on $ y\equal{}x$  :?:

@keyree10:

no problem at all.

i ONLY said in my previous post that IF $ f(x)$ intersects the line $ y\equal{}x$ then it will have a point in common with its inverse obviously.  (But the converse is not true in general!)   :lol:

And also I did NOT say that those are the ONLY points that they will have in common.  They MAY have other points in common aside from the points where $ f(x)$ intersects the line $ y\equal{}x$.  Let me give you a very simple example.

Let $ f\equal{}\{(2,5),(6,6),(5,2),(3,4)\}$.  This is a obviously a simple function.
Its inverse is $ f^{\minus{}1}\equal{}\{(5,2),(6,6),(2,5),(4,3)\}$.  They have $ 3$ points in common, namely $ \{(2,5),(6,6),(5,2)\}$

So there is really no paradox here.  I will leave it up to you to determine how we can get ALL the intersection points of $ f(x)\equal{}a^x$ and $ g(x)\equal{}log_{a}x$   :lol:

OK, Thanks. I thought you were giving that instance as an explanation to resolve the paradox completely. I understand now, that you were only pointing out one case where $ f$ and $ f^{\minus{}1}$ must necessarily intersect.
BTW, In my opinion there [i]is[/i] no paradox  :)

If $ f(x)$ is an increasing function then solution set of $ f(x) \equal{} x\Longleftrightarrow$ solution set of $ f(x) \equal{} f^{ \minus{} 1}(x)$
But in case of $ f(x)$ is decreasing,$ f(x) \equal{} x\implies$ solution set of $ f(x) \equal{} f^{ \minus{} 1}(x)$ converse may not be true.

Explanation:

Note regardless of $ f(x)$ increasing or decreasing we have
$ f(x) \equal{} x\implies f^{ \minus{} 1}(f(x)) \equal{} f^{ \minus{} 1}(x)\implies x \equal{} f^{ \minus{} 1}(x)\implies f^{ \minus{} 1}(x) \equal{} f(x)$

for converse,
Note $ f(x)$ is an increasing $ \Longleftrightarrow f^{ \minus{} 1}(x)$ is an increasing
Let $ x_{1}$ be a solution of $ f(x) \equal{} f^{ \minus{} 1}(x)$ and $ f(x_{1})\neq x_{1}$
suppose $ f(x_{1}) < x_{1}\implies f^{ \minus{} 1}(f(x_{1})) < f^{ \minus{} 1}(x_{1})$          (as $ f^{ \minus{} 1}(x)$ is an increasing)
$ \implies x_{1} < f^{ \minus{} 1}(x_{1}) \equal{} f(x_{1})$
Which is a contradiction

[size=150]Hence in case of f is an increasing function we have solution set of $ f(x) \equal{} x\Longleftrightarrow$ solution set of $ f(x) \equal{} f^{ \minus{} 1}(x)$[/size]

Now in case of a decreasing function if their is any point of intersection of the function with its inverse below the line $ y\equal{}x$ then its mirror image will be also a point of intersection.
e.g. consider $ f(x)\equal{}1\minus{}x^{3}$ and its inverse $ f^{\minus{}1}(x)\equal{}(1\minus{}x)^{\frac {1}{3}}$; Clearly they are intersecting at $ (1,0)$ and at $ (0,1),$ both the points are not on the line $ y\equal{}x$
Observe why it is happening?
Because in our example function $ f(x)$ contains two points $ (1,0)$ and $ (0,1)$ which are mirror image of each other in the line mirror $ y\equal{}x$.

[size=150]Hence whenever curve $ y\equal{}f(x)$ contains two points symmetric to the line $ y\equal{}x,$ its inverse will pass through these points which are not the points of $ y\equal{}x$ [/size]

For my solution  please follow
http://www.mathlinks.ro/Forum/viewtopic.php?p=1633590#1633590


abstract algebra
algebra
function
domain
Ring Theory
superior algebra
superior algebra unsolved

Let $ D$ a principal ideal domain,  $ F$ the field of fraction of $ D$. Show for any $ D\minus{}$module $ A$ of finite rank, not free but squarefree that
\[ card(\hom(A,F)) \leq card(\hom(A,F/D)).\]

remark: the rank of a module $ A$ is the maximal number of linearly indipendent element in $ A$.

[quote="Mr B"]Let $ D$ a principal ideal domain,  $ F$ the field of fraction of $ D$. Show for any $ D \minus{}$module $ A$ of finite rank, not free but square free [/quote]What do you mean by "squarefree"? That the elementary divisors of the torsion sub-module of $ A$ are square free?


Have you figured it out or are you just giving us really tedious numbers to work with on purpose?

ok ok ill make it easier

make a chart of the distance traveled in minutes

[quote="akalra1"]I fell like posting a problem in this forum here goes:

Amelia and Bamelia are friends. They live only 50 miles away from each other. Amelia has a pet hamster that can run 37 miles an hour. Bamelia has a pet molerat that can run 43 miles an hour. Assuming that Amelia and Bamelia let their pets loose and the pets run to each other's houses, how far will they be away from Bamelia's house when the two pets meet?

NO QESTIONS ABOUT THE NAMES![/quote]
[hide=" :rotfl: "][hide="Why cant we ask questions about the names..."]Combined, the hamster and molerat run 80 miles an hour. Since they have to run 50 miles, it will take them $\frac{5}{8}$ of an hour. $43 \cdot \frac{5}{8}=\frac{215}{8}$, which is $\boxed{26\frac{7}{8}}$ miles away from Bamelia's house.[/hide][/hide]


algebra
polynomial
calculus
calculus computations

If $Q(x)$ is the quotient when  $x^{60}-8x^3+4x^2+2x+1$ is divided by $(x-1)$, then find the sum of the coefficients of $Q(x)$

[quote="Altheman"]If $Q(x)$ is the remainder when  $x^{60}-8x^3+4x^2+2x+1$ is divided by $(x-1)$, then find the sum of the coefficients of $Q(x)$[/quote]

$x^{60}-8x^3+4x^2+2x+1=x^{60}-x^3-7(x^3-x^2)-3(x^2-x)-(x-1)$ is divisible by $x-1$ so $Q(x)=0$ :?

$Q(x)$ is [b]not[/b] the initial equation, but it is the equation after you divide by $(x-1)$

This can be done purely algebraically, but there's a reason this was posted in the Calculus section. Look for a nice way to do it...

[hide="What jmerry was talking about. "]For any polynomial $P(x)$ the sum of the coefficients of $P$ is $P(1).$ In this particular case, $P(1)=1-8+4+2+1=0$ so $P(x)$ is divisible by $x-1.$(as more or less promised and as hardsoul rediscovered). That's not even terribly important; we could have a remainder and it wouldn't change much of anything below.

Write $P(x)=(x-1)Q(x).$ Differentiate.

$P'(x)=Q(x)+(x-1)Q'(x).$ Now set $x=1.$

$P'(1)=Q(1)+0=Q(1).$

We were asked to compute $Q(1).$ We now know it is the same as $P'(1).$

$P'(x)=60x^{59}-24x^2+8x+2.$

$Q(1)=P'(1)=60-24+8+2=46.$[/hide]

Does it mean that whenever we divide a polynomial by [b]x-1[/b], we are basically differentiating it? :huh:

(gee, I feel so stupid)

EDIT: What am saying???
i guess brains do not work well by midnight

please disregard this post  :oops:

I am absolutely confused with your definitions! Altheman asked for [b]remainder[/b] and ended up finding coefficients of the [b]quotient[/b]??? I'm too confused and need to be clarified the difference between these two terms  :stretcher:

The short version: The problem was stated wrong, and should have said quotient all along.

I'll edit Altheman's original post so it says "quotient" instead of "remainder." I think both jmerry and read it as if it said "quotient," and that's what it should say for the problem to make much sense.

Wow, this problem was deceivingly simple, I like it.


function
limit
real analysis
real analysis unsolved

Let $ f: \mathbb{N}^* \to \mathbb{N}^*$ be a bijective function so that there exists $ \lim_{n \to \infty}{\frac{f(n)}{n}}$. Evaluate $ \lim_{n \to \infty}{\frac{f(n)}{n}}$.

I am not sure how to make the following rigorous, but note that if $ f$ is a bijection then so is $ f^k(n)$ for every $ k$, and since $ f(n)$ must be unbounded in order to be a bijection one has

$ \lim_{n \to \infty} \frac {f(f(n))}{f(n)} \frac {f(n)}{n} \equal{} \left( \lim_{n \to \infty} \frac {f(n)}{n} \right)^2$

and similarly

$ \lim_{n \to \infty} \frac {f^k(n)}{n} \equal{} \left( \lim_{n \to \infty} \frac {f(n)}{n} \right)^k$.

Thus if the limit exists, one can repeatedly compose the function with and construct a sequence of bijections from $ \mathbb{N}$ to $ \mathbb{N}$ such that $ \lim_{n \to \infty} \frac {f^k(n)}{n}$ can either grow without bound, be arbitrarily close to zero, or be $ 1$.  The first implies that $ f^k$ cannot be onto, the second that $ f^k$ cannot be one-to-one, so the third is the only correct answer.

Let $ N > 1$.
First, suppose $ f(\{1,2,3,\cdots,N \minus{} 1\}) \equal{} \{1,2,3,\cdots,N \minus{} 1\}$.
Then $ f(N) \equal{} N$ or
$ f(N) > N$.
When $ f(N) > N$, there is an $ M > N$ with $ f(M) \equal{} N < M$.
So, in this case, there is always an $ M \ge N$ with $ f(M)\le M$ and an $ M$ with $ f(M)\ge M$.

Second, if $ f(\{1,2,3,\cdots,N \minus{} 1\}) \ne \{1,2,3,\cdots,N \minus{} 1\}$.
Then  $ f(\{1,2,3,\cdots,N \minus{} 1\})$ contains a maximal element $ k\ge N$ and misses a positive integer $ m$ with $ 1 \le m < N$.
Then there is an element $ M \ge N$ with $ f(M) \equal{} m < M$
and there is an $ P \ge N$ with $ f(P) \ge P$, because if $ f(p) < p$ for all $ p \ge N$, then the maximal element of the set
$ f(\{1,2,3,\cdots,k\})$ is $ k$. So $ f(k \plus{} 1) \ge k \plus{} 1$.
Then, in this case, there is always an $ M \ge N$ with $ f(M)\le M$ and an $ M$ with $ f(M)\ge M$.
Ergo, The only possible limit is 1


exterior angle
algebra
system of equations

Angle BAC and the exterior angle CBD of triangle ABC are each bisected. If the angle formed by these bisectors when they meet is 50 degrees, find the number of degrees of Angle C of triangle ABC

Because it's hard to do this without a picture,

http://artpad.art.com/gallery/?ivn96ciz3a4

Is my answer.

(Probably want to watch it at medium speed).

or you can go to this old page...

http://www.artofproblemsolving.com/Forum/viewtopic.php?&t=30021

OMFG THATS SOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOO AWESOME

wow talk about not being to scale :rotfl:

Anyway I did it by letting $m\angle C=y$ and half of $m\CBD=x$ then solve a system of equations like you did


analytic geometry
graphing lines
slope
function
algebra
linear equation

[quote="soccernut15"]1.  The cost for 7 dance lessons is $\$82$. The cost for 11 lessons is $\$122$.  Write a linear equation to find the total cost C for L lessons.  Then use the equation to find the cost for 4 lessons.[/quote]
[hide]$f(x)=10x+12$

For 4 dance lessons, the price is $\$52.$[/hide]

[hide]

Using the slope intercept formula u find that the slope is 10 and the y intercept is 12. Therefore, your linear equation is C = 10L + 12. To solve the second part, substitute 4 for L in the equation so you get 52 dollars.

[/hide]

[quote="soccernut15"]1.  The cost for 7 dance lessons is $\$82$. The cost for 11 lessons is $\$122$.  Write a linear equation to find the total cost C for L lessons.  Then use the equation to find the cost for 4 lessons.[/quote]

[hide=" :juggle: ][color=violet]$f(x)=10x+12$
plug in 4 for x, and u get, $f(4)=10(4)+12$
which is equal to $52 dollars$[/color][/hide]

[hide]f(x)= 10x+2

the answer should  :roll:  be 52

-soccerstar123[/hide]

Ilikepie, can you please explain the f(x) thingy?

$f(x)$ is a function of $x.$ You plug the value of $x$ into the equation and get your output.

If $f(x)=3x+4,$ then for each $x,$ you mupltiply it by 3 and add 4.

$f(2)=3\cdot2+4=10$

[url]http://www.artofproblemsolving.com/Wiki/index.php/Function[/url]


linear algebra
group theory
linear algebra unsolved

Let A and B be $n\times n$ matrices such that AB is invertible.  Prove that A and B are invertible.

Depends on what you know about matrices: for example, the identity $\det(AB)=\det A\det B$ would tell you that $\det A\ne 0$ and $\det B\ne 0$, hence the matrices are invertible.

You can just write down inverses of $A$ and $B$ in terms of $A$, $B$, and $(AB)^{-1}$. owk

I don't think that's enough. You can either do it mlok's way or [url=http://www.mathlinks.ro/Forum/viewtopic.php?p=780970#780970]this way[/url] (it's a proof from a linear algebra book - well, I tweaked it a little).

You could go group theory way. If $A$ and $B$ are invertible then $A,B\in M_{n}(k)$ is a group (choose the appropriate field $k$). So $AB\in M_{n}(k)$, and since $M_{n}(k)$ is a group $AB$ has an inverse... or I think maybe a variant on what owk was going for is $(AB)^{-1}=B^{-1}A^{-1}$ and since $B^{-1}$ exists and $A^{-1}$ exists, the product exists, therefore $(AB)^{-1}$ exists. Neither of these are really $\emph{proofs}$ though. They both rely on knowing something else.

The calculations $A\cdot B(AB)^{-1}=1$ and $(AB)^{-1}A\cdot B=1$ show that $A$ has a right inverse and $B$ has a left inverse. In the finite-dimensional case, this implies invertibility. (Use the equivalence of injectivity and surjectivity which follows from $n=\dim\ker+\dim\mathrm{im}$.) owk


inequalities
geometric series
inequalities unsolved

While solving one of the problems from algebra section i got this kind of inequality: prove or disprove: let $n$ be an even number then for any $x>0$ we have \[ \frac{x^n+1}{2}\geq \sqrt{\frac{x^2+x^4+x^6+...+x^{2(n-1)}}{n-1}} \] and I don't know what to do with it  :(

Sum the RHS by geometric series?  We get $\sqrt{ \frac{x^{2n} - x^2}{(n-1)(x^2 - 1)} }$ and I'm not really sure what to do with that because $x$ can be less than one...


Solve equation :$ \sqrt {x \minus{} 2} \plus{} \sqrt {4 \minus{} x} \plus{} \sqrt {2x \minus{} 5} \equal{} 2x^2 \minus{}5x$

Because of the square roots $ 2.5 \le x \le 4$. You can check that $ x\equal{}3$ is a solution.

[quote="Lamb Biryani"]Because of the square roots $ 2.5 \le x \le 4$. You can check that $ x \equal{} 3$ is a solution.[/quote]

Actually, $ x\equal{}3$ is the only solution.

[hide="Solution"]First we check that $ 2x\minus{}5\equal{}0$ does not provide a solution, so we divide by $ \sqrt{2x\minus{}5}$ (under the assumption that $ 2x\minus{}5>0$) and get
\[ x\sqrt{2x\minus{}5}\equal{}\sqrt{\frac{x\minus{}2}{2x\minus{}5}}\plus{}\sqrt{\frac{4\minus{}x}{2x\minus{}5}}\plus{}1\]
\[ \equal{}\sqrt{\frac12\plus{}\frac1{2(2x\minus{}5)}}\plus{}\sqrt{\minus{}\frac12\plus{}\frac3{2(2x\minus{}5)}}\plus{}1.\]
Note that $ x\sqrt{2x\minus{}5}$ is increasing on $ (2.5,4]$ while $ \sqrt{\frac12\plus{}\frac1{2(2x\minus{}5)}}\plus{}\sqrt{\minus{}\frac12\plus{}\frac3{2(2x\minus{}5)}}\plus{}1$ is decreasing, so there can be at most one solution, which can easily be found to be $ \boxed{x\equal{}3}$.[/hide]


So the Titan Invitational is January 24th at Northview High School. Make sure you guys all register and come out to participate. Thanks.

Can you tell us the format?

Norcross should be going.

Is Titan Invitational another name for the Northview Tournament?

I'll try to come as part of the Parkview math team, though I need to check my schedule first.

Where do we register? Also, how much is it for individual registration? And, what is the format/schedule?

Hmm, well I'm only a test writer, but here's the format:

Written portion: 25 multiple choice, 5 free response
Ciphering: 10 ciphering
Relay: 3 relay rounds with 4 parts apiece

I think invitations were sent out way back, so I dunno.  You'd have to contact Mr. Stone or FrisbeeBaller425, whose name I cannot utter for fear of....uh.....FrisbeeBaller425...

I, being merely a test writer and not an awesome officer like FrisbeeBaller425, cannot remember the scoring system, so do not think that the rest of our math team is as ignorant and/or disorganized.  It's just me... :roll:

Wow Sitan...you're actually afraid of Heng?

What's the difficulty level of the tourney?  AMC?  AIME?  USAMO?

The difficulty is around the AMC & AIME level.
So far, Heng has finished picking the ciphering round questions. The written questions will be done by Thursday, and I have no idea when the relays will be done.

There will be written 30 MC, 5 FR for 200 points
There will be ciphering for 100 points
There will be 3 relays for another 300 points

Yes, it is the northview tournament, contact stone@fultonschools.org for registration.

[quote="Math Geek"]Hmm, well I'm only a test writer, but here's the format:

Written portion: 30 multiple choice, 5 free response
Ciphering: 10 ciphering
Relay: 3 relay rounds with 4 parts apiece

I think invitations were sent out way back, so I dunno.  You'd have to contact Mr. Stone or FrisbeeBaller425, whose name I cannot utter for fear of....uh.....FrisbeeBaller425...

I, being merely a test writer and not an awesome officer like FrisbeeBaller425, cannot remember the scoring system, so do not think that the rest of our math team is as ignorant and/or disorganized.  It's just me... :roll:[/quote]

Also there will be a laser tag tournament. Only with real lasers. And lightsabers. :starwars:

OK, maybe that was just an excuse to use that smiley.

But yay titan invitational! And yea, invitations were sent in late October i think.

Whoops, sorry for the double post.  Dunno how that happened...


$ \sum\frac{ab}{(a\plus{}b)^2}<\equal{}\frac14\plus{}\frac{4abc}{(a\plus{}b)(a\plus{}c)(b\plus{}c)}$    $ a,b,c>0$

[quote="Kamyar1991"]$ \sum\frac {ab}{(a \plus{} b)^2} < \equal{} \frac14 \plus{} \frac {4abc}{(a \plus{} b)(a \plus{} c)(b \plus{} c)}$    $ a,b,c > 0$[/quote]
It is equivalent to $ (a\minus{}b)^2(a\minus{}c)^2(b\minus{}c)^2\geq0.$

The problem is equivalent to

\[ \sum_{cyc}(a\minus{}b)^2 (\frac{(a\minus{}c)(b\minus{}c)}{a\plus{}b})\ge 0\]

Now you can use $ SOS$ or $ SCHURE$ inequalities :wink:

very good and intelligence solution. my solution is very bad i multiple all of them and then i proved it. :blush:

It's one my old ineq (since 2006), has posted on toanthpt.net
My solution (also the way to creat it):
Let $ x \equal{} \frac{a\minus{}b}{a\plus{}b}$ ... $ \Rightarrow x\plus{}y\plus{}z\plus{}xyz\equal{}0\[ \Rightarrow x^2\plus{}y^2\plus{}z^2\plus{}2(1\plus{}x)(1\plus{}y)(1\plus{}z)\equal{}2\plus{}(x\plus{}y\plus{}z)^2  \geq\ 2\] \Rightarrow \sum\ \frac{(a\minus{}b)^2}{(a\plus{}b)^2}  \plus{} \frac{16abc}{(a\plus{}b)(b\plus{}c)(c\plus{}a)} \geq\ 2$$ \Rightarrow \sum\ \frac{ab}{(a\plus{}b)^2}  \leq\ \frac{4abc}{(a\plus{}b)(b\plus{}c)(c\plus{}a)}  \plus{} \frac{1}{4}$
Solution 2 is same to arquady:
Give $ b\equal{}c$, ineq is always trues. So, it equivalent to:
$ (a\minus{}b)^2(b\minus{}c)^2(c\minus{}a)^2 \geq\ 0$
 :)


Find all the integers m,n such that $E(2^{1/2}m)=E( (2^{1/2}+2)n )$.
Please help me!!

See the title, it's floor function. Don't know who spreads this strange symbol, but that's not the first time someone here uses it.

Sorry but is a easy problem is just a special case of a theorem, 1/a+1/b=1 then A={[ma]/ m integer-{0}} and B={[mb]/ m integer-{0}}
are disjoint, it's easy to prove......

sorry but we dont have as function in high-school so explaine please
i think that it has no solution  :o

Suppose that exist $q=[ma]=[nb]$ then $q< ma<q+1$ and $q<nb<q+1$ then $q/{m}<a<(q+1)/{m}$ and $q/{n}<b<(q+1)/{n}$ then $m/{q}>1/a>m/{q+1}$ and $n/{q}>1/b>n/{q+1}$ adding up....
$(m+n)/{q}>1>(m+n)/{q+1}$ which is equivalent a q+1>m+n>q Impossible...!!!!!! :lol:
When m, n are negative is the same!!!!!!


function
real analysis
real analysis unsolved

Let $q$ be continiously differentiable function such that there exists $x_0>0$ s.t. $q(x)>0$ and $q'(x)>0$ for all $x>x_0$.
Prove that all the solutions of the differential equation $y''+q(x)y=0$ are bounded near the $+\infty$

already discussed: look at $y^2+\frac{(y')^2}{q}$


Let x,y,z>0 such that xyz=1
Prove that :
((x+y+z)/3) 3  \geq  \sqrt ((x 2 +y 2 +z 2 )/3)

Quite easy, but nice. First, it is clear that ( \sum x) 2  \geq 6+ \sum x 2 . Thus, it suffices to prove that ((6+ \sum x 2 )/9)^3 \geq  \sum x 2 /3. Put  \sum x 2 =3+3x, with x>=0 and reduce it to (1+x/3)^3>=1+x, obvious.


let $m$ be natural and odd number and $a$ be an intiger and$(a,m)=1$ prove that $x^2\equiv a\ (\mod\ m)$

has $\prod_{p|m}\left(1+\left(\frac{a}{p}\right) \right)$ answers.

There are solutions iff there are some modulo all prime divisors:
If at least one of the Legendre symbols is $-1$, there are no solutions and the term also gives $0$.
If all are $=1$ (since $0$ is excluded), the number of solutions is clearly identical to that of $x^2 \equiv 1 \mod m$, and the $\mathrm{si}$-formula gives the disired result (see http://www.mathlinks.ro/Forum/viewtopic.php?t=15033). Alternativly you can observe that there are exactly $2$ solutions modulo every prime power dividing $m$, so there are $2^k$ solutions by the chinese remainder theorem ($k$ is the number of different prime divisors of $m$), which is what the term gives.


Αν $ x,y,z> 0$ και $ x\plus{}y\plus{}z\equal{}1$ να αποδείξετε ότι:
$ \frac{x}{y^2\plus{}z^2}\plus{}\frac{y}{x^2\plus{}z^2}\plus{}\frac{z}{x^2\plus{}y^2}>\frac{9}{4}$

Πιστεύω ότι το φράγμα αυτό είναι αρκετά αδύναμο αλλά προς το παρόν δεν μπορώ να το βελτιώσω...

Με μία (πολύ) πρόχειρη ματιά, μου φαίνεται ότι το $ 4$ είναι το πιο σφιχτό φράγμα που μπορείς να πετύχεις...

Cheerio,

Durandal 1707

Μία απόδηξη με κάτω φράγμα $ 4$....
$ \sum\frac {x^2}{xy^2 \plus{} xz^2} \ge \frac {1}{\sum_{sym}x^2y}$
Etsi exoume ndo:
$ \frac {1}{\sum_{sym}x^2y}\ge 4 \Longleftrightarrow x^3 \plus{} y^3 \plus{} z^3 \plus{} 3\sum_{sym}x^2y \plus{} 6xyz \ge 4\sum_{sym}x^2y\Longleftrightarrow x^3 \plus{} y^3 \plus{} z^3 \plus{} 6xyz \ge \sum_{sym}x^2y$
Apo Schur $ x^3 \plus{} y^3 \plus{} z^3 \plus{} 3xyz \ge \sum_{sym}x^2y$.Telos..
Tha koitaksw kai egw argotera mipos vrw ena kalitero katw fragma.....

[quote="Dimitris X"]Tha koitaksw kai egw argotera mipos vrw ena kalitero katw fragma.....[/quote]

Μην παιδεύεσαι: για $ x \equal{} y \equal{} 1/2, z \equal{} 0$, η LHS δίνει $ 4$ οπότε, από συνέχεια... :D

Cheerio,

Durandal 1707

[quote="Durandal"]
Μην παιδεύεσαι: για $ x \equal{} y \equal{} 1/2, z \equal{} 0$, η LHS δίνει $ 4$ οπότε, από συνέχεια... :D

Cheerio,

Durandal 1707[/quote]

Πρόσεξε όμως ότι $ x,y,z>0$ :)

[quote="Dimitris X"][quote="Durandal"]
Μην παιδεύεσαι: για $ x \equal{} y \equal{} 1/2, z \equal{} 0$, η LHS δίνει $ 4$ οπότε, από συνέχεια... :D

Cheerio,

Durandal 1707[/quote]

Πρόσεξε όμως ότι $ x,y,z > 0$ :)[/quote]

Epektinei tin anisotita gia $ x,y,z \geq 0$ kai gia tin dothisa anisotita leei oti to infimum einai to 4 logo sinexeias...

I'm very sorry for butting in this forum where I don't belong to.
[hide="Bilot4"]Bilot4(dimitri): This is Tim Lyoo(RSI) I've sent you a private message on mathlinks so check your message box :)  Its been a long time[/hide]
Another solution:
We will show the following.
$ \frac {x}{y^2 \plus{} z^2}\geq \frac {4x^3}{x^3 \plus{} y^3 \plus{} z^3}$
 Adding up the three fractions implies the result.
This is equivalent to$ \frac {x(x \plus{} y \plus{} z)}{y^2 \plus{} z^2}\geq \frac {4x^3}{x^3 \plus{} y^3 \plus{} z^3}$
or
$ (x \plus{} y \plus{} z) \times (x^3 \plus{} y^3 \plus{} z^3)\geq 4x^2y^2 \plus{} 4x^2z^2$
by Cauchy Schwartz we have
$ (x \plus{} y \plus{} z) \times (x^3 \plus{} y^3 \plus{} z^3)\geq (x^2 \plus{} y^2 \plus{} z^2)^2$
so we only need to show
$ (x^2 \plus{} y^2 \plus{} z^2)^2\geq 4x^2y^2 \plus{} 4x^2z^2$
which is equivalent to
$ (x^2 \minus{} y^2 \minus{} z^2)^2\geq 0$ which is obviously true.

I've sent the message (I didn't know you were online yet) doesn't have much in it though ...
 :)  :)  I'm not sure but did you have an MSN id? (I've been too lazy these days)

[quote="Durandal"]
Μην παιδεύεσαι: για $ x \equal{} y \equal{} 1/2, z \equal{} 0$, η LHS δίνει $ 4$ οπότε, από συνέχεια... :D
Cheerio,
Durandal 1707[/quote]

Ωραίος! Και ωραία λύση Δημήτρη! Τελικά πράγματι η ανισότητα ήταν πολύ χαλαρή :blush:

@Tim: Marvelous solution my friend!! I can't believe that you found me here! I am so glad that I'll see you (and Bob) in IMO :)

να δωσω και μια δικια μου λυση στο θεμα.
 (θεωρω x,y,z μη αρνητικους πραγματικους)

αρκει να δειξουμε οτι $ \sum xy(x \plus{} y) \leq \frac {1}{4}$, ευκολα απο B-C-S

παρατηρουμε ομως οτι το αριστερο μελος της προς αποδειξη ανισοτητας μας γραφεται,
$ \sum xy(x \plus{} y) \equal{} (x \plus{} y \plus{} z)(xy \plus{} yz \plus{} zx) \minus{} 3xyz \equal{} \sum xy \minus{} 3xyz$
$ \equal{} xy(1 \minus{} 3z) \plus{} z(x \plus{} y)$
Θα διακρινουμε τωρα 2 περιπτωσεις,
-για $ z > \frac {1}{3}$ ισχυει,
$ xy(1 \minus{} 3z) \plus{} z(x \plus{} y) \leq z(1 \minus{} z) \leq \frac {1}{4}$, το ζητουμενο

-για $ z \leq \frac {1}{3}$, ισχυει
$ xy(1 \minus{} 3z) \plus{} z(x \plus{} y)\leq \frac {(1 \minus{} z)^2(1 \minus{} 3z)}{4} \plus{} z(1 \minus{} z) \equal{} A$,

(χρησιμοποιησα οτι $ xy \leq \frac {(x \plus{} y)^2}{4} \equal{} \frac {(1 \minus{} z)^2}{4}$
,ισοτητα για x=y)

Συνεπως αρκει να δειξουμε οτι,

$ A \leq \frac {1}{4}$

το οποιο γραφεται ισοδυναμα

$ z(3z^2 \minus{} 3z \plus{} 1) \geq 0$,που ισχυει.  :)

Η ισοτητα παρατηρειται στην πρωτη περιπτωση για $ (x,y,z) \equal{} (\frac {1}{2},0,\frac {1}{2})$ ή $ (x,y,z) \equal{} (0,\frac {1}{2},\frac {1}{2})$
και στην δευτερη περιπτωση για $ (x,y,z) \equal{} (\frac {1}{2},\frac {1}{2},0)$


modular arithmetic
number theory
relatively prime

Find the smallest positive integer k such that 1^2+2^2+3^2+...+k^2 is a multiple of 200.

I'll post a bit of the solution: note I haven't excluded other cases, I'll just show how to get to our case.

[hide]$k(k+1)(2k+1) = 1200 = 3(5^{2})(2^{4})$

Note all three of those terms are relatively prime, So:

1.) $k$ is even so $16 | k$
2.) $k+1$ is even so $16 | (k+1)$

The first case yields the result we want: I'll use $k \equiv a\pmod{b}\equiv a(b)$ to save space (and effort).
$k \equiv 0 (16) \Rightarrow k = 16t$
And $(k+1)(2k+1) \equiv 0(3) \Rightarrow 2k^{2}+3k+1 \equiv 0(3)\Rightarrow k^{2}\equiv 1(3) \Rightarrow t^{2}\equiv 0(3) \Rightarrow t \equiv 1,2 (3)$.
A quick check shows that $t \equiv 1(3)$ is best by a little CRT with $2(16t)^{2}+3(16t)+1 \equiv 0(25) \Rightarrow 12t^{2}-2t+1\equiv 0(25)$, if we substitute $t = 3s+1$. We find that $s = 2$, $t = 7$, and $k = 112$.
Of course, I think this much thinking isn't necessary. It's 1 AM, so I think rather you can divide the case up easier since the terms are relatively prime and one has to divide only $3$ and the other only $25$, but whatever.[/hide]


Is there any contest on math or science research where you can enter without school, by that mean you can attend by just filling your own qualification and stuffs?

I know USAMTS does that and I'll probably enter that one.

But is there any other ones?

With prizes and scholarships... If possible. :D  :D  :D  :D

Take a look at http://www.mathlinks.ro/Forum/contest.php :D :P

For a more comprehensive list, look at [url]http://www.artofproblemsolving.com/Resources/AoPS_R_Contests.php[/url]

If you are particularly interested in research, the Siemens Westinghouse is the one to shoot for.

here's another list where you might find something:
[url]http://www.academiccompetition.org/contests.html[/url]

Do you program? If so, be sure to  check out [url]http://www.usaco.org[/url]


trigonometry
calculus
calculus computations

show that
sin(cosx)=cos(sinx) has no solution

hello, use that
$ \sin(x)\minus{}\cos(x)\equal{}\minus{}2\sin(\pi/4\minus{}x/2\minus{}y/2)\sin(\pi/4\minus{}x/2\plus{}y/2)$
Sonnhard.

using $ \sin (\frac{\pi}{2}\plus{}x)\equal{}\cos x$
then $ \sin (\cos x) \equal{} \cos ( \sin x) \to$
$ \sin( \cos x)\equal{} \sin (\frac{\pi}{2} \plus{}\sin x)$

since $ \frac{\pi}{6} < \frac{\pi}{2}\plus{}\sin x < \frac{5\pi}{6}$ and $ \minus{}\frac{\pi}{3} < \cos x < \frac{\pi}{3}$

and sine is positive in the first interval, we know $ \cos x \equal{} a\in [0.1]$

since $ \sin a \equal{} \sin (\pi \minus{}a)$ we know

$ \frac{\pi}{2}\plus{}\sin x \equal{} a$ or $ \pi\minus{}a$ so $ \pm \sin x \equal{} \frac{\pi}{2}\minus{}a$

$ \sqrt{1\minus{}a^2}\equal{} \frac{\pi}{2}\minus{}a$

$ 1\minus{}a^2\equal{} \frac{\pi^2}{4}\minus{}a\pi\plus{}a^2\to 2a^2\minus{}a\pi\plus{}\frac{\pi^2\minus{}4}{4}\equal{}0$

$ a\equal{}\frac{\pi \pm \sqrt{ \pi^2\minus{}2(\pi^2\minus{}4)}}{4}\equal{}\frac{\pi \pm \sqrt{8\minus{}\pi^2}}{4}$

since there are no real solutions, we cannot have such an x.

Or just note that for $ |x|<\frac\pi 2$, one has $ \cos(\sin x)\ge \cos x>\sin(\cos x)$ and that for $ \frac\pi 2\le |x|\le\pi$, one has $ \cos(\sin x)>0\ge\sin(\cos x)$.


function
induction
algebra proposed
algebra

[quote="khanhtrung"]Let positive integer k. N : set of all positive integers n >= k
Find all functions from N to itself satisfy: f(m 2 +f(n)) = (f(m)) 2  +n  for all m,n[/quote]
IMO92: Find all function f:R->R:f(x 2 +f(y))=y+(f(x)) 2
i think everyone can solve IMO92, but noone can solve this pro similar it :D
this pro very easy for khanhtrung and me :P

i think we should solve this equation on Q
it's very difficult???
goodluck


[quote="vishalarul"][quote="sharkman"]A car rental agency charges 35 dollars for the first day and 22 dollars for each additional day. If John's car rental bill was 167 dollars, for how many days did John rent the car?

MY WORK:

first day = 35 dollars

22 dollars times 6 days = 132 dollars

I then added 132 dollars to 35 dollars = 167 dollars

My answer is first day + 6 days = 7 days.

The book's answer is 6 days.

How can this be?[/quote]

Maybe the book doesn't count the first day.

But I think that you are right...[/quote]
167-35=132
132/22=6
6+1=7
you are right in funcia world at least. :huh:
@sharkman:yeah maybe it said "additional days" and you didn't notice


My son just finished one of his class this week, since I knew he would have a tournament on last day, so I kind of take a peek once in a while just to see how was he doing, oh, God, he was so nerves that he messed up even the questions that he knows very well.......

I think one of the reasons for this may be : Being in a school that is not interested in Math competing (they don't have Mathcount, Math League or AMC ...etc), that  he rarely got chances to participate Math contests.

He is extremely happy that he found this site, and learned so much from the class and forum, he is also hoping when he finally finishes his middle school, he will get opportunity in high school to participate some math contests....

However, I think doing well in contests not only requires solid knowledge, it also ask for the ability to concur the stresses and nerveness on contest.....which mostly comes from experiences......and I believe most beginners will be lacking of this kind of experiences....

I see there are a lot of people with advanced math knowledge here posting tournament test  or  answering questions for new learners, you are just great.  Is it possible that you guys also set up something just like the last day of the class, make it live competition?  it can be done in a chat room, preannounce a scheduled time, then when a question is posted, whoever answered the question first and right got the highest points, or something like that.........I truly believe all  the new beginners can benefit from it.....Thanks in advance.

Keep an eye out for some of our MATHCOUNTS Math Jams; we'll be holding several throughout the year.  We occasionally run these just as you suggest - we make a little contest out of it.

If there's interest, we may try something similar with slightly harder problems.

The weekly math chats on friday might be good, although so far no one has kept score

We have done this before, I think.... joml88 should host this, and I would be glad to keep score or make use of myself in withever way suits everyone...

I can do one at the Mathcounts level.  I just need to put some problems together.  I would also like to know how many ppl will participate.  I don't really want to do this if no one participates.  What date and time do you wanna do it?  Tomorrow is too soon.  So maybe at the end of next week.

Thanks for all your reply and consideration.

I will  mention this on Mathcount board so more people will be aware about this and sign up.

Thanks again.


How many real solutions does the following system of equations have? Justify

x+y=3
3xy-z^2=9

$ z^2\ge 0$ so let $ z\equal{}0$ to maximize the equation $ 3xy\minus{}z^2$

From AM-GM $ \frac{x\plus{}y}{2}\ge \sqrt{xy} \Longrightarrow 3xy \le \frac{27}{4}$

Since $ \frac{27}{4} \le 9$ there are no real solutions

Do not double post:
[url]http://www.artofproblemsolving.com/Forum/viewtopic.php?t=249446[/url]


inequalities
function
inequalities unsolved

[i]Second Day – December 13, 1997[/i]
4. Let $e > 0$be a given real number. Find the least value of $f(e)$(in terms
of $e$ only) such that the inequality
$a^3 + b^3 + c^3 + d^3 \le e^2(a^2 + b^2 + c^2 + d^2) + f(e)(a^4 + b^4 + c^4 + d^4)$
holds for all real numbers $a, b, c, d$.

[quote="Lovasz"]I don't understand, if f(e) is a number?[/quote]

I think it is a function of  $e$ .e.g  $f(e)=e^2+1$


What does everyone here normally get (aka on average) when taking a past AIME for practice?
I typically get 8-12, depending on my stupid mistake count.

This year,I usually get like a 9.  The best I've done was an 11, the worst a 7.

[quote="georgiamathdude"]what's a woot aime?[/quote]

practice AIMEs given in the AOPS class [url=http://www.artofproblemsolving.com/Classes/AoPS_C_WOOT.php]WOOT[/url]

for me, i usually get 7-9, with a 13 on 2000 AIME II. and a... uh 5 on 1994 AIME.

I've never gotten more than a 3 (except a complete fluke 8 when I kept looking at answers)

Haven't done AIME for a while...

Last year I got around 8-9 on practice AIMEs (all recent years, typically solved 10-11) and ended getting a 6 on the real thing.


quadratics
number theory solved
number theory

Show that for each prime p there exists an integer x s.t. p | x8-16.

If 2 is quadratic residue mod p then  it's obvious since we can even find a such that p|a^2-2. The same if -2 is a quadratic residue mod p. Now observe that x^8-16=(x^2-2)(x^2+2)((x-1)^2+1)((x+1)^2+1). Suppose also that (2/p)=(-2/p)=-1. Then (-1/p)=1 which means that we can find x such that p| (x-1)^2+1. And we are done.

That's so cool! I did something slightly different: if (2/p)=(-2/p)=-1, it means that p=8k+5, and then I showed that x8-16<>0(p) for all x is equivalent to the fact that there is an x s.t. p|x4+1, which means that x has order 8 in the multiplicative group of Zp, which means that p=8k+1, so we have a contradiction. This is the short form, the actual solution is a bit longer. Anyway, I like yours much better :).

Why from (-1/p)=1we know that we can find x such that p| (x-1)^2+1?

if it's spam sorry....We know that exist x: $ x^2\equal{}\minus{}1 modp$I should prove that 2x=1 or 2x=-1 modp and this finishes the problem but I can't.

Hello grobber, will u explain me how u arrived at p divides $ x^{4}\plus{}1$

Slightly different than the above:
If $ p \not\equiv 1 \mod 8$, then being a $ 8$-th power is equivalent to being a $ 4$-th power, thus $ 16$ is a $ 8$-th power.
So $ p \equiv 1 \mod 8$, giving that $ 2$ is a square $ \mod p$, giving again the result.

Note that this is a very simple counterexample to Grunwalds theorem which he "proved" about 75 years ago. It was reproven some years later, and it lasted about a decade before the error was spotted (by Wang i believe). It can be corrected, giving the Grunwald-Wang-theorem.
See http://www.mathlinks.ro/viewtopic.php?p=1271603#1271603 and http://www.mathlinks.ro/viewtopic.php?p=1194384#1194384 .

matex (L)(L)(L): Where is your problem¿ Find $ y$ with $ p|y^2 \plus{} 1$ and set $ x \equal{} y \plus{} 1$.


geometry
geometric transformation
reflection
perpendicular bisector
similar triangles
Pythagorean Theorem

the basic idea is that the crease is a perpendicular bisector to the diagonal $AC$.
then you can solve with similar triangles :)

I was able to find the length of the crease but i was quick to notice that the fold is perpindicular to the length of AC.

[hide]Suppose $AB=CD=4$ and $AD=BC=3$.  Draw $AC$, and let $M$ be the midpoint of $AC$.  Draw the perpendicular bisector of $AC$.  Let $X$ be the intersection point of $AB$ and the perpendicular bisector.

By the Pythagorean Theorem on triangle $\bigtriangleup ABC$, $AC=5$.  Therefore, $AM=\frac{5}{2}$.  By $AA$, triangles $\bigtriangleup ABC$ and $\bigtriangleup AMX$ are similar.  Therefore, $\frac{BC}{AB}=\frac{MX}{AM}$.  Substituting, we obtain $\frac{3}{4}=\frac{MX}{\frac{5}{2}}$.  Therefore, $MX=\frac{15}{8}$, and the length of the crease is $\boxed{\frac{15}{4}}$.[/hide]

Sorry, i didn't need to find the length of the crease, i found that already what i'm trying to prove is, the length of ac is perpindicular to the fold which coincides the opposite corners a and c.

I think the basic idea is that we want $C$ to be the reflection of $A$ over the crease.

conserning this problem, where do A and C meet.  They could meet almost anywhere and the creases would be different lengths.  Am i just reading something wrong?

[quote="king_23"]A rectangular sheet of paper ABCD is 3 inches in width and 4 inches in length. The paper is folded so that the 2 diagonal opposite corners A and C coincide. Determine the length of the creases in the paper.[/quote]

[hide]
Let $x$ be the distance from the length to the crease itself.
\[ x^2 + 3^2 = (4-x)^2 \implies x = 7/8 \] the length of the crease is $(4-2x)^2 +3^2 = l^2$ so $l = 15/4$[/hide]


When i tried to register for the AMC 10, the program told me to look up my CEEB numba, but since I go to a middle school, I don't have one. What should I do?

The first page of the on-line registration process says:
"PLEASE NOTE:  This registration process is intended for teachers, principals, school officials and district employees authorized to commit a school to participating in the AMC contests. Parents, students, and other individuals not directly associated with the school administration are urged to contact a teacher, or the AMC office."

I suggest that you contact a teacher, principal, school official or district employee authorized to commit a school to participating in the AMC contests.  We have a procedure for assigning an internal code number for schools which do not have a CEEB number.  This internal number helps us distinguish the various kinds of registering schools.

Steve Dunbar
MAA Director, American Mathematics Competitions


Let $ A$ be a $ m\times n\ (m\neq n)$ matrix with the entries $ 0$ and $ 1$. Suppose that if $ f$ is injective such that $ f: \{1,\ 2,\ \cdots ,\ m\}\longrightarrow \{1,\ 2,\ \cdots ,\ n\}$, then there exists $ 1\leq i\leq m$ such that $ (i,\ f(i))$ element is $ 0$.

Prove that there exist $ S\subseteq \{1,\ 2,\ \cdots ,\ m\}$ and $ T\subseteq \{1,\ 2,\ \cdots ,\ n\}$ satisfying the condition:

$ i)$ if $ i\in{S},\ j\in{T}$, then $ (i,\ j)$ entry is $ 0$.

$ ii)$ $ card\ S \plus{} card\ T > n$.

Just use Hall's theorem.
We only need to consider a graph $G=(X,Y;E)$ : $X={1,2,...,m}$, $Y={1,2,...n}$, $E={(x,y)|(x,y) \text{element is} 1}$.
Because there is no X-saturating matching. We know that there exist a subset $I \subseteq X$ such that $|N(I)| < |I|$.
So $I  \times (Y \setminus N(I))$ satisfy.


[hide]
EDIT: I read the problem wrong
So the maximum number of rounds would be 6 turns because each time half are removed. Then, the minimum number of rounds is 1 because all the players can get more than 500,000 away from the leader. This makes 6-1=5.
[/hide]

[quote="Captain Sugar"]an arcade is hosting a pinball tournament. 64 people enter. after every round, everyone who is more than 500,000 points away from the leader gets eliminated [b]unless[/b] more than half of the contestants are closer than 500,000 points away from the leader. in that case, the bottom half is eliminated. find the difference between the maximum number of rounds and the minimum number of rounds.[/quote]
[hide]Umm.."in that case, the bottom half is eliminated" ...

So doesn't that mean in one round they can all be eliminated if they're all 500,000 points away ...? so 6-1=5? Not sure if i understood the problem correctly[/hide]

But if all of the people are more than 500,000 points away, that means less than half of the contestants are closer than 500,000 points away...

Basically I think you (chinese chess person) misread the problem

[quote="Captain Sugar"]...more than half of the contestants are closer than 500,000 points away from the leader...[/quote]

I think you read the question wrong, Treething. Not sure though.  :?

[hide]The way I see it, if the leader is 500,000 points ahead of everyone after the first round, he wins, so the minimum number of rounds is 1.  If everyone is less than 500,000 points away from the leader, then half go away each round, so it takes 6 rounds.  The difference is therefore $6-1=\boxed{5}$.

What if everyone has the same score?[/hide]

[quote="mathgeniuse^ln(x)"][quote="mathgeniuse^ln(x)"]Well at least you guys have a chance of answering the question.  I don't even get it. :mad:[/quote]

[quote="b-flat"]The way I see it, if the leader is 500,000 points ahead of everyone after the first round, he wins, so the minimum number of rounds is 1. If everyone is less than 500,000 points away from the leader, then half go away each round, so it takes 6 rounds. The difference is therefore $6-1=5$.

What if everyone has the same score? [/quote]

Exactly my point[/quote]

he should really add 'assuming no two people can get the same score in one round'

[quote="Captain Sugar"][quote="binonunquineist"]Actually I can file a protest on this problem. It could be infinity, since maybe everyone besides the leader is exactly 500,000 points away from the leader. Then no one is more then 500,000 points away, so they are not eliminated, and no one is actually closer then 500,000 points away, so in consequence, no is eliminated. The same thing can happen forever and ever and ever...

If everyone must have different scores, this still works. We can normally allow 62 of the participants to be eliminated. Then, the last two people can endlessly have a difference of exactly 500,000 iin their scores, forever and ever and ever...[/quote]
you know what i mean-500,000 or closer, including 500,000!

actually, 5 is the answer. the trick was that someone could win the tournament in the first round. Yes! my trick worked![/quote][/quote]

But people have realized your trick though.

[quote="Ignite168"][quote="Captain Sugar"][quote="binonunquineist"]Actually I can file a protest on this problem. It could be infinity, since maybe everyone besides the leader is exactly 500,000 points away from the leader. Then no one is more then 500,000 points away, so they are not eliminated, and no one is actually closer then 500,000 points away, so in consequence, no is eliminated. The same thing can happen forever and ever and ever...

If everyone must have different scores, this still works. We can normally allow 62 of the participants to be eliminated. Then, the last two people can endlessly have a difference of exactly 500,000 iin their scores, forever and ever and ever...[/quote]
you know what i mean-500,000 or closer, including 500,000!

actually, 5 is the answer. the trick was that someone could win the tournament in the first round. Yes! my trick worked![/quote][/quote]

5 is not the answer.[/quote]

Then what is the answer?

[quote="SplashD"][/quote][quote="Ignite168"][quote="Captain Sugar"][quote="binonunquineist"]Actually I can file a protest on this problem. It could be infinity, since maybe everyone besides the leader is exactly 500,000 points away from the leader. Then no one is more then 500,000 points away, so they are not eliminated, and no one is actually closer then 500,000 points away, so in consequence, no is eliminated. The same thing can happen forever and ever and ever...

If everyone must have different scores, this still works. We can normally allow 62 of the participants to be eliminated. Then, the last two people can endlessly have a difference of exactly 500,000 iin their scores, forever and ever and ever...[/quote]
you know what i mean-500,000 or closer, including 500,000!

actually, 5 is the answer. the trick was that someone could win the tournament in the first round. Yes! my trick worked![/quote][/quote][quote="SplashD"]

5 is not the answer.[/quote]

Then what is the answer?[/quote]
Half is not clearly defined, what if there are 3 people left? Also, they could tie, or someone could be exactly 500,000 away. So the tournament could go on forever, or you could just say this is a poorly worded question with no answer.


[quote="akma327"]solve for all solutions of the equation:

4sinx - 3cosx = 2[/quote]
Just use $ \sin x\equal{}\pm \sqrt{1\minus{}\cos^2 x}$

[quote="akma327"]solve for all solutions of the equation:

4sinx - 3cosx = 2[/quote]

$ \sin x \equal{} \pm \sqrt {1 \minus{} \cos^2 x}$, so by substitution, the equation becomes $ \pm 4 \sqrt {1 \minus{} \cos^2 x} \minus{} 3\cos x\equal{}2$

Let $ \cos x\equal{}k$.

The equation is now $ \pm 4 \sqrt{1 \minus{} k^2} \minus{} 3k \equal{} 2.$

Squaring the whole equation, we get $ 16\cdot(1\minus{}k^2)\equal{}(2\plus{}3k)^2.$

We get $ 16\minus{}16k^2\equal{}4\plus{}12k\plus{}9k^2$.

This equals $ 25k^2\plus{}12k\minus{}12.$

The solutions to this equation are $ \dfrac{\minus{}12 \pm 8\sqrt{21}}{50}$,

So $ k$ is approximately $ .49321$, or $ \minus{}.9372$.

Therefore, $ \cos x\equal{}.49321$

$ \cos x\equal{}\minus{}.9372$

That's as far as I got.  You could use a calculator from here?  Maybe I made a mistake.

Edit: Sorry I didn't see it , you're both right.

If you square the equation, doesn't the sign become positive?  I am now officially confused.   :huh:

AdithyaGanesh is right; when you square, the sign does not matter but you may get extraneous solutions. When you square root, you have to consider both positive and negative.

Simple example: $ \sqrt{x} \equal{} 1$. Squaring, we get $ x \equal{} 1$, but $ \minus{}x \equal{} 1$ is nonsensical.

$ \pm4\sqrt{1\minus{}\cos^2x} \equal{} 2 \plus{} 3\cos x$

$ \pm\sqrt{1\minus{}\cos^2x} \equal{} \frac12 \plus{} \frac34\cos x$

$ 1 \minus{} \cos^2x \equal{} (\frac12 \plus{} \frac34\cos x)^2$

$ 1 \minus{} \cos^2x \equal{} \frac14 \plus{} \frac34 \cos x \plus{} \frac9{16} (\cos x)^2$

$ 0 \equal{} \minus{}\frac34 \plus{} \frac34 \cos x \plus{} \frac9{16} (\cos x)^2$

$ \frac9{16} (\cos x)^2 \plus{} \frac34 \cos x \minus{} \frac34 \equal{} 0$

For the meanwhile, we'll let $ \cos x \equal{} a$.

$ \frac9{16} a^2 \plus{} \frac34 a \minus{} \frac34 \equal{} 0$

Finding the roots of this, we get:

$ a \equal{} \frac{\minus{}\frac34 \pm \sqrt{\frac9{16} \minus{} (4)\left(\frac9{16}\right)\left(\minus{}\frac34\right)}}{2 \times \frac9{16}}$

$ a \equal{} \frac{\minus{}\frac34 \pm \sqrt{\frac9{16} \plus{} (4)\left(\frac{27}{64}\right)}}{\frac98}$

$ a \equal{} \minus{} \frac23 \pm \frac{8 \sqrt{\frac{36}{16}}}{\frac98}$

$ a \equal{} \minus{} \frac23 \pm \frac{12}{\frac98}$

$ a \equal{} \minus{} \frac23 \pm \frac{96}{9}$

$ a \equal{} \minus{} \frac23 \pm \frac{32}{3}$

$ a \equal{} \{\minus{}\frac{34}{3},10\}$

So we have $ \cos x \equal{} \minus{}\frac{34}{3}$ and $ \cos x \equal{} 10$.

Thus, there are no real solutions for the equation?

(After going to Wolfram Alpha, the solutions were about $ \minus{}2.9096$ and $ 1.0550$...)

[quote="ernie"]$ 1 \minus{} \cos^2x \equal{} \frac14 \plus{} \frac34 \cos x \plus{} \frac9{16} (\cos x)^2$

$ 0 \equal{} \minus{} \frac34 \plus{} \frac34 \cos x \plus{} \frac9{16} (\cos x)^2$
[/quote]

You forgot the $ \cos^2 x$ term on the LHS.

hello, rewriting your equation in the form
$ \frac{4}{5}\sin(x)\minus{}\frac{3}{5}\cos(x)\equal{}\frac{2}{5}$ then you can set
$ \sin(\phi)\equal{}\frac{4}{5}$
$ \cos(\phi)\equal{}\frac{3}{5}$
and we get
$ \cos(\phi\plus{}x)\equal{}\minus{}\frac{2}{5}$
Sonnhard.

hello, this can be done without squaring something.
Sonnhard.


1. Let $ a,b,c,d \ge 0$ prove that $ \frac{1}{a}\plus{}\frac{1}{b} \plus{} \frac{4}{c}\plus{}\frac{16}{d} \ge \frac{64}{a\plus{}b\plus{}c\plus{}d}$

2. Let $ x,y,z \ge 0$ with $ xyz \equal{} 1$. Find the minimum value of $ \frac{x}{y\plus{}z} \plus{} \frac{y}{x\plus{}z} \plus{} \frac{z}{x\plus{}y}$

3. Let $ a,b,c \ge 0$. Prove that $ \sqrt{3(a\plus{}b\plus{}c)} \ge \sqrt{a} \plus{} \sqrt{b} \plus{} \sqrt{c}$

These are classic and easy...

1)
From c-s we can easily prove that:
$ LHS \ge \frac{(1\plus{}1\plus{}2\plus{}4)^2}{a\plus{}b\plus{}c\plus{}d}\equal{}\frac{64}{a\plus{}b\plus{}c\plus{}d}.$

2)
$ \sum \frac{x}{y\plus{}z}\equal{}\sum\frac{x^2}{xy\plus{}xz} \ge \frac{(x\plus{}y\plus{}z)^2}{2(xy\plus{}yz\plus{}zx)} \ge \frac{3}{2}.$

Thats the nebsit's inequality.

3)
After squaring it suffices to prove that:

$ a\plus{}b\plus{}c \ge \sqrt{ab}\plus{}\sqrt{bc}\plus{}\sqrt{ca} \Longleftrightarrow (\sqrt{a}\minus{}\sqrt{b})^2\plus{}(\sqrt{b}\minus{}\sqrt{c})^2\plus{}(\sqrt{c}\minus{}\sqrt{a})^2 \ge 0$
which is clearly true.

thank you so much!

Yeah I just started inequalities so they're pretty hard for me...


1)Να δειξετε οτι καθε ακεραιος>6 μπορει να γραφτει ως αθροισμα δυο πρωτων μεταξυ τους αριθμων(2 τροποι καθοτι ευκολο)                                                                                2)Καθε κυβος ακεραιου αριθμου ισουται με τη διαφορα 2 τετραγωνων 2 ακεραιων εστω (x,y) για τους οποιους ισχυει 9|x ή y  .

Ας λυσω κατι και γω!Για το α ερωτημα:Ισχυει οτι 4ν=(2ν-1)+(2ν+1) ,4ν+1=2ν+(2ν+1) ,4ν+2=(2ν+3)+(2ν-1) ,4ν+3=(2ν+1)+(2ν+2) αρα αποδειχθηκε.Για το β ερωτημα:     αποδυκνειεται ευκολα με την βοηθεια της ταυτοτητας   ν^3=[ν(ν+1)/2]^2-[ν(ν-1)/2]^2


AMC USA(J)MO USAMO AIME algebra polynomial

How do/did you guys prep for USAMO?

If u have considerable amount of time (say, a year) to prep, what

would u do?

How do u work thru ACoPS and PSS or other olympiad books?

Thanks,

Yea, and make sure you go through AoPS volume 2 too.

Actually, you've probably already done that.  But yea, that's something I still need to do.  Maybe that's why I only got a 7 on the AIME?

I have some notes, too ,but they don't cover in detail, do they?

[quote="DPopov"]To tell you the truth, I wouldn't directly go through those books. There are plenty of online notes that cover all of the material that is tested on the USAMO and they are FREE! :) I am an owner of those books and though they are good resources, I think I could get along just fine with the online resources I have compiled for myself over the course of the last year.[/quote]

I would greatly appreciate it if you could post the links to these free online notes you have compiled. Thanks.

Hmm, well I have :
[url=http://www.tjhsst.edu/~tmildorf/math/Adv.htm]Mildorf's[/url]
[url=http://web.mit.edu/rwbarton/Public/mop/]Reid Barton's[/url]
[url=http://eng.mathdb.org/]A Chinese site, "Mathematical Database", available here in English[/url]
[url=http://homepage.smc.edu/kennedy_john/POLYTHEOREMS.PDF]Polynomial Theorems (pdf)[/url]

Possibly some others; Professor Reiter's website might have some good notes, but I haven't looked around there much.  [url=http://www.upal.ca/adeel/]Chess64's website[/url] is also great.


how do i get line-spacing in .tex?

also,
if my paper is double-spaced, and I wish to include a few lines for mathematica code, how do I single-space the code, yet keep the paper double-spaced?

Use the setspace package with\usepackage{setspace}. Put \doublespacing in the preamble. Then put \singlespacing before the paragraph(s) that should be single spaced and then turn it off afterwards with \doublespacing.


function
trigonometry
abstract algebra
real analysis
real analysis unsolved

$a_{k}>0$, $k \in \mathbb{Z}$ a sequence of real numbers such that:
1/ $a_{-k}= a_{k}$
2/ $a_{k+1}\leq a_{k}$ for $k \geq 0$
3/ $a_{k}\leq \frac{a_{k-1}+a_{k+1}}{2}$ for $k>0$

Prove that there exists a $2 \pi$-periodic function $f \in L^{1}([0;2 \pi])$ with $(a_{k})_{\mathbb{Z}}$ as fourier coefficient

and $a_{k}$ tends towards $0$ of course  :blush:

Assume we have found $f$, then it must be……

is it a solution ? I don't see the obvious thing ...

Once upon a time, I knew this theorem. I've been trying to reconstruct it without consulting any sources.

My first attempt was to sum the series by parts twice, summing the exponentials and differencing the coefficients. Someone probably should double-check this ...

I got $f(x)=a_{0}+2\sum_{n=1}^{\infty}(a_{n}-2a_{n+1}+a_{n+2}) \cdot\frac{1-\cos((n+1)x)}{2\sin^{2}\left(\frac{x}{2}\right)}$

Unfortunately, to make any use of this, I appear to need that

$\sum_{n=1}^{\infty}(n+1)(a_{n}-2a_{n+1}+a_{n+2})<\infty,$

which we clearly do not have.

As far as I remember, the assumptions imply $a_{k}=\sum_{j\ge 1}{c_{j}}(1-|k|/j)^{+}$ with nonnegative $c_{j}$ such that $c_{j}\ge 0$ and $\sum_{j}c_{j}<\infty$. This brings the Fejer kernel into the picture. Since the $L^{1}$ norm of the Fejer kernel is $1$ (as opposed to the Dirichlet kernel, which blows up in $L^{1}$), we get convergence in $L^{1}$.


[quote="ehehheehee"][hide]A number if a perfect number if and only if the sum of all its factors add up to twice itself.

The sum of all the factors of 2^k(2^(k+1)-1) is equal to by sum of a geometric series:

(2^(k+1)-1)/(2-1)*(2^(k+1)-1+1)
(2^(k+1)-1)*(2^(k+1))
(2^(k+1))(2^(k+1)-1)

Which is conviently twice of 2^k(2^(k+1)-1)
[/hide][/quote]
Thanks, ehehheehee. That helps a lot. :)

[quote="Gyan"][quote="Inspired By Nature"]prove that all perfect numbers are of the form $2^{k}(2^{k+1}-1)$, if $(2^{k+1}-1)$ is a prime number.[/quote]
hmm .. It is very easy to prove that all numbers of the form:
$2^{k}(2^{k+1}-1)$,  (and if   $(2^{k+1}-1)$ is a prime number )
are perfect numbers.

It is also easy to prove that all [b] even [/b] perfect numbers are of the form:
$2^{k}(2^{k+1}-1)$,  ( if   $(2^{k+1}-1)$ is a prime number )

But no one has yet proved that: [quote][b]all [/b]perfect numbers are of the form $2^{k}(2^{k+1}-1)$, if $(2^{k+1}-1)$ is a prime number.[/quote]

(And no one has yet found one which is not of this form .. question is still open!)[/quote]
That is a good point Gyan. I just needed to know this for all numbers. But the last ones sound interesting and much harder. As Gyan stated, the question is still OPEN. :idea:

[quote="srulikbd"]yes the first is easy but how do you prove the second? It is also easy to prove that all [b] even [/b] perfect numbers are of the form:
$2^{k}(2^{k+1}-1)$,  ( if   $(2^{k+1}-1)$ is a prime number )  thanks[/quote]

Let $\sigma(n)$ means sum of all divisors  of $n$ and  we know:
1)  $n$ is a perfect number iff $\sigma(n) = 2n$
2) If $a$ and $b$ have no common factors, $\sigma(ab)=\sigma(a)\sigma(b)$

So if $n$ is an even number, write it as $n=2^{k}p$ where $k \ge 1$ and $p$ is an odd number.
For $n$ to be perfect:
$\sigma (n) = \sigma(2^{k}) \sigma(p) = 2n = 2^{k+1}p$
 obviously $\sigma(2^{k})= (2^{k+1}-1)$
 let us call this ($2^{k+1}-1) = q$ (IOW $2^{k+1}= q+1$)

So we have
$q \sigma(p) = (q+1) p$
or $q(\sigma(p)-p) = p$
So $q$ must divide $p$  Or $p=kq$ ($k$ is positive integer)
Or $\sigma(p)-p = kq/q =k$
If $k>1$  we know divisors of $p$ are (atleast )  (1,k,p  etc ...
so  we get $1+k+etc = k$ gives contradiction,  this implies $k=1$
Now if  $p$ is prime, we do have $\sigma(p)=p+1$ but if it is not we again reach contradiction  then because LHS  would be greater than 1.
(Because if there is a factor $x>1$ we would have $\sigma(p)-p \ge 1+x$ etc)

Hence, $q=p$  has to be prime.
(And if an even number  $n$  is perfect, it has to be of the form $2^{k}p$ where $p=2^{k+1}-1$
QED.

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

It seems like we don't know yet...that is only true for even perfect numbers...

[quote]It seems like we don't know yet...that is only true for even perfect numbers...[/quote]
Yes, and that was the point of the post(s). we can prove that all [i]even [/i]perfect numbers are of the form given above but we don't know  that all perfect numbers are even numbers only.


Find the volume of the solid bounded by the planes:
$ x\equal{}1 \qquad y\equal{}0 \qquad z\equal{}0$
$ x\equal{}y  \qquad x\plus{}y\plus{}z\equal{}0$

Any takers?

[hide="Hint"]Consider the image of the solid on the x-y plane.[/hide]


Trapezoid ABCD is inscribed in a circle. The lengths of the bases AB and CD are 10 and 20, respectively,, and AD =$ \sqrt{61}$. What is the area of triangle BCD?

[hide="Very Vague Solution"]The height of the trapezoid is $ \sqrt {61 \minus{} 5^2} \equal{} 6$. Now, the triangle can be split into two triangles. Then, the total area of triangle $ BCD$ is:
\[ \frac {15 \cdot 6}{2} \plus{} \frac {5 \cdot 6}{2} \equal{} \boxed{60}
\]
[/hide]


group theory
abstract algebra
superior algebra
superior algebra solved

Find all the subgroups of $(Z \times Z, +)$.

already discussed in the forum. Buty the answer is not all the product of subgroup of $Z$. (a line for example ?)

alekk is right.
The subgroups of $\mathbb{Z} \times \mathbb{Z}$ are the 'sublattices' of it.
So they are generated by two elements $a,b$ (some of them can be zero) as  $a\mathbb{Z} + b\mathbb{Z}$.
Hope it's correct now  :blush:


function
integration
calculus
real analysis
real analysis unsolved

If $ f$ is a continuous differentiable function such that there exist $ m$ so that $ f'(x) \geq m$ for all $ x$, then prove that $ \int f(x) dx \leq \frac {4}{m}$, where the upper limit of the integral is $ b$ and the lower limit of the integral is $ a$, $ b$ and $ a$ is a real number.

This is how you write definite integrals: $ \int_a^b$.  (you can use reply to see the code, or click on it)  Also, this problem is obviously false without more conditions on $ f$, $ a$, $ b$, and $ m$.

$ f$ is a decreasing function on the interval $ [a,b]$. Sorry if I have missed the statement. Thanks in advance.

Now it makes even less sense than before. :roll:


inequalities
logarithms
limit
calculus
integration
real analysis
real analysis unsolved

Let $ x_n>0$ be an increasing divergent sequence and $ a\le1$. Prove that $ \sum_{n\equal{}1}^{\infty}(\frac{x_n\minus{}x_{n\minus{}1}}{x_n})^a$ is divergent.

" is positive and increasing "  means $ \frac {x_{n \minus{} 1}}{x_{n}} < 1$ divergent means that there exist subsequence $ n_{k}$ and $ \delta < 1$ such that
 $ \frac {x_{n_{k} \minus{} 1}}{x_{n_{k}}} < \delta < 1$
Thus we have $ \sum \left( 1 \minus{} \frac {x_{n \minus{} 1}}{x_{n}} \right)^{a} \geq \sum_{k} \left( 1 \minus{} \frac {x_{n_{k} \minus{} 1}}{x_{n_{k}}} \right)^{a}\geq \sum_{k} (1\minus{}\delta)^{a} \equal{} \infty$

Extremal, that proof is incorrect.

$ n$ is a positive increasing divergent sequence. There is no $ \delta<1$ such that any subsequence of $ \frac{n\minus{}1}n$ remain less than $ \delta.$

You were told that $ x_n\to \infty.$ You're not entitled to concluding that it grows geometrically fast; you must allow for it growing arbitrarily slowly.

I do have an argument. The rough outlines: let $ n_k$ be such that $ x_{n_k}$ is approximately $ k$ (this will take some cleanup of the definition).

Look at the sum of $ \frac{(x_n\minus{}x_{n\minus{}1})^a}{x_n^a}$ for $ n$ running (approximately) from $ n_k$ to $ n_{k\plus{}1}.$ The denominators of those fractions are all approximately $ k^{a}$ The sum of the numerators can be bounded below by $ \sum (x_n\minus{}x_{n\minus{}1}),$ which telescopes and sums to approximately 1. (I used that $ u^a>u$ for $ 0<u<1.$)

This is going to lead to an overall estimate for the series that it is unfavorably comparable to $ \sum_k\frac1{k^a},$ which diverges for $ a\le 1.$

There are some annoying details to be cleaned up in that proof. (For one thing, I may have to do something to assure that $ x_{n_{k\plus{}1}}\minus{}x_{n_k}\ge 1.$) But it should be fixable.

:blush:  Oops. I'm sorry for my stupid mistakes..

I don't see: why there exist such $ n_{k}$ where $ x_{n_{k}} \approx k$ ?
How do you choose such $ n_{k}$ for example for $ x_{n} \equal{} e^{n^{2}}$ ?

Ok  :lol: .
Let me try to correct my stupid mistakes.
1) if $ \frac {x_{n \minus{} 1}}{x_{n}}$ don't tend to $ 1$ then my  (first "stupid") argument about existence $ \delta$ shows that the serie diverge.
Suppose $ \frac {x_{n \minus{} 1}}{x_{n}} \to 1$
then our serie  (by bernulli inequality) is greater that :
$ \sum (1 \minus{} \frac {x_{n}}{x_{n \plus{} 1}})$ and convergence or divergence of this is same as to study convergence or divergence of this serie (since there is no big difference between $ 1 \minus{} x_{n}/x_{n \plus{} 1}$ and $ \ln x_{n}/x_{n \plus{} 1}$) :
$ \sum \ln \frac {x_{n}}{x_{n \plus{} 1}} \equal{} \lim_{N} \ln \frac {x_{1}}{x_{N}} \to \minus{} \infty$.

Yes, I was coming around to about that. My first argument did assume slow growth; more rapid growth is less of a problem, but still has to be dealt with. Here's my current phrasing, which is fairly similar to what Extremal just said.

We're looking at $ \sum\left(\frac{\Delta x_n}{x_n}\right)^a.$

Since $ a<1$ and $ \frac{\Delta x_n}{x_n}<1,$ we have $ \left(\frac{\Delta x_n}{x_n}\right)^a>\frac{\Delta x_n}{x_n}.$

So we might as well be looking at $ \sum \frac{\Delta x_n}{x_n}$ (the thing we'd have if $ a\equal{}1.$)

If we were facing the integral $ \int \frac{x'(t)}{x(t)}\,dt,$ we'd just say the antiderivative was $ \ln x(t).$ By analogy, look at $ y_n\equal{}\ln(x_n).$

The $ y_n\minus{}y_{n\minus{}1}\equal{}\ln\left(\frac{x_n}{x_{n\minus{}1}}\right)\equal{}\minus{}\ln\left(1\minus{}\left(\frac{x_n\minus{}x_{n\minus{}1}}{x_n}\right)\right)\le\frac{x_n\minus{}x_{n\minus{}1}}{x_n}.$

So $ \sum\frac{x_n\minus{}x_{n\minus{}1}}{x_n}\ge \sum y_n\minus{}y_{n\minus{}1}.$

That last sum telescopes to give $ y_n$ (plus or minus a constant) as the sequence of partial sums, and $ y_n\to\infty$ since $ x_n\to\infty.$

No need to deal with different cases regarding the rate of growth for $ x_n.$


You know on the left hand side there are links that lead you to various places on the site?
And under those links there is some interesting formula or something?
There is a proof or something about the Law of Cosines.
I try to follow it, but I don't understand.
Can someone help?
Thanks.

make it into 2 right angle triangle and then find the height of the triangle. after that subsitute to find the side. Try...
thx

Explaining the 'proof without words' on the site:

Consider a triangle with sides a,b,c, and angle $ \theta$ on point C. Draw a circle about B with radius a. Draw the diameter extended from AB; you get the radius a and the length a-c, since we subtract c from the radius in the opposite direction. Now, extend BC in the original triangle to make the diameter; we get a right triangle when we extend AC. By the definition of cosine, the length from C to where the extension of AC meets the circle again is $ 2a\cos{\theta}$. We subtract b from this to get the length of the extension that is not on ABC. Now we use Power of a Point to get
$ (2a\cos{\theta}\minus{}b)(b)\equal{}(a\plus{}c)(a\minus{}c)$
$ 2ab\cos{\theta}\minus{}b^2\equal{}a^2\minus{}c^2$
$ c^2\equal{}a^2\plus{}b^2\minus{}2ab\cos{\theta}$.

The animation is a way more concise way of putting this. Hope this made any sense at all... :maybe:

Hi. There are some proofs of the cosine rule here: http://en.wikipedia.org/wiki/Law_of_cosines

Hope this helps you.


$(\sqrt{2})^x+(\sqrt{3})^x=(\srt{13})^{\frac{x}{2}}$
Find number of solutions for x

i dont think there are any. since x can not be odd. and if it were even it seems as if the RHS would always be larger than the LHS>
i could be wrong.  :?

[hide]This can be written $f(x)=1$, where $f(x)=\left( \frac{2}{13} \right)^{x/2}+\left( \frac{3}{13} \right)^{x/2}$.

$f(x)$ is a decreasing function (a sum of exponents with bases less than 1) so it is one-to-one, meaning at most one solution.

This solution exists because $f(0)=2>1$ and $f(2)=\frac{5}{13}<1$ so there is a solution $x$ with $0<x<2$ by Intermediate Value Theorem.

Therefore there must be $\boxed{1}$ solution.[/hide]


It is true that if a function $ f$ is strictly monotonic on its domain $ D$ then $ f$ is invertible on its domain. But I want to develop a method to prove or disprove that a function has an inverse:

Is it true that a function $ f$ is strictly monotonic on its domain $ D$ iff $ f$ is invertible on its domain?

Also, let us consider $ f$ again. For each interval in $ D$ partitioned by the values of $ x$ for which $ f'(x)\equal{}0$, we need to consider the sign of $ f'(x)$. Is it true that $ f'(x)$ is either positive on all intervals OR $ f'(x)$ is negative on all intervals IF AND ONLY IF $ f(x)$ is monotonic?

Sorry that there is a bit of calculus here, but I feel this should go into pre-calculus.

[quote="ThetaPi"]

Is it true that a function $ f$ is strictly monotonic on its domain $ D$ iff $ f$ is invertible on its domain?

[/quote]
No, an example $ f: \{1,2,3\}\to\{1,2,3\}$ is definite by $ f(2)\equal{}2,f(1)\equal{}3,f(3)\equal{}1$.

I would like an algebraic way of showing that a function has an inverse / does not have an inverse because graphing it and seeing whether it passes the horizontal line test is not always feasible.


limit
logarithms
ratio
analytic geometry
calculus
calculus computations

We need a lemma related to the ratio test. Suppose that $a_n$ is a sequence of positive numbers and we have that $\lim_{n\to\infty}\frac{a_{n+1}}{a_n}=r,$ where $r>0.$ Then for any $0<s<r$ and for any $t>r,\, \exists\,N$ such that for $n>N,\,s^n<a_n<t^n.$

Now we note that clearly $u_n\to\infty$ as $n\to\infty$ (and quite rapidly). We write:

$u_{n+1}=u_n^2\left(1+\frac1{u_n}\right)$ so that

$\ln(u_{n+1})=2\ln(u_n)+\ln\left(1+\frac1{u_n}\right)$

$\frac{\ln(u_{n+1})}{\ln(u_n)}=2+\frac{\ln\left(1+\frac1{u_n}\right)}{\ln(u_n)}$

The last term clearly goes to zero, so $\lim_{n\to\infty}\frac{\ln(u_{n+1})}{\ln(u_n)}=2.$

By the lemma above, for all $\delta>0,\,(2-\delta)^n<\ln(u_n)<(2+\delta)^n$ for large enough $n.$

Hence, for large enough $n,\,n\ln(2-\delta)<\ln\ln(u_n)<n\ln(2+\delta),$ or

$\ln(2-\delta)<\frac{\ln\ln(u_n)}n<\ln(2+\delta).$

This is enough to prove the result.

Suppose $s<q<r.$ Since $\lim_{n\to\infty}\frac{a_{n+1}}{a_n}=r,$ there exists $N$ such that for $n\ge N,\,\frac{a_{n+1}}{a_n}>q.$

Thus $a_{N+1}>qa_N,\,a_{N+2}>qa_{N+1}>q^2a_N,$ and so on by induction. The general estimate is that for $n>N,\,a_n>q^{n-N}a_N.$

So $\frac{a_n}{s^n}>\frac{a_N}{s^N}\cdot\frac{q^{n-N}}{s^{n-N}}.$

Take the limit as $n\to\infty.$ Since $q>s,$ this limit is $\infty.$ Hence, eventually, $\frac{a_n}{s^n}>1,$ meaning $a_n>s^n.$

The argument for the other side is similar.

As I said, this is related to the ratio test for the convergence of series; check how various analysis books prove the ratio test.

Anyway, Kent you could have also applyed Cesaro-Stolz theorem after you deduced that $\lim_{n\to\infty}\frac{\ln(u_{n+1})}{\ln(u_{n})}=2.$ ;)

[quote="erdos"]show that :  $\lim_{n\to+\infty}\frac{\ln(\ln(u_n))}{n} = \ln(2)$
where $(u_n)$ is a sequence defined by : $u_0 \in ]1,+\infty[$ and $\forall n \geq 1: u_{n+1}=u_n+u_{n}^{2}$[/quote]

In other words, if $u_n = A^{2^n}$ then $A$ converges to a limit.  This is easy to show if you change from "$u$ coordinates" to "$A$ coordinates".


function
integration
calculus
real analysis
real analysis solved

Let f:[0,1]->[0,1] be a function of class C1, strictly increasing, f(1)=1 and f'(1)=/=0. What's the limit of n\int {0->1}fn(x)dx as n->00 ?

My proof seems to be very nasty. In fact, it might be wrong, since I never took the time to write it down :). Does ne1 have a nice prof?

For all u s.t. 0 < u < 1 we can write :
Integral(0 =< x =< 1){fn(x).dx}  =  Integral(0 =< x =< u){fn(x).dx} + Integral(u=< x =< 1){fn(x).dx} <  fn(x).u + fn(1).(1-u) = fn(u).u + (1-u)
But |f(u)| < 1, so fn(u).u --> 0, so the the asked constant limit, called L, satisfies  0 =< L =<  1-u  for all 0 < u < 1, so L = 0

    :cool:

See [url]http://www.artofproblemsolving.com/Forum/viewtopic.php?t=40029[/url] for the solution. (Diogene didn't notice the factor $n$ in front of the integral)


Solve the following system of equations:

{
x + y + xy = 11
1/x +1/y + 1/(xy) = 1
}

[hide="Solution"]Multiply second equation by $xy$
$y+x+1 = xy$
Substitute that expression for $xy$ into the first equation and we get $x+y = 5$, thus $xy = 6$

So $y = \frac{6}{x}$

$x+\frac{6}{x}= 5$
$x^{2}-5x+6 = 0$
$(x-3)(x-2) = 0$
So $x = 3$ or $x = 2$. Thus $y = 2$ or $y = 3$.
Solutions are $(x,y) = \{(3,2),(2,3)\}$[/hide]


let a,b,c,d be positive real numbers such than abcd=1
prove that
(1+ab)/(1+a)+(1+bc)/(1+b)+(1+cd)/(1+c)+(1+da)/(1+d)>=4


A custom made window is a single pane of glass consisting of a semicircular sheet of glass on top of the narrow end of a rectangular sheet of glass. The rectangle has dimensions of 4 feet by 3.5 feet. How many feet of weather-stripping will be needed to go around the glass? Express your answer to the nearest integer.

Okay, so we have to find the perimeter of this shape.  This is simply $ 4 \plus{} 4 \plus{} 3.5 \plus{} \frac{3.5}{2}\pi$, which (rounded) equals $ \boxed{17}$.


khob hala ke marhale 2 maloom shodo kesayi ke ghabool shodan ke beheshon tabrik migiam kesayi ke nashodanam (mesle khodam) dobare shoroo konim ba ye masale sade ;) pas dobare bacheha shoroo konim
$n\geq 3$ yek adade sahih ast.sabet konid baraye adade mosbate $x_{1}\leq x_{2}\leq \dots \leq x_{n}$:
$\frac{x_{n}x_{1}}{x_{2}}+\frac{x_{1}x_{2}}{x_{3}}+\dots+\frac{x_{n-1}x_{n}}{x_{1}}\geq x_{1}+x_{2}\dots+x_{n}$

in soal ba hardi-poolia ya ba holder sade hal mishe .be nazare man hale injoor soala ziad komaki be movafaghiat to marhale 2 nemikone ke mesale vazehesh marhale 2 emsale ba oon vaze soalash.(makhsoosan ke dige be nodrat gabr midan)

sunset u be khodet negah nakon mesle har sal yeseri bachehaye aval dabirestani mian too  mathlinks to age baladi nemikhad fekr  koni nazaram lazem niss bedi


floor function
modular arithmetic
algebra solved
algebra

Let $\lambda$ the positive root of the equation $t^2-1998t-1=0$.  It is defined the sequence $x_0,x_1,x_2,\ldots,x_n,\ldots$ by $x_0=1,\ x_{n+1}=\lfloor\lambda{x_n}\rfloor\mbox{ for }n=1,2\ldots$ Find the remainder of the division of $x_{1998}$ by $1998$.

Note: $\lfloor{x}\rfloor$ is the greatest integer less than or equal to $x$.


I have problems with d and e.... anyone can explain these to me??

A long, uniform rod of mass M and length l is supported at the left end by a horizontal axis into the page and perpendicular to the rod, as shown above. The right end is connected to the ceiling by a thin vertical thread so that the rod is horizontal.   The moment of inertia of the rod about the axis at the end of the rod is Ml2/3.   Express the answers to all parts of this question in terms of M, l, and g.
a.	Determine the magnitude and direction of the force exerted on the rod by the axis.

The thread is then burned by a match. For the time immediately after the thread breaks, determine each of the following:
b.	The angular acceleration of the rod about the axis
c.	The translational acceleration of the center of mass of the rod
d.	The force exerted on the end of the rod by the axis

The rod rotates about the axis and swings down from the horizontal position.
e.	Determine the angular velocity of the rod as a function of q, the arbitrary angle through which the rod has swung.

Answers:
a: ma/2
b:3g/2l
c:3g/4
d:mg/4
e:...long enough...

solution:
d:
F = ma = mg-F(r)
Therefore, F(r) = mg/4 -----answer

e:
delta K = delta U
(1/2) I W^2 = (mg l/2) sin(theta)

where did sin(theta) come from??

If that is the entire solution, I have a feeling it's a typo since the variable $ \theta$ is never defined.  Seems like it should be $ q$.

(i) as shown above ??? I couldn't see anything above!!!
(ii)the moment of inertia is $ \; {{Ml^2}\over 3} \;$  instead of Ml2/3.
(iii)For d) you already have the solution; just isolate F(r), with [b]a[/b] given by c).
(iv)JRav already told you.


I can't think of any solutions:

Find all groups of three regular polygons with side length 1 that can surround a single point such that each polygon shares a side with the other two.

One such arrangement is 1 hexagon, 1 square, and 1 dodecagon.

[hide="Hint"]The three polygons must have the same apothem.[/hide]

The distance between the center of a polygon and any one of its sides.

so that means you can form an equilateral triangle with the centers of the three polygons right?


[quote="buzzer11"]Bob has an regular, octagonical garden. His garden has a side length of 6 feet. The radius is 5 feet. Bob wants to plant as many Bob Bushes in his garden. one Bob Bush takes up 2 square feet of space. How many Bob Bushes can Bob plant in his garden?[/quote]
[hide]
$\tan{\frac{180}{8}}=\frac{3}{x}$

$\frac{3}{0.4142}=x$

$x\approx=7.242877837$

Area of the garden is

$8*\frac{6*7.242877837}{2}$

$8*21.72863351$

$\approx174$

So he can fit $\frac{174}{2}=87$ bushes.[/hide]

Isn't this a little complicated for basics?

maybe there's another way to solve it. :?

I think the answer is 48. Is that correct? The side measurements are sort of contradicting.


Given $ a,b,c >0$ and $ a\plus{}b\plus{}c \ge \frac{a}{b}\plus{}\frac{b}{c}\plus{}\frac{c}{a}$
Prove that:
$ 27(a^3\plus{}1)(b^3\plus{}1)(c^3\plus{}1) \ge (a\plus{}b\plus{}c\plus{}\frac{1}{a}\plus{}\frac{1}{b}\plus{}\frac{1}{c})^3$
Good luck  :)

[quote="quykhtn-qa1"]Given $ a,b,c > 0$ and $ a \plus{} b \plus{} c \ge \frac {a}{b} \plus{} \frac {b}{c} \plus{} \frac {c}{a}$
Prove that:
$ 27(a^3 \plus{} 1)(b^3 \plus{} 1)(c^3 \plus{} 1) \ge (a \plus{} b \plus{} c \plus{} \frac {1}{a} \plus{} \frac {1}{b} \plus{} \frac {1}{c})^3$
Good luck  :)[/quote]
Potla and Peine,no one interested in this inequality? It very nice :)

LOL , Are you trying to sell this inequality

[quote="peine"]sorry quykhtn-qa1! I have not see this topic, give me a chance for trying to prove it,
P.S: my friend potla cannot log in for the moment.[/quote]
Thanks.I hope you wii like it :lol:

[quote="peine"]I like much your inequality but It's so difficult! I cannot find any way to prove it,  :wink:[/quote]
No,This inequality nice and easy :lol: .I think you can solve it :)

[quote="Potla"]Sorry that I was offline for several days, and I got to connect the internet today..... :|
I am thinking for a proof, and your inequalities seem to be simple application of AM GM which I cannot find! :([/quote]
I hope you can solve it :)

[quote="aadil"]it is quite a tough  problem.shouldnt this be moved to the olympiad section?[/quote]
this inequality easy.You can solve it? :maybe:

[quote="enndb0x"]This is growing older , can you just post your solution  :oops:[/quote]
You can use AM-GM and HOLDER inequalities :oops:

[quote="Agr_94_Math"]Yes quykhtn-qa1 my friend, you are right. This is a nice and easy inequality and I too have got a solution using  Cauchy Bunchiacopski and AM GM.
I am waiting for someone to post a solution, especially the ones to whom u have delicated this:Potla and Piene.[/quote]
I see you wrote:many inequalities easy,But you can't post solution :wink:
This inequality very easy.Please post solution of you :oops:

[quote="arqady"][quote="quykhtn-qa1"]Given $ a,b,c > 0$ and $ a \plus{} b \plus{} c \ge \frac {a}{b} \plus{} \frac {b}{c} \plus{} \frac {c}{a}$
Prove that:
$ 27(a^3 \plus{} 1)(b^3 \plus{} 1)(c^3 \plus{} 1) \ge (a \plus{} b \plus{} c \plus{} \frac {1}{a} \plus{} \frac {1}{b} \plus{} \frac {1}{c})^3$
Good luck  :)[/quote]
It's easy even for me.  :)
By Holder we obtain $ \prod_{cyc}(3a^3 \plus{} 3)\geq(ab \plus{} ac \plus{} bc \plus{} a \plus{} b \plus{} c)^3.$
Hence, it remains to prove that $ ab \plus{} ac \plus{} bc\geq\frac {1}{a} \plus{} \frac {1}{b} \plus{} \frac {1}{c},$ which is equivalent to $ abc\geq1.$
But $ a \plus{} b \plus{} c \ge \frac {a}{b} \plus{} \frac {b}{c} \plus{} \frac {c}{a} \equal{} \frac {1}{abc}(a^2c \plus{} b^2a \plus{} c^2b) \equal{} \frac {1}{3abc}\sum_{cyc}(2a^2c \plus{} b^2a)\geq$
$ \geq\frac {1}{3abc}\sum_{cyc}3\sqrt [3]{a^5b^2c^2} \equal{} \frac {a \plus{} b \plus{} c}{\sqrt [3]{abc}},$ which gives $ abc\geq1.$[/quote]
Nice proof :lol: Thanks.


If $ \frac{2}{3}$ of a ton of hay costs $ \frac{4}{5}$ of an eagle, how many eagles will $ \frac{1}{6}$ of a ton cost?  What is the number of eagles in the answer to this
problem? Express your answer as a common fraction.

$ \frac{2}{3}\equal{}\frac{4}{6}$ so we need a fourth of the original, $ \boxed{\frac{1}{5}}$

Or, you could set up a proportion.

Let $ x$ = the number of eagles

$ \frac{\frac{2}{3}}{\frac{4}{5}} \equal{} \frac{\frac{1}{6}}{x}$

Now let's just do a bit of simplifying.

$ \frac{5}{6} \equal{} \frac{\frac{1}{6}}{x}$

And then we cross multiply:

$ 5x \equal{} 6 \cdot \frac{1}{6}$

$ 5x \equal{} 1$

$ x \equal{} \frac{1}{5}$

So our answer is $ \implies \boxed {\frac{1}{5}}$

[quote="JRav"]My way is faster.  :lol:[/quote]

Yes, it is, but for me, isabella2296's way was better because it helped to understand the problem.


How to prove :

1) $ Ext(B_1[l^1])\equal{}\{\pm e_n : n\in \mathbb N \}$

2) $ Ext(B_1[l^2])\equal{}\{x: \parallel{}x\parallel{}\equal{}1\}$

3)$ Ext(B_1[l^\infty])\equal{}\{x_n : |x_n|\equal{}1, \forall n \}$

4) $ Ext(B_1[C[0,1])\equal{}\{|f(x)|\equal{}1\}$

Where Ext : Extreme points and $ B_1$ : the unit ball

I know may all of them have the same idea, but how we can prove that ??

 :)

What do you mean by "extreme points"? You seem to have written down the definition of the unit ball in each case...

The definition :

Let $ E\subseteq (X , \parallel{}.\parallel{})$ be a convex set, then $ x\in E$ is called extreme points of E if $ x \equal{} \frac 12 (y \plus{} z)$  then $ x \equal{} y \equal{} z$ , ( i.e. $ x$ is not the middle of any two points ) .

I found for  unit ball $ B_1$ there is a nice theorem for the extreme point which is :

We say $ x$  is an extreme point of $ B_1[X]$ iff there is no $ y$ in $ B_1[X]$ such that $ \parallel{} x \minus{} y\parallel{}\leq 1$ and $ \parallel{} x \plus{} y\parallel{} \leq 1$

I thought about (1) as follows :

a) I need to prove $ \pm e_n$ is an extreme point ( Easy ).

b) I need to prove if $ x$ is any other extreme point , it must be of the set $ \{\pm e_n \}$ .

depend on the theorem I wrote , it was very easy to get a contradiction if I choose $ y \equal{} e_n$. ??? Is this true ?? I don't think so ??

Does my thinking true ?

If yes , I could do (a)  in all other questions , but I still have problem in doing (b).

 :)  :)


AMC USA(J)MO USAMO algebra polynomial function blogs

I saw this problem, which is from the 1975 USAMO:

A polynomial $ P(x)$ of degree $ n$ satisfies $ P(0) \equal{} 0$, $ P(1) \equal{} 1/2$, $ P(2) \equal{} 2/3$, ... , $ P(n) \equal{} n/(n\plus{}1)$. Find $ P(n\plus{}1)$.

Why can't we just plug $ n\plus{}1$ into $ P(x)$?

[quote="123456789"]I saw this problem, which is from the 1975 USAMO:

A polynomial $ P(x)$ of degree $ n$ satisfies $ P(0) \equal{} 0$, $ P(1) \equal{} 1/2$, $ P(2) \equal{} 2/3$, ... , $ P(n) \equal{} n/(n \plus{} 1)$. Find $ P(n \plus{} 1)$.

Why can't we just plug $ n \plus{} 1$ into $ P(x)$?[/quote]

Because then you just get $ \frac{n\plus{}1}{n\plus{}2}$ and you cant really do anything else from there...

Oh..stupid me. I thought $ \frac {n \plus{} 1}{n \plus{} 2}$ could be considered an answer...lol
So what would be the answer?

[quote="jli"]oooh wait...oops

its becaues P(n) is a number not a function. so you cant plug n+1 in[/quote]What do you mean by "P(n) is a number not a function, so you cant plug it in"?

On Tenoreoz blog, he posted the answer as $ \displaystyle P(n\plus{}1)\equal{}\frac{(\minus{}1)^{n\plus{}1}\plus{}(n\plus{}1)}{n\plus{}2}$. Is that any better than $ \displaystyle \frac{n\plus{}1}{n\plus{}2}$?

[quote="123456789"]I saw this problem, which is from the 1975 USAMO:

A polynomial $ P(x)$ of degree $ n$ satisfies $ P(0) \equal{} 0$, $ P(1) \equal{} 1/2$, $ P(2) \equal{} 2/3$, ... , $ P(n) \equal{} n/(n \plus{} 1)$. Find $ P(n \plus{} 1)$.

Why can't we just plug $ n \plus{} 1$ into $ P(x)$?[/quote]

Where do you see a $ P(x)$ to plug into? $ P(x)$ is a certain, unknown polynomial that happens to give those values for those $ x\equal{}0...n$. $ P(x)\equal{}x/(x\plus{}1)$ isn't a polynomial of degree n.

Oh, I get it; so $ P(x)$ happens to match up with those values for $ 1$ through $ n$, but what comes after that is unknown, right?

for example, f(x) is a function but f(5) is just a number so similarly, f(n) is just a number.

like if f(5)=5, you cant say f(5+1)=5+1=6...

Why is this considered a polynomial?

$ P(x) \equal{} \frac {x(x \minus{} 1)(x \minus{} 2)\cdots(x \minus{} n)\cdot \frac {( \minus{} 1)^{n \plus{} 1}}{(n \plus{} 1)!} \plus{} x}{x \plus{} 1}$

http://www.artofproblemsolving.com/Forum/weblog_entry.php?t=196728

I'm kinda confused...

A similar question:  given a polynomial $ P(x)$ of degree $ n$ such that $ P(k) \equal{} 2^k, k \equal{} 0, 1, 2, ... n$, find $ P(n\plus{}1)$.

The answer is [b]not[/b] $ 2^{n\plus{}1}$.  :)

[b]123456789:[/b]  The numerator is divisible by $ x \plus{} 1$.  Plug in $ x \equal{} \minus{}1$.

[quote="t0rajir0u"][b]123456789:[/b]  The numerator is divisible by $ x \plus{} 1$.  Plug in $ x \equal{} \minus{} 1$.[/quote]I still don't see it...sorry...

[quote="123456789"]I still don't see it...sorry...[/quote]

Since the top polynomial is divisible by x+1, when you divide, what you get is a polynomial. If you read the "Factor Theorem" in the blog entry, it should tell you that since -1 is a root of the top, then the top in factored form has an (x+1) somewhere.

Also I'd answer t0rajir0u's question but it wouldn't be fair since I read it on his blog before :)

I don't see how the numerator is divisible by $ x \plus{} 1$...I don't see how to factor $ x\plus{}1$ out of the numerator...

[b]@ 123456789:[/b] [hide="Explanation"]By the factor theorem, $ x \minus{} k$ is a factor of $ P(x)$ iff. $ P(k) \equal{} 0$

So, to show that $ x \plus{} 1$ is a factor, we must plug in $ x \equal{} \minus{} 1$ into the numerator and show that this makes the numerator zero.

First, let's simplify this section: $ x(x \minus{} 1)(x \minus{} 2)\cdots(x \minus{} n)$ when $ x \equal{} \minus{} 1$. We have $ n \plus{} 1$ terms, and we get $ ( \minus{} 1)( \minus{} 1 \minus{} 1)( \minus{} 1 \minus{} 2) \ldots ( \minus{} 1 \minus{} n)$ $ \equal{} ( \minus{} 1)( \minus{} 2)( \minus{} 3) \ldots ( \minus{} (n \plus{} 1)) \equal{} ( \minus{} 1)^{n \plus{} 1} (n \plus{} 1)!$.

So now, the numerator is $ ( \minus{} 1)^{n \plus{} 1} (n \plus{} 1)! \cdot \frac {( \minus{} 1)^{n \plus{} 1}}{(n \plus{} 1)!} \plus{} ( \minus{} 1) \equal{} 1 \plus{} ( \minus{} 1) \equal{} 0$, and we are done.[/hide]

So while there is a way to see that is a factor, we can't write it in factorized form?

So P(x)=(x)/(x+1), for x=0,1,2...,n-1,n take all the terms to one side to get

(x+1)P(x)-x=0.

Since that equals 0 for x=0,1,2,...,n-1,n, we can set that equal to another polynomial Q(x) such that Q(x) has those exact same zeros.  Since (x+1)P(x)-x has degree of n+1, Q(x) has degree n+1 so Q(x) is just simply equal to Q(x)=C(x)(x-1)...(x-n), where C is a constant.

So since $ (x \plus{} 1)P(x) \minus{} x \equal{} C(x)(x \minus{} 1)...(x \minus{} n)$, we want to find $ C$.  Plug in $ x \equal{} \minus{} 1$ (because the left side has to be a constant) so we get $ 1 \equal{} C( \minus{} 1)^{(n \plus{} 1)}(n \plus{} 1)!$.  Rearrange that to get $ C \equal{} \frac {( \minus{} 1)^{(n \plus{} 1)}}{(n \plus{} 1)!}$

Plugging in $ x \equal{} n \plus{} 1$, we get $ (n \plus{} 2)P(n \plus{} 1) \minus{} n \minus{} 1 \equal{} \frac {( \minus{} 1)^{(n \plus{} 1)}(n \plus{} 1)!}{(n \plus{} 1)!}$.
$ (n \plus{} 1)!$ cancels out so we get $ P(n \plus{} 1) \equal{} \frac {( \minus{} 1)^{n \plus{} 1} \plus{} n \plus{} 1}{n \plus{} 2}$

Too lazy to type the rest in latex.

[b]t0rajir0u's problem:[/b]

[hide="Solution"]Let $ \Delta_0 [f](k) \equal{} f(k)$ and $ \Delta_{m \plus{} 1} [f](k) \equal{} \Delta _m [f](k) \minus{} \Delta _m [f](k \minus{} 1)$ define the family of difference polynomials of $ f(k)$.

Note that $ 2^{y} \minus{} 2^{y \minus{} 1} \equal{} 2^{y \minus{} 1}$, a nice property of the base-2 exponential function. From here, it is easy to see that $ \Delta_m [f](k) \equal{} 2^{k \minus{} m}$ for $ m \leq k \leq n$

Now, as we go through the samples, we see that $ \Delta_{n} [f](n) \equal{} 2^{n \minus{} n} \equal{} 1$ is the only known value of $ \Delta_{n} [f](x)$. To minimize degree, we assume that $ \Delta_{n} [f](k) \equal{} 1 \; \forall k$. Thus, $ \Delta_{n} [f](n \plus{} 1) \equal{} 1$.

Recall that $ \Delta _m [f](n \plus{} 1) \equal{} \Delta_{m \plus{} 1} [f](n \plus{} 1) \plus{} \Delta _m [f](n) \equal{} \Delta_{m \plus{} 1} [f](n \plus{} 1) \plus{} 2^{n\plus{}1 \minus{} m}$

This is a telescoping sum, and we find that $ f(n \plus{} 1) \equal{} \Delta_0 [f](n \plus{} 1) \equal{} 2^0 \plus{} 2^1 \plus{} \ldots \plus{} 2^{n} \equal{} 2^{n \plus{} 1} \minus{} 1$[/hide]

Yep.  :)

[hide="The cheap solution"] If you're familiar with binomial coefficients, you might realize that

$ P(x) \equal{} {x \choose 0} \plus{} {x \choose 1} \plus{} ... \plus{} {x \choose n}$.

This gives $ P(n\plus{}1) \equal{} {n\plus{}1 \choose 0} \plus{} ... \plus{} {n\plus{}1 \choose n} \plus{} {n\plus{}1 \choose n\plus{}1} \minus{} {n\plus{}1 \choose n\plus{}1} \equal{} 2^{n\plus{}1} \minus{} 1$.  :) [/hide]
See [url=http://www.artofproblemsolving.com/Forum/weblog_entry.php?t=178381]the blog post to which caffeineboy refers[/url] for an explanation of the technique behind both solutions.