Inequality
Summation
rearrangement inequality
IMO
algebra
IMO 1978

[i][b]Let's solve the problem, which is still unsolved:[/b][/i]

We know that all the unknowns are integers, so the smallest one must greater or equal to 1.

Let me denote the permutations of $(k_1,k_2,...,k_n)$ with $(y_1,y_2,...,y_n)=y_i (*)$.

From the rearrangement's inequality we know that $Random Sum\geq Reversed Sum$.

We will denote we permutations of $y_i$ in this form $y_n \geq ...\geq y_1$.

So we have $\frac{k_1}{1^2}+\frac{k_2}{2^2}+...+\frac{k_n}{n^2} \geq \frac{y_1}{1^2}+ \frac{y_2}{2^2}+...+ \frac{y_n}{n^2} \geq 1+\frac{1}{2}+...+\frac{1}{n}$.

Let's denote $\frac{y_1}{1^2}+ \frac{y_2}{2^2}+...+ \frac{y_n}{n^2}=T$ and  $1+\frac{1}{2}+...+\frac{1}{n}=S$.

We have $T \geq S$. Which comes from $y_1 \geq1, y_2 \geq2, ...,y_n \geq n$.

So we are done.

It is easy proved by induktion. Let $f_0(x)=f(x)$ and $f_i(x)=f_{i-1}(x)$ if $f_{i-1}(i)=i$, else exist k>i $f_{i-1}(k)=i,f_{i-1}(i)=m>i$ we denote $f_i(x)=f_{i-1}(x),x\not=k,i, f_i(i)=i,f_i(k)=m$. Let \[ S_n(f)=\sum_{k=1}^n \frac{f(k)}{k^2} \]. It is easy to show, that $S_n(f_i)\ge S(f_{i+1})$.

We just have to use the Rearrangement Inequality.

[quote="orl"]Let $ f$ be an injective function from $ {1,2,3,\ldots}$ in itself. Prove that for any $ n$ we have: $ \sum_{k \equal{} 1}^{n} f(k)k^{ \minus{} 2} \geq \sum_{k \equal{} 1}^{n} k^{ \minus{} 1}.$[/quote]

Let $ f(k)\equal{}a_k$, where $ a_k \in \{1,2,3,...\}\equal{}\mathbb{N}$.

\[ \sum_{k\equal{}1}^{n} f(k)k^\minus{}2 \equal{}\sum_{k\equal{}1}^n {\frac{a_k}{k^2}} \geq \sum_{k\equal{}1}^{n}{\frac{1}{k}}\]

By AM-GM, for each $ k\in \mathbb {N}$ we have:

\[ \frac{a_k}{k^2} \plus{}\frac{1}{a_k} \geq \frac{2}{k}\]

\[ \implies \sum_{k\equal{}1} ^n {\frac{a_k}{k^2}} \plus{}\sum_{k\equal{}1}^n {\frac{1}{a_k}} \geq 2\sum_{k\equal{}1}^n {\frac{1 }{k}}\]

Therefore, it is enough to prove that:

\[ \frac{1}{1} \plus{}\frac{1}{2} \plus{}... \plus{}\frac{1}{n} \geq \frac{1}{a_1} \plus{}\frac{1}{a_2} \plus{}...\plus{}\frac{1}{a_n}\]

But this is obviously true because all $ a_i$ numbers are different natural numbers.

Very simple question. The only trick is in realising that the rearrangement inequality can be used and then seeing that positive integers are greater than 1.

Hmmm ... your [b]lemma[/b] is just Abel's summation formula. Let's call a rabbit a rabbit.

[quote="orl"]Let $f$ be an injective function from ${1,2,3,\ldots}$ in itself. Prove that for any $n$ we have: $\sum_{k=1}^{n} f(k)k^{-2} \geq \sum_{k=1}^{n} k^{-1}.$[/quote]
Assume to sequence $f(x_1),f(x_2),\cdots,f(x_k)$ where $x_k$'s are the permutation of $\{1,2,\cdots,n\}$                  and   $\left\{\frac{1}{k^2}\right\}$.They are similar.So applying chebyshev inequality,
\begin{align*}\sum_{k=1}^{n} \frac{f(k)}{k^2}& \ge \frac{1}{n} \sum_{k=1}^{n} f(k). \sum_{k=1}^{n} \frac{1}{k^2}\\ & \ge \frac{1}{n}.\frac{n(n+1)}{2}. \frac{1}{n} \left(\sum_{k=1}^{n} \frac{1}{k}\right)^{2}\\ & \ge \frac{n(n+1)}{2n^2}.\sum_{k=1}^{n}\frac{1}{k}.\frac{n^2}{1+2+\cdots+n}\\ & \ge \sum_{k=1}^{n}\frac{1}{k}\end{align*}

Also we can use: $\frac{f(k)}{k^2}\geq \frac{2}{k}-\frac{1}{f(k)}$

Let $a_1,a_2,\dots,a_n$ be a sequence of distinct positive integers such that $f(i)=a_i$.

Note that $\sum\limits_{k=1}^n a_k^{-1}\le\sum\limits_{k=1}^n k^{-1}$ since all $a_k$ are distinct positive integers.

Cauchy Schwarz yields \[\left(\sum\limits_{k=1}^n a_k^{-1}\right)\left(\sum\limits_{k=1}^n a_k k^{-2}\right)\ge\left(\sum\limits_{k=1}^nk^{-1}\right)^2.\] Dividing by $\sum\limits_{k=1}^n a_k^{-1}$ yields \[\sum\limits_{k=1}^n a_k k^{-2}\ge\frac{\left(\sum\limits_{k=1}^nk^{-1}\right)\left(\sum\limits_{k=1}^nk^{-1}\right)}{\sum\limits_{k=1}^n a_k^{-1}}\implies \sum\limits_{k=1}^n a_k k^{-2}\ge\left(\sum\limits_{k=1}^nk^{-1}\right)\cdot\frac{\left(\sum\limits_{k=1}^nk^{-1}\right)}{\sum\limits_{k=1}^n a_k^{-1}}.\] Since $\sum\limits_{k=1}^n a_k^{-1}\le\sum\limits_{k=1}^n k^{-1}$, then $\frac{\left(\sum\limits_{k=1}^nk^{-1}\right)}{\sum\limits_{k=1}^n a_k^{-1}}\ge 1$. Thus, we have $\sum\limits_{k=1}^n a_k k^{-2}\ge\sum\limits_{k=1}^nk^{-1}$ as desired.

We can simply put the R.H.S as $1/k$ to $k/k^2$ and use rearrangement inequality it's done in like two lines.
The R.H.S is a scalar product of two sequences sorted in the opposite way.
The R.H.S is a scalar product of rearranged sequences. Q.E.D

[quote=al_wang]What's an injective function?[/quote]
Let's say there is a function going from $X->Y$
Then, imagine arrows flying from the $X$ to $Y$
It's an injective function when the arrow from Xs is shot at different Ys, which mean than one Y cannot be shot two times
Thus, if $f(x1)$ is unequal to  $f(x2)$, then also $x1$ and $x2$ is different

Sorry for reviving this thread, but couldn't resist posting my solution to this problem.
[b]Solution[/b]
Clearly $f$ being injective implies that:
$\sum \dfrac{f(k)}{k^2} = \sum \dfrac {a_k}{k^2} $ where $a_k = f(k) $.
Thus, the inequality to prove is :
$ \sum \dfrac {a_k}{k^2} \ge \sum \dfrac {1}{k^2} $
Now consider a permutation $(b_1,b_2, \dots , b_n)$ of ${a_k}$ such that $b_i \le b_j   \forall i \le j $
Therefore by rearrangement inequality $ \sum \dfrac {a_k}{k^2} \ge\sum \dfrac {b_k}{k^2}$
[b]Claim[/b]: $\sum \dfrac{b_k}{k^2} \ge \sum \dfrac {1}{k^2} $
Proof: this is trivial since $ b_1 < b_2 < \dots < b_n$ and $b_1 \ge 1 $
Thus, $b_k \ge k$ for any $ k \in { 1,2,\dots,n} $

[quote=Electron_Madnesss]
Proof: this is trivial since $ b_1 \le b_2 \le \dots \le b_n$ and $b_1 \ge 1 $
[/quote]

It should be  $< $, not  $\leq $, but other than that good formalisation.

This is the easiest IMO I have ever seen! It's just Rearrangement.

We may use induction as well. If for all permutations of $S=\{1,2,...,n\}$  we have the inequality is satisfied then we can show that for all permutations of $S_1=S\cup\{n+1\}$ it will be satisfied.

   Let, $\sigma_{f,n}=\sum_{k=1}^{n}\frac{f(k)}{k^2}\geq \sum_{k=1}^n \frac{1}{k}=\sigma_{0}(n)$ for all bijective functions $f:S\rightarrow S$.

So, $\sigma_{f,n}+\frac{n+1}{(n+1)^2}\geq \sigma_{0}(n+1)$

Now, we interchange any $f(k)$ with $n+1$. $\sigma_{f,n}$ will be changed by $\Delta=(n+1-f(k))(\frac{1}{k^2}-\frac{1}{(n+1)^2}) >0$ as $f(k)<n+1$. Hence, $\sigma_{f,n+1}\geq \sigma_{0}(n+1)$.

 As, the inequality is true for all permutations for $n=2$, ($\sigma_{f,2}=\{9/4, 3/2\}$ and $\sigma_{0}(2)=3/2$ ) so is true for all $n$.

Let $(b_1,b_2,\ldots,b_n)$ be the permutation of $(a_1,a_2,\ldots,a_n)$ such that $b_1\le b_2\le\ldots\le b_n$. Thus by the rearrangement inequality we get that
$$\sum_{k=1}^{n}\frac{a_k}{k^2}\ge\sum_{k=1}^{n}\frac{b_k}{k^2}$$
Clearly $b_1\ge 1,b_2\ge 2,\ldots,b_n\ge n$, so
$$\sum_{k=1}^{n}\frac{b_k}{k^2}\ge\sum_{k=1}^{n}\frac{k}{k^2}=\sum_{k=1}^{n}\frac{1}{k}$$


.for each positive integer $n$.let $f(n)$ denote the number of ways of representing $n$ as a sum of powers of 2  with nonnegative integer exponents.representations which differ only in th ordering of their summands are considered to be the same .for instance,$f(4)=4$ because $4=4+0=2+2=2+1+1=1+1+1+1$
 Prove that for $n\Rightarrow 3$
                                       $2^{\frac{n^2}{4}}<f(2^n)<2^{\frac{n^2}{2}}$

It's [url=http://www.mathlinks.ro/Forum/resources.php?c=1&cid=16&year=1997]IMO 1997[/url] Problem 6.
You can find a solution at many places in the internet, e.g., http://imo.math.ca/Sol/97/solutions.html


[quote="rustam"]Let the polynom P(x) has a distinct n=deg(P(x)) roots         $a_1,a_2,...,a_n$ and let $b_1,b_2,...,b_{n-1}$ be the roots of the polinom P'(x).
Find the value of the sum:
$\displaystyle\sum_{i,k} \frac{1}{a_i-b_k}$[/quote]
Since $\frac{P'(x)}{P(x)}=\sum_{i=1}^n \frac{1}{x-a_i}$ we have that for any $k=1,...,n-1$ the sum $\displaystyle\sum_{i=1}^{n} \frac{1}{b_k-a_i} =0$, so the sum $\displaystyle\sum_{i,k} \frac{1}{a_i-b_k}=0$.


function
limit
calculus
calculus computations

1 .
if : $ f\circ g$ is differentiable at $ a$ , and $ f$ is differentiable at $ g\left( a\right)$; then : $ g$ is differentiable at $ a$

2 .
if : $ f\circ g$ is differentiable at $ a$ , and $ g$ is differentiable at $ a$; then : $ f$ is differentiable at $ g\left( a\right)$

                                                       thank you.

Use Caratheodory's Lemma

I'm assuming $ f,g: \mathbb{R}\rightarrow\mathbb{R}$
1. $ f\circ g$ is differentiable at $ a$ $ \implies$ there exists a function $ \phi : \mathbb{R}\rightarrow\mathbb{R}$ that is continuous at $ a$; such that
$ f(g(x)) \minus{} f(g(a)) \equal{} \phi (x)(x \minus{} a)$
We require that $ \phi (a)\neq 0$

Similarly $ f$ differentiable at $ g(a)$ $ \implies$ there exists a function $ \psi : \mathbb{R}\rightarrow\mathbb{R}$ that is continuous at $ g(a)$; such that
$ f(g(x)) \minus{} f(g(a)) \equal{} \psi (x)(g(x) \minus{} g(a))$
Thus $ \lim_{x\rightarrow a}\frac {g(x) \minus{} g(a)}{x \minus{} a} \equal{} \lim_{x\rightarrow a}\frac {\phi (x)}{\psi (x)} \equal{} \frac {\phi (a)}{\psi (a)}$

Similarly for (ii)

[color=green] Unnecessary quote (it never makes sense to quote the previous post in full) removed by moderator[/color]

Example :
$ f = |x|$ and  $ g(x) = \{^{1 , x = rational}_{ - 1,x = irrational}$

So
1.$ f'(g(a))$ must not be $ 0$
2. $ g$ must be continuous at $ a$

Both statements obviously false. Let f be constant in the first one, let g be constant in the second one.

@Comment below: Ok, I did not understand what you guys were trying to do.

Of course both statements are false as given, [quote="shreyas_patankar"]We require that  $ \phi (a)\neq 0$[/quote]
Or, equivalently,
[quote="ICQ-2008"]$ f'(g(a))$ must not be $ 0$[/quote]


I flip a coin which is designed so that it has a $\frac{3}{4}$ chance of landing heads. What is the probability that in 2 flips, only one comes up heads?

[hide]
If you get tails on the first flip, that was a $\frac{1}{4}$ chance of happening, then you get a heads on the second flip which had a $\frac{3}{4}$ chance of happening. $\frac{1}{4} * \frac{3}{4} = \frac{3}{16}$ But since it can happen the other way around, the answer is $\frac{6}{16} = \boxed{\frac{3}{8}}$
[/hide]

[hide]
say you get a tails then heads
the chance of getting tails is 1/4 and the chance of getting heads is 3/4
1/4*3/4=3/16
but you can also get a heads then tails so
3/16*2=$\boxed{\frac{3}{8}}$[/hide]

[quote="buzzer11"][hide]3/8?[/hide][/quote]
how did you get that?


it is \sqrt{}
read carefully and again the instructions

$\sqrt{something} $

[quote="mysmartmouth"]Sorry, I am new to Latex, but what is the command for the square root symbol? How would I TeX rt(x)?[/quote]

[code]\sqrt{What is inside the square root}[/code] or for cube roots, [code]\sqrt[3]{what is inside the cube rooot}[/code]

Also, I am working on a latex guide for beginners.

When you get that Latex guide, could you email me a copy?  My email address is mysmartmouth@gmail.com

Thanks!


Given: two chords that intersect at right angles within a circle.
One of the chords has lengths 3 and 4 (meaning it has length 7, but it is intersected, and the segments that it is "split" into are 3 and 4)
The other chord has length 6 and 2 (meaning it has length 8, but it is intersected, and the segments that it is "split" into are 6 and 2)
Find the length of the radius

[hide="solution"]
All right, my solution seems messy, but it might be right, I guess. Using Power Of A Point, we get that the PA*PB for this point is 12. Then, using the 2 and 3 sides, we construct a right triangle. The length from the point P to the midpoint of the hypotenuse is equal to half the hypotenuse, which happens to be $ \frac{\sqrt{13}}{2}$. Using Power of a Point and adding to the other length, then dividing by 2, we get that the radius is $ \frac{61}{52} * \sqrt{13} $.
[/hide]

Got something different.
[hide]One of the triangles inscribed in the circle has sides of $\sqrt {13}$, $3 \sqrt 5$ and $8$. Using the formula $\frac {abc}{4A}$ we get that the radius is $\frac {\sqrt {65}}{2}$. Or, using root-mean-square, $\sqrt {\frac {a^2 + b^2 + c^2 + d^2}{4}}$, we again get $\frac {\sqrt {65}}{2}$.[/hide]

When I did it yesterday with an unsure method, I came up with tarquin's results

[hide]
I just drew two other perpendicular chords opposite the two already draw, with the same lengths, and created a rectangle and found the diagonal of that (which would be the diameter of the circle) and then divided by two to get

$\frac{\sqrt65}{2}$
[/hide]


Source: AHSME 1973 #19

Define $n_a!$ for $n$ and $a$ positive to be
\[ n_a! = n(n-a)(n-2a)(n-3a)\dots(n-ka)  \]
where $k$ is the greatest integer for which $n>ka$.  Then the quotient $72_8!/18_2!$ is equal to
$\text{(A)} \ 4^5 \qquad \text{(B)} \ 4^6 \qquad \text{(C)} \ 4^8 \qquad \text{(D)} \ 4^9 \qquad \text{(E)} \ 4^{12}$

[hide="Solution"]
(D)
List the first few terms.  We quickly see that we can factor out 8's in the numberator and 2's in the denominator and the quotient becomes
\[ \frac{8^9(9!)}{2^9(9!)} = 4^9  \]
[/hide]


geometry
similar triangles
geometry solved

Can you please help me with the following questions:

#1 Is it possible to prove The angle sum of a triangle without using EPA?

#2 Is it possible to prove that restangles exist without using EPA? if yes, how?

Thank you so much

Dina

In hyperbolic geometry, the axiom of parallel lines does not hold, and more than one parallel to a given line passes through a given point not on this line. The sum of internal angles of a hyperbolic triangle is less than 180 deg. The larger is the triangle, the bigger is its angular deficit, so that all similar triangles (with equal angles) must be congruent. All hyperbolic quadrilaterals having all 4 internal angles equal (their sum being less than 360 deg) have all 4 sides congruent and they are hyperbolic squares. Moreover, 2 ultraparallel lines (non-intersecting non-parallel lines) have a unique common normal. If they had more than one, these 4 lines would form a quadrilateral with all 4 internal angles right, their sum equal to 360 deg, which is impossible (in hyperbolic geometry).


function
calculus
integration
algebra
domain
Ring Theory
superior algebra

Let $ C\left(\mathbb{R},\mathbb{R}\right)$ be the set of continuous functions from $ \mathbb{R}\rightarrow\mathbb{R}$.

I know how to show that $ C\left(\mathbb{R},\mathbb{R}\right)$ is not a field under $ \plus{},\times$, but how do you show that it's not an integral domain?  Are there any other idempotents besides $ f\left(x\right)\equal{}0,1$?

An appropriate product of two [url=http://mathworld.wolfram.com/BumpFunction.html]bump functions[/url] is the zero function.

Edit:  Is the idempotent question a separate question?  If $ f(x)^2 \equal{} f(x)$ for a particular value of $ x$, then $ f(x) \equal{} 0, 1$, and the only such continuous functions are $ f(x) \equal{} 0$ and $ f(x) \equal{} 1$.

Yeah they're two separate questions.  I originally thought that this was an integral domain, but read that it wasn't and I'm not really sure why not.

Simpler than general bump functions:
Take
$ f_1(x) \equal{} \begin{cases} x \ \text{ for } x \leq 0 \\ 0 \ \text{ for } x>0 \end{cases}$
$ f_2(x) \equal{} \begin{cases} x \ \text{ for } x \geq 0 \\ 0 \ \text{ for } x<0 \end{cases}$
then their product is $ 0$.

Sorry, I read the original question as "smooth functions."   :oops:


geometry
perimeter
ARML
3D geometry
sphere
trigonometry

A formula for the area of any regular polygon is (ap)/2. where a is the length of the apothem and p is the perimeter.

Does anyone have a proof for this?

Draw lines from the center of the polygon to each vertex and it should become obvious.

you can easily come up with this formula. consider an n-gon with an apothem a and a perimeter p. Each side is p/n. find the area of one of the n triangles that make up the n-gon. (1/2)(b)(h) or (1/2)(a)(p/n) since there are n triangles, multiply by n to get the total area which is (1/2)(a)(p)

Here is a problem that is similar from a problem from ARML:

What is the radius of the largest sphere inscribed in a teetrehedron if the volume of the tetrehedron is V and the surface area is S

[quote="EunuchOmerta"]I prefer
\[ A = \frac{n*e^2*\tan(\frac{90*(n - 2)}{n})}{4}  \][/quote]

why is there an e^2?


geometry
geometric transformation
rotation
reflection
geometry unsolved

Let $ABC$ be a triangle .Let $A'BC,B'CA,C'AB$ be equalteral triangles ,out side of the triangle .
Three points are given in the plane ,construct a triangle like $ABC$ s.h these points become its $A',B',C'$.
By construction ,I mean construction with Compass and ruler.

The composition of the rotations around $A',B',C'$, each one of angle $\frac\pi 3$, is the reflection in $A$, so just take any point $X$ in the plane, find its image $X'$ through this transformation, and take $A$  to be the midpoint of $XX'$.

Good lord,Man how do you remember every singel way of problem solving from when you were studing for competitions?
Well ,quite right.
I by myself find the $A$ directly .
Let define $R^{\alpha}_O$ ,be a rotation around $O$ by $\alpha$ degree .

[color=blue]Lemma:If $R^{\alpha}_{O_1}\circ R^{\beta}_{O_2}=R^{\alpha+\beta }_{O} $then $O_1O_2O=\frac{\beta}{2} $ and $O_2O_1O=\frac{\alpha}{2}$[/color]

Clearly $R^{60}_{B`} \circ R^{60}_{A`} \circ R^{60}_{C`}(A)=A$ but we know that
Composition of three rotations is a rotaion by the sum of the angles by a different Center ,But these rotations has no effect on $A`$ so ,$A`$ is the center of the composition of the rotations .
Lets find the $A$:

Let $R^{60}_{A`} \circ R^{60}_{C`}=R^{120}_{O}$,according to lemma we can find $O$.
Now $R^{60}_{B`} \circ R^{120}_{O}=R^{180}_{A`}$ and by using lemma we can find $A$.
To be precise
Let $O$ be a point such that $B`C`O=C`B`O=30    , O \in \ in \ side \ the A`B`C`$ .
then $AOA`=60,AA`0=30$ ,so $A$ is found.
Well grobber solution is much more easier .

I shall re-word the problem:
[quote]If, in the given configuration, with $ BCA', CAB'$ and $ ABC'$ equilateral triangles erected outwardly on the sides of the triangle $ ABC$ one erases the vertices $ A, B$ and $ C$, find their position again, knowing the points $ A', B'$ and $ C'$ only.[/quote]

Proof: We know that $ AA' \equal{} BB' \equal{} CC'$ and every two of these lines make an $ 120^\circ$ angle between them; if, the three equilateral triangles are erected inwardly (i.e. $ A$ and $ A'$ on the same side of $ BC$, the property still holds, with $ AA' \equal{} BB' \equal{} CC'$, but this time two lines only make an $ 120^\circ$, the third one being the bisector of this angle.

Next: we construct, outwardly on the sides of the $ \triangle A'B'C'$ the equilateral triangles $ B'C'A", C'A'B"$ and $ A'B'C"$ and claim that the midpoints of $ A'A", B'B"$ and $ C'C"$ are the required points $ A, B$ and $ C$.

Proof of this claim: we considerate known the position of the points $ A, B$ and $ C$. According to the previous statements, $ AA' \equal{} BB' \equal{} CC'$ and their angles are $ 60^\circ$. Applying the same property to the $ \triangle AB'C'$ and equilateral triangles $ B'C'A", ACB'$ and $ ABC'$, we get $ BB' \equal{} CC' \equal{} AA"$ and their angles are $ 60^\circ$ , hence $ AA' \equal{} AA"$ and both of them lie on the bisector of the angle between $ BB'$ and $ CC'$, similarly for $ B$ and $ C$, done.

[b]Remark (own):[/b] [i][color=darkblue]The center of $ \triangle ABC'$ is the midpoint of the line segment joining the centers of $ \triangle B'C'A"$ and $ \triangle A'C'B"$, and other two similar[/color][/i]

Best regards,
sunken rock


topology
real analysis
real analysis unsolved

Let $ X$ be a compact Hausdorff space, and let $ f: X \to \mathbb{R}$. The graph of $ f$ is the set $ G(f) \equal{} \{(x,f(x)): x\in X\} \subset X\times\mathbb{R}$. Show that $ f$ is continuous iff $ G(f)$ is compact.

Consider the bijective map $ F: X\to G(f),x\mapsto (x,f(x))$ for both implications.

If $ f$ is continuous, so is $ F$, hence it has compact image.

If $ G(f)$ is compact, $ F^{ \minus{} 1}$ is continuous (a priori) and closed (i.e. $ F^{ \minus{} 1}$ sends closed subsets to closed subsets), thus a homeomorphism. Then $ f \equal{} p\circ F$ is continuous with the (continuous) projection $ p: G(f)\times\mathbb R\to\mathbb R$ on the second factor.


Please post your predictions:
Winner
Loser
# of games

[b]RAYS OVER PHILLIES IN 6 GAMES[/b]

I'll agree with that, but the Rays will beat the Phillies in the other four games.   :D
[b]Rays over Phillies in 5 games[/b]

Hang on...don't put the Rays in the Series just yet.  Boston just won Game 5 of the ALCS 8-7.

ALCS: RAYS OVER RED SOX 4-2, MVP: UPTON

WORLD SERIES: RAYS OVer phillies 4-1, MVP: SONNANSTINE or CRAWFORD

Well, looks like I'm going to be right with either the Phillies winning or the 7 games.

go brad lidge. phillies win 4-1!!!!!!!the world series champions!

Funny sign of the day: Phillies in 5 1/2 games!
Lol: Game 5 had a 48 hour rain delay!

Each side had a top player with a strong Long Beach connection.

The Rays had Evan Longoria, who attended my university. (Known as CSULB to us academics but as Long Beach State University to the Athletic Department.)

The Phillies had Chase Utley, who attended Long Beach Polytechnic High School - the same school that both jmerry and my other child went to.

As you may have noticed, Utley had the better World Series.


probability
combinatorics unsolved
combinatorics

Find the probability of a total of $ 9$ when three fair dice are tossed once.

It is $ \frac{25}{216}$. I hope someone will show why is that. :(


I just had a math test today, and there was a equation that simplified to likee cot x=0. However, I do not think that csc, cot, or sec can be zero because they are 1/sin,cos,tan, and 1/x cannot be zero. I simply put "undefined". Was I right that cotx=0 isn't true?

Well technically, cot x = (1/tan x) and that equals zero if and only if (tan x)=infinity, which in some sense is undefined, unless you are working with extended real numbers, in which 1/(infinity)=0.

However, another way to put it is (cot x) approches 0 as x approaches (pi)/2, 3(pi)/2, 5(pi)/2 etc, since at these values
(tan x )= infinity.

Lizzie

$\cot x = \frac{1}{\tan x}$ however, it's more like....

$(\tan x)^{-1} = (\frac{\sin x}{\cos x})^{-1} = \frac{\cos x}{\sin x}$

So for values such that $\cos x = 0$, $\cot x = 0$

This occurs at $\frac \pi2$ and $\frac{3\pi}{2}$ where $0 \leq x < 2\pi$

EDIT: another thing: we define the range of $\cot$ to be all real numbers so there have to be solutions.


geometry
inequalities
calculus
derivative
inequalities proposed

[color=purple]Let an acute triangle ABC with the side length a,b,c and its area S.
        Prove that
 $\frac{a^{2}+b^{2}+c^{2}}{4\sqrt{3}S}\ge\frac{3}{2(cosA+cosB+cosC)}$
  Dear pvthuan do you remember the origin of this problem ?  :lol:
 And if you (=mathlinker) want the stronger :D :
$\frac{ab+bc+ca}{4\sqrt{3}S}\ge\frac{3}{2(cosA+cosB+cosC)}$
   Have fun ! :D [/color]

[color=purple]No one like this problem ? [/color]

I think here is a beautiful problem but any one can post proff ?

[quote="evarist"][color=purple]Let an acute triangle ABC with the side length a,b,c and its area S.
        Prove that
 $\frac{a^{2}+b^{2}+c^{2}}{4\sqrt{3}S}\ge\frac{3}{2(cosA+cosB+cosC)}$
   [/color][/quote]
Put $x=\sqrt{\frac{b^{2}+c^{2}-a^{2}}{2}},y=\sqrt{\frac{c^{2}+a^{2}-b^{2}}{2}},z=\sqrt{\frac{a^{2}+b^{2}-c^{2}}{2}}$. We have got the algebraic inequality!.

[color=purple]Can you (N.T.TUAN) or other mathlinker can write more clearlier ? I think it's a nice problem :)  but you are don't like it  :wink: [/color]

My solution is very ugly.It uses derivative method
 We need prove:
 $(sin^{2}a+sin^{2}b+sin^{2}c)(cosa+cosb+cosc)\geq3\sqrt{3}sinasinbsinc$
 $\leftrightarrow F(a)=(sin^{2}a+sin^{2}b+sin^{2}(a+b))(cosa+cosb-cos(a+b))-3\sqrt{3}sinasinbsin(a+b)\geq0$
 $F(a)'=(sin2a-sin2c)(cosa+cosb+cosc)+(sin^{2}a+sin^{2}b+sin^{2}c)(sinc-sina)+3\sqrt{3}sin(a-c)sinb$
Assume that $a\geq b\geq c$
 $(sin2a-sin2c)(cosa+cosb+cosc)\geq 3(sin\frac{c-a}{2}cosb$(1)
  $sin^{2}a+sin^{2}b+sin^{2}c\geq 2sina+2sinb$(2)
From(1),(2),it's easy to prove F(a)'$\geq 0$
->F(a)$\geq F(b)$
   or F(a)$\geq F(\pi-2b)$
Now,it only has two variables,We use derivative method again....

 Who has a better solution because this way is complicated!


analytic geometry
geometry
geometric transformation
reflection
geometry open

I'm new at barycentric coordinates. I'd like to know wheter there exists a formula which enable us to calculate the coordinates of the centroid of an arbitrary  triangle. Can someone help me? Thanks :)

A good article by Paul Yiu:

P.S Do you know Gabriel Alexander Reyes? Maybe it is you?

Thanks for your help guys!

[quote="prowler"] P.S Do you know Gabriel Alexander Reyes? Maybe it is you?[/quote]

Sure!!! I'm Gabriel!! :D Why the question?? :)

[quote="prowler"]I know you from Mathematical Reflections :)
I am Ivan Borsenco, whom you send your solutions.[/quote]

Pleased to meet you Ivan!! :D Actually I had no idea about who you are, because your nickname did not give me any information!


[hide="hint 1"]  Try expressing them as improper fractions.[/hide]

[hide="hint 2"]  Try simplifying common terms. [/hide]

Well, I got it before I posted it. Last year was my last MathCounts year.  :(

Anyways, I posted the problem because the solution (which G-Unit's hints point towards) is pretty cool... I hadn't seen a _____ like this before (I avoid putting the type of problem here, since it could give it away...but you know what fills in the blank if you've done these types of problems before).[/code]

[hide]If you simplify each term, you get
(3/2)(4/3)(5/4)(6/5)(7/6)(8/7)
Just cancel the terms into 8/2
>>4<<[/hide]

[hide]Make improper fractions to get $ \frac 32 * \frac 43 * \frac 54 * \frac 65 * \frac 76 * \frac 87$
All but the $2$ and $8$ cancel leaving $ \frac 82 = \boxed 4$ [/hide]

[quote="chesspro"]I found this problem when I was randomly flipping through the AoPS book. It's a pretty old problem, so I'm pretty sure people haven't seen it yet. Not sure what level this is, but it shouldn't be too hard...

Express in simplest form:

\[\left(1 + \frac{1}{2}\right)\left(1 + \frac{1}{3}\right)\left(1 + \frac{1}{4}\right)\left(1 + \frac{1}{5}\right)\left(1 + \frac{1}{6}\right)\left(1 + \frac{1}{7}\right)\][/quote]
 Ilove asking my friends these kinds of problems they're like huh? lemme go get a calculator! anyways the asnwer is :[hide="answer"] 4 because you expand the fractions and then cancel out till you get 8/2 which equals 4[/hide]


geometry
geometric transformation
rotation

with what initial velocity $v_{0}$ must a wheel of radius $r$ mass $m$ and moment inertia $I = \frac{1}{2}mr^{2}$ have to be able to climb onto a step of height $h$

I don't know if it is corrected.
   Use the law of conservation of energy. The system is your wheel, the earth.
  So the potential energy is converted from the kinetic energy and rotation energy. Hence
$mgh = \frac{1}{2}mv_{0}^{2}+\frac12 I w^{2}$ but $w= \frac{v}{r}$
so $mgh= \frac{1}{2}mv_{0}^{2}+\frac12 \frac12 mr^{2}\frac{v_{0}^{2}}{r^{2}}$
    $mgh= \frac{1}{2}mv_{0}^{2}+\frac14 mv_{0}^{2}$
    $gh= 0.75v_{0}^{2}$
    $v_{0}= \sqrt{\frac{4}{3}gh}$ :)


$ u\equal{}\sqrt {z_1z_2}$ , prove that $ \left|z_1 \right|\plus{}\left|z_2 \right|\equal{}\left|\frac {z_1 \plus{}z_2}{2}\plus{}u \right|\plus{}\left|\frac {z_1 \plus{}z_2}{2}\minus{}u \right|$

Think geometrically. $ sqrt{z_{1}z_{2}}$ and $ \frac {z_{1} \plus{} z_{2}}{2}$ are on the same line.  It can be proved by simple construction.  I'm a beginner of LaTex, sorry.  :)

I was wrong.  One of the right way is to square the identity you want to proof, and use Pappus's theorem twice on the right hand side.


geometry
inequalities
trigonometry
inequalities unsolved

[quote="Jianxing113725"]how to think about....
Prove that $ \sin\frac{A}{2}\plus{}\sin\frac{B}{2}\plus{}\sin\frac{C}{2}\geq1\plus{}\sqrt{\frac{r}{2R}}.$[/quote]
Let $ b\plus{}c\minus{}a\equal{}x,$ $ a\plus{}c\minus{}b\equal{}y$ and $ a\plus{}b\minus{}c\equal{}z.$
$ \sin\frac{A}{2}\plus{}\sin\frac{B}{2}\plus{}\sin\frac{C}{2}\geq1\plus{}\sqrt{\frac{r}{2R}}\Leftrightarrow$
$ \Leftrightarrow\sum_{cyc}\sqrt{\frac{1\minus{}\frac{b^2\plus{}c^2\minus{}a^2}{2bc}}{2}}\geq1\plus{}\sqrt{\frac{\frac{S}{p}}{2\cdot\frac{abc}{4S}}}\Leftrightarrow$
$ \Leftrightarrow\sum_{cyc}\sqrt{\frac{(a\plus{}b\minus{}c)(a\plus{}c\minus{}b)}{4bc}}\geq1\plus{}\sqrt{\frac{(a\plus{}b\minus{}c)(a\plus{}c\minus{}b)(b\plus{}c\minus{}a)}{4abc}}\Leftrightarrow$
$ \Leftrightarrow\sum_{cyc}\sqrt{(x\plus{}y)xy}\geq\sqrt{(x\plus{}y)(x\plus{}z)(y\plus{}z)}\plus{}\sqrt{2xyz}\Leftrightarrow$
$ \Leftrightarrow\sum_{cyc}\sqrt{x(x\plus{}y)(x\plus{}z)}\geq2\sqrt{xyz}\plus{}\sqrt{2(x\plus{}y)(x\plus{}z)(y\plus{}z)}.$
$ \sqrt{(x\plus{}y)(x\plus{}z)}\geq x\plus{}\sqrt{yz}.$
Let $ x\equal{}u^2,$ $ y\equal{}v^2$ and $ z\equal{}w^2,$ where $ u,$ $ v$ and $ w$ are non-negative numbers.
Id est, it remains to prove that
$ u^3\plus{}v^3\plus{}w^3\plus{}uvw\geq\sqrt{2(u^2\plus{}v^2)(u^2\plus{}w^2)(v^2\plus{}w^2)},$ which is true by SOS.

Thanks so much...But have to use Schur...

I have a new idea.We have that $ r\equal{}4Rsin\frac{A}{2}sin\frac{B}{2}sin\frac{C}{2}$,so the ineq is equivalent to:
$ x\plus{}y\plus{}z$≥$ 1\plus{}\sqrt{2xyz}$ where x=sin(A/2),y=sin(B/2),z=sin(C/2)
<=> $ (x\plus{}y\plus{}z\minus{}1)^2$≥$ 2xyz$
<=> $ x^2\plus{}y^2\plus{}z^2\plus{}2xy\plus{}2yz\plus{}2zx\minus{}2(x\plus{}y\plus{}z)\plus{}1$≥$ 2xyz$
Since $ x^2\plus{}y^2\plus{}z^2\plus{}2xyz\equal{}1$,it remains to prove that:
$ xy\plus{}yz\plus{}zx\plus{}1$≥$ x\plus{}y\plus{}z\plus{}2xyz$
<=> $ 1\minus{}(x\plus{}y\plus{}z)\plus{}xy\plus{}yz\plus{}zx\minus{}xyz$≥$ xyz$
<=>$ (1\minus{}x)(1\minus{}y)(1\minus{}z)$≥$ xyz$  (*)
We have to prove (*) if $ x^2\plus{}y^2\plus{}z^2\plus{}2xyz\equal{}1$and$ x,y,z>0$(**)
Since x,y,z satisfy (**),we can let x=cosM,y=cosN,z=cosP where MNP is an acuted-angle triangle.
(*)<=>$ (1\minus{}cosM)(1\minus{}cosN)(1\minus{}cosP)$≥$ cosM cosN cosP$(this is very easy to prove)

[quote="SUPERMAN2"]I have a new idea.We have that $ r \equal{} 4Rsin\frac {A}{2}sin\frac {B}{2}sin\frac {C}{2}$,so the ineq is equivalent to:
$ x \plus{} y \plus{} z$≥$ 1 \plus{} \sqrt {2xyz}$ where x=sin(A/2),y=sin(B/2),z=sin(C/2)
<=> $ (x \plus{} y \plus{} z \minus{} 1)^2$≥$ 2xyz$
<=> $ x^2 \plus{} y^2 \plus{} z^2 \plus{} 2xy \plus{} 2yz \plus{} 2zx \minus{} 2(x \plus{} y \plus{} z) \plus{} 1$≥$ 2xyz$
Since $ x^2 \plus{} y^2 \plus{} z^2 \plus{} 2xyz \equal{} 1$,it remains to prove that:
$ xy \plus{} yz \plus{} zx \plus{} 1$≥$ x \plus{} y \plus{} z \plus{} 2xyz$
<=> $ 1 \minus{} (x \plus{} y \plus{} z) \plus{} xy \plus{} yz \plus{} zx \minus{} xyz$≥$ xyz$
<=>$ (1 \minus{} x)(1 \minus{} y)(1 \minus{} z)$≥$ xyz$  (*)
We have to prove (*) if $ x^2 \plus{} y^2 \plus{} z^2 \plus{} 2xyz \equal{} 1$and$ x,y,z > 0$(**)
Since x,y,z satisfy (**),we can let x=cosM,y=cosN,z=cosP where MNP is an acuted-angle triangle.
(*)<=>$ (1 \minus{} cosM)(1 \minus{} cosN)(1 \minus{} cosP)$≥$ cosM cosN cosP$(this is very easy to prove)[/quote]
Nice solution  :)


For those who don't know about it, N is a flash game in which you control a ninja who has to traverse 5 rooms within 90 seconds.

http://www.harveycartel.org/metanet/n.html

It is here that we talk about it.

I've been playing for about a year and a half, although I'm trying to cut down on my playing time to study for the AIME. Have you heard that N just got released for xbox live? As soon as my 360 finishes getting repaired, I'm probably going to be spending far too much time playing that, it's supposed to be awesome.

[url=http://numa.notdot.net/browse?sort=rating&order=desc&q=author:miyoser]My N Maps[/url]
To play:
Open up N.
Press "~".
Copy the data from the text box in the map you want to play.
Paste it into the upper text box in N.
Hit "L" to load.
Hit "P" to play.
To restart, hit "~", "R", "P".

[quote="andersonw"]I play, but I really stink at it. But I did beat this crazy custom level called DoorEerie5.

One of my friends has beat every single group from 00-99 and has every color ninja, he's crazy at it.[/quote]Most people on the Forums have.  I don't really try to, though.   I prefer maps by the community to the MetaNet maps.

I've played it. I'm not very good at it, though.

[quote="7h3.D3m0n.117"]Hm for the Xbox is it supposed to be from the Arcade or is it an actual game?[/quote]

Arcade[/i]

Nothing shows when I click on the link...

I can't get past these:
Episode 69, Level 2
Episode 85, Level 3
Episode 96, Level 4
70s column: haven't tried yet

70s is my favorite of all the columns.  Have you downloaded NReality and played Legacy?

I've played it and beat the entire game (well the original 100 at least).

ZOMG N game N game N game. Thats all our entire grade did during school last year it was a freaking competition. We had a few guys beat every level and I think I got most of the colors I dont remember. Not many have played since.

THAT GAME IS AWESOME!!!

I'm serious..but I have only beaten 2 columns (I have the olive green flavor I think) and my records are horrible :) .

So back to playing N to beat my record. :D

So I was going to make this thread only to discover that it had already been created, hence the revival.

Looks like there's a sizeable proportion of people here who play N... It has been recently announced that after half a decade or so [url=http://forum.ninjarobotyeti.com/viewtopic.php?f=7&t=6040]N v1.5 is coming out[/url] so those long waits of anticipations should hopefully be coming to an end. Woot.

If there are any N fans still here then I'd like to attract your attention to [url=http://www.diygamer.com/2010/06/torchlight-tournament/]this thingy[/url] which seems to be intensifying at the moment. Vote for N, you know you want to.

~ 999_Springs (Destroyer of the N highscore board with [url=http://n.infunity.com/playerLvTop20.php]901[/url] Nreality level highscores, [url=http://n.infunity.com/playerEpTop20.php]162[/url] NReality episode highscores and 38 metanet highscores. Formerly a forum moderator.)

Reviving threads is all right as long as the message is pertinent, as 999Springs' was.

[quote="andrewjjiang97"]Please do not bump 2-year-old threads.[/quote]
You have obviously not played N.
[quote="999Springs"][url=http://forum.ninjarobotyeti.com/viewtopic.php?f=7&t=6040][size=200]N v1.5 is coming out[/size][/url][/quote]
Any member of the N community would agree that [i]this[/i] official announcement by M&R is sufficiently significant to warrant a revival of the 2-year-old thread.

PWNED.

[hide=" "]why u looking at this?[/hide]

Pwned? I don't think andrewjjiang97 has a point.

[quote="andrewjjiang97"][quote="PhireKaLk6781"]According to Valentin Vornicu, Fun Factory Threads should never be revived unless for emergency reasons, and that instead a thread marked as a revival thread should be made. Since this thread has already been well revived, I'll let it be.[/quote][/quote]

Dude stop backseat modding.

I'm not. I'm merely stating a hard defense by a respected person.

[quote="metainf"]N is really fun.
Being a very curious person, I try to hack the code used to create maps in the computer version.
I have found out some pretty odd stuff.
But I can't really remember it all.
I remember that doing something to a twamp code makes it stay still, and the shock pad some where else.[/quote]
Hack the code!? Wow. The fact that your username only differs in the last letter from a well-known highscorer (who is better than me) is very disturbing.


group theory
abstract algebra
vector
superior algebra
superior algebra unsolved

Let $ K/F$ be any finite extension and let $ \alpha \in K$. Let $ L$ be a Galois extension of $ F$ containing $ K$ and let $ H \leq \text{Gal}(L/F)$ be the subgroup corresponding to $ K$. Define the norm of $ \alpha$ from $ K$ to $ F$ to be $ N_{K/F}(\alpha)\equal{}\prod_{\sigma}\sigma(\alpha)$, where the product is taken over all the embeddings of $ K$ into an algebraic closure of $ F$ (so over a set of coset representatives for $ H$ in $ \text{Gal}(L/F)$ by the Fundamental Theorem of Galois Theory). This is a product of Galois conjugates of $ \alpha$. In particular, if $ K/F$ is Galois this is $ \prod_{\sigma \in \text{Gal}(K/F)}\sigma(\alpha)$

(a) Prove that $ N_{K/F}(\alpha \beta)\equal{} N_{K/F}(\alpha)N_{K/F}(\beta)$, so that the norm is a multiplicative map from $ K$ to $ F$.

[This may be trivial, but I don't know how to show this.]

Actually it is trivial. Does the set of embeddings depend on the element $ \alpha$?

Also, an equivalent formulation is that since $ K$ is a vector space of $ F$, we have that $ t_\alpha: K\to K$ by $ u \mapsto \alpha u$ is a linear transformation. Then $ N_{K/F}(\alpha)\equal{}\det (t_\alpha)$.

So now that is just saying that the determinant of products is the product of determinants, which is true.

(I learned this definition, and then had to derive that the one you gave is equivalent).


Let $ n$ be an integer divisible by $ 4$. Prove that $ n^2 \plus{} 2$ can be written as $ a^4 \plus{} b^4 \plus{} c^4\plus{}d^4 \minus{} 4abcd$ for some non-negative integers $ a$, $ b$, $ c$, and $ d$.

(Even though you don't need to do this for a rigirous proof, it would be helpful for the sake of discussion to show how you came up with your solution).

Um, which one is it? $ c^{d4}$ or $ c^4\plus{}d^4$?

Altheman meant $ a^4 \plus{} b^4 \plus{} c^4 \plus{} d^4 \minus{} 4abcd$.

[hide]
I don't know the solution to this one...  :maybe:  :oops:
[/hide]

[hide="Go Excel"]
$ (0,0,1,1)$ for $ 2$, and $ n\equal{}4k, k\ge 1 \Longrightarrow (k\minus{}1,k,k,k\plus{}1)$.   :P
[/hide]

[quote="azjps"][hide="Go Excel"]
$ (0,0,1,1)$ for $ 2$, and $ n \equal{} 4k, k\ge 1 \Longrightarrow (k \minus{} 1,k,k,k \plus{} 1)$.   :P
[/hide][/quote]

I think you mean $ |n|\equal{}4k$, since $ n$ can be negative.

You want to concentrate on the content problem instead of getting hung up on trivialities. It is obvious that I made a typo. This reminds me of another example that I just saw was that somebody had a problem to prove that ABCD or whatever is cyclic. Then somebody comments; hey its ABDC or whatever. Its obvious that is the same thing.

@azjps, (-1,0,0,1) works for $ 2$...$ 0$ is certainly an integer divisible by $ 4$ so your formula still holds.

@distracted523, you don't need absolute value bars, you can say n=4k where k is an integer.

Not to be argumentative, but he did say $ k \geq 1$...unless I'm misinterpreting something. :whistling:

Whoops, I changed my post after I noticed that it said non-negative $ a,b,c,d$ and thought all the variables were non-negative (but I guess it’s kind’ve clear anyway).  :oops:

I know that but I'm saying that if you are correcting him that the better way to correct it is to say that $ k$ is an integer instead of taking absolute value signs...

Anyway, the fact that $ n$ can be negative is not important because $ n^2\equal{}(\minus{}n)^2$ or whatever.

Note that $ a^4 \plus{} b^4 \plus{} c^4 \plus{} d^4 \minus{} 4abcd \equal{} (a^2 \minus{} b^2)^2 \plus{} (c^2 \minus{} d^2)^2 \plus{} 2(ab \minus{} cd)^2$.  We want $ ab \minus{} cd \equal{} \pm 1$ and $ c^2 \equal{} d^2$, hence $ ab \equal{} c^2 \pm 1$ and we can take $ a \equal{} c \plus{} 1, b \equal{} c \minus{} 1$.  (Incidentally, the above proves AM-GM in four variables.  There is a lovely generalization in a thread that I can never find.)

http://www.artofproblemsolving.com/Forum/viewtopic.php?t=115712

It is pretty crazy how he came up with that.


function
induction
number theory unsolved
number theory

Fìind all function$ f: Q\to Q$ such that
$ f(x\plus{}y\plus{}z)\plus{}f(x\minus{}y)\plus{}f(y\minus{}z)\plus{}f(z\minus{}x)\equal{}3f(x)\plus{}3f(y)\plus{}3f(z)$ for all $ x,y,z\in Q$

$ x\equal{}y\equal{}z\equal{}0$ gives $ f(0)\equal{}0$.
$ y\equal{}0,z\equal{}0$  gives $ f(x)\equal{}f(\minus{}x)$.
$ y\equal{}1 ,z\equal{}0$ gives $ f(x\plus{}1)\plus{}f(x\minus{}1)\equal{}2f(x)\plus{}2f(1)$ ,
by induction for every natural number $ n ,f(n)\equal{}f(1)n^2$.
for $ x\equal{}(n\minus{}1)y ,z\equal{}0$ we have $ f(ny)\equal{}2f((n\minus{}1)y)\plus{}2f(y)\minus{}f((n\minus{}2)y)$. by induction again we obtain $ f(ny)\equal{}n^2f(y)$.
now let $ x\equal{}p/q$ then $ f(1)p^2\equal{}f(p)\equal{}f(qx)\equal{}q^2f(x)$.


if u qualify for AIME, can u take both AIME 1 and 2?

[quote="gauss1181"]Yeah. AIME 1 and 2 are just there in case if a person can't attend on one date due to schedule conflicts, so he/she can just attend on the other date. It doesn't make a difference, but you could do it on both dates and they judge your best score for your USAMO qualification.[/quote]

No, that's not true. You may take the AMC on both dates, but you may only take the AIME once. The second date is only open to those who have conflicts for the first date.

Twiz is correct.  If you qualify, you don't even get a choice of which one to take.  The only way you can take the AIME-II is if you're unable to take AIME-I for some legitimate reason, like you're sick that day, or your school is on spring break.

2007 AIME Teachers' Manual, page 5, section VII, last sentence:
"Under no circumstances can a student take both AIME's."

http://www.unl.edu/amc/d-publication/d1-pubarchive/2006-7pub/07-AIME-TM/TM-%20AIME,2007.pdf

Steve Dunbar
MAA Director, American Mathematics Competitions.

Oh, now i get it. you can take AMC 10/12 twice but not AIME...


What is the value of $ \displaystyle{(2^3)}^{\frac{4}{3}}$?

The two $ 3's$ cancel out since the exponents multiply so you get $ 2^4\equal{}\boxed{16}$.

Darn beaten


How large should be the pressure $ P$, pumped by a pump to supply water with velocity $ v$ to a part of a building at the height $ H$, where it is consumed. The liquid may be considered as incompressible and inviscid and the flow, as steady and irrotational. (Hint: take into account that the water inside the pump has zero velocity and that it is consumed at the atmospheric pressure $ P_{0}$).

Use equation of continuity, conservation of energy, Bernoulli's equation, etc.

CANT U GIVE THE PROBLEM WITH SOME ACTUAL DATA???LIKE HEIGHT AND CSA OF THE PIPE etc??????

That isn't necessary. In my opinion, we have all that we require.


[i]Find all functions $ f \ : \ N \ \to \ N$ that satisfy two conditions :

$ 1 /  f \left( xf(y) \right) \ = \ y f(x)  \  \forall \  x \ ; \ y \in \mathbb N$

$ 2/ f(p) = p$  for all primes $ p$[/i]

There fre not suth functions.
I think $ f: Q_\plus{}\to Q_ \plus{}$.

[quote="Rust"]There fre not suth functions.
I think $ f: Q_ \plus{} \to Q_ \plus{}$.[/quote]

Obviously f(x)=x is a satisfactory function.

[quote="nguyenvuthanhha"]$ f(xf(y)) \equal{} yf(x) \\ f(p) \equal{} p$[/quote]
1.clearly$ f(f(y)) \equal{} yf(1)$.
Hence $ fof$ is bijective. so is $ f$.
2.$ f(f(1)) \equal{} f(1)$. so $ f(1) \equal{} 1.$
3.if $ x \equal{} f(z), f(f(x)f(y)) \equal{} yx$ or
$ f(xy) \equal{} f(x)f(y).$
4.rest follows from multiplicative character of $ f$.

I think
$ \frac{1}{ f \left( xf(y) \right) \ }= \ y f(x) \ \forall \ x \ ; \ y \in \mathbb N$

$ \frac{2}{ f(p)} = p$  for all primes $ p$

[quote="Rust"]I think
$ \frac {1}{ f \left( xf(y) \right) \ } = \ y f(x) \ \forall \ x \ ; \ y \in \mathbb N$

$ \frac {2}{ f(p)} = p$  for all primes $ p$[/quote]
Is it so, Hmm... :maybe:  :maybe:


geometry
arithmetic sequence
number theory proposed
number theory

Let a, b, and c be the sides of a Pythagorean triangle, with c the hypotenuse.
[b](a)[/b] Show that c-a and c+a cannot both be sides of single Pythagorean triangle.
[b](b)[/b] Show that  none of $c^2+4ab$, $c^2-4ab$, or $c^2-9a^2$  can be square

This is going to be long and fatiguing, and I wouldn't have posted it if you hadn't posted it in the geometry section (I will later move it into number theory)... anyway, I'm glad to have a solution (hope that there are no mistakes), so I leave simplifying it to others.

[color=blue][b]Theorem.[/b] Let a, b, c be the sides of a Pythagorean triangle, with c being the hypotenuse.
[b](a)[/b] Prove that the numbers c - a and c + a cannot be two sides of one Pythagorean triangle.
[b](b)[/b] Show that $c^2+4ab$ is not a perfect square.
[b](c)[/b] Show that $c^2-4ab$ is not a perfect square.
[b](d)[/b] Show that $c^2-9a^2$ is not a perfect square.[/color]

[i]Proof.[/i] We will use some well-known facts:

[color=green][b](1)[/b][/color] No four perfect squares can form an arithmetic progression (with nonzero difference).
(References: http://www.mathlinks.ro/Forum/viewtopic.php?t=252 , http://www.mathlinks.ro/Forum/viewtopic.php?t=18624 .)

[color=green][b](2)[/b][/color] The equation $x^4-y^4=z^2$ has no solutions in positive integers x, y, z.
(Reference: http://www.mathlinks.ro/Forum/viewtopic.php?t=528 .)

[color=green][b](3)[/b][/color] The equation $x^4+y^4=z^2$ has no solutions in positive integers x, y, z.
(References: http://www.mathlinks.ro/Forum/viewtopic.php?t=47238 , http://www.mathlinks.ro/Forum/viewtopic.php?t=29429 , http://www.mathlinks.ro/Forum/viewtopic.php?t=3542 .)

Also, we assume that the Pythagorean triangle with sides a, b, c is primitive, i. e. that the numbers a, b, c are pairwisely coprime. (In fact, multiplying the three numbers a, b, c with a common factor k leaves the assertions of Theorem [b](a)[/b], [b](b)[/b], [b](c)[/b] and [b](d)[/b] invariant.)

Now to the actual proof of the Theorem:

Since a, b, c are the sides of a Pythagorean triangle, with c being the hypotenuse, we have $a^2+b^2=c^2$.

[b](a)[/b] Assume that the numbers c - a and c + a are two sides of one Pythagorean triangle. Let q be its third side. Then, either $\left(c-a\right)^2+q^2=\left(c+a\right)^2$, or $\left(c+a\right)^2+q^2=\left(c-a\right)^2$, or $\left(c+a\right)^2+\left(c-a\right)^2=q^2$.

- The first of these cases, $\left(c-a\right)^2+q^2=\left(c+a\right)^2$, yields $q^2=\left(c+a\right)^2-\left(c-a\right)^2=4ca$. Hence, 4ca is a perfect square. Thus, since 4 is a perfect square, ca must also be a perfect square. Since c and a are coprime, this means that both numbers c and a are perfect squares. In other words, $c=C^2$ and $a=A^2$ for some positive integers C and A. Then, $a^2+b^2=c^2$ becomes $A^4+b^2=C^4$, so that $C^4-A^4=b^2$. But this contradicts with [color=green][b](2)[/b][/color].

- The second of these cases, $\left(c+a\right)^2+q^2=\left(c-a\right)^2$, is impossible since $\left(c+a\right)^2+q^2>\left(c+a\right)^2>\left(c-a\right)^2$.

- The third of these cases, $\left(c+a\right)^2+\left(c-a\right)^2=q^2$, rewrites as $2\left(c^2+a^2\right)=q^2$. Hence, $q^2$ is even, so that q is even, and thus q = 2t for some positive integer t. Thus, $2\left(c^2+a^2\right)=q^2=\left(2t\right)^2=4t^2$, so that $c^2+a^2=2t^2$. Since $c^2=a^2+b^2$, this becomes $\left(a^2+b^2\right)+a^2=2t^2$, so that $2a^2+b^2=2t^2$, and $b^2=2t^2-2a^2$. Thus, $b^2$ is even, so that b is even. Now, since the numbers a, b, c are the sides of a primitive Pythagorean triangle with c being the hypotenuse and b being even, they can be written in the form $a=m^2-n^2$, b = 2mn and $c=m^2+n^2$ for some coprime positive integers m and n. Hence, the equation $2\left(c^2+a^2\right)=q^2$ becomes $2\left(\left(m^2+n^2\right)^2+\left(m^2-n^2\right)^2\right)=q^2$, or, equivalently, $4\left(m^4+n^4\right)=q^2$. Hence, $q^2$ is divisible by 4, so that q is even, and thus q = 2r for some positive integer r. Hence, $4\left(m^4+n^4\right)=q^2=\left(2r\right)^2=4r^2$, so that $m^4+n^4=r^2$. But since m, n and r are positive integers, this contradicts with [color=green][b](3)[/b][/color].

Hence, all three cases lead to an absurdum, so that our assumption was wrong, and it follows that the numbers c - a and c + a cannot be two sides of one Pythagorean triangle. Theorem [b](a)[/b] is proven.

[b](b)[/b] Assume that $c^2+4ab$ is a perfect square. Then, we have four perfect squares forming an arithmetic progression with difference 2ab (what is obviously nonzero):

$c^2-2ab=\left(a^2+b^2\right)-2ab=\left(a-b\right)^2$;
$c^2$;
$c^2+2ab=\left(a^2+b^2\right)+2ab=\left(a+b\right)^2$;
$c^2+4ab$.

This contradicts with [color=green][b](1)[/b][/color]. Hence, $c^2+4ab$ is not a perfect square, and Theorem [b](b)[/b] is proven.

[b](c)[/b] Assume that $c^2-4ab$ is a perfect square. Then, we have four perfect squares forming an arithmetic progression with difference 2ab (what is obviously nonzero):

$c^2-4ab$;
$c^2-2ab=\left(a^2+b^2\right)-2ab=\left(a-b\right)^2$;
$c^2$;
$c^2+2ab=\left(a^2+b^2\right)+2ab=\left(a+b\right)^2$.

This contradicts with [color=green][b](1)[/b][/color]. Hence, $c^2-4ab$ is not a perfect square, and Theorem [b](c)[/b] is proven.

[b](d)[/b] Assume that $c^2-9a^2$ is a perfect square, i. e. that we have $c^2-9a^2=u^2$ for some integer u. Thus,

$u^2=c^2-9a^2=\left(c-3a\right)\left(c+3a\right)$.

Hence, (c - 3a) (c + 3a) is a perfect square.

Since the numbers a and b are coprime, they cannot both be divisible by 3; hence, either one of them, or both of them are not divisible by 3. The square of a number divisible by 3 is congruent to 0 mod 3, and the square of a number not divisible by 3 is congruent to 1 mod 3. Hence, the number $a^2+b^2$ must be either congruent to 1 or to 2 mod 3. In other words, $a^2+b^2$ is not divisible by 3. Since $a^2+b^2=c^2$, this yields that $c^2$ is not divisible by 3, so that c is not divisible by 3.

On the other hand, it is well-known that a primitive Pythagorean triangle has an odd hypotenuse and one odd and one even catet. Thus, c is odd. Now, we distinguish two cases: whether a is even and b is odd, or b is even and a is odd.

[i]Case 1:[/i] The catet a is even and the catet b is odd. Then, we can write the sides a, b, c of our primitive Pythagorean triangle in the form a = 2mn, $b=m^2-n^2$ and $c=m^2+n^2$ for two coprime positive integers m and n.

Now, if the numbers c - 3a and c + 3a had a common divisor > 1, this divisor would also divide (c - 3a) + (c + 3a) = 2c and (c + 3a) - (c - 3a) = 6a. But on the other hand, this common divisor could not be divisible by 3 (in fact, it must divide c - 3a, but c - 3a is not divisible by 3, since c is not divisible by 3), and it could not be divisible by 2 (in fact, it must divide c - 3a, but c - 3a is not divisible by 2, since c is odd and a is even). Hence, this common divisor would be coprime with 2 and with 6, so that, from the fact that it divides 2c and 6a, we could conclude that it divides c and a, what contradicts with the fact that the numbers c and a are coprime. Hence, the numbers c - 3a and c + 3a have no common divisor > 1; hence, they are coprime. Thus, from the fact that (c - 3a) (c + 3a) is a perfect square, we conclude that both numbers c - 3a and c + 3a are perfect squares.

Hence, we get four perfect squares forming an arithmetic progression with difference 2a (what is obviously nonzero):

$c-3a$;
$c-a=\left(m^2+n^2\right)-2mn=\left(m-n\right)^2$;
$c+a=\left(m^2+n^2\right)+2mn=\left(m+n\right)^2$;
$c+3a$.

This contradicts with [color=green][b](1)[/b][/color]. Hence, we obtain a contradiction in Case 1.

[i]Case 2:[/i] The catet b is even and the catet a is odd. Then, since the numbers a, b, c are the sides of a primitive Pythagorean triangle with hypotenuse c, we can write them in the form $a=m^2-n^2$, b = 2mn and $c=m^2+n^2$ for two coprime positive integers m and n. Hence, $c-3a=\left(m^2+n^2\right)-3\left(m^2-n^2\right)=2\left(2n^2-m^2\right)$ and $c+3a=\left(m^2+n^2\right)+3\left(m^2-n^2\right)=2\left(2m^2-n^2\right)$. Thus,

$\left(c-3a\right)\left(c+3a\right)=2\left(2n^2-m^2\right)\cdot 2\left(2m^2-n^2\right)=4\left(2n^2-m^2\right)\left(2m^2-n^2\right)$.

So, instead of saying that (c - 3a) (c + 3a) is a perfect square, we can say that $4\left(2n^2-m^2\right)\left(2m^2-n^2\right)$ is a perfect square. Since 4 is a perfect square, we can thus conclude that $\left(2n^2-m^2\right)\left(2m^2-n^2\right)$ is a perfect square.

Now, if the numbers $2n^2-m^2$ and $2m^2-n^2$ had a common divisor > 1, then this divisor would also divide $\left(2n^2-m^2\right)+\left(2m^2-n^2\right)=m^2+n^2=c$ and $\left(2m^2-n^2\right)-\left(2n^2-m^2\right)=3\left(m^2-n^2\right)=3a$; but it could not divide 3 (since it divides c, and c is not divisible by 3), and thus, from the fact that it divides 3a, it would follow that it divides a, and thus it would be a common divisor of the numbers c and a, what contradicts with the fact that the numbers c and a are coprime. Hence, the numbers $2n^2-m^2$ and $2m^2-n^2$ have no common divisor > 1; hence, they are coprime. Therefore, from the fact that $\left(2n^2-m^2\right)\left(2m^2-n^2\right)$ is a perfect square, we can conclude that $2n^2-m^2$ and $2m^2-n^2$ are perfect squares.

Now, we get four perfect squares forming an arithmetic progression with difference $m^2-n^2=a$ (what is obviously nonzero):

$2n^2-m^2$;
$n^2$;
$m^2$;
$2m^2-n^2$.

This contradicts with [color=green][b](1)[/b][/color]. Hence, we obtain a contradiction in Case 2.

So we have obtained contradictions in both cases, and thus our assumption was wrong, so we obtain that $c^2-9a^2$ is not a perfect square. Theorem [b](d)[/b] is proven.

  Darij


use these numbers:  15, 4, 10, 19, and 22 one time with any combination of the operations *,+,-, and / to make a true equation.  u can use parenthesis.

example    19-4-15+22-10=12

can u make the numbers 0 through 7?

I could only get 2, 4, and 6

HELP!!!! PLEASE!!!!!!

:) Were you able to figure out how to get 5? I haven't figured it out yet, and will keep working on it, but if you find it, I'd like to know.

Thanks.  :lol:

i couldn't find it, but i'll have soon! :)

Thx a million times a million to the millionth power!
 :D  :D  :D  :D  :)  :D  :D  :D  :D

Can u tell me how u did that?
is there some kind of method?
i used guess and check.

I just played around with the numbers

I thought of ways to make five

20/4
4+1
3+2
6-1
etc....

And tried to break it down...

Wow. It all seems so simple to me now.  :)  :)  :)

thank you for showing me what you did! :D


A rectangular room measures 30 feet in length and 12 feet in height, and the ends are 12 feet in width. A spider rests at a point one foot up from the floor at the middle of one end. The center of the spider's web is located one foot down from the ceiling at the middle of the other end of the room. What is the shortest distance the spider can walk to get to her web?

I solved this, but my way took too long. I am hoping one of you guys knows a great way to simplify this problem.

The simplest way appears to be casework on what faces the spider traverses.  I don't think it's simpler than this because the answer will in general depend both on the dimensions of the room and on the positions of the spider and the web.

I agree with @t0rajir0u; but since this problem is not so unpleasant, which gives "partially symmetric" conditon, we can do less casework though...

Case 1: $ A_1$ to $ B_1$, passing through $ EFGH$. $ D_1 \equal{} \sqrt {42^2 \plus{} 10^2}$
Case 2: $ A_2$ to $ B_2$, passing through $ EHJI$. $ D_2 \equal{} 42$
Case 3: $ A_1$ to $ B_2$, passing through $ EFGH$, $ EHJI$. $ D_3 \equal{} \sqrt {17^2 \plus{} 37^2}$
Case 4: $ A_2$ to $ B_1$, passing through $ EHJI$, $ EFGH$. $ D_4 \equal{} \sqrt {7^2 \plus{} 35^2}$

$ D_1$ is obviously greater than $ D_2$, as is $ D_3$ than $ D_4$. $ D_4 \equal{} 7\sqrt 26$, $ D_2 \equal{} 7*6$; because $ 6^2 > 26$, $ D_4$ is the minimum value, so the answer is $ 7\sqrt 26$.

Just for the sake of curiousity.

Assume the room is 4-dimensional (30x12x12x12). (hypercube)

There is a spider.
Assume his initial position is (1,0,0,0).
His destination is (30,12,12,11)

If he can only walk on the walls of the hypercube.

What is the shortest path possible?

[hide="third dimension problem"]I'm certain that we can just use the pythagorean theorem to get

$ \sqrt{42^2\plus{}10^2}\equal{}2\sqrt{466}$[/hide]

4 dimensions.... ugh.


Let a,b,c are positive numbers such that $ a\plus{}b\plus{}c\equal{}3$.Prove that
$ \sum_{cyc}\frac {a}{b^2\plus{}2c} \ge 1$
 :D

$ \bigstar$ .Lemma :$ a\plus{}b\plus{}c\equal{}3;a,b,c \ge 0$
=>$ ab^2\plus{}bc^2\plus{}ca^2\plus{}abc \le 4$ .

$ LHS \ge \frac{9}{\sum ab^2 \plus{}2\sum ab}$
I need to prove that :
$ 9 \ge (\sum ab^2   \plus{}abc) \plus{} 2\sum ab \minus{}abc$
Use this lemma and schur  ($ abc \ge \frac{ 4(ab\plus{}bc\plus{}ca) \minus{}9}{3}$).
Setting :$ x\equal{}ab\plus{}bc\plus{}ca$
$ 5\plus{}abc \ge 5\plus{}\frac{ 4(ab\plus{}bc\plus{}ca) \minus{}9}{3}$ .
We have to prove:
$ 3  \ge (\sum ab)$ Which is obvious true

[quote="mitdac123"]$ \bigstar$ .Lemma :$ a \plus{} b \plus{} c \equal{} 3;a,b,c \ge 0$
=>$ ab^2 \plus{} bc^2 \plus{} ca^2 \plus{} abc \le 4$ .
[/quote]
excuse me.can you write the steps that we get to this?
thank you.

[quote="thegod277"]Let a,b,c are positive numbers such that $ a \plus{} b \plus{} c \equal{} 3$.Prove that
$ \sum_{cyc}\frac {a}{b^2 \plus{} 2c} \ge 1$
 :D[/quote]
$ \sum_{cyc}\frac {a}{b^2 \plus{} 2c} \ge ( a\plus{}b\plus{}c)^2\frac{1}{ab^2\plus{}bc^2\plus{}ca^2 \plus{}2(ab\plus{}bc\plus{}ca)}\ge1 <\equal{}>$
$ a^2\plus{}b^2\plus{}c^2\ge ab^2\plus{}bc^2\plus{}ca^2$ - which is obviously true.

[quote="Sergic Primazon"][quote="thegod277"]Let a,b,c are positive numbers such that $ a \plus{} b \plus{} c \equal{} 3$.Prove that
$ \sum_{cyc}\frac {a}{b^2 \plus{} 2c} \ge 1$
 :D[/quote]
$ \sum_{cyc}\frac {a}{b^2 \plus{} 2c} \ge ( a \plus{} b \plus{} c)^2\frac {1}{ab^2 \plus{} bc^2 \plus{} ca^2 \plus{} 2(ab \plus{} bc \plus{} ca)}\ge1 < \equal{} >$
$ a^2 \plus{} b^2 \plus{} c^2\ge ab^2 \plus{} bc^2 \plus{} ca^2$ - which is obviously true.[/quote]
Your last inequality is wrong,try $ a\equal{}\frac{1}{2},b\equal{}\frac{3}{2}$ and $ c\equal{}1 .$

[quote="rachid"][quote="Sergic Primazon"]

$ a^2 \plus{} b^2 \plus{} c^2\ge ab^2 \plus{} bc^2 \plus{} ca^2$ - which is obviously true.[/quote]
Your last inequality is wrong,try $ a \equal{} \frac {1}{2},b \equal{} \frac {3}{2}$ and $ c \equal{} 1 .$[/quote]
I think, for $ a \equal{} \frac {1}{2},b \equal{} \frac {3}{2}$ and $ c \equal{} 1$ the last inequality is obviously true. :wink:

[quote="rachid"][quote="Sergic Primazon"][quote="thegod277"]Let a,b,c are positive numbers such that $ a \plus{} b \plus{} c \equal{} 3$.Prove that
$ \sum_{cyc}\frac {a}{b^2 \plus{} 2c} \ge 1$
 :D[/quote]
$ \sum_{cyc}\frac {a}{b^2 \plus{} 2c} \ge ( a \plus{} b \plus{} c)^2\frac {1}{ab^2 \plus{} bc^2 \plus{} ca^2 \plus{} 2(ab \plus{} bc \plus{} ca)}\ge1 < \equal{} >$
$ a^2 \plus{} b^2 \plus{} c^2\ge ab^2 \plus{} bc^2 \plus{} ca^2$ - which is obviously true.[/quote]
Your last inequality is wrong,try $ a \equal{} \frac {1}{2},b \equal{} \frac {3}{2}$ and $ c \equal{} 1 .$[/quote]
 :D  :D
$ a^2 \plus{} b^2 \plus{} c^2\ge ab^2 \plus{} bc^2 \plus{} ca^2$ - obviously true then $ a\plus{}b\plus{}c \equal{}3$

$ 1/4\plus{}9/4\plus{}1> 9/8\plus{}3/2\plus{}1/4$ :D :D

Yes, very simple  :wink:
The last is equivalent to:
\[ (a \plus{} b \plus{} c)(a^2 \plus{} b^2 \plus{} c^2) \minus{} 3(ab^2 \plus{} bc^2 \plus{} ca^2)
\]

\[ \iff \sum a^3 \plus{} \sum a^2b\ge 2\sum ab^2
\]

which is always true ;)


In the adjoining plane figure, sides $AF$ and $CD$ are parallel, as are sides $AB$ and $EF$, and sides $BC$ and $ED$. Each side has length of 1.  Also, $\measuredangle FAB = \measuredangle BCD = 60^\circ$. The area of the figure is

[asy]
size(200);
defaultpen(linewidth(0.8));
pair A = dir(145), F = A + (0,-1), E = (0,-1), C = dir(35), D = C + (0,-1), B = origin;
draw(A--B--C--D--E--F--cycle);
label("$A$",A, dir(100));
label("$B$",B,2*N);
label("$C$",C,dir(80));
label("$D$",D,dir(0));
label("$E$",E,S);
label("$F$",F,W);
label("$60^\circ$",A, 6*dir(295));
label("$60^\circ$",C, 6*dir(245));
[/asy]

$\displaystyle \textbf{(A)} \ \frac{\sqrt 3}{2} \qquad \textbf{(B)} \ 1 \qquad \textbf{(C)} \ \frac{3}{2} \qquad \textbf{(D)} \ \sqrt{3} \qquad \textbf{(E)} \ 2$

[hide]
sqrt[3]

As by cutting off FED and putting it into ABC we get a rectangle. We have the width, 1, but we need the height. 2 * sin 60 = a = sqrt[3].
[/hide]

[color=green][size=75]edited by mod[/size][/color]

You forgot to put the end [/hide] tag. ;)

[hide="Answer"]Because of the given parallels, we must have $4$ equilateral triangles of side length $1$. Therefore, we have $\frac{4\sqrt{3}}{4}=\sqrt{3}\Rightarrow \boxed{D}$.[/hide]

I'm having a hard time constructing the given figure with the 6 given points... could anyone tell me what figure it is... seriously this type of problem should definitely include a figure!


we can rewrite it to

(x+y)ˇ2+(y+1)ˇ2+(x-1)ˇ2=0

hence x=1 y=-1

[quote="stergiu"]Solve the equation :

   $2x^{2}+2y^{2}+2xy-2x+2y+2 = 0$

 that is ,  find reals $x,y$ wich satisfy the above equation.

 [u]Babis[/u][/quote]

[hide="solution"]$(x+y)^{2}+(x-1)^{2}+(y+1)^{2}=0$

Since a square is at least 0, $x+y=x-1=y+1=0$ so $x=1$ and $y=-1$.[/hide]

[quote="surshu"][hide]we can rewrite it to

$(x+y)^{2}+(y+1)^{2}+(x-1)^{2}=0$

hence $x=1$ $y=-1$[/hide][/quote]
The funny thing is, if you graph it, it is just a point a $(1,-1)$.

By the way, I know you are new here, but you should hide your solutions. Just highlight whatever you want to be hidden and then hit the "hide" button above.

[hide]blah[/hide] is really [code][hide]blah[/hide][/code]
[hide="Look, nothing is hidden"]nothing[/hide] is [code][hide=Look, nothing is hidden]nothing[/hide][/code]

(these were made with codes, which are basically the same)


[hide]The answer is 55^2 or 3025.[/hide]

I agree with Inspired by Nature that you shouldn't give hints ( but answers stay :D). Anyway the solution is:
[hide]$1^3+2^3+3^3+...+10^3=(1+2+4+5+...+10)^2=$
$=\frac{(10\times11)^2}{2}=55^2=3025$[/hide] :|
Actually I kinda like this new $\LaTeX$ :roll:

[quote="Inspired By Nature"][quote="pgpatel"]Alex is playing with cubes. First he takes a unit cube. Then he takes a 2 by 2 by 2 cube and breaks it up into unit cubes and takes them. He does so for cubes up to a 10 by 10 by 10 cube. How many unit cubes does he have?
Countdown level :D
[hide="hint"]$1^3+2^3+3^3...+x^3=(1+2+3+...+x)^2$[/hide]
[hide="bigger hint"]$(55+5)(55-5)=55^2-25$[/hide]
[hide="answer"]$3025$[/hide]
Is it just me or does the latex look different?[/quote]
You shouldn't give any hints. That way its a little more challenging.  :idea:[/quote]
Don't look at them then, they're hidden for a reason.

[quote="Treething"]Not everybody is as smart as you think you are.[/quote]
You are making an assumption. I never said I was smart. I simply made a remark. I meant no offence.  :|

[hide]$1^3+2^3+3^3...+10^3=(1+2+3...+10)^2=55^2=3025$[/hide]

This identity is on one of the flash videos on the AoPS home page.


For positive real numbers $a,b,c,d$ such that $abcd=1$ prove the inequality

$ a^2+b^2+c^2+d^2 +ab+ac+ad+bc+bd+cd \geq 10$

Sorry, but I don't remember if it was already posted !!!

[quote="grobber"]Isn't this just a trivial application of AM-GM for the ten numbers?[/quote]

I guess you are only allowed to use $x^2\ge 0$ for this particular problem. :-)


inequalities
Functional Analysis
topology
advanced fields
advanced fields unsolved

let $E,F$ be two Banach spaces. Prove that a linear operator $T: E\to F$ is compact if, and only if, $T^{*} : F^{'} \to  E^{'}$ is compact.

$x,y$ denote the typical elements of $E,F$ respectively, while $x^*,y^*$ denote the elements of $E^*,F^*$ respectively. $E(0,1)$ is the closed unit ball of $E$, and we use similar notations for $F,E^*,F^*$. Also, in general, for a Banach space $X$ and $x\in X,\ x^*\in X^*,\ (x,x^*)$ will denote $x^*(x)$.

$T\Rightarrow T^*$

Fix $\varepsilon>0$. We want to find $y_i^*,\ i=\overline{1,n}$ such that for every $y^*\in F^*(0,1)$ there is $i\in\overline{1,n}$ such that $\forall x\in E(0,1),\ |(x,T^*y^*-T^*y_i^*)|<\varepsilon$. The last inequality can be rewritten as $\forall y\in T(E(0,1)),\ |(y,y^*-y_i^*)|<\varepsilon$. The Banach-Alaoglu Theorem states that the unit ball of $F^*$ is weak$^*$ compact, so if we manage to show that $\Omega=\{\omega^*\in F^*(0,1)\ |\ |(y,\omega^*)|<\varepsilon,\ \forall y\in T(E(0,1))\}$ contains a weak$^*$ open set (open in $F^*(0,1)$, that is) around the origin of $F^*$ then we're done, because from the cover of $F^*(0,1)$ with sets of the form $y^*+\Omega$ we'll be able to extract a finite subcover $y_i^*+\Omega,\ i=\overline{1,n}$, and this is exactly what we want.

Let $y_j,\ j=\overline{1,m}$ be a system of points in $T(E(0,1))$ such that for all $y\in T(E(0,1))$ there is a $j\in\overline{1,m}$ such that $\|y-y_j\|<\frac\varepsilon 2$. Suppose $\|\omega^*\|\le 1$ and $|(y_j,\omega^*)|<\frac\varepsilon 2,\ \forall j$. Then, given $y\in T(E(0,1))$, we have $|(y,\omega^*)|\le|(y_j,\omega^*)|+|(y-y_j,\omega^*)|<\frac\varepsilon 2+\frac\varepsilon 2=\varepsilon$. This means that the weak$*$ open subset of $F^*(0,1)\ \{\omega^*\in F^*(0,1)\ |\ |(y_j,\omega^*)|<\frac\varepsilon 2,\ \forall j\}$ is contained in $\Omega$, which is what we wanted to prove.

$T^*\Rightarrow T$

By the above, $T^{**}: E^{**}\to F^{**}$ is compact. It's easy to see that the image through $T^{**}$ of $E(0,1)$ ($E,F$ are regarded as subspaces of $E^{**},F^{**}$ respectively through the canonical embedding) is the image through $T$ of the same unit ball, and it's a subset of the image through $T^{**}$ of $E^{**}(0,1)$, which is relatively compact in $F^{**}$. Since $F$ is embedded in $F^{**},\ T(E(0,1))$ is relatively compact.


AMC
AIME
trigonometry
analytic geometry
AMC 12
AMC 12 B
AMC 10

How do you prevent yourself from procrastinating? How do you motivate yourself to work when you've temporarily lost the drive to do so?

[hide="Story"]Today I picked up Aops volume 2 and tried to go through the chapter on complex numbers. I was really intimidated by the chapter because I hadn't seen any of the topics in it before, and I ended up giving up and putting the book down after just an hour or so. I had planned to spend the whole day studying from the book because the AIME is coming in a few weeks. Instead, I sat around in a dazed state, overate, and watched a video. Suddenly after a whole day wasted, I became interested in reading the chapter again. Where would be some places that I went wrong? How do you get over initial discouragement?

[hide="Digression"]On a side note, supposing I know just enough information to theoretically be able to do all the problems on an AMC 10, what topics should I learn about before the AIME? The AMC forum is blocked now... [/hide][/hide]

Ah, the joys of procrastinating. In elementary school, you do that as you please. But as you enter AIME, MC, High School and Middle School, it does change your ways..I usually procrastinate for huge projects like the Science Fair. I'm doing it right now. But, I do give in and stress so much I start doing it the week before it's due. I hate not to get the full credit I deserve for procrastinating, and teachers just under appreciate it.

[hide]My Story:

Once I don't feel like adding the final touches to something, I leave it where I left it, and pick up my iPod. For a few (hours, days, minutes, weeks, insert time period here) I just lay around, not doing anything or reading...After that, I pick up the homework, but not like you said, I find no point whatsoever in doing it, but common sense and the picture of me being beaten (okay, not beaten, but maybe yelled at) motivates me. A little.[/hide]

I was also recently reading the Complex Numbers chapter in AoPS vol 2, and I finished everything right up to the hyperbolic sines and cosines. I guess what motivated me through was a drive to understand how to solve problems like "Find all 6th roots of i."

Procrastination is a difficult ideal to go through, especially with schoolwork. The general idea is to start a significant portion of your work, and then procrastinate to the point where making up the work in a short time will not be as dire.

Initial discouragement is also tough when you have to learn something. Either the material is far unlike anything you've seen, or you're just not understanding the concept fully enough to practice some simple questions.

If the material is a bit hazy (like the trig part in complex numbers), then I would review the concepts used in that chapter. For the complex numbers chapter, I reviewed polar coordinates and basic polar trig.

If you can't grasp the concepts then go slowly. There's no need to rush (unless you're procrastinating that is   :lol: ). Just because they give $ x \equal{} r \cos x$ you don't have to take it for their word. Figure out why that makes sense rather than go on blind faith.

If you can't grasp the context, I'd ask someone. Not procrastinate. Challenging things [i]will[/i] get you motivated enough.

[quote="D3m0n Shad0w"]If you can't grasp the context, I'd ask someone. [/quote]

Well that too. But by figuring out the hard stuff by yourself, you'll improve in your work habits. Otherwise, you might end up yelling for help everytime a question mark appears over your head, which is [i]not[/i] a good way to learn.

But, if you [i]truly[/i] did not understand, researched around, asked here on AoPS, or just are the impatient type, I would still ask. Example: We are currently doing Science Fair, and I'm just starting my research report. I go for the basic stuff on BrainPop for information about photosynthesis, which is very overall and I still do not grasp the concept. Then, I ask my brother for his biology text. That is much clearer, and the rest of the websites I scour through do not make any sense. I read the biology text, and the only thing I understand is Light waves. And only partially. Therefore, I ask my brother. Yes, he did explain things to me, yes, the rumors are true, I did say "Phossy",  :blush: , but still. Explaining things made me get the base of a truly great research report, and even more knowledge to base off of.

Hmmm...
I seem to have a completely different problem. I don't work as efficiently as I should because I get distracted not by other things but by the work itself.

Ya, I think I qualify as an enriched learner rather than accelerated, but I don't know a ton about either.

I think "I better do (insert task here).", and then I don't do it. Then I get really annoyed because I wasted the whole day and I go back to what I should have done, and do it. It gets really annoying.

The power of procrastination... If something doesn't interest me. It dies off and I won't do it.  Unless someone else forces me to  :wink: What's really stupid is that in Language Arts, we have to make a Mandala that shows our inner and outer selves (LOL)... yep.  It's. STUPID.

I used to procrastinate a lot when it comes to tasks I don't consider too productive, such as packing up, waking up in the morning (this counts, right?) and sending a whole bunch of necessary emails. Now I find joy in completing everything early as fast as I can, though sometimes I do it TOO fast. Only problem is, it feels like running in a hamster cage sometimes: new tasks pile up when I finish the old.

Well, tracing back to the main question here, I think nothing can fully motivate you to work unless the reward at the end is worth all the pain.  Some of this comes through maturity, and knowing what's right and what's wrong, and a lot of it also comes through how [i]important[/i] your work is to you.

To stress this,

[hide="My Story of Success"]

A month ago I really wanted an iPod Touch, but I knew that nobody trusted me enough to buy one for me  :rotfl:  and nobody was willing to pay so much money on behalf of somebody else.  And I was DYING for this iPod Touch (I would think about it 24/7 and how glossy the screen would look if it were in my hand.  i dreamt about it, whined about it and argued about going to the Apple Store EVERY DAY... etc. you get the point), so I was willing to do anything to get it, including making every cent of the money myself.

Only two weeks later I had accumulated over 400 bucks through money earned from placing in piano competitions and some previous savings money that I had kept stowed away in my Piggy Bank  :D .  Albeit I got this money in checks of 100 dollars at a time, I was still motivated enough to work for my money even if it meant dollar by dollar.

My point being, that if scoring well on the AIME matters as much to you as getting an iPod Touch was for me, there's motivation for you.  However, if you don't [i]want[/i] it very badly and aren't [b]willing[/b] enough to work for it, then good luck on getting it.
[/hide]

[hide="My Story of Failure"]
Last night, I was prompted to start studying for the AMC12B (which is in 2 days. yikes.).  I had all the materials I needed, a whole binder of problem sets, a book on fundamental AMC topics, and AoPS.  However, I found it extremely hard to concentrate on my work.  As soon as my parents left the room, I logged onto Gmail and AIM and had conversations for the rest of the hour.  I also plugged myself into iTunes and started listening to some random music.

I felt *kind of* bad inside that I wasn't studying hard enough and I wasn't even willing enough to [i]cram[/i] for the AMC12 in two days.  I talked to my friends about how I always felt bad that I was feeling lazy, but even that wasn't enough to make me log off and get down to business.  The rest of the hour flew by and suddenly I heard my parents coming downstairs.  I had no excuse for my not-even-procrastination.

I often wish that acing the AMC12 would be as important to me as getting an iPod Touch, but I have no way to FORCE myself to LIKE preparing for the AMC12.

Currently I am wallowing in my own self-pity trying to find out how those successful people manage to concentrate so well!  :(
[/hide]

[quote="D3m0n Shad0w"]Mandala? What's that?  :wink:  Over the summer, I went to a Student Writer's Workshop, and we had to trace ourselves (well, someone did it for us) and shadow stream. Shadow streaming is just writing down whatever you're thinking (that you want to expose, thank you) and do "inner, deep thoughts". I don't get why every teacher thinks that what you write is so "intimate". Well, I ended up volunteering for me to go last, (procrastination) and I was the first one to be finished. Whether it was because I was the smallest, or shadow streaming comes easily to me, I don't know...[/quote]

Oh, and one of the teachers praised me highly for it.

Hah, I did the exact same thing this past week.

I cram and hope for the best.

It's worked fairly well for every single math contest I've done as of yet ;)

[quote="1=2"][quote="sailingdog"][quote="infinity4ever"]
[hide="My Story of Failure"]
Last night, I was prompted to start studying for the AMC12B (which is in 2 days. yikes.).  I had all the materials I needed, a whole binder of problem sets, a book on fundamental AMC topics, and AoPS.  However, I found it extremely hard to concentrate on my work.  As soon as my parents left the room, I logged onto Gmail and AIM and had conversations for the rest of the hour.  I also plugged myself into iTunes and started listening to some random music.

I felt *kind of* bad inside that I wasn't studying hard enough and I wasn't even willing enough to [i]cram[/i] for the AMC12 in two days.  I talked to my friends about how I always felt bad that I was feeling lazy, but even that wasn't enough to make me log off and get down to business.  The rest of the hour flew by and suddenly I heard my parents coming downstairs.  I had no excuse for my not-even-procrastination.

I often wish that acing the AMC12 would be as important to me as getting an iPod Touch, but I have no way to FORCE myself to LIKE preparing for the AMC12.

Currently I am wallowing in my own self-pity trying to find out how those successful people manage to concentrate so well!  :(
[/hide][/quote]

I feel exactly the same way! I'm supposed to be studying for AMC 12B right now.  :blush:[/quote]

OH MY GOD do not cram. Study before hand and relax when the test is near. If you cram, you would be trying desperately to remember what you just read, so you wouldn't get much sleep.[/quote]

Actually, cramming works quite well and can earn you a critical extra 6 points  :D

Ya, I was 5<x<7 points higher than on the A because of cramming.  :lol:

[quote="Xevarion"]Cramming is not a good strategy. It works in the sense that it probably will get you a couple more questions, especially when the topics covered on the tests you take are predictable and crammable and not too diverse. It's also your only choice if you have a few days until the test and you're not fully prepared.

But eventually one (or both) of the following things will happen:
1) You will be taking tests where the topics that are hard to predict, or for which you need understanding rather than memorized knowledge, and cramming will be ineffective.
2) You will be taking tests on subjects that you don't already know well enough so that you'll do decently on the tests without any studying at all. However, you'll have learned from past experience that you can do well on tests by cramming, and when you apply this strategy you will be pushing a bad grade to a mediocre grade, whereas good studying would allow you to earn a good grade (and still sleep well the night before).

My point is not that you should never cram. I'm only saying that you should be aware that one day (when you're in college, or maybe not until you're in graduate school or trying to get some kind of certification later on), you will need to be able to use other methods of studying besides just cramming, or you won't be too happy with the results.[/quote]

Yes, I realize that now from last year's Science Fair. I did a sloppy job, and even though I still got an A for participating, my teachers were NOT happy. I'm working a bit more efficiently this year (but still a fragment of cramming, but it won't come back to bite me) and my strict teacher said my mini-backboard was good to go, even though I haven't completed my experiment. He said that my hypothesis, research report, etc, etc, looked like I had put forth all of the effort, which I did. Spent a few hours on it. Also, I got a good example of photosynthesis from my brother, which he explained to me.

Between being able to do the work right on, versus procrastinating even the littlest assignments, I like being able to relax and not have my stomach fluttering around with butterflies.

In addition to Xevarion:

Interesting, neither of those two reasons are the main reason I consider cramming a bad idea. The main reason is that the brain doesn't work that way. Cramming can get you a higher score on a test, but you will forget all that material rather quickly. Studying a little everyday for a long time is how you actually learn and remember.

This is especially bad for cumulative finals (which most math classes have). If you have crammed for all the tests before the final, you will probably have to relearn everything again instead of just leisurely brushing up on the details.

My point is that in a lot of American middle and high schools, smart students can get by via the "pump and dump" studying technique. You cram before the tests, dump out the material, and forget it all. Of course the fact that you don't learn the material is bad from the point of view where it's good if you learn the material, but it is an effective strategy in terms of getting good grades with minimal effort. And this strategy will not continue to be effective forever...

In swedish grades 6-9 the principal told my father that they had students before with similarly good grades (I went to a private school so grades were not that easy), but no one who worked as little as I did. He described that I got maximal grades with minimal work and expressed worry that I refused to do any sort of extra-curricular material.

I turned out just fine. These days I find that if I am not motivated to work I need the break. I end up studying about 6 days a week.

Edit: Actually now that I think about it I remember that once I got the worst passing grade on a math test because I was studying the material of a grade above me on my own, and I was not responsible enough to do that. My teacher asked if I was disappointed, and I had to hold back the tears. Sheesh.


geometry
geometric transformation
rotation
trigonometry
linear algebra
matrix
complex numbers

I worked through this problem for a while and I can't seem to find any insight into it.

The sequence $ \{ a_n \}^{\infty}_{n\equal{}1}, \{ b_n \}^{\infty}_{n\equal{}1}$ are defined through $ a_1 \equal{} \alpha, b_1 \equal{} \beta,$ and $ a_{n\plus{}1} \equal{} \alpha a_n \minus{} \beta b_n, b_{n\plus{}1} \equal{} \beta a_n \plus{} \alpha b_n$ for all $ n \geq 1.$ How many pairs $ (\alpha, \beta)$ of real numbers are there such that

\[ a_{1997} \equal{} b_1, \ \text{and} \ b_{1997} \equal{} a_1 \text{?}\]

I started to work on it but didn't get far. The terms weren't bad in terms of $ \alpha$ and $ \beta$ but I didn't see any special thing about them. Help!  :blush:

[hide="Humongous hint"] Let $ z_n \equal{} a_n \plus{} i b_n$.  ;) [/hide]
[hide="Motivation"] The rotation of the point $ (x, y)$ an angle $ \theta$ counterclockwise is given by

$ x' \equal{} x \cos \theta \minus{} y \sin \theta$
$ y' \equal{} x \sin \theta \plus{} y \cos \theta$;

in other words, it is given by the matrix $ \left[ \begin{array}{cc} \cos \theta & \minus{} \sin \theta \\
\sin \theta & \cos \theta \end{array} \right]$.  This is an immediate consequence of writing a point in polar form $ (r \cos t, r \sin t)$ and applying the angle-summation identities, but once you've seen this transformation written out explicitly the proper interpretation of this problem should jump out at you.  :)

Generally, when we analyze a system of two first-order linear recurrences, we are looking at a sequence of repeated applications of the same matrix transformation.  In this case, we have the composition of a rotation and a scaling, and the structure of the sequences becomes evident immediately.  :)  I offered the complex-number interpretation merely as the simplest way to work with this structure algebraically. [/hide]
[hide="Edit:  Further motivation"] If you play around with the sequences for awhile, you might realize that

$ a_{n\plus{}1}^2 \plus{} b_{n\plus{}1}^2 \equal{} (\alpha^2 \plus{} \beta^2) (a_n^2 \plus{} b_n^2)$.

In order for the problem condition to be satisfied, we therefore require that $ \alpha^2 \plus{} \beta^2 \equal{} 1$ (or that $ a_n^2 \plus{} b_n^2 \equal{} 0$), which should suggest the correct approach immediately.  :) [/hide]

Thanks, I never realized the problems like this before in that manner! I'm not especially good with rotation so thanks even more for that explanation.

[hide="Well...."]
I have to go to sleep now so I don't have too much time but...

To solve for problem and your equation, I got

$ \beta^2 \plus{} \alpha^2 \equal{} (\alpha^2 \plus{} \beta^2)(a_{n}^2 \plus{} b_{n}^2)$

which implied to me that there are only five possible pairs (I didn't check them yet) ---- (-1,0), (1,0), (0,-1), (0,1), and (0,0) ----.

But as I'm typing this, I realized that the numbers don't have to be integers... Hmm.. Am I doing something wrong here (that is not going to correct method?  :oops: ) [/hide]

[hide="Solution"] Let $ z \equal{} \alpha \plus{} \beta i$.  Then $ z_1 \equal{} z$ and $ z_{n\plus{}1} \equal{} z \cdot z_n \implies z_n \equal{} z^n$.  We therefore wish to solve

$ z^{1997} \equal{} i \bar{z}$.

[b]Case:[/b]  $ |z| \equal{} 0$.  Then $ z \equal{} 0$.

[b]Case:[/b]  $ |z| \equal{} 1$.  Then $ z^{1998} \equal{} i$, which has $ 1998$ complex roots for a total of $ \boxed{1999}$ pairs.  :) [/hide]

I understand your solution little bit but I don't get how:

$ z_{n\plus{}1} \equal{} z \cdot z_n \rightarrow z_n \equal{} z^{n}$ (I'm really rusty on complex numbers blah).

Mmm... Definitely need to review complex numbers..

I haven't looked at the problem, but the issue isn't that of complex numbers but of notation. The recursion $ a_n \equal{} k \cdot a_{n\minus{}1}$ and $ a_1 \equal{} k$ gives an explicit formula of $ a_n \equal{} k^n$ (just replace them with $ z$s).  :)

So, t0r has pretty thoroughly covered the beautiful and insightful approach to this problem.  One can also solve it by more conventional but less wonderful methods.  (I'm going to use $ a$ and $ b$ instead of $ \alpha$ and $ \beta$ to save some typing.)

$ a_{n + 1} = a\cdot a_n - b\cdot b_n$ implies \begin{align*}a_{n + 1} - a\cdot a_n & = a \cdot (a_n - a \cdot a_{n - 1}) - b\cdot (b_n - a\cdot b_{n - 1}) \\
& = a\cdot (a_n - a\cdot a_{n - 1}) - b\cdot(b\cdot a_{n - 1}) \\
& = a\cdot a_n - (a^2 + b^2) a_{n - 1},
\end{align*}
so
\[ a_{n + 1} - 2a\cdot a_n + (a^2 + b^2) a_{n - 1} = 0.\]
We can do something similar for $ b_n$.  Then we use the characteristic equations to find the exact solutions of the recursions (straightforward, with tiresome arithmetic) and solve the resulting equations.  They are exactly the same as t0r's equations once we've separated out the real and imaginary part, and so the end is completely equivalent.


function
combinatorics unsolved
combinatorics

let $A=\{1,2,...,k\}$.how many sequences are there like  $(B_1,B_2,...,B_n)$ from subset of $A$  such that:$B_1\subseteq B_2\subseteq ...\subseteq B_n$

this is rather straightforward.
 every element $x\in A$ either belongs to $B_n$ or $B_n^C$.Now if it belongs to $B_n$ then there is also a unique least $i$ such that $x\in B_i$ and so also in $B_j$ for all $j\geq i$. so to each such sequence $(B_1\subseteq B_2\ldots\subseteq B_n)$ you can associate a function $f:A\rightarrow \{1,2,\ldots,n+1\}$, with $f(x)=i$ if $i$ is the smallest such integer where $x\in B_i$ and $f(x)=n+1$ if no such $i$ exists and it is easy to see that this correspondence is invertible.
thus the number of such sequences is $(n+1)^k$.


Can anybody recommend a beginner proof book for people who score around 5 on AIME?

What specifically are you looking for? How to approach proof problems? How to write rigorous proofs? What facts and theorems are helpful for proofs?

For the first and last, this site and some of the books in the "Ready for more" and "For the pros" sections [url=http://www.artofproblemsolving.com/Resources/AoPS_R_Books.php]here[/url] are good places to start. For the second, take a look at this site's "How to write a solution", seen [url=http://www.artofproblemsolving.com/Resources/AoPS_R_Articles.php]here[/url]. I don't know of detailed books that explain the process, but maybe someone else does.

thank you. i am looking for a book that covers different strategies and when to use them. Also, common tricks to use in proofs.


let the sequence $ a_{n}$ with $ a_(1)\equal{}a$ and $ a_(2)\equal{}b$ and $ a_(n\plus{}1)\equal{}a^{2}_{n}\plus{}c/a_(n\minus{}1)$ with $ ab\equal{}0$ and ($ a,b,c$ real numbers). Prove that every term is integer if and only if the $ a,b, a^2\plus{}b^2\plus{}c/ab$  are integers. :?:  :?:

[quote="Γιώργος"]Let the sequence $ a_{n}$ with $ a_{1} \equal{} a$ and $ a_{2} \equal{} b$ and $ a_{n \plus{} 1} \equal{} a^{2}_{n} \plus{} \frac {c}{a_{n \minus{} 1}}$ with $ ab \equal{} 0$ and ($ a,b,c$ real numbers). Prove that every term is integer if and only if the $ a,b, a^2 \plus{} b^2 \plus{} \frac {c}{ab}$  are integers. :?:  :?:[/quote]

I'm not sure how this works but

Since $ ab \equal{} 0$, either $ a \equal{} 0$ or $ b \equal{} 0$

Either way

$ a_3 \equal{} b^2 \plus{} \frac {c}{a}$ (If this is the case then $ a \not \equal{} 0$)
$ a_4 \equal{} (b^2 \plus{} \frac {c}{a})^2 \plus{} \frac {c}{b}$ (If this is the case then $ b \not \equal{} 0$)

Either way, $ ab \not \equal{} 0$

[quote="Γιώργος"]let the sequence $ a_{n}$ with $ a_(1) \equal{} a$ and $ a_(2) \equal{} b$ and $ a_(n \plus{} 1) \equal{} a^{2}_{n} \plus{} c/a_(n \minus{} 1)$ with $ ab \equal{} 0$ and ($ a,b,c$ real numbers). Prove that every term is integer if and only if the $ a,b, a^2 \plus{} b^2 \plus{} c/ab$  are integers. :?:  :?:[/quote]
It is not true.
Conditional is
1) $ a,b\in Z$
2)$ ab|a^2\plus{}b^2\plus{}c$


vector
linear algebra
matrix
search
conics
ellipse
induction

FIRST MATHEMATICS EXAMINATION
Duration: 4 hours
     Throughout the questions, V is supposed to be a n-dimensional vector space on the Field K (which will always be R or C). A subset F of $ L(V)$, set of endomorphisms of V is said to be triangulable if there exist a basis in V in which the matrix of ANY element of F is upper triangular. It is reminded that a subspace W of V is said stable by F if, for all $ u \in F$, W is stable by u, i.e. $ u(x) \in W$ for all $ x \in W$

     The main topic of this problem consists in looking for eigenvectors that are common to all elements of a subset F of L(V), endowed with suitable properties, the main application being to obtain some sufficient conditions for triangulation.

PART I
In the first five questions of this part, K=C.
1°) Show that, for a subset F of L(V) to be triangulable, it is necessary that the elements of F have at least one common eigenvector.

     We assume in the remaining of this part, that F is a subset of L(V) such that, for any $ u \in F$ and $ v \in F$, one have u° v = v ° u. The purpose of the following questions is to show that F is triangulable.
2°) Let $ u \in F$, $ \lambda$ is a eigenvalue of u and let $ V_{u}(\lambda)$ be the corresponding eigenspace. Show that $ V_{u}(\lambda)$ is stable by F.
3°) Show that the elements of F have a common eigenvector.
4°) Show that F is triangulable.
5°) We additionaly assume that any element of F is diagonalizable. Can we find a basis in V in which the matrix of any element of F is diagonal?
6°) Re-do the problem raised in question 2°, replacing C by R.

PART II
In this part, K = C.

     Given $ u \in L(V)$ and $ v \in L(V)$, let $ [u,v] \equal{} uo v \minus{} vo u$. A subset F of L(V) is called a Lie algebra (made of endomorphisms of V) if the following conditions are fulfilled :

(i)  F is a vector subspace of L(V)
(ii)  For any $ u \in F$ and $ v \in F$, $ [u,v] \in F$.
     We will call dimension of a Lie algebra F, which will be denoted by dim(F), its dimension as a subspace over the field K.

     Let F be a Lie albegra, we call ideal of F any subspace I of F such that $ [u,v] \in I$ for any $ u \in F$ and $ v \in I$
1°) Let F be a Lie algebra of dimension 2, such that there exists $ u_{0} \in F$ and $ v_{0} \in F$ obeying $ [u_{0},v_{0}] \neq 0$ . Let also F' ba another Lie algebra of dimension 2, obeying the same property. Show that there exists an isomorphism (vector space isomorphism) f from F to F' such that f([u,v])=[f(u),f(v)] for any $ u \in F$ and $ v \in F$

     Let F be a Lie algebra and I an ideal of F. Given a linear form l on I, we denote by W the subspace of V of vectors x such that v(x)=l(v)x for any $ v \in I$. The purpose of questions 2° to 5° is to show that W is stable by F.

     Let $ u \in F$, and x be a non-vanishing element of W ; we define by recursion a sequence $ (x_{k})$ in the following way: $ x_{0} \equal{} x$ and $ x_{k} \equal{} u(x_{k \minus{} 1})$ for any integer $ k \geq 1$.
2°) Show that, for any $ k \in N$ and any $ v \in I$, $ v(x_{k}) \minus{} l(v)x_{k}$ lies in the subspace generated by $ (x_{0},x_{1},...,x_{k \minus{} 1})$.
3°) Let U be the subspace of V generated by the vectors $ x_{k}$, where k is any positive integer. Show that U is stable by $ I\cup u$
4°) Give a relation between $ l([u,v])$ and the trace (i.e. the sum of eigenvalues) of the restriction to U of the endomorphism [u,v].
5°) Show that W is stable by F.

     A Lie algebra F is said to be solvable if their exists an increasing sequence
$ (0) \equal{} F_0 \subset F_{1} \subset ... \subset F_{p} \equal{} F$
of subspaces F such that, for any integer k obeying $ 1 \leq k \leq p$, one has :
$ [u,v] \in F_{k \minus{} 1}$ for any $ u \in F_{k}$ and $ v \in F_{k}$
6°) Show that any Lie algebra of dimension $ \leq 2$ is solvable.

     The purpose of the following questions is to proove the "Lie Theorem" which states that any solvable Lie algebra is traingulable. We consequently let F be a solvable Lie algebra.
7°) Let d=dim(F). Show that there exists an ideal I of F, of dimension d-1. Show that I is also a solvable Lie algebra.
8°) Show that the elements of F have a common eigenvector.
9°) Show that F is triangulable.
10°) Show that, conversely, any triangulable Lie algebra is solvable.
11°) Show that the result of question I.4° is a corrolary of the "Lie Theorem".

PART III
     In this part, the field K can be either R or C. For any $ u \in L(V)$, we will denote by adu the element of L(L(V)) defined by $ ad_{u}(v) \equal{} [u,v]$ for any $ v \in L(V)$.
1°) Show that, for $ u \in L(V)$ and $ v \in L(V)$, we have $ ad_{[u,v]} \equal{} [ad _{u},ad_{ v}]$.
2°) Show that, if u is a nilpotent operator in L(V), then $ ad_{u}$ is a nilpotent operator in L(L(V)).
3°) Let F and G be two Lie algebras (made of endomorphisms of V) such that $ G \subset F$. Let H be a complementary subspace of G in F, and let q be the projector on H parallel to G, i.e. the application from F to H which maps any $ u \in F$ to the unique $ v \in H$ such that $ u \minus{} v \in G$. Show that there exists one and only one linear application
$ p$: $ G \rightarrow L(H)$
such that, for any $ g \in G$ and any $ u \in F$,$ p(g)(q(u)) \equal{} q([g,u])$

     We now denote by F a Lie algebra (made of endomorphisms of V), such that any element of F is a nilpotent operator in V. The purpose of the following questions is to demonstrate the "Engel's theorem", which states that their exists a non vanishing vector $ x \in V$ such that u(x)=0 for any $ u \in F$.
4°) Let G be a second Lie algebra (made of endomorphisms of V), such that $ G \subset F$ and $ G \neq F$. We keep the same notations as in the previous question and we let $ F' \equal{} p(G)$, $ V' \equal{} H$. Show that $ F'$ is a Lie algebra ( made of endomorphisms of V'), that dim(F') < dim(F) and that any element of F' is nilpotent.
5°) Let d=dim(F). We assume that, for any finite dimensional vector space W on the field K, and any Lie algebra B ,made of nilpotent endomorphisms of W, and such that $ dim(B) \leq d \minus{} 1$, there exists a non-vanishing vector $ x \in W$ such that u(x)=0 for any $ u \in B$. We also keep the notations and hypothesis of question 4°. Show that there exists a Lie algebra $ G_{1}$,made of endomorphisms of V, obeying the following properties: $ G \subset G_{1} \subset F$
$ dim(G_{1}) \equal{} dim(G) \plus{} 1$
$ G_{1}$ is an ideal of G.
Conclude that there exists an ideal $ F_{1}$ of F, such that $ dim(F_{1}) \equal{} d \minus{} 1$.
6°) Prove Engel's theorem.
7°) Show that any Lie algebra made of nilpotent operators of de the vector space V is triangulable.

I wonder who will start to write solutions for this.  :D  Yet, what is even more strange is that they used to give this theorem in oral examination!!! Yes, at Ulm this was quite common. But hopefully there are much easier proofs of this theorem (as usual, in written exams they search for the most complicated and long solution).

Who are supposed to take this exam?
High school graduates?

It is entrance exam for undergraduates?
Or qualification exam for graduates?

It is entrance exam for undergraduates. After high-school (which I had the luck not to do there  :D ) frechmen have 2 years of preparation to enter ENS and other big schols in France and this is the entrance exam for ENS after these 2 years.

A detailed and complete solution of this problem can be found in the book:

Problèmes corrigés de Mathématiques posés aux concours des ENS

by Laurent Rosaz and Marc Zeitoun, Ellipses, Paris 1990. ISBN: 2-7298-9085-8 :lol:  :lol:  :lol:

[quote="Michael"]A detailed and complete solution of this problem can be found in the book:

Problèmes corrigés de Mathématiques posés aux concours des ENS

by Laurent Rosaz and Marc Zeitoun, Ellipses, Paris 1990. ISBN: 2-7298-9085-8[/quote]

Is there any change at all this book circulates on the internet these days? I can't order it... Links are very welcome in a private message or in this thread if the moderators allow that ;)

I don't have that book either, but why don't we try solving this problem instead? :)

Let's begin with part I then.
The 1st question is completely clear.
For the 2nd one, take some element $ x$ in $ V_{u}(\lambda),$ then $ uv(x) = vu(x) = \lambda v(x),$ we're done.
For the 3rd, 4th and 5th questions, proceed by induction on $ \dim V.$ (of course I can detail that, in case you can't manage them by yourself)
As for the 6th question, when doing the 3rd, 4th and 5th you'll notice that the hypothesis $ K = \mathbb{C}$ is essential for the 3rd and 4th ones, while it doesn't matter if $ K = \mathbb{R}$ for the 5th one.

My real interest was not the problem actually; I was searching for (harazi's) sources or where I could find those ENS exam questions. Via google I ended up on mathlinks again :D Perhaps I just can't find it on those french sites since I don't speak french. But I guess they aren't available online, and only in the private library of ENS.

Aha. :)

Well, for the entrance exam problems, you can go and check their site, namely here : http://www.ens.fr/concours/Rapports/2006/MP/ for the subjects from 2001 till 2006 (and soon 2007, I guess).
As for the oral exams, indeed there's no such thing as a link gathering them all (or at least I'm not aware of it :)), but there are some books with them, however.
So all in all, MathLinks does sound like the better place to get them online, yeah.

The post depresses me.  I almost finish my undergraduate degree and I can't understand the entrance exam problem for undergraduate.  I'm really really depressed now.  :lol: