analytic geometry
graphing lines
slope
rotation
function

You plot weight $ (y)$ against height $ (x)$ for three of your friends and obtain the points $ (x_1,y_1)$, $ (x_2,y_2)$, $ (x_3,y_3)$.  If $ x_1 < x_2 < x_3$ and $ x_3\minus{}x_2 \equal{} x_2\minus{}x_1$, which of the following is necessarily the slope of the line which best fits the data?  "Best fits" means that the sum of the squares of the vertical distances from the data points to the line is smaller than for any other line.

$ \textbf{(A)}\ \frac{y_3\minus{}y_1}{x_3\minus{}x_1} \qquad \textbf{(B)}\ \frac{(y_2 \minus{} y_1)\minus{}(y_3\minus{}y_2)}{x_3\minus{}x_1}$

$ \textbf{(C)}\ \frac{2y_3 \minus{} y_1 \minus{} y_2}{2x_3\minus{}x_1\minus{}x_2} \qquad \textbf{(D)}\ \frac{y_2\minus{}y_1}{x_2\minus{}x_1} \plus{} \frac{y_3\minus{}y_2}{x_3\minus{}x_2} \qquad \textbf{(E)}\ \text{none of these}$

After guessing around, it is not too difficult to prove that the answer is $ \boxed{\textbf{(A)}}$. But how would one actually find this legitimately?

[hide="Official Solution"]
The best fit line goes through some point $ (x_2,z)$.  Claim: the correct slope for this line makes the directed vertical distances to it from $ (x_1,y_1)$ and $ (x_3,y_3)$ the same.  To see this, imagine any line through $ (x_2,z)$ and rotate it until the two directed vertical distances to it are the same.  Call this distance $ d$.  For any other line through $ (x_2,z)$ the directed distances to it from $ (x_1,y_1)$ and from $ (x_3,y_3)$ will be $ d\plus{}e$ and $ d\minus{}e$ for some $ e \neq 0$. Thus, the sum of the squares will increase from $ 2d^2\plus{}(x\minus{}y_2)^2$ to $ (d\plus{}3)^2\plus{}(d\minus{}e)^2 \plus{} (z\minus{}y_2)^2\equal{}2d^2\plus{}2e^2\plus{}(z\minus{}y^2)^2$.

This proves the claim.  Finally, the slope from $ (x_1,y_1\plus{}d)$ to $ (x_3,y_3\plus{}d)$ is $ \frac{y_3\minus{}y_1}{x_3\minus{}x_1}$, whatever $ d$ and $ z$ are.
[/hide]

That's pretty much what I just said. :wink:  But still, how would you think of the claim in the first place?

Well you didn't really show what you were thinking...

[hide]Let the line of best fit be $ y \equal{} mx \plus{} b$ for some set of points $ (x_i, y_i)$

Let some other line be $ y \equal{} px \plus{} q$. Notice that $ y \equal{} (m \minus{} p)x \plus{} (b \minus{} q)$ is the line of best fit for the set of points $ (x_i, y_i \minus{} px_i \minus{} q)$.

In other words, if we subtract some amount from each of the y coordinates equal to some function of the x-coordinate at each point, then the new line of best fit is the original line of best fit minus the function we subtracted.

In this way, you can just transform the data so that $ y_1 \equal{} y_3 \equal{} 0$ and notice that the new slope should be zero by AM-QM.[/hide]


[quote="silver"]What is the absolute value of m for which the equation 4x2 - mx + 9 = 0 has equal roots?[/quote]

$4x^2-mx+9=0$?
[hide]
Find the perfect square...so that it becomes $(2x\pm 3)(2x\pm 3)$. $m=12, -12$, $|m|=12$

Or you could use the quadratic formula... [/hide]

please stop posting the same problem in two different threads.

continue discussion of this topic in

http://www.artofproblemsolving.com/Forum/viewtopic.php?t=97797


$ ABCD$ is a parallelogram and $ X$ is any point on $ BC$ produced. $ AX$ cuts $ DC$ at $ Y$. Prove $ |AYB| \equal{} |DAX|$   (where $ |..|$ represents area eg $ |ABC|$ is area of triangle $ ABC$)

Y and D lie on the parallel DC to AB ---> [ABY]=[ABD] .1

X and C lie on the parallel BC to AD ---> [ABD]=[ADX] .2

From 1 we get [ABY]=[ADX]

I don't have a paper or pencil so I proved it in my head  :D

[hide="Solution"]X is the same distance to AD as any other point on BC so [DAX]=[DAC] (since C and X are the same distance to AD)=(1/2)[ABCD].
Similarly [AYB]=[ACB]=(1/2)[ABCD][/hide]


Using the digits 1 through 9, how many different 9-digit numbers can be formed so that all two adjacent digits are consecutive numbers? Repetitions of digits are allowed.

(Example: - - 34 - - - - - , 5456 - - - - -, 2121 - - - - -)

Is it
[hide]
$9 * 2^8$ ?
Since there are 9 possible digits which could go in the first position and then from there on out there is a selection of 2 digits which will be consecutive to the first digit.  this is probably wrong though because if 0 or 9 is chosen as the next digit there is only 1 possible selection.  I'm not quite sure how to include those.
Maybe someone could help?
Thanks
[/hide]

[quote="SnowStorm"]Is it
*****
[/quote]

Your answer is too large.

Hmm... SnowStorm's solution can't be too small. It can be only large.
Do you have a solution, sick?

Sorry, I did a mistake.
The answer is 1370.


primero lo que dice el problema:

Considerar 5 puntos A, B,C ,D y E tal que ABCD es un paralelogramo y BCED es un cuadrilatero cilico. sea l una linea que pasa por A. suponer que l intersecta en el interior del segmento DC en F e intersecta la linea BC en G. Suponer tambien que EF = EG = EC. Provar que l es la bisectriz del angulo DAB.

Solucion:[hide]sea P, Q y R los pies de las perpendiculares hacia BD, FC y CG desde. como E es el circuncentro del triangulo FCG (debido a que EF = EG = EC) entonces Q y R son los puntos medios de los segmentos CG y FC, respectivamente. ademas por el teorema de la linea de simson tenemos que P, Q, R son colineales (ya que E esta en el circuncirculo del triangulo BCD), trazemos un punto T como la interseccion de la paralela por D a AG, y BG, luego AD = BC = GT (debido a los paralelogramos ABCD y AGTD) luego R es el punto medio de BT, y como QR paralela a AG (ya que son puntos medios de FC y CG) entonces PR es paralela a DT luego como R es punto medio de BT, P es punto medio BD, por lo tanto el triangulo BED es isosceles con DE = EB, por propiedades de cuadrilateros cilicos tenemos: <ECD = <EBD = <BDE = <ECE por lo tanto <CFG = <FGC entonces <BAG = <CFG = <FGC = <FAD por lo que l es la bisectriz del < BAD.[/hide]
la puse aqui por que no se hablar bien ingles y queria postear esta solucion de este problema por que fue el que mas me gusto de los problemas de esta imo.

Muy bien tony!! Mi solucion fue practicamente lo mismo, use la recta de Simson sin darme cuenta (como k probe la recta de simson), y ya visto que ED=EB use algo de trigonometria para llegar a que era bisectriz. Un problema muy bonito!
Como vas con los demas problemas de la IMO de este ano??

pues como me dijo manuel, tome la imo e hice los problemas 1,2,4,5 el que mas me gusto obviamemnte fue el 2, pero dime como van los puntajes, si me lo puedes enviar.los problemas 3 y 6 el 3 no lo entendi (ta en ingles y es combi) el 6 de plano no lo toque no me gusta, jejeje si me pudieras poner el 3 en espàñol te lo agradeceria muxo por que quiero intentarlo, desde aqui les mandan saludos todos los yucatecos!!!


let $ x, y, z$ be real number in the interval $ [\frac12;2]$ and $ a, b, c$ a permutation of them. Prove the inequality:

$ \dfrac{60a^2\minus{}1}{4xy\plus{}5z}\plus{}\dfrac{60b^2\minus{}1}{4yz\plus{}5x}\plus{}\dfrac{60c^2\minus{}1}{4zx\plus{}5y}\geq 12$

[quote="every"][quote="sandu2508"]
$ 60(x^2 \plus{} y^2 \plus{} z^2) \minus{} 3\geq 60(x \plus{} y \plus{} z) \minus{} 48$ or
$ 4(x^2 \plus{} y^2 \plus{} z^2) \plus{} 3\geq 4(x \plus{} y \plus{} z)$
[/quote]
Why do you have this,please?[/quote]

form the that fact that $ 4xy \plus{} 5z\leq5(x \plus{} y \plus{} z) \minus{} 4$ we get that $ \frac {60a^{2} \minus{} 1}{4xy \plus{} 5z}\geq\frac {60a^2 \minus{} 1}{5(x \plus{} y \plus{} z) \minus{} 4}$
Similarly we get that $ \frac {60b^{2} \minus{} 1}{4xy \plus{} 5z}\geq\frac {60b^2 \minus{} 1}{5(x \plus{} y \plus{} z) \minus{} 4}$ and $ \frac {60c^{2} \minus{} 1}{4xy \plus{} 5z}\geq\frac {60c^2 \minus{} 1}{5(x \plus{} y \plus{} z) \minus{} 4}$

So adding that relation we get: $ \frac {60a^{2} \minus{} 1}{4xy \plus{} 5z} \plus{} \frac {60b^{2} \minus{} 1}{4yz \plus{} 5x} \plus{} \frac {60c^{2} \minus{} 1}{4zx \plus{} 5y}\geq\frac {60(a^2 \plus{} b^2 \plus{} c^2) \minus{} 3}{5(x \plus{} y \plus{} z) \minus{} 4}$

Because $ a, b, c$ are $ x, y, z$ in different order we get that $ a^2 \plus{} b^2 \plus{} c^2 \equal{} x^2 \plus{} y^2 \plus{} z^2$

So it remains to prove that $ \frac {60(x^2 \plus{} y^2 \plus{} z^2) \minus{} 3}{5(x \plus{} y \plus{} z) \minus{} 4}\geq12$

Which is equivalent to $ 60(x^{2} \plus{} y^{2} \plus{} z^{2}) \minus{} 3\geq 60(x \plus{} y \plus{} z) \minus{} 48$

Hope it's more understandable now.

[quote="sandu2508"]nobody posted a solution so I'll post it.

$ (x \minus{} 2)(y \minus{} \frac12)\leq0$ which is equivalent with $ xy \minus{} 2y \minus{} \frac12x \plus{} 1\leq0$

$ (x \minus{} \frac12)(y \minus{} 2)\leq0$ which is equvalent with $ xy \minus{} 2x \minus{} \frac12y \plus{} 1\leq0$

Add last 2 inequalities:

$ 2xy\leq\frac52(x \plus{} y) \minus{} 2$

Multiply by $ 2$ and add $ 5z$ to both sides:

$ 4xy \plus{} 5z\leq5(x \plus{} y \plus{} z) \minus{} 4$

Similarly we get: $ 4yz \plus{} 5x\leq5(x \plus{} y \plus{} z) \minus{} 4$,   $ 4zx \plus{} 5y\leq5(x \plus{} y \plus{} z) \minus{} 4$. So your inequality becomes:

$ 60(x^2 \plus{} y^2 \plus{} z^2) \minus{} 3\geq 60(x \plus{} y \plus{} z) \minus{} 48$ or

$ 4(x^2 \plus{} y^2 \plus{} z^2) \plus{} 3\geq 4(x \plus{} y \plus{} z)$ now just group to form squares:

$ (2x \minus{} 1)^2 \plus{} (2y \minus{} 1)^2 \plus{} (2z \minus{} 1)^2\geq0$

It may be solved also using Cauchy's Inequality.[/quote]

Wow!
That's really nice! Thanks for sharing your nice ideas.  :)


trigonometry
function
geometry unsolved
geometry

[quote][color=darkred]Let $ w$ be the circumcircle of a given triangle $ ABC$ and let $ M\in w$ be a mobile point.

Ascertain the maximum value of the product $ p\equal{}MA\cdot MB\cdot MC$ .[/color][/quote]

$ a\leq c \leq b$   $ P\in AC$   --> $ BM*AC=CM*AB+AM*BC$ ---->

$ BM*AC=CM*AB+AM*BC\geq2\sqrt{CM*AB*AM*BC}$  --->

$ CM*AM\leq\frac{(BM*AC)^{2}}{4AB*BC}=\frac{(b*BM)^{2}}{4ac}$ ---->

$ P=MB\cdot MA\cdot MC%Error. "leqMB" is a bad command.
\cdot\frac{(b*BM)^{2}}{4ac}=\frac{b^{2}BM^{3}}{4ac}\leq\frac{b^{2}(2R)^{3}}{4ac}=\frac{2b^{2}R^{3}}{ac}$

$ P_{max}=\frac{2b^{2}R^{3}}{ac}$

$ BM=2R$

$ CM=\frac{bR}{c}$

$ AM=\frac{bR}{a}$

If I am not mistaken dear [b][size=100]zaya_yc[/size][/b], you mean that the antidiametric (= diametric opposite) point of $ B,$ has the property as the problem states.

But I think this is not correct. We can see in my drawing, where $ BC \equal{} \alpha \equal{} 1,90 < AB \equal{} c \equal{} 7,36 < AC \equal{} b \equal{} 7,48$, that the midpoint $ M$ of the arc $ AC,$ not containing the vertex $ B$ and another point $ P,$ between $ A$ and $ M,$ have equal products as the problem states and bigger than the one with respect to the point $ Q,$ where $ BQ \equal{} 2R.$

That is we have $ MA\cdot MB\cdot MC \equal{} PA\cdot PB\cdot PC > QA\cdot QB\cdot QC$ $ ,(1)$

I conjecture that the midpoint $ N$ of the arc $ PM,$ has the property of the maximum product as the problem states.

That is, I think $ p_{max} \equal{} NA\cdot NB\cdot NC,$ but I have not in mind any proof about it.

Kostas Vittas.

PS. I don’t know the way to determine the point $ P,$ which is the key of the solution I thought, when the triangle $ \bigtriangleup ABC$ is given.
I made the configuration below, considered as fixed the point $ P,$ on a circle $ (O)$ taken as cord the given segment $ AC,$ and the point $ B,$ is determined based on the problem [url=http://www.mathlinks.ro/Forum/viewtopic.php?t=191359]Point’s determination 6[/url].

$ 2[\triangle MAB] = MA \cdot MB \cdot \sin C$

$ 2[\triangle MBC] = MB \cdot MC \cdot \sin A$

$ 2[\triangle MAC] = MA \cdot MC \cdot \sin B$

$\Longrightarrow 8[\triangle MAB][\triangle MBC][\triangle MAC] = (MA \cdot MB \cdot MC)^2 \cdot  \sin A \cdot \sin B \cdot \sin C$

Let $(x:y:z)$ be the barycentrics of $M$ WRT $\triangle ABC.$ Dividing the latter equation into $ [\triangle ABC]^3$ yields

$ \Longrightarrow 8xyz = \frac {(MA \cdot MB \cdot MC)^2 \cdot \sin A \cdot \sin B \cdot \sin C}{[\triangle ABC]^3}$

Setting $\frac {1}{k} = \frac {\sin A \cdot \sin B \cdot \sin C}{8[\triangle ABC]^3}=\text{const}$

${ \Longrightarrow f(x,y,z) = \sqrt {kxyz} \ \ (1) \ \ , \ \ x + y + z = 1} \ \ (2)$

Since $ M \in \omega \Longrightarrow \ \frac {a^2}{x} + \frac {b^2}{y} + \frac {c^2}{z} = 0 \ \ (3)$

Now, we might use Lagrange multipliers on the fuction $f(x,y,z)$ subject to constraints $(2)$ and $(3),$ but the computations seem pretty heavy, so I will leave my proof at this point.


limit
trigonometry
calculus
calculus computations

hi!
calculate this limit:

$ \lim_{x\to0}\frac{1\minus{}\prod_{k\equal{}1}^n\cos{kx}}{x^{2}}$

thanx.

$ \prod_{k\equal{}1}^n\cos(kx)\equal{}\prod_{k\equal{}1}^n\left(1\minus{}\frac{k^2x^2}2\plus{}O(x^4)\right)$

$ \equal{}1\minus{}\sum_{k\equal{}1}^n\frac{k^2x^2}2\plus{}O(x^4)$

$ \equal{}1\minus{}\frac{n(n\plus{}1)(2n\plus{}1)}{12}x^2\plus{}O(x^4)$

Subtract that from $ 1,$ divide by $ x^2,$ and take the limit to get that the answer is $ \frac{n(n\plus{}1)(2n\plus{}1)}{12}.$

another method

$ \lim_{x\to0}\frac{1\minus{}\prod_{k\equal{}1}^{n}\cos{kx}}{x^{2}}
 \equal{} \lim_{x\to0}\frac{\sum_{k\equal{}1}^{n}{k}\sin(kx)\prod_{m\equal{}1\\m\neq{k}}^{n}\cos(mx)}{2x}
 \equal{}\frac{\sum_{k\equal{}1}^{n}k^2}{2}
 \equal{}\frac{n(n\plus{}1)(2n\plus{}1)}{12}.$

hi [b]Jolie[/b] :)
can you explain how you get : $ \lim_{x\to0}\frac{1\minus{}\prod_{k\equal{}1}^{n}\cos{kx}}{x^{2}}\equal{}\lim_{x\to0}\frac{\sum_{k\equal{}1}^{n}{k}\sin(kx)\prod_{m\equal{}1\\m\neq{k}}^{n}\cos(mx)}{2x}$ ?
thanx.

another method:
we have: $ 1\minus{}\prod_{k\equal{}1}^{n}\cos{kx}$=$ \sum_{k\equal{}1}^{n}(1\minus{}\cos{kx})\prod_{k\equal{}1}^{n\minus{}1}\cos{(k\plus{}1)x}$

so we get immediately that: $ \lim_{x\to0}\frac{1\minus{}\prod_{k\equal{}1}^{n}\cos{kx}}{x^{2}}$=$ \frac{\sum_{k\equal{}1}^{n}k^{2}}{2}$=$ \frac{n(n\plus{}1)(2n\plus{}1)}{12}$.

[quote="leshik"]I'm almost sure that this is Hospital rule in Jolie solution  :wink:[/quote]Yes, it is, two applications actually.


ratio
number theory unsolved
number theory

Prove that for k>0,m>n>3 (all integers), F_m/F_n=2^k has no solutions.

(Where F_0=0, F_1=1 and F_(i+2)=F_(i+1)+F_i.)

I am just going to rewrite it becuase it seems very interesting:

Let $F$ be the fibbonacci sequence, prove there doesnt exist $m,n>3$ such that $f_m/f_n$ is a power of $2$.

It may be related to a problem posted by Harazi...however I dont have a solid solution for this.

well the first thing is that $n|m$, next step is to prove $m=n2^t$. suppose $m=npr$ with $p$ an odd prime, then we have that:

$(1+\sqrt{5})^m-(1-\sqrt{5})^m=((1+\sqrt{5})^{nr}-(1-\sqrt{5})^{nr})A\sqrt{5}$ where $A$ is odd, since $(1+\sqrt{5})^n-(1-\sqrt{5})^n=2^n\sqrt{5}k$ (*).

so we get an odd number would divide $f_m/f_n$, contradiction.

Now we must have that $f_{2m}/f_{m}$ must be a power of $2$, so we must have that $2^t=(1+\sqrt{5})^n+(1-\sqrt{5})^n=4k+2$ using (*) so we get $(1+\sqrt{5})^n+(1-\sqrt{5})^n=2$ and from here we get all good values or possible values for $n$. and easily see there is no solution with $m,n>3$

your solution seems to be wrong... (forgive me if *i'm* wrong...)

for instance, your last claim... $ (1+\sqrt 5)^n + (1-\sqrt 5)^n = 2 $ you say gives you all good values of n.  but what about n = 3?  (for the case $ 2^2 F_3 = F_6 $.)  n = 3 doesn't satisfy that relation...

aside from that...

earlier in the proof you said

"$ (1+\sqrt{5})^m-(1-\sqrt{5})^m=((1+\sqrt{5})^{nr}-(1-\sqrt{5})^{nr})A\sqrt{5} $ where $ A $ is odd"

this can't be true because the LHS is equal to $ \sqrt 5 $ times an integer, while the RHS is an integer

Fact: For $k\ge3$, $2^k|F_n \iff 3\cdot2^{k-2}|n.$
For $2^k$ to divide $\frac{F_n}{F_m}$, we must have that either $2^{n-1}|\frac nm$ or $3|\frac nm$. Thus $n-m\ge (3-1)\cdot2^{k-2}=2^{k-1}$ and $\frac{F_n}{F_m}\ge\left(\frac{\sqrt{5}+1}{2}\right)^{2^{k-1}}.$ For $k>3$, this quotient is too big, and we are done.
Similar reasoning shows that for any prime $p$, there are finitely many powers $p^k$ which can be the ratio of Fibonacci numbers. This finite collection is almost always contained in $\{1,p\}$, with exceptions beyond 2 possible only if every term of the Fibonacci sequence divisible by $p$ is divisible by a higher power of $p$.

To correct my previous post: we have that $f_{2n}/f_n=L_n$ where $L_n$ is the nth Lucas number. So we must have $L_n$ is a power of $2$, seing this sequence modulo 8, we see none of its terms is a multiple of 8, so we get
 $L_n=1$, $L_n=2$, $L_n=4$,  from here we get $n=1$, $n=3$, the rest is easy!


[quote="DangChienbn"]I am sorry if I have mistakes in English. This is an inequality on serec in inequalities :D
Let a, b, c>0. Prove that
$ \frac {a^4 \plus{} b^4 \plus{} c^4}{ab \plus{} bc \plus{} ca} \plus{} \frac {3abc}{a \plus{} b \plus{} c} \geq \frac {2}{3(a^2 \plus{} b^2 \plus{} c^2)}$
By pqr is very easy but AMGM ?[/quote]
this ineq. is wrong, you can try with $ a\equal{}b\equal{}c\equal{}1$ :)
p/s: Chiến ku qua đây àh  :lol:

Mr.Bikrone, when a=b=c=1, the inequality still [size=150]true[/size]
The inequality becomes:
$ \frac{1^{4}\plus{}1^{4}\plus{}1^{4}}{1\plus{}1\plus{}1}\plus{}\frac{3}{1\plus{}1\plus{}1}\geq\frac{2}{3(1^{2}\plus{}1^{2}\plus{}1^{2})}$
$ \Leftrightarrow 2\geq\frac{2}{9}$
I think it's true.

[quote="DangChienbn"]I am sorry if I have mistakes in English. This is an inequality on serec in inequalities :D
Let a, b, c>0. Prove that
$ \frac {a^4 \plus{} b^4 \plus{} c^4}{ab \plus{} bc \plus{} ca} \plus{} \frac {3abc}{a \plus{} b \plus{} c} \geq \frac {2}{3(a^2 \plus{} b^2 \plus{} c^2)}$
By pqr is very easy but AMGM ?[/quote]
try $ a\equal{}b\equal{}c\equal{}\frac{1}{2}$

Oh, you're right. This inequality is wrong.
Hic!
[hide] :( No more discussion  :mad: [/hide]

This one is the correct one :lol:

$ \frac {a^4 \plus{} b^4 \plus{} c^4}{ab \plus{} bc \plus{} ca} \plus{} \frac {3abc}{a \plus{} b \plus{} c} \geq \frac {2(a^2 \plus{} b^2 \plus{} c^2)}{3}$

Equality if and only if $ a \equal{} b \equal{} c$ :)

We can also prove the stronger:

$ 5\frac {a^4 \plus{} b^4 \plus{} c^4}{ab \plus{} bc \plus{} ca} \plus{} 9\frac {3abc}{a \plus{} b \plus{} c} \geq 14\frac {(a^2 \plus{} b^2 \plus{} c^2)}{3}$

But the slightly stronger is wrong:

$ 5\frac {a^4 \plus{} b^4 \plus{} c^4}{ab \plus{} bc \plus{} ca} \plus{} 10\frac {3abc}{a \plus{} b \plus{} c} \geq 15\frac {(a^2 \plus{} b^2 \plus{} c^2)}{3}$


vector
linear algebra
matrix
algebra
system of equations

Show that there do not exist scalar $ c_1, c_2, c_3$ such that $ c_1(1, 2, \minus{}3) \plus{} c_2(5, 7 ,1) \plus{} c_3(6, 9, \minus{}2) \equal{} (4, 5, 0)$

That is equivalent to say that the corresponding system of equations does not have a solution, and that means the determinant of (1,2,-3), (5,7,1), and (6,9,-2) is zero, which indeed is.

Computationally: don't use determinants. Use Gaussian elimination to attempt to solve these equations. If they're inconsistent, that will be be made clear in the calculation.

[quote="darij grinberg"][quote="Carcul"]that means the determinant of (1,2,-3), (5,7,1), and (6,9,-2) is zero, which indeed is.[/quote]

It is,  but that is not enough for the system to have no solution.

  darij[/quote]

It is. By Cramer x_i = det(A_i)/det(A).

[quote="andib"]It is. By Cramer x_i = det(A_i)/det(A).[/quote]
That is the wrong conclusion to draw.

Consider these two systems:

$ \begin{pmatrix}1 & 0 & 0 \\
0 & 1 & 1 \\
0 & 1 & 1\end{pmatrix} \begin{pmatrix}x \\
y \\
z\end{pmatrix} \equal{} \begin{pmatrix}1 \\
2 \\
3\end{pmatrix}$

and

$ \begin{pmatrix}1 & 0 & 0 \\
0 & 1 & 1 \\
0 & 1 & 1\end{pmatrix} \begin{pmatrix}x \\
y \\
z\end{pmatrix} \equal{} \begin{pmatrix}1 \\
2 \\
2\end{pmatrix}$

The same $ 3\times 3$ matrix appears in both problems, and it is a matrix with determinant zero. The first system has no solutions; it is inconsistent. And yet, the second system does have solutions - infinitely many solutions, as a matter of fact - which Cramer's rule is completely unable to find.

What darij was pointing out is that your statement that the coefficient matrix has determinant zero does not distinguish between the two cases I have just presented.

Please do not accept Cramer's rule as either the best method or the last word in finding solutions of linear systems. Gaussian elimination is a robust workhorse of a method, and perfectly able to cope both both of the systems I just presented.

Normally, three vectors in space will generate the whole space. So saying that one vector cannot be generated by three given ones had better mean the three vectors are linearly dependent. So we should be able to "make" one from the other two.

Let's show that $ (6,9,\minus{}2)$ is indeed a linear combination of $ (1,2,\minus{}3)$ and $ (5,7,1)$. Solving the system $ p\plus{}5q\equal{}6,2p\plus{}7q\equal{}9$ to equate $ x,y$ components gives $ p\equal{}1,q\equal{}1$. Of course, you might have also noticed that the third is the sum of the first two.

Now we just need to show that $ (4,5,0)$ cannot be generated by $ (1,2,\minus{}3)$ and $ (5,7,1)$. Again trying to equate $ x,y$ components shows that the only way to do so is to take $ (5,7,1)\minus{}(1,2,\minus{}3)$. But this doesn't give 0 for the $ z$ component. So $ (4,5,0)$ indeed can't be generated.

In the first part of scorpius119's proof (the middle paragraph of his post) he shows that the three vectors in question form a linearly dependent set. This can be done in dozens of different but equivalent ways. One of those equivalent ways would be to show that the $ 3\times 3$ matrix formed by these vectors has determinant zero. Hence at the end of that paragraph we are exactly where andib was when he noted that that determinant was zero.

But that's still not enough. scorpius119 could not have stopped there; his last paragraph was required, which is what Darij and I have been saying.


trigonometry
quadratics
algebra
trig identities
Law of Cosines

All agles A, B, C, D and diagonals e=AC and f=BD are given. Draw the quadrilateral.

I know there exists an ugly solution using law of cosines leading to graphical solution of quadratic equation, but I'm looking for a nice synthetic solution.
For inscribed quadrilaterals it has obviously infinitely many solutions, so suppose $A+C\not=\pi$.
Thanks for any ideas.

i have been seen a solution with homotechy  :!:

[quote="Arne"]Why don't you post it then? :)[/quote]

hmmm....

ok dear Arne...

i dont know, why i dont have time .

as soon as possible i will do it ;) .

it seems to me that all you have to do is draw a quadrilateral with those angles, and create another similar one with the correct diagonal length, also i think that the 2nd side length is unnecessary

I'm afraid not all quadrilaterals with the same angles are similar...
Altheman, your solution doesn't work.
I'm really curious how to solve this...


({{1}. {2}, {4}, {1,2}, {1,4}, {3,4}, {1,3,4}, {2,3,4}}, ).
For the given poset

a. Find all maximal elements.
b. Find all minimal elements.
c. Is there a greatest element?
d. Is there a least element?
e. Draw the Hasse diagram.

What's the ordering?  (It is probably inclusion, but one has to be sure.)

You seem to be spending a lot of time posting your homework from a discrete math class of some kind.  Instead of asking us to do your homework for you, please attempt the problems first and tell us your progress and where you've gotten stuck; you'll find that people are much more willing to help if you've at least put genuine effort in.


Does anyone know any cheap hosting that supports $\LaTeX$? Just wondering, could be handy to make a list of them or so...

I don't think there are any that offer such a thing. Only on request basis, but I think that most will not do it for a cheap price, because it may involve security holes.

To put it simple, using latex and php, a hacker (or you, the one who's renting the place) can do a lot of damage to the server (which usually hosts other sites as well).

You might be lucky with having a dedicated server, as you would be only "risking" your site. But that, as you might know, is pretty expensive.


inequalities
calculus
integration
real analysis
real analysis unsolved

Suppose $ f$ is continuous on $ \mathbb{R}$, $ f(x)>0$ for $ 0<x<1$ and $ f(x)=0$ otherwise. Let $ h_c(x)=\sup\{n^cf(nx) : n=1,2,3,\ldots\}$.

b) Prove that $ h_1$ is in weak $ L^1$.

So I call $ h=h_1$, and $ g(\lambda)=\lambda m\{x: |h(x)|>\lambda\}$. If $ g$ is bounded, then we have weak $ L^1$.

Fix $ n\in\mathbb{N}$. Then

\begin{eqnarray*}
h(\lambda) & = & \lambda m\{x : |h(x)|>\lambda\} \\
 & \leq & \lambda m\{x : nf(nx) > \lambda\} \\
 & \leq & \frac{\lambda}{n}\\
\end{eqnarray*}

Then right as I was about to get excited since I have no experience working with weak $ L^1$, I realized that this doesn't prove it, since the next part is to show that for $ c>1$ it is not in weak $ L^1$ and if this proof showed anything, then that would work the same way.

Woops. I found the fix.

That last inequality can be taken a step further to be relevant. We know that $ n>\frac{\lambda}{M}$ where $ M\equal{}\max f(x)$, meaning that $ \frac{\lambda}{n}\leq M$. This no longer poses problems, since if $ c>1$, then we only get $ \leq M^{1/c}\lambda^{1\minus{}1/c}$ which is not bounded. Of course this doesn't show that it is unbounded, but it gets rid of my previous concern.

but showing that it's unbounded isn't that hard once you see what's going on.  say that $ \|f\|_{\infty}$ is attained at $ \ell\in (0,1)$ and fix $ 0 < \varepsilon < \|f\|_{\infty}$, and pick $ N$ large enough that $ |f| > \varepsilon$ on $ (\ell - \ell/N,\ell)$.  then on $ (0,\ell/N)$, $ h_c$ is at least $ (\|f\|_{\infty} - \varepsilon)N^c$.

taking $ \lambda_n\triangleq (\|f\|_{\infty} - \varepsilon)n^c$, for every $ n > N$ we have
\begin{eqnarray*} \lambda_n m(\{x: |h_c(x)| > \lambda_n\}) & = & (\|f\|_{\infty} - \varepsilon)n^cm(\{x: |h_c(x)| > \lambda_n\}) \\
& \geq & (\|f\|_{\infty} - \varepsilon)n^c\left(\frac {\ell}{n}\right) \\
& = & (\|f\|_{\infty} - \varepsilon)\ell n^{c - 1} \end{eqnarray*}
but this is not bounded as $ n\to\infty$, so $ h_c$ is not in weak $ L^1$.

as an aside, $ h_1$ is indeed in weak $ L^1$, but it is not in $ L^1$.  an easy way to see this is that the integral of $ |h_1|$ over $ X_n\triangleq(0,\ell/n)$ for any $ n > N$ (where we are using the same notation as above) is at least $ (\|f\|_{\infty} - \varepsilon)\ell$.  but for every $ f\in L^1$ and $ t > 0$, there exists $ \delta_{f,t} > 0$ such that $ m(X) < \delta_{f,t}$ implies $ \int_X |f|dm < t$.  letting $ n\to\infty$ in the above leads to a contradiction if we assume that $ h_1\in L^1$.

actually upon rereading your posts i don't necessarily see where you are getting your estimates from in the first post.  for example, if $ n = 2$, then we can have $ g(\lambda)$ arbitrarily close to $ \lambda$ by choosing $ \lambda$ appropriately.

this (i think) works, on the other hand.  we may assume that $ \lambda \geq \|f\|_{\infty}$.  then there exists an integer  $ k = k(\lambda)\geq 2$ such that $ (k - 1)\|f\|_{\infty}\leq\lambda < k\|f\|_{\infty}$.  if the supremum of $ f$ is achieved at $ \ell\in (0,1)$, then $ |h_1(x)| > \lambda$ implies $ 0 < x < 1/(k - 1)$.

then we have
\begin{eqnarray*} \lambda m(\{x: |h_1(x)| > \lambda\}) & < & k\|f\|_{\infty} m(\{x: |h_1(x)| > \lambda\}) \\
& \leq & k\|f\|_{\infty} \left(\frac {1}{k - 1}\right) \\
& = & \frac {k}{k - 1} \|f\|_{\infty} \\
& \leq & 2\|f\|_{\infty} \end{eqnarray*}


For every positive integer $ n$, let $ f(n)$ denote the sum of the decimal digits of $ n$. Prove that there are infinitely many positive integers $ k$ such that $ f(3^k)\geq f(3^{k\plus{}1})$.

Obviosly? if only finite number k satisfyed $ f(3^k)\ge f(3^{k\plus{}1})$, then $ f(3^k)>9(k\minus{}m)$ for big k. Obviosly $ f(3^k)<9\plus{}9klog_{10}3<9(k\minus{}m)$ - contradition.


Which next generation console are you anxious about. If you bought the 360, and plan to stick with only a 360, then vote for that. I left out "unsure" because I know tons of people would select that, thus making the poll less interesting. If your unsure, just post it here.

Man, PC > all consoles.

PC has games that are actually competitive. Much easier to find other players and find leagues. Much larger multiplayer base.

I basically oppose the idea of different consoles and always new generations.
For instance i have a couple of games for super nintendo , nintendo 64 and ps2

You always have to get a new console, either because there is a new generation or because the game is distributed by competing companies.

What irks me even more is that for instance a game for super nintendo cannot be played on a nintendo 64.

Yeah I do not fancy sim games or realistic shoot'em ups, I like to play Super Mario 64 etc

i agree, PC is not much more expensive and has about infinity times more capabilities.

I won't be getting any  :D

First of all, my parents wouldn't let me get any, and I don't really want one that badly.  I think that they are a rip-off compared to a nice TI-84+ SE and a decent computer.  You can program both the calculator, and the PC, so they are way better already.  Besides, PCs have better resolution, are faster, more memory (don't waste money on save cartridges).

But most importantly, computers have a mouse, the ultimate input tool, which is not something that comes on game counsels.

I think if I get one it will be in the distant future. Too much money when they come out! In my eyes, the Sony is making a mistake by putting too much trust in high-def and blue-ray. It adds more to the cost than buyers are willing to cough up. I'm not sure about Wii either. We'll have to see how it holds up. The 360 seems Ok, but I've only played it a few times.

they're all hugely expensive. the games are too. :(

I voted for the Xbox 360. I know that the PC has more capabilities and whatnot, so theoretically there is no reason to invest in consoles. I personally prefer consoles over the PC for several reasons. First off, i SUCK at using the keyboard and mouse to play games. I just can't get coordinated with it, and i hate it. Secondly, my computer SUCKS and it would cost a fortune to make it a good gaming rig. That, to me, isn't worth it because i don't care enough about computer games to spend that kind of money. I only use the computer for internet and AIM pretty much. Thirdly, ALL computer games require a good internet connection whereas on consoles, only the online games require a good internet connection (well not ALL computer games, but pretty much any PC game that is worth buying these days requires online capabilities to get the full enjoyment out of the game). Even the online games for consoles usually have a single player mode or something that does not require the user to play online. So with PC's, lag is much more of an issue than with consoles.

I prefer Xbox360 to the other consoles because i like shooters, and Xbox has the widest variety of shooters. It also has Halo: Combat Evolved <3 (Halo 2 sucks). Yes, i know it is on the PC as well, but i hate PC gaming, as explained above. The playstation consoles are more suited for people who prefer RPG's, and the gamecube's games just flat out suck. It took me 2 years to realize that before i got the Xbox.

[quote="Totally Zealous"]The video game industry is going to collapse one day. Look, graphics and gameplay are getting better. They're getting more expensive. Back then a super nintendo mario bros was only $\$20$ compared to the $\ge$$\$60$ prices today. We will reach a day where video games will become so expenisive no one will be able to buy them. And graphics and gameplay are getting better, but one day they will reach the limit of real life level graphics and real life level gameplay. Then the graphics can't get any better. Also, real life level gameplay = real life = not staring at a screen. So drop that controller and go outside and run around like idiots. The best video game ever is playing your own life. Video games are improving fast but they can't improve any further when they reach a certain point (real life). Any more improvements are just paradoxing itself.

I also believe people who play video games all day are those stupid ones that ruined their lives (by playing video games) and are afraid to correct it so they sit at a screen and pretend they're in a magical land where they have the perfect life.

and NO I don't play much video games, just some gameboy on long car trips where there is nothing else to do.

Did that make any sense because it now sounds confusing to me (how i wrote it)?[/quote]

Congratulations, you just described inflation.

[quote="Totally Zealous"]The video game industry is going to collapse one day. Look, graphics and gameplay are getting better. They're getting more expensive. Back then a super nintendo mario bros was only $\$20$ compared to the $\ge$$\$60$ prices today. We will reach a day where video games will become so expenisive no one will be able to buy them.[/quote]

No, SNES games never did cost 20 dollars, new.  Most SNES games were at least 50 dollars; the SNES game Final Fantasy III (which, oddly enough, is the sixth game in the series; long story) originally cost 80.

[quote="dts"][quote="Ignite168"]I highly doubt there will be a good SSM for the Wii because of its controller. While it may be good for games like Madden where you can pretend to throw the Wii's controller, how would you do games like SSM? While Nintendo's idea looks good on paper, I don't think it will be practical.[/quote]

Actually, the Wii version of SSB won't use the regular controller.  Nintendo's making a normal, dual-analog controller for older games and games also released on other systems, and SSBB is going to use that one.

(also, they're letting non-Nintendo characters in this time...)[/quote]

Well that seems smart, I thought the Wii was just going to become an over glorified Eyetoy like object.

b-flat is one clueless little kid, he likes computers for all the wrong reasons. IF YOU GET MY DRIFT.

[quote="Ignite168"][quote="Totally Zealous"]text[/quote]

Congratulations, you just described inflation.[/quote]

this post actually deserves a smilie: :rotfl:

[quote="eryaman"]I probably won't get any, possibly the Wii, if the reviews are good.  The XBox 360 and PS3 seem basically like the older versions, maybe with a little better internet connectivity, better graphics, etc.

If Wii can get the controller right, I might buy it.  After all, it will most likely be much cheaper than the other consoles, plus it will have Super Smash Bros... :D[/quote]
I would buy wii just for the ssb brawl, and probably mario party 8 :)

computer graphics are better? i don't know about that.  the best pc graphics cards costs millions of dollars, and ain't much or at all better than a tv and bluray.

Wii is definitely doing great, bunch of my friends have it (more than when the 360 was released).  Haven't heard of anyone with a PS3 yet.

But i can assure the 600 tag for ps3 is cheap, and people will shell out money.  For people in Asian, 600 is nothing.


I noticed a larger percentage of left handed people in math areas than in the general human population (almost 10%) .  Let's see if it is the case in this forum.  Can anyone explain that?

I think if anyone knew there would be huge articles and what not on it.  I am a left handed person though, but I only use my left hand for writing and eating (sometimes).  SO I guess I'm sorta ambidextrous in the fact that I do all sports with my right side.

[quote="BHorseMath"]Theres a theory that all left handed people are the result of a twin that never happened.[/quote]

How does that work, BHorseMath? It sounds like the right hand of the twin that didn't form takes the place of the left hand of the twin that did... or something like that? I thought development of organs happens after the point where there clearly are or aren't twins.

I read somewhere that right/left-handedness is caused by the direction embrionic fluid washes across the embryo.

By the way, I'm right-handed, and trying to become ambidextrous through practice

[quote="tetrahedr0n"] I thought development of organs happens after the point where there clearly are or aren't twins.[/quote]

Twinning generally happens within about 3 weeks of conception.  So, yes, well before organs form.

Off topic - I like your avatar Danbert :)

you'll never write as well when forcing yourself to use your opposite hand. why would you want to be ambidextrious if you are already right-handed? the world is MADE for right-handed people! lefties have to have special desks, special scissors... have you ever tried writing in marker with your left hand? you smear all over the entire page and cramp your hand trying to keep out of the ink. when i made posters, i always had to write out the message in pencil and then write over it, backwards, in marker- and i would STILL smear!

[quote="bleumoose"]appr. 20% of the population is naturally left-handed.[/quote]Seems to be the case here too... btw I am left handed also.  :D

ACE!

hm, question for the lefties- did you experience similar problems with poster-making, etc? did anyone have the experience of being asked by a teacher to change writing hands, or are we past all that now?
and for the righties- ever wanna be left-handed?
and for the ambis by practice- WHY?

[quote="bleumoose"]hm, question for the lefties- did you experience similar problems with poster-making, etc?[/quote]what do you mean "poster-making" ?

[quote="bleumoose"]hm, question for the lefties- did anyone have the experience of being asked by a teacher to change writing hands, or are we past all that now?[/quote]
Yes I had this. Back in the communist dictatorial times in Romania, students were beaten to write with the right hands. In these days, this treatment is not allowed anymore (thank God!).

[quote="bleumoose"]and for the ambis by practice- WHY?[/quote]I can write with my right hand (but very very ugly), but I assume the main reason is
"just in case" (something happens with your other hand just before a math competition ;) ).

[quote="bleumoose"]and for the ambis by practice- WHY?[/quote]

Being ambi helps you do two things at once, a very useful skill.

i meant, when you make signs, or something that requires to write with either a marker or a dark pen. i always smear the entire page because my hand rests on what i have just written (which isnt quite dry).

ah, yes, the ambi thing "just in case" makes sense now (i am only slightly ambi, because of breaking a finger on my left hand and being forced to use my right- it was not a pleasant time, but luckily WAS after all math competitions had been held for the year  ;) ).

[quote="bleumoose"]i meant, when you make signs, or something that requires to write with either a marker or a dark pen. i always smear the entire page because my hand rests on what i have just written (which isnt quite dry).
[/quote]I also had this problem, untill I discovered how to  write without smearing ... ;) (with my left hand of course :P).

The ambi thing can be a lot just for fun.  Some sports make right easier (stupid baseball), and I've always felt more comfortable playing hockey w/a right-handed stick even though I'm a lefty--and had a left-handed stick.
I've also noticed that while writing with the opposite hand is not as precise, it works pretty well with madly-fast scribbling and especially writing upside-down or sideways (eg, on someone else's paper).
Sometimes, just for a kick when I'm trying to get through mc math tests in a third of the given time or something, it helps to hold a second pencil to mark the answers, etc.  But I guess there's no real practical application there; just like learning a second language for the fun of being able to communicate 2 ways/push yourself  ;)

[quote="fedwinri"]But I guess there's no real practical application there; just like learning a second language for the fun of being able to communicate 2 ways/push yourself  ;)[/quote]I don't agree with these statements. Partially because there is a real application in being ambi (in case of your writing hand breaking you can still write - do your job, and drive!) and partially because knowing a second language is not just for fun / push yourself, it's a key element in understanding, dealing and working with people which don't speak your language. Of course being a native English rules out 1/6 of the globe, but still around 4 billion people remain, that don't know English :) So learning a second language is a must in our times, but this belongs perhaps to another topic.

I'm right handed but for some reason, I like to write with my left

[color=green]I'm naturally ambidextrous, but forced to use my left hand b/c of an injury.

I think we're past that in America right now, as well as probably most other countries, but there still is some minour things going on. The hardest thing to do is left-handed calligraphy![/color]

[quote]Maybe because the right brain controls the left side, and the right brain is used for maths?[/quote]

WHOA!!!! you got that mixed up man!!!

left brain is used for math or literature or writings.. like that

but right brain is used for art or music

okay..

I'm ambidextrous by practice because I used to do mind-map and one of my mind-map tutor said it helps you work both brains at the same time, somehow....

so I'm ambidextrous by practice

[quote]Did you hear about that psychology study about Left-handers living like 20 years less than Right-handers? I think they said one possibility was because a lot of dangerous items like chainsaws are designed for right-handers, so left-handers may be more prone to accidents. Then after a lot of controversy another study came out with normal results. But yeah, that's crazy stuff. Anyone look into this one? I read it in my psych book.[/quote]

I don't think it's psychologically correct ;)

though...
many smart peeps are left handed

Sir Isac Newtans

and

Einstein were left handed

there are more famous people who are lefthanded but I can't remember all of them

Hi,

[quote="peter"]Maybe because the right brain controls the left side, and the right brain is used for maths?[/quote]

As far as I know, that is taken as an old fashion theory. I think I read somewher that most of the neurologist and psicologist think that this is not necessarly true (left part of the body controlled by the right side of the brain and viseversa). I tend to think that there should be around a 25% of people should be naturally left-handed, because it is a resecive gen.

[quote="JBL"]Twinning generally happens within about 3 weeks of conception. So, yes, well before organs form.[/quote]

I saw a program in Discovery Health may be a month ago, that said, according to the researh of some doctors, around of 15% of us were in some moment one of a twin pair, but the other fetus was reabsorved for some reason. Does anybody knows what about the percentage of deliveries that are multiples.

Ahhh...!! I am righthanded (unfortunatelly not ambidextrus, neither by practice or naturally) but given that my father is left handed, ambidextrus by practice, there is a chance that a son of mine would be left handed.

Best regards,

[quote="shinwoo"]there are more famous people who are lefthanded but I can't remember all of them[/quote]How about Bill Clinton?

ive been trying to be ambidextrous by practice for three months now (largely unsuccessful), but since i cant even write "properly" (as in, the way most people write) and i have my own writing style, it is incredibly difficult to try left handed

[quote="bleumoose"][quote="djimenez"]I tend to think that there should be around a 25% of people should be naturally left-handed, because it is a resecive gen.[/quote]

incorrect assumption. many recessive genes (i cannot name any specific ones right now, but the ones i am thinking of have to do with looks- like if your earlobes are attached to your head... and if you can roll your tongue or not) are actually dominant IN THE POPULATION (that is, there are many more recessive alleles than dominant alleles, so there are more people exhibiting the recessive trait). just thought i'd point that out.[/quote]

For example, Huntington's disease is dominant, but it is [b]very[/b] rare.


function
floor function
algebra
system of equations

CAn you hepl me sovle this system equation ?

$ \{\begin{array}{l}  x + [y] + {z} = 3,9  \\  y + [z] + {x} = 3,5  \\  z + [x] + {y} = 2 \end{array}$
với {A} = A - [A]

Is it easy for you ? So can you prove this quickly ?  :)

first define numbers $ x \equal{} \overline{a.b}$ $ y \equal{} \overline{c.d}$ and $ z \equal{} \overline{e.f}$
from three exations we get
$ f \plus{} d \equal{} \overline{k.0}$
$ d \plus{} b \equal{} \overline{l.5}$
$ b \plus{} f \equal{} \overline{m.9}$

When we solve this easy system of equations we get that $ b \equal{} 2$,$ d \equal{} 3$, and $ f \equal{} 7$
So $ x \equal{} \overline{a.2}$ $ y \equal{} \overline{c.3}$ and $ z \equal{} \overline{e.7}$

Since this is defined one all real numbers the solution are $ a \plus{} b \plus{} c \equal{} 3$ for any $ a,b,c\in N$(cant have negatives beacuse $ \overline{a.2} \minus{} \overline{c.3} \neq \overline{a.2} \plus{} \overline{c.3}$.
So all solutions for a,c,e valuse are permutations of (0,0,3),(0,1,2),(1,1,1)
Too get solutions for x,y,z add b,d,f respectably.
Im too lazy to post all the solutions  :D

There's no solution. Subtract the third equation from the first to get $ \{ x\}\equal{}1.9\plus{}\{ y\}$, where the brackets denote fractional part and lie in $ [0,1)$.

If you'll quote the text of the original poster, you'll see that s/he intended for several of the expressions on the left to be curly-bracketed.  This largely explains the comment "với {A} = A - [A] " that the O.P. included under the equations.  It doesn't explain why people don't make use of the preview function, however.

Oh, I see, looking at the code, the third expressions should be bracketed. In that case...

Add the three equations to get $ 2(x\plus{}y\plus{}z)\equal{}9.4\Rightarrow x\plus{}y\plus{}z\equal{}4.7$. Subtracting off each of the equations from this gives
\[ \{ y\}\plus{}\lfloor z\rfloor \equal{}0.8\]
\[ \{ z\}\plus{}\lfloor x\rfloor \equal{}1.2\]
\[ \{ x\}\plus{}\lfloor y\rfloor \equal{}2.7\]
So the unique solution is $ \boxed{(1.7,2.8,0.2)}$ (it's easy to see that $ \{ u\}\plus{}\lfloor v\rfloor \equal{}w$ if and only if $ \{ u\}\equal{}\{ w\},\lfloor v\rfloor \equal{}\lfloor w\rfloor$).


[quote="linzk3"]Complete the square to find the center and the radius of hte circle  3x^2 + 3y^2 - 6x + 30y + 15 = 0[/quote]

[hide="Solution"]
$ 3x^{2}\plus{}3y^{2}\minus{}6x\plus{}30y\plus{}15 \equal{} 0$

$ 3x^{2}\minus{}6x\plus{}3y^{2}\plus{}30y \equal{}\minus{}15$

$ 3(x^{2}\minus{}2x)\plus{}3(y^{2}\plus{}10y) \equal{}\minus{}15$

$ 3((x\minus{}1)^{2})\plus{}3((y\plus{}5)^{2}) \equal{}\minus{}15\plus{}3\plus{}75$

$ (x\minus{}1)^{2}\plus{}(y\plus{}5)^{2}\equal{} 21$

$ \text{Center}\equal{}\boxed{(1,\minus{}5)}$

$ \text{Radius}\equal{}\boxed{\sqrt{21}}$
[/hide]

[quote="7h3.D3m0n.117"]$ 3(x^{2}\minus{}2x)\plus{}3(y^{2}\plus{}10y) \equal{}\minus{}15$

$ 3((x\minus{}1)^{2})\plus{}3((y\plus{}5)^{2}) \equal{}\minus{}15\minus{}1\plus{}25$[/quote]

The first line is correct, but the second one is wrong for two reasons.  The second line should be this:
$ 3((x\minus{}1)^{2})\plus{}3((y\plus{}5)^{2}) \equal{}\minus{}15\plus{}3\plus{}75$

Your center is correct, but the radius is actually [hide]EDIT:$ \sqrt{21}$[/hide]

[quote="7h3.D3m0n.117"]Where do you get $ 75$? $ \left(\frac{10}{2}\right )^{2}\equal{} 25$ not $ 75$.[/quote]

You have to multiply its by 3 because of the parentheses.

Wait... My radius answer is wrong also.  Anaxerzia explained why you have to multiply by three, but my radius answer is wrong.  From my second line, we get:
$ 3(x\minus{}1)^{2}\plus{}3(y\plus{}5)^{2}\equal{}63$
$ (x\minus{}1)^{2}\plus{}(y\plus{}5)^{2}\equal{}21$
Radius$ \equal{}\sqrt{21}$...

Sorry, I had forgotten to divide by three...


Harvard college Yale Princeton Columbia MIT

Hello everyone,
I know I have posted similar topic, but this is a new(and improved) one.

I am an international student from Europe, having the following EC:
Member of the business club, its president for 1 year
Vice-president of the webdesign club
founder of one club, which aimed to help people in mathematics in 8th grade, Founder&President
Member of the We Care Club
Member of the MUN club, participated in 5 conferences
Member of the basketball team - a few second places
Member of the snowboard team - one second place
Member of the math team of my school - about 10 awards
Member of honor roll for excellent gpa
Average GPA: 3.7/4

I have excellent essay, SAT results and recommendations.

What do you think - am i going to get into one of those universities ( I know there isn't any formula, I just want your opinion)
Harvard
Cambridge
Yale
Princeton
Brown
Columbia
Cornell
and the other Ivys

Thank you !
Excellence :)

Why isn't MIT in your list?...just out of curiousity  :roll:  How much did you score on the SAT subject test and on the SAT Reasoning test?

[quote="exceLLence"]I have accidentally forgot it... MIT is on my list.. My sat scores  M: 790, CR: 740, W: 740 (Total: 2270)
SAT II Math: 740, Physics: 740, Chemistry 740

:)[/quote]

you did pretty well, I believe, you have a plausible chances to get accepted at least to the universities at the bottom of your list due to your high SAT 1 score.

Good luck!

A very high amount of weight is given to your essays and recs. Without seeing at least your essays, we can't say anything. Even with an essay, there is a chance that we will misjudge it, but without the essay, no reasonable prediction can be made. At least, not for the Ivy's. I'm not sure how Cambridge rolls.

You have an impressive list of EC's. When you write your Curriculum Vitæ, put the emphasis on interesting things you did, initiatives you took, and the effect of what you accomplished. Separate the "Activities", which detail things that you did, with a short, concise description (1-2 lines) of what you did and what it accomplished, from an "Honors and Awards" section, which details what you have won. And obviously, place the "Activities" section first.

If you have undergone any major projects, such as research, science fair projects, engineering projects, etc, use your best discretion as to whether you want to place that under a separate Projects and Research category or place them under activities.

Presentation is very important.

comment on my ec's too tzf

there are in the same forum

I already have. But my comments would be exactly the same as the comments I gave him. Pretend I wrote my last post here to you.

this means i should emphasize more about my intel isef project

Sure. I would also place some emphasis on debate... I'm not sure how involved you were, but if I recall, you went to another country to compete. That should be interesting.

Also note that for many schools in the US, if you send them an abstract of your ISEF paper (make it a bit longer than what you would use for a science fair, since the paper is not present... but not too long), they will take a look at it.

thanks;
and yes i went to france to compete in a debate

also where in my extra cirricular should i place my inter school activities?

Do you have a resume? If so, PM it to me as an attachment and I'll be able to give you more meaningful feedback.


[quote="gerrard"]Find all polinomials P(x) with real coefficients  satisfying the condition :
$P(x).P(2x^{2})=P(x^{3}+x), x /in R$[/quote][/quote]
x=0, we have P(0)=0 or 1.

[quote="gerrard"]Find all polinomials P(x) with real coefficients  satisfying the condition :
$P(x).P(2x^{2})=P(x^{3}+x), x /in R$[/quote][/quote]

maybe you mean $P(2x^{3}+x)$ in rhe RHS?

I have seen this problem in my training.Old but very good problem.
I am sorry but i will give you only [hide="hint"]try to analyse its roots.[/hide]


Can anybody help me with optimizing that code?? I did my best but I still got timeout.
Algorithm is simple DP.
[code]#include <stdio.h>
int main()
{
      FILE *fin=fopen("inflate.in", "r");
      FILE *fout=fopen("inflate.out", "w");
      int M, N, m, i;
      fscanf(fin, "%d %d", &M, &N);
      int best[M+1];
      int times[N+1];
      int points[N+1];
      for(i=1; i<=N; i++)
            fscanf(fin, "%d %d\n", &points[i], &times[i]);
      for(m=1; m<=M; m++){
          best[m]=0;
          for(int n=1; n<=N; n++)
              if(times[n]<=m&&best[m]<best[m-times[n]]+points[n])
                          best[m]=best[m-times[n]]+points[n];
          if(best[m]<best[m-1])
              best[m]=best[m-1];

      }
      fprintf(fout, "%d\n", best[M]);
      return 0;
}[/code]

I know this is mentioned in a later section, but I remember using the Knapsack algorithm for this one.

Yeah that is Knapsack problem. I 've already finished that. It is an integer knapsack problem and I used DP. If anybody is interested in code I can post it.

i have the same code but still runs in 1.1 sec in the last case...

is there anything i need to do for this problem


Take any number in base 5.  Rearrange the digits however you like.  Subtract the rearranged number from the original number.  What is the largest integer that will ALWAYS divide the difference?

Can you generalize your answer to an arbitrary base?

[hide]The largest integer is 4

For any number y in base x, if you subtract a number with the digits of y scrambled, the difference will be divisible by x-1.[/hide]

[hide]If the digits in a number in base x added up are divisble by x-1, then the whole number is divisble by x-1.  Since the number with the rearanged digits has the same digits, the rearanged number is divisble by x-1.  If a (mod x-1) =  0 and c (mod x-1) = 0 and a+b = c, then b (the difference) (mod x-1) =0.[/hide]

[quote="h_s_potter2002"]If the digits in a number in base x added up are divisble by x-1, then the whole number is divisble by x-1.  [/quote]
Why?

The rest of your argument doesn't follow, I think.   For example, take 213 in base 5.  This is 58 in base 10, which is not divisible by 4, nor is 132 in base 5 (which is 42 in base 10).  But their difference is 31 in base 5 (or 16 in base 10), which *is* divisible by 4.

So in your argument, I think you want that a = c (mod x-1), not that they're both 0 (mod x-1).  But this then begs the question: why are a and c equal (mod x-1)?

You have the right answer and I think the right intuition; I'm just trying to get more precise reasoning.


geometry
3D geometry
octahedron
tetrahedron
LaTeX
ratio
symmetry

(1) A regular octahedron is placed on a horizontal rest. Draw the plan of top-view  for the regular octahedron.

(2) Let $ G_1,\ G_2$ be the barycentres of the two faces of the regular octahedron parallel to each other. Find the volume of the solid by revolving the regular tetrahedron about the line $ G_1G_2$ as the axis of rotation.

I'm very sorry, I forgot to write that the side length of the octahedron is $[color=red]1[/color]$.

Could you show us your solution?

kunny

Hi Kunny,

The solution for this particular problem is too lengthy and requires too many mathematical syntaxes (LaTex) and graphics for me to put it in here.

Dien

[quote="Vo Duc Dien"]Here goes

V = [pi x (side length)^3 / (18sqrt(6)]x[7 + 2xsqrt(3)]

Vo Duc Dien

Solution is in my book.[/quote]

The answer is wrong.

First, we find the distance $G_1G_2$, which joins the centers of two opposite faces of the octahedron.  This is also the long diagonal of the cube whose edges are the segments joining the centers of adjacent octahedral faces.  Due to similarity of triangles and the fact that the center of an equilateral triangle divides the altitude into a ratio of 1:2, it follows that the cube edge is one-third the distance between opposite vertices of the octahedron--i.e., the diagonal of the square joining four octahedral vertices in a plane.  If the side length of the octahedron is $s$, then this distance is $s\sqrt{2}$, and consequently the cube edge has distance $s\sqrt{2}/3$, and therefore $G_1 G_2 = s\sqrt{2/3}$.

Now consider horizontal slices of the octahedron perpendicular to $\overline{G_1 G_2}$.  These are equiangular hexagons, which degenerate into equilateral triangles when the cross section coincides with one of the octahedral faces.  (To see this, it helps to have drawn the diagram in the first part of the question.)  When rotating the octahedron about the axis $\overline{G_1 G_2}$, these hexagons become circles, whose radii equal the distance from the hexagon's center to a vertex--i.e., its circumradius.

Because of symmetry, place the octahedron's center at the origin and consider only $0 \le z \le s/\sqrt{6}$, half the vertical height.  For $z$ in this interval, the circumradius of a hexagon at height $z$ varies from $s/2$ to $s/\sqrt{3}$.  Thus, consider a separate diagram in which one has drawn a regular hexagon whose inradius (apothem) is $s/2$ and whose circumradius is $s/\sqrt{3}$.  Then half the side length is simply $s/\sqrt{12} = s/(2 \sqrt{3})$.  Hence the circumradius as a function of $z$ is given by the function \[ r(z) = \sqrt{(s/2)^2 + z^2/2}, \quad 0 \le z \le s/\sqrt{6}. \]  Therefore the volume of the desired solid of revolution is \begin{align*} V(s) &= 2 \int_{z=0}^{s/\sqrt{6}} \pi r(z)^2 \, dz \\ &= 2\pi \int_{z=0}^{s/\sqrt{6}} \frac{s^2}{4} + \frac{z^2}{2} \, dz \\ &= \pi \left[ \frac{s^2}{2} z + \frac{z^3}{3} \right]_{z=0}^{s/\sqrt{6}} \\ &= \frac{5\pi s^3}{9 \sqrt{6}}. \end{align*}  When $s = 1$, we have $V(1) = \frac{5\pi}{9 \sqrt{6}}$.


The number of People in a closed room is proportional to the number of poeple outside the room.

Proof that if there is one person in the room, and if he leave the room, all people in the world will be killed.

And now another question.

Is he a murder?!?

ok. let's see.

suppose a generic squared room [b]A[/b], and it's perfect, on where we assume that it' width is [/b]m[/b], with it's basis area equal to [b]a[/b]².

as it's square, we can strongly assume that:
[b]m[/b]² = [b]a[/b]²
[b]m[/b] = [b]a[/b]

ok, now we are getting to the point. the afirmation above represents that the room [b]A[/b] isn't perfect at all, because it's widht and other sides varies from [b]m[/b] (and its's multiples) to [b]a[/b] (and it's multiples).

now, let's assume that Earth is square and plane and the area outside the room's walls is represented as [b]e[/b] (do not make confusion with Euler's constant). For instance, we can assume that:

Earth = [b]e[/b]² - [b]a[/b]²  (assuming that the side of the generic room [b]A[/b] is [b]a[/b])

or

Earth = [b]e[/b]² - [b]m[/b]² (assuming that the side of the generic room [b]A[/b] is [b]m[/b])

now, let's see. we are almost there.
assume that there are [b]x[/b] people inside the room [b]A[/b]. as the question was said, we have to assume that there are [b]x[/b] people, or we lose the game. let's put into tables:

__________________________________
|time  |  people inside  |  people outide|
|   1   |          x          |           x       |
``````````````````````````````````
([i]Table 1: relates the people outside and people ouside the room on the assumed time [/i])

and now let's assume one person inside the room got to [b]pee[/b] (not pi). it's not polite to pee insides, as the room [b]A[/b] got no bathrooms neither toilets. And we are assuming that the generic person [b]z[/b], that needs to pee, is polite and is going to leave the room to do so. let's see what happens with the world on the next table:

__________________________________
|time  |  people inside  |  people outide|
|   2   |          x-1       |           x+1    |
``````````````````````````````````
([i]Table 2: relates the people outside and people ouside the room on the assumed time {again} [/i])

it was said that we [b]must[/b] assume that people outside is equal to people inside. or, mathmatically:

pi = po

(pi: do not confunde with 3,14...)

on time 2, we have:

x - 1  =  x + 1
-1 = 1
2 = 0

this result only proves that, when someone leaves to [b]pee[/b], or other stuff, the balance will broken down and 2 other generic persons will be 0, as they will annulate each other.

and, the other question. the generic person [b]z[/b], he/she/it is not a murder for 3 motives:

1 - he didnt mention to do that;
2 - everybody died, the police too (no one would arrest him/her/it);
3 - the famou dictation: "peeing is not a crime"

Salutes. [/list][/quote]

by the way

Earth = e² + a²

       or

Earth = e² + m²

(not minus)

I miswrote it

I'm going to take a guess here and assume Fabiomr1 = Thales418.

what?! what r u talking about?!

Fabiomr1 = Thales418

Fabiomr/Thales = 418
Fbiomr/Thles = 418
Fm/T*io/he*r/l = 418s

Force x mass/period * initial image/height x euler's constant * radius/log = 418 seconds?!

no way

u' r nuts

Fabiomr1 is dunce

he/she/it dont know latex!

Thales418 know latex

are all the world dunce, watson!?

:huh: What exactly is the point of this again?
Sorry, I'm kinda confused here about what's so funny in this thread and all the others.
Maybe I just dont get the joke

ra5000

stupid things are funny....

math is a stupid thing funny

we need do laught

Math is not stupid.
Stupid doesn't mean funny.
We need to laugh. But not at pointless math.
Combining these, I still don't get it.
And you spelled my name wrong. Congrats.

ra5001

math is funny, yes!!!

i dont want to be a stupid teacher that tell to students

$x=7$ know by core this!!
it will be on the test .. bla bla lba lba bla.....

if they make math more funny, dear ra5001, people like you would be more happy

I'm happy enough. Thanks.
And I'll be more happy when people can spell my name right.
Oh, and math doesn't have to be funny for me to be happy. It's OK, I think I get it now.

So... Math is funny for some, and then it makes them happy.
But for others (Like me) Math isn't funny, but it still makes them happy.
And others just hate math.
OK, thanks for answering my question! (No sarcasm, seriously, I get it now. Thanks.)

One more thing; my name isn't ra5000, ra5001, ra5002, or anything else of the sort, OK?

ok  But induction your name is raInfinity?!

hehe

its just a joke...

you never understand!?

math is happy

the only true thing in the world

the others are all false for $n>0$

yah yah


ARML
analytic geometry
graphing lines
slope
logarithms
ratio
geometry

1.

Let $S=\{1,2,3,\ldots,24,25\}.$  Compute the number of elements in the largest subset of $S$ such that no two elements in the subset differ by the square of an integer.

2.

Starting at the origin, a beam of light hits a mirror (in the form of a line) at $A(4,8)$ and is reflected to the point $B(8,12).$  Compute the exact slope of the mirror.

3.

If $a\cdot b\cdot c=240$, compute the number of distinct ordered triples $(a,b,c)$ such that $a$ is a multiple of 2, $b$ is a multiple of 3 and $c$ is a multiple of 5.

4.

Compute the largest real value for $b$ such that the solutions to the following equation are integers:

\[\left(\log_{2^{10}}{x^{2b}}\right)^2=\log_{2^{10}}{x^4}.\]

5.

Compute the ratio of the volume of a sphere to the volume of the largest regular octahedron that will fit inside it.

5:
[hide]
If the sphere has radius r, then we can find the volume of the octahedron like this: split it into 2 pyramids, each pyramid has a base in the shape of a square with side length $ \sqrt{2}r $ and height r. So the volume is $ 2(\frac{1}{3})(r)(2r) = \frac{4r^3}{3} $.

The volume of the sphere is $ \frac{4 \pi r^3}{3} $.

So the ratio is $ \pi $.  :lol:
[/hide]

It can also be easily proven that the largest octohedron is the regular one.

[hide]
simplify to $4b^2(logx) = 40 log2$
$b^2 = \frac {10log2}{logx}$
when $x=1$ it is undefined
when $x=2$ then $b = \sqrt {10}$
[/hide]

3. $abc=240$.  So
$abc=2^4 * 3 * 5$
We know that there must be a 3 in the prime factorization of b, a 5 in c, and a 2 in a.  We can first assign those.  There are then 3 2s left, and there are $3^3$ ways to distribute them.  But wait: there are some possibilities which are counted twice.  The 3 possibilities where the three twos are all on one number are each counted once.  So those are fine.  The ones where one number has 2 2s and another has one are each counted 3 times.  There are 6 of those.  Therefore, we must subtract 12.  So we now have 15.  Then, there is one possibility where each number has one of the three 2s.  This is counted 6 times.  So the final answer is
$10$
Indeed, these are
$(2^4, 3, 5)  (2^3, 2*3, 5)  (2^3, 3, 2*5)  (2^2, 2^2*3, 5)  (2^2, 2*3, 2*5)  (2^2, 3, 2^2*5)  (2, 2^3*3, 5)  (2, 2^2*3, 2*5)  (2, 2*3, 2^2*5)  (2, 3, 2^3*5)$

2. Well, the segment from 0 to (4,8) has an angle of $tan^-1 2$...so then you'd need a calculator to find it...
Here's one way to do it...extend the second line segment so it's the same length as the first: 4sqrt[5].  Then it reaches the point (4+2sqrt[10], 8+2sqrt[10])
Then the slope of this line is clearly (8+2sqrt[10])/(4+2sqrt[10]).  This is simplified to(sqrt[10]+1)/3

I think that works...

2. The silver line is the mirror. Let the slope equal $m$. Since angle of incidence equals angle of reflection, $\angle EAO = \angle BAH$.  We have
$90-\arctan(4/8)-\arctan(m)=90-\arctan(1)-\arctan(1/m)
\\ \arctan(1/2)+\arctan(m)=\arctan(1)+\arctan(1/m)
\\ \tan(\arctan(1/2)+\arctan(m))=\tan(\arctan(1)+\arctan(1/m))
\\ \frac{1/2+m}{1-m/2}=\frac{1+1/m}{1-1/m}
\\ \frac{1+2m}{2-m}=\frac{m+1}{m-1}
\\ 3m^2-m-1=0
\\ m=\frac{\sqrt{10}+1}{3}
$


pigeonhole principle
number theory unsolved
number theory

Here's the challenge...
Show that every positive integer which is not divisble by 2 or 5 has a multiple which consists entirely of 1's.

We have $(10,9n)=1$ ($n$ is our number), so $9n|10^{\varphi(9n)}-1$, meaning that $n|\frac{10^{\varphi(9n)}-1}9$. The number on the right consists only of $1$'s.

pigeonhole principle on n numbers of the form 111,111,... consinsting of 1,2,..,n 1's each. 2 will have the same residue modulo n(if we suppose that no one is divisible by n) and the rest follows easily.


If $ \frac{p^{a}-1}{p-1}=\frac{q^{b}-1}{q-1}$ then there are primes $ a_{1}|a$ and $ b_{1}|b$ such that $ p^{a_{1}}-1|q^{b_{1}}-1$ or vice versa.

[quote="mathpk"]
Is follow statement true?
If $ \frac{p^{a}-1}{p-1}=\frac{q^{b}-1}{q-1}$ then there are primes $ a_{1}|a$ and $ b_{1}|b$ such that $ p^{a_{1}}-1|q^{b_{1}}-1$.[/quote]
It is difficult to answer this question, because you did not say whether $ p,q,a,b$ are integers, positive, primes, etc.  If $ p$ and $ q$ are primes and $ a$ and $ b$ are integers such that $ a,b\ge 3$, then we have \[ 1+p+\cdots+p^{a-1}=\frac{p^{a}-1}{p-1}=\frac{q^{b}-1}{q-1}=1+q+\cdots+q^{b-1}\] which yields that \[ p(1+p+\cdots+p^{a-2})=q(1+q+\cdots+q^{b-2})\] which happens iff $ p=1+q+\cdots+q^{b-2}$ and $ q=1+p+\cdots+p^{a-2}$.
But, anyways before looking into this more, what are $ p,q,a,b$?  Thanks.

Oops sorry :blush:
 p and q are different odd primes and a,b are positive integers


Seven segment display for a calculator doing arithmetic base 4
Proof by mathematical induction of extension of distributive law for sets ( intersection over union)

How should I get started? Thanks!

Can some mod please move or delete this...


I have read that the planet Mars will be the most shining object in the day 27 of this month. It seems that Mars will appear so big as a full moon, and so in the night of that day the sky will appear to have two moons! According to the source, Mars will be at 34.65 miles from the Earth, and Mars will be so close again only on the year 2287. Is this true? If it is, then I think no one should miss this rare and probably beautifull fenomenon.

great catch if this is true then we will have two moons :wow:

:| I was going to type topic about this in this forum to WARN people against this year's Mars Hoax.
No the Mars will NEVER get that close, in the fore-seeable future  :wink: .
From an reliable email
[quote="SpaceWeather.com"]
Space Weather News for August 4, 2007
http://spaceweather.com

BEWARE THE MARS HOAX: It's August, which means it's time for the annual Mars Hoax. An email is going around claiming that Mars will approach Earth on August 27th; the encounter will be so close, the email states, that Mars will rival the full Moon in size and brightness. (Imagine the tides!) Don't believe it. The Mars Hoax email first appeared in 2003. On August 27th of that year, Mars really did come historically close to Earth. But the email's claim that Mars would rival the Moon was grossly exaggerated. Every August since 2003, the email has staged a revival.

Here's something that is true: Mars is having a close encounter with the Pleiades star cluster, easily seen in the eastern sky before sunrise. Especially good mornings to look are August 6th and 7th when the crescent Moon joins the planet and the cluster to form a pretty celestial triangle. Set your alarm!
[/quote]
:maybe:  As the above stated, if the moon has such control on the tides imagine two celestial bodies doing so.
The event with Mars the mornings of yesterday and today would have been exciting though.
For those who were or are interested the sky map for this morning was
[img]http://www.spaceweather.com/images2007/07aug07/skymap_north.gif[/img].

 :wink: There is another exciting event happening with the moon, I will post a new topic about it.

Thanks for clearing the issue, AstroPhys. Reading again the message, its indeed ridiculous to claim that Mars will approach Earth so closely.

If Mars was actually that close, it would be in the earth's atmosphere and collide with the earth.

Aww, and I was so looking forward to seeing the polar caps without a telescope...

Vamathletes, I don't think that Mars would be in Earth's atmosphere if it were that close. After all, Mars is much larger than the moon, and thus would be farther away if it appeared to be the same size as the moon. Since the moon isn't in our atmosphere, Mars wouldn't be, either.

It would be physically possible too. For if mars was visible as the size of the moon, our oceans would be affected, and so far they aren't.

[quote="KrazyNerd"]Vamathletes, I don't think that Mars would be in Earth's atmosphere if it were that close. After all, Mars is much larger than the moon, and thus would be farther away if it appeared to be the same size as the moon. Since the moon isn't in our atmosphere, Mars wouldn't be, either.[/quote]
I meant that [quote="Carcul"]Mars will be at 34.65 miles from the Earth[/quote].
Sorry about the confusion.


[quote="Inequalities Master"]Let $ a,b,c,d > 0$ s.t. $ abcd \equal{} 1$.Prove that:
$ \frac {1}{1 \plus{} ab \plus{} bc \plus{} ca} \plus{} \frac {1}{1 \plus{} bc \plus{} cd \plus{} db} \plus{} \frac {1}{cd \plus{} da \plus{} ac} \plus{} \frac {1}{1 \plus{} da \plus{} ab \plus{} bd}\leq 1$.
[/quote]
Let $ a\equal{}x^2,$ $ b\equal{}y^2,$ $ c\equal{}z^2$ and $ d\equal{}t^2,$ where $ z,$ $ y,$ $ z$ and $ t$ are positive numbers.
Hence, we need to prove that $ \sum_{cyc}\frac{xyzt}{xyzt\plus{}x^2y^2\plus{}x^2z^2\plus{}y^2z^2}\leq1.$
But $ \sum_{cyc}\frac{xyzt}{xyzt\plus{}x^2y^2\plus{}x^2z^2\plus{}y^2z^2}\leq\sum_{cyc}\frac{xyzt}{xyzt\plus{}xyz(x\plus{}y\plus{}z)}\equal{}1.$


does anyone no any jokes about pi. Here is one

a farmers boy came home from his first day at middle school. His father said
"say something in algebra!"
"Pi R Squared" The boy replied
"what are they teaching you in that school?" the farmer asked
"everyone knows that cornbread is square, pies are round!"

Not exactly on pi, more on the lines of 'Soak a toe' :rotfl:  :rotfl:
[url=http://www.pen.k12.va.us/Div/Winchester/jhhs/math/lessons/trig/legend.html]www.pen.k12.va.us/Div/Winchester/jhhs/math/lessons/trig/legend.html[/url]

A joke that's pretty stupid --

Q:What do you get when you take the sun and divide its circumference by its diameter?
A: Pi in the sky.


Man I am really confused about this
will someone help me out

WRITING ABOUT MATH
Finding patterns inpolynomial Division

Complete the following polynomial divisions

a. (x^2 - 1) divided by (x-1) =

b. (x^3 - 1) divided by (x-1) =

c. (x^4 - 1) divided by (x-1) =

Write a brief description of the pattern that you obtain, and use your result to find a [b]formula[/b] for the polynomal division (x^[i]n[/i] - 1)/(x - 1). Create a numerical example to test your formula

i know how to do the polynomial division....but what is the formula in the paragraph part

dang
that helped a lot

thank you
(do you use aim)


vector
calculus
function
calculus computations

This came up in my class today: in $ \mathbb{R}^3,$ assuming $ f$ and $ g$ are sufficiently smooth scalar functions, show that $ \text{div}\,(\nabla f\times \nabla g)\equal{}\nabla\cdot(\nabla f\times\nabla g)\equal{}0.$

It is, of course, and you can show that by just bashing it out. But I was trying to come up with some more elegant way of explaining why it had to be true, or some motivation for it, and didn't come up with anything I found particularly convincing. Any ideas?

Hmm,

isn't

$ \nabla f \times \nabla g$ the same as $ df \wedge dg$ using differential forms (requires a check but it looks right), and
so

$ d (df \wedge dg) \equal{} 0$?


geometry
3D geometry
tetrahedron
geometry proposed

[quote="shasha"]Four swallows are catching a fly. At first, the swallows are at the four
vertices of a tetrahedron, and the fly is in its interior. Their maximal speeds
are equal. Prove that the swallows can catch the fly.[/quote]

This version of the problem is false.  The fly could start near the center of one of the
faces and fly outward perpendicular to the face.

[quote="statistical"]
This version of the problem is false.  The fly could start near the center of one of the
faces and fly outward perpendicular to the face.[/quote]

Why do you think that swallows cannot catch the fly in this case?
I think its obvious they CAN do it in this special case.


inequalities calculus derivative algebra function domain quadratics

Any relationship for $ x,y,z$. I think it would be $ \frac{k}{3}$. I got this intutively. I will have to check that!

[quote]some body del this post  :)  i have an error in posting[/quote]

What error?

[quote="Kouichi Nakagawa"]You use Cauchy - Schwarz ineq and AM $ \ge$ GM.[/quote]

Can you solve it for me ?  :?:

[quote="Kouichi Nakagawa"]\begin{eqnarray*} (1^2 + 1^2 + 1^2)((a + x)^2 + (b + y)^2 + (c + z)^2) & \ge & (a + b + c + x + y + z)^2 \\
& \ge & (3\sqrt [3]{abc} + x + y + z)^2 \\
& = & (9\sqrt [3]{k} + x + y + z)^2 \end{eqnarray*}

Thus, $ (a + x)^2 + (b + y)^2 + (c + z)^2 \ge \frac {1}{3}(9\sqrt [3]{k} + x + y + z)^2$[/quote]

 :rotfl: You'r wrong  :rotfl:
when does equality  occur ? I think you will write $ a = b = c = > x = y = z$ :roll:

$ (a+x)^2+(b+y)^2+(c+z)^2 \ge$$ {(a+b+c+x+y+z)^{2}/{3}}$by the Engel's extension of C-S or the Andreescu lemma and then apply AM-GM

[quote="manjil"]$ (a + x)^2 + (b + y)^2 + (c + z)^2 \ge$$ {(a + b + c + x + y + z)^{2}/{3}}$by the Engel's extension of C-S or the Andreescu lemma and then apply AM-GM[/quote]
like Kouichi Nakagawa's ?

[quote="foraus"]
like Kouichi Nakagawa's ?[/quote]
I think mine is lower than his!

[quote="Kouichi Nakagawa"]\begin{eqnarray*} (1^2 + 1^2 + 1^2)((a + x)^2 + (b + y)^2 + (c + z)^2) & \ge & (a + b + c + x + y + z)^2 \\
& \ge & (3\sqrt [3]{abc} + x + y + z)^2 \\
& = & (\color{red}3\color{black}\sqrt [3]{k} + x + y + z)^2 \end{eqnarray*}

Thus, $ (a + x)^2 + (b + y)^2 + (c + z)^2 \ge \frac {1}{3}(\color{red}3\color{black}\sqrt [3]{k} + x + y + z)^2$

EDIT : oops! I find a tiny typo.[/quote]

change 9 by 3 ?! try again !
I think your sollution before you edit :
$ P_{max} = \frac {1}{3}(\color{red}3\color{black}\sqrt [3]{k} + x + y + z)^2$

but the result isn't like that

I want to see the right sollution for this problem ; i thinks after use cauchy schwarz inequality ; you have to use AM-GM inequality but not easy to use it .

[quote="Akashnil"][quote="Kouichi Nakagawa"]I revised a typo some time ago.
Sorry.
I wanted you to supplement the slightest mistake like that within the brain.[/quote]
No, that's not the problem. Your solution is still wrong. Because the minimum is wanted, and there must be a equality case. But according to your solution there isn't an equality case for all $ x,y,z$.[/quote]

OK.

C-S ineq equality case : $ \frac {a \plus{} x}{1} \equal{} \frac {b \plus{} y}{1} \equal{} \frac {c \plus{} z}{1}$.
$ \therefore a \plus{} x \equal{} b \plus{} y \equal{} c \plus{} z$.
AM $ \ge$ GM equality case : $ a \equal{} b \equal{} c( \equal{} \sqrt [3]{k})$

Thus, $ a \equal{} b \equal{} c$ and $ x \equal{} y \equal{} z$.

[quote="Kouichi Nakagawa"](equality case : $ a \equal{} b \equal{} c( \equal{} \sqrt [3]{k})$, $ x \equal{} y \equal{} z( \equal{} \minus{} \sqrt [3]{k})$)[/quote]
$ x,y,z$ are [b]constants[/b]. That can't be in the equality case. Only $ a,b,c$ are variables.
[hide]I've'nt read rest of your post, sorry if I'm stupid[/hide]

[quote="Kouichi Nakagawa"][quote="foraus"]Yes $ x;y;z$ are constants and $ x,y,z$ aren't const

$ P_{min} \equal{} 0$ isn't right[/quote]

Yes. $ P_{min} \equal{} 0$ is $ x$, $ y$, $ z$ is special.[/quote]

If I change that:
$ x \equal{} 2; y \equal{} 3;z \equal{} 4 ;k \equal{} 5$ (random value  :D )

what's your sollution  :lol:

あぁ～そういうことね。
$ a\plus{}x\equal{}b\plus{}y\equal{}c\plus{}z$ を満たしながら, $ a\equal{}b\equal{}c$ にするためには $ x$, $ y$, $ z$ も変数でないといけないということね。
しかし, $ x$, $ y$, $ z$ は定数である。
ここで矛盾が生じたのか…。
なるほどね。
おかしかった部分が分かったよ。

Lagrangeの未定乗数法を使って Mathematica で計算してみました。
計算結果が余りにも凄いことになっているので, ファイルを添付します。

この問題って本当に綺麗な形で答えが出るの？

[quote="foraus"]Find minimum value of $ P \equal{} (a \plus{} x)^2 \plus{} (b \plus{} y)^2 \plus{} (c \plus{} z)^2$
and a,b,c are positive real numbers ;$ abc \equal{} k$ with x;y;z;k are constants[/quote]
Nice inequality.
I rewrite it as follows (just in a simpler form)
"If $ a,b,c$ are positive real numbers such that $ abc \equal{} 1$. For any given constant $ x,y,z>0,$ determine the minimum value of the following expression
\[ P(a,b,c) \equal{} (a \plus{} x)^2 \plus{} (b \plus{} y)^2 \plus{} (c \plus{} z)^2
\]
"
Now, the weighted AM-GM will help us ;)
Notice that the equality case is $ a \equal{} \frac {\sqrt {x^2 \plus{} k} \minus{} x}{2},b \equal{} \frac {\sqrt {y^2 \plus{} k} \minus{} y}{2},c \equal{} \frac {\sqrt {z^2 \plus{} k} \minus{} z}{2}$ where $ k$ is the unique positive proof of the following equation
\[ (\sqrt {x^2 \plus{} k} \minus{} x)(\sqrt {y^2 \plus{} k} \minus{} y)(\sqrt {z^2 \plus{} k} \minus{} z) \equal{} 8
\]
:)

[quote="Akashnil"][quote="can_hang2007"]Now, the weighted AM-GM will help us ;)[/quote]
Please show us how...
[quote="can_hang2007"]Notice that the equality case is $ a \equal{} \frac {\sqrt {x^2 \plus{} k} \minus{} x}{2},b \equal{} \frac {\sqrt {y^2 \plus{} k} \minus{} y}{2},c \equal{} \frac {\sqrt {z^2 \plus{} k} \minus{} z}{2}$ where $ k$ is the unique positive proof of the following equation
\[ (\sqrt {x^2 \plus{} k} \minus{} x)(\sqrt {y^2 \plus{} k} \minus{} y)(\sqrt {z^2 \plus{} k} \minus{} z) \equal{} 8
\]
:)[/quote]
I know, that's easily obtained in my Lag Mult approach, But I simply can't find the required minimum value :([/quote]

Thanks a lot. :)

Edit: So we can't have the minimum value in closed form without $ k$...

[quote="Akashnil"]My approach with Lagrange Multipliers:

We need to find the extrema of $ f(a,b,c) \equal{} (a \plus{} x)^2 \plus{} (b \plus{} y)^2 \plus{} (c \plus{} y)^2$
With the condition that $ g(a,b,c) \equal{} abc \minus{} k \equal{} 0$
Let $ \Lambda(a,b,c,\lambda) \equal{} f(a,b,c) \plus{} \lambda g(a,b,c) \equal{} (a \plus{} x)^2 \plus{} (b \plus{} y)^2 \plus{} (c \plus{} y)^2 \plus{} \lambda (abc \minus{} k)$
According to Lagrange Multiplier Theorem extrema occurs when the derivative $ d\Lambda \equal{} 0$
This gives the system of equations:
(I used $ \delta$ for partial derivatives.)

$ \frac {\delta \Lambda }{\delta a} \equal{} 2a \plus{} 2x \plus{} \lambda bc \equal{} 0\cdots (I)$

$ \frac {\delta \Lambda }{\delta b} \equal{} 2b \plus{} 2y \plus{} \lambda ca \equal{} 0\cdots (II)$

$ \frac {\delta \Lambda }{\delta c} \equal{} 2c \plus{} 2z \plus{} \lambda ab \equal{} 0\cdots (III)$

$ \frac {\delta \Lambda }{\delta \lambda} \equal{} abc \minus{} k \equal{} 0\cdots (IV)$

But the system seems too complex :( . If it is solved, then our problem is solved as well.

@foraus: Do you have the answer or the equality cases?[/quote] In this case you cant use Lagrange multipliers because you have open subset of variables

[quote="dima ukraine"] In this case you cant use Lagrange multipliers because you have open subset of variables[/quote]
What is open subset of variables? The domain of $ a,b,c$ is positive I know. But then I will find all the extrema cases in the real domain and then exclude those which fall outside the domain. And possibly check for the bounds of the domain(some of $ a,b,c\equal{}0$) which won't give minimum.

I think Lagrange Multipliers work very nicely here(very little calculation). All I was worrying about was a nice closed form minimum without other variables($ k$ in can's solution or  $ \lambda$ in my solution) and equation, which appears not to be available even in other solutions(can_hang2007's solution). We just need a few more steps to arrive at the answer he gave.

Multiply $ (I)$ by $ a$, replace $ abc$ by $ k$ and use the quadratic formula to solve for $ a$ in terms of $ x,\lambda,k$ After obtaining the expressions for $ a,b,c$ like this, define $ \lambda$ in terms of $ (IV)$. Now we have a equality case like the one can_hang2007 gave(he assumed the original problem's $ k \equal{} 1$ which isn't much problem)

[quote="Akashnil"][quote="dima ukraine"] In this case you cant use Lagrange multipliers because you have open subset of variables[/quote]
What is open subset of variables? The domain of $ a,b,c$ is positive I know. But then I will find all the extrema cases in the real domain and then exclude those which fall outside the domain. And possibly check for the bounds of the domain(some of $ a,b,c \equal{} 0$) which won't give minimum.

I think Lagrange Multipliers work very nicely here(very little calculation). All I was worrying about was a nice closed form minimum without other variables($ k$ in can's solution or  $ \lambda$ in my solution) and equation, which appears not to be available even in other solutions(can_hang2007's solution). We just need a few more steps to arrive at the answer he gave.

Multiply $ (I)$ by $ a$, replace $ abc$ by $ k$ and use the quadratic formula to solve for $ a$ in terms of $ x,\lambda,k$ After obtaining the expressions for $ a,b,c$ like this, define $ \lambda$ in terms of $ (IV)$. Now we have a equality case like the one can_hang2007 gave(he assumed the original problem's $ k \equal{} 1$ which isn't much problem)[/quote] Sorry for my English :blush:  I mean open domain


[quote="mathmen"]let $ a,b,c$ sides of triangle
prove that

[img]http://www.freeimagehosting.net/uploads/370df079b1.jpg[/img][/quote]
I think, $ |\frac {a \minus{} b}{a \plus{} b} \plus{} \frac {b \minus{} c}{b \plus{} c} \plus{} \frac {c \minus{} a}{c \plus{} a}| < \frac {1}{22}$ is also true, but
$ |\frac {a \minus{} b}{a \plus{} b} \plus{} \frac {b \minus{} c}{b \plus{} c} \plus{} \frac {c \minus{} a}{c \plus{} a}| < \frac {1}{23}$ is wrong already.

$ \left|\frac {a \minus{} b}{a \plus{} b} \plus{} \frac {b \minus{} c}{b \plus{} c} \plus{} \frac {c \minus{} a}{c \plus{} a}\right| < k$

the best $ k$ is $ k\equal{}\frac{8\sqrt{2}\minus{}5\sqrt{5}}{3}$...

kuing, thank you for the marking \left|. :lol:


A very general question: are there any techniques to find out the approriate substitutes needed to solve a differential equation?

Especially. second order differential equations! Come on, let's have a discussion about it.

There are many different types of differential equations and substitutions, do you have any examples?

any second order linear differential equations.

Here's a review sheet I made for Differential Equations (there might be a few typos), but you can see some examples of second order equations.  If anything doesn't make sense, let me know.

Gosh, this is good! Thank you very much! Did you make it yourself?

Thanks, and yes I wrote it over the summer.  All of the examples aren't mine though (some are).  I didn't have a book to go off of, so I got most of the info off the internet / memory, but most of the derivations are my own.


geometry circumcircle inequalities trigonometry calculus derivative integration

請大家在這裡一切分享problem of the day 的解答.
(也許現時的題目並沒有太大難度)

[b]The problem of 10/10/2005:[/b]

Find all sets $A$ and $B$ that satisfy the following conditions:

a) $A \cup B= \mathbb{Z}$;

b) if $x \in A$ then $x-1 \in B$;

c) if $x,y \in B$ then $x+y \in A$.

[b]The problem of 11/10/2005: [/b]

Find all $f: \mathbb{N^*}\to\mathbb{N^*}$ satisfying:
\[ (f^2(m)+f(n)) \mid (m^2+n)^2  \]

[quote="Soarer"]what is N star? does it include 0?[/quote]
I'm not sure. I just know that in main land China,
$\mathbb{N}=\{0,1,2,\cdots\}$ and $\mathbb{N}^*=\{1,2,\cdots\}$.

[b]The problem of 12/10/2005:[/b]

Six points are chosen on the sides of an equilateral triangle $ABC$: $A_1, A_2$ on $BC$, $B_1, B_2$ on $CA$ and $C_1, C_2$ on $AB$, such that they are the vertices of a convex hexagon $A_1A_2B_1B_2C_1C_2$ with equal side lengths.

Prove that the lines $A_1B_2, B_1C_2$ and $C_1A_2$ are concurrent.

[quote="dingdongdog"]Is it IMO2005-1?[/quote]
ya

我觉得最好每个问题一个topic吧.

There will soon be a page with all these displayed, so don't worry about it. Until now the random system was not that efficient, starting from now, problems should be displayed "more randomly" (if that's even a word :P).

[b]The problem of 13/10/2005:[/b]

You are at a point (a,b) and you need to reach another point (c,d). Both points are below the line x = y and have integer coordinates. You can move in steps of length 1, either upwards of to the right, but you may not move to a point on the line x = y. How many different paths are there?

This problem is related to Catalan's number.

[b]The problem of 14/10/2005:[/b]

Two circles $k_1$ and $k_2$ intersect at two points $A$ and $B$. Some line through the point $B$ meets the circle $k_1$ at a point $C$ (apart from $B$), and the circle $k_2$ at a point $E$ (apart from $B$). Another line through the point $B$ meets the circle $k_1$ at a point $D$ (apart from $B$), and the circle $k_2$ at a point $F$ (apart from $B$). Assume that the point $B$ lies between the points $C$ and $E$ and between the points $D$ and $F$.

Finally, let $M$ and $N$ be the midpoints of the segments $CE$ and $DF$.

Prove that the triangles $ACD$, $AEF$ and $AMN$ are similar to each other.

[b]The problem of 15/10/2005:[/b]

For positive integer $a \geq 2$, denote $N_a$ as the number of positive integer $k$ with the following property: the sum of squares of digits of $k$ in base a representation equals $k$. Prove that:

a.) $N_a$ is odd;

b.) For every positive integer $M$, there exist a positive integer $a \geq 2$ such that $N_a \geq M$.

[quote="ifai"][b]The problem of 16/10/2005:[/b]

Segments AC and BD meet at P, and |PA| = |PD|, |PB| = |PC|. O is the circumcenter of the triangle PAB. Show that OP and CD are perpendicular.[/quote]
Suppose $PQ$  is the diameter. Extend $OP$ to $R$, where $R$ is the intersection of $OP$ and $CD$.
Then $\angle BAP=\frac 12\widehat{PB}$, and $\angle OPB=\frac 12\widehat QB$, hence $\angle BAP+\angle OPB = \frac 12(\widehat{PB}+\widehat{QB}) = \frac 12\widehat{PQ} = \frac{\pi}{2}$. And it is easy to see that $\angle BAP+\angle OPB =\angle PDC+\angle DPR=\angle PRC$. Hence $\angle PRC=\frac \pi 2$.

[quote="liyi"][quote="ifai"][b]The problem of 17/10/2005:[/b]

Prove the trigonometric inequality $\cos x < 1 - \frac{x^2}{2} + \frac{x^4}{16}$, when $\dsp x \in \left(0, \frac{\pi}{2} \right)$.[/quote]

Are we allowed to use differentiation or integration?[/quote]
If not, then how would you do?

[quote="ifai"][b]The problem of 17/10/2005:[/b]

Prove the trigonometric inequality $\cos x < 1 - \frac{x^2}{2} + \frac{x^4}{16}$, when $\dsp x \in \left(0, \frac{\pi}{2} \right)$.[/quote]
It's the same as $\cos x <\left(1-\frac{x^2}4\right)^2, x\in\left(0, \frac{\pi}2\right)$
But I don't think it's very useful.

[b]The problem of 18/10/2005:[/b]

An arbitrary point [i]M[/i] is selected in the interior of the segment [i]AB[/i]. The square [i]AMCD[/i] and [i]MBEF[/i] are constructed on the same side of [i]AB[/i], with segments [i]AM[/i] and [i]MB[/i] as their respective bases. The circles circumscribed about these squares, with centers [i]P[/i] and [i]Q[/i], intersect at [i]M[/i] and also at another point [i]N[/i]. Let [i]N '[/i] denote the point of intersection of the straight lines [i]AF[/i] and [i]BC[/i].

a) Prove that [i]N[/i] and [i]N '[/i] coincide;

b) Prove that the straight lines [i]MN[/i] pass through a fixed point [i]S[/i] independent of the choice of [i]M[/i];

c) Find the locus of the midpoints of the segments [i]PQ[/i] as [i]M[/i] varies between [i]A[/i] and [i]B[/i].

[b]The problem of 19/10/2005:[/b]

In an $n$-gon all of whose interior angles are equal, the lengths of consecutive sides satisfy the relation
\[ a_1 \geq a_2 \geq \dots \geq a_n \]
Prove that $a_1=a_2= \ldots= a_n$.

[b]The problem of 20/10/2005:[/b]

We can obviously put 100 unit balls in a 10x10x1 box.
How can one put 105 unit balls in? How can we put 106 unit balls in?

[b]The problem of 21/10/2005:[/b]

Let $a, b, c$ be sides of a triangle whose perimeter does not exceed $2 \cdot \pi$., Prove that $\sin a, \sin b, \sin c$ are sides of a triangle.

ifai as I told you there is no need to re-post the problems on the board. Here's a link with the automatic generation of the collection: http://www.mathlinks.ro/Forum/resources.php?potd=1&week=-1 (will be also upgraded soon).

If you still wish to put the problems here, maybe it's a good idea to provide a translation in Chinese for them ;)

[b]The problem of 24/10/2005:[/b]

設 $\triangle ABC$ 內接於圓 $\Gamma$。圓 $\Theta$ 位於 $\triangle ABC$ 的內角 $A$ 的一側並分別交 $AB, AC$ 於 $M_1, N_1$、內切 $\Gamma$ 於 $P_1$（原句：(Circle $\Theta$) touches internally $\Gamma$ at $P_1$）。點 $M_2, N_2, P_2$ 與 $M_3, N_3, P_3$ 類似地定義於內角 $B, C$ 上。證明 $M_1N_1, M_2N_2, M_3N_3$ 共點，且該點分別是三線的中點。


consider p is a prime number and n and a are natural numbers and (p^a)-1/p-1=2^n.
find all divisors of n.a

see here http://www.mathlinks.ro/viewtopic.php?p=1251798#1251798
$ a$ should be a power of two.but you cant determine the prime divisor of $ n$(for example put$ a \equal{} 2$)


algebra
polynomial
difference of squares
special factorizations

I have worked on this question for over half an hour but I can't get it!
Find all integer solutions $(m,n)$ to:
$n^4+2n^3+2n^2+2n+1=m^2$.
Could someone please point me in the right direction? (Please don't tell me the answer, just a couple of hints). Thanks!

[quote="Gyan"]Sorry this hint almost gives it away .. but peek if you like:

[hide]IOW if , you assume $k=n^2+n$  and thus there is a  quantity which is greater than $k^2$ (or $n^4+2n^3+n^2$) but[b] less than [/b]$(k+1)^2$ (Or
$n^4+2n^3+3n^2+2n^2+1$  .. Can it be a perfect square? [/hide]
[/quote]

[hide]or equal [/hide]

[hide="my hint"]what is $n^4+2n^3+n^2+n^2+2n+1$ ?[/hide]

Yes  $n^4 + 2n^3+3n^2+2n+1$ is $(n^2+n+1)^2$ so the original equation becomes:
$(n^2+n+1)^2 - n^2 =m^2$
After doing that I'm kind of stuck...
I tried difference of squares and it gave me:
$(n+1)^2(n^2+1)=m^2$ and I got stuck here.

- And Gabriel, after following your method I arrived at:
$n+1/n=k$ so:
$k^2+2k=(m/n)^2$ and I didn't know what to do afterwards.

Is there some factoring trick that I don't know of???
My first try was to change this equation into the form $n^2-m^2=k$ where $k$ is a constant, but I don't know how to arrive at that.

[quote="10000th User"][quote="Gyan"]Sorry this hint almost gives it away .. but peek if you like:

[hide]what is  $(n^2+n)^2$  and what is $(n^2+n+1)^2$ ?

[hide]Can a perfect square exist between $k^2$ and $(k+1)^2$ ?
[hide]IOW if , you assume $k=n^2+n$  and thus there is a  quantity which is greater than $k^2$ (or $n^4+2n^3+n^2$) but less than $(k+1)^2$ (Or
$n^4+2n^3+3n^2+2n^2+1$  .. Can it be a perfect square? [/hide]
[/hide][/hide][/quote]
Those are not just hints, that's like 90% of the solution...  :rotfl:[/quote]
And I still don't get it... :(

You have to learn to let go of ideas coolly because your mind keeps thinking of difference of squares.

Gyan was NOT mainly suggesting difference of squares

$(n^2+n)^2=n^4+2n^3+n^2$

$n^4+2n^3+2n^2+2n+1=(n^2+n)^2+(n+1)^2$

$(n^2+n+1)^2=n^4+2n^3+3n^2+2n+1$

Compare.

[hide="Super hint OMG"]Think of consecutive squares...[/hide]

[quote="10000th User"]You have to learn to let go of ideas coolly because your mind keeps thinking of difference of squares.[/quote]

[hide]But couldn't you use difference of squares?

Because like Hellstar said, you'd get $(n+1)^2(n^2+1) = m^2$ after using that.  Taking the square root of both sides gives us $(n+1)\sqrt{n^2+1} = m$, and we can see that $n^2+1$ has to be a perfect square.  And then go from there.[/hide]

Many have already pointed it out,  but if it is still not clear ... (Excluding the case when n=0)

 ...t the point is there can not be a perfect square [b]between [/b]   $n^2$ and $(n+1)^2$  simply
because there is NO integer between two integers $n$ and $n+1$ .. (Or for matter any two consecutive integeres) .

The expression given is [b]greater [/b]than $(n^2+n)^2$    and[b] less [/b] than $(n^2+n+1)^2$  so $m$  has to be greater than $(n^2+n)$  and less than  $n^2+n$ +[size=150][color=red]1[/color][/size])  ...  (Notice that these two numbers are consecutive!) .. And that's impossible    Hope this helps.

$(n^2+n+1)^2$ can also be equal to $n^4+2n^3+2n^2+2n+1=m$ ...
dont forget that

Peeta -  yes, but  only if n=0  (and if one excludes this  case - [b]As I have noted in my post above [/b] - there are no other solutions...)

Thanks, the last three posts really helped. I didn't realize that $n$ had to be 0.
So the solutions $(m,n)$ are $(0,1)$ and $(0,-1)$?

When n = -1, the polynomial on the left has a value of 0. It seems to me that there are three solutions for (m,n). Namely, (0, -1), (1,0) and (-1,0).


We are given 4 lattice points A,B,C,D which forms a convex quadrilateral and is not a parallelogram. We start with those 4 points. We can obtain new points by taking the intersection of two lines, each of which passes through 2 of our given points. Can we obtain all rational points using this method?


invariant
induction
complex numbers
combinatorics solved
combinatorics

Hi everyone,
A quick problem that should be very easy....
Given n lamps in a circle, the initial configuration is that one is on and the rest are all off. At any stage you may take a lamp that is on, turn it off and at switch the states of the two adjacent lamps. For which n is it possible to have all lamps off after some number of moves?

I still maintain that this must be a relatively easy problem. It is obvious that the final configuration of all lamps off can be attained for n congruent to one mod 3.. .and apparently not for n 2,0 mod 3. It seems difficult to find an invariant so I believe induction may be the best method of approach.... any ideas on how to show that given that k will not give a final configuration with all lamps off, then neither will k +3?? Any help would be very, very much appreciated!!

That's exactly what I was trying to prove, but there's one problem (it took me LONG to see this): we can turn off all the lamps for n=5 and 5=2 (mod 3) so maybe there's another answer? Or maybe this is just an exception?

Im not very sure, but  id say this problem is in the book "Mathematical Chalenges" (Romanian Authors). The invariant was with complex numbers in their exponential form e^ki.
I'll check in the book and i will post the sol.

A complex proof does seem like it would work. Just to clarify, switching the states does not mean interchanging, but taking each in turn and switching its individual state.... just in case I had been unclear.
I'll look forward to a solution!

I claim that n works iff n-3 works. consider the lamps to be A!,...An around a circle in clockwise direction and lets assume that a1=1, ai=0 for i not 1. we know that we can reach the configuration: an=a1=a2=0,
and an-1=a3=1 and others zero. our aim is to show that the operations can now be performed on thesystem a3,...,an-1 with out any problem. the only problem that may be caused is when we want to perform the operation on the triple an-1,a3,a4 or an-2,an-1,a3 lets consider the first one.  a3 should be 1. Now notice that an=a1=a2=0 apply the operation on the triples (a2,a3,a4) then (a1,a2,a3) then (an,a1,a2) ,( an-1,an,a1) then
finally (a1,a2,a3) the result is just like performing the operation on (an-1,a3,a4) so lets consider a3,...,an-1 apply operation on an-1,a3,a4 so that we will end up with only a4 being on. Now we see that n works iff n-3 works.
n=4 works so  all n=1 mod 3 works.
n=5 works so all n=2 mod 3 works.
n=3 doen't work so n=0 mod 3 doesn't work.
so except multiples of 3 any n works.


geometry
circumcircle
cyclic quadrilateral
geometry proposed

Let $ ABCD$ be a cyclic quadrilateral. Let $ M\equal{}AC\cap BD, N\equal{}AB\cap CD, P\equal{}AD\cap BC$. Prove that $ MN$ is the polar of $ P$.

[quote="nguyentrang"]Let $ ABCD$ be a cyclic quadrilateral. Let $ M \equal{} AC\cap BD, N \equal{} AB\cap CD, P \equal{} AD\cap BC$. Prove that $ MN$ is the polar of $ P$.[/quote]
$ MN \cap AD \equal{}S$ and $ MN \cap BC \equal{}T$
Easy to prove $ (PSAD)\equal{}(PTBC)\equal{}\minus{}1$
so S,T is in the polar of P
so MN is the polar of P

Actually there's a nice proof... using a little of inversion help and Pascal's theorem. Try to find it  :)

It is enough to prove that $ OP\perp MN,$ where $ O$ is the circumcenter of $ ABCD.$

But it is true, if we consider the non convex cyclic quadrilateral $ ABDC.$

As well known: [b][size=100][color=DarkBlue]In every cyclic quadrilateral (convex or non convex), the line connecting the intersection point of its diagonals,
with the center of its circumcircle, is perpendicular to the line connecting the intersection points, of its opposite sidelines[/color][/size][/b].

An elementary proof of this result, has already been presented in the topic [url=http://www.mathlinks.ro/Forum/viewtopic.php?t=110887]Triangle, perpendicular lines[/url].

Kostas Vittas.


Prove that if $\alpha + \beta + \gamma = \pi$ and $\alpha_1 + \beta_1 + \gamma_1 = \pi$ then the folowing inequality is true: \[ \frac{cos \alpha_1}{sin \alpha} + \frac{cos \beta_1}{sin \beta} + \frac{cos \gamma_1}{sin \gamma} \le cot \alpha +cot \beta + cot \gamma \]

Use the following fact

[i]Fact[/i]: If $x,y,z$ are real and $\alpha_1 + \beta_1 + \gamma_1 = \pi$, then

$\sum yzcos{\alpha_1} \leq \frac{x^2+y^2+z^2}{2}$.

By the way, I believe your problem lacks the condition $sin{\alpha}.sin{\beta}.sin{\gamma} > 0$