abstract algebra
vector
algebra
polynomial
superior algebra
superior algebra theorems

let  $R$  be commutative ring and  $M$  is finitely generated  $R$ module
   $\varphi: M\longrightarrow M$  is a  surjective  $R$ module homomorphism
 prove that  $\varphi$  is injective

Let $k$ be the minimal number of generators of $M$. Let $e_{1},e_{2},...,e_{k}$ be these generators of $M$. Then since $\phi$ is surjective we have that $\phi(e_{1}),...,\phi(e_{k})$ also generate $M$. If there were $x= r_{1}e_{1}+...+r_{k}e_{k}$ such that $\phi(x) = 0$, then it would imply that $\phi(e_{1}),...,\phi(e_{k})$ aren't linearly independent and, thus, $M$ could be generated by $k-1$ elements - contradiction with our choic of $k$.

@eugene: this works only if $R$ is a division ring. it's right that the images are linearly dependant, but this does not guarantee, that one image depends on the others ...

... however I can't manage to find a proof either. does anyone has an idea for this interesting statement? I mean, this seems to be familiar only for vector spaces ...

Let $R_{n}=\ker\varphi^{n}$ i.e. $\varphi^{2}=\varphi\circ\varphi$, don't think of something else. It is obvious that $R_{n}$ form an increasing sequence of submodules of $M$. We can put $R_{0}=0=\ker id_{M}$.

Assume there exists $x_{0}$ a nonzero element of $\ker\varphi$. There exists $x_{1}\in M$ such that $\varphi(x_{1})=x_{0}$. Then $x_{1}\not\in\ker\varphi$, otherwise $0=\varphi(x_{1})=x_{0}$, which contradicts the choice made for $x_{0}$. Inductively we construct a sequence $x_{n}\in R_{n+1}\setminus R_{n}$, so $R_{n}$ is a strictly increasing sequence of submodules of $M$.

If $M$ were a noetherian module, than we would have a contradiction right here, but unfortunately, if $R$ is not noetherian, this hardly ever happens.

Let's try to go round this "small" impediment.

$\varphi$ is an endomorphism of a finitely generated module, so it satisfies a Hamilton-Cayley type polynomial relation:
$\varphi^{m}+a_{m-1}\varphi^{m-1}+\ldots+a_{0}=0$.
By applying the above identity to $x_{0}$, we get $a_{0}=0$.
Let $\rho=\varphi^{m-1}+a_{m-1}\varphi^{m-2}+\ldots+a_{1}$.
Then $\rho\circ\varphi(x)=0$ for all $x\in M$.
Since $\varphi$ is surjective it admits a set-theoretic (not necessarily an endomorphism) section $g: M\to M$ such that $\varphi\circ g(x)=x$ for all $x\in M$. We can take $g(0)=0$.

We have $\rho\circ\varphi\circ g(x)=0(g(x))=0$ for all $x\in M$. We see $0$ as the null endomorphism of $M$. So $\rho(x)=0$ for all $x\in M$.
Taking $\rho(x_{0})$ proves $a_{1}=0$. Continuing this process we will finally obtain $\varphi=0$ which contradicts its property of being surjective.

[Small note]: We didn't really need the sequence $x_{n}$ constructed above. Nor have we really used $R_{n}$. They are reminiscence of a previously bad idea I had for the problem. No, I didn't fall the noetherian trap. I thought I could apply the polynomial relation to $x_{n}$ to prove that all the coefficients are $0$. That didn't quite worked out as smoothly as it seamed, but I like the construction though :).

Ok. The above proof is flawed (Sorry) as somebody pointed out to me.

Attempt no2:

$ M$ is an $ R[f]$ module because $ f$ is linear (define $ f\cdot m\equal{}f(m)$ and extend), so we can say that multiplication by $ f$ on $ M$ is surjective, which means $ M\equal{}(f)M$
Then there exists an element $ 1\plus{}kf\in R[f]$ such that $ (1\plus{}kf)M\equal{}0$.
If $ f(x)\equal{}0$, then $ fx\equal{}0$. But this and $ (1\plus{}kf)x\equal{}0$ imply $ x\equal{}0$, so $ f$ is injective.

ah, you apply nakayama to $ M$ over $ R[f]$ ... nice

I hope this proof is correct. In the earlier version I said that $ f(x_0)\equal{}0$ implies $ a_0\equal{}0$, but it implied $ a_0x_0\equal{}0$ and I found no easy way around.


Find all function $f$:$R$ to $R$ such that:

     $f^{2}(x)$ + $2yf(x)$ + $f(y)$ = $f(y+f(x))$
for all $x,y$ in $R$

I think we have that :
$i,f(f(x))=f(-f(x))=f^{2}(x)+f(0)$
$ii,\forall x\in R$ there exits $y,z \in R$ such as : $x=f(y)-f(z)$

Then we have $f(y)=f(f(y)-f(z))=f^{2}(y)-2f(y)f(z)+f^{2}(z)+f(0)=x^{2}+f(0)$

I think we have 2 solution :

$f(x)=0$ and $f(x)=x^{2}+a$

Yes, I solve, when $f^{2}(x)$ mean $f(f(x))$. In this case y=f(x) give $3y^{2}+f(y)=f(2y).$

Let $g(x)=f(x)-x^{2}$. It give $g(y+x^{2}+g(x))=g(y)$.
1. $g(x)=-x^{2}\Longrightarrow f(x)=0.$
2. If $z=x^{2}+g(x)=f(x)\not =0$ then z is period $g(y+z)=g(y)$.
If $g(x)$ is continiusly it give $g(x)=const, f(x)=x^{2}+a$.
But it had infinetely many non continiusly solutions. For example:
$f(x)=0,x-\ irrational, \ f(x)=r(r=const), \ x,r-\ rational$.


Let $ a,$ $ b,$ $ c,$ $ d$ and $ e$ are positive numbers
such that $ abc\plus{}abd\plus{}abe\plus{}acd\plus{}ace\plus{}ade\plus{}bcd\plus{}bce\plus{}bde\plus{}cde\equal{}10.$ Prove that
\[ 1\plus{}\frac{13}{a\plus{}b\plus{}c\plus{}d\plus{}e}\geq\frac{36}{ab\plus{}ac\plus{}ad\plus{}ae\plus{}bc\plus{}bd\plus{}be\plus{}cd\plus{}ce\plus{}de}\]

Denote $ a \plus{} b \plus{} c \plus{} d \plus{} e \equal{} A$, $ ab \plus{} \dots \equal{} B$. Then $ f(x) \equal{} x^5 \minus{} Ax^4 \plus{} bx^3 \minus{} 10x^2 \plus{} \dots$ has five positive roots $ a,\,b,\,c,\,d,\,e$, hence $ f''$ has three positive roots, so $ 20x^3 \minus{} 12Ax^2 \plus{} 6Bx \minus{} 20$ has three positive roots $ u,\,v,\,w$, we have $ uvw \equal{} 1$, $ \alpha \equal{} u \plus{} v \plus{} w \equal{} 3A/5$, $ \beta \equal{} uv \plus{} vw \plus{} uw \equal{} 3B/10$. So, we have to prove $ 5 \plus{} 39/\alpha\ge 54/\beta$. Ok, remember Schur inequality for numbers $ W \equal{} uv,\,U \equal{} vw,\,V \equal{} uw$. It says that $ U^3 \plus{} V^3 \plus{} W^3 \plus{} 3UVW\ge U^2(W \plus{} V) \plus{} V^2(W \plus{} U) \plus{} W^2(U \plus{} V)$, i.e. $ (U \plus{} V \plus{} W)^3 \plus{} 9UVW\ge 4(U \plus{} V \plus{} W)(UV \plus{} VW \plus{} UW)$, i.e. $ \beta^3 \plus{} 9\ge 4\alpha\beta$, i.e. $ 9\ge \beta(4\alpha \minus{} \beta^2)$. Assuming the contrary, i.e. that $ 5 \plus{} 39/\alpha < 54/\beta$, we get $ \alpha > 39\beta/(54 \minus{} 5\beta) > 0$. Substitute this to $ 9\ge \beta(4\alpha \minus{} \beta^2)$ and after simplifications get $ (\beta \minus{} 3)(5\beta^3 \minus{} 39\beta^2 \plus{} 39\beta \plus{} 162)/(54 \minus{} 5\beta) < 0$, hence $ 5\beta^3 \minus{} 39\beta^2 \plus{} 39\beta \plus{} 162 < 0$, what is impossible for positive $ \beta$.

Yes, 	Fedor Petrov! :lol:
My proof is similar, but instead of Schur, I used $ (u \minus{} v)^2(u \minus{} w)^2(v \minus{} w)^2\geq0.$  Schur is easier here.

[quote=arqady]
I used $ (u \minus{} v)^2(u \minus{} w)^2(v \minus{} w)^2\geq0.$  [/quote]
Dear Mr Michael Rozenberg
Could you please post your full solution ?

Now, I see much more easier proof by $uvw$.
We need to prove that
$$\frac{1}{w^2}+\frac{13}{5uw}\geq\frac{18}{5v^2}.$$
Now, by the Rolle's theorem we can see in the last inequality the inequality of three variables $x$, $y$ and $x$,
where $x+y+z=3u$, $xy+xz+yz=3v^2$ and $xyz=w^3$.
We see that it's enough to prove the last inequality for the maximal value of $u$, which happens for equality case of two variables.
After replacing $x$ on $x^3$ we can assume $y=z=1$, which gives
$$(x-1)^2(10x^4-34x^3+59x+10)\geq0,$$ which is obvious.


Do there exist powers of 2 that begin with 2,7,4,5 in this order?

can there be numbers inbetween them? so 2745567 would be legal, but would 207465 be?

I think you should rephrase it a little :)

[quote]...that begin with 2,7,4,5 in this order?[/quote]
I think he meant your
[quote]...so 2745567 would be legal...[/quote]

It is obvious that log10 2 is irrational. Hence the set {n log10 2} (where n runs through all integer values, and {} denotes fractional part) is dense on [0, 1]. (This is called Jacobi's theorem, or also Kronecker's theorem.) You should be able to derive the result from this consideration :): infinitely many such powers of 2 must exist.

well it looks like a difficult problem anyway. I think the way to solve it is the way a calculator counts it in notation of x,xxx . 10^a

so some force of two resulting 2,745...   .   10^a

2^38 comes close to it, but is a little too much, but I guess the solution will be that there is none for some reason :D

There are even infinitely many of them Peter :)

:( strange excercice then

btw does anyone know how to calculate 2logs with a calculator???

Yea well it was so simple that I overlooked it :redface:


$ x_0\equal{} a$
$ x_{n\plus{}1}\equal{}x_n^2\minus{}x_n$
Which value of a that $ x_n$ have limit.

[quote="Rijul saini"]I am assuming $ x_n \in \mathbb{N}$
We have the required answer as $ x_n \equal{} 1 \ or \ 2$
For all otther values $ x_n \longrightarrow \ infinity$[/quote]
You are wrong.a=0 is also lead $ x_n$ to have the lim too :lol:

Please read my assumption carefully...
Natural numbers does not mean 0

I'm going to post two graphs as a hint to solving the problem under the assumption $ a \in \mathbb{R}$.
The first provides all the information you might need. The second is there for convenience.


What is ''Problems from the Book'' forum for?

For harazi and Titu's book http://www.artofproblemsolving.com/Books/AoPS_B_Item.php?item_id=5003


[quote="daermon"]If $x^{2}+px+8$ has real, distinct roots, prove that $\mid{r_{1}+r_{2}}\mid>8$[/quote]

are you sure the 8 isn't $4\sqrt{2}$?

[quote="daermon"]If $x^{2}+px+8$ has real, distinct roots, prove that $\mid{r_{1}+r_{2}}\mid>8$[/quote]
not true... take $r_{1}=-4$ and $r_{2}=-2$
[hide="but"]
from the quadratic formula,
$p^{2}-32>0$
$|p|>4\sqrt{2}$
and since $p=-(r_{1}+r_{2})$
$|r_{1}+r_{2}|>4\sqrt{2}$
so yea, spalshd is right[/hide]

oops sry I meant $4\sqrt{2}$  :oops:
you guys are right


[quote="pureblack2003"]find $k \in \mathbb{Z}$ so as the system is compatible
$[x]+y+z=5$
$x+k[y]+z=3$
$x+y+3[z]=7$[/quote]
Obviosly $y+z,x+z,x+y$ are integers, therefore $\{x\}=\{y\}=\{z\}=0 \ or \ \frac{1}{2}$.
In first case $z=1, (k-1)y=-2,x=4-y$. it give $k=-1,0,2,3.$
In second case $z=1.5, \ (k-1)[y]=-2, y=\frac{-2}{k-1}+\frac{1}{2}, x=3.5-[y]$, therefore $k=-1,0,2,3$.


induction
inequalities proposed
inequalities
BPSQ
algebra
High school olympiad

[hide]Hint:prove for n=3,and apply induction[/hide]

Let $a$, $b$ and $c$ be non-negatives. [url=https://artofproblemsolving.com/community/c6h2072698p14870889]Prove that[/url]$$ab^2+bc^2+cd^2+da^2\leq\frac{4}{27}(a+b+c)^3.$$
Let $d=\min\{a,b,c,d\}$.Thus,$$ab^2+bc^2+cd^2+da^2=ab^2+bc^2+ca^2+abc+cd^2-abc+da^2-ca^2$$$$\leq  ab^2+bc^2+ca^2+abc\leq\frac{4}{27}(a+b+c)^3.$$


Find all functions $f : \mathbb{R} \to\mathbb{R}$ such that $f(x +y) = f(x) f(y) f(xy)$ for all $x, y \in \mathbb{R}.$

First let's note that if $f(x) = 0$ for some $x$ then $f(x) = 0$ for all $x$ by translation.
So let's suppose $f(x) \neq 0 \forall x$
If $f$ is a solution  so is $-f$

$x=y=0 : f(0)^2 = 1$
If $f$ is a solution  so is $-f$ so we take for the moment $f(0) = 1$

Taking successively values for $x, y$: $1,-1;-2,1;-1,-1$ we get that $f(1) = f(-1) = f(-2) = f(2) = 2$
Then taking $x,-x$, $x-1,1$, we get $f(-x) = f(-x^2)$ or $f(x) = f(-x)$

Looking again $x,-x$ gives $f(x)^3= 1$ i.e $f(x) = 1$

Only solutions are $-1, 0, 1$

[color=#f00][Mod edit: see pco's solution [url=http://artofproblemsolving.com/community/c6h146995p832434]here[/url].][/color]

\[f(x+y)=f(x)f(y)f(xy)\]         .....(1)
We note that if f(r)=0 for some real r then f(x)=0 for all reals x,
We thus work now fully on the assumption that f(r) not equals zero for any r.
 Taking x=a+b and y=c in (1) we get
  \[f(a+b+c)
    =f(a+b)f(c)f(ac+bc)
    =f(a)f(b)f(ab)f(c)f(ac)f(bc)f(abc^2)\]
Thus we get
\[f(a+b+c)=f(a)f(b)f(c)f(ab)f(bc)f(ca)f(abc^2)\]     .....(2)
Again taking x=a and y=b+c in (1) we get
\[f(a+b+c)
=f(a)f(b+c)f(ab+ac)
=f(a)f(b)f(c)f(bc)f(ab)f(ac)f(a^2bc)\]
Thus we get
\[f(a+b+c)=f(a)f(b)f(c)f(ab)f(bc)f(ca)f(a^2bc)\]    ......(3)
Comparing (2) and (3) we get
\[f(abc^2)=f(a^2bc)\]            ...(4)
If a,b,c are not equal to 0, then we put b=1/ac in (4) to get
\[f(c)=f(a)\]
Since a and c are arbitrary reals, it follows that f is a constant function, say
\[f(x)=c\] for all reals x.
Then substitution in (1) leads to
\[c=c^3\]
giving values c=0,1 and -1
Thus we get three solutions:
\[ f(x)=0\]
\[f(x)=-1\]
\[f(x)=1\]
each extending over all reals x.

We claim that the only solutions are $f \equiv 0, \pm 1~\forall x \in \mathbb{R}$. Obviously, they satisfy the original condition. Let us now show these are the only ones.

Let $P(x,y)$ denote the assertion. We start off with the crux claim.

[color=#f00][b]Claim.[/b][/color] We have, $f(x) = f(x^2)$ for all $x \in \mathbb{R}$.
[i]Proof.[/i] Combine $P(x,x)$ and $P(x,2)$ to get $f(x+2)=f(x)^3 \cdot f(2) \cdot f(x^2)$. Again, combining this along with $P(x,1)$ and $P(x+1,1)$, we get $$f(x)^3 \cdot f(2) \cdot f(x^2) = f(x+2) = f(x+1)^2 \cdot f(1) = f(x)^4 \cdot f(1)^3$$ It follows that $f(x) \cdot f(1)^3 = f(2) \cdot f(x^2)$. But $P(1,1)$ implies that $f(2)=f(1)^3$. Therefore, we must have $f(x)=f(x^2)$, as desired. $\square$

Replacing $x$ with $-x$ in the previous claim yields $f(x) = f(-x)$. So, $f$ must be an even function. To finish off, from $P(x,-x)$, we have $$f(0) = f(x) \cdot f(-x) \cdot f(x^2) = f(x)^2 \cdot f(x) = f(x)^3$$
It follows that $f$ must be a constant function. Plugging $f(x) = c$ back in the original equation, we get $c=0,\pm 1$. This finishes the problem. $\blacksquare$

Let $P(x,y)$ be the assertion $f(x+y)=f(x)f(y)f(xy)$.
If $f(k)=0$ for some $k$:
$P(x-k,k)\Rightarrow\boxed{f(x)=0}$ which is a solution.
If $f$ is a solution, then $-f$ is also a solution.
We can now assume that $f(x)\ne0$ for all $x$ and $f(0)>0$.
\begin{align*}
P(x,y+z)&\Rightarrow f(x+y+z)=f(x)f(y+z)f(xy+xz)\\
&\Rightarrow f(x+y+z)=f(x)f(y)f(z)f(yz)f(xy+xz)\\
&\Rightarrow f(x+y+z)=f(x)f(y)f(z)f(xy)f(yz)f(xz)f(x^2yz)\\
&\Rightarrow Q(x,y,z):f(x^2yz)=f(xy^2z)=f(xyz^2)
\end{align*}
$Q\left(\frac1x,x,1\right)\Rightarrow f(x)=f(1)\forall x\ne0$
$P(0,0)\Rightarrow f(0)=1$
$P(1,1)\Rightarrow f(1)=f(1)^3\Rightarrow f(1)^2=1$
$P(1,-1)\Rightarrow f(1)f(-1)^2=1\Rightarrow f(1)=1$
Thus $\boxed{f(x)=1}$. The solution $-f$ corresponds to $\boxed{f(x)=-1}$.


set theory
advanced fields
advanced fields unsolved

Could you give an example of an uncountable well ordered chain ,where the natural ordering is considered(please give a concrete example in the special case of real numbers also).

That was my question. In general, check out [url=http://en.wikipedia.org/wiki/Ordinal_number]ordinals[/url]. You can pick them to be as large as you like, and they are well-ordered by construction.

As for reals. Are there well-ordered uncountable subsets of $ \mathbb{R}$? I think I remember reading somewhere that there aren't, but my set-theory isn't in great shape...

With respect to what ordering?  As far as I can tell, by "natural ordering" you mean the field order on $ \mathbb{R}$, but no uncountable subset of $ \mathbb{R}$ is well-ordered with respect to this ordering (sketch: uncountable subsets of $ \mathbb{R}$ that are well-ordered contain no intervals and no limit points, so every point is isolated and can be covered by an interval, but the real line can only contain countably many disjoint intervals).  The existence of a well-order of an arbitrary set ([i]not[/i] one compatible with field operations) is equivalent to the Axiom of Choice.

On the other hand, I'm not sure you're asking the question you want to ask since this question, as stated, has nothing to do with posets.


group theory
abstract algebra
superior algebra
superior algebra unsolved

suppose that Cis a family of subsets of a group G which forms a partition of G,and suppose further that XY belongs C for any X,Y belongs C.show that exactly one of the sets belonging to C is a subgroup of G,that this subgroup is normal in G,and that C consists of its cosets.

Nice ;). But we have to assume, that the Ci are non-empty.

[hide="solution"]
We have an equivalence relation of $G$, whose classes have the property

$\forall a,b \in G ~\exists c \in G : [a] [b] = [c]$

But since $a*b \in [a] [b]$, this means

$\forall a,b \in G : [a] [b] = [ab]$

We show that $[1]$ is a normal subgroup:

$1 \in [1]$ is trivial. $a,b \in [1]$ implies

$[1] = [a a^{-1}] = [a] [a^{-1}] = [1] [a^{-1}] = [a^{-1}]$

$[1] = [1] [1] = [a] [b] = [ab]$

hence $a^{-1} , ab \in [1]$. If $g \in G$ and $a \in [1]$, then

$[g a g^{-1}]=[g] [a] [g^{-1}] = [g] [1] [g^{-1}] = [g g^{-1}] = [1]$

i.e. $g a g^{-1} \in [1]$. So, $[1]$ is a normal subgroup. It is the only subgroup of the form $~ [a]$, because $1 \in [a]$ implies $[a] = [1]$.

Now for all $a \in G$, we show $[a] = a [1]$:

$g \in [a] \Rightarrow [a^{-1} g] = [a^{-1}] [g] = [a^{-1}] [a] = [1] \Rightarrow a^{-1} g \in [1] \Rightarrow g \in a [1]$

$g \in a [1] \Rightarrow a^{-1} g \in [1] \Rightarrow [g] = [a] [a^{-1} g] = [a] [1] = [a] \Rightarrow g \in [a]$

So, each $~ [a]$ is a coset of the normal subgroup $[1]$.
[/hide]


floor function
induction
logarithms
integration
limit
real analysis
real analysis unsolved

I proposed this "small" problem: Let a be the sequence, a: [b]N[/b]->[b]N[/b] such that lim a(n)/n=0. Prove that the series whose n-th term is ((-1)^n)a(n)/n converges. Do you have any ideas or suggestions which way to go and which methods to use in proving that hypothesis? Or can you disprove it by constructing the counterexample?

[hide="counterexample"]$ \begin{cases}a_{2k} \equal{} \lfloor\sqrt{k}\rfloor \\
a_{2k \plus{} 1} \equal{} 0\end{cases}$[/hide]

[quote="ifk"][hide="counterexample"]$ \begin{cases}a_{2k} \equal{} \lfloor\sqrt {k}\rfloor \\
a_{2k \plus{} 1} \equal{} 0\end{cases}$[/hide][/quote]

That is not the counterexample. The sequence a does not take the value zero anywhere. It is strictly positive, a: N->N, not                    a: N -> N U {0}

[quote="ifk"]I assumed $ \mathbb{N} \equal{} \{0, 1, 2, \ldots \}$, but you can put your favourite (fixed) natural number instead of $ 0$ :)[/quote]

What happens if I do that? Does it diverges than? I can not SEE that. Can you actually prove it or describe some steps that would constitute a proof?

[hide="proof"]Suppose that
$ \begin{cases}a_{2k} \equal{} \lfloor\sqrt {k}\rfloor \\
a_{2k \plus{} 1} \equal{} n_{0}\end{cases}$
Let $ S_{n} \equal{} \sum_{k \equal{} 0}^{n}\frac {( \minus{} 1)^{n}a_{n}}{n}$.
We have $ \frac {( \minus{} 1)^{2n}a_{2n}}{2n} \equal{} \frac {a_{2n}}{2n} > \frac {\sqrt {n} \minus{} 1}{2n} \equal{} \frac {1}{2\sqrt {n}} \minus{} \frac {1}{2n}$
By induction:
$ \sum_{k \equal{} 1}^{n}\frac {1}{\sqrt {k}} \ge \sqrt {n}$
We have also:
$ \sum_{k \equal{} 2}^{n} \frac {1}{n} \le \ln n$
(left side is the Lower Darboux Sum of $ x\mapsto \frac {1}{x}$ on $ [1, n]$, for partition $ (1, 2, \ldots, n \minus{} 1, n)$ of that interval, and right side equals $ \int_{1}^{n}\frac {1}{x}\, dx$)
Now we can estimate:
$ S_{2n} \equal{} \sum_{k \equal{} 1}^{n}\frac {a_{2k}}{2k} \minus{} \sum_{k \equal{} 0}^{n\minus{}1} \frac {a_{2k \plus{} 1}}{2k \plus{} 1} \ge \frac {\sqrt {n}}{2} \minus{} \frac {1}{2}(1 \plus{} \ln n) \minus{} \frac {n_{0}}{2}(3 \plus{} \ln (n \minus{} 1)) \to \infty$
since $ \lim_{n\to \infty}\frac {\ln n}{\sqrt {n}} \equal{} 0$.
Thus there cannot exist finite limit $ \lim_{n\to \infty}S_{n}$ and the series diverges.[/hide]


Rationalize the denominator:

$\frac{1}{3+5 \sqrt[3]{2}+7 \sqrt[3]{4}}$

I would post this in Intermediate, but too many Pre-Olympiad-level problem solvers answer questions there (myself included  :rotfl: ) and I intend for this to be good food for thought for those of you who already know how to rationalize square roots  :wink:

[quote="kevyy"][hide]
could u rationalize it by thinking of $3+5\sqrt[3]{2}$ as one quantity and $7 \sqrt[3]{4}$ as another?[/hide][/quote]

Some of my fellow campers at PROMYS thought this, but you'll find that the idea doesn't work.

Farenhajt:  Tricky.  Try the general case ;)

$\frac{1}{a+b \sqrt[3]{2}+c \sqrt[3]{4}}$

[hide="Another spoiler! :)"]Here's a more general method.

We must find rational numbers $p,q,r$ such that

$\frac{1}{3+5\sqrt[3]{2}+7\sqrt[3]{4}}=p\sqrt[3]{4}+q\sqrt[3]{2}+r$

Multiplying and simplifying, we get the following:

$\sqrt[3]{4}(3p+5q+7r)+\sqrt[3]{2}(14p+3q+5r)+(10p+14q+3r)=1$

Since $\sqrt[3]{4}$ and $\sqrt[3]{2}$ are irrational, this can only hold if the following system has a solution:

$\begin{cases}3p+5q+7r = 0\\ 14p+3q+5r=0\\ 10p+14q+3r=1\end{cases}$

It can be solved by any known method, and as it is, the solution is $(\frac{4}{1019},\frac{83}{1019},-\frac{61}{1019})$

Hence $\frac{1}{3+5\sqrt[3]{2}+7\sqrt[3]{4}}=\frac{4\sqrt[3]{4}+83\sqrt[3]{2}-61}{1019}$[/hide]

This "spoiler" contains an outline of an answer to the [b]t0rajir0u[/b]'s generalization, but I really can't be bothered to post step-by-step solution of the system with the parameters $a,b,c$ - forgive me my laziness ;) Therefore I give only the

[hide="general formula"]$\frac{1}{a+b\sqrt[3]{2}+c\sqrt[3]{4}}=\frac{(b^{2}-ac)\sqrt[3]{4}+(2c^{2}-ab)\sqrt[3]{2}+(a^{2}-2bc)}{a^{3}+2b^{3}+4c^{3}-6abc}$[/hide]

and also

[hide="even more general formula"]$\frac{1}{a+b\sqrt[3]{n}+c\sqrt[3]{n^{2}}}=\frac{(b^{2}-ac)\sqrt[3]{n^{2}}+(nc^{2}-ab)\sqrt[3]{n}+(a^{2}-nbc)}{a^{3}+nb^{3}+n^{2}c^{3}-3nabc}$ for $\sqrt[3]{n}\notin\mathbb{Q}$[/hide]

And after all this, I should give the real crucial hint, and

[hide="here it is"]$(x+y+z)(x^{2}+y^{2}+z^{2}-xy-yz-zx)=x^{3}+y^{3}+z^{3}-3xyz$[/hide]

Thank you very much.

Especially that last factoring trick.

Now, I believe this problem is solvable...

An even more general statement can be found [url=http://www.mathlinks.ro/Forum/viewtopic.php?t=83135]here[/url].


Finds all integer roots of the equation $ x^2 \plus{} y^2 \equal{} 458\equal{}2(2^2\plus{}15^2)$

All squares mod $ 8$ are $ 0,1,4$. $ 446$ is $ 6$ mod $ 8$. No two of $ 0,1,4$ can be added to get $ 6$, so no solutions.

Sorry, my exact problem is
"Find all integer roots of the equation:
$ x^2\plus{}y^2\equal{}458\equal{}2(2^2\plus{}15^2)$"

$ 458\equal{}2.229$ where $ 229$ is a prime in the form $ 4k\plus{}1$ so it has solution.
Other way you can suppose $ x>y$ and $ 2y^2<458$ and try to find with computer.

[hide="The relevant lemma"] $ (a^2 \plus{} b^2)(c^2 \plus{} d^2) \equal{} (ac \pm bd)^2 \plus{} (ad \mp bc)^2$ [/hide]
[hide="Solution"] $ x^2 \plus{} y^2 \equal{} (1^2 \plus{} 1^2)(2^2 \plus{} 15^2) \equal{} (2 \pm 15)^2 \plus{} (2 \mp 15)^2 \equal{} 17^2 \plus{} 13^2$

Hence $ (x, y) \equal{} (\pm 13, \pm 17), (\pm 17, \pm 13)$ (four cases each).  We know there are no other solutions from some lemmas in $ \mathbb{Z}[\imath]$, or we can search exhaustively. [/hide]

Hi all,
We haved 4 solutions (13,17),... But they're all solutions? Are there anymore solution?

[quote="t0rajir0u"]We know there are no other solutions from some lemmas in $ \mathbb{Z}[\imath]$, or we can search exhaustively. [/quote]

Specifically, $ \mathbb{Z}[\imath]$ is a UFD and $ (x \plus{} yi)(x \minus{} yi) \equal{} (1 \plus{} i)(1 \minus{} i)(2 \plus{} 15i)(2 \minus{} 15i)$ is a factorization into primes, unique up to multiplication by a unit.


trigonometry
algebra
polynomial
algebra unsolved

can you prove that if sinx=3/5 then x is not rational multiple of pi,if it is true

I think there are different approaches to this problem.

Note that $\cos x=\frac{4}{5}$. Suppose $x=\frac{\pi p}{q}$ and consider Tchebychev's polynomial $T_{2q}(t)=\cos(2q\arccos(t))$. We see that $1=T_{2q}(\cos x)=T_{2q}(4/5)$, so $4/5$ is a root of $T_{2q}-1=0$. But it is well known that $T_n(t)\in\mathbb{Z}[t]$ has a power of $2$ as a leading coefficient. Therefore it is impossible $T_n(4/5)-1=0$, since $5$ doesn't divide powers of $2$ (we use ratonal root theorem).

Cool! :)

Another way would be proving once more (this would be like the fourth time it's done on the forum :)) that $\cos x$ can be rational when $x$ is a rational multiple of $\pi$ iff it takes the values $0,\pm 1,\pm\frac 12$. We do this by observing that $2\cos x$ is an algebraic integer, which is also rational (it's an algebraic integer because it's equal to $z+\bar z$, with $z=\cos x+i\sin x$, and $z,\bar z$ are algebraic integers), meaning that it's an integer, and, since it takes values between $-2$ to $2$, we get $2\cos x\in\{0,\pm 1,\pm 2\}$.


I like the following problem very much :D

Two given circles C1 and C2 intersect at A and B.
Let C be the second intersection of C2 and the diameter of C1 passing through B.
Let D be the second intersection of C1 and the diameter of C2 passing through B.
Let C3 be the circle passing through B, C and D.
Prove that the center of C3 is on the line joining A and B.

:)

Join the centers of the circles C1,C2,C3 Let the centers be C1,C2,C3 respectively.

Now C1C3 is the perpendicular bisector of BC and similarly C2C3 is the perpendicular bisector of BD.

Hence in the triangle C1C2C3. C1D & C2C are the altitudes and B is the orthocenter of the triangle.

So C3B extended on to C1C2 is the altitude on to C1C2. So C3B extented is perpendicular to C1C2.

But C1C2 is the perpendicular bisector of AB hence C3 lies on AB extended.


Ascertain the real numbers $x,y,z$ so that
$8yz\sqrt{xyz(x+y+z)}\le (x+y)(x+z)(y+z)^2\ .$
and establish when is the equality.

We can see:
$4yz\leq (y+z)^2$;
$2\sqrt{xyz(x+y+z)}=2\sqrt{yz*(x^2+xy+xz)}\leq yz+x^2+xy+xz=(x+y)(x+z)$
Multiplied we get what we need.
Equality hodls iff $y=z$ and $x^2+xy+xz=yz$, i.e. $x^2+2xy-y^2=0$, i.e $x=(\sqrt{2}-1)y$(I consider $x,y,z>0$), i.e $(\sqrt{2}+1)x=y=z$.


Do there exist real numbers $a$ and $b$ such that [list=a][*] $a+b$ is rational and $a^n +b^n $ is irrational for all $n \in \mathbb{N}$ with  $n \ge 2$? [*] $a+b$ is irrational and $a^n +b^n $ is rational for all $n \in \mathbb{N}$ with $n \ge 2$?[/list]

a) There exist such numbers: take $a$ to be any transcendental number and $b=1-a$ then $a^n+(1-a)^n=q$ for $q\in \mathbb{Q}$ and $n\geq 2$ would mean $a$ is algebraic, contradiction

b) answer is no, proof ad absurdum. $-\frac{a^3+b^3}{a+b}+a^2+b^2=ab \not \in \mathbb{Q}$.
But we also have
$a^{n+2}+b^{n+2}=(a^n+b^n)(a^2+b^2)-a^2b^2(a^{n-2}+b^{n-2})$ and
$a^{n+3}+b^{n+3}=(a^n+b^n)(a^3+b^3)-a^3b^3(a^{n-3}+b^{n-3})$
so both $(ab)^2, (ab)^3$ are rational and hence $ab$ is rational, contradiction.


$ A_{1},....A_{k}$are some subset of set $ X$. for each $ i$ we have$ |A_{i}|$=3.for  every $ a,b \in X$there are only $ p A_{i}$ such that $ {a,b}\subset A_{i}$(p is constant)
and also we know that there exists a partition of $ X$ like $ X_{1}$and$ X_{2}$ such that both $ X_{1}$and$ X_{2}$ have common elements with each $ A_{i}$
prove that $ |X|\leq 4$

Let $ |X_1| \equal{} m$, $ |X_2| \equal{} n$ and $ |X| \equal{} \ell$.  ($ \ell \equal{} m \plus{} n$)

For each pair $ \{a,b\}$ from $ X_1$ we have $ p$ different subsets $ A_i$

Then there are in total $ {m\choose 2}\cdot p$ different subsets $ A_i$ (beacuse each $ A_i$ cuts both $ X_1$ and $ X_2$)

Also there is $ {n\choose 2}\cdot p$ (beacuse of $ X_2$)

So we have at least $ {m\choose 2}\cdot p \plus{} {n\choose 2}\cdot p$ subsets $ A_i$:

$ k \geq {m\choose 2}\cdot p \plus{} {n\choose 2}\cdot p \geq {\ell ^2 \minus{} \ell\over 4}\cdot p$

-------------------------------------------------------------------------------------------------

On the other side each pair $ \{a,b\}$ from $ X$ is ''connected'' with $ p$ diffrent subsets $ A_i$
and each subset $ A_i$ is connected with exactly $ 3$ pairs $ \{a,b\}$, so we have
\[ {\ell \choose 2}\cdot p \equal{} 3k\]
-------------------------------------------------------------------------------------------------
Joining both equations we get $ \ell\leq 4$.


system of equations
combinatorics
algebra
IMO
IMO 1966

In a mathematical contest, three problems, $A,B,C$ were posed. Among the participants ther were 25 students who solved at least one problem each. Of all the contestants who did not solve problem $A$, the number who solved $B$ was twice the number who solved $C$. The number of students who solved only problem $A$ was one more than the number of students who solved $A$ and at least one other problem. Of all students who solved just one problem, half did not solve problem $A$. How many students solved only problem $B$?

$ a\equal{}\text{the number of participants who solved only problem A},\\b\equal{}\text{the number of participants who solved only problem B},\\c\equal{}\text{the number of participants who solved only problem C},\\d\equal{}\text{the number of participants who solved problem B and problem C but not problem A}$

[quote="DPopov"]Of all the contestants who did not solve problem $ A$, the number who solved $ B$ was twice the number who solved $ C$.[/quote]$ b\plus{}d\equal{}2(c\plus{}d)\quad\ldots(1)$

[quote="DPopov"]The number of students who solved only problem $ A$ was one more than the number of students who solved $ A$ and at least one other problem.[/quote]$ a\equal{}1\plus{}(25\minus{}a\minus{}b\minus{}c\minus{}d)\quad\ldots(2)$

[quote="DPopov"]Of all students who solved just one problem, half did not solve problem $ A$.[/quote]$ a\equal{}b\plus{}c\quad\ldots(3)$

(2) and (3): $ 3b\plus{}3c\equal{}26\minus{}d\quad\ldots(4)$.

(1): $ b\minus{}2c\equal{}d\quad\ldots(5)$.

(4) and (5): $ 4b\plus{}c\equal{}26\quad\ldots(6)$

(5): $ b\minus{}2c\equal{}d\ge0\implies b\ge 2c\quad\ldots(7)$

(i) $ b\le 5$
(7): $ c\le 2$
(6): $ c\ge 6$
Contradiction

(ii) $ b\ge 7$
(6): $ c\le \minus{}2$
Contradiction

$ \boxed{b\equal{}6}$

i didnt get how did you get B = 6 ...
can you please explain how did you compare at the end of your solution....

I did some algebraic manipulations, but without inequalities. Here's what I did:

Define the following variables:

$ a$: Number of students who only solved A
$ b$: " " B
$ c$: " " C
$ w$: Number of students who solved A and B but not C
$ x$: " " A and C but not B
$ z$: " " B and C but not A
$ y$: Number of students who solved all 3 problems

Then, based on the given information, we can get the following equations:

$ a \plus{} b \plus{} c \plus{} w \plus{} x \plus{} y \plus{} z \equal{} 25$

$ b \plus{} z \equal{} 2(z \plus{} c)$
$ \Rightarrow b \plus{} z \equal{} 2z \plus{} 2c$
$ \Rightarrow z \equal{} b \minus{} 2c$

$ a \equal{} 1 \plus{} w \plus{} x \plus{} y$

$ a \equal{} b \plus{} c$

Using substitution, the first equation becomes

$ 2b \plus{} 2c \plus{} a \minus{} 1 \plus{} b \minus{} 2c \equal{} 25$
$ 3b \plus{} 3c \minus{} 1 \plus{} b \minus{} 2c \equal{} 25$
$ 4b \plus{} c \equal{} 26$

Now the possible ordered pairs $ (b,c)$  of positive integers $ b$, $ c$ that satisfy this equation are:

$ b \equal{} 1, c \equal{} 22$
$ b \equal{} 2, c \equal{} 18$
$ b \equal{} 3, c \equal{} 14$
$ b \equal{} 4, c \equal{} 10$
$ b \equal{} 5, c \equal{} 6$
$ b \equal{} 6, c \equal{} 2$

By doing some calculations, we see that the first four ordered pairs listed above are not possible because that will resut in $ a \plus{} b \plus{} c \equal{} 2(b \plus{} c) > 25$, which is not possible. Also, $ b \equal{} 5$ and $ c \equal{} 6$ won't work because that will result in $ z \equal{} 5 \minus{} 12 \equal{} \minus{} 7$, which isn't possible.

Now let's try $ b \equal{} 6$ and $ c \equal{} 2$. Substituting and calculating gives us $ a \equal{} 8$, $ z \equal{} 2$, and $ w \plus{} x \plus{} y \plus{} z \equal{} 7$, all of which are possible.

Therefore $ \boxed{6}$ students in the olympiad solved only problem $ B$.


calculus
integration
trigonometry
symmetry
geometry
geometric transformation
function

$1.\blacktriangleright\ x\in R\Longrightarrow\int_{0}^{1}|t^{2}+x|\ dt\ge \frac{1}{4}\ .$
$2.\blacktriangleright\ x\in (-1,1)\Longrightarrow \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}}\frac{1}{x^{2}+2x\sin t+1}\ dt\ge \pi\ .$

On #2 I'd prefer to use symmetry to make the make the integral over a whole interval of length $2\pi$ followed by a translation so we have a cosine rather than a sine:

$I(x)=\frac12\int_{-\pi}^{\pi}\frac{dx}{x^{2}-2x\cos t+1}$

$I(x)$ is an even function of $x,$ so it suffices to assume that $x\in[0,1).$ We know that

$\sum_{k=-\infty}^{\infty}x^{|k|}e^{ikt}= \frac{1-x^{2}}{1-2x\cos t+x^{2}}$

So $I(x)=\frac1{2(1-x^{2})}\int_{-\pi}^{\pi}\sum_{k=-\infty}^{\infty}x^{|k|}e^{ikt}\,dt$

$=\frac{2\pi}{2(1-x^{2})}$ since all terms of the sum except the $k=0$ term integrate to zero.

This is clearly an increasing function of $x$ on $[0,1)$ hence it has its minimum value at $I(0)=\pi.$

[A Fourier analyst like me does tend to have instant recognition of the Poisson kernel.]

Number 2 can be tackled in the following way too. Use $\sin t = \frac{2\tan \frac{t}{2}}{1+\tan^{2}\frac{t}{2}}$ and put $u = \tan\frac{t}{2}$. Thus $du = \frac{1}{2}\sec^{2}\frac{t}{2}dt$ and the integral can be written as $\int_{-1}^{1}\frac{2}{(x^{2}+1)(u^{2}+1)+4xu}du = \frac{2}{x^{2}+1}\int_{-1}^{1}\frac{du}{(u+\frac{2x}{x^{2}+1})^{2}+(\frac{1-x^{2}}{1+x^{2}})^{2}}= \frac{\pi}{1-x^{2}}$. The result follows.


I got confused into thinking the registration deadline was June 13, and panicked when I learned otherwise. While I see from the other thread that I could get the legal forms in on time by scanning the documents and E-mailing them, payment is another question. Is there some way to pay over the Internet, is it okay if the check arrives a little late, or do I need to run over to Fed-Ex?

Send the check by regular mail and drop an email to us to let us know that it's on the way.


trigonometry
integration
calculus
calculus computations

$ \frac{dx}{dt} \minus{}4x \minus{}\cos t\equal{}0$

 :D  :o

Write this as
$ \frac{dx}{dt}\minus{}4x(t)\equal{}\cos{t}$.
Set $ \mu(t)\equal{}\exp\left(\int \minus{}4\ dt\right)\equal{}e^{\minus{}4t}$
Then $ x(t)\equal{}e^{4t}\int e^{\minus{}4t}\cos{t}\ dt\equal{}e^{4t}\frac{1}{17}e^{\minus{}4t}(\sin{t}\minus{}4\cos{t})\equal{}\frac{\sin{t}\minus{}4\cos{t}}{17}$.
Let's check it:
$ \frac{dx}{dt}\equal{}\frac{\cos{t}\plus{}4\sin{t}}{17}\Rightarrow\frac{dx}{dt}\minus{}4x(t)\equal{}\frac{\cos{t}\plus{}4\sin{t}}{17}\minus{}\frac{4\sin{t}}{17}\plus{}\frac{16\cos{t}}{17}\equal{}\cos{t}$
So we're good.

[quote="vinskman"]$ \frac {dx}{dt} \minus{} 4x \minus{} \cos t \equal{} 0$

 :D  :o[/quote]
[hide=" This may a solution"]
Suppose $ x(t)\equal{}A\cos t\plus{}B\sin t$ , where $ A,B$ are constants.

$ \frac {dx}{dt} \equal{}\minus{}A\sin t\plus{}B\cos t$

$ \implies  4x \plus{} \cos t\equal{}\minus{}A\sin t\plus{}B\cos t$

$ \implies  4( A\cos t\plus{}B\sin t)\plus{} \cos t\equal{}\minus{}A\sin t\plus{}B\cos t$

$ \implies  4A\plus{}1\equal{}B ...(1) , 4B\equal{}\minus{}A ....(2)$

$ \implies  \minus{}16B\plus{}1\equal{}B \implies B\equal{}\frac{1}{17} , A\equal{}\minus{}\frac{4}{17}$

$ \therefore x(t)\equal{}\minus{}\frac{4}{17}\cos t\plus{}\frac{1}{17}\sin t$

[/hide]

Either way works.  You used undetermined coefficients, I used integrating factors.

$ x(t) \equal{} e^{4t}\int \left(e^{ \minus{} 4t}\cos t\ dt \plus{} C\right) \equal{} e^{4t}\left(\frac {1}{17}e^{ \minus{} 4t}(\sin t \minus{} 4\cos t) \plus{} C\right)$

$ \equal{} \boxed{\frac {1}{17}(\sin t \minus{} 4\cos t) \plus{} Ce^{4t}}$.


geometry
limit
symmetry
integration
real analysis
real analysis unsolved

Let $ S_n$be the area of the figure on the plane, bounded with the curve $ x^{2n}\plus{}y^{2n}\equal{}1$. Show that $ \displaystyle\lim_{n\to\infty}S_n\equal{}4$.

Due to symmetry reasons we have that

$ A\equal{}4\int\limits_{0}^{1}\left(1\minus{}x^{2n}\right)^{1/2n}dx\equal{}4\frac{1}{2n}\int\limits_{0}^{1}(1\minus{}x)^{1/2n}x^{1/(2n)\minus{}1}dx$

$ \equal{}\frac{4}{2n}\beta\left(\frac{1}{2n}\plus{}1,\frac{1}{2n}\right)\equal{}4\frac{1}{2n}\frac{\Gamma(1\plus{}\frac{1}{2n})\Gamma(\frac{1}{2n})}{\Gamma(1\plus{}\frac{1}{n})}\rightarrow 4$.

I used the fact that $ \frac{1}{2n}\Gamma\left(\frac{1}{2n}\right)\equal{}\Gamma\left(1\plus{}\frac{1}{2n}\right)\rightarrow 1.$

Let $ E_n\equal{}\{(x,y): x^{2n}\plus{}y^{2n}< 1\}$

I claim that $ E_1\subset E_2\subset E_3\subset \cdots$ and $ \bigcup_{n\equal{}1}^{\infty}E_n\equal{}\{(x,y): \max(|x|,|y|)<1\}.$

For an increasing sequence of measurable sets, the limit of the measures equals the measure of the union, and the union is a square of area $ 4.$

It shouldn't be that hard to verify my claims.


algebra
polynomial
blogs
quadratics
function
number theory proposed
number theory

(1) Let $ f(x)$ be a polynomial with real coefficients. Prove that  the necessary and sufficient condition such that  $ f(x)$ is integer for any integers $ x$ is that $ f(0)$ is integer and $ f(k) \minus{} f(k \minus{} 1)$ is integer for any integers $ k$.

(2) Find the necessary and sufficient condition such that $ ax^2 \plus{} bx \plus{} c$ is integer for any integers $ x$ in terms of $ a,\ b,\ c$.

how come people who want to study humanities are examined in maths? :o

1. Necessity is immediate. for sufficiency: $ f(k)\equal{}\sum_{i\equal{}1}^{k} f(i)\minus{}f(i\minus{}1) \plus{}f(0) \in N$ for k>0, $ f(k)\equal{}\minus{}\sum_{i\equal{}1}^{\minus{}k} f(1\minus{}i)\minus{}f(\minus{}i) \plus{}f(0) \in N$, for k<0.

2. by (1), it is necessary and sufficient that $ f(0)\equal{}c$ is an integer and that $ a_{k}\equal{}f(k)\minus{}f(k\minus{}1)\equal{}b\plus{}a(2k\minus{}1)$ is an integer for every integer k. So $ a_{k}\minus{}a_{k\minus{}1}\equal{}2a$ is also an integer. Hence $ a_{k}$ is an integer iff b-a is an integer.
The conditions are: c, 2a (or a+b) and b-a are integers

Repeatedly applying the first result gives the following

[b]Proposition:[/b]  $ \left( x \in \mathbb{Z} \implies f(x) \in \mathbb{Z} \right) \Leftrightarrow f(x) \equal{} \sum_{k \equal{} 0}^{n} a_k {x \choose k}, a_k \in \mathbb{Z}$.

See [url=http://www.artofproblemsolving.com/Forum/weblog_entry.php?t=178381]my blog post[/url].  For $ n \equal{} 2$ this gives $ f(x) \equal{} a {x \choose 2} \plus{} b {x \choose 1} \plus{} c$; in other words, the quadratic and linear coefficients are either both integer or both half-integers.  (Interestingly enough, the above holds for any function from the integers to the integers if we allow $ n \equal{} \infty$.)


We know that $k$ is a positive integer and the equation \[ x^3+y^3-2y(x^2-xy+y^2)=k^2(x-y) \quad (1) \] has one solution $(x_0,y_0)$ with
$x_0,y_0\in \mathbb{Z}-\{0\}$ and $x_0\neq y_0$. Prove that

i) the equation (1) has a finite number of solutions $(x,y)$ with $x,y\in \mathbb{Z}$ and $x\neq y$;

ii) it is possible to find $11$ addition different solutions $(X,Y)$ of the equation (1) with $X,Y\in \mathbb{Z}-\{0\}$ and $X\neq Y$ where $X,Y$ are  functions of $x_0,y_0$.

i)$(1)\Leftrightarrow x^2-xy+y^2=k^2.$  (look $x\neq y$)
Hence, $(x-\frac{y}{2})^2+\frac{3y^2}{4}=k^2$ and $-\frac{2k}{\sqrt3}\leq x,y\leq\frac{2k}{\sqrt3}.$
ii)???

[quote="Arne"][quote="silouan"]ii) It is possible to find $11$ addition different solutions $(X,Y)$ of the equation (1)

with $X,Y\in Z*$ and $X\neq Y$ where $X,Y$ are  functions of $x_0,y_0$[/quote]

This part of the question isn't clear to me.[/quote]

We can say ii) It is possible to find $11$ addition different solutions $(X,Y)$ as functions of $x_0,y_0$ of the equation (1)
 It is better now ?

(i) $k^2=x^2-xy+y^2=\frac{1}{2}(x-y)^2+\frac{1}{2}(x^2+y^2)\geq\frac{1}{2}(x^2+y^2)$
(ii) Note that if $(x,y)$ is a solution to $x^2-xy+y^2=k^2$, then $(x-y,x)$ is another solution. This linear transformation gives an orbit of solutions.
$(x,y),(x-y,x),(-y,x-y),(-x,-y),(-x+y,-x),(y,-x+y)$
Taking negative gives a double of the solutions.

P.S. Please refer also to Turkey 2005 Problem 5. That problem is stated as follows.
(a) Prove that for each positive integer $n$, the number of ordered pairs $(x,y)$ of integers satisfying $x^2-xy+y^2=n$ is finite and divisible by $6$.
(b) Find all ordered pairs $(x,y)$ of integers satisfying $x^2-xy+y^2=727$.

exactly leepachin nice solution. The problem from Turkey is similar.

Thx. :)
What about your solution and the official one?


Let $ a,b,c,d\geq 0, a\plus{}b\plus{}c\plus{}d\equal{}1$. Prove that

$ (a^2\plus{}b^2\plus{}c^2)(b^2\plus{}c^2\plus{}d^2)(c^2\plus{}d^2\plus{}a^2)(d^2\plus{}a^2\plus{}b^2)\leq \frac{1}{64}$

[quote="outback"]Anybody?  :)[/quote]
Because it is too old,
My hint ;Use mixing variables (the SMV of hungkhtn) and the genaral one is same to
Let try again by your self

[quote="outback"]Let $ a,b,c,d\geq 0, a \plus{} b \plus{} c \plus{} d \equal{} 1$. Prove that

$ (a^2 \plus{} b^2 \plus{} c^2)(b^2 \plus{} c^2 \plus{} d^2)(c^2 \plus{} d^2 \plus{} a^2)(d^2 \plus{} a^2 \plus{} b^2)\leq \frac {1}{64}$[/quote]
It is a special case of my problem:
"If $ a,b,c,d$ are nonnegative real numbers such that $ a\plus{}b\plus{}c\plus{}d\equal{}1$. For each $ n>0$, determine that maximum value of
\[ P(a,b,c,d)\equal{}(a^n\plus{}b^n\plus{}c^n)(b^n\plus{}c^n\plus{}d^n)(c^n\plus{}d^n\plus{}a^n)(d^n\plus{}a^n\plus{}b^n)\]
For $ n\equal{}2$, we can solve it by using this (a trivial fact)
\[ P(a,b,c,d) \le P(a\plus{}\frac{d}{3},b\plus{}\frac{d}{3},c\plus{}\frac{d}{3},0)\]
where $ d\equal{}\min\{a,b,c,d\}$.
:)

[quote="can_hang2007"][quote="outback"]Let $ a,b,c,d\geq 0, a \plus{} b \plus{} c \plus{} d \equal{} 1$. Prove that

$ (a^2 \plus{} b^2 \plus{} c^2)(b^2 \plus{} c^2 \plus{} d^2)(c^2 \plus{} d^2 \plus{} a^2)(d^2 \plus{} a^2 \plus{} b^2)\leq \frac {1}{64}$[/quote]
 we can solve it by using this (a trivial fact)
\[ P(a,b,c,d) \le P(a \plus{} \frac {d}{3},b \plus{} \frac {d}{3},c \plus{} \frac {d}{3},0)
\]
where $ d \equal{} \min\{a,b,c,d\}$.
:)[/quote]
I think the following reasoning easier.
Let $ a\geq b\geq c\geq d.$
Hence, $ P\left(a\plus{}\frac{c\plus{}d}{2},b\plus{}\frac{c\plus{}d}{2},0,0\right)\minus{}P(a,b,c,d)\geq0.$ :wink:

Is there a "classical" solution to this?

well $RHS=\frac{(a+b+c+d)^8}{64}$

If you want to try a classical one


Last year GIL sold their books during IMO(may be some other as well,what i learnt from an IMO contestants last year)
are they going to sell their books during IMO this year.Is any other publication going to sell?If anyone knows about it,please inform.
(i only know that the book of Hung is going to be sold)

Yes, I am really happy since Gil chooses my book for IMO. But please don't post any question about the book publicly in Mathlinks, just PM for me or Mircea Lascu. Thank you,

Could someone post or pm me some information on these books and GIL, please?

Here is the list of books published by GIL in English

1. Authors: T. Andreescu, D. Andrica
Title: An Introduction to Diophantine Equations
ISBN: 973-9238-88-2
2. Authors: T. Andreescu, D. Andrica
Title: 360 Problems for Mathematical Contests
ISBN: 973-9417-12-4
3. Authors: T. Andreescu, V. Cartoaje, G. Dospinescu, M. Lascu
Title: Old and New Inequalities
ISBN: 973-9417-35-3
4. Author: Negut A.
Title: Problems for the Mathematical Olympiads
ISBN: 973-9417-52-3
5. Author: V. Cartoaje
Title: Algebraic Inequalities. Old and New Methods
ISBN: 973-9417-66-3
6. Authors: M. Becheanu, B. Enescu
Title: Balkan Mathematical Olympiads 1984-2006
ISBN: 973-9417-80-8
7. Authors: P. Boyvalenkov, E. Kolev, O. Mushkarov, N. Nikolov
Title: Bulgarian Mathematical Competitions 2003-2006
ISBN: 973-9417-86-0
8. Author: P. K. Hung
Title: Secrets in Inequalities (volume 1)
ISBN: 973-9417-88-4

In case you are interested in any of these books and will come in Vietnam (Hanoi) at the IMO visit http://www.gil.ro/en and click the Order button. Complete the form there and the books will be delivered to you personally at IMO! Special discounts are offered with this. Best regards!


Let $ a,b,c \geq 1$. Prove or disprove

 $ \sqrt{\frac{a^2\plus{}b^2\plus{}c^2}{3}} \leq \left(\frac{\sqrt{a\plus{}b\plus{}c}\plus{}\sqrt{a^2\plus{}b^2\plus{}c^2\minus{}a\minus{}b\minus{}c}}{\sqrt{3}\plus{}\sqrt{a\plus{}b\plus{}c\minus{}3}} \right)^2$

I have no solution. Do anyone can solve it?

I tested your numbers. But I get the LHS is 7.79376640376898, RHS is 8.05380126525657. It still right.
Please check in advanced.


Let $a,b,c$ be sidelengths of a triangle and $S$ be its area, prove or disprove $S(\frac{1}{a}+\frac{1}{b}+\frac{1}{c})^2 > \frac{15}{4}$
Edit: It must be wrong. It's nonsense..  :blush:


floor function
Euler
calculus
calculus computations

konnichiwa  :D

Can you do this sum ?

$ \sum_{p \equal{} 2}^{\infty} \frac {p \minus{} \lfloor\sqrt {p}\rfloor^2}{p^2\lfloor\sqrt {p}\rfloor^2}$

[hide]$ \text{answer: }\qquad \boxed{\gamma}$
i.e. it is an expression for euler's constant[/hide]

I changed it to
$ \sum_{p \equal{} 2}^{\infty} \frac {1}{p\lfloor\sqrt {p}\rfloor^2} \minus{} \frac {\pi^2}{6} \plus{} 1$
then what?

$ \sum _{p \equal{} 2}^{\infty } \frac {p \minus{} \left\lfloor \sqrt {p}\right\rfloor ^2}{p^2 \left\lfloor \sqrt {p}\right\rfloor ^2}$

Using:
$ n^2\leq p\leq (n \plus{} 1)^2 \minus{} 1\Longleftrightarrow \left\lfloor \sqrt {p}\right\rfloor \equal{} n$

$ \sum _{p \equal{} 2}^{\infty } \frac {p \minus{} \left\lfloor \sqrt {p}\right\rfloor ^2}{p^2 \left\lfloor \sqrt {p}\right\rfloor ^2} \equal{} \sum _{n \equal{} 1}^{\infty } \sum _{p \equal{} n^2}^{(n \plus{} 1)^2 \minus{} 1} \frac {p \minus{} \left\lfloor \sqrt {p}\right\rfloor ^2}{p^2 \left\lfloor \sqrt {p}\right\rfloor ^2} \equal{} \sum _{n \equal{} 1}^{\infty } \sum _{p \equal{} n^2}^{(n \plus{} 1)^2 \minus{} 1} \frac {p \minus{} n^2}{n^2 p^2}$

Use the definition of Harmonic Number and get:
$ \sum _{n\equal{}1}^{\infty } \frac{H_{(n\plus{}1)^2\minus{}1}\minus{}H_{n^2\minus{}1}}{n^2}\minus{}\frac{\pi ^2}{6}$

And now we need Euler  :!:  :!:

This series doesn't converge against $ \gamma$ .

I ran the partial sums out to $ n\equal{}1024,$ and then used the sums for $ n\equal{}4,16,64,256,1024$ as the base of an extrapolation table. My best estimate for the sum is $ 0.5508775.$

That number also appeared in my post [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=159891]here.[/url]

It's not $ \gamma.$

doe john and eulerintegral are right (and i guess kent beat me too)


linear algebra
matrix
inequalities
linear algebra unsolved

Let the Matrix $A=(a_{ij})$ be simmetric and all the main minors be positive.
Prove the inequality $a_{ik}^2\leq a_{ii}\cdot a_{kk} \: \forall i,k=1,...,n$

This matrix is thus a positive symetric matrix, and you can use Cauchy-Schwarz ..


e^x-x=20.

The answer is very close to pi (look [url=http://www.xkcd.com/c217.html]here[/url]), but what is it?

You can use Newton's method: $g(x)=x-(e^{x}-x-20)/(e^{x}-1)=(xe^{x}-e^{x}+20)/(e^{x}-1)$, starting at $\pi$ or $3$, but is there a better solution.[/url]

Most arbitrary equations don't have nice solutions.


MATHCOUNTS
AMC
AMC 12
AIME
USA(J)MO
USAMO
geometry

Alright, the time has come for the annual tourney to finally begin. These are the contestants for the month of January;

pythag011
lotsofmath
AIME15
Bugi
johnnybeard
boredtodeath
BOGTRO
poincare
Bachukas

This will be the new sign-up thread, but you are not allowed to sign up yet until I say so.

Me, pythag011, Yongyi781, and 1=2 are the graders for January - April.
@pythag, Yongyi, and 1=2 - If you no longer want to be a grader, please tell me and I will replace you with the next person that would like to grade. Sign-ups for grading are up right now and are always up, unless I say so. There are already two other people that would like to grade, but they are still waiting to.

Each monthly contest will consist of forty questions. Twenty of these will be on this thread, and the other twenty will be on a different section of the forum, and I will tell you the location later. I will post one question per day, and if I am offline one day, then I will post extra questions to make up for those days.

The first ten questions will be MOEMS lvl, and they will each be worth one point.
The next ten questions will be hard MathCounts State level, or easy nationals level, and they will each be worth five points.
The next ten questions will be medium-hard AMC12 level, and they will each be worth ten points.
The next six questions will be AIME level, and they will each be worth twenty points.
The next three questions will be USAMO level, and they will each be worth thirty points.
The final question will be IMO level, and it will be worth forty points.

Of course, just because you get the answer right, doesn't mean you will get full score. That is the purpose of the graders. They will decide how much credit you deserve for each question.

For MOEMS lvl questions, you get a point as long as you get the right answer, regardless of the solution.
For the rest of the rounds the minimum score you can get is half the highest possible (rounded up) if you get the answer right. If you are close but incorrect then you can still make some credit. That is the graders' decision. I do not need the contestants to confirm, but I do need confirmations from the graders.

Here is question one;

1. Find the value of the following;

    $ 3 times 5 \plus{} (2^3 / 2^2) \minus{} 5$

2. Solve for $ x$;

    $ 2x \plus{} 5 \equal{} 11$

Lol, it'll get harder, I promise.

All contestants must pm their answers to me, and at the end of the month, I will pm your answers to the rest of the graders. Pythag011, will be grading everyone, but himself.

Oh, then I have them swapped.

[hide="questionsanswered"]pythag011 - hasn't answered any questions
lotsofmath - hasn't answered any questions
AIME15 - hasn't answered any questions
Bugi - hasn't answered any questions
johnnybeard - hasn't answered any questions
boredtodeath - hasn't answered any questions
BOGTRO - hasn't answered any questions
poincare - hasn't answered any questions
Bachukas - hasn't answered any questions [/hide]

[hide="confirmations"]Wickedestjr - Confirm
pythag011 - Confirm?
Yongyi781 -
1=2 - [/hide]

pythag011, you confirm, right?

I Confirm.

Well... instead of saying USAMO level, you could be like

Replace USAMO with USAMO/IMO #1/4

Replace IMO with USAMO/IMO #2/5

[hide="questionsanswered"]pythag011 - answered questions 1 and 2
lotsofmath - answered questions 1 and 2
AIME15 - answered questions 1 and 2
Bugi - answered questions 1 and 2
johnnybeard - hasn't answered any questions
boredtodeath - hasn't answered any questions
BOGTRO - hasn't answered any questions
poincare - hasn't answered any questions
Bachukas - hasn't answered any questions [/hide]

[hide="confirmations"]Wickedestjr - Confirm
pythag011 - Confirm
Yongyi781 -
1=2 - [/hide]

3. How many numbers are there in this pattern;

    $ 19/3, 36/3, ... 189/3$    ?

[hide="questionsanswered"]pythag011 - answered questions 1 and 2
lotsofmath - answered questions 1, 2, and 3
AIME15 - answered questions 1 and 2
Bugi - answered questions 1, 2, and 3
johnnybeard - hasn't answered any questions
boredtodeath - hasn't answered any questions
BOGTRO - hasn't answered any questions
poincare - hasn't answered any questions
Bachukas - hasn't answered any questions [/hide]

[hide="confirmations"]Wickedestjr - Confirm
pythag011 - Confirm
Yongyi781 -
1=2 - [/hide]

4. If the area of a rectangle is 45 sq. cm. and the length of the rectangle is 9 cm, then what is the height of the rectangle?

Alright, there are some rules that I forgot to mention.

After I have submitted a total of 20 questions, only the top six contestants will be aloud to proceed. Then after the next ten questions have been submitted, the top five will move on, and then there will be a winner at the end. I have just realized that there are not forty days in the month of January, so I am going to have to pack in more than one problem a day. In order to catch up, I will have to ask two questions today.

Some of the contestants have been doubting that I will be capable of presenting (problems>AMC10), so I am allowing any people that are not participating (including other judges that aren't playing) to pm me problems that I might choose to use.

Anyway;

5. How many factors does 12360 have?

6. What is the value of 2^(2^1 - 2^0 - 2^(-1) - 2^(-2) - ...)  ?

[hide="questionsanswered"]pythag011 - 1, 2, 3, 4, 5, and 6.
lotsofmath - answered questions 1, 2, 3, 4, 5, and 6.
AIME15 - answered questions 1, 2, 3, 4, 5, and 6.
Bugi - answered questions 1, 2, 3, and 4.
johnnybeard - hasn't answered any questions
boredtodeath - hasn't answered any questions
BOGTRO - has answered questions 1, 2, 3, and 4.
poincare - hasn't answered any questions
Bachukas - has answered questions 1, 2, 3, 4, 5, and 6[/hide]

[hide="confirmations"]Wickedestjr - Confirm
pythag011 - Confirm
Yongyi781 -
1=2 - [/hide]

7.  Joe and Gary found some money in the snow one day. Joe had found it first, so he took the majority of the money. Gary took the rest. As they walked home, Joe got attacked by a bunch of pigeons, and he lost $ \$$4. Then he noticed that he had three times as much money as Gary. Gary went to go buy a snack. It cost $ 20$. This was 50% of Gary's money. Gary ended up buying it. What percent of the money that they found in the snow do they have now?

8. There are two committees called the AoPS is Awesome committee, and the AoPS Rocks committee. The AoPS is Awesome committee will have four members, and the AoPS Rocks committee will have six members. There are twelve people that would like to join one of the committees. How many possible committees are possible?

9. A square is inscribed inside a circle. The circle has an area of 64pi. What is the area of the square?

10. Find the sum of all possible values of $ a$, such that the three digit integer $ 34a$ is divisible by three.

11. What is the value of $ 1 \plus{} 2 \plus{} 3 \plus{} 4 \plus{} 5 \plus{} 6 \plus{} 7 \plus{} 8 \plus{} 9 \plus{} 10 \plus{} 11 \plus{} 12 \plus{} 13 \plus{} 14 \plus{} 15 \plus{} 16 \plus{} 17 \plus{} 18 \plus{} 19 \plus{} 20$?

12. What is the sum of all integers $ x$, such that $ 14 < x \plus{} 4$, and $ x \plus{} 10 < 34$ ?

[size=200][b]Your solutions to questions 11 and 12 are required to achieve a high score. [/b][/size]

Sorry for the long delay.

For the last question, the contestants certainly shouldn't get as much as 15 points for simply getting the correct answer! (Same for the second and 3rd last, the answer alone shouldn't be worth 10 points).

And plus, many USAMO/IMO problems do not even have an "answer," just simply a proof.

[quote="Wickedestjr"]10. Find the sum of all possible values of $ a$, such that the three digit integer $ \overline{34a}$ is divisible by three. [/quote]

More clear now. :wink:

Alright, for the last seven questions, you get at least one fifth of the maximum possible, just by proving it. You may get less by proving half of it.

For number ten $ a$ is one of the digits of $ 34a$.

13. How many four digit numbers are there, such that the sum of the digits is more than the product of the digits?

1=2 didn't want to be a grader, so now I am asking james4l.

This is the current list for being a grader.

Graders;
Bugi

If anybody would like to add their names to this list please say so on this thread.

14. If the average of the three sides of a triangle is 4 inches, and two of the side lengths are 3 inches, and 5 inches, then what is the area of the triangle?

15. At a really nice pet store. Dogs cost 5 dollars, and cats cost 3 dollars. If I have 80 dollars, then what is the number of possible animal arrangements can I take home, if I must pay all the money on dogs and cats? (one example would consist of ten dogs, and ten cats.)

16. Simplify;

$ (1/8) \plus{} 3(1/12) \plus{} 3(1/18) \plus{} (1/27)$

17. There is a store with 40 vehicles in it. All forty vehicles are either cars or motorcycles. If there are a total of 140 wheels on the forty vehicles, then how many cars are there at the store?

18. A man is standing on the coordinate (0, 0), and he is planning on walking from (0,0) to (6, 6). If the man can only go up or to the right, then how many possible paths are there?

19.  A circle is inscribed inside a square with a side length of 10 cm. What is the area of this circle?

20. What is the largest possible sum of $ x$ and $ y$, if they are both integers, and $ (7/x \plus{} y) \equal{} 14$ ?

21. If $ 5^r$ is equal to $ 30$, then what is the value of 5^(2r + 1)?

22. Joe goes to a store to buy bread. Each loaf of bread costs three dollars, and it also costs an additional 2 dollars to make the purchase. If Joe has 50 dollars, then what is the maximum number of loaves of bread that he can buy?

Questions 1 - 20 are due in a week.

Can you post who's done which questions?  If not, at least PM me the ones that I've done, and the ones that I still need to do.

How do you know if a pm that you sent to somebody has been opened up yet? (or is that not possible)

Here;

pythag011 - completed first nine questions
lotsofmath - completed first 20 questions
AIME15 - completed first 12 questions
johnnybeard - completed nothing
boredtodeath - completed nothing
BOGTRO - completed first four questions
bachukas - completed first six questions

Click "Outbox", and if it's in the Outbox, then it hasn't been opened yet.

wait, other way around, right?

if it is NOT in the outbox, it has NOT been opened...i thought?

No, if it's in Outbox, it's unopened.  If it's in Sentbox, then it's been sent...:P

Okay, well everyone has opened up there pms apparently, so that means they have just completely ignored it, without even replying, so I am taking johnnybeard, boredtodeath, and poincare out of the tourney.

Well, I am still waiting on answers from everybody. Nobody has answered all the questions I have up.


function
complex analysis
complex analysis unsolved

Hi,
is there any non-constant rational function $ f: \mathbb C\to \mathbb C\cup \{\infty\}$ that is periodic?

A nonconstant rational function can only take the value $ c$ a number of times equal to its degree.


If you qualified for the AIME, will they send the AIME, the alternate AIME, or can you choose?

We will be making announcements about AIME qualification later this week, at which point we will let you know all the details.


I was just wondering how other ppl did on the RCML.

Please state:
Your name
What Test ur takin
Grade level
Score!

thx yall

Are you talking about the math league?
(Different regions, different names, same robs though)

uhh... the rocket city math league at least :)

I wanted to take the RCML test, but our school doesn't do it.

that sucks.... what would test would you have gone 4?

Isn't Apollo the one for Algebra II students?

yeah... its the hardest one. first rnd geometry and alg 2 were mixed 2getha i think

Too bad they don't have one for Pre-Calc students, 'cause that's what I'm in next year.

Yes! I feel happy today because i just found out that NWCS got 1st place in the nation for the RCML Alg 1!!! (need a little ice 4 that burn?) ooooohhhhhh yeah..... Good thing they r giving out trophies 4 the top 20 individuals... not that its the trophies that counts.

I did horrible still compared 2 what I thought that I would have gotten. 25 18 18. The 25 was good but the 18s were really too low. Surprising how 2 questions can lower your score to an 18. 10/12 (5/6) is way more than 18/25.

4th place in the nation (mercury test)    :first:
Scores:21,19,22
Sorry if im being too arrogant

Are there any awards?  :D

Oh yea
I'm from tennessee
Jefferson Middle School in Oak Ridge

yeah i think there are trophies for the top 20 in each level. I got sixth 4 alg 1. (25 18 18) Another person from Northwest Chinese School also tied with me. You beat me! Unfair!!! jk. I did reeaaaaallly bad last 2 rounds. Good job anyhow. ;)

RCML will have a Comprehensive division this year for Pre-Calculus and above students, so now all students who are in Pre-Algebra and above can participate. The competition will be free again this year. For more information, please visit our subforum at [url=http://www.artofproblemsolving.com/Forum/index.php?f=305]www.artofproblemsolving.com/Forum/index.php?f=305[/url].


find all real x such that

  x^2+9x^2/(x+3)^2=16

This is just a small typo to this problem:

[quote]Find all real $ x$ such that

$ x^{2}\plus{}\frac{9x^{2}}{(x\plus{}3)^{2}}\equal{} 16$[/quote]

I think that this one should be proposed in algebra forum.

It uses the same idea as Problem 4 from Canada 1992.
Step 1. You may prove that $ x^{2}\plus{}\frac{9x^{2}}{(x\plus{}3)^{2}}\equal{} (\frac{x^{2}}{x\plus{}3})^{2}\plus{}\frac{6x^{2}}{x\plus{}3}$
Step 2. Substitute: $ u \equal{}\frac{x^{2}}{x\plus{}3}$
Step 4. Solve the equation for $ u$.
Step 5. Solve the equation(s) for $ x$.
Step 6. Check if the solutions satisfies starting equation.
You are ready...

$ x^{2}\plus{}\frac{9x^{2}}{(x\plus{}3)^{2}}\equal{} 16$
$ x^{2}(1\plus{}\frac{3}{x\plus{}3}^{2})\equal{}16$
trigonometric substitution: $ \frac{3}{x\plus{}3}\equal{}\tan{\alpha}$
$ 9\frac{(\cot{\alpha}\minus{}1)^{2}}{\cos{\alpha}^{2}}\equal{}16$
$ \frac{\cot{\alpha}\minus{}1}{\cos{\alpha}}\equal{}\pm\frac{4}{3}$
$ \frac{1}{\sin{\alpha}}\minus{}\frac{1}{\cos{\alpha}}\equal{}\pm\frac{4}{3}$
$ \cos{\alpha}\minus{}\sin{\alpha}\equal{}\pm\frac{2}{3}\sin{2\alpha}$
after squaring:
$ 1\minus{}\sin{2\alpha}\equal{}\frac{4}{9}\sin{2\alpha}^{2}$
$ \sin{2\alpha}\equal{}\minus{}3>1$, or $ 0.75\to \sin{2\alpha}\equal{}0.75$
$ \cot{2\alpha}\equal{}\pm\frac{\sqrt{7}}{3}$
$ x\equal{}3(\cot{\alpha}\minus{}1)\equal{}3(\cot{2\alpha}\plus{}\frac{1}{\sin{2\alpha}}\minus{}1)\equal{}1\pm\sqrt{7}$

It is just factorable into two quadratics...

$ x^{2}\plus{}\frac{9x^{2}}{(x\plus{}3)^{2}}\minus{}16\equal{}\frac{(x^{2}\minus{}2x\minus{}6)(x^{2}\plus{}8x\plus{}24)}{(x\plus{}3)^{2}}$

This can easily be done by the method of undetermined coefficients.


Are points $ A  \equal{} (\minus{}4, 6, 0)$, $ B \equal{} (8, 2, 1)$ and $ C \equal{}(2, \minus{}3, 1)$ collinear of not? Explain.

If you want to know if they are [i]collinear[/i], see if they are scalar multiple of a certain vector.  :wink:

Of course, these are points and not vectors. How would you get vectors from these points?

The question from a calculus class might be "What is the area of $ \triangle ABC.$" That's overkill, but it is a related question.


function
calculus
integration
real analysis
real analysis unsolved

Let x_n defined as in the previous problem. Prove that the sequence  \sqrt n* (x_n-9/16) tends to 0.
  Open question:
   The sequence n(x_n-9/16) is convergent.

Sorry, but which is the previous problem?

I think that the limit asked by Harazi is 1/2. This appears as a particular instance of this general problem.

Let $f$ be a positive (strictly) decreasing function of class $C^{1}$, and let $F$ be such that $F'=f$. Let us assume that $F(1)-F(0)>1$.

Let us denote by $x_{\infty}$ the unique solution of
\[F(x+1)-F(x)=1\]
and by $x_{n}$ the unique solution of
\[\frac{1}{n}\sum_{k=1}^{n}f\left(x+\frac{k}{n}\right)=1,\]
which exists for every large enough $n$.

Then
\[x_{n}=x_{\infty}-\frac{1}{2n}+o\left(\frac{1}{n}\right),\]
and in particular $x_{n}\to x_{\infty}$ and $n(x_{\infty}-x_{n})\to 1/2$.

If this is true, then Harazi's problem is just the case where $f(x)=x^{-1/2}$ (after some algebra).

What is really surprising for me is that $x_{\infty}$ depends on $f$, while the factor 1/2 is completely independent on $f$!!!

The main trick in the proof of the general result is to interpret the sum as an approximation of the integral from 0 to 1 of $f(x+t)$ (with respect to the variable $t$).


function
calculus
derivative
inequalities
geometry
3D geometry
algebra

If a,b,c are real numbers such that 0<=a,b,c<=1
prove that $ a/1 \plus{} bc$+$ b/1 \plus{} ac$+$ c/1 \plus{} ab$ <=2.

I know there are many solutions to this problem.

But i want to know if this function is convex or concave or not coz i dont know how to take the derivative of functions in 3 variables.

P.S. i want some practice in concavity or convexity of functions so plz post a solution using concavity or convexity only if possible

Thx,
Befuddlers

http://www.mathlinks.ro/viewtopic.php?p=1355440&sid=505442c42ad16dabc7907b37ba9247ac#1355440

It's not something new to take derivative of functions in 3 variables.

It's called [url=http://en.wikipedia.org/wiki/Partial_derivative]partial derivative[/url]

You just need to consider two variables as constants and then differentiate with respect to one variable.

If $ f(a,b,c)\equal{}\frac{a}{1\plus{}bc}\plus{}\frac{b}{1\plus{}ca}\plus{}\frac{c}{1\plus{}ab}$

then $ \frac{\delta f(a,b,c)}{\delta a}\equal{}\frac{1}{1\plus{}b c}\minus{}\frac{b c}{(1\plus{}ca)^2}\minus{}\frac{b c}{(1\plus{}ab)^2}$

And $ \frac{\delta^2 f(a,b,c)}{\delta a^2}\equal{}\frac{2b c^2}{(1\plus{}ca)^3}\plus{}\frac{2b^2 c}{(1\plus{}ab)^3}$ which is positive.

So the function is convex with respect to $ a$. And similarly for $ b,c$

Thus the function attains it's maximum at the endpoints of the range of $ a,b,c$

So
$ \max(f(a,b,c))\equal{}\max(f(0,0,0),f(1,0,0),f(1,1,0),f(1,1,1))\equal{}$
$ \equal{}\max(0,1,2,1.5)\equal{}2$.

[hide]I'll explain the last 2 sentences more if you ask. But I guess it suffices to state it in a contest.[/hide]

[quote="Conjurer"][b]Akashnil:[/b] Please explain the last 2 sentences.[/quote]
I'll try. Hope it's understandable.

[hide]
[b]Theorem 1[/b]
Suppose a single variable real function $ f$ is convex in the interval $ [a,b]$. We claim that for all $ c\in [a,b]$
\[ f(c)\le \max(f(a),f(b))
\]
It' is intuitively evident from the graphical representation of the convex function $ f$. But I'll include a rigorous proof:

[hide="Click to reveal Rigorous proof of Theorem 1"]
By definition of convex function: $ f''(x)\ge 0\ \forall \ x\in [a,b]$. Hence $ f'(x)$ is increasing in the interval $ [a,b]$.

[u]Case I[/u]: $ f'(c)\ge 0$

$ f'(c)$ increasing in the interval $ [c,b]$ implies $ f'(x)\ge 0\ \forall \ x\in [c,b]$
Hence $ f$ is increasing in the interval $ [c,b]$.
Since $ b > c$, $ f(b)\ge f(c)$
Proved.

[u]Case I[/u]: $ f'(c) < 0$

$ f'(c)$ increasing in the interval $ [a,c]$ implies $ f'(x) < 0\ \forall \ x\in [a,c]$
Hence $ f$ is decreasing in the interval $ [a,c]$.
Since $ a < c$, $ f(a)\ge f(c)$
Proved.

$ \boxed{}$[/hide]

$ f(a,b,c)$ behaves as a single variable function of $ a$ if $ b,c$ are constant and $ a$ is variable.

Since $ f(a,b,c)$ is convex wrt $ a$ when $ b,c$ constant, by Theorem 1: $ f(a,b,c)\le \max(f(0,b,c),f(1,b,c))$ for all $ b,c$.
But $ f$ is convex with respect to $ b,c$ also.
Thus for all $ a,b,c$:
$ f(a,b,c)\le \max(f(0,b,c),f(1,b,c))\le$
$ \max(\max(f(0,0,c),f(0,1,c)),\max(f(1,0,c),f(1,1,c))) \equal{}$
$ \max(f(0,0,c),f(0,1,c),f(1,0,c),f(1,1,c))\le$

$ \cdots$

$ \equal{} \max(f(0,0,0),f(0,0,1),f(0,1,0),f(0,1,1),f(1,0,0),f(1,0,1),f(1,1,0),f(1,1,1))$
(since $ f$ is a cyclic function of $ a,b,c$)
$ \equal{} \max(f(0,0,0),f(0,0,1),f(0,1,1),f(1,1,1))$

So for all $ a,b,c$:
$ f(a,b,c)\le \max(f(0,0,0),f(0,0,1),f(0,1,1),f(1,1,1))$[/hide]

Nice.

Thanks a lot Akashnil.

Guess i can try some problems on convexity now.

Well, another exampple where convexity in 3 variables.

COnsider the following problem from the US Olympiad f 1980:

For $ 1\ge a,b,c \ge 0$, prove that $ \sum_{cyclic} \frac {a}{b \plus{} c \plus{} 1} \plus{} (1 \minus{} a)(1\minus{}b)(1\minus{}c)  \le 1$.

A manipulative solution requires excellent skills in inequalities,  but a solution is possible without any manipulation.

Let us consider the $ LHS$ of the inequality by $ f(a,b,c)$.  Partial differentiation with respect to the variables $ a,b,c$ will give us that $ f$ is convex as its second derivative in all the 3 cases of partial differentiation{for $ a,b,c$} come out to be greater than $ 0$. {In case you dont know partial differentiation, please go through Akashnil's post. He has written it excellently}.

So, since is $ f$ is a ocnvex function, it assumes its maximum on the extremal points forming 8 vertices $ (0,0,0),.....(1,1,1)$ forming a closed cube with these vertices and since they are the boundary points, they are the only points of the closed cube whhc ar enot midpoint sof two other points of the cube. Now, by citing the THeorem of Weirstrass, we complete our proof saying that  a continuous function on a bounded and closed domain assumes its maximum and minimum , here for us, at $ (1,1,1)$ and $ (0,0,0)$.

Or simply finsh the proof by saying that $ max [ f(0,0,0), f(0,0,1), ....f(1,1,1)]$ is $ 1$ and hence we prove the inequality.


function
trigonometry
limit
calculus
calculus computations

Is $f(x) = x{\sin{\frac{1}{x}}}, x\ne0, 0, x=0$ differentiable at $x=0$ or not? What about $f(x) ={x^{2}}{\sin{\frac{1}{x}}}, x\ne0, 0, x = 0$? Is that differentiable at $x=0$?

Since we have $f(0)=0,$ in both cases we're looking at

$f'(0)=\lim_{x\to0}\frac{f(x)}x,$ if that limit exists.

Hence if $f(x)=x\sin\left(\frac1x\right),$ the limit becomes

$\lim_{x\to0}\sin\left(\frac1x\right).$

That limit does not exist; $\sin\left(\frac1x\right)$ equals both $1$ and $-1$ (and every value in between) infinitely often in any neigbhorhood of zero.

But if $f(x)=x^{2}\sin\left(\frac1x\right),$ the limit becomes

$\lim_{x\to0}x\sin\left(\frac1x\right).$

By the "pinching" or "squeeze" theorem, that is trapped between $-|x|$ and $|x|$ as $x\to0$ and hence has a limit of zero. In that case, $f$ is differentiable at zero, and $f'(0)=0.$

Amusing variant: Let $g(x)=x^{2}\sin\left(\frac1x\right)+\frac{x}4$ for $x\ne 0$ with $g(0)=0.$

Then $g$ is differentiable everywhere and $g'(0)>0,$ but there is no interval containing zero on which it is an increasing function.


[quote="MysticTerminator"]SEASON-ENDING LOSSES TO MARYLAND, CAROLINA

FIRST-ROUND ACC TOURNAMENT LOSS TO 10TH SEEDED NCSU

FIRST-ROUND NCAA TOURNAMENT LOSS TO 11TH SEEDED VCU[/quote]

LOL! DUKE STINKS!  :rotfl: GO VCU. NICE FACTS THERE MysticTerminator! DUKE=SUCKY!!!

EVEN AS A GROSSLY OVERRATED 6 SEED, DUKE IS THE FIRST UPSET OF THE 2007 NCAA TOURNAMENT

It was just bad luck, although they were ranked higher than they should have been.


calculus
derivative
function
logarithms
calculus computations

Is that really the derivative? I thought that just a different way of expressing it.

no that is just a different way to express the function.
In this form it is much easier to see what the derivative is.
This of course assumes that you know
$\frac{d}{dx}\ln(x)=\frac{1}{x}$

Yeah, so you just apply the quotient rule here? So I'm getting it as
$\frac{\frac{\ln{a}}{x}-\frac{\ln{x}}{a}}{2\ln{a}}$.
Is that right?

no. you are differentiating with respect to $x$. Thus $\ln(a)$ is a constant.


quadratics
number theory
greatest common divisor
LaTeX
number theory unsolved

Prove that the equation $3y^2 = x^4 + x$ has no positive integer solutions.

Congratulations, you succeed to post each of you topics in the wrong section... :D
Please try to wait a little before posting them into the solved problems area.

Pierre.

Clearly GCD(x,y)=1 and then x is a power of 3 (x=3^L) with (L>0) odd and  then

3^(3L)=A^2-1=(A-1)(A+1)   A integer

so A=2 and then 3L=1 Contradiction (L integer).

Best regards.

This can be complicated a little to:

[i]The equation:

$3z^3=x^3+y^3$

has no solution in integers except for $z=0$ [/i]

Well, a simple calculation gives us that x+y=3L  and easily again we get that L=1, so x+y=3., so your equation transforms into

Z^3=9-3X+X^2     and      X+Y=3

then if try to solve the quadratic form (in x) we get that D=4z^2-33 wich is only a square iff |D|=4 or |D|=16  then x is NOT an integer, then No integer solution if |Z|>0  hence  Z=0, and Y=-X.

Best regards.

Sorry for reviving this OLD problem, but can somebody explain to me why it's "clear" that gcd (x, y) = 1 ?

since $x(x^3+1)=3y^2$,$(x,x^3+1)=1$
hence $x=3t^2,x^3+1=s^2$ or $x=t^2,x^3+1=3s^2$(nod 3,contradiction!)
so $x^3+1=s^2$,since $3|x$ hence $(x+1,x^2-x+1)=1$
hence $x^2-x+1=l^2$
then $x=1$,but it;s easy to check that it's not a solution.

[quote="RobertuX"]Well, a simple calculation gives us that $x+y=3L$  and easily again we get that $L=1$, so x+y=3., so your equation transforms into

\[Z^3=9-3X+X^2\]     and      \[X+Y=3  \]

then if try to solve the quadratic form (in x) we get that $D=4z^2-33$ wich is only a square iff |D|=4 or |D|=16  then x is NOT an integer, then No integer solution if |Z|>0  hence  Z=0, and Y=-X.

Best regards.[/quote]
Please use $\LaTeX$.

My solution:
$3y^2=x^4+x=x(x^3+1)=x(x+1)(x^2-x+1)$
Because $3|3y^2 \rightarrow 3|x(x+1)(x^2-x+1)$ <1>
We know that $gcd(x,x+1)=1, gcd(x,x^2-x+1)=1$ <2>
Let $gcd(n+1,x^2-x+1)=d \rightarrow d|n+1 \rightarrow d|n^2+2n+1 \rightarrow d|n^2+2n+1-(n^2-n+1) \rightarrow d|3 \rightarrow d=1,3$ <3>
Case 1: $3|x+1 \rightarrow 3|x^2-x+1 \rightarrow x^2-x+1=3k \rightarrow gcd(x+1,k)=1$ (from <3>)
Therefore $y^2=x(x+1)k$ we see that $gcd(x,x+1)=1,gcd(x,k)=1,gcd(x+1,k)=1$ (from <1><2><3>)
So $x,x+1,k$ are perfect squares so $x=a^2,x+1=b^2 \rightarrow b^2-a^2=1 \rightarrow b=1,a=0 \rightarrow x=y=0$
Case 2: $3|x \rightarrow gcd(x+1,x^2-x+1)=1$
Let $x=3q \rightarrow y^2=q(x+1)(x^2-x+1)$
We know that $gcd(q,x+1)=1,gcd(d,x^2-x+1)=1,gcd(x+1,x^2-x+1)=1$
So $q,x+1,x^2-x+1$ are perfect squares thus $x^2-x+1=c^2$
If $x=0 \rightarrow y=0$
If $x=1$ FALSE!
If $x>1$ thus $(x-1)^2<x^2-x+1<x^2 \rightarrow (x-1)^2<c^2<x^2 \rightarrow x-1<c<x$ FALSE! because $c$ is an natural number.
Answer: $\boxed{(x,y)=(0,0)}$


[quote="21 7 15"]hi geniuses, hows the preparation for imo,cool right.sum it up 7!+8!+9!+...+2006! :roll:[/quote]

Please read your PMs....I would love to be your apprentice.

to all divide both sides by 7!.To Funcia how does it go

oh yeah, posting this ridiculously annoying sum definitely goes in HSB :roll:

lets see...[hide="is the answer"]a very large number?[/hide]
Ok but seriously, I would like to know how to do this. (How hard is it actually?)
I'll give it my best shot and post any progress, but I'm not making any promises about how correct it will be!  :P


induction
absolute value
graph theory
combinatorics proposed
combinatorics

We have $ 3$ chess clubs. Each club contains $ n$ players. Assume that every player has played with at least $ n \plus{} 1$ players from the other two clubs. Prove that there are three players from different clubs who have all played against each other.

Is it necessary the condition of "at least n+1"... I mean, I know that the result is true for the case in which each player has played with exactly n+1 players. Check this out (the problem is from 1977 Kürschák competition):

"Three schools have n students each. Each students know a total of n+1 students from the other two schools. Prove that there exist three students, one from each school, who know each other"

My proof is as follows. Draw a graph with vertices representing students, and edges representing the relation "they know each other". So we have a graph with 3n vertices, and obviously 3n(n+1) edges. Now we use the following graph theory result: If a graph has e edges and v vertices, and e is greater than or equal to v, then it must contain a cycle. But since 3n(n+1) is obviosly greater than 3n, we are done.

You've just shown that this graph has a cycle. You want it to have a triangle, i.e. a cycle of length $3$.. ;)

[quote="Jutaro"]Now we use the following graph theory result: If a graph has e edges and v vertices, and e is greater than or equal to v, then it must contain a cycle.[/quote]

well, how do you prove it? (i'm very, very new to graphs)

You can prove it by induction on $v$, for instance. First, you show that if the graph has no cycles, then there must be a vertex of degree at most $1$, and after that, you delete such a vertex and the edge adjacent to it (or just the vertex, if it has degree $0$, i.e. it's an isolated vertex), and you use the induction hypothesis on the remaining graph.

[quote="costica"]Source: Andrei Negut - ''Problems for the Mathematical Olympiads'' [/quote]

Is that a book?  :)

[quote="grobber"]I think I've even posted this one some time ago, but I'm sure I had a nastier solution back then :). Anyway, it looks very familiar. It's a bit hard to explain though.

Assume there are no triangles in our graph, which is a tripartite $n+1$-regular graph, consisting of classes $A,B,C$ of vertices.

For each vertex $x$, consider the absolute value $d_x$ of the difference between the number of vertices in one class to which it is connected and the number of vertices in the other class to which it's connected. Let $t$ be the maximum among all these $d_x$. It means that there is a vertex $x$, which we may assume WLOG belongs to $A$, that is connected to $p$ vertices in $B$ and $q$ vertices in $C$ s.t. $p+q=n+1,\ p\le q,\ q-p=t$. A vertex $y$ among the $p$ neighbours of $x$ from $B$ can be connected to at most $n-q$ neighbours from $C$ (because of the assumption that there are no triangles), so it's connected to at least $q+1$ vertices in $A$, meaning that $d_y\ge t+2$, which is a contradiction, because $t$ was maximal.[/quote]

As what mentioned above, I let $n=100, p=100$ and $q=1$. Then we have $d_x=q-p=99$. Vertex $y$ can be connected to at most $n-q=99$ neighbours from $C$, and it's connected to at least $q+1=2$ vertices in $A$. Then we have $d_y \le 97$. It doesn't yield any contradiction... Is it okay with my statement? :roll:  :roll:  :roll:

grobber's solution in his post above (post #2 in this topic) is not entirely correct, since he assumes that every player knows [i]exactly[/i] $n+1$ players, whereas the problem only states that every player knows [i]at least[/i] $n+1$ players. But it is easy to correct his solution: Instead of choosing a vertex $x$ maximizing $d_x$, we have to choose a vertex $x$ maximizing the number

$\max\left\{\text{number of vertices in one class connected to }x,\text{number of vertices in the other class connected to }x\right\}$.


topology
real analysis
real analysis unsolved

Let $A\subset \mathbb{R}^{n}$ be measurable. $f;A\to \mathbb{R}$ is such that for every $\varepsilon>0$ there exists a closed set $F\subset A$ such that $m(A\setminus F)<\varepsilon$ and $f|F$ is continuous. How to prove that $f$ is measurable?

You want to prove that $\{x\in A\colon f(x)\le t\}$ is measurable for any $t\in\mathbb R$. Set $\epsilon=1/n$ and find $F=F_{n}$ as above: then $\{x\in A\colon f(x)\le t\}=\bigcup_{n}\{x\in F_{n}\colon f(x)\le t\}$ + a null set.


ARML
trigonometry
Stanford
college
geometry
3D geometry
Mafia

AoPS users (with more than about 80 posts) who will be on one of the Southern California ARML teams:

apusey
easyas3.14159...
JAM
jli
kimby_102
kohjhsd
litoxe2718281828
pkerichang
shbmsc90
ShoeFactory
Teddy
Zellex

Coaches:

Kent Merryfield
zmli

(And plenty of people with fewer AoPS posts, of course.)

[quote="13375P34K43V312"]easyas3.14159... and jli are both in 8th grade, am I wrong?[/quote]
As are also apusey and kohjhsd. The 8th graders probably won't be on the A team.

[quote="Altheman"]woot i will be there too (Chicago ARML)[/quote]
How about a list of AoPS regulars on any of the Chicago teams? Or at least a guess at a few of them?

South Carolina:

mysmartmouth
Iversonfan2005/2007 (possibly...he might have something else on those dates)

And I should be there as an assistant coach.

We'll either be going to Penn State or Vegas.  I'm guessing we'll stick with going to Penn State but it has yet to be decided.

[quote="daermon"]for SFBA ARML:
daermon (duh)
junggi
i think i managed to recruit discreetfouriertransform
bluestalk (i think)
...
there are lots more that im not sure of
...
maybe i'm not the one to be asking :P[/quote]
erm i guess I should add some even though I don't know a lot of SFBA ARML people

ccy
uberl33tmage(I think)
ubemaya
Paul christiano and Adrian Sanborne are coming probably(I forgot their aops usernames...)

Sly Si is coming as a coach I think?

(ya I'm a noob I barely know anyone)

Washington:

paladin8
t0rajir0u (me)
Kamior (okay, so Adam doesn't post [i]that[/i] often...)
xxreddevilzxx
ch1n353ch3s54a1l
seungjun
seungsoo

North Carolina...

jb05
MysticTerminator
n^4+4
not_trig
b-flat
yasha
teki-teki

Possibly some of the other people whom I don't know on the NC forum.

[quote="roadnottaken"]Ohio:

tjhance
Xaero
roadnottaken (me)

maybe DanK[/quote]

To add to the list:

ehehheehee
h_s_potter2002
nr1337

Upstate NY:

Myself
Perfectnumber628
TimReyn
EmilGouliev
qqwref (sp?)
Jazzy200084

If anyone knows the list of AAMC AoPSers who are going, feel free to continue the list

Yep, I'll be there with SFBA again.  I could never pass up a chance to go to ARML :)

[quote="xxreddevilzxx"]is cowpi and kerry xing the same person?[/quote]

I don't think so; they don't go to the same school for one thing.

[quote="13375P34K43V312"]hmmm...Massachusetts?

13375P34K43V312
Phelpedo (?)
yunorman
vfiriou (sp?)

anyone else? I'm in 8th grade and have a decent shot at the A team I think.[/quote]

Oh, so now Eastern Mass is the "Massachusetts" team? What's Western Mass then? The [i]other[/i] Massachusetts team? Haha

(even though that's totally true  :( ... at least we're in division A this year)

So there'll be myself, and I think cosinator (unless I'm confusing you with someone else). And possibly some others who I don't know on AoPS.

There's been a "Massachusetts" team for at least ten years now; I assume this one used to be "Eastern Mass".  "Wester Mass" still exists under the same name.

[quote="The Zuton Force"]
So there'll be myself, and I think cosinator (unless I'm confusing you with someone else). And possibly some others who I don't know on AoPS.[/quote]
I am indeed on Western Mass A.

Thanks for noticing me.  I don't think there are many people on Aops on Western Mass ARML.

some more NC ARMLers (although not necessarily A-teamers)

not_trig
b-flat
blueboat (Hyunji)
misha?
nemosupport?

and morpheus, who is a former NC ARMLer (noah) with a grand total of about 2 posts so far... :P

Actually, all of those but 2 only have about 2 posts :). And morpheus only has 1 post, last time I checked.

[quote="i_like_pie"][quote="daermon"]for SFBA ARML:
daermon (duh)
junggi
i think i managed to recruit discreetfouriertransform
bluestalk (i think)
...
there are lots more that im not sure of
...
maybe i'm not the one to be asking :P[/quote]
I'm going! :D[/quote]

I'm 75% sure I'm going as well.

daermon, davidyko, [i]et al.[/i] - how are you handling transportation this year? Overnight bus? Are you flying?

[quote="Sly Si"]We're taking a bus again; I don't think Ted even looked into flying this year.  It should be almost the exact same arrangement as last year: leave Thursday evening, get in Friday morning and chill/practice until the competition that night, then leave after awards on Saturday to get back late that night.  I thought it worked quite well last year, and no one really seemed to mind the bus ride that much.[/quote]
totally!  the bus ride was awesome
silas taught me how to cube :)

We have a substantially shorter bus ride than you, so we'll leave Friday morning and arrive in Las Vegas at perhaps 1:00.

Size calculation: 1 bus = a maximum of three teams. We'll have three teams, and maybe a couple of alternates, and a collection of coaches, and we'll have zero or one empty seat. So either SFBA limits itself to three teams or brings two buses.

Anyone on our bus who doesn't already cube would know whom to ask.

Indiana ARML Aopser's... jeffdshen (who is kinda like... not on aops ever) lol
fanatic
me...
stupidkid
artofowna
usaha
pretty much it

forgot to add mr. indianamath -_-... that's all i know


I simply think you don't have an avatar and you either don't have or haven't activated your signature...
Check
http://www.mathlinks.ro/Forum/usercp.php?mode=editprofile&sub=avatar
and
http://www.mathlinks.ro/Forum/usercp.php?mode=editprofile&sub=sig
carefully.


function
induction
inequalities proposed
inequalities

It's morning rigth before school, so I'm not sure if the RHS is really correct due to function. I think this is right.  Using AM-GM:
$\\\frac{f(x_1)+...+f(x_n)}{n}\ge f(\sqrt[n]{x_1x_2...x_n})\\f(x_1)+...+f(x_n)\ge nf(\sqrt[n]{x_1x_2...x_n})$

I'll be back.

Masoud Zargar

I think you are wrong. You solve in functions not in numbers

are you referring to backward induction?
When did I have a solution of a question like this?

You use the first at a the ineq of UK ST 2005


AMC AIME MATHCOUNTS quadratics trigonometry USAMTS geometry

I will include a poll for the scores but don't restrict your discussion to just your scores you may also discuss the questions.

I got an 8

I did numbers 1-6, 8-9.

I thought this test was comparable to the 2006 AIME

numbers 1-3 can be Mathcounts problems
4 was a bit harder but the trick is used alot (completing the square or quadratic formula to and then set it to a perfect square)
5 was very easy if you understood the problem
6-7 were probalby the hardest of the first 10, they required some additional steps
8 is a very famous kind of problem, infact like 20 years ago (on like 1984 AIME?, there was a probelm like it but with cotangents)
9 just required some algebra but i thought it was easy for 9
10 was quite brute forceable-or assumable (too bad i didnt brute force it correctly)
11-15 i didnt even read any of the problems (expcet 11 but i gave up once i couldnt find a pattern)

you beast... hey does anyone know what mark got???

[quote="uldivad9"]I got a 10, with problems 1,2,3,4,5,7,9,10,12,and 14. I mad a dumb mistake on 6, and didn't know how to do 8 because I have no trig knowledge.[/quote]

number 8 was essentially variation on #3 first round usamts right?

I did it for practice and i got 8.

# 1, 2, 3, 4, 5, 7, 8, 11.

didn't see the pattern on 6 and forgot that 3-3=0 on number 9.

i didin't really try that much on the others. i couldnt' get anything out of the trapezoid and i put 300 for number 12.

[quote="davidyko"]Semi-retaliation spam from California.
I got a 9. (#1-9), later failed when half-heartedly doing #10 and #12.[/quote]
I have great luck, I fail AIME on the year its really easy heh 7. Oops this is Indiana :oops: (no I'm not that dumb, same reason as david)

Dont worry about it you can spam as much as you want
[hide][hide][hide]only 8 times for my score[/hide][/hide][/hide]

Im predicting Indiana will have 25 USAMO qualifiers

last year it only had like 14 or so.

omg wthhh got 5 out of 5 answered in 2:30 (yeah below normal time limits, but whatev it was practice)

did # 1, 2, 3 within the first 15 minutes, spent lots of time attempting almost all of them before I solved #9 and 15 (I heart geometry :D) in 40 minutes -_-

looking over the solutions, I could've pretty easily gotten four more. Depressing.

oh yeah cognos, you're probably gonna get 5 + (7 - 2) = [b]10[/b] on the AIME next year if we both keep working at the same rates (my rate = 0.00001)


Q: It takes three days for four people to paint five houes. How long will it take two people to paint six houses?

Thanks so much if you help, I have been struggling with this problem forever, and i want to know how to solve it. I am just looking for how to solve it, answer isn't nessasary.

Since you know that 4 people can paint 5 houses in 3 days, now you know that 2 people can paint 2.5 houses in 3 days. Therefore, set up the proportion: 2.5/3=6/x

then you get the answer 7.2 days

Just think logically and use proportions. These are the facts:

1. The more people we have, the less time it'll take, and vice versa - and proportionally at that. Hence 2 people vs. 4 people is $2\mathrm{\ times\ slower}$.

2. The more houses we have, the more time it'll take, and vice versa - and proportionally at that. Hence 6 houses vs. 5 houses is ${6\over 5}\mathrm{\ times\ slower}$.

Therefore, $N=3\mathrm{\ days}\cdot\left({2\cdot{6\over 5}}\right)={36\over 5}$ days


[hide][quote="anirudh"]If x − 2x + 3x = 100, what is the value of x?[/quote]

2x=100
x=50[/hide]

[hide]x - 2x + 3x = 100
2x=100
x=50

hehe. I think I wrote the exact same thing as...five other people.[/hide]

2x=100
x=50

I did the same as 6 people!!! :wow:

[quote="anirudh"]If x − 2x + 3x = 100, what is the value of x?[/quote]

[hide] 2x=100
          x=50[/hide]

Please don't use complex numbers in this forum...thank you.

Eitherway, the answer is indeed 50

[hide]
Simplifying, we get $2x=100$
So, x=50.[/hide]

[hide]x-2x+3x becomes 4x-2x. that makes 2x. 2x=100 makes x=50.
x=50.[/hide]

[hide]

2x=100
x=50[/hide]

easy! i think i did the same thing as the 14 people ahead of me

Yes, 14 people have answered 50 and your post isn't exactly useful. Worse is that this topic is 2.5 months old.

[quote="ProtestanT"]wut's so bad about a 2.5month old topic?[/quote]
you don't awaken old topics. You just don't. Especially when what you say is pointless.

14 people made a point that x=50, let's just stop now and get rid of this thing once and for all.

why is this in MOEMS?

it should be in Classroom math

No, it should be in the graveyard. Somebody lock it.


function
calculus
integration
limit
real analysis
real analysis unsolved

$f_n$ sequence of functions on $[0,1]$, $f_0(x)>0$ Riemann integrable

Let $p\geq 2$

$f_{n+1}(x)= ( \int_{0}^{x}f_n(t)dt )^{1/p}$

What is $\lim_{n \rightarrow + \infty} f_n(x)$  ?
-----------------------------------------------------------
liyi posted here the case $p=2$
http://www.artofproblemsolving.com/Forum/viewtopic.php?t=20994

Assuming convergence, the only possible answer is

[tex]f(x) = (\frac{p-1}{p}x)^\frac{1}{p-1}[/tex]

using [tex]y = \lim\limits_{n\to\infty} f_n(x)[/tex],

[tex]\Rightarrow y^p = \int_{0}^{x}y dt[/tex]

[tex]\Rightarrow p\cdot y^{p-1}\cdot\frac{dy}{dx} = y[/tex]

and solving the differential equation.

y=0 is not a solution, because clearly the sequence of functions moves away from values close to 0, and f_0(x) > 0 forbits it as a starting point.  The arbitrary constant disappears due to y(0) = 0.

I leave the proof of convergence to others  :)

[quote="jacobi"]
I leave the proof of convergence to others  :)[/quote]

This is the most difficult part  :)
Thank's for your answer


geometry
ratio
geometric transformation
homothety
circumcircle
projective geometry
cyclic quadrilateral

let the incircle of a triangle ABC touch BC,AC,AB at A1,B1,C1 respectively. M and N are the midpoints of AB1 and AC1 respectively. MN meets A1C1 at T . draw two tangents TP and TQ through T to incircle. PQ meets MN at L and B1C1 meets PQ at K . assume I is the center of the incircle .

prove IK is parallel to AL

I've managed to reduce it to showing that AT and AL are perpendicular:

K is the intersection of B1C1 and PQ, so the polar of P with respect to the incircle of ABC passes through the poles of B1C1 and PQ, which are A and T respectively, so the polar of K with respect to the incircle is AT, but it's well-known that the polar of a point with respect to a circle is perpendicular to the line joining the point to the center of the circle, so AT perpendicular to IK, so in order to show that IK || AL "all" (:)) we need to show is that AT is perpendicular to AL as well.

This might not be significantly easier, but it's still progress, isn't it? :).

I'm sort of stuck right now, but here are some things I observed (I drew a dynamic sketch with Euklides):

Let T' be the analogous of T (i.e. let A1B11 cut MN at T'). Let MN cut BC at S. Then the following hold:

(1) ATA1T' is cyclic;
(2) <SAT = <TAB;

If we manage to prove these, then I think it's kind of done.

I think I can prove (1) from the previous post:

Let the line through T parallel to AB cut AA1 at R and let the line throguh T' parallel to AC cut AA1 at R'.

N is the midpoint of AC1 and TR and AB are parallel, so the lines TR, TA, TN, TC1 form a harmonic cross-ratio, so the points R, A, X, A1 also form a harmonic cross-ratio (X is the intersection of AA1 and TT'). We prove the same for R', A, X, A1 and we get R=R', so A1TRT' is obtained from A1C1AB1 by a homothety, so RT and RT' are tangent to the circumcircle of A1TT', so A1R is a symmedian in the triangle A1TT', and from the fact that A is the harmonic conjugate of A1 with respect to (R, X), we easily find that A is on the circle A1TT'.

I guess the other part isn't tht hard either (part (2) from 2 posts ago):

First of all, the tangents to the circumcircle of ATT' through T and T' and the diagonal AA1 are concurrent. This is called a hramonic cyclic quadrilateral and it follows that the tangents to its circumcircle through A and A1 and its diagonal TT' are also concurrent, so SA is tangent to the circumcircle, so <SAT=<AT'T=<TAB (this is true because: <TNA=\pi-<ANM=\pi/2+A/2=<TAT'=><TAB=<TAN=<AT'T).

My friends and I have the proof below:

After the arguments Grobber just states.

we know that,AT=QT=PT(TQ,TP are tangents to O)
Draw a tnagent line to circle T at point A,intersect PQ at S.Let A1C1 interesect circle T at U(outside the circle O).
Because circle T and O are orthogonal,
Consider the inversion with point Aand radius AB1,under which
 the circle O does not move.As we can see,the circle T changes to a line
passes through O.And U' the inversion of U
lies on the circumcircle of AB1C1,so AU'O=AC1O=90, since the angle doesnt change under the inversion,AU must be the diameter of circle T.

Next, we draw a circle with center S and radius AS,therefore the circle T
is orthogonal to the circles O and S.And the radical axis of O and T pass
through the center of circle S.It can be proved that,(this is not difficult so i
just dont type them),circle S is also orthogonal to circle O. So,if we know
already the circle S is orthogonal to circleO.By the same argument
above,if circle S intersect B1C1 at V,we can claim that S is the midpoint
of AV.
By now,we conclude that,ST is parallel to UV,then ST bisects segments
AC1 and AB1,the complete the proof.

This problem is just so cool that we thought about it all afternoon！

Best regards,
chao

Consider $A$ as a degenerate circle, then we will have $MN$ is the radical axis of $A$ and $\odot I$. So we know that $T$ is the circumcenter
of $\triangle APE$, furthermore we have $L$ the radical center of $A, \odot I$ and $\odot T$. Therefore $AL$ is tangent to $\odot T$ at $A$, which means $AL \perp AT$.
On the other hand, it is not hard to see that $AT$ is the polar of $K$ with respect to $\odot I$, hence $IK \perp AT$.
Finally, we have $AL \parallel IK$.

Dear Mathlinkers,

http://www.mathlinks.ro/Forum/viewtopic.php?search_id=1137818265&t=161416

Sincerely
Jean-Louis


Let be $ G\equal{}\{(x,y)\in \mathbb{R}^2\ |\ x^2\minus{}y^2\equal{}3x\minus{}4y\}$ and $ H\equal{}\{(x,y)\in \mathbb{R}^2\ |\ 2xy\equal{}4x\plus{}3y\}$ . Determine :

$ M\equal{}\{z\equal{}x^2\plus{}y^2\ |\ (x,y)\in G\cap H\}$.

If $ y\equal{}0$ , then  $ (0,0)\in G\cap H$ . Suppose $ y\ne 0$ . Thus, $ (x,y)\in G\cap H\Longleftrightarrow\left\|\begin{array}{ccc}
x^2\minus{}y^2 & \equal{} & 3x\minus{}4y\\\
2xy & \equal{} & 4x\plus{}3y\end{array}\right\|\implies$ $ \frac {x^2\minus{}y^2}{2xy}\equal{}\frac {3x\minus{}4y}{4x\plus{}3y}\ \ (*)$ .

Denote $ t\equal{}\frac xy$ . The relation $ (*)$ becomes $ \frac {t^2\minus{}1}{2t}\equal{}\frac {3t\minus{}4}{4t\plus{}3}\Longleftrightarrow 4t^3\minus{}3t^2\plus{}4t\minus{}3\equal{}0\Longleftrightarrow$ $ (4t\minus{}3)\left(t^2\plus{}1\right)\equal{}0\Longleftrightarrow t\equal{}\frac 34$ .

Therefore, $ \left\|\begin{array}{c}
x\equal{}3\lambda\\\
y\equal{}4\lambda\end{array}\right\|$ $ \implies$ $ 24\lambda^2\equal{}24\lambda\ ,\ \lambda\ne 0$ $ \implies$ $ \lambda\equal{}1$ $ \implies$ $ \left\|\begin{array}{c}
x\equal{}3\\\
y\equal{}4\end{array}\right\|$ . In conclusion, $ G\cap H\equal{}\{(0,0)\ ;\ (3,4)\}$ and $ M\equal{}\{0,25\}$ .

From the 2 equations we deduce that $ (x^2 \minus{} y^2)^2 \plus{} (2xy)^2 \equal{} (3x \minus{} 4y)^2 \plus{} (4x \plus{} 3y)^2 \implies (x^2 \plus{} y^2)^2 \equal{} 25(x^2 \plus{} y^2)$ and the conclusion follows immediately.


advanced fields
advanced fields unsolved

[b]Which of the following is a topological property?[/b]
(A) boundedness
(B) being a Cauchy sequence
(C) completeness
(D) being an limit point
(E) length

When you see something like this, think of the definitions. If the definition uses open sets, but nothing extra, then it is topological. This means that D) limit points are the answer.

Boundedness needs some sense of distance. Cauchy sequence needs some sense of distance. Completeness needs a distance. Length...


[b]basics[/b]
(1)why is magnetic field generated when current is passed through a conductor?
and (2)why is current induced in a conductor by change in magnetic flux? :?:

This isn't exactly "Introductory Physics".  I would move this to the advanced physics forum.


inequalities
modular arithmetic
induction
factorial
number theory

Prove that, for all positive integers $ n$,

$ \left(\sum\limits^\infty_{k \equal{} 1} k^\frac{2n}{1\minus{}2n}\right)^{2n\minus{}1} > \dfrac{2\pi^{2n}(2n)!\prod\limits_{p\minus{}1 | 4n} p}{144^n}$

A rather abstract problem, but... :ninja:

Of all the things to intimidate me on this fine young night...

So I'm guessing the product thingy is the product of all primes s.t. $ a^{4n}\equiv 1\pmod{p}$.

Man this entire problem looks overly intimidating. And i really suck at inequalities.

Would Lagrange multipliers bash it properly enough?

kevin hu

you should try to be more helpful than posting random stuff like this

this is NOT helpful

man if you could've solved a problem like this on team contest, now that's another story :wink:

ok so the LHS probably telescopes or something... well actually it's a zeta thing. The 1-2n and 2n-1 look suspicious. Blargh on the RHS, which, I must say, looks incredibly taylor-series-like, (probably because it is  :wink: ), is pretty silly because of the pi and nt-product thing. Meh.

ok so man nvm it is not actually taylor series due to the lack of sum involved

darn

induction

but induction is probably TOO WEAK

where'd you get this overly yucky problem anyway

err also lagrange multipliers? rofl... TRY LAGRANGE MULTIPLIERS, PLEASE

yeah so I'm not sure where you found this - is it supposed to be tractable without the use of analytic number theory? (some things seem to hint applying the zeta / gamma functional equations but I'm really not sure what exactly that gets you)

Differentiating both sides looks appealing. If you differentiate the LHS 2n times you get a factorial, so thats a hint. I don't see how that helps though. Then, one can conceivably use induction on the infinite part.

Taylor series might help too...

Also, taking the equation modulo $ \prod\limits_{p\minus{}1 | 4n} p$ might work, since then the residues on the right side are $ \sqrt{p\plus{}1}^{2n\minus{}1}\pmod{p}$, and that would repeat, right?

I don't know

I suck at inequalities.

Hehe. Actually, if you plug in $ n\equal{}1$, it's false.  :rotfl:  :rotfl:  :rotfl:

err I am pretty sure that it is equality for n=1 or something w/e

oh wait lawl

what is up with that stray 5 XD

Yeah you would have to divide by 5 to get equality at 1.

Even so, maybe by other hints would work...

Aryth, what the heck is up with the number-theoretic product on the RHS?  Everything else works, and is differentiable on the interval $ (1,\infty)$.  Except for that.  Seriously, that's pretty freaking lame.  And it's wrong.

True.   :spam:  :oops_sign:
Oops, i sorry fer invadeing