linear algebra
matrix
vector
quadratics
inequalities
linear algebra unsolved

This is the matrix forum, after all. Can you restate this using matrices and vectors?

At first glance, I see a symmetric $2\times2$ matrix $A$ and the first three terms of your expression seem to be the quadratic form $x^{T}AX.$ The rest of it is less obvious, however.

There is no need to reformulate anything: the statement is incorrect without additional assumptions. Just take $a_{11}=a_{22}=0$, $a_{12}=1$, $b_{1}=b_{2}=-2$. Then the given inequality is trivial: $2x_{1}x_{2}\ge 0$ but the conclusion $0\ge 4$ is false.


Can someone do #17 from http://www.kalva.demon.co.uk/russian/rus96.html

It seems not possible, but i can't prove it.  What a great problem though.

The problem is: is there any n such that when all the integers up to n without break, form a palindrome.  (such as if n = 9, then the number is 123456789, which is not a palindrome)

Is there any such n?

Ah, I can't resist posting the answer to this one.

[hide]

Suppose the last number we wrote down has x+1 digits. Consider the point where we had written the number 10^x. This part looks like:

...99 100...00 100....01

Now lets see what happens when we reverse this. We still have that group of x zeroes in the middle - and since 0 cannot be the start of one of the numbers we wrote down, that must come from writing down 10^x also. So those 0s must be the middle of our palindrome. But of course thats impossible, as there are 9s to the left and 0s to the right. (x = 0 or 1 also doesn't work).

[/hide]

I don't think you can assume there is only one block of x 0s in the final string of digits - there could be up to 9 of them (if the final number is between 9x10^x and 10^(x+1)-1).

But the proof is easily fixed to handle this.

I'm not assuming there is only one block of x zeroes. I thought the same as you and that it was going to be a little problem, but it isn't: the next number after 10..00 must be a 1 (as 10..00 + 1 = 10..01 :))

Let the last number that we write in the sequence be n.
Let the 1st digit of the last number be m.
Then the last digit of (n-1)th number has to be m+1.
But m+1 = 0 (because last digit of last number is 1, so last digit of n-1th number should be 0)

but m+1 = 0 is impossible, because m cannot be nagative.

Hence we cannot obtain a palindrome just by writing down numbers from 1 through n.

Any questions on this proof?

[quote="Krishna Kothoor"]Let the last number that we write in the sequence be n.
Let the 1st digit of the last number be m.
Then the last digit of (n-1)th number has to be m+1.
[/quote]

Uh.. how do you get this?

TripleM - I thought you implicitly assumed there could only be one block of x 0s when you wrote "those 0s must be the middle of our palindrome". Certainly you can only conclude this if there are an odd number of blocks of x 0s in the supposed palindrome.

How does your proof work if the last number is between 2.10^x and 3.10^x ? There are now two blocks of x 0s in the string (one from 10^x, the other from 2.10^x), and neither is in the middle.

Those 0s must be in the middle of our palindrome since they not only come from writing the number 10^x, but when we flip it around since the digit preceding those 0s is a 1, they must also come from writing the number 10^x backwards. So that must be the middle of the palindrome.

(in fact, if there was a second block of 0s, 2000...0 200..01 then that must ALSO be the middle of a palindrome. In fact anywhere where theres a block of those 0s must be the middle of the palidrome, all leading to contradictions.)

Look at it this way. Suppose we start from the last number n. let the last number have m digits.  There must be a number with m-1 zeroes before we wrote the last number. Now for this to be a palindrome we must have m-1 zeroes somewhere in the beginning too. But this is impossible because all the numbers in the beginning(or first half) are smaller that the number with m-1 zeroes. So we cannot have m-1 zeroes in the first half of palindrome.


If the complex number $ z$ satisfy the relation $ \frac{z\minus{}2}{z}\equal{}i\tan(arg z)$.Then $ z$ lies on
a)straight line
b)circle
c)parabola
d)ellipse


In this moment of the prize-giving, the three embraced people (me, Pietro and Maria) just know that they are first, second or third...

The first three contestants: from the left, Maria (second), Pietro (first), and Andrea (me, second as Maria).

From the left: unknown, Pietro (1st, piever on mathlinks), Andrea (2nd, edriv on mathlinks), Maria (2nd, leblanc on mathlinks), Roberto Dvornicich (Italian team leader at the IMO), the president of the italian mathematical society.


geometry
inequalities
trigonometry
inequalities proposed

Let $A,B,C$ are three angles of one triangle. Prove that
 $(cosA+cosB+cosC)^{2}\geq 2\sqrt{3}sinAsinBsinC$

[hide]
i know that this is an ugly aproach, but i just love multiplying :)
ok.let $P$ denote the area of the triangle, $a,b,c$ are sidelengths.
$cosA= \frac{a(b^{2}+c^{2}-a^{2})}{2abc}$ and cyclic.
$sinA=\frac{a}{2R}$ and cyclic. so we have
$\left(a^{2}(b+c-a)+b^{2}(c+a-b)+c^{2}(a+b-c)\right)^{2}\geq \left(\frac{abc}{R}\right)^{3}\sqrt{3}=P^{3}\cdot64\sqrt{3}$
$a=y+z,b=x+z,c=x+y$ gives us
$\left((y+z)^{2}(2x)+(z+x)^{2}(2y)+(x+y)^{2}(2z)\right)^{2}\geq 64\left(\sqrt{xyz(x+y+z)}\right)^{3}\sqrt{3}$ squaring gives us
$(\sum_{sym}x^{2}y+6xyz)^{4}\geq 3\cdot2^{8}(xyz(x+y+z))^{3}$
$(\sum_{sym}x^{4}y^{2}+2\sum_{cyc}x^{4}yz+2\sum_{cyc}x^{3}y^{3}+14\sum_{sym}x^{3}y^{2}z+42x^{2}y^{2}z^{2})^{2}\geq$
$3\cdot2^{8}(xyz(x+y+z))^{3}$
now we have that the $LHS\geq$
$(2\sum_{cyc}x^{4}yz+2\sum_{cyc}x^{4}yz+12\sum_{sym}x^{3}y^{2}z+24x^{2}y^{2}z^{2}+36x^{2}y^{2}z^{2})^{2}=$
$=(xyz(4(x+y+z)^{3}+36xyz))^{2}$
denote $f=(x+y+z)^{3}$ and $g=xyz$. we know $f\geq 27g$
we need to prove
$(g(4f+36g))^{2}\geq 3\cdot2^{8}(xyz(x+y+z))^{3}=3\cdot2^{8}g^{3}f$
$(f+9g)^{2}\geq 48fg$
$f^{2}-30fg+81g^{2}\geq 0$
$(f-27g)(f-3g)\geq 0$ which is obviusly true...[/hide]
hope to see a nice solution too ;)

Maybe using $\sum cos \alpha = \frac{R+r}{R}$, $\prod sin \frac{\alpha}{2}=\frac{r}{4R}$, $\prod cos \frac{\alpha}{2}=\frac{s}{4R}$, so the inital inequality comes down to:

$(\frac{R+r}{R})^{2}\geq 2 \sqrt 3*8 \prod sin \frac{\alpha}{2}\prod cos \frac{\alpha}{2}\leftrightarrow$

$\leftrightarrow (R+r)^{2}\geq \sqrt 3 rs$

[quote="goc"]$ \sqrt{3}rs=\sqrt{3}rR(sin\alpha+sin\beta+sin\gamma)\leq \sqrt{3}rR(\frac{3\sqrt{3}}{2})=\frac{9rR}{2}$ so you have
$ (r+R)^{2}\geq \frac{9rR}{2}$ which is true because
$ (2R-r)(R-2r)\geq 0$ so this is a much shorter and nicer solution. the onlly thing is that i don't know the products of the sines and stuff...[/quote]

i tried that way but obviously i made a mistake somewhere... :blush:


Hi! Is this problem open yet?

CRUX 1356 Proposed by Walther Janous,
Show that
$\frac{x_{1}}{\sqrt{1-x_{1}}}+\frac{x_{2}}{\sqrt{1-x_{2}}}+...+\frac{x_{n}}{\sqrt{1-x_{n}}}\geq\frac{ \sqrt{x_{1}}+...+\sqrt{x_{n}}}{\sqrt{n-1}}$
for positive real numbers $x_{1}, . . . , x_{n}(n \geq 2)$ satisfying $x_{1}+...+x_{n}= 1$.

No, see here (page 28, problem 2) http://www.animath.fr/cours/inegalites.pdf  :wink:

Thanks a lot.  :)
Which of the problems in CRUX are open now? How can I know this?

Send your question to an Editor, this is the list
http://journals.cms.math.ca/CRUX/editorial.html#EDITORIAL

I think so.


$x,\ y,\ z$ が $0\leq x\leq 1,\ 0\leq y\leq 1,\ 0\leq z\leq 1,\ x+y+z=1$ を満たすとするとき, $x^{3}+y^{3}+z^{3}-3xyz$ のとりうる値の範囲を求めよ。

Porte さん、こんばんわ！1時間ほど待ってください。 :oops:
その間、こちらでもどうぞ。
[url]http://www.mathlinks.ro/Forum/viewtopic.php?t=109292[/url]

kunny

[quote="porte"]

yについて整理して、
$3y^{2}-3(1-x)y+3x^{2}-3x+1$
$=3\left( y-\frac{1-x}{2}\right)^{2}+\frac{(3x-1)^{2}}{4}$

となるので、$y=\frac{1-x}{2}$、$3x-1=0$ のとき、すなわち $x=y=z=\frac{1}{3}$ のとき、与式は最小値 $\boxed{0}$ をとる。

最大値は $y=0 or 1$ のとき、すなわち $x,y,z$ のいずれかが $1$ のとき$\boxed{1}$ をとる。[/quote]

$y$ について2次関数とみなしたようですが, 軸の位置を確認する必要がありますね。

P.S. $x^{3}+y^{3}+z^{3}$ の場合はどうなるでしょうか？ :lol:

うーん…
グラフをかいてみました :lol:

$0\leq \frac{1-x}{2}\leq \frac{1}{2}$　であることに注意すれば、

最大値は $y=1$ のとき、$\boxed 1$

最小値は $y=\frac{1-x}{2}$ 、$3x-1=0$ 即ち $x=y=z=\frac{1}{3}$ のときで $\boxed 0$

$x^{3}+y^{3}+z^{3}$ も…同じやり方でやってしまいそうです :(
それか $x+y+z=1$ $(0\leq x,y,z\leq 1)$ が与えられているので、$f(X)=X^{3}$　$(0\leq X\leq 1)$ のグラフを考えて、Jennsen の不等式でしょうか(でもそれだと変域が全部求まらないですよね :blush: )


geometry
circumcircle
geometric transformation
homothety
reflection
geometry proposed

Let $ H$ , $ O$ be the orthocenter and the circumcenter of $ \triangle ABC$ respectively. Denote $ \{\begin{array}{c}
E\in BH\cap AC\\\
F\in CH\cap AB\end{array}$ ,

$ \{\begin{array}{c}
X\in EF\cap AH\\\
Y\in AO\cap BC\end{array}$ and $ \{\begin{array}{c}
M\in BC\ ,\ MB=MC\\\
N\in XY\ ,\ NX=NY\end{array}$ . Prove that $ A\in MN$ . [url=http://www.mathlinks.ro/Forum/viewtopic.php?t=195986]See here an its extension ![/url]

Nice!

Let $ A'$ be the point diametrically opposite $ A$ on the circumcircle of $ \triangle{ABC}$.  Notice that quadrilaterals $ AEHF$ and $ ABA'C$ are similar.  Therefore, the homothecy about $ A$ which maps $ X$ to $ H$ also maps $ Y$ to $ A'$.  Therefore, the same homothecy maps the midpoint of $ XY$ (which is $ N$) to the midpoint of $ HA'$ (which is $ M$).  This yields our result.

Just couldn't help posting a picture :lol:
[asy]size(350);
fullbreak=true;
pointfontpen=fontsize(8);
void f()
{
pair A=MP("A",D((0,0))),B=MP("B",D((1,2))),C=MP("C",D((-1,0))),
     H=MP("H",D(orthocenter(A,B,C))),O=MP("O",D(circumcenter(A,B,C))),
     E=MP("E",D(IP(D(L(B,H,0)),D(L(A,C,1.3,0))))),
     F=MP("F",D(IP(D(L(C,H,0)),D(L(A,B,0.4,0))))),
     X=MP("X",D(IP(D(L(E,F,0)),D(L(A,H,0))))),
     Y=MP("Y",D(IP(D(L(A,O,0)),D(L(B,C,0))))),
     M=MP("M",D(WP(D(B--C)))),
     N=MP("N",D(WP(D(X--Y))));
D(A--B--C--A,black+linewidth(1));
D(L(M,N),blue+dashed);
}
f();pathflag=false;f();[/asy]

$ 1)$  From above, we deduce that $ PO//XY//HA'$, where $ P$ is the mid point of $ AH$. A more metric proof of the fact $ HA'//XY$ without this elegant use of homothety is to see that if $ H'$ is the reflection of $ H$ across $ BC$ and $ D\equal{}AH\cap BC$ then $ \frac{AY}{YA'}\equal{}\frac{AD}{DH'}\equal{}\frac{AD}{DH}\equal{}\frac{XA}{XH}$, because $ (A,X,H,D)\equal{}\minus{}1$.

A similar metric argument gives us $ PO//XY$. Moreover, a similar homothetic argument gives us the same result. Indeed, since $ \triangle AFE \approx \triangle ABC$ and $ P,O$ are their circumcenters, respectively, then this similarity(-homothety) implies $ \frac{AO}{AP}\equal{}\frac{AY}{AX}$ so $ PO//XY$.

$ 2)$ Another even more metric solution is by using Menelaus' theorem for the $ \triangle XDY$ for the points $ A,N,M$. Indeed, it suffices to prove that $ \frac{YM}{YD}\equal{}\frac{AX}{AX\plus{}AD}$. But we have $ \frac{YM}{YD}\equal{}\frac{HA}{2AD}$  therefore we have to prove that $ \frac{AH}{2AD}\equal{}\frac{AX}{AX\plus{}AD} \Leftrightarrow \frac{DA}{DH}\equal{}\frac{XA}{XH}$, which is true.


Ya u are the all know ones and u shall bestow thy knowledge on me lol  :D

[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=53947]go here to see some usefull formulas. :)[/url]


function
algebra
polynomial
inequalities
number theory unsolved
number theory

For function $f(x)$ define in $R$ such that for all  $c\in R^+$ then there exists the polynomial $P_c(x)\in R[x]$ s.t.:
$|f(x)-P_c(x)|\le cx^{1998} \forall x\in R$
Show that: $f(x)\in R[x]$

Sketch of a proof: Assume at least one of the coefficients is complex. Then there exists $y \in \mathbb{R}$ such that $Im\{f(y)\}\not = 0$. Plug $y$ into inequality and take $c$ small enough - we obtain contradiction because $P_c (x) \in \mathbb{R}[x]$  so $f(y)=a+bi$ then $|f(y)-P_c(y)|=\sqrt{(a-P_c(y))^2+b^2}$ where $b^2>0$ so there exist such a $c$ with left side larger.


Consider the convex quadrilateral $ ABCD$ with $ A(0,\ 0),\ B(3,\ 0),\ C,\ D$ which are on the curve $ y \equal{} x(3 \minus{} x)\ (0 < x < 3)$ in the $ xy$ plane.
Find the maximum area of the quadrilateral.

That doesn't make sense as written. Do you mean the maximum area?


geometry
geometric transformation
reflection
probability

In how many distinct ways can I arrange my five keys on a keychain, if I want to put my house key next to my car key? Two arrangements are not considered different if the keys are in the same order (or can be made to be in the same order without taking the keys off the chain).

Since reflections are possible, we can treat this as a four key keychain without reflections, which there are $ \frac {4!}4 \equal{} \boxed6$ arrangements of.

Another way to do this is to note that the number of ways to arrange $ 5$ keys on a keychain is $ \frac{5!}{5\cdot2}\equal{}12$, and since the probability that the house key is next to the car key is $ \frac12$, there are $ \frac12\cdot12\equal{}\boxed6$ ways.

[hide=Oops…Probably should have hid this.]On a circle with five points, there are $5$ ways to place the car key, then there are $2$ ways to place the house key, since the house key has to go on either side of the car key. Then, there are $3!$ ways to order the other keys for a total of $5\cdot2\cdot3!=60$ ways.

Wait! There is still rotational and reflective symmetry. So, divide by $5$ for rotational symmetry and $2$ for reflective symmetry for a total of $\frac{60}{10}=\boxed6$ ways.[/hide]

[hide=SOL]First, put the house key and car key next to each other on the keychain. It doesn't matter where they are put, because the keychain can be rotated and reflected to move them to any other pair of adjacent locations. The remaining three keys can be put on the keychain in $3!=\boxed{6}$ ways.[/hide]

This topic shouldn't be named ''Counting with symmetry'', It should be named ''problems with keychains'' -_-

Why 2 for reflective symmetry? I could see a horizontal line but where is the other one? Shouldn't there be also two diagonal lines and a vertical line?

[quote=dlu200774][hide = "What?"] Shouldn't the answer be 3, because with a keychain you can literally view it from both sides. The Alcumus solution treats the keychain like a table, but a keychain has 1/2 the number of rotations as a table since no one flips over the table. [\hide][/quote]

No, because they already got rid of all of the reflections and rotations in the process. Plus, does three ways make sence? For five keys?

You get 5 choices for the house key. Then only 2 choices for the car key because the car key must be next to the house key. After that the next key has 3 options. Then the next has 2 and then 1.
$5\cdot 2 \cdot3 \cdot 2 \cdot1=60$
Now you have to divide by 5 for rotational symmetry 60/5=12 then divide by 2 for reflectional symmetry 12/2=6


I am trying to prove that the eigenvalues of $A^{T}A$ are non-negative.

I have no idea where I'm going with this. I'm trying to use the fact that $|A|$ = $|A^{T}|$, and that the diagonal of $A^{T}A$ is positive, but I don't know what I'm doing. Any ideas?

Use the inner product: if $x$ is an eigenvector of $A$ with eigenvector $\lambda$, what's $x^{T}A^{T}Ax$?

The complex version is almost identical; substitute $A^{*}A$ for $A^{T}A$.

Sorry for bringing this topic to light. Continuing from jmerry idea:

$x^{T}A^{T}Ax$

If $x$ is an eigenvector of A we know that $A^{T}Ax=\lambda x$ so $x^{T}A^{T}Ax=x^{T}\lambda x$.

I don't know how to move on from here. Is there something I should know to prove this?

Are we talking about eigenvectors of $A$ or $A^{T}A$?

If $A^{T}Ax=\lambda x$, then after taking inner products of both sides with $x$, you get $\langle A^{T}Ax,x\rangle=\lambda \langle x,x\rangle$. But $\langle A^{T}Ax,x\rangle=\langle Ax,Ax\rangle$, hence $\parallel Ax\parallel ^{2}=\lambda\parallel x\parallel ^{2}$ and $\lambda\ge 0$.

if A is square and nonsingular then the eigenvalues of A^T * A are positive since eigenvectors are nonzero.


I need some help. Which of the following has the same value as $\frac{1}{\log_{2}50!}+\frac{1}{\log_{3}50!}+\cdots+\frac{1}{\log_{50}50!}$?
a. $\frac{1}{50}$ b. $1$ c. $\frac{1}{50!}$ d. $50$ e. $51$

thanks

[hide="Super hint"] $\frac{1}{\log_{a}b}= \log_{b}a$ [/hide]

ARGH! I should have seen that! AAAAAAAAAHHHHHHHH! I'm such an idiot! :wallbash_red:


Two cars are approaching an intersection. One is 2 miles south of the intersection and is moving at a constant speed of 30 mph. At the same time, the other car is 3 miles east of the intersection and is moving at a constant speed of 40 mph. Express the distance d between the cars as a function of time t.

The distance $d_{1}$ of the south car is given by $d_{1}= 2-v_{1}t$, where $v_{1}= 30mph$. Similarly for the east car: $d_{2}= 3-v_{2}t$, where $v_{2}= 40mph$. Then the distance $d$ between them is given by

$d = \sqrt{d^{2}_{1}+d^{2}_{2}}= \sqrt{(2-v_{1}t)^{2}+(3-v_{2}t)^{2}}= \sqrt{13-t(4v_{1}+6v_{2})+t^{2}(v^{2}_{1}+v^{2}_{2})}$
$\Rightarrow d = \sqrt{13-360t+2500t^{2}}$.

Notice that the equation $13-360t+2500t^{2}= 0$ doesn't have real solutions, which means that the cars will not arrive at the intersection at the same time (the distance between them will never be zero if they keep their speed constant).

In other words, there is no solution for the final quadratic equation given?


geometry
rectangle
modular arithmetic
floor function
rotation

First Question:(English)
By the time a party is over, $ 28$ handshakes have occurred. If everyone shook everyone else's hand once, how many people attended the party?

(Spanish)
Al terminar una fiesta un total de $ 28$ dadas de mano se dieron. Si todos le dieron la mano a todos una sola vez, ¿cuántas personas había en la fiesta?

Second Question:(English)
The last three digits of $ N$ are $ x25$. For which values of $ x$ can $ N$ be the square of an integer?

(Spanish)
Los últimos tres dígitos de $ N$ son $ x25$. ¿Para cuántos valores de $ x$ puede ser $ N$ el cuadrado de un entero?

Third Question:(English)
On an arbitrary triangle $ ABC$ let $ E$ be a point on the height from $ A$. Prove
that $ (AC)^2\minus{}(CE)^2\equal{}(AB)^2\minus{}(EB)^2$.

(Spanish)
En cualquier triángulo $ ABC$ sea $ E$ un punto sobre la altura desde $ A$. Probar que $ (AC)^2\minus{}(CE)^2\equal{}(AB)^2\minus{}(EB)^2$.

Fourth Question:(English)
Find all integers $ b$ and $ c$ such that the equation $ x^2\minus{}bx\plus{}c\equal{}0$ has two real roots $ x_1,x_2$ satisfying $ x_1^2\plus{}x_2^2\equal{}5$.

(Spanish)
Encuentra todos los enteros $ b$ y $ c$ tales que la ecuación $ x^2\minus{}bx\plus{}c\equal{}0$ tenga  dos soluciones reales $ x_1,x_2$ con $ x_1^2\plus{}x_2^2$.

Fifth Question:(English)
Let $ ABCD$ be a quadilateral inscribed in a circle. The diagonal $ BD$ bisects $ AC$. If $ AB\equal{}10, AD\equal{}12$ and $ DC\equal{}11$, find $ BC$.

(Spanish)
Sea $ ABCD$ un cuadrilátero inscrito en un círculo. La diagonal $ BD$ biseca a $ AC$. Si $ AB\equal{}10, AD\equal{}12$ y $ DC\equal{}11$, hallar $ BC$.

Sixth Question:(English)
The entries on an $ N$ by $ N$ board are colored black and white like it is usually done in a chessboard, and the upper left hand corner is black. We color the entries on the chess board black according to the following rule: in each step we choose an arbitrary $ 2$ by $ 3$ or $ 3$ by $ 2$ rectangle that still contains $ 3$ white entries, and we color these entries black. For which values of $ n$ can the whole board be colored black in a finite number of steps?

(Spanish)
Las casillas de un tablero de $ n$ por $ n$ se están coloreadas blanco y negro como en el tablero usual de ajedrez, donde la casilla de la esquina superior izquierda es negra. Coloreamos de negro las casillas blancas de la siguiente manera: en cada paso escogeremos un rectángulo arbitrario de tamaño $ 2$ por $ 3$ ó $ 3$ por $ 2$ que todavía contenga tres casillas blancas y coloreamos de negro estas tres casillas blancas. ¿Para qué valores de $ n$ se puede colorear todo el tablero de un número finito de pasos?

[hide="1"]
The number of handshakes is $ \binom{n}{2}$. Therfore,

$ \binom{n}{2} \equal{} 28 \implies n^2 \plus{} n \minus{} 56 \equal{} 0 \implies (n \plus{} 8)(n \minus{} 7) \equal{} 0$.

Thus, there were $ \boxed{7}$ people that attended the party.
[/hide]
[hide="2"]
We can rewrite it as $ 100x \plus{} 25 \equal{} k^2$ for some integer $ k$. Factoring, we have $ 5^2(4x \plus{} 1) \equal{} k^2$. So $ 4x \plus{} 1$ must also be a perfect square.

$ 4x \plus{} 1 \equal{} 9 \implies x \equal{} 2 \\
4x \plus{} 1 \equal{} 25\implies x \equal{} 6$

Therefore, $ N$ is a perfect square for $ \boxed{x \in \{2,6\}}$.
[/hide]
[hide="3"]
Let $ H$ be the foot of the altitude from $ A$. We have

$ EH^2 \equal{} BE^2 \minus{} BH^2 \equal{} CE^2 \minus{} CH^2 \implies CH^2 \minus{} BH^2 \equal{} CE^2 \minus{} BE^2.$

From $ \triangle ABC$, we have

$ AH^2 \equal{} AB^2 \minus{} BH^2 \equal{} AC^2 \minus{} CH^2 \implies CH^2 \minus{} BH^2 \equal{} AC^2 \minus{} AB^2.$

Therefore, $ AC^2 \minus{} AB^2 \equal{} CE^2 \minus{} BE^2 \implies AC^2 \minus{} CE^2 \equal{} AB^2 \minus{} BE^2$. $ \Box$
[/hide]
[hide="4"]
Because $ x_1$ and $ x_2$ are roots, we have

$ x_1^2 \minus{} bx_1 \plus{} c \equal{} 0 \implies x_1^2 \equal{} bx_1 \minus{} c \\
x_2^2 \minus{} bx_2 \plus{} c \equal{} 0 \implies x_2^2 \equal{} bx_2 \minus{} c$

Substituting that into the given equation, we have

$ x_1^2 \plus{} x_2^2 \equal{} 5 \implies b(x_1 \plus{} x_2) \minus{} 2c \equal{} 5 \implies b^2 \equal{} 2c \plus{} 5.$

Taking mod 2 on both sides, we have

$ b^2 \equiv 2c \plus{} 5 \pmod{2} \implies b^2 \equiv 1 \pmod{2}.$

The solution to that congruence is $ b \equiv 1 \pmod{2}$. Therefore, $ b \equal{} 2k \plus{} 1$ for some integer $ k$. Substituting that back into the equation from before, we have

$ b^2 \equal{} 2c \plus{} 5 \implies (2k \plus{} 1)^2 \equal{} 2c \plus{} 5 \implies c \equal{} \frac {2k^2 \plus{} 2k \plus{} 1 \minus{} 5}{2} \implies c \equal{} k^2 \plus{} k \minus{} 2.$

Thus, we have $ \boxed{(b,c) \equal{} (2k \plus{} 1, k^2 \plus{} k \minus{} 2)}, \forall k \in \mathbb{Z}$.
[/hide]

Your solution for (1) is wrong. Recheck the equation.

Solution for (4) is incomplete. Review all the initial conditions.

Um actually the first one is $ 8$ people, if there were $ 7$ people there would be $ \dfrac{6\cdot 7}{2} \equal{} 21$ handshakes.

On number two your missing $ \{0\}$, number $ 4$, the solutions are $ b \equal{} \pm 3, c \equal{} 2$ and $ b \equal{} \pm 1,c \equal{} \minus{} 2$. If you look at the discriminant $ \sqrt {b^2 \minus{} 4c} \equal{} \sqrt {(b^2 \minus{} 2c) \minus{} 2c} \equal{} \sqrt {5 \minus{} 2c}$ and this has to be greater or equal to $ 0$. So

$ \sqrt {5 \minus{} 2c}\ge 0$
$ 5 \minus{} 2c\ge 0$
$ 5\ge 2c$
$ c\le 2$

Also
Notice that $ 2b^2 \minus{} 4c \equal{} 10$ which means that $ 2b^2 \minus{} 10 \equal{} 4c$ substituting this into the discriminant:
$ \sqrt {b^2 \minus{} 4c} \equal{} \sqrt {b^2 \minus{} (2b^2 \minus{} 10)} \equal{} \sqrt {10 \minus{} b^2}$, and again this is greater or equal to $ 0$. So:
$ \sqrt {10 \minus{} b^2}\ge 0$
$ 10\ge b^2$
$ \minus{} \sqrt {10}\le b\le \sqrt {10}$
Since b is an integer:
$ \minus{} 3\le b\le 3$

On problem number $ 6$, the only way that the $ n$ by $ n$ board can be fully covered in black is if the number of white squares is a multiple of $ 3$, or mathmematical words: \[ \lfloor\dfrac{n^2}{2}\rfloor \equiv 0\pmod{3}\]. The values that satisfy this are in the form of $ 6k,6k \plus{} 1,6k \plus{} 5$ if I remember correctly.

Yes marcos, or at least that's what I heard everyone say after they took the test.

Yeah Alan told me it was that but sunehra posted 2 and 6.

Yea I kinda zipped through reading the problem and forgot that they were the last three digits and not a number, so I skipped the case of $ x \equal{} 0$.

1:

   let $ x\plus{}1$ be the number of people. so $ \frac{x(x\plus{}1)}{2}\equal{}28$.Easy to see that $ x\equal{}7$.
so there were $ 8$ people at the party.

2:

    easy to see that $ 25|N$.so $ \sqrt{N}$ends in $ 25 or 75$.easy to see that only the last $ 3$ digits of $ \sqrt{N}$ have "effect" at the last 3 digits.So to find the possibles $ x$ just compute.

3: this is the carnot´s theorem.use pithagoras in $ ABH,ACH,BEH$ and $ CEH$, where H is the foot of the perpendicular.

[hide="RE: v235711"]1. Nope: Assume that people shake hands in order. The first person shakes $ n \minus{} 1$  people, the second shakes $ n \minus{} 2$ (he already shook person one), and so on until person $ n \minus{} 1$ shakes person $ n$. If you add it all up, you get

$ S(n) \equal{} (n \minus{} 1) \plus{} (n \minus{} 2) \plus{} \dots \plus{} 2 \plus{} 1 \equal{} \frac {n}{2}(n \minus{} 1)$.

We know that $ S(n) \equal{} 28$, so we just find some $ n$ that works: $ n^{2} \minus{} n \minus{} 56 \equal{} 0 \Longleftrightarrow n \equal{} 8$.

2. Um, what? You just expanded $ (10^{2}a \plus{} 10b \plus{} 5)^{2}$ in $ \mod 1000$, and everything disappeared except $ 10^{2}(b^2 \plus{} b) \plus{} 25 \equal{} 10^{2}[b(b \plus{} 1)] \plus{} 25$, and you just had to find triangular numbers $ \leq 5$, including $ 0$.[/hide]

[b]tonypr[/b] — the number of white squares has to be of the form $ 3k$, so $ n^{2} \in \{ 6k, 6k \plus{} 1 \ |\ k \in \mathbb{R} \} \Longrightarrow n \equiv 0, 1, \minus{} 1 \mod 6$. When $ n^{2} \equal{} 6k$, you just cover up the whole board with $ k$ such rectangles (they always fit neatly); otherwise, you cover the board with $ k$ rectangles in a way such that the middle black square is uncovered: to do this, you start on the upper left corner and color $ \frac {n \minus{} 3}{2}$ such rectangles to the right, then you rotate the board by a quarter turn and color $ \frac {n \minus{} 3}{2}$ rectangles... as you continue rotating the board, the amount of rectangles you can color before rotating will decrease, but you will always end up with only the center black square left "uncovered" by such coloring triangles.


[hide]
I agree. The answer is $-\frac{5}{4}$[/hide]

me 2, now whats the largest product in that set?

[hide="da anwer"]$\frac{5}{4} \cdot 1=\frac{5}{4}$
i'm not sure if that's right[/hide]

i meant largest with three terms multiplied

Now what's the largest number obtained from a product of two numbers divided by another number?

fast way to think

[hide="fast stuff"]

this kind of stuff came along in many practice tests I took before

one way to do it fast is

find 2 smallest numbers,
(if negatives are involved, find the smallest largest negative and the largest poitive to multiply )

[/hide]

[quote="MCrawford"]What is the smallest possible product of three distinct members of the set $\{ -1, -\frac{2}{3}, \frac{2}{3}, 1, \frac{5}{4} \}$?[/quote]

[hide]$\frac{5}{4}\cdot 1\cdot -1=-\frac{5}{4}$[/hide]

[quote="math92"][hide]$\frac{15}{8}$?[/hide][/quote]
yeah, that corect

[hide="Click here for free money"]

$\frac{5}{4} * 1 *-1=-\frac{5}{4}$

[/hide]


[color=cyan]Given a cube of side length four.  The midpoints of each edge are chosen, and they are used to define eight tetrahedra, each of which shares a vertex with the cube, and has three other vertices at the midpoints of the three edges which meet at that vertex.  If the eight tetrahedra are cut off, what is the volume of the resulting solid?[/color]

[hide]the volume of a tetrahedron is 1/3 bh... the base is a right triangle with legs of 2, so the area is 2. The height is also 2; therefore the volume is 4/3. There are 8 tetrahedra; their total volume is 32/3; therefore the volume of the remaining figure is 64-32/3= 160/3 [/hide]


Find the maximum value of b-1/(a^2+b^2) when a,b>0 verify a^3+b^5<=a^2+b^2.

[quote="harazi"]Find the maximum value of $ b\minus{}\frac{1}{a^2\plus{}b^2}$ when $ a,b>0$ verify $ a^3\plus{}b^5\leq a^2\plus{}b^2.$[/quote]$ b\minus{}\frac{1}{a^2\plus{}b^2}\equal{}\frac{1}{2}\minus{}\frac{a^2\plus{}b^2\minus{}a^3\minus{}b^5}{a^2\plus{}b^2}$

$ \minus{}\frac{a(2a\minus{}b\minus{}1)^2}{4\left(a^2\plus{}b^2\right)}\minus{}\frac{\left(4a\minus{}b^2\minus{}2b\minus{}1\right)^2}{32\left(a^2\plus{}b^2\right)}\minus{}\frac{(b\minus{}1)^2\left(32b^3\plus{}63b^2\plus{}58b\plus{}31\right)}{32\left(a^2\plus{}b^2\right)}\leq\frac{1}{2},$

with equality if and only if $ a\equal{}b\equal{}1.$


vector geometry calculus integration inequalities function research

If $ v_{1}$, $ v_{2}$, ..., $ v_{n}$ are $ n$ vectors in $ \mathbb{R}^{m}$ and $ \lambda_{1}$, $ \lambda_{2}$, ..., $ \lambda_{n}$ are $ n$ reals such that $ \sum\limits_{i \equal{} 1}^{n}\lambda_{i}\left\vert w^{T}v_{i}\right\vert \geq0$ for every $ w\in\mathbb{R}^{m}$, then $ \sum\limits_{i \equal{} 1}^{n}\lambda_{i}\left\vert \left\vert v_{i}\right\vert \right\vert \geq0$, where $ \left\vert \left\vert \cdot\right\vert \right\vert$ is the Euclidean norm on $ \mathbb{R}^{m}$.

For $ m \equal{} 2$, this is well-known (see, e. g., Prasolov's geometry book) and can be proven in two ways: one is by integration over the unit circle (just remark that $ \left\vert \left\vert v_{i}\right\vert \right\vert \equal{} \frac {1}{4}\int\limits_{B\left( 0,1\right) }\left\vert w^{T} v_{i}\right\vert dw$), the other uses the weighted Karamata inequality (write $ v_{i} \equal{} \left( \begin{array}[c]{c} a_{i} \\
b_{i} \end{array}\right)$; then, the condition that $ \sum\limits_{i \equal{} 1}^{n}\lambda_{i}\left\vert w^{T}v_{i}\right\vert \geq0$ for every $ w\in\mathbb{R}^{m}$ yields $ \sum\limits_{i \equal{} 1}^{n}\lambda_{i}\left\vert b_{i}\right\vert\left\vert \frac {a_{i}}{b_{i}} \minus{} t\right\vert \geq0$ for any $ t\in\mathbb{R}$, what yields $ \sum\limits_{i \equal{} 1}^{n}\lambda_{i}\left\vert b_{i}\right\vert f\left( \frac {a_{i}}{b_{i}}\right) \geq0$ for any convex function $ f: \mathbb{R} \rightarrow\mathbb{R}$, and in particular, $ \sum\limits_{i \equal{} 1}^{n}\lambda_{i}\left\vert \left\vert v_{i}\right\vert \right\vert \geq0$ because the function $ f\left( x\right) \equal{} \sqrt {1 \plus{} x^{2}}$ is convex).

I don't know whether any of these two arguments can be generalized to $ m > 2$; as for the first argument, I am missing the basics in analysis, while for the second one, I don't see how to generalize it.

I am also wondering whether the Euclidean norm can be replaced by an arbitrary norm here - at least for $ m \equal{} 2$, this works (with a similar proof by weighted Karamata). If this works for every $ m$, then the above fact could be a source of inequalities.

  Darij

[quote="Soarer"]This question may be stupid, but does karamata inequality work for possibly negative weights?[/quote]

Actually, when I said weighted Karamata inequality, I meant Theorem 8b in [url=http://www.cip.ifi.lmu.de/~grinberg/PopoviciuFormal.pdf]my Popoviciu text[/url]. But I am not that sure of the proof anymore because we actually need the condition $ \sum\limits_{i = 1}^{n}\lambda_{i}\vert b_{i}\vert = 0$ additionally, which is generally not satisfied. I'll think about it once again.

  darij


trigonometry
combinatorics theorems
combinatorics

hi-
use sum or diffenrece identiteis to solve the equation exactly

cos 3x cos x= sin 3x sin x

how do i start/do this problem. i have the list of identites in front of me but i dont know what do to.
can you help me?
thanks

Problems like this belong in [url=http://www.artofproblemsolving.com/Forum/index.php?f=149]High School Basics[/url].

On your list of identities is the identity $ \cos (a \plus{} b) \equal{} \cos a \cos b \minus{} \sin a \sin b$.  Do you think this identity might help?

[hide="Specifically, what if..."] $ a \equal{} 3x, b \equal{} x$? [/hide]

thank you. next time i will post it in high school basics.


trigonometry
MATHCOUNTS
ratio
Pythagorean Theorem
geometry
similar triangles
trig identities

7. Wall A and B are 15 ft apart. On e rope with length 25 ft. is attached to the top of wall A and the bottom wall B. Another rope with length 39 ft. is attached to the top of wall B and the bottom of wall A. How far above the ground is the point where the two points intersect?

I get the height of the wall, then I could do nothing. The solution says that you use the formula $ \frac{xy}{x\plus{}y}$ for $ x$ and $ y$ are the heights of the wall. I am not familiar with this formula. Can you guys perhaps explain and how the formula was derived?

Heh, on these I always try it the hard way (analytic geometry).
But I'm pretty sure they want you to use similar triangles. (Which isn't really a formula per se).
If you follow along, DRAW A DIAGRAM! I don't know LaTeX/Asymptote, so I'm explaining everything
[hide]
First, use the Pythagorean Theorem to find the heights of the walls (ground to rope). You should get 20 feet for the 25-rope and 36 for the 39-rope.
Draw an altitude from the intersection point to the ground. Call the length [i]h[/i].
The altitude divides the 15 feet ground distance into two parts: I will call them [i]a[/i] (close to the 25 feet) and [i]b[/i] (close to the 39 feet).
Now by AA, you have two pairs of similar triangles, so b/h=(a+b)/20 and a/h=(a+b)/36. Add the equations to get:
(a+b)/h=(a+b)/20+(a+b)/36. We defined a and b so that a+b=15.
15/h=15/20+15/36. Multiplying by 180h/15 gives
180=h(9+5).
h=[b]90/7 feet[/b].
[/hide]

EDIT: The formula does work! I think from this solution, you can pretty much see the derivation of it.

Worst comes to worst, if the formula doesn't work, you can use trig:

first, separate triangle A, which includes Wall A, the ground (15 ft. apart), and the string, which is 25 ft.

next, make traingle B, which includes wall B, the ground (15 ft. apart), and the string, which is 39 B.

In the whole picture, you can see triangle C, which includes the ground, and portions of both strings, cut off where the two points meet.  Draw a picture, and you'll see what I mean.

Take triangle B, and find the base angle by using you calculator (I use a TI-84)

cos-1 (2nd + cos on your calculator) of 15/39 (looks like cos-1(15/39))

you should get about 67 (~67.38013505)

now, you know that the other part of the angle (again, look at you figure) is about 23 degrees.

Now, you can find the other angle, by looking at triangle A

Same thing: cos-1(15/25) comes up to about 53 (~53.13010235)

OK, now you know that the other part of this angle (see your picture) is about 37 degrees.

We can now use AAS congruence to solve the triangle (or ASA)

First of all, look at triangle C, and figure out the third angle.

It's 60 degrees.

use the law of sines (sinA/a = sinB/b = sinC/c) to figure out c:

sin67/15 = sin53/c

isolate c:

c*sin67/15 = sin53

c*sin67 = sin53*15

c = (15*sin53)/sin67

c is about 13.

Now, you can change to right triangles...

cut triangle C in half, and concentrate on the left side

first, find the other angle

angle A is 67, so angle B must be 23

Then, use sines: sin67 = h/13 (hypotenuse)

13*sin67 = h

h is about 12 (~11.97953265)

It's as easy as that.

 :|

1. The formula does work. (Why?) [hide="hint"]Use similar triangles[/hide]
2. Don't post trig in the MathCounts forum.

once you use similar triangles, you'll see that the length between the two ropes. doesn't even matter.

[hide="hint"]It a classic problem, get 2 ratios with the same denominator and add them[/hide]


Twelve couples are seated around a circular table such that all of men are seated side by side, and every women are seated to opposite of  her husband. In every step, a woman and a man next to her are swapping. What is the least possible number of swapping until all couples are seated side by side?

$\textbf{(A)}\ 36  \qquad\textbf{(B)}\ 55  \qquad\textbf{(C)}\ 60  \qquad\textbf{(D)}\ 66  \qquad\textbf{(E)}\ \text{None}$

Let $M_i$ be the man, and $W_i$ be the woman where $(M_i,W_i)$ are married with each other.
Let $|M_iW_i|$ be the distance of the couple. Initially, all $|M_iW_i| = 12$.
So initially $\sum_{i=1}^{12} |M_iW_i| = 12\times 12 = 144$.
At the end, we want each $|M_iW_i| = 1$, so $\sum_{i=1}^{12} |M_iW_i| = 12\times 1 = 12$.
At each step, suppose $M_x$ and $W_y$ are being swapped. So $|M_xW_x|$ can ve decreased at most by $1$. Similarly $|M_yW_y|$ can be decreased at most by $1$.
So at each step $\sum_{i=1}^{12} |M_iW_i|$ can be decreased by at most $2$.
So at least $\dfrac {144-12}{2} = \dfrac {132}2 = 66$ steps are required to get $12$ from $144$.
We showed that  the least possible number of swapping is equal to or greater than $66$. Now we give an example by $66$ swappings:
$M_1 \rightarrow W_{12}, W_{11}, \dots, W_2$ : $11$ times
$M_2 \rightarrow W_{12}, W_{11}, \dots, W_3$ : $10$ times
$\vdots$
$M_{10} \rightarrow W_{12}, W_{11}$ : $2$ times
$M_{11} \rightarrow W_{12}$ : $1$ time
There are $1+2+\dots + 11 = \dfrac{11\cdot 12}{2} = 66$ swappings.


Let $\frac{x^2+y^2}{x^2-y^2} + \frac{x^2-y^2}{x^2+y^2} = k$. Compute the following expression in terms of $k$:
\[ E(x,y) = \frac{x^8 + y^8}{x^8-y^8} - \frac{ x^8-y^8}{x^8+y^8}.  \]
[i]Ciprus[/i]

I think this problem has post in Batic Way.

[quote="hrithikguy"]Would the graders want this to be in the same form as the one above, or would this be fine?[/quote]
In other words, "Hey dear grader, it's your turn to simplify and give me full marks!"  :roll:

[hide="Solution?"]
$k=\dfrac{x^2+y^2}{x^2-y^2}+\dfrac{x^2-y^2}{x^2+y^2}$

$=\dfrac{x^4+2x^2y^2+y^4+y^4-2x^2y^2+y^4}{x^4-y^4}$

$=2\left(\dfrac{x^4+y^4}{x^4-y^4}\right)$

Note that we can see that $\dfrac{x^k+y^k}{x^k-y^k}+\dfrac{x^k-y^k}{x^k+y^k}=2\left(\dfrac{x^{2k}+y^{2k}}{x^{2k}-y^{2k}}\right)$ for all even $k$.  This is easy to prove.

This means that $\dfrac{k}{2}=\dfrac{x^4+y^4}{x^4-y^4}$.

Now we can see that $\dfrac{k}{2}+\dfrac{1}{k/2}=\dfrac{k}{2}+\dfrac{2}{k}=\dfrac{k^2+4}{2k}$

$=\dfrac{x^4+y^4}{x^4-y^4}+\dfrac{x^4-y^4}{x^4+y^4}=2\left(\dfrac{x^8+y^8}{x^8-y^8}\right)$.  This means $\dfrac{k^2+4}{4k}=\dfrac{x^8+y^8}{x^8-y^8}$.  We can finally say that $\dfrac{k^2+4}{4k}+\dfrac{4k}{k^2+4}=\dfrac{x^8+y^8}{x^8-y^8}+\dfrac{x^8-y^8}{x^8+y^8}$, which is our desired.

Now we just simplify to get $\dfrac{k^2+4}{4k}+\dfrac{4k}{k^2+4}=\dfrac{(k^2+4)^2+(4k)^2}{4k(k^2+4)}$$=\dfrac{k^4+8k^2+16+16k^2}{4k(k^2+4)}=\boxed{\dfrac{k^4+24k^2+16}{4k(k^2+4)}}$.

EDIT: The distribution properties failed in my first solution.  Still, I'm getting an expression that is not equivalent to the other solutions but according to Wolfram|Alpha works....[/hide]

My  solution :

[b]Lemma I used[/b] :
if $\frac{a}{b} = \frac{c}{d} \text{then} \frac{a+b}{a-b} = \frac{c+d}{c-d} $
this lemma is also know'n as componendo and dividendo and the proof of this lemma is very simple.

So

$ \frac{x^2+y^2}{x^2-y^2}+\frac{x^2-y^2}{x^2+y^2}= k $
Simplifying gives
$\frac{k}{2}=\frac{x^4+y^4}{x^4-y^4}$
Using [b]Lemma[/b]

$\frac{2x^4}{2y^4} = \frac{k+2}{k-2} $
$(\frac{x}{y})^4 = \frac{k+2}{k-2} $

Now we have to find $\frac{x^8+y^8}{x^8-y^8}+\frac{x^8-y^8}{x^8+y^8} $

which is equivalent to $ 2\frac{x^{16}+y^{16}}{x^{16}-y^{16}} $

$ 2\times\frac{(\frac{k+2}{k-2})^4+1}{(\frac{k+2}{k-2})^4-1}$

$2\times\frac{(k+2)^4+(k-2)^4}{(k+2)^4-(k-2)^4} $

[quote="JBL"]Why should a grader care whether denominators have been cleared?  It's a correct expression, and it satisfies the necessary condition that it contains only $k$ and no other instances of $x$ or $y$.[/quote]

This is not a fair statement IMO, considering how many graders do not accept fractions that are not in lowest terms or surds that are not simplified (for example) ;)

[quote="djmathman, April 2012"]
We can finally say that $\dfrac{k^2+4}{4k}+\dfrac{4k}{k^2+4}=\dfrac{x^8+y^8}{x^8-y^8}+\dfrac{x^8-y^8}{x^8+y^8}$, which is our desired.
[/quote]
Oops.  Well now I know why this solution was wrong....


Give three positive number $x,y,z$ satisfying $x+y+z = 1$. Prove that:
$(\frac{x+1}{y+1})^n+(\frac{y+1}{z+1})^n+(\frac{z+1}{x+1})^n \leq 1+2^n+2^{-n}$ where $n$ is real number

I post this one just because this topic is dropped down. I think firstly we want to solve this one in the case $n = 1$
$\frac{x+1}{y+1}+\frac{y+1}{z+1}+\frac{z+1}{x+1} \leq \frac{7}{2}$
Do someone want to post a solution ?


Let $*$ be an operation on a non-empty set $S$. Prove that if $1$ is in $S$ and $(x*1)*x = x$ for all $x$ in $S$, then there exists $a$ in $S$ such that $a*(a*x)=x$ for all $x$ in $S$

Let $S=\{2n+1\mid{n}\in\mathbf{Z}\}$ and define the operation $*$ by $x*y=x$ for all $x,y\in{S}$. Then $1\in{S}$ and $(x*1)*x=x$, but $a*(a*x)=a$ is a constant, so the proposition fails. Is this a valid counterexample? I think there's something wrong with this problem since we want \[{\forall{x}\in{S},(x*1)*x=x\Longrightarrow\exists{a}\in{S}\text{ s.t. }\forall{x}\in{S},a*(a*x)=x,}\]but that can't be true, can it?

Well, it depends on the meaning of the phrase "there exists a in S such that relation for all x in S."  You wrote it explicitly as, "$\exists{a} \in S \forall{x}\in{S}$" which is also how I interpreted it and for which your counterexample is valid.  However, ambiguity in English leads to the possibility that this phrase might also mean "$\forall{x}\in{S}\exists{a} \in S$" in which case your counterexample is not valid, as the proof is fairly simple.

If $a$ doesn't have to be constant, you can make $a = x * 1$.   Then we get $a * (a * x) = (x * 1) * ((x* 1) * x) = (x * 1) * x = x$.


Sean $a,b$ y $c$ los lados de un triangulo cualquiera. Pruebe que:
$\frac{\sqrt{b+c-a}}{\sqrt{b}+\sqrt{c}-\sqrt{a}}+\frac{\sqrt{c+a-b}}{\sqrt{c}+\sqrt{a}-\sqrt{b}}+\frac{\sqrt{a+b-c}}{\sqrt{a}+\sqrt{b}-\sqrt{c}}\leq 3$

Tenemos que $\sqrt a+\sqrt b > \sqrt{a+b}\ge \sqrt c$ y lo mismo para las ciclicas. Esto quiere decir que $\sqrt a , \sqrt b , \sqrt c$ son los lados de un triangulo también. Entonces podemos hacer la sustitución de Ravi es decir que podemos poner:
$\sqrt a =x+y$
$\sqrt b =y+z$
$\sqrt c =z+x$

Con $x,y,z$ reales positivos.  Por lo tanto la desigualdad es equivalente a:
$\sum_{cyc}\frac{\sqrt{(x+y)^{2}+(y+z)^{2}-(z+x)^{2}}}{2y}\le 3$
Abriendo los cuadrados y multiplicando por $2xyz$ a ambos lados obtenemos que la desigualdad es igual a:

$\sum_{cyc}\sqrt{2x^{2}yz^{2}(x+y+z)-2x^{3}z^{3}}\le 6xyz$

Usando Jensen (con $f(a)= \sqrt{a}$ que es concava) obtenemos que es suficiente probar que:

$\sum_{cyc}\sqrt{2x^{2}yz^{2}(x+y+z)-2x^{3}z^{3}}\le 3\sqrt{\frac{2(x+y+z)(\sum_{cyc}x^{2}y^{2}z)-2(\sum_{cyc}x^{3}y^{3})}{3}}\le 6xyz$

Esto ultimo se tiene si y solo si:
$\frac{(x+y+z)(\sum_{cyc}x^{2}y^{2}z)-(\sum_{cyc}x^{3}y^{3})}{3}\le 2x^{2}y^{2}z^{2}$

Que es equivalente (después de expandir y poner todo del mismo lado) a:

$\frac{1}{2}\sum_{sym}x^{2}y^{2}z^{2}+\frac{1}{2}\sum_{sym}x^{3}y^{3}-\sum_{sym}x^{3}y^{2}z = \sum_{cyc}xy(xy-yz)(xy-zx) \ge 0$

Que no es otra cosa que la conocida desigualdad de Schur.

QED

[color=blue]Disculpa mi ignorancia.
¿Cómo se efectúa la sumatoria simétrica?[/color]

$\sum_{sym}x^{a}y^{b}z^{c}= x^{a}y^{b}z^{c}+x^{a}y^{c}z^{b}+x^{b}y^{a}z^{c}+x^{b}y^{c}z^{a}+x^{c}y^{a}z^{b}+x^{c}y^{b}z^{a}$

Es decir todas las combinaciones posibles de exponentes. Es importante que cuando hay dos exponentes iguales también hay seis terminos. Una notación conocida y util es (como dato curioso): $\sum_{sym}x^{a}y^{b}z^{c}= [a,b,c]$

Naturalmente toca definir ese simbolo si se va a usar, pero este facilita bastante la notación.

[quote="jasamper88"]$\sum_{sym}x^{a}y^{b}z^{c}= [a,b,c]$[/quote]
[color=blue]Esta notación vi ya en algunas publicaciones, mas no logré comprender bien cómo se aplica en el campo de las desigualdades. ¿Quién podría tomarme unas clases particulares?[/color]

realmente las sumas simetricas son importantes no solo para ahorrar espacio al escribir, sino tambien por estar relacionadas con muirhead, de hecho el enunciado formal del teorema se escribe usando esa notacion!

supongo q ya sabes q para indicar una suma simetrica se suele escribir "sym" abajo de la sigma, yo acostumbro escribir "sim" por adaptarlo al español. en cuanto al uso de la notacion en si, basicamente la sumatoria indica q las letras se "permutan" de todas las maneras posibles, lo mejor q podrias hacer es revisar varios ejemplos (las sumas se efectuan respecto a tres variables $a$, $b$, $c$):
\[\sum_{\mathrm{sim}}a^{3}b=a^{3}b+b^{3}c+c^{3}a+ab^{3}+bc^{3}+ca^{3}\]

\[\sum_{\mathrm{sim}}a^{2}bc=2(a^{2}bc+b^{2}ca+c^{2}ab) \]

\[\sum_{\mathrm{sim}}a^{2}=2(a^{2}+b^{2}+c^{2}) \]

\[\sum_{\mathrm{sim}}abc=6abc \]
como te digo, si pensas q las letras se permutan respecto a los exponentes, eso explica q hayan siempre 3!=6 terminos en todas las sumas, aunq algunos se repitan; en general una suma simetrica respecto a $n$ variables tiene $n!$ terminos (no necesariamente distintos)

espero q esto te sirva de algo  :)


In a certain cross-country meet between two teams of five runners each, a runner who finishes in the $ n$th position contributes $ n$ to his team’s score. The team with the lower score wins. If there are no ties among the runners, how many different winning scores are possible?

$ \textbf{(A) } 10 \qquad \textbf{(B) } 13 \qquad \textbf{(C) } 27 \qquad \textbf{(D) } 120 \qquad \textbf{(E) } 126$

[hide] There are a total of $ \frac{10(11)}{2}\equal{}55$ points available in the contest. Hence, the highest score that can win is $ 27$, and the lowest $ 15$ (since $ 1\plus{}2\plus{}3\plus{}4\plus{}5\equal{}15$). This is a total of $ \boxed{13}$ socres. Thus,the answer is B.[/hide]


[quote="happyme"]How many positive integers $x$ are there such that $3x$ has three digits and $4x$ has four digits?[/quote]

[hide]Minimum for $4x$ to have four digits is $4x=1000,$ where $x=250$. ($3x=750$ and has three digits)

Maximum for $3x$ to have three digits is when $3x=999,$ where $x=333$. ($4x=1332$ and has four digits)

So we count from 250 to 333. There are $333-250+1=\boxed{84}$ such numbers.[/hide]


AMC
AIME
geometry
vector
analytic geometry
graphing lines
slope

The angle bisector of angle A in the triangle A (-8, 5), B (-15, -19), C (1, -7) is ax + 2y + c = 0. Find a and c.

can someone help me with this? i don't know exactly how to approach this..

can this be done with vectors?

thanks

You [i]could[/i] use vectors, but it would be very algebra intensive. A better way is to draw the triangle and find the distances AB and AC. Then use the Angle-Bisector Theorem to find how far along side BC the bisector from A crosses (call this crossing D). Using this you can find the coordinates of D.

Then you have the coordinates of two points on the angle bisector (A and D), so you can find it's slope and intercept.

ok, i had started doing that.. but i thought it might get a little messier than necessary. i'll try it out, thanks.

My first instinct would to be what Gauss suggested.  It might also be possible to use that facts that the slope of a line is the tangent of the angle it makes with the x axis, and try to solve it using trigonometry.  I'm not sure this will be nice either, though.

Ah, figures -- those aren't any randomly selected points you've got there.  I don't know in general what the nicest method is, but I'll put money on its not being vectors.  My method is almost certainly worse here, although I haven't actually tried it out.

Also, triangle ABC is a right 3-4-5 triangle


geometry
3D geometry
octahedron
Asymptote

How can I make an octahedron with asymptote with one of the "square" vertexes labled A and one of the "points" labeled B?

Thank you so much Dojo!!!!!!

Just wondering, how could you use it without showing the edges that are blocked from view? And maybe shading the faces?

But this works out just ifne.

hmm but that's a bit lopsided..

[asy]import three;
size(250);
currentprojection=perspective(9,3,4);
pair A=(1,0,0), B=(0,0,1), C=(0,1,0), D=(-1,0,0), E=(0,-1,0), F=(0,0,-1);
filldraw(A--E--F--cycle,paleblue);
filldraw(A--C--F--cycle,paleblue);
filldraw(A--C--B--cycle,paleblue);
filldraw(A--E--B--cycle,paleblue);
draw(C--D--E^^B--D--F,dashed);
D("A",A,W); D("B",B,N);[/asy]

I think what he wanted is:
[asy]import three;
size(250);
currentprojection=perspective(9,3,4);
pair A=(1,0,0), B=(0,0,1), C=(0,1,0), D=(-1,0,0), E=(0,-1,0), F=(0,0,-1);
filldraw(A--E--F--cycle,paleblue);
filldraw(A--C--F--cycle,paleblue);
filldraw(A--C--B--cycle,paleblue);
filldraw(A--E--B--cycle,paleblue);
D("A",A,W); D("B",B,N);[/asy]

Which is without the back ends.


In an integer of two digits is $k$ times the sum of its digits, the number formed by interchanging the digits is the sum of the digits multiplied by:
\[ \text{(A)}\ (9-k) \qquad \text{(B)}\ (10-k) \qquad \text{(C)}\ (11-k) \qquad \text{(D)}\ (k-1) \qquad \text{(E)}\ (k+1) \]

[hide]the answer is $C$
lets call what we want $m$ then:
$10a+b=k(a+b)$
and
$10b+a=m(a+b)$ so
add them up you get:
$11(a+b)=(m+k)(a+b)\Rightarrow m+k=11\Rightarrow m=11-k$[/hide]

[hide]$(C)$

I was going to go and do a bunch of algebra (EDIT: thought it would be more than what amir used, ah well, both work) but I decided to just use a real number- $21$.

So we know (from the first part of the problem)

$2(10) + 1 = k(2+1)$

$21 = 3k$

$7 = k$

Then switching the digits

$1(10) +2 = (2+1)(x)$

$12 = 3x$

$4 = x$

Where x is one of the answers given. We can see that the only answer that words is $(11 - k)$ which is $(11 - 7)$ which is $4$

Giving us the answer of $(C)$[/hide]


A house has three smoke detectors. In a fire, each one has an equal chance of going off and thus alerting the family (the fire alarms are indepenent of each other). If even one alarm goes off, the family is alerted of the fire. In a fire, there is a 93.6% chance that the family is alerted. In a fire, given that the family survived, what is the probability that all three fire alarms went off? Express your answer as a common fraction.

[hide="My answer"]
I believe it is $\frac{27}{117}$, not completely sure though. As always.

EDIT: yeah. It simplifies to $\frac{3}{13}$. I knew I was missing something, lol...
[/hide]

Oh, by the way, assume that if the fire alarms do not alert the family, they [b]die[/b], and assume that if they are alerted, they survive.

this is what i got
[hide]the chance that they die is 0.064

therefore, the chance a fire alarm will go off is 0.6

((0.6)^3)/(0.936)=

(3/13)[/hide]

[hide]
There is a 6.4% or $\frac{8}{125}$ chance that the family will not be alerted. Find the cube root to get $\displaystyle\frac{2}{5}$. That is the chance an individual alarm will not go off and that means there's a $\displaystyle\frac{3}{5}$ chance it will go off. Now cube that to get $\frac{\frac{27}{125}}{\frac{117}{125}}$ or $\frac{27}{117}$ which simplifies to $\boxed{\frac{3}{13}}$.[/hide]

[hide]solving $x^3-3x^2+3x=.936$ you get $x=\frac{3}{5}$, the probability that an alarm goes off. $\frac{(\frac{3}{5})^3}{.936}$ gives you $\frac{3}{13}$.[/hide]

[hide="ONLY for fireemblem13"]Nice signature!!![/hide]


geometry
inequalities proposed
inequalities

Prove that, for any triangle with sides a,b,c and area A
         [b]a^2+b^2+c^2 => 4*3^1/2*A[/b]

See here.

[url]http://www.mathlinks.ro/viewtopic.php?t=14388[/url]

let $ x\plus{}y\equal{}a ,y\plus{}z\equal{}b,z\plus{}x\equal{}c$  and $ x\plus{}y\plus{}z\equal{}3$
 We have : $ x^2\plus{}y^2\plus{}z^2 \plus{}9 \geq 12 \sqrt{xyz}$
 By Cauchy's  ineq  for 12 positive real numbers  we have :
 $ LHS \geq 12 \sqrt[6]{xyz} \geq RHS$
because of : $ xyz \leq 1$


calculus
derivative
inequalities
real analysis
real analysis unsolved

without using derivatives (looking for minimum), prove that $ x^3\plus{}9y^3  \ge 8xy^2$ for all $ x,y\ge 0$

Young's Inequality says that for positive $ a,b,p,q$ such that $ \frac1p \plus{} \frac1q \equal{} 1,$ then $ ab\le\frac1pa^p \plus{} \frac1qb^q.$

What can you do here with $ p$ and $ q$ being $ 3$ and $ \frac32\,?$

wow, less then 15 min and two replies :)
thank you very much


The sum of two numbers $ x$ and $ y$ is 399, and the value of the fraction $ \frac{x}{y}$ is 0.9. What is the value of $ y \minus{} x$?

hello, with $ x\equal{}\frac{9}{10}y$ we get $ \frac{19}{10}y\equal{}399$, from here we have $ x\equal{}189$ and $ y\equal{}210$, so
$ y\minus{}x\equal{}21$.
Sonnhard.


Here is another picture! :P
Again from left to right
Sailor,Tiks,Iura :wink:

Here is another one :wink: .
In the middle [b]Albanian Eagle[/b],[b]Tiks[/b] :P (I don't know the nickes of the other two guys :( )

I hope to go to IMO in 2008.That means that maybe the only I can meet in IMO is Valentin Vornicu. :(  :(  :( ZetaX is so tall!You and Darij and  Albanian Eagle is so cool!

hey that's Son of Skanderbeg(from my team) in the left and Vitalie (from Moldova) in the right!

[quote="Tiks"]Hey guys here are some pictures from IMO 2006
 :wink:
From left to right
Darij Grinberg,Tiks,Valentin Vornicu,Z[size=0]spammer[/size]etaX,(and a short boy from our team :D )

[edit by $\zeta X$: it's ZetaX, not Zatex ;) ]
[edit by darij: ...][/quote]
I am waiting for Valentin to edit this post :P  :blush: .


I was looking at who's online, and I found this...

Guest   Fri Nov 20, 2009 11:25 am   Forum index

This was the same for 5 other guests, too.

[quote="hunter34"]Guests can view the forum index.[/quote]

I believe myyellowducky82 meant he saw the Guests classified under the registered users.

[quote="PowerOfPi"]I see "Fri Nov 20, 2009 1:25 pm" for the date/time. Maybe different time zones. (Central)[/quote]

That still does not explain the November 20, 2009 "future" posting time.

[quote="caffeineboy"][quote="r15s11z55y89w21"]
I believe myyellowducky82 meant he saw the Guests classified under the registered users.[/quote]

No... I'm pretty sure the problem is the date/time. Unless you are posting from the future, and Nov 20, 2009 already happened?

I confirm the same thing; I see the same date/time (Fri Nov 20, 2009 11:25 AM ) for a few guests too.[/quote]
Sorry I wasn't clear. I was talking about the "future" time. All of the other guests have "Today" on their date/time.

[quote="Dojo"]Both parts don't make sense.

1) The fact a gues would be shown on the registered users.
2) The date is so far in the future.[/quote]
For me, the guests weren't shown in the registered users section.

All of the registered users are on the top section, and all of the guests are in the bottom section.

See here: http://www.artofproblemsolving.com/Forum/viewonline.php

You can set your time on the AoPS, Probalbly some people With messed up times, cause if your a guest, it goes with the time set for your computer.

5 people having the exact same time, and it stays like that? Unlikely. (actually I don't know)

Sorry, I didn't make myself clear, I meant ONE person, browsing this forum in FIVE different windows.

Oh, I get it now. :) Is it possible to see guest's IP addresses? If they're the same, we'll know if this is true or not.

I don't think this really makes a difference, cause only admins can do that, and thats not enough to bother them with.


Bar Mitzvah? what's that?

And yeah, I get to miss the last week of school (all those exams) :) :D

If you don't know what a Bar Mitzvah is, go look it up.

...and I don't even get to miss finals. What a pity. (But most of them are AP classes anyway, so whatever.)

Yeah, I was thinking on the side of MOP too. I can't miss this opportunity (and I might not get in next year, seeing as I'll be a senior and you need to be in the top 12).

[quote="Kamior"]It's not a choice between three days of MOP and that: this year, you can't come for only part of it. I don't know how significant Bar Mitzvahs are (or even if that's the correct plural form), but if you were having trouble deciding between three days of MOP and the Bar Mitzvah, you'd probably want MOP.[/quote]
I think its Bar Mitzvot  :)

Wow Blue MOP seems pretty big.... but a Bar Mitzvah only happens once in your brother's lifetime right?  Hard choice...

[quote="jb05"][quote="Phelpedo"]No it's definitely Bar Mitzvah (or Bat Mitzvah, depending on if it's a boy a girl). Anyway... if you guys don't know what a Bar Mitzvah is, you'd better stop doing math and go outside.[/quote]

I think he was referring to the plural.

I honestly don't see why the 13th birthday is any different than the 12th or 14th birthday. Unless you actually don't [b]want[/b] to go to MOSP, you should definitely skip the Bar Mitzvah. But then, I'm not religious, so I wouldn't understand.[/quote]Look on [url=http://en.wikipedia.org/wiki/Bar_Mitzvah]Wikipedia[/url].  I agree with Phelpedo; you should know something about this.  It is definitely a lot more than just the thirteenth birthday.  Often, the subject has been preparing for months in advance.

Your brother's Bar Mitzvah will only happen once in his life, and in yours, and since it's such an important ceremony, don't you think you should be there to support him, and to congragulate him on his coming of age?  MOSP might be fun, but your family, and your brother, should come first.  Maybe you should ask your brother what he thinks; try to get an honest answer from him, and see how much it'd mean to him if you were there, and whether he'd be hurt if you weren't.

Getting the opportunity to go to MOP is extraordinary.  Many people, myself included, never made it that far in the US IMO selection process.  I know the basics of what a Bar Mitzvah is but not being Jewish and not having been around other Jewish people I don't really know just [i]how[/i] important it is.  You are the best person to judge whether or not you need to/want to go to the Bar Mitzvah.

Would you say that Bar Mitzvah's and weddings are roughly of the same importance?  That would serve as a good method of comparison for me (and other people here who aren't Jewish).  Hm...hypothetically choosing between my sister's marriage and MOP would be TOUGH indeed :P

[quote="krustyteklown"][quote="Phelpedo"]No it's definitely Bar Mitzvah (or Bat Mitzvah, depending on if it's a boy a girl). Anyway... if you guys don't know what a Bar Mitzvah is, you'd better stop doing math and go outside.[/quote]

ummm, not everyone is Jewish or hangs around a lot of Jews...[/quote]
Evidently. I'm not Jewish myself, but I'd imagine that most people would still have heard of stuff like Bar Mitzvahs, the Old Testament, and Judaism.

[quote="jb05"]I think he was referring to the plural.

I honestly don't see why the 13th birthday is any different than the 12th or 14th birthday. Unless you actually don't want to go to MOSP, you should definitely skip the Bar Mitzvah. But then, I'm not religious, so I wouldn't understand.[/quote]
Correct me if I'm wrong, but Bar Mitzvot is not the plural of Bar Mitzvah. As for the 13th birthday being special, well that's just a cultural/religious thing. This is just an analogy, but by your reasoning December 25th is no different than December 24th or 26th.

Err... you guys do not get it.  A Bar Mitzvah is not just some random 13th birthday party.  It is a ceremony to celebrate one's becoming a Jewish adult, which involves said person reading from the Torah and perform other services he has been practicing.  Like the USAMO, one studies very hard to have a limited grasp of Hebrew and understanding of the Jewish religion before his Bar Mitzvah.  So yeah, it is nothing like December 25th.

By the way, I know all that and I'm not Jewish either.  So if you don't know what a Bar Mitzvah is, you do indeed need to realize that there's this thing called the world out there :? ... pretty profound I know :P

[quote="qqwref"]This is quite annoying... my brother's having his bar mitzvah during the MOSP, but going back home to attend would take three days (of MOSP...) and I'd probably be really tired during the ceremony. Personally, I think that the MOSP is more important... but what do you guys think?[/quote]

Don't be selfish. This is a really important time in his life, and not having you there to share it with him, you whom he looks up to as a role model, would really hurt him, in my opinion. But it's your choice.

Considering the plural of Mitzvah in hebrew is Mitzvot, the plural of Bar Mitzvah is Bar Mitzvot.

But, likely knowing my dis-knowledge-ness of Hebrew grammar, I'm probably horrendously incorrect.

EDIT: That was the case  :P

[quote="krustyteklown"][quote="Phelpedo"]No it's definitely Bar Mitzvah (or Bat Mitzvah, depending on if it's a boy a girl). Anyway... if you guys don't know what a Bar Mitzvah is, you'd better stop doing math and go outside.[/quote]

ummm, not everyone is Jewish or hangs around a lot of Jews...[/quote]
Hey, half our math team is Jewish, and you should know.

[hide]The other half is Asian...[/hide]

[quote="sirorange"]Considering the plural of Mitzvah in hebrew is Mitzvot, the plural of Bar Mitzvah is Bar Mitzvot.[/quote]

It is interesting to note that although Mitzvot is the hebrew plural for Mitzvah, saying Bar Mitzvot would probably not make gramatical sense.  Since "Bar Mitzvah" is refering to the boy, not the ceremony, the correct plural would be "B'nei Mitzvah", as "B'nei" is plural for "Bar".

MOP is a huge opportunity. There may be things that are worth missing it for, but I have a hard time imagining your brother's ceremony being *so* important. But I'm not Jewish, so the usual disclaimers apply. As has been said, you should ask your brother.

[quote="Phelpedo"]
Evidently. I'm not Jewish myself, but I'd imagine that most people would still have heard of stuff like Bar Mitzvahs, the Old Testament, and Judaism. [/quote]

Umm....Old Testament is a Christian idea. In Judaism, there is no "New Testament," so what Christians call the "Old Testament" is just simply, to put it in English terms, the bible to Jews.

As for the Bar Mitzvah, I could understand going to MOP instead. Your brother will still have his Bar Mitzvah regardless of whether or not you're there, and there are various other members of the family of course.

[quote="DPatrick"]I think many of you are missing the point.  You're not talking about missing MOP, you're talking about missing [i]three days[/i] of MOP.  There's a big difference.

In my opinion, MOP is not so special that missing 3 days of it is a big deal.  (Unless they're the first 3 days, but that's another story.)  When I was at MOP, a couple of the seniors took a couple of days off to go to their high school graduations.  (I didn't, because I didn't care to.)  You can take the MOP problems that you'll miss on the plane with you.

What exactly are you losing by taking 3 days out of MOP to attend what is literally a once-in-a-lifetime event?  Very little in my opinion.  If the choice were missing all of MOP or the IMO or something, that's another story.  But 3 days of a 3-week-long program is fairly insignificant.[/quote]

I thought MOP was an all-or-nothing deal this year. Correct me if I'm wrong, but I remember reading in other threads that, for example, people who have already accepted RSI, which overlaps about a week with MOP, cannot attend MOP at all.

[quote="paladin8"]I thought MOP was an all-or-nothing deal this year. Correct me if I'm wrong, but I remember reading in other threads that, for example, people who have already accepted RSI, which overlaps about a week with MOP, cannot attend MOP at all.[/quote]That's correct, and it was that way too when I was in high school (I went to RSI my junior year even though I was HM on the USAMO and was invited to MOP).  That doesn't (I believe) prohibit short absences in the middle of the MOP for compelling personal reasons (as I said, seniors were allowed to miss a couple of days to go to their high school graduation).  But I would check with the AMC to be sure.

If the decision really does come down to missing MOP entirely, then what I said in my last post goes out the window.

Here are my basic positives/negatives so far:

Reasons to attend the Bar Mitzvah:
- It's a once-in-a-lifetime event. (For him.)
- My brother's been studying hard for it for a year or so.
- It would be nice to show support for my brother.
- It's customary for the entire family to come. Most of the family will be there.
Reasons not to attend:
- My family's really not all that observant in the Jewish religion.
- I'll be a senior next year so I may not get into MOP again.
- It might not be possible to skip the three days (by the rules, or because that weekend is pretty early and I might miss some important stuff).
- I'd probably be tired/annoyed during the ceremony due to having to come home (and anticipating the having to fly back to Nebraska).
- My brother's very sensitive and if I accidentally say something that he doesn't want to hear he gets sulky.
- I don't know very many of my brother's friends.

It's still a hard choice.  :|

Perhaps if you talk to Zuming or the big-wigs up at MOP, you guys can arrange something where they will let you leave for a few days.

[quote="qqwref"]Here are my basic positives/negatives so far:

...

It's still a hard choice.  :|[/quote]

I think you need more information. Email someone at MOSP and ask them what your options are. I would start with Donita Bowers. She is the person who sent the email. Tell them the dates you will come and go. You will be leaving on a Friday, and coming back on Sunday, right? Do Moppers get weekends off? In which case you are only missing Friday.

Talk to your family. Ask them how they feel about it. Let them feel that you are taking their feelings into account. Will they care terribly if you don't come? Are you the only sibling? I think the point that most struck me from your list is that you are not a particularly observent family.

I agree with DPatrick, if you cannot get off those 3 days, then you should probably go to MOPS and make it up to your brother when you get home. Take him somewhere special, and spend the day with him.  Make sure he realizes that you are doing this to commerate his Bar Mitzvah. That way when he looks back on it, he wont think you missed his special day, he will remember the effort you made to commemorate the event.

Good luck!

[i]Do Moppers get weekends off?[/i]

Good question... anyone know?

Not completely. There are still some classes, albeit they are shorter, usually.

[quote="chess64"]Well it's somebody else's Bar Mitzvah so I don't see why you are even obligated to attend, but if you're his close friend or something, I guess you should go.[/quote]

Oh woops I didn't realize it was your brother, I thought you said it was your friend. That definitely changes everything! :oops:

My advice - If you cannot miss those three days of MOP, tell your brother that you are very sorry, but you may only have this opportunity once in your life.  Then, talk with your parents and arrange to send him a very fancy and formal card with a very personal and heartfelt message from you on it, for his bar mitzvah, and have your parents surprise him with the card at the appropriate time during the ceremony.  This way, you get to go to MOP, he gets a pleasant surprise and knows you are thinking of him, although he will still probably wish that you could have come.  If you can miss those three days, then by all means GO.  If you can't send mail, I can't help you there.  Figure it out.

That's my advice.

Edit: :blush:  Wow I am not sure how I got to this thread but I thought it was recent! did not realize that I was reviving something really old, I'm sorry (thanks for pointing out boy soprano)

Uh, MOP ended about four and a half months ago, and qqwref was there... :maybe:   I can't recall how the bar mitzvah worked out, but in the future, wouldn't a PM be better than reviving an eight-month-old thread? :wink:

[quote="sapphyre571"]put yourself in your brother's place
you know you'd want him there at your bar mitzvah
are you a junior or senior?[/quote]
Oh my god, read the post above yours.


Let $ p,\ q$ be real numbers. Define a sequence $ \{a_n\}$ defined by $ a_1 \equal{} p,\ a_2 \equal{} q,\ a_{n \plus{} 2} \equal{} 2a_{n \plus{} 1} \plus{} a_n\ (n \equal{} 1,\ 2,\ \cdots).$ Find the necessary and sufficiently condition in regarding $ p,\ q$ for which $ \sum_{n \equal{} 1}^{\infty} a_n$ converges.

hello kunny, for the sequence i have got
$ a_n\equal{}\frac{1}{4}(1\plus{}\sqrt{2})\left(\minus{}((1\minus{}\sqrt{2})^n(2\plus{}\sqrt{2})\plus{}(1\plus{}\sqrt{2})^n(\minus{}10\plus{}7\sqrt{2}))p\plus{}(\sqrt{2}(1\minus{}\sqrt{2})^n\plus{}(1\plus{}\sqrt{2})^n(\minus{}4\plus{}3\sqrt{2}))q\right)$
Sonnhard.

I have examined $ a_1$ to get $ a_1\equal{}\frac{2\sqrt{2}}{4}p.$

[quote="Dr Sonnhard Graubner"]hello kunny, are you sure, i got $ p$.

Sonnhard.[/quote]

Very sorry, your result was correct. :blush:

kunny

Therefore equal the coefficient of $ (1\plus{}\sqrt{2})^n$ to $ 0$ in order to obtain convergence condition.


Βάζω στο συνημμένο μια άσκηση(με το σχήμα)  που ελπίζω να σας αρέσει. Επειδή φεύγω για διακοπές, θα δω τις απαντήσεις σας μέσα Αυγούστου. Αν κάποιος θελήσει να με ρωτήσει κάτι , για οποιοδήποτε θέμα ,  ας με πάρει στο 697 9382042.

Καλές διακοπές σε όσους δεν έφυγαν ακόμα και καλό καλοκαίρι σε όλους !

Μπάμπης

[b]ΑΣΚΗΣΗ [/b]
Οι παρεγγεγραμμένοι κύκλοι που αντιστοιχούν στις γωνίες Α και Β ενός ορθογωνίου τριγώνου ΑΒΓ με υποτείνουσα ΒΓ εφάπτονται με την ευθεία ΑΓ στα σημεία Ε και Ζ αντίστοιχα. Αν Μ είναι το μέσο της ΑΓ και η ΔΖ τέμνει τον περι-γεγραμμένο κύκλο του τριγώνου ΑΔΓ στο σημείο Θ , να αποδειχθεί ότι :
  α) Το τετράπλευρο ΔΜΘΕ είναι εγγράψιμο.
  β) Το Γ είναι έγκεντρο του τριγώνου ΔΘΕ.

ΑΔ είναι το ύψος του δοσμένου τριγώνου και προβλέπω ότι μέχρι αύριο, η άσκηση αυτή θα έχει λυθεί.

( Από αυτούς που δεν έχουν φύγει ακόμα, γιατί οι φευγάτοι είναι γάτοι... :)  )

Καλά να περνάτε όπου κι αν βρίσκεστε, Κώστας Βήττας.

Βρισκομαι ακομη σε διακοπες αλλα μου αρεσε η ασκηση.
Για να μπορεσω να χρησιμοποιησω latex, στην λυση μου θα αλλαξω το Θ με το G.
Θεωρω $ s$ την ημιπεριμετρο.

Ι)

Αρκει $ ZM.ZE \equal{} ZD.ZG$ ομως $ ZD.ZG \equal{} ZA.ZC \equal{} MC^2 \minus{} ZM^2$.

$ MC^2 \minus{} ZM^2 \equal{} \frac {b^2}{4} \minus{} (\frac {b}{2} \minus{} s \plus{} a)^2 \equal{} 2as \plus{} bs \minus{} s^2 \minus{} a^2 \minus{} ab$

$ ZM.ZE \equal{} (MC \minus{} ZC)(ZC \plus{} CE) \equal{} [\frac {b}{2} \minus{} (s \minus{} a)].[(s \minus{} a) \plus{} (s \minus{} b)] \equal{} (\frac {b}{2} \minus{} s \plus{} a).(2s \minus{} b \minus{} a)$

το οποιο μετα απο πραξεις καταληγει ομοιως στο $ 2as \plus{} bs \minus{} s^2 \minus{} a^2 \minus{} ab$.

Αρα $ ZM.ZE \equal{} ZD.ZG$.

ΙΙ)

Στο παραπανω εγγραψιμο τα τοξα $ MG,MD$ ειναι ισα αρα $ \angle{CED} \equal{} \angle{CEG}$.
Απο τον περιγεγραμμενο κυκλο εχουμε οτι $ 2.\angle{CDG} \equal{} \angle{CMG}$
και απο το εγγραψιμο οτι $ \angle{CMG} \equal{} \angle{GDE}$.
Αρα $ \angle{CDG} \equal{} \angle{CDE}$.

Αρα $ C$ εγκεντρο του $ GDE$.

Χαιρετισμους σε ολους.

Συγχωρέστε με που για ευκολία στο [b][size=100]Latex[/size][/b], χρησιμοποιώ και γράμματα που υπάρχουν στο Λατινικό αλφάβητο.

[quote="stergiu"][b][size=100][color=DarkBlue]Οι παρεγγεγραμμένοι κύκλοι που αντιστοιχούν στις γωνίες $ \angle A$ και  $ \angle B$ ενός ορθογωνίου τριγώνου $ %Error. "bigtriangle" is a bad command.
ABC$  με υποτείνουσα $ BC,$ εφάπτονται με την ευθεία $ AC$ στα σημεία $ E$ και $ Z$ αντίστοιχα. Αν $ M$ είναι το μέσο της $ AC$ και η ευθεία $ DZ,$ όπου $ AD$ είναι το ύψος του $ \bigtriangleup ABC,$ τέμνει τον περιγεγραμμένο κύκλο του τριγώνου $ \bigtriangleup ADC$ στο σημείο $ F,$ να αποδειχθεί ότι :
  α) Το τετράπλευρο $ DEFM$ είναι εγγράψιμο.
  β) Το σημείο $ C$ είναι το έγκεντρο του τριγώνου $ \bigtriangleup DEF$.[/color][/size][/b][/quote]
Ας δούμε μία απόδειξη με αρμονικά συζυγή που δίνουν όμορφες λύσεις χωρίς πολλές πράξεις.

$ \bullet$ Είναι γνωστό ότι τα σημεία $ A,\ Z,\ C,\ E,$ αποτελούν αρμονική σημειοσειρά.

$ ($ Από τα όμοια ορθογώνια τρίγωνα $ \bigtriangleup EI_{a}C,\ \bigtriangleup ZI_{b}C$ $ \Longrightarrow$ $ \frac {CE}{CZ} = \frac {I_{a}E}{I_{b}Z}$ $ ,(1)$ και από όμοια ορθογώνια

τρίγωνα $ \bigtriangleup EI_{a}A,\ \bigtriangleup ZI_{b}A$ $ \Longrightarrow$ $ \frac {AE}{AZ} = \frac {I_{a}E}{I_{b}Z}$ $ ,(2)$ και από $ (1),\ (2)$ $ \Longrightarrow$ $ \frac {CE}{CZ} = \frac {AE}{AZ}\ ).$

Έτσι, η δέσμη $ D.AZCE$ είναι επίσης αρμονική και επειδή $ AD\perp BC,$ συμπεραίνουμε ότι $ \angle ZDC = \angle CDE$ $ ,(3)$

$ \bullet$ Από ορθογώνια τρίγωνα $ \bigtriangleup DAC,\ \bigtriangleup FAC,$ έχουμε ότι $ MD = MA = MF$ και άρα $ \angle MDF = \angle MFD$ $ ,(4)$

Επειδή τώρα, $ \angle EMF = 2\angle CAF = 2\angle CDF = \angle EDF,$ προκύπτει ότι το τετράπλευρο $ DEFM$ είναι εγγράψιμο και το πρώτο ζητούμενο της πρότασης έχει αποδειχθεί.

$ \bullet$ Όπως λέει ο Σάκης, στον κύκλο $ (K)$ που περιγράφει το τετράπλευρο $ DEFM,$ από $ MD = MF$ έχουμε ότι $ \angle MED = \angle MEF$ $ ,(5)$

Από $ (3),\ (5)$ προκύπτει ότι η κορυφή $ C$ του $ \bigtriangleup ABC,$ είναι το έγκεντρο του τριγώνου $ \bigtriangleup DEF$ και το δεύτερο ζητούμενο έχει αποδειχθεί.

Κώστας Βήττας.


inequalities
function
algebra
polynomial
inequalities theorems

[quote="barcelona"]Hi,
Can anyone give me some examples with the Newton inequality,because i can't understand it.Thanks.[/quote]

First of all we start from the theory. Newton says that we define by $ s_{k} \equal{} \frac {\sigma_{k}}{\binom{n}{k}}$ the symmetrical mean of the numbers $ a_{1},a_{2},...,a_{n}$, where $ \sigma_{k}$ is the symmetrical function (of order $ k$) of the numbers $ a_{1},a_{2},...,a_{n}$ of the coefficient of the term $ x^{n \minus{} k}$ of the polynomial $ (x\plus{}a_{1})(x\plus{}a_{2})\cdot \cdot \cdot (x\plus{}a_{n})$.

That means, that if for example we have $ n \equal{} 3$ then we have the polynomial $ (x \plus{} a_{1})(x \plus{} a_{2})(x \plus{} a_{3}) \equal{} x^{3} \plus{} (a_{1} \plus{} a_{2} \plus{} a_{3})x^{2} \plus{} (a_{1}a_{2} \plus{} a_{2}a_{3} \plus{} a_{3}a_{1})x \plus{} a_{1}a_{2}a_{3}$ then

$ \sigma_{0} \equal{} 1$ since the coefficient of $ x^3$ equals to $ 1$
$ \sigma_{1} \equal{} a_{1} \plus{} a_{2} \plus{} a_{3}$ since the coefficient of $ x^2$ equals to $ a_{1} \plus{} a_{2} \plus{} a_{3}$
$ \sigma_{2} \equal{} a_{1}a_{2} \plus{} a_{2}a_{3} \plus{} a_{3}a_{1}$
$ \sigma_{3} \equal{} a_{1}a_{2}a_{3}$

So, Newton's inequality says $ s_{k \minus{} 1}s_{k \plus{} 1}\leq s^{2}_{k}$.

Now examples: For $ n \equal{} 3$ we have $ s_{0}s_{2}\leq s^{2}_{1}\Longleftrightarrow \frac {ab \plus{} bc \plus{} ca}{3}\leq \left(\frac {a \plus{} b \plus{} c}{3}\right)^{2}$

$ s_{1}s_{3}\leq s^{2}_{2}\Longleftrightarrow \frac {a \plus{} b \plus{} c}{3}\cdot abc\leq \left(\frac {ab \plus{} bc \plus{} ca}{3}\right)^{2}$

Hi,
can anyone explain me in more detail the proof of Newton inequality,because i can't understand it.Thanks.


I've seen two covers pictures, one, behind mario as he prepares to drive, and another, with a ball view af mario chipping (i think).

Is there a difference.  I am thinking of buying one.  The one with the 2nd cover is the one i am thinking of.  I need NTSC.

help?

[url]http://en.wikipedia.org/wiki/Mario_Golf_Toadstool_Tour[/url]
vs.
search ebay, and then it should be 6th or 7th on the list.

Also, do they reissue games with different covers when WII came out?


Find the number of the connected graphs with 6 vertices. (Vertices are considered to be different)

Labeled graphs of graphs up to isomorphism? (I think the second one...)

can anybody post  answer (i mean just the number)

[url=http://www.research.att.com/~njas/sequences/A001349]A001349[/url] in the [url=http://www.research.att.com/~njas/sequences/]OEIS[/url] is connected graphs up to isomorphism.  Probably you can figure out how to find connected labeled graphs on your own.

i think you could complet the graph to a complete one and then color the edges red if the exist initialy and blue if the didn't exist initaly , and then count the pemutations...

oh its not so hard the number of graph that are connected with 6 vertices is the number of graph with 6 vertices $-$the graph that are not connected!!
and the graph that are not connected with $6$vertices can find with Recursive:)

...from the booklet Turkish Mathematical Olmypiad 2006:

[b]Problem:[/b]
An airline company is planning to run two-way flights between some of the six cities $A$, $B$, $C$, $D$, $E$ and $F$. Determine the number of ways these flights can be arranged so that it is possible to travel between any two of these six cities using only the flights of this company.

[b]Solution:[/b]
Using the inclusion-exclusion principle we find that the number of ways the flights can be arranged so that each city is connected to at least one other city by a flight is

\[2^{\binom 62} - \binom 61 2^{\binom 52} +\binom 62 2^{\binom 42} -  \binom 63 2^{\binom 32} + \binom 64 2^{\binom 22} - \binom 65 + \binom 66 = 27449\]

Among these, the configurations that do not satisfy the condition of the problem, and the corresponding number of ways the flights can be arranged are shown in the following figure:

[asy]
unitsize(30);
pair A, B,C,D,E,F;
real x, y;
A=(0,0); B=(2,1); C=(2.5, 0.8); D=(2.3, -1); E=(1.9, -1.1); F=(0, -0.4);
draw(A--B);
draw(C--D);
draw(E--F);

dot(A);dot(B);dot(C);dot(D);dot(E);dot(F);
label("$\frac{6!}{2^3 3!}$", (3, 0));

x=5;
A=(x+0,0); B=(x+0.2,1); C=(x+1.5, 0.8); D=(x+1.6, 0.3); E=(x+1.5, -1.1); F=(x+0, -0.4);
draw(A--B--C--cycle);
draw(D--E--F--cycle);
dot(A);dot(B);dot(C);dot(D);dot(E);dot(F);
label("$\frac{1}{2}{{6} \choose {3}}$", (x+2.5, 0));

x=10;
A=(x+0,0); B=(x+0.2,1); C=(x+1.5, 0.8); D=(x+1.6, 0.3); E=(x+1.5, -1.1); F=(x+0, -0.4);
draw(A--B--C--cycle);
draw(D--E--F);
dot(A);dot(B);dot(C);dot(D);dot(E);dot(F);
label("$3 \cdot {{6} \choose {3}}$", (x+2.5, 0));

x=0; y=-4;
A=(x+0,y+0); B=(x+0.2,y+1); C=(x+1.5, y+0.8); D=(x+1.6, y+0.3); E=(x+1.5, y+-1.1); F=(x+0, y+-0.4);
draw(A--B--C);
draw(D--E--F);
dot(A);dot(B);dot(C);dot(D);dot(E);dot(F);
label("$\frac{3 \cdot 3}{2} {{6} \choose {3}}$", (x+2.5, y+0));

x=5; y=-4;
A=(x+0,y+0); B=(x+1.7,y+1); C=(x+2.0, y+0.8); D=(x+1.8, y+-1); E=(x+0.6, y+-1.1); F=(x+0, y+-0.4);
draw(A--B);
draw(C--D--E--F--cycle);
draw(C--E);draw(F--D);
dot(A);dot(B);dot(C);dot(D);dot(E);dot(F);
label("${{6} \choose {2}}$", (x+2.5, y+0));

x=10; y=-4;
A=(x+0,y+0); B=(x+1.7,y+1); C=(x+2.0, y+0.8); D=(x+1.8, y+-1); E=(x+0.6, y+-1.1); F=(x+0, y+-0.4);
draw(A--B);
draw(C--D--E--F--cycle);
draw(C--E);
dot(A);dot(B);dot(C);dot(D);dot(E);dot(F);
label("${{6} \choose {2}}{{4} \choose {2}}$", (x+2.5, y+0));

x=0; y=-8;
A=(x+0,y+0); B=(x+1.7,y+1); C=(x+2.0, y+0.8); D=(x+1.8, y+-1); E=(x+0.6, y+-1.1); F=(x+0, y+-0.4);
draw(A--B);
draw(C--D--E--F--cycle);
dot(A);dot(B);dot(C);dot(D);dot(E);dot(F);
label("$3\cdot {{6} \choose {2}}$", (x+2.5, y+0));

x=5; y=-8;
A=(x+0,y+0); B=(x+1.7,y+1); C=(x+1.8, y+0.8); D=(x+1.5, y+-1); E=(x+1.6, y+-1.8); F=(x+0, y+-0.4);
draw(A--B);
draw(C--D--F--cycle);
draw(E--D);
dot(A);dot(B);dot(C);dot(D);dot(E);dot(F);
label("$4 \cdot 3 \cdot {{6} \choose {2}}$", (x+2.5, y+0));

x=10; y=-8;
A=(x+0,y+0); B=(x+1.7,y+1); C=(x+1.8, y+0.8); D=(x+1.5, y+-1); E=(x+1.6, y+-1.8); F=(x+0, y+-0.4);
draw(A--B);
draw(C--D--F);
draw(E--D);
dot(A);dot(B);dot(C);dot(D);dot(E);dot(F);
label("$4 \cdot {{6} \choose {2}}$", (x+2.5, y+0));

x=0; y=-12;
A=(x+0,y+0); B=(x+1.7,y+1); C=(x+2.0, y+0.8); D=(x+1.8, y+-1); E=(x+0.6, y+-1.1); F=(x+0, y+-0.4);
draw(A--B);
draw(E--F--C--D);
dot(A);dot(B);dot(C);dot(D);dot(E);dot(F);
label("$\frac {4!}{2} {{6} \choose {2}}$", (x+2.5, y+0));

[/asy]

Therefore, the answer is $27449 - 745 = 26704$

An involved solution, I think, is good to pair up with the booklet answer above,

Note that, the preamble requires, isomorphic graphs are counted multiple times, {\em e.g.}, consider two graphs, with two edges; where, the first is, $(A-B,A-C)$; and, the second is, $(D-E,D-F)$. While these two graphs are isomorphic, they must be counted differently, due to the presence of labels.

Our strategy, is to count, through the cardinality of the largest connected subgraph. Let, $M_i$, be the total number of graphs on $6$-vertices, whose maximal connected subset, consists of $i$-vertices (for instance, a graph in $M_3$, must contain a triangle; while any subset with $4$, $5$, or $6$ vertices, must be unconnected). Also note that, $M_0=1$ (simply, a graph with no edges), and $M_1=0$ (if there is an edge, then, at the very least, we can find $2$ connected vertices).

Note that, there are, a total of, $2^{\binom{6}{2}}=32768$ graphs, that can be constructed on, $6$-vertices. Hence, the answer we seek is, $32768 - \sum_{i=2}^5 |M_i|-1$.

$a)$ $|M_2|:$ Suppose that, the maximal connected subset, consists of $3$-vertices, only. Therefore, the possibilities are, $2+2+2$, $2+2+1+1$; and $2+1+1+1$ (think of these summations, as, partitioning $6$-vertices into equivalence classes, depending on the connectivity).

$\bullet$ $2+2+2$ gives us, the following graph, and isomorphisms, $(A-B,C-D,E-F)$, hence, a total of,
$$
\frac{\binom{6}{2}\binom{4}{2}\binom{2}{2}}{3!}=15.
$$
$\bullet$ $2+2+1+1$ gives us, $(A-B,C-D)$, and isomorph graphs (namely, graphs with two edges, where, each edge, has different endpoints). Therefore, the total is,
$$
\frac{\binom{6}{2}\binom{4}{2}}{2}=45.
$$
$\bullet$ Finally, $2+1+1+1+1$ gives us a graph with only one edge, namely, there are, $\binom{6}{2}=15$ such graphs, on $6$-vertices.

Hence, $|M_2|=75$.
$b)$ Next, we continue by $|M_3|$. The possibilities are, $3+3$, $3+2+1$, and $3+1+1+1$. Also, as a quick exercise, let us count, the total number of connected graphs, on $3$-vertices, where, the vertices are fixed. We can either have a triangle, or, $V$-shapes, and there are, $3$ $V$-shaped structure. Hence, fixing $3$-vertices, gives us, $4$ connected graphs (notice that, the total number of graphs, on $3$-vertices, is, $2^{\binom{3}{2}}=8$.

$\bullet$ $3+3$ can be obtained, either, through, $(T,T)$, $(T,V)$, or $(V,V)$; where, $T$ is a triangle, and $V$ is a $V$-shaped structure. There are, $\displaystyle\frac{\binom{6}{3}}{2!}=10$ $(T,T)$-configurations; $\displaystyle\binom{6}{3}\cdot 3 =60$, $(T,V)$-configurations; and finally, $\displaystyle\frac{\binom{6}{3}\cdot 3 \cdot 3}{2}=90$, $(V,V)$-configurations. The total is, $90+60+10=160$ configurations.
$\bullet$ $3+2+1$ can be obtained in, a total of,
$$
\binom{6}{3}\cdot 4 \cdot \binom{3}{2} = 240,
$$
diferent ways; where, we first select, three vertices, that will be connected, and count $4$ valid arrangements; and then, select $2$ more vertices, among the remaining $3$.
$\bullet$ Finally, $3+1+1+1$ can be obtained in, $\binom{6}{3}\cdot 4 =80$ different way.

Thus, $|M_3|=480$.

$c)$ $|M_4|$ is a little bit more involved. Let us first, begin by, a fact, that will be very useful in the remaining.
[b]Lemma:[/b] On $4$-vertices, there are, $38$ connected graphs.
[i]Proof:[/i] Argue via the total number of edges. Notice that, any graph, on $4$-vertices, with $4$ or more edges, is connected, giving us, $\displaystyle\binom{6}{4}+\binom{6}{5}+\binom{6}{6}=22$. Furthermore, there are configurations on three edges, that are, also, connected; and these are simply, the shapes, without any triangles. Hence, a total of, $\binom{6}{3}-\binom{4}{3}=16$; where, $\binom{6}{3}$ is number of different ways, in which, we can select $3$-edges, out of $6$; and the subtraction, is to eliminate the triangles. Therefore, the answer is, as desired, 38.

With the lemma in our hands, now, there are two candidate arrangements, $4+2$, and $4+1+1$; and therefore, the answer is,
$$
|M_4|=\binom{6}{4}\cdot 38 \cdot 2 = 1140,
$$
where, four connected vertices, can be selected in $\binom{6}{4}$ different ways; having selected those $4$ vertices, there are $38$ arrangements; and finally, we have, either, the remaining $2$ connected; or disconnected, thus, a multiplication by $2$.

$d)$ $|M_5|$ is hard to work with, directly. Hence, we instead, take a convoluted approach, and begin by, computing the total number of connected graphs, on $5$-vertices. First, note that, there are, a total of, $2^{\binom{5}{2}}=1024$ graphs, that live on $5$-vertices. Among these;

$\bullet$ the ones, where the maximal connected subset, has a size $4$, are, $\displaystyle\binom{5}{4}\cdot 38=190$,
$\bullet$  the ones, with a maximal connected subset of size $3$, are, $\displaystyle\binom{5}{3}\cdot 4 \cdot 2 =80$, where, the multiplication by $2$, is to take care of, $3+2$, or $3+1+1$;
\item and finally, the ones, with a maximal connected subset of size $2$, are, $\displaystyle \frac{\binom{5}{2}\binom{3}{2}}{2}+\binom{5}{2}=25$, where the first, takes care of, $2+2+1$, and the second, for, $2+1+1+1$.

Thus, the total number of connected graphs, that can be constructed on $5$-vertices, are,
$$
1024-190-80-25-1=728,
$$
where, the last item, is to take care of a graph, without any edges.

With this, we can now, clinch $|M_5|$ via,
$$
\binom{6}{5}\cdot 728 = 4368.
$$

$e)$ Now, we will finish up the answer,
\begin{align*}
|M_6| &= 2^{\binom{6}{2}} - |M_5|-|M_4|-|M_3|-|M_2|-1\\
&=32768 - 4368 - 1140 - 480 - 75 - 1 \\
&=26704.
\end{align*}

[quote=Math-Ninja]That was a 3 year revival btw[/quote]

Post necessary stuff, don't spam. There is no rule preventing you from posting, if you think you have a detailed/different approach.


[quote="fjwxcsl"]let  $ {x,y,z}\ge0,$ prove
\[{ \sum_{cyc}{\sqrt {y^2 + z^2}(y - z)^2}\geq{\frac {2\sqrt {5}}{5}}(x^3 + y^3 + z^3 - 3xyz)}.
\]
[/quote]
This inequality can be rewritten as: $ A(y-z)^2+B(z-x)^2+C(x-y)^2 \ge 0,$ where
$ A= \sqrt{5(y^2+z^2)} -(x+y+z)$
and $ B,C$ are similar.
Without loss of generality, we can assume that $ x \ge y \ge z \ge 0,$ then $ A \le B \le C$. Moreover, we have
$ A+B=\sqrt{5(y^2+z^2)}+\sqrt{5(x^2+z^2)} -2(x+y+z) \ge \sqrt{5[(x+y)^2+4z^2]}-2(x+y+z) =\sqrt{(4+1)[(x+y)^2+4z^2]} -2(x+y+z)\ge 2(x+y)+2z-2(x+y+z)=0.$
Thus $ C \ge B \ge 0$, and with note that $ (z-x)^2 \ge (y-z)^2$, we obtain
$ A(y-z)^2+B(z-x)^2+C(x-y)^2 \ge (A+B)(y-z)^2+C(x-y)^2 \ge 0.$
This completes our proof. :)

Very nice  proof!

And we have another   nice  inequality:

\[{ \sum_{cyc}{\sqrt {y^2 + z^2}(y - z)^2}\leq2(x^3 + y^3 + z^3 - 3xyz)}.\]

[quote="fjwxcsl"]Very nice  proof!

And we have another   nice  inequality:
\[{ \sum_{cyc}{\sqrt {y^2 + z^2}(y - z)^2}\leq2(x^3 + y^3 + z^3 - 3xyz)}.
\]
[/quote]
It is equivalent to
$ \sum (x+y+z-\sqrt{y^2+z^2})(y-z)^2 \ge 0,$
which is trivial as $ y+z \ge \sqrt{y^2+z^2}.$ :)


Let $X$ be an arbitrary set and $f$ a bijection from $X$ to $X$. Show that there exist bijections $g,\ g':X\to X$ s.t. $f=g\circ g',\ g\circ g=g'\circ g'=1_X$.

really nice!
so.. choose a subset $R\subset X$ such that $X = \bigcup_{x\in R}\text{orb}(x)$, where, if $x, y \in R$, then $\text{orb}(x)$ and $\text{orb}(y)$ are disjoint iff $x\neq y$.
lets divide $R = R_\infty \cup R_{0}\cup R_{1}$, where $\forall x \in R_\infty$ $\text{orb}(x)$ is infinite, $\forall x \in R_{j}$ $|\text{orb}(x)| \equiv j \mod 2$, for $j=0,1$.
now define, for every $x\in R_\infty$, and any $k$ integer, $g(f^{k}(x)) = f^{1-k}(x)$, and $h(f^{k}(x)) = f^{2-k}(x)$.
for every $x\in R_{0}$, let $2n_{x}$ be the length of its orbit, and define then $g(x) = f^{2}(x), g(f(x)) = f(x), g(f^{2}(x)) =x$, and, for $k\ge3$, let $g(f^{k}(x)) = f^{2n_{x}+2-k}(x)$; $h(x) = f^{3}(x), h(f(x)) = f^{2}(x), h(f^{2}(x)) = f(x), h(f^{3}(x)) = x$, and for $k\ge 4$, let $h(f^{k}(x)) = f^{2n_{x}+3-k}(x)$.
finally, given $x\in R_{1}$, let $2m_{x}+1$ be its orbit length, and define: $g(x) = f^{2}(x), g(f(x)) = f(x), g(f^{2}(x)) =x$, and, for $k\ge3$, let $g(f^{k}(x)) = f^{2m_{x}+3-k}(x)$; $h(x) = f^{3}(x), h(f(x)) = f^{2}(x), h(f^{2}(x)) = f(x), h(f^{3}(x)) = x$, and for $k\ge 4$, let $h(f^{k}(x)) = f^{2m_{x}+4-k}(x)$.
everything gets easy to verify...

anyway, just to give the idea (since everything i wrote is really boring, the key is working inside every single orbit, just to be sure everything is bijective.. the construction inside every orbit is forced as soon as you ask for $g(x) = f^{2}(x)$.. it's just worth making a sketch (a graphical sketch) of what really happens here...
nice problem, though..

Ugh..., what is orbit ???  :(
Does anyone have easier solutions to read....

uh, sorry :)
you call $\text{orb}(x)$ the set $\{f^{k}(x) | k \in \mathbb{Z}\}$... just a matter of vocabulary, no concepts involved...
anyway, just read the sketch of the solution, since the previous part, as i already sad, is boring details, just to put a simple idea in "mathematical language"..


What is $\infty^\infty$?

Answer: [hide] I think that it is infinity [/hide]

[quote="mathcnt2005"]What is $Infinity{}^I{}^n{}^f{}^i{}^n{}^i{}^t{}^y$?
 I mean the sign thing for infinity. The answer is not Infinity^Infinity if Infinity is a giant variable.
By the way, how do you do the sign for infinity?

Answer: [hide] I think that it is infinity [/hide][/quote]

Eh.. I don't think this belongs here...

$\$$\infty$\$$ -> $\infty$

I was wondering if they could post it on another mathcounts futuristic thing.

$\infty^{\infty}$ is an indeterminite form i think...

If $\lim_{x\to \infty}{f(x)}=\infty$ and $\lim_{x\to\infty}{g(x)}=\infty$ then $\lim_{x\to\infty}{f(x)^{g(x)}}=\infty.$

And this should go in Calculus Computations and Tutorials.


A "rectangular" parallelpiped has rectangles for all of its faces. The sum of the lengths of the four diagonals of a "rectangular" parallelepiped is $28$. The diagonal of one face has length $2\sqrt{10}$, while the diagonal of another face has length $3\sqrt{5}$. Find the number of cubic units in the volume of the "rectangular" parallelepiped.

Also, what is a parallelepiped?

[hide]A parallelepiped is a rectangular prism.  My solution to this problem involves some guessing.  The value 2sqrt(10) is the hypotenuse of a 2*6 right triangle.  Also, 3sqrt(5) is the hypotenuse of a 3*6 right triangle.  This yields a diagonal of sqrt(49), which is 7.  The four diagonals have a sum of 7+7+7+7, or 28.  The volume is then 2*3*6 or 36.[/hide]


[quote="kritikos"]Έχουμε ένα τυχαίο τετράπλευρο και πάνω στις δύο πλευρές του παίρνουμε τα σημεία E,F στις AD , BC αντίστοιχα τέτοια ώστε $ \frac {AE}{ED} \equal{} \frac {BF}{FC} \equal{} \frac {AE'}{E'D} \equal{} \frac {BF'}{F'C} \equal{} \frac {AB}{CD}$. Τα σημεία $ E'\in{AD}$ και $ F'\in{BC}$ . Να αποδειχθεί ότι οι ευθείες E'F' και EF είναι κάθετες .[/quote]

[b] Υπόδειξη[/b]
Για να μην πάρω τη μπουκιά από τα παιδιά , λέω μόνο μια υπόδειξη:

 [hide]Η κάθε κόκκινη ευθεία είναι παράλληλη με τη διχοτόμο που σχηματίζουν προεκτεινόμενες οι δύο απένταντι πλευρές που ''περιέχουν '' αυτήν την κόκκινη ευθεία.Αυτό είναι μια πολύ βασική πρόταση.Αρκεί να πάρετε μια διαγώνιο και σ'αυτήν ένα σημείο που την διαιρεί σε λόγο ΑΒ/ΓΔ .Από το  θεώρημα Θαλή τα δύο τμήματα που ενώνουν το σημείο αυτά με τα άλλα αντίσοιχα σημεία σημεία των δύ απένταντι πλαυρών είναι παράλληλα με τις πλευρές και σχηματίζουν ισοσκελές τρίγωνο. Αλλά στο ισοσκελές τρίγωνο η διχοτόμος της εξωτερικής γωνίας είναι παράλληλη με τη βάση. Από την άλλη μεριά η εξωτερική γωνία αυτού του ισοσκελούς τριγώνου έχει παράλληλες πλευρές με τη γωνία που σχηματίζουν οι απένταντι πλευρές του τετραπλεύρου.
 ( Η παραπάνω πρόταση βγαίνει και με εφαρμογή του Θ. Μενελάου , αλλά θέλει πιο πολύ δουλειά. Για αυτό αναγκάστηκα και έκανα άλλη απόδειξη - αυτή που περιέγραψα). Τώρα , η άσκηση μετασχηματίζεται σε άλλη επίσης γνωστή κλπ[/hide]

Νομίζς οτι με μιγαδικούς πάει περίπατο :wink:

Έστω $ M$ το συμμετρικό σημείο του $ A,$ ως προς την ευθεία $ EF$ και $ P\equiv EF\cap DM.$

Είναι φανερό ότι $ \angle APE \equal{} \angle DPE$ $ ,(1)$

Σύμφωνα με το παρακάτω Λήμμα, συμπεραίνουμε ότι είναι και $ \angle BPF \equal{} \angle CPF$ $ ,(2)$

Επειδή τώρα, το σημείο $ E'$ είναι το αρμονικό συζυγές του $ E,$ ως προς τα σημεία $ A,\ D$ και ισχύει η $ (1),$ προκύπτει ότι  $ E'P\perp EF$ $ ,(3)$

Ομοίως, επειδή το σημείο $ F'$ είναι το αρμονικό συζυγές του $ F,$ ως προς τα σημεία $ B,\ C$ και ισχύει η $ (2),$ προκύπτει ότι  $ F'P\perp EF$ $ ,(4)$

Από $ (3),$ $ (4),$ συμπεραίνουμε ότι $ E'P\equiv F'P\equiv E'F'\perp EF$ και η πρόταση έχει αποδειχτεί.

[b][size=100][color=DarkBlue]ΛΗΜΜΑ. - Δίνεται ένα τυχαίο κυρτό τετράπλευρο $ ABCD$ και έστω $ E,\ F,$ δύο σημεία επί των πλευρών $ AD,\ BC$ αντιστοίχως, τέτοια ώστε να είναι $ \frac {EA}{ED} \equal{} \frac {FB}{FC} \equal{} \frac {AB}{CD}.$ Εάν $ M$ είναι το συμμετρικό σημείο του $ A$ ως προς την ευθεία $ EF$ και $ P\equiv EF\cap DM,$ αποδείξτε ότι $ \angle BPF \equal{} \angle CPF$[/color][/size][/b].

Θα βάλλω αργότερα την απόδειξη που έχω υπόψη μου για το Λήμμα αυτό και ευχαριστώ θερμά τον [b][size=100]Ανδρέα Βαρβεράκη[/size][/b], για την όμορφη αυτή πρόταση.

Κώστας Βήττας.

[quote="vittasko"][b][size=100][color=DarkBlue]ΛΗΜΜΑ. - Δίνεται ένα τυχαίο κυρτό τετράπλευρο $ ABCD$ και έστω $ E,\ F,$ δύο σημεία επί των πλευρών $ AD,\ BC$ αντιστοίχως, τέτοια ώστε να είναι $ \frac {EA}{ED} \equal{} \frac {FB}{FC} \equal{} \frac {AB}{CD}.$ Εάν $ M$ είναι το συμμετρικό σημείο του $ A$ ως προς την ευθεία $ EF$ και $ P\equiv EF\cap DM,$ αποδείξτε ότι $ \angle BPF \equal{} \angle CPF$[/color][/size][/b].[/quote]
[b][size=100]Απόδειξη[/size][/b]. – Λαμβάνουμε το σημείο $ N,$ μεταξύ των $ P,\ C,$ έτσι ωστε να είναι $ MN\parallel CD$ και έστω τα σημεία $ K\equiv EF\cap AM$ και $ L\equiv EF\cap BN.$

Από $ \angle APE \equal{} CPE$ και $ MN\parallel CD,$ έχουμε ότι $ \frac {PN}{PC} \equal{} \frac {MN}{CD} \equal{} \frac {PM}{PD} \equal{} \frac {PA}{PD} \equal{} \frac {EA}{ED} \equal{} \frac {AB}{CD} \equal{} \frac {FB}{FC}$ $ \Longrightarrow$ $ \frac {PN}{PC} \equal{} \frac {FB}{FC}$ $ ,(1)$ και $ MN \equal{} AB$ $ ,(2)$

Η ευθεία $ PLF,$ είναι διατέμνουσα στο τρίγωνο $ \bigtriangleup BCN$ και σύμφωνα με το  [b][size=100]θεώρημα του Μενελάου [/size][/b],

ισχύει η ισότητα $ \frac {FB}{FC}\cdot \frac {PC}{PN}\cdot \frac {LN}{LB} \equal{} 1,$ από την οποία προκύπτει ότι $ LB \equal{} LN$ $ ,(3)$ λόγω της $ (1).$

Από $ (2),$ $ (3)$ και από $ LA \equal{} LM$ $ ($ γιατί η $ EF$ είναι μεσοκάθετη στο $ AM$ $ ),$ συμπεραίνουμε ότι τα τρίγωνα $ \bigtriangleup ABL$ και $ \bigtriangleup MNL$ είναι ίσα και άρα έχουμε $ \angle ALB \equal{} \angle MLN$ $ ,(4)$

Από $ (4)$ και επειδή είναι $ \angle ALK \equal{} \angle MLK,$ προκύπτει ότι $ \angle KLB \equal{} \angle KLN \equal{} 90^{o}$ και άρα έχουμε ότι $ BN\perp EF$ $ ,(5)$

Από $ (3)$ και $ (5),$ συμπεραίνουμε ότι $ \angle BPF \equal{} \angle CPF$ και η πρόταση έχει αποδειχθεί.

Κώστας Βήττας.

[quote="stergiu"][quote="vittasko"][quote="stergiu"]....Η κάθε κόκκινη ευθεία είναι παράλληλη με τη διχοτόμο που σχηματίζουν προεκτεινόμενες οι δύο απέναντι πλευρές που ''περιέχουν '' αυτήν την κόκκινη ευθεία....[/quote]
Αυτό νομίζω ότι αληθεύει μόνο όταν το δοσμένο τετράπλευρο $ ABCD$ είναι εγγράψιμο, γιατί μόνο τότε οι διχοτόμοι των γωνιών που σχηματίζουν οι απέναντι πλευρές του, είναι κάθετες μεταξύ τους.

Η πρόταση αυτή, υπάρχει ως [b][size=100]θεώρημα 131[/size][/b], στη σελίδα 302 του βιβλίου  [b][size=100]Ασκήσεις Γεωμετρίας Ιησουϊτών[/size][/b] (από τις εκδόσεις Α.ΚΑΡΑΒΙΑ - Αθήνα παλιότερα, ή σε ανατύπωση του 2007, από τις εκδόσεις ΜΑΤΙ - Κατερίνη.

Οι ευθείες $ EF,\ E'F'$ όπως ορίζει το πρόβλημα, ταυτίζονται με τις διχοτόμους των γωνιών που σχηματίζουν οι διαγώνιες του τετραπλεύρου $ ABCD$  όταν αυτό είναι εγγράψιμο και το σημείο τομής των συμπίπτει, με το σημείο τομής αυτών των διαγωνίων.

Κώστας Βήττας.

ΥΓ. Ήθελα μόνο να ξεκαθαρίσω αυτό το σημείο, αν το κατάλαβα καλά και δεν έχω λύση γι’ αυτό το όμορφο πρόβλημα.[/quote]

Κώστα , πιθανόν να μου έχει ξεφύγει κάτι. Αλλά θυμίσου και μια τελευταία άσκηση  που στο τετράπλευρο οι απέναντι πλευρές ήταν ίσες και είχαμε τα μέσα των δύο άλλων πλευρών. Πάλι κάτι παρόμοιο ίσχυε. Ο Σιλουανός είχε  βάλει λύση και σχήμα.
 Θα το ξανακοιτάξω κάποια στιγμή που θα είμαι ξεκούραστος.Οι ισουίτες δεν την έχουν λυμένη ;
 Μπάμπης
ΧΧΧΧΧΧΧΧΧΧΧΧΧΧΧΧΧΧΧΧΧΧΧΧΧΧΧΧΧΧΧΧΧΧΧΧΧΧΧχ[/quote]

 Κώστα , έχεις δίκαιο. Σε άλλο πράγμα όντως πήγε το μυαλό μου :( . Εκτός αυτού , όσα έγραψα αναφέρονταν σε άλλη άσκηση. Τώρα βλέπω ότι τα τέσσερα σημεία δεν βρίσκονται ανα δύο σε απένταντι πλευρές όπως νόμισα.Δεν διάβασα προσεκτικά την άσκηση και βασίστηκα σε μια γρήγορη ματιά στο σχήμα. Καλά να πάθω !
 Γράψε  λάθος  :)  και πάμε παρακάτω.

 Μπάμπης

[quote="vittasko"][quote="stergiu"]....Η κάθε κόκκινη ευθεία είναι παράλληλη με τη διχοτόμο που σχηματίζουν προεκτεινόμενες οι δύο απέναντι πλευρές που ''περιέχουν '' αυτήν την κόκκινη ευθεία....[/quote]
Αυτό νομίζω ότι αληθεύει μόνο όταν το δοσμένο τετράπλευρο $ ABCD$ είναι εγγράψιμο, γιατί μόνο τότε οι διχοτόμοι των γωνιών που σχηματίζουν οι απέναντι πλευρές του, είναι κάθετες μεταξύ τους.

Ως [b][size=100]θεώρημα 131[/size][/b], υπάρχει τέτοια πρόταση στη σελίδα 302 του βιβλίου  [b][size=100]Ασκήσεις Γεωμετρίας Ιησουϊτών[/size][/b], από τις εκδόσεις Α.ΚΑΡΑΒΙΑ - Αθήνα παλιότερα, ή σε ανατύπωση του 2007 από τις εκδόσεις ΜΑΤΙ - Κατερίνη.

Οι ευθείες $ EF,\ E'F'$ όπως ορίζει το πρόβλημα, ταυτίζονται με τις διχοτόμους των γωνιών που σχηματίζουν οι διαγώνιες του τετραπλεύρου $ ABCD$  όταν αυτό είναι εγγράψιμο και το σημείο τομής των συμπίπτει, με το σημείο τομής αυτών των διαγωνίων.

Κώστας Βήττας.

ΥΓ. Ήθελα μόνο να ξεκαθαρίσω αυτό το σημείο, αν το κατάλαβα καλά και δεν έχω λύση γι’ αυτό το όμορφο πρόβλημα.[/quote]

  Κώστα , αυτή την άσκηση είχα στο μυαλό μου και τη νέα δεν την κοίταξα καν  τι έλεγε  :(  :

 http://www.mathlinks.ro/viewtopic.php?t=200396

   Μπάμπης

[quote="stergiu"]....

Κώστα , αυτή την άσκηση είχα στο μυαλό μου...  http://www.mathlinks.ro/viewtopic.php?t=200396

Μπάμπης[/quote]
Μπάμπη σε ευχαριστώ, το είδα και εγώ σήμερα το πρωϊ.

Αυτό το πρόβλημα αφορά στο εγγράψιμο τετράπλευρο και αποδεικνύεται πολύ απλά όπως μας το δίνει ο [b][size=100]sunken rock[/size][/b].

Ο [b][size=100]Gabriel[/size][/b] όμως, μας παραπέμπει σε ένα άλλο πρόβλημα (παραλλαγή του δικού μας νομίζω), που αφορά σε τυχαίο τετράπλευρο και όπου ο [b][size=100]yetti[/size][/b] δίνει μία λύση βασισμένη στο [b][size=100]Θεώρημα του Μενελάου[/size][/b], η οποία νομίζω ότι μπορεί να εφαρμοστεί και σε μας εδώ.

Θα το ξανακοιτάξω πάλι και αν δεν έχω κάποιο λάθος, θα βάλω την απόδειξη με βάση το σκεπτικό του [b][size=100]yetti[/size][/b].

Κώστας Βήττας.

Υ.Γ. Ευτυχώς για μένα που δεν είχα διαβάσει τη λύση του [b][size=100]yetti[/size][/b], γιατί μπορεί να μην είχα ασχοληθεί και έχει ενδιαφέρον να μάθουμε (ίσως ο Δημήτρης ξέρει να μας πει), αν ο [b][size=100]Ανδρέας Βαρβεράκης[/size][/b], έχει υπόψη του διαφορετική (τρίτη) λύση, να τη βάλουμε κι αυτή στο φόρουμ.


Show that there exists a composite number $n$ such that $a^n \equiv a \; \pmod{n}$ for all $a \in \mathbb{Z}$.

Take $n=561= 3 \cdot 11 \cdot 17$, checking is trivial.

My proposal (not easy): prove that there is an infinite number of composite $n$ with $a^n \equiv a \mod n$ for all integers $a$.

I think the real problem here is: find a way to come up with the number 561. ;)

Also, do you have a solution for your variant? If it has a nice solution, we may add it.

Any such number must be a [url=http://mathworld.wolfram.com/CarmichaelNumber.html]Carmichael Number[/url], but I don't know if a problem Erdos couldn't resolve can be described as "not easy"  :P

We want those squarefree $ n$ such that $ p \minus{} 1 |n \minus{} 1$ for all primes $ p|n$.
We can try $ n \equal{} pqr$ a product of $ 3$ distinct primes, then we want $ p \minus{} 1|qr \minus{} 1,q \minus{} 1|rp \minus{} 1,r \minus{} 1|pq \minus{} 1$.
Choose $ p \equal{} 3$, the first property becomes trivial for odd $ q,r$ and we want $ q \minus{} 1 | 3r \minus{} 1, r \minus{} 1|3q \minus{} 1$.
Now write $ 3q \minus{} 1 \equal{} k(r \minus{} 1)$, so $ q \equal{} \frac {k(r \minus{} 1) \plus{} 1}{3}$, we are left with $ \frac {k(r \minus{} 1) \minus{} 2}{3} | 3r \minus{} 1$ or in other words $ 3| k(r \minus{} 1) \minus{} 2 | 9r \minus{} 3$. For big $ r$, this requires $ k \leq 9$. Lets try $ k\equal{}3$:
We want a prime $ r$ such that $ 3r \minus{} 5 | 9r \minus{} 3$. This requires  $ 3r\minus{}5 | 12$, simply giving $ r\equal{}17$ and we end up with $ q\equal{}11$.