Prove that $ \sum\frac{a^5}{b^2c^2\plus{}ab^2c}\geq \frac{1}{2(a\plus{}b\plus{}c)}$ $ \forall a, b, c>0$.
Happy Bengali New Year to all mathlinkers! :lol:
Hope you will enjoy it(though it is too easy), dear friend! :oops:

[quote="Potla"]Prove that $ \sum\frac {a^5}{b^2c^2 \plus{} ab^2c}\geq \frac {1}{2(a \plus{} b \plus{} c)}$ $ \forall a, b, c > 0$.
Happy Bengali New Year to all mathlinkers! :lol:
[/quote]
Thank you! But your inequality is wrong. :(  Try $ a\equal{}b\equal{}c\equal{}0.1.$  :wink:

[quote="arqady"][quote="Potla"]Prove that $ \sum\frac {a^5}{b^2c^2 \plus{} ab^2c}\geq \frac {1}{2(a \plus{} b \plus{} c)}$ $ \forall a, b, c > 0$.
Happy Bengali New Year to all mathlinkers! :lol:
[/quote]
Thank you! But your inequality is wrong. :(  Try $ a \equal{} b \equal{} c \equal{} 0.1.$  :wink:[/quote]
Yes,you are right.I think it must be: $ LHS \ge \frac{a\plus{}b\plus{}c}{2}$
This inequality very weak and we can find the inequalities than stronger :)
@potla: Thank you,my friend.But this inequality is false.You can edited it.Regards :wink:

[quote="quykhtn-qa1"][quote="arqady"][quote="Potla"]Prove that $ \sum\frac {a^5}{b^2c^2 \plus{} ab^2c}\geq \frac {1}{2(a \plus{} b \plus{} c)}$ $ \forall a, b, c > 0$.
Happy Bengali New Year to all mathlinkers! :lol:
[/quote]
Thank you! But your inequality is wrong. :(  Try $ a \equal{} b \equal{} c \equal{} 0.1.$  :wink:[/quote]
Yes,you are right.I think it must be: $ LHS \ge \frac {a \plus{} b \plus{} c}{2}$
This inequality very weak and we can find the inequalities than stronger :)
@potla: Thank you,my friend.But this inequality is false.You can edited it.Regards :wink:[/quote]
Sorry for the typo in my first post..... :blush:
The inequality should be
]Prove that $ \sum\frac {a^5}{b^2c^2 \plus{} ab^2c}\geq \frac {3}{4(a \plus{} b \plus{} c)}$ $ \forall a, b, c > 0$.

It might be weak, but i hope you will try to solve it now, Quy! :lol:

[quote="Potla"][/quote][quote="quykhtn-qa1"][quote="arqady"][quote="Potla"]Prove that $ \sum\frac {a^5}{b^2c^2 + ab^2c}\geq \frac {1}{2(a + b + c)}$ $ \forall a, b, c > 0$.
Happy Bengali New Year to all mathlinkers! :lol:
[/quote]
Thank you! But your inequality is wrong. :(  Try $ a = b = c = 0.1.$  :wink:[/quote]
Yes,you are right.I think it must be: $ LHS \ge \frac {a + b + c}{2}$
This inequality very weak and we can find the inequalities than stronger :)
@potla: Thank you,my friend.But this inequality is false.You can edited it.Regards :wink:[/quote][quote="Potla"]
Sorry for the typo in my first post..... :blush:
The inequality should be
]Prove that $ \sum\frac {a^5}{b^2c^2 + ab^2c}\geq \frac {3}{4(a + b + c)}$ $ \forall a, b, c > 0$.

It might be weak, but i hope you will try to solve it now, Quy! :lol:[/quote]

Sorry, but your second inequality is wrong again. Try a=b=c=0.01.

[quote="Potla"][/quote][quote="quykhtn-qa1"][quote="arqady"][quote="Potla"]Prove that $ \sum\frac {a^5}{b^2c^2 + ab^2c}\geq \frac {1}{2(a + b + c)}$ $ \forall a, b, c > 0$.
Happy Bengali New Year to all mathlinkers! :lol:
[/quote]
Thank you! But your inequality is wrong. :(  Try $ a = b = c = 0.1.$  :wink:[/quote]
Yes,you are right.I think it must be: $ LHS \ge \frac {a + b + c}{2}$
This inequality very weak and we can find the inequalities than stronger :)
@potla: Thank you,my friend.But this inequality is false.You can edited it.Regards :wink:[/quote][quote="Potla"]
Sorry for the typo in my first post..... :blush:
The inequality should be
]Prove that $ \sum\frac {a^5}{b^2c^2 + ab^2c}\geq \frac {3}{4(a + b + c)}$ $ \forall a, b, c > 0$.

It might be weak, but i hope you will try to solve it now, Quy! :lol:[/quote]
$ RHS$ must be : $ \frac{a+b+c}{2}$ ,or must be condition,my friend.

[quote="quykhtn-qa1"][/quote][quote="Potla"][/quote][quote="quykhtn-qa1"][quote="arqady"][quote="Potla"]Prove that $ \sum\frac {a^5}{b^2c^2 + ab^2c}\geq \frac {1}{2(a + b + c)}$ $ \forall a, b, c > 0$.
Happy Bengali New Year to all mathlinkers! :lol:
[/quote]
Thank you! But your inequality is wrong. :(  Try $ a = b = c = 0.1.$  :wink:[/quote]
Yes,you are right.I think it must be: $ LHS \ge \frac {a + b + c}{2}$
This inequality very weak and we can find the inequalities than stronger :)
@potla: Thank you,my friend.But this inequality is false.You can edited it.Regards [/quote][quote="Potla"]
Sorry for the typo in my first post.....
The inequality should be
]Prove that $ \sum\frac {a^5}{b^2c^2 + ab^2c}\geq \frac {3}{4(a + b + c)}$ $ \forall a, b, c > 0$.

It might be weak, but i hope you will try to solve it now, Quy! :lol:[/quote][quote="quykhtn-qa1"]
$ RHS$ must be : $ \frac {a + b + c}{2}$ ,or must be condition,my friend.[/quote]
OH! Sorry for my nuisance! I am really sorry as I was incorrect.......
Yeah, RHS = $ \frac{a+b+c}{2}$ and I am jumbled up in my inequalities.
Hope you will solve it now.... :oops:  :(

[quote="Potla"][/quote][quote="quykhtn-qa1"][/quote][quote="Potla"][/quote][quote="quykhtn-qa1"][quote="arqady"][quote="Potla"]Prove that $ \sum\frac {a^5}{b^2c^2 + ab^2c}\geq \frac {1}{2(a + b + c)}$ $ \forall a, b, c > 0$.
Happy Bengali New Year to all mathlinkers! :lol:
[/quote]
Thank you! But your inequality is wrong. :(  Try $ a = b = c = 0.1.$  :wink:[/quote]
Yes,you are right.I think it must be: $ LHS \ge \frac {a + b + c}{2}$
This inequality very weak and we can find the inequalities than stronger :)
@potla: Thank you,my friend.But this inequality is false.You can edited it.Regards [/quote][quote="Potla"]
Sorry for the typo in my first post.....
The inequality should be
]Prove that $ \sum\frac {a^5}{b^2c^2 + ab^2c}\geq \frac {3}{4(a + b + c)}$ $ \forall a, b, c > 0$.

It might be weak, but i hope you will try to solve it now, Quy! :lol:[/quote][quote="quykhtn-qa1"]
$ RHS$ must be : $ \frac {a + b + c}{2}$ ,or must be condition,my friend.[/quote][quote="Potla"]
OH! Sorry for my nuisance! I am really sorry as I was incorrect.......
Yeah, RHS = $ \frac {a + b + c}{2}$ and I am jumbled up in my inequalities.
Hope you will solve it now....   ([/quote]
It very weak and easy,my friend :wink: .
$ LHS \ge \frac{(a^3+b^3+c^3)^2}{2abc(ab+bc+ca)} \ge RHS$


Andy and Bethany have a rectangular array of numbers with $ 40$ rows and $ 75$ columns. Andy adds the numbers in each row. The average of his $ 40$ sums is $ A$. Bethany adds the numbers in each column. The average of her $ 75$ sums is $ B$. What is the value of $ \frac{A}{B}$?

$ \textbf{(A)}\ \frac{64}{225}     \qquad
\textbf{(B)}\   \frac{8}{15}    \qquad
\textbf{(C)}\    1   \qquad
\textbf{(D)}\   \frac{15}{8}    \qquad
\textbf{(E)}\    \frac{225}{64}$

[hide="Solution"]We don't have any restrictions on the numbers themselves, so we can just let them all be $ 1$. Thus, Andy gets $ 75$ for each of his sums, so the average $ A\equal{}75$. Bethany gets $ 40$ for each of her sums, so the average $ B\equal{}40$. Finally, $ \frac{A}{B}\equal{}\frac{75}{40}\equal{}\frac{15}{8}$, or answer choice $ \boxed{\mathbf{(D)}}$.[/hide]

[hide="Solution"]
Denote the sum of all the numbers in the array by $ S$.  Then, the average of Andy's sums is $ \dfrac{S}{40}$, while the average of Bethany's sums is $ \dfrac{S}{75}$.  Thus, $ \dfrac{A}{B}\equal{}\dfrac{\frac{S}{40}}{\frac{S}{75}}\equal{}\dfrac{75}{40}\equal{}\dfrac{15}{8}$, or $ \boxed{\textbf{(D)}}$.
[/hide]


calculus
integration
floor function
real analysis
real analysis solved

In the following,   $e$    denotes Napier's constant and let  [ . ]  be the integer part. Prove or disprove that if $n$ is a positive integer, then \[\int_{0}^{\infty}e^{-t}(t-1)^{n}dt = \left[ \frac{n!}{e}+\frac{1}{2}\right] \; .\]

I think it's true, but I don't have the proof. If we denote that by $a_n$, we have $a_0=1,\ a_n=(-1)^n+na_{n-1}$. Amazingly, this seems to give $\displaystyle a_n=\left\lfloor \frac {n!}e+\frac 12\right\rfloor,\ \forall n>0$ (for $n=0$ the equality doesn't hold).

Divide both sides of your recurrence by n!, and sum.  You will then get an exact expression for $a_n$ involving the finite power series of $e^{-1}$.

I wonder why I didn't see that.. There's something cool which seems to be going on as well: I think that whenever $n$ is odd, $\{\frac {n!}e\}<\frac 12$, and whenever $n$ is even, $\{\frac{n!}e\}>\frac 12$. Is this actually true? I've computed the approximations for a few values with Mathematica, and it seems correct.

it is...actually the bounds can be sharpened to $\{\frac{n!}{e}\}\geq \frac{n}{n+1}$(or something very close to it) for even n and analogous for odd n. but this is clear since $\frac{1}{e}=1-\frac{1}{2}+\frac{1}{3!}\pm...$, after multiplication with n!, all terms up to $\frac{1}{n!}$ are integral and the next one becomes $\pm \frac{1}{n+1}$ depending on whether n is even or odd. now, with a little work one can show that the other terms have opposite sign in sum(very easy if it is true...i didn't check :) ), so we get the inequality.

Peter

let's [tex]a_n[/tex] be the sequence of integrals!
notice that n!=integral(0, :inf: ,exp(-t)t^ndt)
so n!/e=integral(1, :inf: ,exp(-t)(t-1)^ndt)  "changing variables"
so the difference between [tex]a_n[/tex] and n!/e equals
 integral(0,1,exp(-t)(t-1)^ndt)=(-1)^n*integral(0,1,exp(-t)(t-1)^ndt)
so tha abs of the difference equals <1/(n+1)<=2
 because exp(-t)<1 and integral(0,1,(1-t)^n)=1/(n+1)
so the equality holds for all n>=1
bye

The file attached by [b]sam-n [/b]([i]Mehdi Hassani)[/i] is a paper published in
Journal of Integer Sequences. This paper includes a list of wonderful
equalities. Interesting paper ! Alex


Let $A=\{a_n\}_n$ and $B=\{b_n\}_n$ be two disjoint closed countable sets in a regular space $X$. Prove that $A$ and $B$ are separable using open sets in $X$.

Take $U_1$ around $a_1$ s.t. $B,\overline{U_1}$ are disjoint, then take $V_1$ around $b_1$ s.t. $\overline{V_1},A\cup\overline{U_1}$ are disjoint, then $U_2$ around $a_2$ s.t. $\overline{U_2},B\cup\overline{V_1}$ are disjoint, then $V_2$ around $b_2$ s.t. $\overline{V_2},A\cup\overline{U_1}\cup\overline{U_2}$ are disjoint, and so on (the $U$'s and the $V$'s are open). The desired open sets separating $A$ and $B$ will be $\bigcup U_i$ and $\bigcup V_i$.


In the vertexes of a regular $ 31$-gon there are written the numbers from $ 1$ to $ 31$, ordered increasingly, clockwise oriented.
We are allowed to perform an operation which consists in taking any three vertexes, namely the ones who have written $ a$,$ b$, and $ c$ and change them into $ c$, $ a\minus{}\frac{1}{10}$ and $ b\plus{}\frac{1}{10}$ respectively ( $ a$ becomes $ c$, $ b$ becomes $ a\minus{}\frac{1}{10}$ and $ c$ turns into $ b\plus{}\frac{1}{10}$
Prove that after applying several operations we can reach the state in which the numbers in the vertexes are the numbers from $ 1$ to $ 31$, ordered increasingly,anti-clockwise oriented.

We can get $ (c,b,a)$ from $ (a,b,c)$:
$ (a,b,c)\rightarrow(c,a\minus{}\frac{1}{10},b\plus{}\frac{1}{10})$,
$ (b\plus{}\frac{1}{10},a\minus{}\frac{1}{10},c)\rightarrow(c,b,a)$

We can see that we are able to change places of any two numbers. So we change 2 with 31, 3 with 30, 4 with 29, ... , 16 with 17

Done  :)

I actually got (a,b,c) from (a,b,c)  :oops_sign:

[quote="Bachukas"]We can get $ (c,b,a)$ from $ (a,b,c)$:
$ (a,b,c)\rightarrow(c,a \minus{} \frac {1}{10},b \plus{} \frac {1}{10})$,
$ (b \plus{} \frac {1}{10},a \minus{} \frac {1}{10},c)\rightarrow(c,b,a)$

We can see that we are able to change places of any two numbers. So we change 2 with 31, 3 with 30, 4 with 29, ... , 16 with 17

Done  :)[/quote]
Wrong!
$ (a,b,c)\rightarrow(c,a \minus{} \frac {1}{10},b \plus{} \frac {1}{10})$,
$ (b \plus{} \frac {1}{10},a \minus{} \frac {1}{10},c)\rightarrow(c,b,a)$
They follow that $ (a,b,c)\rightarrow (a,b,c)$ (arrange positions in triples correctly). And nothing in your post is helpful.  :wink:


Harvard
college
induction
combinatorics proposed
combinatorics

A graph $ G$ is called divisor graph if vertex $ x\in V(G)$ is connected to vertex $ y$ iff $ x|y$ or $ y|x$. The following result was proposed and solved by Le Anh Vinh (my homelander, who has been at Harvard for his masters degree)  :) :
  Prove that: for any positive interger $ n$ and for all $ 0< m\leq{n\choose 2}$ we can always find out a divisor graph of $ n$ vertices and $ m$ edges !!!
   By the way, when I said vertex $ x$ above I meant a vertex assigned to an integer $ x$.

I think mathematical induction works here.

Let $ D(n, m)$ be divisor graph with n vertices and m edges. Let $ X = {a_1, a_2, ..., a_n} (a_i >1)$ be the set of vertices of $ D(n, m)$. Then, if we add $ X' = {a_1, a_2, ..., a_n, a_1...a_n + 1)}$ to get new divisor grap with n + 1 vertices and number of edges is still m. If we take $ X' = {a_1, a_2, ..., a_n, a_1...a_n)}$ then new divisor graph has m+n edges.

Of course, we have to work a bit for small n.

Call $ t$ the integer, which satisfies $ t(t \minus{} 1)/2\le m < t(t \plus{} 1)/2.$, $ k \equal{} m \minus{} t(t \minus{} 1)/2$. We have $ k\le t$.

Let $ p_1,..,p_t,q_1,...q_{n \minus{} t}$ be $ n$ distinct prime numbers.

Define $ a_i \equal{} p_1...p_i$ for $ i \equal{} 1,2...,t$ and $ b \equal{} p_1...p_kq_1$ and. $ c_i \equal{} q_{i \plus{} 1}$ for $ i \equal{} 1,2,...,n \minus{} t \minus{} 1$

Consider the graph  $ G$ made of $ a_1,a_2,..,a_t,b,c_1,..,c_{n \minus{} t \minus{} 1}$. $ G$ has exactly $ n$ vertices and the number of edges is:

$ t(t \minus{} 1)/2 \plus{} k \equal{} m$ .

[quote="keira_khtn"]@ Namdung: Your method doesn't work. Note that: the result states that there always exists a graph of arbitrary order and size! I think the construction of the desired graph is much more sophisticated![/quote]

Why not? I try to construct D(n, m) for all n and m by induction:

For example: D(3,1): (2, 4, 5)
                    D(3,2): (2, 3, 6)
                    D(3,3): (2, 4, 8)
Then by proposed construction we have
                    D(4,1): (2, 3, 5, 31)   (ofcourse, we can take (2, 4, 5, 7)!)
                    D(4,2): (2, 3, 6, 37)
                    D(4,3): (2, 4, 8, 65)
                    D(4,4): (2, 3, 5, 30)
                    D(4,5): (2, 3, 6, 36)
                    D(4, 6): (2, 4, 8, 64)

Please, check it again and find where I do the misteake?


geometry perimeter rectangle calculus integration

[hide]18 is the Per.[/hide]

You beat me to it ;)

Although I have a solution, could others please share theirs :).

[hide]Um, from the problem we have

$2a+2b=ab$

$a=(2b)/(b-2)$

a must be an integer. b-2 cannot divide b unless it is really small so we try some small numbers and it turns out that 3*6 and 4*4 are the only possibilities. It isn't a square, so...[/hide]

[quote="bpms"]The area and perimeter of a rectangle are the same. The rectangle isn't a square. The sides are all whole numbers. What is the perimeter?[/quote]

[hide=":)"]It's 18 because 3*6=18 and 2(3+6)=18.

For some reason the first number I thought of was 18, but this could have also been solved with algebra.

[/hide]

(sry I forgot the = in my hide  :oops: )

[quote="nonie"][quote="bpms"]The area and perimeter of a rectangle are the same. The rectangle isn't a square. The sides are all whole numbers. What is the perimeter?[/quote]

[hide=":)"]It's 18 because 3*6=18 and 2(3+6)=18.

For some reason the first number I thought of was 18, but this could have also been solved with algebra.

[/hide][/quote]

Could you please hide your answer?

[hide]Let the sides of the rectangle be $x$ and $y$. Then:
2x+2y=xy trying some numbers, we get 3 and 6 as the side lengths, giving a perimeter of 18.[/hide]

[quote="mathgeniuse^ln(x)"][hide]I don't know any clever methods, but if you let x and y be two of the sides then 2x+2y=xy.  then we try some numbers and we get x=4, y=4 or x=6,y y=3 as two integral solutions, and six is the first number that shows that it is not a square.

$18$[/hide][/quote]
[quote="mathgeniuse^ln(x)"][quote="robinhe"]A non-square rectangle has integer dimensions.  The number of square units in its area is numerically equal to the number of units in the perimeter.  What is the number of units in the perimeter of this rectangle?

Does anyone know of a way that's not guess and check (which I used and didn't like)?[/quote]

[hide]Let the dimensions be $x$ and $y$

From the conditions of the problem

$2(x+y)=xy$

So $xy-2(x+y)=0$

Therefore $(x-2)(y-2)=4$ (Clever Manipulation)

So either $(x-2)=2$ and $(y-2)=2$, $(x-2)=4$ and $(y-2)=1$, and vice versa

So $(x,y)=(4,4)$, $(x,y)=(6,3)$, and $(x,y)=(3,6)$

So perimeter/area is 18 of the non-square rectangle

$18$[/hide][/quote]
What????
The way mathgenius did it is known as Simon's Favorite Factoring Trick. That's the way I did it. That way, and binouniquineist's way are the only ways that seem to be fool proof.
Any other ways???

OMG, I forgot about that.  Now I will remember it.  Thank you very much.

I am such an idiot.

[hide]
We can write an equation:
$2x+2y=xy$
If x=3 and y=6, the equation works.
So, 18.[/hide]

[hide]
[quote="bpms"]The area and perimeter of a rectangle are the same. The rectangle isn't a square. The sides are all whole numbers. What is the perimeter?[/quote]

The two sides of the rectangle are 6 and 3.

2(6+3)=2(9)=18
6*3=18
[/hide]

[quote="mathgeniuse^ln(x)"][quote="bpms"]Knowing Simon's favorite factoring trick is very important by the way.[/quote]

Thank You Captain Obvious.

But yes, that is a good trick.

Moving from the problem

Find all rectangles such that the area is 3 times the perimeter.[/quote][hide]$(x-6)(y-6)=36$ from complex zeta's favorite factoring trick
42,7
8,24
9,18
10,15
12,12
[/hide]


trigonometry
geometry
circumcircle
inequalities proposed
inequalities

Let be $ ABC$ an arbitrary triangle . Prove that :

$ \sin^4\ A\plus{}\sin^4\ B\plus{}\sin^4\ C\ge \frac{27}{2}\left(\frac{r}{R}\right)^3$ ;

$ \cos^4\ A\plus{}\cos^4\ B\plus{}\cos^4\ C\ge \frac{3}{8}\left(\frac{r}{R}\right)^4\left(5\minus{}\frac{r_a}{r}\right)\left(5\minus{}\frac{r_b}{r}\right)\left(5\minus{}\frac{r_c}{r}\right)$

where $ R$ is the radius of the circumcircle circle , $ r$ is the radius of the inscribed circle , $ r_a,r_b,r_c$ are the radius of the exinscribed circle .

[quote="arqady"][quote="alex2008"]Let be $ ABC$ an arbitrary triangle . Prove that :

$ \sin^4\ A \plus{} \sin^4\ B \plus{} \sin^4\ C\ge \frac {27}{2}\left(\frac {r}{R}\right)^3$

[/quote]
The stronger $ \sin^4\ A \plus{} \sin^4\ B \plus{} \sin^4\ C\ge \frac {27}{4}\left(\frac {r}{R}\right)^2$ also true and easy.[/quote]

$ LHS \ge \frac{(\sin^2 A\plus{}\sin^2 B \plus{} \sin^2 C)^2}{3}\equal{} \frac{(s^2\minus{}r^2\minus{}4rR)^2}{12R^4} \ge$ $ 3\frac{r^2(2R\minus{}r)^2}{R^4}\ge \frac {27}{4}\left (\frac {r}{R}\right)^2$

[quote="alex2008"]Let be $ ABC$ an arbitrary triangle . Prove that :

$ \cos^4\ A \plus{} \cos^4\ B \plus{} \cos^4\ C\ge \frac {3}{8}\left(\frac {r}{R}\right)^4\left(5 \minus{} \frac {r_a}{r}\right)\left(5 \minus{} \frac {r_b}{r}\right)\left(5 \minus{} \frac {r_c}{r}\right)$

where $ R$ is the radius of the circumcircle circle , $ r$ is the radius of the inscribed circle , $ r_a,r_b,r_c$ are the radius of the exinscribed circle .[/quote]

$ LHS \ge \frac{(cos^2 A \plus{} cos^2 B \plus{} cos^2 C)^2}{3}\equal{}$ $ \frac{1}{3}(3\minus{}\frac{s^2\minus{}r^2\minus{}4rR}{2R^2})^2 \ge \frac{1}{3}(3\minus{}\frac{4R^2\plus{}2r^2}{2R^2})^2\equal{} \frac{1}{3}(1\minus{}\frac{r^2}{R^2})^2$

$ RHS \le \frac {3}{8}\left(\frac {r}{R}\right)^4(5\minus{}\frac{r_a\plus{}r_b\plus{}r_c}{3r})^3\equal{}\frac {1}{9}\left(\frac {r}{R}\right)^4(7\minus{}2\frac{R}{r})^3$

$ \Leftrightarrow (7\minus{}2\frac{R}{r})^3 \le 3(\frac{R^2}{r^2}\minus{}1)^2$

Let :  $ \frac{R}{r}\equal{}t \ge 2$

$ \Leftrightarrow (7\minus{}2t)^3 \le 3(t^2\minus{}1)^2$ which is true for $ t \ge 2$


Is there a way to prove for vectors a, b, and c, that
\[a\times(b\times c) = (a\cdot c)b-(a\cdot b)c \]
without actually multiplying everything out?

Yes, use index notation [url=http://web.mit.edu/8.07/www/IndexNotation.pdf]see here[/url].

A full solution can be found in [url=http://www.maths.qmul.ac.uk/~mm/MAS204/chap5.pdf]this[/url] document on page 4.


inequalities
calculus
integration
LaTeX
algebra
polynomial
inequalities proposed

Prove that for any real numbers $a_{1},a_{2}, \ldots,a_{n}$ the following inequality holds:

\[\displaystyle \left( \sum_{i=1}^{n} a_{i} \right)^{2} \leq \sum_{i,j=1}^{n} \frac{ij}{i+j-1} a_{i}a_{j}.\]

This one can be done by induction.
First for $n=2$,it is clearly.
Assume that it is true for $n=k$.For $n=k+1$,we must prove that:
[tex] \frac{(n-1)^2}{2n-1}\cdot a_n^2+\frac{(t-1)^2}{2t-1}\cdot a_t^2+2\frac{(n-1)(t-1)}{n+t-1}\cdot a_n\cdot a_t \geq 0 [/tex] and the proof will complete.
But this is clearly by Cauchy and [tex] (2n-1)(2t-1) \leq (n+t-1)^2 [/tex] :)

Could you please be more explicitely? The idea is interesting, but again I don't see obvious things. I have a three line solution, but yours seems more interesting. Unfortunately, I do not understand it very well.

hmm...am i misunderstanding something or does it suffice to show $\frac{ij}{i+j-1]\geq 1}$, which is trivial? the right side then becomes the same as the left side...but it can't be that trivial...

Peter

Trivial is too much, they are REALS, Peter, not positive numbers!

I think, Harazi, that your solution is with integrals.
Take f(x)=sum{a_i*x^i} g(x)=f'(x).
denote S(f) be integral from 0 to 1 of f ( I'm stupid can't write it in LaTeX)
The inequality rewrites S(g)^2<=S(g^2).
But is this true?
We all know (x_1+..x_n)^2 is not always smaller than x_1^2+..x_n^2

Yes, it is a combination between integrals and Cauchy. The solution is very simple, but the problem isn't simple at all.

Can such problem has another solution?

iura.
Integrals in Latex is quite simple:
\left(\int_{0}^1 g(x)dx\right)^2 \leq \int_0^1 f^2(x)dx
will produce
$\left(\int_{0}^1 g(x)dx\right)^2 \leq \int_0^1 f^2(x)dx$

Well, another solution would possibly be that of Anh Cuong. But I see no answer of him.

Oh yes! How could I forget CBS! (A stupid excuse is that I don't use integrals since I never need them)
Another problem can be solved by this method is Poland (1990-1992 I think):
Prove sum{a_ia_j/(i+j)}>=0 for REALS a_i.
At this problem I have an elegant solution using Polynomial fractions.
So you can try it, Myth!

Another stunning result:
\[
\sum_{i,j=1}^\infty \frac{a_ia_j}{i+j}\leq \pi\sum_{i=1}^\infty a_i^2.
\]

It is Hilbert's inequality.

For your problem, iura, we define $f(x)=\sum\limits_{k=1}^n a_ke^{kx}$.
Then $\sum\limits_{i,j}\dfrac{a_ia_j}{i+j}=\int\limits_{-\infty}^0f^2(x)dx$

Oh,I am soory,but I must be wrong,because I don't understand what I wrote too :blush:

had another integral approach on Poland problem:
Let f(x)=sum{a_ix^i}
Then LHS is just the integral of f^2(x)/x

i knew there had to be something wrong...thanks
Peter


calculus
integration
logarithms
Euler
Gauss
calculus computations

Can anybody, please, tell me the value of the integral
$\int_0^\infty (\ln x)^2 e^{-bx^2}\,dx$ ($b>0$ is a parameter).
Thanks in advance  :)

Here a way to get the value of your integral

$\int_{0}^{+\infty}e^{-x^2}(\ln x)^2dx=\frac{\sqrt{\pi}}{8}\left(\frac{\pi^2}{2}+(\gamma +2\ln 2)^2\right)$ where $\gamma$ Euler constant

Do the change variable$x=\sqrt{b}y$ and use

$\int_{0}^{+\infty}e^{-by^2}\ln ydy=\frac{-1}{4}(\gamma +\ln(4b))\sqrt{\frac{\pi}{b}}$

with  Gauss integral $\int_{0}^{+\infty}e^{-by^2}dy$


Ian has $ 24$ more pencils than Sylvia. Ian has $ 4$ times as many pencils as she has. How many pencils should Ian give to Sylvia so they will have an equal number of pencils?

[hide]Let Sylvia have $ p$ pencils and $ Ian$ have $ p\plus{}24$ pencils. \[ 4p\equal{}p\plus{}24\]By subtracting $ p$ from both sides, \[ 3p\equal{}24\]Now we have \[ p\equal{}8\]. Then, we need Ian to give Sylvia $ 12$ pencils for them to have an equal number.[/hide]

Alternatively, since the amount of pencils they have differs by 24, our answer is simply $ 24 \div 2 \equal{} 12$.


induction
number theory proposed
number theory

Prove if $u>v>1$ then the equation
\[u^x \equiv 1(mod v^x)\]
has finite natural answers for x.
____________________________________
No pain, no gain

See solution to problem 3, mathlinks contest , round 1. And by the way, I don't find this problem easy.

I remember that I have posted it before and If I'm right one of american member solved it btw with the lemma which is used in contest 3 round one it's pretty obvious

Yes, but it's not very easy to prove that lemma and to think of it. The fact that problem 3 was the most difficult in that round shows that the problem isn't easy at all.

I was looking through some older problems and I found this one. I'll just sketch it.

First of all, we can assume $v$ is actually a prime $p$. Then, if $d|p-1$ is the order of $u$ modulo $p$ and $p^t\|u^d-1$, then we can easily show by induction on $k\ge 0$ that $dp^k$ is the order of $u$ modulo $p^{t+k}$ and $p^{t+k}\|u^{dp^k}-1$. This means that if $p^n|u^n-1$, then, for large enough $n$, we will have $dp^{n-t}|n$, and this can't happen infinitely many times, as the left side increases much faster than the right side.


a, b, c are not less than 1. Show that: $ a^{2}\plus{}b^{2}\plus{}c^{2}\plus{}abc(a\plus{}b\plus{}c)\leq(a^{2}b^{2}\plus{}b^{2}c^{2}\plus{}c^{2}a^{2})\plus{}(ab\plus{}bc\plus{}ca)$

if i'm not wrong , the inequality is equivalent to
$ (a\minus{}b)^{2}(c^{2}\minus{}1)\plus{}(b\minus{}c)^{2}(a^{2}\minus{}1)\plus{}(c\minus{}a)^{2}(b^{2}\minus{}1)\geq 0$
which is true

nmtv, don't post problem from M&Y again  :mad:


analytic geometry
AMC
AIME
Asymptote
Mafia
FTW

Welcome to the new game AOPS Planet! Basically you are the leader of a "country". More to follow! Sign up here!

EDIT 1: Once we have enough people, I will randomly assign a "country" to everyone which they can name. The size and population of the country is random.

People:

Eggylv999

Note: To discourage inactivity it will be easier for your country to be conquered when you are absent....

I prefer a country that is in good shape to crush supermathboy's.

People:

Eggylv999
sup3rcrash3r
SuperMathBoy
math154
isabella2296

JUst add your name to the list.

I guess I will sign up too. But if I get any third world country, or any Soviet country, I will drop out.

READ TOPIC!!

You get to name your country so it's obviously fake.

People:

Eggylv999
sup3rcrash3r
SuperMathBoy
math154
isabella2296
jain_harshi
chenhsi
bugi
aime15
pythag011

[size=200]
JUst add your name to the list.[/size]

10 people is good for now. Signups closed! However if you would like to replace, or possibly lead a rebelling province, please post here or pm me!

Here is the map of planet Mewto. I am God. What I say goes. Anyone who disobeys will be smote. Got it? Good. Here's where random.org says you live:
[geogebra]12d3bd247d08aa4393e987ee1deed0a2dd787ea9[/geogebra]

OK. If 9 people sign up here :P
And please no mewtos in my planet(classic war game)

I sign up. Just dont kill me at the beginning. :) :lol:

I won't jain_harshi!
It's all about war and alliances as all countries are equal, also as I am the chief, you should win me on your side so I can help you against others. How?

[quote="coolts"]hey! hey! can i be in this one? mewto was obviously such a retard that he didnt notice my posts asking him about it for his game...[/quote]

No such words, even if you are right.

[quote="coolts"]hey! hey! can i be in this one? mewto was obviously such a retard that he didnt notice my posts asking him about it for his game...[/quote]

Coolts if you are to call me such nasty names, please be justified in doing so. You took over jain_harshi's country following his assassination. Please read the thread before accusing me unjustly.

i join i will give you ftw rating lol jk

1 more, but as I don't have much time, don't expect this will be a nice game.


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

Um, I wanna prepare for the AMC 12 and the AIME and I was wondering whether the AOPS books were helpful. I dont mean AOPS V.1 and V.2 I mean the Intro books like Intro to Counting and Intro to Geometry and Intro to Number Theory.  Would those three cover most of the material on the AMC 12? Assuming you've already got Algebra down. And what about the AIME problem solving series class? like after you complete those three books would that help?

Well, this would really help, but if you were to study them all, you'd be even better prepared  :wink: I personally found that mastering volume 1 and reading through volume 2 (not the weird geometry stuff) and just doing a whole bunch of practice AMC 12 and AIME problems really helps boost your score.

What "weird geometry stuff" are you talking about? Hopefully not chapters 21 or 22, which are the most important for Olympiad geometry. Vectors are important too. Also, I notice you use the word "reading" instead of "mastering" for Volume 2. Volume 2 material is important and common at the AIME level (Volume 1 is more like MathCounts/AMC10).

In short, yes, they are helpful.

Volume I is useful for the geometry and the basic number theory stuff.  Volume II is where the majority of the material for AMC12/AIME comes from.  You'd do well to get the Intro to NT or Intro to Geometry books.  Algebra is actually a relatively tricky topic for AMC.

Needless to say, books are useful, but practice is the most important.  Doing old tests/learning the concepts behind problems will raise your score quite a bit for each.

i just saw the book First Steps for Math Olympians
by J. Douglas Faires in my visit back to school....
its a great source of prep for AMC test....

check it out:
http://www.artofproblemsolving.com/Books/AoPS_B_Item.php?page_id=52

but last yaer, my prep was just doing past tests...haha, and it paid off pretty well  :D

I guess, if you don't think you can get yourself to study without one.

AMC prep is kinda like MathCounts. You want to do well enough to get into the USAMO, but really it's not a big deal.

whats a good text for brushing up on algebra besides the aops texts? and does ne1 know the breakdown of subjects on the AMC and the AIME? like how much of the exam will be Algebra, how much will be geometry, and so on? what's the most heavily tested subject?

uh..do old tests? there is a contest directory on this site...

[quote="amirhtlusa"]i have to disagree, a high amc score is a huge factor in qualifying through floor score in my opinion.[/quote]

Uh...if you qualify through floor the AMC doesn't matter at all. If you're capable of qualifying, don't bother with doing the AMC too much, because it's not really important.

Amir got a perfect on the AMC 12 and that's what got him into USAMO... I wouldn't be saying that :D

sorry,
what i meant to say was:
getting a high score on AMC is a huge factor from qualifying through cut off score for usamo qualification.  as last year it applied to me when i could have qualified for usamo with a 7. (but i did with a 10 instead :)  )

the previous statement is also somewhat true, given the past "tie-brakes" among 10 and below graders.

Well amir would have gotten into the USAMO with a 117 on the AMC12, so the perfect score didn't actually matter that much. And as it turned out, the tiebrakers made no difference for grades 10 and below.

um amirthusa? you got a perfect score on the AMC? how did you do it? like what kinda prep did you do? did you just do old tests? i mean didn't you have to learn the concepts behind the questions first? or did you just know it? what did you use to learn the concepts?

[quote="Boy Soprano II"][quote="Karth"]Ah, well I was wrong about discrete math, I guess. Statistics would be considered discrete math, so yes, there is discrete math on the AMC and AIME. However, don't go diving into extreme matrix action with eigenvalues and vector spaces  :wink:[/quote]
The only statistics on the AIME have to do with how many people got various scores :wink: .  But combinatorics is usually "discrete", as is number theory.[/quote]
Yes, that's what I meant by discrete math (as in combinatorics, etc.)
In fact, according to Wikipedia, discrete math includes logic, set theory, number theory, combinatorics, graph theory (which comes up occasionally), functions, probability, and even proofs, so discrete math is [i]definitely[/i] included in AMC/AIME/USAMO.

[quote="Boy Soprano II"][quote="Karth"]Ah, well I was wrong about discrete math, I guess. Statistics would be considered discrete math, so yes, there is discrete math on the AMC and AIME. However, don't go diving into extreme matrix action with eigenvalues and vector spaces  :wink:[/quote]
The only statistics on the AIME have to do with how many people got various scores :wink: .  But combinatorics is usually "discrete", as is number theory.[/quote]

There is probability though, which is probably encompassed in Wiki's idea of statistics.

[quote="nebula42"][quote="simulacrum"]aw that sucks i just signed up for the linear algebra course in my school[/quote]
I don't think that would make much of a difference though, because the discrete math course at most schools probably doesn't go in depth enough to enable you to solve AIME level problems anyway.  :wink:  Just about every person who does well on AIME, etc. does math outside of school math.[/quote]

Agreed  :) Take geometry for example... if you ask the average high schooler (or even college student for that matter) what Ceva's Theorem was, this would be an example of their facial expression:  :|  :huh:  :huuh:

A bit of an aside... these smiles are awesome.  :P

[quote="simulacrum"]my school's pretty advanced, i go to the best math science high school in nyc, but yeah ur probly rite most of the kids in our school don't become usamo qualifiers, i think there were like nine last yr.[/quote]

Wow... that's amazing! From one school?? I think the highest this year were schools like Phillips Exeter Academy (15 ppl?) and Thomas Jefferson (13 ppl or so). Your school must be pretty awesome  :D

It's not like Ceva matters on the AIME anyway.

But it doesn't hurt to use Cevalaus once in a while on the AIME... mass points, which is connected to those two are ridiculously helpful on the AIME when you have a whole bunch of concurrent lines and triangles.

[quote="Karth"][quote="me@home"]Is your school Stuyvesant or something like that? Or Bronx science something...? (My aunt went to one of those, i forget which, and she's a math prof now... lol)[/quote]

No.. my school is Northview (Georgia).[/quote]

Um... I'm pretty sure me@home was talking to simulacrum.

Yup I go to Stuyvesant. And Stuyvesant is a pretty awesome school but its got nothing on you guys    :)


Show that there is a positive number in the Fibonacci sequence which is divisible by $ 1000$

We will show something stronger: for each k there are infinitely many fibonnacci numbers which are divisible by k.
All the equalities below are taken mod k.
Let the fibonnaci sequence start with f1=1 and f2=1.
Consider couples $ c_{1}\equal{} (f_{1},f_{2})$, $ c_{2}\equal{} (f_{2},f_{3})$, etc mod k. There are only n x n possibilities of such a couple, so some couples must appear in the sequence more times. Let $ c_{a}\equal{} c_{b}$. But since $ c_{k\minus{}1}$ is determined by $ c_{k}$, we have
$ c_{a\minus{}1}\equal{} c_{b\minus{}1}$ and by induction
$ c_{1}\equal{} c_{b\minus{}a\plus{}1}$
so $ c_{b\minus{}a\plus{}1}\equal{} (1,1)$
$ c_{b\minus{}a}\equal{} (0,1)$
$ f_{b\minus{}a}\equal{} 0$,
we found a fib. number divisible by k.

In similar way it can be shown that for z in { 1,-1,2,3,5,8,... }  there are infinitely many fib. nmbers yielding theese residues mod k.


number theory
prime numbers
number theory open

Every even number ([b]other than 58[/b]) of the form [b]10n+8 [/b]can be expressed, at least once, as sum of a prime number plus a square :

[b]10n+8= p + k^2[/b]

n>=0
k>0

Example

8=7+1^2
18=17+1^2
28=3+5^2
38=37+1^2=13+5^2
48=47+1^2=23+5^2

and so on

Tested until 10^6
Any counterexample?

[quote="doyunkim"]wow it's interesting... but unfortunately i cant say anything more than this! how sad....[/quote]

Really sad

This is likely to be true... But I highly doubt someone here can find a proof for this.
Every statement involving a sum of a prime is basically really hard to prove or to disprove. Look at Goldbach conjecture,
Sophie-Germain conjecture, etc... I think we are missing a piece of the whole thing in number theory. For fun I just tried to write down
about ten conjectures. It's easy, just wonder if there are infinitely many (prime) numbers with some particular property, not a trivial one, but also not too restrictive...

What is interesting in finding some conjecture is that each conjecture can enlight one or more aspects of the primes.
I had given in another forum  a more restrictive version of Goldbach conjecture :

Every even number 2n can be expressed at least once as sum of two primes such as the two are both > int(sqrt(2n)).

Example

2n=20
int(sqrt(20))=4

20=7+13
7>4
13>4

No one has found a counterexample.


limit
trigonometry
calculus
calculus computations

Evaluate $ \lim_{x\rightarrow 0}(\frac{arcsinx}{x})^{\frac{1}{x^2}}$

$ \lim_{x\rightarrow 0}\;\left(\dfrac{\sin^{ \minus{} 1}x}{x}\right)^{\dfrac {1}{x^2}}$

$ \equal{} \;\lim_{x\rightarrow 0}\;\left(\dfrac{x \plus{} \dfrac{1}{2}\cdot\dfrac{x^3}{3} \plus{} \textbf O(x^5)}{x}\right)^{\dfrac {1}{x^2}}$

$ \equal{} \;\lim_{x\rightarrow 0}\;\left(1 \plus{} \dfrac{x^2}{6} \plus{} \textbf O(x^4)\right)^{\dfrac {1}{x^2}}$

$ \equal{} e^{\dfrac{1}{6}}$

$ \because \lim_{x\to 0} \left(1 \plus{} x\cdot y \plus{} \textbf O(x^2)\cdot z\right)^{\dfrac {1}{x}}\; \equal{} \;e^y\;\;\;\quad\;\;\;(\;\;\;\text{for}\;\;\;y,z\in\mathbb R\;\;\; )$


geometry
circumcircle
analytic geometry
angle bisector
geometry unsolved

given triangle $ ABC$ with angle bisector $ BM$ and $ CN$ ($ M$, $ N$ belong to $ AC$, $ AB$). $ MN$ intersect circumcircle of $ ABC$ at $ D$.
Prove that: $ 1/BD \equal{} 1/AD \plus{} 1/CD$

It's easy to show that line $ MN$ is the geometric locus of the points whose sum of the oriented distances to $ AC$ and $ AB$ equals the oriented distance to $ BC.$ Indeed, trilinear equation of $ MN$ WRT $\triangle ABC$ is given by $ \beta \plus{} \gamma \minus{} \alpha \equal{} 0.$ Therefore if $ X,Y,Z$ denote the projections of $ D$ on $ CB,BA,AC,$ we get $ DX \equal{} DZ \minus{} DY.$ Since $ \angle DBX \equal{} \angle DAZ$ and $ \angle DCB \equal{} \angle DAB,$ we have $ \triangle DBX \sim \triangle DAZ$ and $ \triangle DCX \sim \triangle DAY.$

$ \frac {BD}{AD} \equal{} \frac {DX}{DZ} \ , \ \frac {CD}{AD} \equal{} \frac {DX}{DY}  \Longrightarrow  \frac {1}{BD} \minus{} \frac {1}{CD} \equal{} \frac {DZ \minus{} DY}{AD \cdot DX} \equal{} \frac {DX}{AD \cdot DX} \equal{} \frac {1}{AD}$

[quote="hqthao"]Oh, great. thanks very much. I google to see trilinear coordinate and know the basic knowledge. But can you tell me more about this coordinate (or barycentric is ok because two coordinate are equivalent)  (for example, does it strong or easy to use, please? (or give me a link to view it).[/quote]
Your welcome, [url=http://mathworld.wolfram.com/]Mathworld[/url] is a good source to learn about trilinear coordinates. Nevertheless, the first part can be proved synthetically by considering a couple of similar triangles.


factorial
superior algebra
superior algebra unsolved

i)Prove that the set of all algebraic integers in Q is  Z

ii) Then find all algebraic integers in the field

Q((sqrt2)) = {a + b(sqrt2) | a, b E Q}

sqrt2 = square root of 2.

i.e. show that every number you found is an algebraic integer and every algebraic integer in Q((sqrt2)) is among these numbers.

sorry. the question actually says the following:

ii) Find all algebraic integers in the field

Q((sqrt2)) = {a + b(sqrt2) | a, b E Q}

sqrt2 = square root of 2.

(this requires proving that every number you've found is an algebraic integer, and every algebraic integer in Q((sqrt2)) is among the number you've found).

[quote="ZetaX"]For i) you can show in more general manner: Let $ R$ be a factorial ring. Then $ a \in \mathrm{Quot}(R)$ is integralover $ R$ iff $ a\in R$.

What have you done until know for part ii)¿[/quote]


I'm currently preparing benzoic acid by the alkaline hydrolysis of benzamide.

1. Is NaOH a catalyst in hydrolysis?

2. Why is alkaline hydrolysis preferred to acidic hydrolysis?

Is it that the neutralization of benzoic acid shifts the equilibrium of hydrolysis to the right, so the conversion of benzamide to benzoic acid is more complete?

3. Why shouldn't we turn off the vacuum pump before removing the rubber tubing from the Büchner flask?

Also, I'd like to ask why D2O has stronger intermolecular forces than H2O? (D2O has is higher melting and boiling than H2O)
Isn't the strength of intermolecular forces actually dependent on the size of and number of electrons in the molecules only?

[quote="Puppet"]1. Is NaOH a catalyst in hydrolysis?[/quote]

No. Hydroxide ions act both as nucleophiles and as bases.

[quote="Puppet"]2. Why is alkaline hydrolysis preferred to acidic hydrolysis?[/quote]

Because it is more irreversible: in the end we get a carboxylate anion, which is unreactive towards nucleophilic attack, while in acidic hydrolysis we get a carboxylic acid molecule, which can again be protonated and attacked.

[quote="Puppet"]3. Why shouldn't we turn off the vacuum pump before removing the rubber tubing from the Büchner flask?[/quote]

To what experimental procedure are you referring to?

[quote="Carcul"][quote="Puppet"]1. Is NaOH a catalyst in hydrolysis?[/quote]
To what experimental procedure are you referring to?[/quote]
Thanks!
I'm talking about obtaining benzoic acid crystals by suction filtration.

Well, the problem that might arise is a pressure difference in the sense of air aspirated to the kitasate flask, which may "blow up" the crystals out of the Büchner.

D20 has stronger hydrogen bonding than ordinary water.


Hi I need help in this exercise:
The set {1;2;...;49}is partitioned into three subsets. Show that at least one of the subsets contains three different numbers a,b,c such that a+b=c.

I can't solve it...

Could somebody post a solution for this problem?


Now we know that

$ a_0 \equal{} a_1 \equal{} a_2 \equal{} a_3 \equal{} 1,$
$ a_na_{n \minus{} 4} \equal{} a_{n \minus{} 1}a_{n \minus{} 3} \plus{} a_{n \minus{} 2}^2, (n > \equal{} 4)$

Find a standard form of $ a_n$,and prove that $ a_n (n \equal{} 0,1,2,...)$ are integers.

A standard form of a sequence means that a standard form. For example:

$ a_1 \equal{} 1,a_{n \plus{} 1} \equal{} 2a_n,(n > \equal{} 1)$

Then the standard form is $ a_n \equal{} 2^{n \minus{} 1}$.

This is a rather special type of sequence called a Somos sequence; see the many references [url=http://www.research.att.com/~njas/sequences/A006720]here[/url] or [url=http://mathworld.wolfram.com/SomosSequence.html]here[/url].

But can sombody give its standard formula?
Or prove it always gives integers?

[quote="t0rajir0u"]Several proofs are available in the references, but the closest you're going to get to a closed form is given by certain values of a [url=http://en.wikipedia.org/wiki/Theta_function#Jacobi_theta_function]theta function[/url] (which was Somos' original motivation for studying such sequences).[/quote]

Can you give me some details?I want to get a general proof.

[quote="t0rajir0u"]Then read the references![/quote]
Which reference?
 http://jamespropp.org/somos.html?

I can't open it.Perhaps the website has a problem.


geometry
circumcircle
incenter
angle bisector
perpendicular bisector

ohhhh
I misread

I will work on it now

whoops

[hide="Hint"]If we can inscribe a circle in $DFEG$, then $DF+EG=FE+DG$, and vice-versa.[/hide]

[quote="cincodemayo5590"][quote="chickendude"][hide="uhh"]
$D, F, E$ and $G$ are all on the incircle
doesn't that make the incircle of $\triangle ABC$ the circumscribed circle of quadrilateral $DFEG$ regardless of any of the other values?
[/hide][/quote]

Circumscriptible means that it has both a circumcircle and an [b]incircle[/b] that is tangent to all four sides.[/quote]

no it doesn't

circumscriptible <-> cyclic

[url=http://www.cut-the-knot.org/Curriculum/Geometry/BicentricQuadri2.shtml]CTK[/url] agrees with Altheman, but I'm finding other conflicting resources too from Google.

Alright, sorry. Stupid english language.

I meant to say that the quadrilateral has an incircle that is tangent to all four sides.

[quote="cincodemayo5590"]The incircle with center $I$ of $\triangle ABC$ meets sides $a,b,c$ at $D,E,F$ respectively. Extend $FI$ until it hits the incircle again at $G$. Prove that quadrilateral $DEFG$ has an incircle that is tangent to all four sides if and only if $a=b$.

(EDIT: Removed "circumscriptible", to avoid confusion.)[/quote]
[hide="Well back to the problem"]
I hope I didn't make a mistake because this seems a little trivial...
Consider the angle bisectors of $\angle DGE, \angle DFE$. The incenter of $\triangle EGD, \triangle EFD$ must lie on these lines respectively. $EGDF$ has an incenter iff the incenters of $\triangle EGD, \triangle EFD$ coincide i.e. the angle bisectors of $\angle DGE, \angle DFE$ coincide. However $G$ was defined to be diametrically opposite $F$ so the two angle bisectors lie parallel to each other with the $I$ lying midway between them, so in order for the two to coincide $I$ must be on the bisector of $\angle DFE$. Let $ED\cap GF=H$. Then since $\triangle DFE$ is isosceles and $H$ is the foot of the angle bisector, it is also the foot of the altitude and median so $FH=FI$ is the perpendicular bisector of $ED$. Now the angle bisector of $C$ passes through $I$ thus is the perpendicular bisector of $ED$ so $C,G,H,I,F$ are collinear in that order. Since $AB$ is tangent to the incenter at $F$ and we have proven that $F$ lies on the angle bisector, it follows that $CB=CA$ as was wanted. All of this was if and only if so we're done.
Wow the idea was trivial but not the formal proof :(  :)
[img]8268[/img]
That might help
[/hide]

[color=darkblue][b][u]Proof.[/u][/b] Observe that $\{\begin{array}{ccc}\triangle CGE\sim\triangle CEF & \Longrightarrow & GE=\frac{EC\cdot EF}{CF}\\\\ \triangle CGD\sim\triangle CDF & \Longrightarrow & GD=\frac{DF\cdot DC}{CF}\end{array}$. The quadrilateral $DEFG$ is circumscriptiblely $\Longleftrightarrow$

$GE+DF=GD+EF$ $\Longleftrightarrow$ $EF\cdot (CE-CF)=DF\cdot (CD-CF)\ ,\ CE=CD$ $\Longleftrightarrow$

$EF=DF$ $\Longleftrightarrow$ $\triangle CFD\sim\triangle CFE$ the ray $[CF$ is the $C$- bisector of the angle $\widehat{ACB}$ $\Longleftrightarrow$ $CA=CB$.[/color]


geometry
geometric transformation
rotation
trigonometry

Given $\triangle ABC$ with $\angle B=30^\circ$, $P$ is inside $\triangle ABC$ such that $\angle PAB=36^\circ$, $\angle PBC=24^\circ$ and $\angle PCB=72^\circ$. Find $\angle PCA$.

[hide]the triangle does not exist

if $x=\angle PAC$ and $y=\angle PCA$ I got $x+y=46$ from considering $\triangle APC$ and $x+y=42$ from considering the sum of the angles of $\triangle ABC$.[/hide]

I'm not sure how you got $x+y=46^\circ$... I'm getting $x+y=42$ in both of them. The triangle [i]does[/i] exist.

$x+y=42$.
Would rotation of the triangle work? I'm not sure...

use trig ceva...but i cannot find the solution to the trig equation...


number theory
least common multiple
relatively prime

Q) If $ a_n$ denotes the sum of n terms of the HP $ \frac {1}{2} , \frac{1}{3} , \frac{1}{4} ,...$then prove that $ a_n$ cannot be equal to any integer for $ n$ belonging to N.

Q) Prove that $ m^r$ where $ m,r$ belong to N , $ r >\equal{}2$ can be expressed as sum of $ m$ consecutive odd integers. Find the first odd integer in case of $ 8^5$.

[quote="azn6021023"][hide="solution to 1"]Consider the sequence $ \frac {1}{2}, \frac {1}{4}, \frac {1}{8}...$.  This has a rate of $ \frac {1}{2}$.  But $ \frac {1}{3} > \frac {1}{4}$ and $ \frac {1}{5} \plus{} \frac {1}{6} \plus{} \frac {1}{7} > \frac {1}{8}...$.  Thus, the rate is increasing by more than 1, and the series diverges.[/hide][/quote]
I don't think that's the question. :P

If you consider the largest prime $ p$ (guaranteed to be above $ \frac{n}{2}$ since we know there's a prime between any $ a$ and $ 2a$), and think about combining all the fractions with the lcm of all from $ 2$ to $ n$...

oops.  Thanks for the clarification.

[hide]The harmonic mean series when rewritten yields [b]something[/b] like $ \frac{abc\plus{}bcd\plus{}cda\plus{}dab}{abcd}$, where each part of the numerator has $ n\minus{}1$ terms and the denominator has $ n$ terms.  Notice that all the factors in the denominator must cancel in the numerator, but the numerator has exactly one term that is relatively prime to the denominator.  Thus, the expression cannot be an integer.[/hide]

I wonder if this is rigorous enough...

[hide="2nd"]Let the consecutive integers have mean $ m^{r\minus{}1}$, so that their sum will be $ m^{r\minus{}1}\cdot m \equal{}m^r$. This is possible because when $ m$ is even, the average of $ m$ consecutive integers is even, and when $ m$ is odd, the average of $ m$ consecutive integers is odd. (Think of it as median instead of mean.)[/hide]

[quote="azn6021023"]Notice that all the factors in the denominator must cancel in the numerator, but the numerator has exactly one term that is relatively prime to the denominator.  [/quote]
This is not close to rigorous enough.  (Try to restrict yourself to thinking about one prime - $ 2$ is a good choice here.)

The second question can be generalized significantly.  Note that a sum of consecutive odd numbers is a difference of squares.

@Brut3Forc3: I dont get your method.

@t0rajir0u : You mean diff of squares divided by 2. But what about that?

Where is the division by two?

[hide="Brut3Forc3s method"] As he states it, it is not quite right.  But the basic idea is that an odd number of consecutive odd integers has a middle integer, which is odd and therefore the average of the entire set of consecutive integers.  For example, in the sequence

$ 3, 5, 7, 9, 11$

$ 7$ is the middle term (which exists because there are $ 5$ terms) and therefore also the average of the $ 5$ terms.  If $ m$ is odd, one takes $ m^{r \minus{} 1}$ to be the middle odd integer and extends the sequence in both directions until there are $ m$ terms.

On the other hand, an even number of consecutive odd integers has two middle odd integers, and the average of the entire set will be the average of those two middle odd integers and therefore even.  For example, in the sequence

$ 1, 3, 5, 7, 11, 13$

the middle terms are $ 5$ and $ 7$ and the average is $ 6$.  If $ m$ is even, one takes $ m^{r \minus{} 1} \minus{} 1, m^{r \minus{} 1} \plus{} 1$ to be the middle terms and extends the sequence in both directions until there are $ m$ terms as before. [/hide]
[hide="My method"] A sum of $ m$ consecutive odd integers takes the form $ (n \plus{} m)^2 \minus{} n^2$.

To see why, recall that

$ k^2 \equal{} 1 \plus{} 3 \plus{} 5 \plus{} ... \plus{} (2k \minus{} 1)$

so the sum of the first $ k$ consecutive odd positive integers is $ k^2$ and the sum of the first $ n \plus{} m$ consecutive odd positive integers minus the sum of the first $ n$ consecutive odd positive integers (which is the sum of $ m$ consecutive odd integers) is $ (n \plus{} m)^2 \minus{} n^2$.

This factors as $ m(2n \plus{} m)$.  In other words, a number can be expressed as the sum of $ m$ consecutive odd integers precisely if it is the product of $ m$ and a number with the same parity as $ m$.  Clearly $ m^r$ is in this category. [/hide]


What do high schools look for in a high school student who is applying? Do they look for the commmunity service you've done, grades, SAT scores, competitions and acctivities you've been in, and things of that sort? Which of these would you say are the most important?

It's a little early for me to be think about these, but it never hurts to start early. Thanks.

Take a skim through this whole forum and you'll get a pretty good sense of what colleges are looking for... there are probably several threads on stuff like this already.

Many of the highschools I am considering applying to, look for strong leaders.  Thus, your portfolio has to show you've taken of projects, services, jobs etc. that develop leader-like qualities, and at the same time, show you're a good citizen.


geometry
trigonometry
perpendicular bisector
angle bisector
trig identities
Law of Cosines
AMC

In triangle ABC, side AC and the perpendicular bisector of BC meet in point D, and BD bisects  :ang: ABC.  If AD = 9 and DC = 7, what is the area of triangle ABD?

Umm...should I just tell you what to do?  Or is this for other people to solve, because I just have two hints: [hide]Stewart's theorem and Angle-Bisector theorem[/hide].

Warped, that's way overkill -- [hide]angle chasing and similar triangles work fine.[/hide]

I ought to agree I think your method works more often, most people wouldn't know Stewart's Theorem.

[hide]On a contest, I'd use Angle Bisector and Stewart without second thought. But of course you don't need to know those theorems to solve this problem. You just have to be a little bit more clever to solve it without them.[/hide]

You should all go ahead and post your solutions so people can see them.

Could someone please enlighten me?  If I get another nonreal answer one more time trying to solve this problem I think I might pull my hair out.

Probably the simplest way to do it is to prove and then use the fact that ABC :sim: ADB.  It should not be too hard from there.

WarpedKlown, why don't you post the Stewart's + Angle Bisector proof?

Ok, I'll go first then:

[hide]Triangle BCD is isosceles with legs BD and DC because its perpendicular bisector runs through D.  BD will be 7 units.  From the angle-bisector theorem, AB:BC = 9:7.  Let AB and BC be 9x and 7x.  Using Stewart's theorem, (9x) :^2:(7) + (7x) :^2:(9) = (7 :^2:)(16) + (9)(7)(16).  That simplifies to 567x :^2: + 441x  :^2: = 784 + 1008 which further becomes 1008x :^2: = 1792.  x = 4/3.  To find the area of triangle ABD, use Heron's formula.  AB = 12, BD = 7, and AD = 9.   :sqrt: [(14)(2)(7)(5)] = 14 :rt5:.[/hide]

CBD is isocelis, thus DB=7. Now we can prove that BDA :sim: CBA, by AAA (angle ABD=DBC=BCD=x, let angle BAC=y, thus ADB=x+x, thus the triangles are similar). NOw let AB=a, a^2=9*16=144=12, using herons here we get  :sqrt:  (5*7*14*2) = 14*  :rt5:

....or you could use law of sines and the 1/2absinC triangle area thing...

All of those are good solutions. I especially like the one using similar triangles. It's probably the most time-efficient solution.

Can someone talk a little about Stewart's Theorem?  I read the [url]http://mathworld.wolfram.com/StewartsTheorem.html[/url] site, but it doesn't really explain when to use it or how to remember it.

its useful for finding the lenghts of cevians and the lenghts of the two parts they divide a triangle side into. Consider a triangle with base a, and the other two sides b and c. if you draw a segment down from the top vertex to side a, it splits it into lenghts m and n respectively. let the lenght of the segment be d. then the relationship cnc + bmb = dad + man applies. (i find this is the easier way to remember it.)

[quote]then the relationship cnc + bmb = dad + man applies. (i find this is the easier way to remember it.)[/quote]

I prefer taping "a(p^2+mn) = (b^2)m + (c^2)n" to my homework assignment book so I can stare at it all throughout class.  :mrgreen:

Or you can just use the Law of Cosines and the fact that cos(x) = -cos(180-x) to prove it if you really forget.

Or you can do what a lot of bay area people (myself included) do and point and various sides and say ``that2that+that2that=that2that+thatthatthat. You can figure out which sides I'm referring to if you draw the diagram and practice a bit.

Actually, in that vein, you can do the "Stewart's Theorem Dance," invented by a Stuy math captain in the class ahead of me.  I'm not sure I can really explain it without actually demonstrating it, unfortunately :).


AMC AIME MIT college email percent USA(J)MO

I couldn't remember my stats on the AIME in past years.  I need this information for my MIT application.

My name is Frederick Mako (oh no, the secret is out).

Does anyone have the Summary of Results book from various years to look at how well I did (I live in Virginia) if I am even on there...

I am concerned with my AIME scores, but I'm pretty sure I got a 136 on the AMC 12 and 144 on the AMC 10.

Thanks, Please help me.

Have you tried asking your math teacher to whom the AIME results and the Summary Book were mailed?

Otherwise, write an email to the AMC office at
amcinfo@unl.edu
giving name, high school, city, state, zip code, year, whether it was AIME I or AIME II, any other information that you have that will help us locate your information, e.g. if you were a Visitor.
(By the way, for potential Visitors, this information is hard to locate since it is not stored with a school's records, and is one reason we charge a full registration for being a Visitor.)

If you cannot wait a couple of days to get an answer, then a phone call to our 800-number during business hours will get your information.  Please be patient and have your information ready, we are fielding a large number of calls due to the upcoming AMC 8.

Steve Dunbar
MAA Director for the AMC.

I think you can also check the VMT archives.

Wait were AMC results books mailed to every school that participated in the AMC contests?

Yes, they were.  The National Summary of Results and Awards was completed in August and mailed to the contest manager of every participating school.

Steve Dunbar
AMC Director

That is what I meant. Obviously, they didn't send one to all 200-some thousand contestants so...

[quote="Boy Soprano II"][quote="neelnal"][quote="Phelpedo"]Which individuals received one?[/quote]

Umm...you school advisor I think.[/quote]
He means which contestants, I think.  For instance, I received one, but I don't know why—it might be because I am a school winner, or because I am in the top $x$ percent, or because I'm in the top $n$ in my region, etc.[/quote]

Probably since you made MOP.

[quote="Elemennop"]I thought it was all usamo qualifiers, right?[/quote]

I think that's it.

Yeah, mailing it to MOPpers would be kind of strange, because as a group, we would probably be more interested in olympiad stuff than AMC and AIME stuff anyhow.

USAMO qualifiers sounds more plausible, especially since I wasn't really regionally outstanding on the AMCs anyhow.


Is there anyway to write a formal proof in LaTeX?
Also, I have a smaller question?
Just to give it a try since I'm new to LaTeX, I tried making the quadratic formula. For some reason, and can't get make a numerator, only a denominator comes up. What I thought would become the numerator, then became the whole number in front of the fraction.

Be sure to include enough {} to enclose elements.

x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}

That's exactly what mine looked like. When you click on the {, you see the highlight come up, and it highlighted the whole top. It really didn't help. Also, after you type something, say an inequality, how do you get it to type normal again. Lastly, how do you make the appearance more desirable, like margins, indents, and how do you transfer what it says in Adobe Reader to somewhere else?

[quote="mathgeek624"]Is there anyway to write a formal proof in LaTeX?[/quote]Ensure you are using the amsmath, amssymb and amsthm packages. Then use
[code]\begin{proof}
This is the first line of a proof \\
this is the last line of a proof
\end{proof}[/code]
[quote="mathgeek624"]When you click on the {, you see the highlight come up, and it highlighted the whole top. It really didn't help. Also, after you type something, say an inequality, how do you get it to type normal again.[/quote]Have you forgotten to use dollar signs or \ [ ... \ ] to enclose your maths? For example:
[code]% inline mode
$x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}$
% or display mode
\[x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}\][/code]
[quote="mathgeek624"]how do you make the appearance more desirable, like margins, indents[/quote]Use the geometry package - the manual is [url=http://tug.ctan.org/tex-archive/macros/latex/contrib/geometry/manual.pdf]here[/url]
[quote="mathgeek624"]how do you transfer what it says in Adobe Reader to somewhere else?[/quote]Press Select text and then copy to the clipboard. You won't be able to do this for maths only for text.


Νομίζω πως αυτή η άσκηση έχει ενδιαφέρον :

[color=red][b]ΑΣΚΗΣΗ[/b][/color]

 Έστω [b]Β[/b] το μέσο μιας ακτίνας [b]ΟΑ [/b]ενός κύκλου και δύο σημεία [b]Γ, Δ[/b] του κύκλου προς το ίδιο μέρος της

ευθείας [b]ΟΑ[/b] , ώστε οι γωνίες [b]ΑΒΔ[/b] και [b]ΟΒΓ[/b] να είναι ίσες. Αν [b]Ε[/b] είναι το μέσο του [b]ΓΔ[/b] και η

ευθεία [b]ΕΒ[/b] τέμνει τον κύκλο στο [b]Ζ [/b], να αποδειχθεί ότι [b]ΒΕ = ΒΖ[/b]

 Μπάμπης

Αυτή η άσκηση έχει περίεργα πράγματα. Είναι από το φόρουμ. Το ΟΑΔΓ πρέπει να βγει εγγράψιμο και αν η ΓΔ τέμνει την ΟΑ στο Ζ, τότε  ΖΑ=ΑΟ.
 Δεν μπορώ να καταλάβω πώς θα βγουν αυτά με ένα ομαλό τρόπο, μια και η λύση του φόρουμ έχει όλο έμμεσες αποδείξεις(με σημεία δηλαδή που συμπίπτουν κλπ- λύση delegat).
 Μπάμπης

Η αποδειξη του εγγραψιμου γινεται με χρηση νομου ημιτονων στα τριγωνα ΟΒΓ και ΟΒΔ.Επισης με θεωρημα Πτολεμαιου στο ΟΒΔΓ αποδεικνυεται οτι ΒΓ-ΒΔ=ΔΕ=ΓΕ.Ετσι μπορει καποιος να εκφρασει ολα τα μηκη του σχηματος συναρτησει του R και των τριγονομετρικων αριθμων της γωνιας ΔΒΑ.Ομως οι πραξεις μου φαγανε 3 σελιδες.Υπαρχουν πολλες αλλες προτασεις που προκυπτουν απο την ασκηση.Απεδειξα με μετρικες σχεσεις οτι ΑΕ=R.Μια προσεγγιση που μπορει να δινει λυση ειναι οτι αν Ι το αντιδιαμετρικο του Α αρκει το Ο να ειναι βαρυκεντρο στο τριγωνο ΙΕΖ,αλλα δεν εβγαλα ακρη.

Αφού δεν απαγορεύεται να βάλω μια λύση της άσκησης. Ίσως ο κος Στεργίου να αναζητεί μια πιο γεωμετρική λύση, αλλά η λύση μου είναι με αναλυτική γεωμετρία. Έχουμε λοιπόν

Ο = (0,0)
Α = (2,0)
Β = (1,0)
$ \Delta = 2(\cos{\theta},\sin{\theta})$
$ \Gamma = 2(\cos{\phi},\sin{\phi})$
Έχουμε
$ \frac{AB \cdot B \Delta}{|AB| |B \Delta| } =  \frac{OB \cdot B \Gamma}{|OB| |B \Gamma| }$.
Υψώνοντας στο τετράγωνο, λύνουμε το σύστημα και βρίσκουμε ότι
$ \Gamma = \frac{2}{5 - 4\cos{\theta}}(4 - 5 \cos{\theta},3 \sin{\theta})$ (Η άλλη λύση Γ = Δ απορρίπτεται)
Παίρνουμε
$ E = \frac{4}{5 - 4\cos{\theta}}(\sin^2{\theta}, 2\sin{\theta} - \cos{\theta}sin{\theta})$
και
$ Z' = \frac{1}{5 - 4\cos{\theta}}(10 - 8\cos{\theta} - 4\sin^2{\theta},4\cos{\theta}\sin{\theta} - 8\sin{\theta)}$
όπου ΒΕ = ΒΖ'. Ελέγχουμε τώρα ότι το Ζ' ανήκει στον κύκλο.

Ας βάλει κάποιος και μια καθαρά γεωμετρική λύση.

[quote="Demetres"]Αφού δεν απαγορεύεται να βάλω μια λύση της άσκησης. Ίσως ο κος Στεργίου να αναζητεί μια πιο γεωμετρική λύση, αλλά η λύση μου είναι με αναλυτική γεωμετρία. Έχουμε λοιπόν

Ο = (0,0)
Α = (2,0)
Β = (1,0)
$ \Delta = 2(\cos{\theta},\sin{\theta})$
$ \Gamma = 2(\cos{\phi},\sin{\phi})$
Έχουμε
$ \frac {AB \cdot B \Delta}{|AB| |B \Delta| } = \frac {OB \cdot B \Gamma}{|OB| |B \Gamma| }$.
Υψώνοντας στο τετράγωνο, λύνουμε το σύστημα και βρίσκουμε ότι
$ \Gamma = \frac {2}{5 - 4\cos{\theta}}(4 - 5 \cos{\theta},3 \sin{\theta})$ (Η άλλη λύση Γ = Δ απορρίπτεται)
Παίρνουμε
$ E = \frac {4}{5 - 4\cos{\theta}}(\sin^2{\theta}, 2\sin{\theta} - \cos{\theta}sin{\theta})$
και
$ Z' = \frac {1}{5 - 4\cos{\theta}}(10 - 8\cos{\theta} - 4\sin^2{\theta},4\cos{\theta}\sin{\theta} - 8\sin{\theta)}$
όπου ΒΕ = ΒΖ'. Ελέγχουμε τώρα ότι το Ζ' ανήκει στον κύκλο.

Ας βάλει κάποιος και μια καθαρά γεωμετρική λύση.[/quote]

Δημήτρη, σε ευχαριστώ που ασχολήθεκες με την άσκηση. Αύριο στο σχολείο θα την τυπώσω και θα  μελετήσω καλά τη λύση, μήπως πάρω και καμία ιδέα και την προσαρμώσω σε λίγο πιο ...καθαρή γεωμετρία.
 Να πω όμως ότι τα μαθηματικά έχουν μια ενιαία οντότητα και πως αγαπώ κάθε λύση. Υπάρχουν εξαιρετικά θεωρήματα που αποδεικνύονται πιο εύκολα τριγωνομετρικά. Από τη Ρουμανία μάλιστα που στείλανε ένα βιβλίο το οποίο έχει όλο ασκήσεις γεωμετρίας  με τριγωνομετρικές λύσεις. Και επιπλέον , οι καλοί χειριστές της τριγωνομετρίας σπάνια χάνουν γεωεμτρία στις ΙΜΟ.Το ότι ψάχνουμε συχνά καθαρά γεωμετρικές λύσεις γίνεται πιο πολύ σαν ένας αγώνας  να βρούμε αυτές τις  ...όμορφες βοηθητικές γραμμούλες  που έχουν μια ξεχωριστή γλύκα. Για κανένα άλλο λόγο. Αυτό που αξίζει σε ένα πρόβλημα είναι η λύση και μετά πώς έγινε η λύση ή πώς μπορεί να βελτιωθεί κλπ.
 Καλό βράδυ και καλές εκλογές!

                                Μπάμπης

[quote="xp"]Μπαμπη καλησπερα
Λοιπον η επομενη παρεμβαση σου βοηθησε
Αν η ΓΔ τεμνει την ΟΑ στο Η τοτε ο κυκλος ειναι απολλωνιος (ευκολα βγαίνει γιατι απ τις γωνιες που δίνεις ισες βγαίνουν οι ΔΑ και ΔΘ -το Θ ειναι αντιδιαμ.του Α- διχοτομοι του τριγωνου ΒΔΗ)
Απ τα συζυγη αρμονικα πανω στην ΘΗ βγαζουμε ΟΑ=ΑΗ=ρ και αφου ΟΕΗ=1 ορθη ειναι ΕΑ=(ΟΗ)/2=ρ=(ΟΖ)
Αν φερουμε την καθετο  ΑΣ στην ΓΗ και αυτη τεμνει τον κυκλο στο Ζ και το ΟΖΑΕ ειναι παρ/μο......
(Την ειχα πιο πολυπλοκα στο τελος αλλα μια παραπλησια λυση ενος φιλου την συντομευσε στο τελος..)[/quote]

 :clap2: Πολύ ωραία λύση!! Ετσι προσπαθούσα να το κάνω κι εγώ αλλά πραγματικά είχα κολλήσει :blush:
Βασικά έχω και μια απορία... μπορείτε λίγο να μου εξηγήσετε πώς προκύπτει εύκολα ότι "οι ΔΑ και ΔΘ -το Θ ειναι αντιδιαμ.του Α- διχοτομοι του τριγωνου ΒΔΗ", από τη δοσμένη γωνιακή σχέση...

Εγώ αυτό για να το βγάλω απέδειξα πρώτα ότι ΒΗ εξωτερική διχοτόμος του ΒΓΔ και μετά από αυτό προκύπτει ότι αν ΒΚ η εσωτερική διχοτόμος του ΒΓΔ τότε ΒΚ είναι η πολική του Η (επειδή είναι και κάθετη στην ΟΚ) κι από αυτό μου βγήκε ότι Β,Κ συζηγή αρμονικά των Α, Θ κι έτσι έδειξα ότι ΔΑ, ΔΘ είναι διχοτόμοι. Γενικώς δηλαδή αισθάνομαι ότι πήγα "μέσω Λαμίας". Μπορείτε να μου εξηγήσετε σας παρακαλώ εναν πιο σύντομο δρόμο?

Δημητρη η ΘΓΑ γωνια βαινει σε ημικύκλιο αρα η ΓΑ ειναι εσωτερικη διχοτομος............
Σιλ και Δημητρη ευχαριστω για τα καλα σας λογια.......

[quote="silouan"]κ. Χρήστο και Δημήτη εξαιρετικά ! κ. Μπάμπη αξίζει να δείτε αυτή τη λύση . Είναι μακράν καλύτερη από αυτές του αγγλικού  :)[/quote]

Σιλουανέ,
 όχι μόνο θα τις δω και θα τις απολεύσω :)  , αλλά και  θα έλεγα να τις καθαρογράψεις και να τις βάλουμε στο βιβλίο που ελπίζω να φτιάξουμε κάποια στιγμή μαζί(θα βοηθήσουν πιστεύω όλοι) και να βάλεις και τα ονόματα των λυτών(Χρήστου και Δημήτρη , ώστε να τη βλέπουν οι νέες γενιες!)).

              Καλό βράδυ σε όλους (βγαίνω μια βόλτα να ξεσκάσω .Σήμερα άγγιξα και χρονικά(ηλικιακά ) το $ 7^2$  :lol: !)!

 Μπάμπης

Κυριε Στεργιου να σας δωσω και τις δικες μου ευχες.Να ειστε ακμαιος και ψυχη του Forum για πολλα χρονια ακομη.

[quote="bilot4"][quote="stergiu"]
                Καλό βράδυ σε όλους (βγαίνω μια βόλτα να ξεσκάσω .Σήμερα άγγιξα και χρονικά(ηλικιακά ) το $ 7^2$  :lol: !)!
 Μπάμπης[/quote]

Αν κατάλαβα σωστά, χρόνια σας πολλά κ. Στεργίου. Να τα εκατοστήσετε!

Παρεπιμπτόντως στην παραπάνω άσκηση σε καμία περίπτωση δεν μπορώ να θεωρηθώ "λύτης"  :)[/quote]

Χρόνια πολά και από εμένα.

Σας ευχαριστώ ολόψυχα όλους !

 Να ξέρετε πως εσείς είστε η καρδιά αυτού του χώρου και γω μόνο μια μικρή φλεβίτσα!Η χαρά μου είναι να είμαστε όλοι  μια ψυχή , να έχουμε στόχους και να αφήσουμε κάτι καλό στις γενιές που έρχονται.
 Δες σας κρύβω ότι βασική μου επιδίωξη είναι να βοηθήσω αυτούς που επέλεξαν να σπουδάσουν μαθηματικά να αφήσουν πίσω τους και συγγγραφικό έργο πάνω στις ολυμπιάδες.Έκανα ήδη μια νύξη στο προηγούμενο μήνυμα. Από τη δική μου τη γενιά λίγα μπορέσουμε να γράψουμε, αλλά εσείς που δεν  ξεκινάτε από το μηδέν  ,  είμαι σίγουρος ότι θα γράψετε καλά και διδακτικά βιβλία για να προωθήσουν την ολυμπιακή παιδεία και να κρατήσουν ζωντανή τη μαθηματική σκέψη και στην Ελλάδα.
[color=red][b]    Σας  ευχαριστώ ξανά για τις ζεστές ευχές σας ! [/b][/color]
 Μπάμπης
ΧΧΧΧΧΧΧΧΧΧΧΧΧΧΧΧΧΧΧΧΧΧΧΧΧ

[quote="r_boris"]Μπάμπη
Επειδή η άσκηση που πρότεινες καθώς και η λύση του Χρήστου μου φάνηκαν ΘΕΟΚΟΥΚΛΑΡΕΣ σκέφτηκα ότι θα μπορούσε να πλουτιστεί με μερικά ακόμα ερωτήματα όπως   το ΓΤ του Μ μέσου της ΕΘ , το μέγιστο της ΒΕ και ακόμη πολλά άλλα που μπορεί κανείς νά βρει σκαλίζοντας λίγο  το θέμα[/quote]

Ροδόλφε, μια και έχεις ασχοληθεί και έχεις ψάξει πιο βαθειά το θέμα, γράψε αν βρεις λίγο χρόνο όλα τα ερωτήματα με τις λυσούλες και βάλτα στο αρχείο του κλαμπ (Word ή pdf) για να μην ξεχαστούν.

 Καλό μεσημέρι !

 Μπάμπης

Να μερικά ερωτήματα για την εν λόγω άσκηση (με λιγο διαφορετική εκφώνηση

Έστω ΟΑ ακτίνα κύκλου (Ο,ρ) με μέσον το  Β. Χορδή ΔΙ διέρχεται από το Β και Γ είναι το συμμετρικό του Ι ως προς την ΟΑ

1.	Να δείξετε ότι η ΓΔ διέρχεται από σταθερό σημείο
2.	Να δείξετε ότι τα Α,Θ,Β,Γ είναι συζυγή αρμονικά ,οπου Θ είναι το αντιδιαμετρικό του Α
3.	Να δείξετε ότι 2ΒΔ=ΔΗ όπου Η κοινό σημείο των ΓΔ και ΟΑ
4.	Να βρείτε τον Γ.Τ του μέσου Ε της ΓΔ
5.	Να βρείτε τον Γ.Τ του μέσου Κτης ΘΕ
6.	Να δείξετε ότι ΕΒ=ΒΖ όπου Ζ είναι σημείο του κυκλου και της ΕΒ στο άλλο ημιεπίπεδο από οποίο
        βρίσκονται τα Γ και Δ ως προς την ΟΑ
7.	Να δείξετε ότι τα σημεία Ο,Β,Δ,Γ είναι ομοκυκλικά
8.	Αν r η ακτίνα του προηγούμενου κύκλου δείξτε  ότι $ 2r.OE\equal{}\rho^2$
9.	Να βρείτε το μέγιστο του ΕΖ
10.	Να βρείτε την μέγιστη οξεία γωνία που σχηματίζει η ΓΔ με την ΟΑ

Τις λύσεις λίγο αργότερα όταν βρω χρόνο

[quote="r_boris"]Να μερικά ερωτήματα για την εν λόγω άσκηση (με λιγο διαφορετική εκφώνηση

Έστω ΟΑ ακτίνα κύκλου (Ο,ρ) με μέσον το  Β. Χορδή ΔΙ διέρχεται από το Β και Γ είναι το συμμετρικό του Ι ως προς την ΟΑ

1.	Να δείξετε ότι η ΓΔ διέρχεται από σταθερό σημείο
2.	Να δείξετε ότι τα Α,Θ,Β,Γ είναι συζυγή αρμονικά ,οπου Θ είναι το αντιδιαμετρικό του Α
3.	Να δείξετε ότι 2ΒΔ=ΔΗ όπου Η κοινό σημείο των ΓΔ και ΟΑ
4.	Να βρείτε τον Γ.Τ του μέσου Ε της ΓΔ
5.	Να βρείτε τον Γ.Τ του μέσου Κτης ΘΕ
6.	Να δείξετε ότι ΕΒ=ΒΖ όπου Ζ είναι σημείο του κυκλου και της ΕΒ στο άλλο ημιεπίπεδο από οποίο
        βρίσκονται τα Γ και Δ ως προς την ΟΑ
7.	Να δείξετε ότι τα σημεία Ο,Β,Δ,Γ είναι ομοκυκλικά
8.	Αν r η ακτίνα του προηγούμενου κύκλου δείξτε  ότι $ 2r.OE \equal{} \rho^2$
9.	Να βρείτε το μέγιστο του ΕΖ
10.	Να βρείτε την μέγιστη οξεία γωνία που σχηματίζει η ΓΔ με την ΟΑ

Τις λύσεις λίγο αργότερα όταν βρω χρόνο[/quote]
Ζητώ συγνώμη για το 9ο ερώτημα Λόγω βιασύνης ξέφυγε ότι ζητούσα το μέγιστο της προβολής Της ΕΖ στην ΟΑ και όχι το μέγιστο της ΕΖ


inequalities
Euler
geometry
circumcircle
inequalities unsolved

Let a,b,c be three sides of the triangle. Prove that:
   $ 9(a^2\plus{}b^2\plus{}c^2)(ab\plus{}bc\plus{}ca)\geq(a\plus{}b\plus{}c)^4$

[quote="math4y"]Let a,b,c be three sides of the triangle. Prove that:
   $ 9(a^2 \plus{} b^2 \plus{} c^2)(ab \plus{} bc \plus{} ca)\geq(a \plus{} b \plus{} c)^4$[/quote]

substitution  $ a\equal{}x\plus{}y,b\equal{}y\plus{}z,c\equal{}z\plus{}x$($ x,y,z>0$)

The following inequality is equivalent to

$ 2(x^2\plus{}5xy\plus{}y^2\plus{}5zx\plus{}5yz\plus{}z^2)(x^2\minus{}xy\plus{}y^2\minus{}zx\minus{}yz\plus{}z^2)\geq 0$

obvious true.

[quote="math4y"]Let a,b,c be three sides of the triangle. Prove that:
   $ 9(a^2 \plus{} b^2 \plus{} c^2)(ab \plus{} bc \plus{} ca)\geq(a \plus{} b \plus{} c)^4$[/quote]
\[ \Leftrightarrow 9\left( {2s^2 \minus{} 8Rr \minus{} 2r^2 } \right)\left( {s^2 \plus{} 4Rr \plus{} r^2 } \right) \ge 16s^4\]
\[ \Leftrightarrow 2\left( {s^2 \minus{} 12Rr \minus{} 3r^2} \right)\left( {s^2 \plus{} 12Rr \plus{} 3r^2} \right) \ge 0\]
\[ \Leftrightarrow s^2 \ge 3r(4R \plus{} r),\]
but we know that $ s^2 \ge r(16 R \minus{} 5 r)$, so just need to prove

\[ 16 R \minus{} 5 r \ge 4R\plus{}r\]
\[ \Leftrightarrow R\ge 2r,\]
which is true by Euler inequality.

[quote="kuing"]
$ 9\left( {2s^2 \minus{} 8Rr \minus{} 2r^2 } \right)\left( {s^2 \plus{} 4Rr \plus{} r^2 } \right) \ge 16s^4$
[/quote]
kuing,can you tell me turn s=?,R=?,r=?
Why we have :$ s^2 \geq r(16R \minus{} 5r)$?

[quote="math4y"][quote="kuing"]
$ 9\left( {2s^2 \minus{} 8Rr \minus{} 2r^2 } \right)\left( {s^2 \plus{} 4Rr \plus{} r^2 } \right) \ge 16s^4$
[/quote]
kuing,can you tell me turn s=?,R=?,r=?
[/quote]

$ 2s\equal{}a\plus{}b\plus{}c$, $ R$ and $ r$ are circumradius and apothem of triangle...

[quote="math4y"]Let a,b,c be three sides of the triangle. Prove that:
   $ 9(a^2 \plus{} b^2 \plus{} c^2)(ab \plus{} bc \plus{} ca)\geq(a \plus{} b \plus{} c)^4$[/quote]
We have to prove that :

$ \frac {3(a^2 \plus{} b^2 \plus{} c^2)}{(a \plus{} b \plus{} c)^2} \geq \frac {(a \plus{} b \plus{} c)^2}{3(ab \plus{} bc \plus{} ca)}$ ,

Substracting $ 1$ from the two sides,the inequality becomes :

$ \frac {2(a^2 \plus{} b^2 \plus{} c^2 \minus{} ab \minus{} bc \minus{} ca)}{(a \plus{} b \plus{} c)^2} \geq \frac {a^2 \plus{} b^2 \plus{} c^2 \minus{} ab \minus{} bc \minus{} ca}{3(ab \plus{} bc \plus{} ca)}$ ,

Which is clearly true because :

$ a^2 \plus{} b^2 \plus{} c^2 \minus{} ab \minus{} bc \minus{} ca \geq 0$ and $ \frac {2}{(a \plus{} b \plus{} c)^2} \minus{} \frac {1}{3(ab \plus{} bc \plus{} ca)} \equal{} \frac {4(ab \plus{} bc \plus{} ca) \minus{} a^2 \minus{} b^2 \minus{} c^2}{(a \plus{} b \plus{} c)^2} > 0$

[b]Remark :[/b]

I think that the following stronger inequality also holds and may be proved by the same method :

$ (7a^2 \plus{} 7b^2 \plus{} 7c^2 \plus{} 2ab \plus{} 2bc \plus{} 2ca)(ab \plus{} bc \plus{} ca) \geq (a \plus{} b \plus{} c)^4$

For $ a,b$ and $ c$ sides of any triangle.

[quote="math4y"][quote="kuing"]
$ 9\left( {2s^2 \minus{} 8Rr \minus{} 2r^2 } \right)\left( {s^2 \plus{} 4Rr \plus{} r^2 } \right) \ge 16s^4$
[/quote]
Why we have :$ s^2 \geq r(16R \minus{} 5r)$?[/quote]

that's Gerretsen Inequality...


number theory
least common multiple
modular arithmetic

A positive integer is 3 more than a multiple of 4 and 4 more than a multiple of 5. What is the least integer it could be?

The unit's digit must end in a 4 or a 9, since it is 4 more than amultiple of 5. Any multiple of 4 is even, so 3 more than a multiple of 4 would be odd, so the number must have a units digit of 9. Thus, the multiple of 4 must have a units digit of 6, so the smallest possible number is 16+3=19.

simpler solution:
this situation means that the required no. is 1 less than a multiple of 4 and 5. lcm of 4 and 5 =20.
so, the answer is 19.

[hide="Modular Arithmetic"]
$x\equiv 3\mod 4$ and $x\equiv 4\mod 5$.

Adding 1 yields $x+1\equiv 0\mod 20\implies x=\boxed{19}$[/hide]


function
calculus
integration
algebra proposed
algebra

I hope that the following one wasn't given in an olympiad till now. It's "mine", but it's easy. Anyway, I think it's not trivial and I decided to post it.

Let $f$ be defined on the positive integers with positive integers values such that for all $n>1$ we have \[f(n+2)-2f(n+1)+f(n)=f(f(n-1)).\]
Prove that there exist $a$ and $b$ such that whenever $n>a$ we have $f(n)=b$.

[i]Edited by Myth[/i]

When you say 'positive' do you include 0? because if you don't (I, for one, don't include 0 when I say 'positive'), then it's not right: if you take n sufficiently large the thing on the left will be 0, so f(f(n-1))=0 for sufficiently large n, meaning that f will be 0 at some point, but f has positive values.

O, sorry. f takes only nonnegative integer values and it is defined on the positive integral numbers. Sorry again!

We have
\[f(n+2)-f(n+1)=f(n+1)-f(n)+f(f(n-1))\geq f(n+1)-f(n).\] Therefore $f(n+1)-f(n)$ is a nondecreasing sequence. It can't be negative all the time (otherwise $f(n)<0$ for some all large $n$). So thre is a moment s.t. $f(n+1)-f(n)\geq 0$.
Suppose there is a $n_0$ s.t. $f(n_0+1)-f(n_0)\geq 1$. It follows that $f(n)$ increase for $n>n_0$. So there is $n_1>n_0+2$ s.t. $f(n_1-1)>n_0$. It follows that $f(n_1+2)-f(n_1+1)=f(n_1+1)-f(n_1)+f(f(n_1-1))\geq f(n_1+1)-f(n_1)+1\geq 2$, since $f(f(n_1-1))>f(n_0)\geq 0$.
As $f(n+1)-f(n)$ is nondecreasing, it means that $f(n)\geq 2n-c$ for some $c$. In particular, $f(n)\geq n+4$ for all large $n$. Thus if $n$ is sufficiently large,
\[f(n+2)-f(n+1)=f(n+1)-f(n)+f(f(n-1))\geq f(f(n-1))\geq f(n+3)>f(n+2)\]
-- contradiction.

Therefore, $f(n+1)-f(n)=0$ for all $n\geq a$ for certain $a$.


helo
resolve in IR:
$ x^3\plus{}x\plus{}2\equal{}0$

Notice that $ x_1\equal{}\minus{}1$ is a solution then $ (x\plus{}1)| x^3\plus{}x\plus{}2$
so $ x^3\plus{}x\plus{}2\equal{}(x\plus{}1)(x^2\minus{}x\plus{}2)$ then apply the quadratic formula...


inequalities
function
search
inequalities open

I've been trying to solve a few inequalities, including the Bernoulli inequality of the form

pi( 1+xi) >= 1 + sigma (xi) (both sigma and pi are from 1 to n)

xi >= -1, all xi have the same sign ('i' is in subscript)

Anyway, I havent seen the accepted proof because I'm more excited to do it my way. Basically, its come down to this.

f(x) is a monotonic function (actually in ln(1+x) but I want a more general approach). For the same restrictions on xi that I've mentioned above,, I have to find an inequality between f(sigma(x)) vs. sigma(f(x)). After some conjecturing, I'm starting to think that inserting an inequality between the two depends on behaviour of the function relative to y=x ), ie. if f(x)>=x or f(x)<=0 is so far agreeing with the results.

Can someone please help prove this formally. And please nothing too fancy. The best I know is single variable calculus.

well if you know that $ f$ is convex or concave then you can use jensen:
[b]let $ f$ be a convex function and let $ w_{1},...,w_{n}$ be constant numbers with sum equal to $ 1$,then we have that:
$ f(\sum w_{i}x_{i})\leq \sum w_{i}f(x_{i})$
(for the case that $ f$ is concave then the inequality is reversed...)[/b]
now for the case that $ w_{1}=...=w_{n}=\frac{1}{n}$ we get that:
$ nf(\frac{\sum x_{i}}{n})\leq \sum f(x_{i})$

note that $ f$ is convex in the interval $ I$ iff for all $ \lambda \in [0,1]$ we have:
$ f(\lambda x+(1-\lambda)y)\leq \lambda f(x)+(1-\lambda)f(y)$
i.e $ f(w_{1}x+w_{2}y)\leq w_{1}f(x)+w_{2}f(y)$ for positive numbers $ w_{1},w_{2}$ s.t $ w_{1}+w_{2}=1$...
and if the sign was reversed,then $ f$ is concave...
also for the convexity of $ f$ you can compute $ f''$ then $ f$ is convex if $ f''>0$ and concave if $ f''<0$...
you can also search jensen to find a good aplication for it,and note that the proof of jensen is just induction...  :)

[quote="BaBaK Ghalebi"]and note that the proof of jensen is just induction...  :)[/quote]
here is a link:
[url]http://www.mathlinks.ro/Forum/viewtopic.php?t=157255[/url]

That sounds like a wonderful result. I got as far as $ nf(\frac{\sum x_{i}}{n})\leq \sum f(x_{i})$ for a continuous interval by the mean value theorem, so again I was conjecturing that it should work for discrete values too.

Also, I was thinking convexity and concavity might play a part too, in case of Bernoulli's inequality, the function in ln(1+x) so its convex, increasing and always less that or equal to x.

All in all, this looks wonderful, so I guess I'm not far off.

Hmm, I guess this is getting stuck exactly where it did yesterday.

The 'n' in
$ f(\sum w_{i}x_{i})\leq \sum w_{i}f(x_{i})$ is causing all the trouble, I really need a comparison of

$ f(\sum x_{i})$ and $ \sum f(x_{i})$

and I know this about the function ln(1+x):

Monotonically increasing
Convex
<=x

I'm thinking that Jensen uses the first two preoperties, but the third has some part to play to get a further result that compares $ f(\sum x_{i})$ and $ \sum f(x_{i})$ . I just cant prove it, but as I said in my first post, I cant help but feel that its important. Maybe I'm totally wrong, and I've dreamt this up, but please help try to get a more specific case of Jensen using this, or maybe contradict my third point to prove I'm thinking wrong.

One more thing

$ f(x) \leq x$
$ f(x_{i}) \leq (x_{i})$
$ \sum f(x_{i}) \leq \sum (x_{i})$

Looks somewhat helpful to me.

also there is another inequality called [b]karamata[/b] which might help you:
it says that :
[b]if $ x_{1}\geq ...\geq x_{n},y_{1}\geq ...\geq y_{n}$ and for every $ k$ we had $ x_{1}+...+x_{k}\geq y_{1}+...+y_{k}$ then we have:
$ f(x_{1})+f(x_{2})+...+f(x_{n})\geq f(y_{1})+f(y_{2})+...+f(y_{n})$
where $ f$ is a convex function...[/b]

for more details in karamata`s inequality you can use this useful link:
[url]http://www.mathlinks.ro/Forum/viewtopic.php?t=14975[/url]

[quote="BaBaK Ghalebi"]well there is another fact which might help you:
[b]let $ f$ be a convex function,then for numbers $ a_{1},a_{2},...,a_{n}$ we have:
$ f(a_{1})+f(a_{2})+...+f(a_{n})\leq f(a_{1}+a_{2}+...+a_{n})+(n-1)f(0)$[/b][/quote]

Wow, this one does it, since $ f(0)$ is 0. Can I get more information on this one please.

Someone please help me out with this, this inequality is posted here as well:

http://www.artofproblemsolving.com/Forum/viewtopic.php?search_id=536430925&t=160205

But nobody has proven it or provided a link. Is it even a calid inequality. It doesn't seem like it, at least not unless f(x) is monotonic.


[code]      1
    1 1 1
  1 2 3 2 1
1 3 6 7 6 3 1[/code]
makes it look nicer.  Your question seems very easy to me, though: [hide]Each row will clearly contain a 1.  If you mean "other than the 1's," however, it's still easy -- [hide]the number in the middle must always be odd.  I can't guarantee any other odd numbers, since in the next row every number but the central and edge numbers are even.[/hide][/hide]

Perhaps you meant even rather than odd?

If you meant "even," [hide="hint"]Notice that odd-even-odd --> even.. there is a pattern..[/hide]

Edit--oops, I think this doesn't work  :blush: but I think the solution would be along these lines.. maybe...
Edit again--OK, now I think you have to think about [hide]how you would get an all odd row in the first place[/hide].

Sorry... :blush:
I meant even, like 2 etc. But i don't understand the hint from probability1.01... I mean 1 3 6 is not odd - even - odd...hm, maybe without the "one"s, but how do you prove the pattern?

Any other ideas?


Can somebody explain to me the Chinese Remainder Theroem WITHOUT modular arithmetic.

This can be very useful on mathcounts.
Google it if you don't know what type of problems i'm talking about.

Mostly remainder problems

lol your chinese yet you don't know chinese remainder theorem? do you not know modular arthmitic?

and why is it called chinese remainder theorem?

i know the basics of modular arithmetic, but I don't know much.
Oh and even my parents don't know what it is :blush:

If you don't understand a concept, search it on this site.  Most likely you will find at least one thing on it.

[quote="shinwoo"]wow you guys are really helpful....

since someone asked this question, you guys should be answering not telling where you can find the answer of it[/quote]

Yeah i think we should  :oops:

Here is where someone else asked about the Chinese Remainder Theorem, and I posted about it:

[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=65217]www.artofproblemsolving.com/Forum/viewtopic.php?t=65217[/url]

It is based entirely on Modular arithmatic, so if you don't understand it, I will try to explain it to you:
[hide="Explaining Modular Arithmatic"]When you are given an equation in the form $n\equiv x\text{ (mod }y)$, What it really means is that $n-x$ is divisible by $y$.

You can add or subtract integer multiples of $y$ from $x$ and the equation will always remain true.  For example:

$7\equiv2\text{ (mod }5)$
$7\equiv-3\text{ (mod }5)$
$7\equiv7\text{ (mod }5)$

This can also be written in other terms:

$n=2+5x$

This also shows that $n$ is 2 more than a multiple of 5, because $5x$ is always a multiple of 5.

I think that is about it for basics...  If you have any more questions, just ask me and I will try to answer them.[/hide]


Keeping your work clean and understandable  is very impportant when doing math. Word is often to difficult to comprehend when using mathmetical symbols.
[b]If you have always wanted to know how to create those beautiful documents with mathematical formatting and stuff, and don't know  how, then you should read this.[/b] I know from asking ComplexZeta there are many ways, but this one is a 100% free way. First you need to download the Miketex from [url]http://www.miktex.org/[/url]. This is the program that has the complilation programs. Make you follow instructions caferully and install that. Then donwload Texnicenter at [url]http://www.toolscenter.org/[/url]. To use TExnic center you need to know how to create LATEX documents. It is very complicated and I just edit documents in Maple and export it in this Latex doucment (Mathematica works just as good). Once you open the document in Texnic center, all you have to do is click a button to make DVIs, PDFs, PS(Poscript) from the Latex document. Miketex includes only a DVI viewer, so I suggest to view PDFs and PSs you should get Adobe Acrobat and Ghostscript.
HAVE FUN![/url]

At AoPS we make a lot of our documents in TeX.

By the way, you can actually do everything you want just using miktex, notepad and dos :) just type in a line in dos and it'll compile it into a dvi or pdf for you, or you could set up a simple program to make that easier. Though having something like texnicenter may be useful too.

[quote="ComplexZeta"]I downloaded a copy of TeXnicCenter, and I really like it. I would definitely recommend LaTeX for someone who wants to make documents that look professional. You can make anything you want (albeit possibly with a lot of work). One nice online introduction is the [url=http://www.ctan.org/tex-archive/info/lshort/english/lshort.pdf]The Not So Short Introduction to LaTeX2e[/url].[/quote]

Oh, it's a PDF.  Well, that makes sense.  Warning: that thing is about 2 MB, which can be ugly if your connection times out frequently.

LaTeX is an absolutely beautiful way of writing documents for mathematics (and / or research publications).  I am now fairly good at latex for that matter, so if you have any specific questions, you can email me!

Yeah LaTeX is brilliant, I've been reading up on it recently. You are able to do so many complex things so easily.. I don't think I'll ever need to use Word again :)


[quote="delta"][hide]Prove that $1*2+2*3+...+n*(n+1)=\frac{n(n+1)(n+2)}3$
[/hide][/quote]
Very good. We'll kill this thing with induction.
[u]Base step[/u]: for $n = 1$, have $1*2=\frac{1*2*3}3=2$, whcih is obviously true.
[u]Induction step[/u]: assume the formula works for $n = k$. Have:
$1*2+2*3+...+k*(k+1)=\frac{k(k+1)(k+2)}3$
We want to prove the formula works for $n=k+1$. Add $(k+1)(k+2)$ to both sides to get:
$1*2+2*3+...+(k+1)*(k+2)=\frac{k(k+1)(k+2)}3+(k+1)(k+2)$
Simplifying, obtain:
$1*2+2*3+...+(k+1)*(k+2)=\frac{(k+1)(k+2)(k+3)}3$
Hence the formula holds for all n and we are done. :)

we can also prove it using telescoping method

$\sum^n_{i=1}(i+1)(i+2)(i+3)-\sum^n_{i=1}i(i+1)(i+2)=3\sum^n_{i=1}(i+1)(i+2)$

Notice how each sum could be written as n^2-n, e.g 2005^2-2005.

[quote="srulikbd"][quote="shyong"][hide="some interesting pattern"]

$\sum^{n}_{i=1}i=\frac{1}{2}n(n+1)$

$\sum^{n}_{i=1}i(i+1)=\frac{1}{3}n(n+1)(n+2)$

$\sum^{n}_{i=1}i(i+1)(i+2)=\frac{1}{4}n(n+1)(n+2)(n+3)$

So what did you see  :D

[/hide][/quote]
cool! how do you prove it? induction?[/quote]
Yes. Just combine the formulas for $\sum^{n}_{i=1}i$ and $\sum^{n}_{i=1}i(i+1)...(i+(n-1))$ and do some algebra and you are done.


vector
real analysis
real analysis unsolved

Prove that normed vector space $E$ is complete iff all absolute convergent series in $E$ are convergent in $E$.

I will prove just one implication :)
Take $u_n$ a Cauchy sequence. $u_n=\sum_1^n v_k$ where $v_k=u_k - u_{k-1}$. One want to show that the series $\sum v_k$ converges:

For any $n \geq 1$, there exists an integer $N(n)$ such that for any $p,q \geq N(n)$, $|\sum_p^q v_k| \leq 2^{-n}$.
If $w_n=\sum_{N(n)}^{N(n+1)-1}v_k$, the series $\sum w_k$ is absolutely convergent. By our hypothesis, this series converges towards a limit $L$:
From here, it is clear that the sequence $u_n$ converges towards $L$.


algebra
polynomial
calculus
limit
algebra proposed

Suppose the sequence {x_n} satisfy :
  for all P is  polynomial with degree 2 then :
   lim{n-> \infty } (x_n + x_p(n))=0
  Do we have lim x_n=0 ?

Consider the sequence x_n=cos(pi*log2log2(n))(*) we will solve it.
But I think It's difficult to find (*).
Do you have nice sol?

I think the place of this problem should be in
College Playground ->Calulus Analysis ->Proposed & Own Problems

I think you also have the condition for any n \in N, p(n) \in N
to define x_p(n)

The subsequence

x_p(n) + x_p^2(n) --> 0 (2)
since it is a subsequence of x_n + x_p(n) -->0 (1)  where p^2(n)=p(p(n))

Taking again a subsequence
x_p^2(n) + x_p^3(n) --->0 (3)

(1) - (2) implies x_n - x_p^2(n) with (3) we get

x_n - x_p^k(n)-->0  when n->+oo
( where p^k(n)=pop...p(n) )

that's all for now

Hi A1lqdSchool,

log2log2(n) do you mean  (ln2)*(ln2)*n or (ln2)*(ln(2n)) or
log_2(log_2(n)) where log_2 is logarithme base 2 ?

+  " x_n - x_p^k(n)-->0 when n->+oo " (*)
From (*) What thing can you obtain?
+ log2 is logarithme base 2

As I said:sequence {x_n} is x_n= cos(pi*log2log2(n))
You can prove \lim (x_n+ x_p(n)) =0
but |x_2^2^k|=1 for all k is natural , thus x_n can't get to 0


Mostrar que si $f$ es un polinomio  de grado $n$, tal que $f(0) = f(1) = ... = f(n-1) = a$ y $f(n)=x$ (con $x$ entero y tal que $0\le x \le n$),
entonces $f(n+1) = (n+1)(x-1)+a$

[color=blue]¿Puedo asumir que los números 1 ; 2 ; … ; [i]n[/i] - 1 son soluciones del polinomio [i]f[/i]([i]k[/i]) - [i]a[/i]?

Es fácil averiguar que el coeficiente independiente de [i]f[/i]([i]k[/i]) es [i]a[/i]. Parece que me estanco… ¿Cómo hago para dar con [i]f[/i]([i]n[/i] + 1) a partir de funciones de valores anteriores? ¿Debo hacer descomposiciones por binomio de Newton o existe algún camino más práctico?[/color]

ay  tony  ese es el  problma q te di una  vez cuando  veniste a mi  ksa  jeje.

  hint :
 [hide]considera el  nuevo polinomio f(x)-1[/hide]

efectivamente a partir del que me diste puse en lugar de 1 a, y en lugar de 0 puse lo que dice, pero basicamente es la misma solucion.
la sugerencia seria buscar el polinomio f(x) - a, ver que este polinomio es unico ya que n+1 imagenes de este polinomio estan determinados, sustituyes n+1 y ya.

[color=blue]Una pequeña observación que no tengo mucho tiempo…

He notado que si [i]f[/i]([i]x[/i]) se divide por [i]n[/i] +1, me da resto [i]a[/i].[/color]

Tiene un error el enunciado:
$f(n+1) = (x-a)(n+1)+a$
Consideremos el polinomio $g(x)=f(x)-a$.
$g$ tiene el mismo grado que $f$ pues solo cambia el coeficiente independiente y tenemos que:
$g(0) = g(1) = ... =g(n-1)= 0$ son las $n$ racices de $g$.
Entoces podemos escribir $g$ como:
$g(x)=k\cdot x(x-1)(x-2)...(x-n+1)$ con $k$ una constante.
Entonces tenemos que:
$g(n)=k\cdot n!$. Sabemos por otro lado que $g(n)=x-a$
entonces $k=\frac{x-a}{n!}$
Ahora evaluando $g$ en $n+1$ obtenemos que:
$g(n+1) = \frac{x-a}{n!}\cdot (n+1)(n)(n-1)...2 = (x-a)(n+1)$
$\implies f(n+1) = g(n+1)+a = (x-a)(n+1)+a$
QED


I want to create a webpage.  I know a way that will only cost me 3 dollars a year, but my dad won't let me use it (I need his permission to use his credit card).  He says that there is a way to make a webpage uing SBC Yahoo DSL (Which we have), but is having trouble finding how to use it.

Can anyone tell me if what he's saying is true, and how that works?  Their website didn't really mention anyting about it.

I agree with your dad, there is a way to  make a webpage through SBC Yahoo DSL, but I don't know how

[quote="JS1527"]Your ISP will probably let you make a free webpage, but there is Yahoo Geocities which is pretty good.  Just get the google toolbar which blocks pop-ups.  There are MANY services on the web that will give you free webpages...[/quote]

I just want to know if there is a way to create web pages using SBC Yahoo DSL.

What state do you live in? I'm reading through SBC Yahoo! DSL's pages, and the methods seem to differ per state.

[quote="Osiris"][quote="h_s_potter2002"]I live in Ohio.

And could you post a link to the pages you found?[/quote]

Then you're stuck with SBC Yahoo! Geocities:

[url]http://geocities.sbc.yahoo.com/[/url].

FYI, I own my own domain, which costs me $6 US per year; webhosting is free. I have 250mb webspace and a 5gb transfer limit (I think). It comes with 50 mailboxes (50MB on each), PHP extensions, CGI, MySQL, et cetera. I have Invision Board and MovableType running on them with no problem. Very handy. :)[/quote]

Thanks for the help.  Can you show me the web page that said that I have to use geocities since I live in Ohio?

And I already have another way to have a website with only 3 dollars a year.  It's just that I [b]have[/b] to use what SBC Yahoo DSL allows me to, if it allows me create a web page without using Geocities.

But [b]Yahoo![/b] owns Geocities, and since you're a SBC [b]Yahoo![/b] customer, you have to use the webhosting service that [b]Yahoo![/b] provides, which is Geocities.

[quote="confuted"][quote="Nikolas"]I don't know what r u talking but let me give u another tip... U can use http://www.geocities.com to build a new 25 MB website. They even provide u with online tools. Then u can go to http://www.dot.tk were u can register (FREE!) for your http://www.your-name.tk ending... End u can also find a free forum or a chatroom to place...If u need any other info's about that contact me... I have built 2 websites based on these tools and i haven't paid even 1$ ! If u like u can visit tgem to see an example of how you can build your website... The URL's are : [url]http://www.imo2004.tk[/url]   and  [url]http://www.mathhacker.tk[/url][/quote]
Why you would use Geocities to host a site when options such as 0catch.com (and others) exist is well beyond my ability to comprehend.[/quote]

because your 0catch doesn't support php perhaps? the age of html websites is over dude!

Nor does Geocities. I don't think that's the point we're trying to make - it's that Geocities offers less space and has a very stringent bandwidth limit. (not to mention the ads.)

I still consider my $6 per year a good investment. :)[/tex]

[quote="nr1337"]h_s_potter2002:  In the event that you want a host instead of SBC Yahoo, then a pretty good host is http://dhost.info .  You can replace the Popup ads (which they don't give you very many of) with Google-Text ads, and you get either 100MB or 150 MB (not sure) webspace, and 300GB bandwidth.  And PHP/MySQL support.

off-topic:  You live in ohio!  Cool!  I do too![/quote]

Actually, I don't want to use SBC Yahoo as a host.  It's just that since everything is free, my dad want to use it.

[quote="nr1337"]h_s_potter2002:  In the event that you want a host instead of SBC Yahoo, then a pretty good host is http://dhost.info .  You can replace the Popup ads (which they don't give you very many of) with Google-Text ads, and you get either 100MB or 150 MB (not sure) webspace, and 300GB bandwidth.  And PHP/MySQL support.

off-topic:  You live in ohio!  Cool!  I do too![/quote]

Thanks! Finally I can upload my >20 mb usefull files. BTW, I suggest use the free and totally gnarley FTP uploader "Filezilla" and use Macromedia Dreamweaver MX 2004 for creating your websites.

Which is available from macromedia.com for the low, low price of USD$399.

[i]This post has been ruthlessly culled by a moderator for advocating illegal activities.[/i]

The free trials of macromedia software can be downloaded even if you don't get the books

Or avoid all the legal bugaboos and code it with notepad.