Now, it seems a little bit more organized if we let

\[ \{\begin{array}{cc}
a=x^2+y^2+z^2 \\ \\
b=xy+yz+zx
\end{array}\]

the inequality turns out into

$ (a-b)(a+3b) \ge 0$

and we are done!

Let $ x^2\plus{}y^2\plus{}z^2\equal{}m$ and $ xy\plus{}yz\plus{}zx\equal{}n$ , then the inequality become

$ \frac{m\plus{}n}{2}\le\sqrt{m\plus{}2n}\sqrt{\frac{m}{3}}$

$ \left(\frac{m\plus{}n}{2}\right)^2\le\frac{m^2\plus{}2mn}{3}$

$ 3(m^2\plus{}n^2\plus{}2mn)\le4(m^2\plus{}2mn)$

$ 0\le m^2\minus{}3n^2\plus{}2mn$

$ 0\le (m\minus{}n)(m\plus{}3n)$ , which it true cause $ m\ge n$ by $ AM\minus{}GM$ and $ m\plus{}3n\ge0$

Let $ x\plus{}y\plus{}z\equal{}p$ and $ xy\plus{}yz\plus{}zx\equal{}q$. Then the inequality turns into $ \frac{p^2\minus{}q}{2}\le p \sqrt{\frac{p^2\minus{}2q}{3}}$. The last is equivalent to $ p^4\minus{}2p^2q\minus{}3q^2\ge 0$. Examining the last as quadratic inequality towards $ p^2$ we get roots $ \minus{}q$ and $ 3q$. So since we want to prove the inequality we need to prove $ p^2 \in \left(\minus{}\infty,\minus{}q\right)$ and $ p^2 \in \left(3q,\plus{}\infty\right)$. Since $ p^2\ge 0$ we only need to prove that $ p^2 \in \left(3q,\plus{}\infty\right)$ which is obvious since $ p^2\ge 3q$ is equivalent after expanding to $ \left(x\minus{}y\right)^2\plus{}\left(y\minus{}z\right)^2\plus{}\left(z\minus{}x\right)^2\ge 0$.

Since given inequality is homogeneous, W.L.O.G. we may assume that $ x^2 \plus{} y^2 \plus{} z^2 \equal{} 3$. Then it is enough to prove that:
\[ 2(x \plus{} y \plus{} z)\geq3 \plus{} xy \plus{} yz \plus{} zx \equal{} \frac {3 \plus{} (x \plus{} y \plus{} z)^2}{2},\]
or
\[ (\lambda \minus{} 1)(\lambda \minus{} 3)\leq0,\]
where $ \lambda \equal{} x \plus{} y \plus{} z$; which is true since $ x^2 \plus{} y^2 \plus{} z^2 \equal{} 3$. :)

[quote=sqing][quote=April]Let $ x$, $ y$ and $ z$ be non negative numbers. Prove that \[ \frac{x^2\plus{}y^2\plus{}z^2\plus{}xy\plus{}yz\plus{}zx}{6}\le \frac{x\plus{}y\plus{}z}{3}\cdot\sqrt{\frac{x^2\plus{}y^2\plus{}z^2}{3}}\][/quote]
[b]Strengthening[/b]
Let $ x$, $ y$ and $ z$ be non negative numbers. Prove that \[ x^2\plus{}y^2\plus{}z^2\plus{}xy\plus{}yz\plus{}zx\le (x\plus{}y\plus{}z)\cdot\sqrt{x^2\plus{}y^2\plus{}z^2+\frac{xy\plus{}yz\plus{}zx}{3}}\][/quote]

let $ x+y+z=3a, xy+yz+xz=3b^2 $ and $a\geq b$
$\iff 9a^2-6b^3+3b^2\leq 3a\sqrt{9a^2-6b^2+\frac{3b^2}{3}} \iff$
$(3a^2-b^2)^2\leq  a^2(9a^2-5b^2) \iff $
$b^4\leq a^2b^2 \iff b\leq a$ this is true.

[quote=sqing][b]KöMaL  2006[/b]
Let $ x$, $ y$  be positive real  numbers. Prove that \[ \frac{x^2\plus{}y^2\plus{}xy}{3}\le \frac{x\plus{}y}{2}\cdot\sqrt{\frac{x^2\plus{}y^2}{2}}\][/quote]

Let $ x+y=2a, xy=b^2 $ we have $a\geq b$
$\iff \frac{(4a^2-2b^2+b^2)}{3}\leq\frac{2a}{2}\sqrt{\frac{4a^2-2b^2}{2}} $
$\iff (4a^2-b^2)^2\leq 9a^2(2a^2-b^2) $
$\iff 2a^4\geq a^2b^2+b^4 $ this is true.

[quote=April]Let $ x$, $ y$ and $ z$ be non negative numbers. Prove that \[ \frac{x^2\plus{}y^2\plus{}z^2\plus{}xy\plus{}yz\plus{}zx}{6}\le \frac{x\plus{}y\plus{}z}{3}\cdot\sqrt{\frac{x^2\plus{}y^2\plus{}z^2}{3}}\][/quote]

By uvw method, we may check the case $z=0$ or $y=z$

[b]case1[/b]:$z=0$
The inequality is equivalent to

$x^4+y^4+2(x^3y+xy^3)\geq x^2y^2$

This holds by AM-GM inequality :D

[b]case2[/b]:$y=z$
The inequality is equivalent to

$x^4+5y^4+4x^3y\geq 6x^2y^2+4xy^3$

By AM-GM,

$x^4+y^4\geq 2x^2y^2$

By Rearrangement,

$y^3+x^3\geq x^2y+xy^2\rightarrow 4y^4+4x^3y\geq 4x^2y^2+4xy^3$

So we're done :D

[quote=sqing]
[b]Strengthening[/b]
Let $ x$, $ y$ and $ z$ be non negative numbers. Prove that \[ x^2\plus{}y^2\plus{}z^2\plus{}xy\plus{}yz\plus{}zx\le (x\plus{}y\plus{}z)\cdot\sqrt{x^2\plus{}y^2\plus{}z^2+\frac{xy\plus{}yz\plus{}zx}{3}}\][/quote]

By uvw method, we may check the case $z=0$ or $y=z$.

[b]case1[/b]:$z=0$
The inequality is equivalent to

$x^3y+xy^3\geq x^2y^2$

This holds by AM-GM inequality :D

[b]case2[/b]:$y=z$
The inequality is equivalent to

$\frac{2}{3}x^3y+\frac{1}{3}y^4\geq x^2y^2$

This holds by AM-GM.So we're done :D

[quote=sqing]
[b]Strengthening[/b]
Let $ x$, $ y$ and $ z$ be non negative numbers. Prove that \[ x^2\plus{}y^2\plus{}z^2\plus{}xy\plus{}yz\plus{}zx\le (x\plus{}y\plus{}z)\cdot\sqrt{x^2\plus{}y^2\plus{}z^2+\frac{xy\plus{}yz\plus{}zx}{3}}\][/quote]

Notice that $ (x+y+z)^2 = x^2 + y^2 + z^2 + 2(xy + yz + zx) $
Also, by AM-GM, $ x^2 + y^2 + z^2 \geq xy + yz + zx $

Let $ x^2 + y^2 + z^2 = a $
and $xy + yz + zx = b $

Now we know that $ (x+y+z)^2 = a +2b $
and we have $ a \geq b $

Now we may convert the whole original inequality into $ a,b $, which is:

$ \sqrt {a+2b} \cdot \sqrt {a+\frac{b}{3}} \geq a + b  $

Since both sides are non-negative we may square the inequality
$ \Rightarrow (a+2b)(a+\frac{b}{3}) \geq a^2 + 2ab + b^2 $
$ \Rightarrow a^2 + \frac{ab}{3} + 2ab + \frac{2b^2}{3} \geq a^2 + 2ab + b^2 $
$ \Rightarrow a + 2b \geq 3b $
$ \Rightarrow a \geq b $

Which we've already stated to be true by AM-GM

Let $s=x+y+z$ and $p=xy+xz+yz$. The inequality becomes $$s^2-p\leq 2s\sqrt{(s^2-2p)/3}$$ $$3(s^2-p)^2\leq 4s^2(s^2-2p).$$ This expands out to $$s^4-2s^2p-3p^2\geq 0.$$ Note that $p\leq \frac{1}{3}s^2$, so $$s^4=\frac{1}{3}s^2(2s^2+s^2)\geq p(2s^2+3p)\rightarrow s^4-2s^2p-3p^2\geq 0,$$ so we are done.


Gauss
geometry
vector
calculus
calculus computations

Anyone willing to help me with this?

http://www.math.sjsu.edu/~hsu/courses/112/p10.pdf

I don't think this is the right forum (it's for algebraic fields, not the physics kind) but anyway:

1. Let $A$ be the area of the base of the cylinder. It's the only part that matters because $E$ is perpendicular to the normal vectors $n$ everywhere else. Then the surface area is $2A$ (since it has two sides). Also the charge is $q = \delta A$. So by Gauss's Law, we have

$E \cdot 2A = 4\pi q = 4 \pi \delta A$

$E = 2 \pi \delta$.

2. Not completely sure how to work it out, but it's the same idea. Find the enclosed charge $q$. It'll be changing for $R \le 1$ and constant for all $R > 1$. Then apply Gauss's Law with surface area $4 \pi R^2$ to get

$E \cdot 4 \pi R^2 = 4 \pi q$

$E = \frac{q}{R^2}$

for whatever $q$ you find by integrating.


geometry
analytic geometry
vector
invariant
geometry proposed

Let ABCDEF is a convex hexagon . $ M , N , P , Q , I , K  $ are the central points of the segments $ AB , BC , CD , DE , EF , FA $ . Prove that :
   $ S(ACE) = S(BDF) <----> MQ , NI , PK $ are concurrent.

I have a solution by calculations using cross products; I have no time to post it now.

  Darij

I think that : to solve this problem , we should  use [b]algebraic area[/b] :

    $ I $ is the central point of the segment AB , $ \overline{s}(AMN) + \overline{s}(BMN) = 2\overline{s}(IMN) $ with any points M ,N .
    Darij , Can you post your solution?

Nttu, can you please post your solution? In fact, what I hope is that it's different from mine. My proof is a real torture to write up, it uses not only cross products but also a bit of barycentric coordinates, and though it has a few interesting ideas it is still connected with some calculations. I won't post it, but here is the main lemma it uses:

[b]Cross-product criterion for concurrent lines.[/b] Let O be an arbitrary point. If A, B, C are three non-collinear points in the plane, and AA*, BB*, CC* are three lines through these points, then these lines AA*, BB*, CC* concur if and only if there exist three vectors $\overrightarrow{a^{\prime }}$, $\overrightarrow{b^{\prime }}$, $\overrightarrow{c^{\prime }}$, parallel to the lines AA*, BB*, CC*, not all being equal to $\overrightarrow{0}$, but summing up to $\overrightarrow{0}$ - in other words, $\overrightarrow{a^{\prime }}+\overrightarrow{b^{\prime }}+\overrightarrow{c^{\prime }}=\overrightarrow{0}$ -, and satisfying the equation

$\overrightarrow{OA}\times \overrightarrow{a^{\prime }}+\overrightarrow{OB}\times \overrightarrow{b^{\prime }}+\overrightarrow{OC}\times \overrightarrow{c^{\prime }}=0$.

I'm using cross products for the first time, and this criterion (which I found in Dan Pedoe, [i]Geometry - A comprehensive course[/i], London 1970, paragraph 10.2) was quite new to me (I just remembered that there was such a criterion somewhere in Pedoe's book, and this is how I found it). The proof of the criterion is not really trivial (it uses a bit of barycentric coordinates anyway), and though it's a quite useful criterion I hope that there is a better solution of your problem.

Some remarks about the problem itself. At first, the convexity of the hexagon ABCDEF is not necessary, if one uses directed areas, of course. Here is a general formulation of the problem:

[b]Theorem 1.[/b] Let A, B, C, D, E, F be any six points in the plane, and let M, N, P, Q, I and K be the midpoints of the segments AB, BC, CD, DE, EF and FA, respectively. Then, the triangles ACE and BDF have the same directed area if and only if the lines MQ, NI and PK concur.

A special case, where the triangles ACE and BDF both have area 0, is:

[b]Theorem 2.[/b] Let A, C, E be three collinear points, and B, D, F three other collinear points. Let M, N, P, Q, I and K be the midpoints of the segments AB, BC, CD, DE, EF and FA, respectively. Then, the lines MQ, NI and PK concur.

Another corollary of Theorem 1 is:

[b]Theorem 3.[/b] Let A, B, C, D, E, F be any six points in the plane, and let M, N, P, Q, I, K, R, S and T be the midpoints of the segments AB, BC, CD, DE, EF, FA, AD, BE and CF, respectively. Then, the lines MQ, NI and PK concur if and only if the lines MT, PS and IR concur, if and only if the lines NR, QT and KS concur. And this holds if and only if the triangles ACE and BDF have the same area.

This, of course, follows from Theorem 1 after applying it to the points A, B, C, D, E, F, to the points A, B, E, F, C, D, and to the points B, C, F, A, D, E.

Thanks, Nttu, for this wonderful problem!

  Darij

I am sure we need geometrical solution, or, at least, solution without hard calculations.
For example consider following simple problem:
Let $ABCD EF$ be convex hexagon, $M$, $N$, $P$, $Q$, $I$ and $K$ are midpoints of corresponding sides. It is known that $MQ$, $NI$ and $PK$ divides area of $ABCD EF$ into two equal parts. Prove $MQ$, $NI$ and $PK$ are concurrent.

[b]Theorem[/b].  Let $a$, $b$, and $c$ be complex numbers.  Then the signed area of the triangle they form is
\[\frac{\bar{a} b + \bar{b} c + \bar{c} a - a \bar{b} - b \bar{c} - c \bar{a}}{4i}.\]

Now, suppose MQ, NI, and PK are concurrent.  WOLOG, assume that the point of concurrency is the origin.  Then
\[\frac{a + b}{d + e}\]
is real.  In other words,
\[{\frac{a + b}{d + e} = \frac{\overline{a + b}}{\overline{d + e}} = \frac{\bar{a} + \bar{b}}{\bar{d} + \bar{e}}},\]
so
\[a \bar{d} + a \bar{e} + b \bar{d} + b \bar{e} = \bar{a} d + \bar{a} e + \bar{b} d + \bar{b}e.\]
Similarly,
\[b \bar{e} + b \bar{f} + c \bar{e} + c \bar{f} = \bar{b} e + \bar{b} f + \bar{c} e + \bar{c} f, \\
c \bar{f} + c \bar{a} + d \bar{f} + d \bar{a} = \bar{c} f + \bar{c} a + \bar{d} f + \bar{d} a.\]

Adding the first and third equation and subtracting the second, we get
\[\bar{a} c + \bar{c} e + \bar{e} a - a \bar{c} - c \bar{e} - e \bar{a} = \bar{b} d + \bar{d} f + \bar{f} b - b \bar{d} - d \bar{f} - f \bar{b}.\]

By the above theorem, triangles ACE and BDF have equal areas.

Haven't yet tackled the converse.

Mr VTT gave one solution :
      We will prove that :   $ \forall O : \overline{s}(OMQ) + \overline{s}(ONI) + \overline{s}(OPK) = \frac{1}{4}[\overline{s}(ACE) - \overline{s}(BDF)] $
      We have :
  $ \overline{s}(OMQ) + \overline{s}(ONI) + \overline{s}(OPK) = \frac{1}{2} [\overline{s}(OMD) + \overline{s}(OME) + \overline{s}(ONE) + \overline{s}(ONF) + \overline{s}(OPA) + \overline{s}(OPF)]  = \frac{1}{4} [\overline{s}(OAD) + \overline{s}(OBD) + .... ]  =  \frac{1}{4} [ \overline{s}(ACE) - \overline{s}(BDF)] $
    So $ MQ , NI , PK $ are concurrent $ <----> $ $ S(ACE) = S(BDF) $

And how do you show that $ \forall O : \overline{s}(OMQ) + \overline{s}(ONI) + \overline{s}(OPK) = 0$ is equivalent to the concurrence of the lines MQ, NI and PK ? This is not difficult using the cross-product criterion for concurrent lines I gave in my former message, but do you have a synthetic proof?

  darij

Yes , I think this problem is really beautiful  .  I've never seen it in anybooks before . It's in my test .
   This problem is still right with [b]inconvex hexagon [/b]or [b]degraded hexagon[/b] . Darij mentioned about it in his post .

[quote="darij grinberg"]And how do you show that $ \forall O : \overline{s}(OMQ) + \overline{s}(ONI) + \overline{s}(OPK) = 0$ is equivalent to the concurrence of the lines MQ, NI and PK ? This is not difficult using the cross-product criterion for concurrent lines I gave in my former message, but do you have a synthetic proof?
[/quote]
Nttu proved that $\overline{s}(OMQ) + \overline{s}(ONI) + \overline{s}(OPK)$ is invariant relatively p. O, so if MQ, NI and PK are concurrent then they common point O' satisfies $\overline{s}(O'MQ) + \overline{s}(O'NI) + \overline{s}(O'PK)=0$ $\Rightarrow$ $ \forall O : \overline{s}(OMQ) + \overline{s}(ONI) + \overline{s}(OPK) = 0$.
If MQ, NI and PK are not concurrent then we can easily find point O'' s.t. equality $\overline{s}(O''MQ) + \overline{s}(O''NI) + \overline{s}(O''PK)=0$ isn't true (more precisely, we can provide all summands are positive).

And myths problem is an old Russian problem wich i posted somewhere in geometry proposed and own problems.

[quote="Myth"][quote="darij grinberg"]And how do you show that $ \forall O : \overline{s}(OMQ) + \overline{s}(ONI) + \overline{s}(OPK) = 0$ is equivalent to the concurrence of the lines MQ, NI and PK ? This is not difficult using the cross-product criterion for concurrent lines I gave in my former message, but do you have a synthetic proof?
[/quote]
Nttu proved that $\overline{s}(OMQ) + \overline{s}(ONI) + \overline{s}(OPK)$ is invariant relatively p. O, so if MQ, NI and PK are concurrent then they common point O' satisfies $\overline{s}(O'MQ) + \overline{s}(O'NI) + \overline{s}(O'PK)=0$ $\Rightarrow$ $ \forall O : \overline{s}(OMQ) + \overline{s}(ONI) + \overline{s}(OPK) = 0$.[/quote]

Myth, thanks! I really wasn't able to come up with this trivial point! Anyway, I wanted to note that the characteristic equation should be $\overline{s}\left( OMQ\right) +\overline{s}\left( OIN\right) +\overline{s}\left( OPK\right) =0$, and not $\overline{s}\left( OMQ\right) +\overline{s}\left( ONI\right) +\overline{s}\left( OPK\right) =0$. And now I also see that the other direction, namely to show that if every point O satisfies the equation $\overline{s}\left( OMQ\right) +\overline{s}\left( OIN\right) +\overline{s}\left( OPK\right) =0$, then the lines MQ, NI and PK concur, is also very easy: Define the point O' as the point of intersection of the lines MQ and NI; then, $\overline{s}\left( O^{\prime }MQ\right) =0$ and $\overline{s}\left( O^{\prime }IN\right) =0$, but on the other hand $\overline{s}\left( O^{\prime }MQ\right) +\overline{s}\left( O^{\prime }IN\right) +\overline{s}\left( O^{\prime }PK\right) =0$, so that $\overline{s}\left( O^{\prime }PK\right) =0$, and the point O' lies on the line PK. In other words, the lines MQ, NI and PK concur.

Another remark: If A, B, C, A', B', C' are six points in the plane, then the following three conditions are equivalent:

[b](1)[/b] The triangles ABC and A'B'C' have the same centroid.

[b](2)[/b] The vectorial identity $\overrightarrow{AA^{\prime }}+\overrightarrow{BB^{\prime }}+\overrightarrow{CC^{\prime }}=\overrightarrow{0}$ holds.

[b](3)[/b] For any point O in the plane, the sum of directed areas $\overline{s}\left( OAA^{\prime }\right) +\overline{s}\left( OBB^{\prime }\right) +\overline{s}\left( OCC^{\prime }\right) $ is constant, i. e. independent of the position of the point O.

Hence, for two such triangles (and in Nttu's problem, the triangles MIP and QNK are two such triangles), the equation $\overline{s}\left( OAA^{\prime }\right) +\overline{s}\left( OBB^{\prime }\right) +\overline{s}\left( OCC^{\prime }\right) =0$ holds for [i]some[/i] point O if and only if it holds for [i]every[/i] point O. This all can be shown using vector products, but it is not necessary to solve Nttu's problem.

  Darij

we can solve this problem using outer product and without Darij's Lemma:
With any $O$
$4(\vec{OM}\wedge\vec{OQ}+\vec{OI}\wedge\vec{ON}+\vec{OP}\wedge\vec{OK}) =(\vec{OA}+\vec{OB})\wedge(\vec{OD}+\vec{OE})+(\vec{OE}+\vec{OF})\wedge(\vec{OB}+\vec{OC})+(\vec{OC}+\vec{OD})\wedge(\vec{OF}+\vec{OA}) =2(\vec{OC}\wedge\vec{OA}+\vec{OA}\wedge\vec{OE}+\vec{OE}\wedge\vec{OC})-2(\vec{OF}\wedge\vec{OD}+\vec{OD}\wedge\vec{OB}+\vec{OB}\wedge\vec{OF}=\overline{S}_{AEC}-\overline{S}_{BFD}$
thus we have
$2(\vec{OM}\wedge\vec{OQ}+\vec{OI}\wedge\vec{ON}+\vec{OP}\wedge\vec{OK})= \overline{S}_{AEC}-\overline{S}_{BFD}$
now if we put $\{O\}=MQ\cap NI\Rightarrow \vec{OM}\wedge\vec{OQ}=0,\vec{OI}\wedge\vec{ON}=0\Rightarrow S_{ACE}=S_{BFD}\Leftrightarrow \vec{OP}\wedge\vec{OK}=0 \Leftrightarrow MQ,PK,NI$ are concurrent

I seek for a cleaner proof to it, so I revive it.


The points on this graph show the end-of-year sales amounts for each year. During what year after 1994 did sales increase the most number of dollars?  Express your answer as a 4-digit year.

[asy]unitsize(101/10);
real textsize = 10, largetext = 12;
draw((10,0)--(0,0)--(0,10),Arrows(4,20));
path verticalmark = (0,-.3)--(0,.3), horizontalmark = (-.3,0)--(.3,0);
draw(verticalmark);
for(int i = 1; i<10;++i)
	draw(shift(i*right)*verticalmark);
for(int i = 1; i<7; ++i)
	draw(shift(9*i/6*up)*horizontalmark);
for(int i = 0; i<3; ++i)
label(string(5(2i+1)),(-.25,9/6 + 9/3*i),left,fontsize(textsize*pt));
for(int i = 0; i<3; ++i)
label("'"+string(94+2i),(2i,-1),fontsize(textsize*pt));
label("'00",(6,-1),fontsize(textsize*pt));
label("'02",(8,-1),fontsize(textsize*pt));
label("\textbf{Year}",(5,-2.5),fontsize(largetext*pt));
label(rotate(90)*"\textbf{Millions of Dollars}",(-2.5,5),fontsize(largetext*pt));
pair A = (0,9/6*2), B = (1,9/6*2.4), C = (2,9/6*3), D = (3,9/6*3.25), E = (4,9/6*4.2), F = (5,9/6*4.5),G = (6,9/6*4.8),H = (7,9/6*5.2), I = (8,9/6*4.75),J = (9,9/6*3.25);
draw(A--B--C--D--E--F--G--H--I--J);
dot(A);dot(B);dot(C);dot(D);dot(E);dot(F);dot(G);dot(H);dot(I);dot(J);[/asy]

Since all the points are at an equal horizontal distance apart, we look for the steepest line segment, which occurs during the year $ \boxed{1998}$.


inequalities
inequalities unsolved
China

Let $ x$ and $ y$ be positive real numbers with $ x^3 \plus{} y^3 \equal{} x \minus{} y.$ Prove that \[ x^2 \plus{} 4y^2 < 1.\]

[quote="orl"]Let $ x$ and $ y$ be positive real numbers with $ x^3 \plus{} y^3 \equal{} x \minus{} y.$ Prove that
\[ x^2 \plus{} 4y^2 < 1.
\]
[/quote]

Since $ x,y \geq 0$ then $ x \minus{} y \equal{} x^3 \plus{} y^3 \geq 0$ or $ x \geq y \geq 0$
Consider $ y \equal{} 0$ separately.
If $ y > 0$ then
\[ x^2 \plus{} 4y^2 < 1
\]
Equivalent
\[ (x \minus{} y)(x^2 \plus{} 4y^2) \leq x^3 \plus{} y^3
\]or
\[ y(y^2 \plus{} (x \minus{} 2y)^2) > 0
\]

Another way:
Note that $x>y$

Let $x=ky$. Then equation becomes $(k^3+1)y^3=(k-1)y$ So $y^2=\frac{k-1}{k^3+1}$ and now $x^2+4y^2=(k^2+4)y^2=\frac{(k^3-k^2+4k-4)}{k^3+1}$ which can be easily verified as $<1$

[quote=sqing]\[ x^2 \plus{} y^2<x^2 \plus{} 4y^2 < 1.\]
The following inequality ( All-Russian Olympiad 1981) is also true.
Let $ x$ and $ y$ be positive real numbers with $ x^3 \plus{} y^3 \equal{} x \minus{} y.$ Prove that \[ x^2 \plus{} y^2 < 1.\]
[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?f=52&p=3223405]Strengthen of CGMO 2005[/url]
[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?f=52&t=558202&hilit=CGMO+2005]Inspired by CGMO 2005[/url]
[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?f=51&t=584040&p=3455944#p3455944]here[/url][/quote]
http://www.artofproblemsolving.com/community/c6h1388075p7728332

[quote=sqing]The following inequality ( All-Russian Olympiad 1981) is also true.
Let $ x$ and $ y$ be positive real numbers with $ x^3 \plus{} y^3 \equal{} x \minus{} y.$ Prove that \[ x^2 \plus{} y^2 < 1.\][/quote]
[url=https://artofproblemsolving.com/community/c893771h1880707p12794608]All Soviet Union MO 1981[/url]

Let $x,y>0 ,x^3+y^3=x-y$ and $x^2+4y^2<1.$ [url=https://artofproblemsolving.com/community/c6h1796634p11911107]Prove that [/url]$$x^2+5y^2 > 4xy$$

24. Let $x$ and $y$ be positive reals satisfying $x^3+y^3=x-y$. [url=https://artofproblemsolving.com/community/c130425h1133049p5274805]Find the maximum [/url]real $k$ such that for all $a<k$ have $x^2+ay^2<1$.

$\mathrm{(A) \ } 4\qquad \mathrm{(B) \ } 3\sqrt[3]{3}\qquad \mathrm{(C) \ } 2+2\sqrt{2} \qquad \mathrm{(D) \ } 2\sqrt{6} \qquad \mathrm{(E) \ } 5$
[url=https://artofproblemsolving.com/community/c6h554797p5244256]2015 Mock AMC10[/url]
Let $ a>b>0$ and $a^5 + b^5 =a-b.$ For $k<\frac{37}{3},$ [url=https://artofproblemsolving.com/community/c6h557847p13331315]prove that[/url]$$a^4 +kb^4 <1.$$


A bag contains 6 red marbles and 4 green marbles.  Joe reaches into the bag and draws out 2 marbles at the same time.  What is the probability that they are both red?  Express your answer as a common fraction.

[quote="math92"][hide]There are 10 marbles total in the bag.  The probablity of drawing a red marble is $\frac{3}{5}$ and the probablity of drawing a red marble again is $\frac{2}{3}$.  $\frac{3}{5} \cdot \frac{2}{3}=\boxed{\frac{2}{5}}$
[/hide][/quote]
He said that he draws teh marbles at the [b]same time[/b], not one by one.

Masoud Zargar

Sorry for double posting.  I think it should be $\left(\frac{3}{5}\right)^2=\frac{9}{25}$.

Masoud Zargar


geometry
3D geometry
rotation
geometric transformation
reflection
vector

For some reason the thread i made earlier was locked, so i started a new one. I was directed to [url]http://en.wikipedia.org/wiki/Pocket_Cube[/url] when i asked how many permutations there are for a Pocket version of the Rubik's Cube.

It says any permutation of the 8 corner cubies is possible (8 positions), and 7 of the cubies can be independently rotated ($ 3^{7}$ positions). There is nothing identifying the orientation of the cube in space, reducing the positions by a factor of 24.

What do they mean when they say 7 of the cubies can be independently rotated? I thought all of them can be rotated independently. And if there is nothing identifying the orientation of the cube in space, why would the number of positions be reduced by a factor of 24?

For the factor of 24: choose a given coloring.  There are 8 places to put a given corner, and 3 rotations of the cube with one fixed corner.  (In other words, the group of symmetries of a cube, not counting reflections, has 24 elements.)

Oh thanks.. i was thinking of 6 faces of a cube and 4 ways to orientate them, does it amount to the same thing? And i'd really like to know why just 7 of the cubies can be independently rotated =P

If you look at the discussion, it's not that 7 can be independently rotated, but rather the rotations are a 7-dimensional vector space of $ \mathbb{Z}/3\mathbb{Z}$. In other words, there are 7 rotations that are all independent of each other which have a period of 3, so there are $ 3^{7}$ rotations. (don't ask me what the 7 basis rotations are - I don't know)

[quote="aidan"]Oh thanks.. i was thinking of 6 faces of a cube and 4 ways to orientate them, does it amount to the same thing?[/quote]

Yes, absolutely.  (It's also twice the number of edges.)

Ok thanks =) But i think i'd have to read up on the 7-dimensional vector thing.


linear algebra
matrix
algebra
polynomial
linear algebra unsolved

Let $ A,B \in M_{n}(R)$ s.t $ A^3\plus{}A^4\equal{}O_{n}$
Prove that $ A^2B^2\plus{}AB\plus{}I_{n}$ is inverbile ,where $ B\equal{}A\plus{}I_{n}$

A matrix is invertible if and only if it does not have zero as an eigenvalue. If $ P(A)\equal{}0$ for some polynomial $ P,$ then the only possible eigenvalues of $ A$ are the roots of $ P$. Finally, for a polynomial $ Q,$ the eigenvalues of $ Q(A)$ as precisely $ Q(\lambda)$ where $ \lambda$ is an eigenvalue of $ A.$ (The latter conclusion may require an algebraically complete field, or at least that the characteristic polynomial of $ A$ splits over the given field.)

Since $ A^4\plus{}A^3\equal{}0,$ the only eigenvalues of $ A$ are $ 0$ and $ \minus{}1.$ Let $ Q(x)\equal{}x^2(x\plus{}1)^2\plus{}x(x\plus{}1)\plus{}1.$ We have $ Q(0)\equal{}1$ and $ Q(\minus{}1)\equal{}1,$ so the only possible eigenvalue of $ Q(A)$ is $ 1.$ Since that's not zero, $ Q(A)$ is invertible.

thx Kent Merryfield for solution...is beautiful

We can calculate the inverse explicitly: $ A^2B^2\plus{}AB\plus{}I\equal{}A^4\plus{}2A^3\plus{}2A^2\plus{}A\plus{}I\equal{}A^3\plus{}2A^2\plus{}A\plus{}I$, and $ (A^3\plus{}2A^2\plus{}A\plus{}I)(\minus{}A^2\minus{}A\plus{}I)\equal{}\minus{}A^5\minus{}3A^4\minus{}2A^3\plus{}I\equal{}I$.

This works over any field- it's just polynomial algebra, and we didn't have to divide by anything.


Determine the smallest positive integer, $n$, which satisfies the equation $n^3+2n^2 = b$, where $b$ is the square of an odd integer.

[hide="Answer"]$n^3+2n^2=n^2(n+2)=b=k^2$ for some odd integer $k$. This means that $n$ must also be odd. Taking the square root of both sides, we have $n\sqrt{n+2}=k$, so we must have $n+2$ a perfect square. Therefore, $\boxed{n=7}$.[/hide]


Demonstrati ca nu exista numere intregi nenule x,y,z astfel ca:

x^3+y^3=z^3  :)

Demonstreaza ca nu exista numere intregi nenule $x,y,z$ si $n\geq 3$ astfel ca $x^n+y^n=z^n$  :rotfl:  :rotfl:

Da-mi-l pe Fermat la telefon....io zic ca si pt puterea 3 e un inceput.... :)

Invocam spiritul lui Fermat poate ne zice solutia lui usoara pe care a gasit-o inaintea trecerii sale in nefiinta

Cred ca ar fi mai bine sa invoci spiritul lui Euler caci el e primul care a demonstrat teorema pentru $n=3$.Se pare ca in realitate Fermat nu avea demonstratia teoremei caci a cautat solutii pentru $n=4$ si $n=5$ plus ca e greu de crezut ca teorema rezolvata de Wiles pe 200 de pagini nemaivorbind de complexitatea solutiei(care anuiesc ca e mare) ar avea o rezolvare elementara. :play_ball:

Intradevar putin probabil sa exista "magica" solutie a lui Fermat. Pt n=3 rezolvarea este cat de cat normala?

pentru n=4 am citit pe undeva ca e o rezolvare elementara cred ca implica de fapt rezolvarea ansamblului x^2+y^2=z^2 care au multe solutii dupa cum pitagora si voi stiti...deja pt n=3 se face la faculta

Am citit în [1] cå Fermat a demonstrat cazul $n=3$, iar Euler cazul $n=4$. O demonstratie mai generala a ecuatiei $x^4+y^4=z^2$ se aflå [url=http://www.mathlinks.ro/Forum/viewtopic.php?t=97533]aici[/url] si nu e chiar asa de grea. Cât despre cazul general, nu se cunoaste nimic "elementar" in afara de cateva "fake-uri" prea putin reusite  :)


A math class of 20 students got their test papers back. The average was 54.0% (it was an easy test!). The professor tells everyone to adjust their mark according to the following formula:

New mark = min(100, old mark/0.9),

with all marks non-negative integers less than or equal to 100. What are the least upper and greatest lower bounds on the new average?

Από το http://www.math.toronto.edu/007/007-news.html#MDC

Αν υποθέσουμε ότι και οι  20 βαθμοί ήταν 54 τότε η μέση τιμή  μετά τη διώρθωση είναι 54. Για κάθε απόκλιση στη βαθμολογία των 20 βαθμών ( από τη μέση τιμή ) εύκολα παρατηρούμε ότι μετά τη διώρθωση έχουμε μεγαλύτερη νεα μέση τιμή, γεγονός που μας κάνει να συμπαιράνουμε ότι η ελάχιστη τιμή ( της μέσης τιμής )μετά τη διώρθωση είναι 54.
Όσο για τη μέγιστη έχω βρεί ότι είναι το 60 αλλά δεν έχω κάνει αυστηρή μαθηματική απόδειξη ... ( πιο πολύ διαισθητικά ).Αν κάποιος έχει κάποια τέτοια άς τη γράψει .

Tip.  Νομίζω ότι επειδή το 54 διαιρείται από το 9 και εμείς κάνουμε adjust διαιρώντας με 0,9 αυτό παίζει ρόλο  :)

Παραθέτω την απόδειξη για το μέγιστο: (χρησιμοποιώ ότι $ min(a,b) \equal{} \frac {a \plus{} b \minus{} |a \minus{} b|}{2}$)

Έστω $ a_1, a_2, ... a_{20}$ οι βαθμοί τότε

$ \frac {a_1 \plus{} a_2 \plus{} ... a_{20}}{20} \equal{} 54 \Leftrightarrow a_1 \plus{} a_2 \plus{} ... a_{20} \equal{} 20 \cdot 54$

$ 2 \cdot 20 \cdot$ (New Average) $ \equal{}$

$ \frac {10}{9}a_1 \plus{} 100 \minus{} |\frac {10}{9}a_1 \minus{} 100| \plus{} \frac {10}{9}a_2 \plus{} 100 \minus{} |\frac {10}{9}a_2 \minus{} 100| \plus{} ... \frac {10}{9}a_{20} \plus{} 100 \minus{} |\frac {10}{9}a_{20} \minus{} 100|$

$ \Leftrightarrow 36 \cdot (New Average) \equal{} a_1 \plus{} 90 \minus{} |a_1 \minus{} 90| \plus{} a_2 \plus{} 90 \minus{} |a_2 \minus{} 90| \plus{} ... a_{20} \plus{} 90 \minus{} |a_{20} \minus{} 90| \equal{}$

$ a_1 \plus{} a_2 \plus{} ... a_{20} \plus{} 20 \cdot 90 \minus{} |a_1 \minus{} 90| \minus{} |a_2 \minus{} 90| \minus{} ... |a_{20} \minus{} 90| \equal{}$

$ 20 \cdot 54 \plus{} 20 \cdot 90 \minus{} |a_1 \minus{} 90| \minus{} |a_2 \minus{} 90| \minus{} ... |a_{20} \minus{} 90| \equal{}$

$ 2880 \minus{} |a_1 \minus{} 90| \minus{} |a_2 \minus{} 90| \minus{} ... |a_{20} \minus{} 90| \Leftrightarrow$

$ 36(New Average) \minus{} 2880 \equal{} \minus{} |90 \minus{} a_1| \minus{} |90 \minus{} a_2| \minus{} ... |90 \minus{} a_{20}|$

$ \le \minus{} |90 \plus{} 90 \plus{} ... 90 \minus{} a_1 \minus{} a_2 ... \minus{} a_{20}| \equal{} \minus{} 20 |90 \minus{} 54| \equal{} \minus{} 20 \cdot 36 \equal{} \minus{} 720 \Leftrightarrow$

$ max(36(New Average) \minus{} 2880) \equal{} \minus{} 720 \leftrightarrow$

$ max(New Average) \equal{} \frac {2880 \minus{} 720}{36} \equal{} 60$

[quote="vangelismak"]Αν υποθέσουμε ότι και οι  20 βαθμοί ήταν 54 τότε η μέση τιμή  μετά τη διώρθωση είναι 54. Για κάθε απόκλιση στη βαθμολογία των 20 βαθμών ( από τη μέση τιμή ) εύκολα παρατηρούμε ότι μετά τη διώρθωση έχουμε μεγαλύτερη νεα μέση τιμή, γεγονός που μας κάνει να συμπαιράνουμε ότι η ελάχιστη τιμή ( της μέσης τιμής )μετά τη διώρθωση είναι 54.
[/quote]

Νομίζω υπάρχει ένα λάθος.
Δεν υπάρχει η δυνατότητα η καινούρια βαθμολογία να είναι 54. Ας υποθέσουμε αυτό γίνεται τότε, επειδή δεν υπάρχει μεταβολή στο σύνολο τον βαθμών, όλοι οι βαθμοί παραμένουν οι ίδιοι, άρα όλοι οι βαθμοί είναι 100 ή 0.

Έστω ότι είναι k, το πλήθος που έχει πετύχει 100 και m το πλήθος που έχει πετύχει 0 τότε:
$ k100 \plus{} 0m \equal{} 20\cdot54 \Leftrightarrow k \equal{} \frac{54}{5} \notin \mathbb{N}$
άρα δεν γίνεται να επιτευχθεί η τιμή 54.

Αυτό που έγινε σε αυτή τη λύση είναι να βρεθεί ένα κάτω φράγμα για τον καινούριο μέσο όρο,
όχι ο κατώτερος καινούριος μέσος όρος.

Έστω 20n το άθροισμα των αρχικών βαθμών των μαθητών που ο βαθμός τους αυξήθηκε τότε:
$ 100k \plus{} 0m \plus{} 20n \equal{} 20\cdot54 \Leftrightarrow k \equal{} \frac{54 \minus{} n}{5}(I)$

η αύξηση θα είναι ελλάχιστη όταν το 20n είναι ελλάχιστο, άρα όταν το k είναι μέγιστο άρα όταν (από $ (I)$) $ k \equal{} 10 \Leftrightarrow n \equal{} 4$

δηλαδή ο ελλάχιστος νέος μέσος όρος είναι ο:
$ \frac{100\cdot10 \plus{} 0 \plus{} \frac{10}{9}(20\cdot4)}{20}\equal{} 50 \plus{} \frac{40}{9} \equal{} 54.44$

Κακή διατύπωση  :(

Η νέα μέση τιμή είναι > 54.  ( δεν είναι το 54 η ελάχιστη τιμή )


NNNNEEEEEEDDDDDD!!!  Help! On these problems get back 2 me ASAP!

Give Explanation how to solve!

Suppose that four circular disks of equal radius are arranged so that we can barely fit a circle of radius 1 into the gap between these four circles.  What is the radius of one of these four circles?
Remember to give an exact answer--- no approximations!!!

The four circular disks are larger than the middle circle!!!

Would it be .5??  The raduis is 2 times the radius of the disc, right?

Sorry...I thought it was a square... :oops:  :blush:

I am not good at asymptote, so I will try to explain as best I can. Bear with me.

Draw the diagram. then draw the centers of the five circles. then draw the square around the four larger circles.

connect the center of the middle circle to the centers of the circle to the upper right and the lower right. This is an isoceles right triangle with sides 1+r, 1+r, and 2r. (r is the radius you are finding) If you are not convinced, extend the legs of the right triangle and you will see that they hit the corners of the square.

So, by the pythagorean thm, you have this equation:

$ 2(1\plus{}r)^2 \equal{} 4r^2$

$ 2r^2\plus{}4r\plus{}2 \equal{} 4r^2$

$ 2r^2\minus{}4r\minus{}2 \equal{} 0$

$ r^2\minus{}2r\minus{}1 \equal{} 0$

$ r\equal{}\frac{4\pm \sqrt{8}}{2}\equal{}\boxed{2\plus{}\sqrt{2}}$

[quote="buzzer11"]I am not good at asymptote, so I will try to explain as best I can. Bear with me.

Draw the diagram. then draw the centers of the five circles. then draw the square around the four larger circles.

connect the center of the middle circle to the centers of the circle to the upper right and the lower right. This is an isoceles right triangle with sides 1+r, 1+r, and 2r. (r is the radius you are finding) If you are not convinced, extend the legs of the right triangle and you will see that they hit the corners of the square.

So, by the pythagorean thm, you have this equation:

$ 2(1 \plus{} r)^2 \equal{} 4r^2$

$ 2r^2 \plus{} 4r \plus{} 2 \equal{} 4r^2$

$ 2r^2 \minus{} 4r \minus{} 2 \equal{} 0$

$ r^2 \minus{} 2r \minus{} 1 \equal{} 0$

$ r \equal{} \frac {4\pm \sqrt {8}}{2} \equal{} \boxed{2 \plus{} \sqrt {2}}$[/quote]
I think you got the problem mixed up, the four circles are SMALLER than the original circle.
Diagram:
[asy]draw(circle((0,0),.5));
draw(circle((0.207106781,0.207106781),0.207106781));
draw(circle((-0.207106781,0.207106781),0.207106781));
draw(circle((0.207106781,-0.207106781),0.207106781));
draw(circle((-0.207106781,-0.207106781),0.207106781));
draw((0,0)--(0.5,0));
label("$\frac{1}{2}$",(0.25,0),N);[/asy]

[quote="mathmath1"]Buzzer 11, The four circles surrounding the middle circle is much larger than the middle circle or they couldn't surround it![/quote]
?
Doesn't the circle with diameter 1 circumscribe the four circles?
Or did I interpret the question wrong...

so you have four circles with radius x and a circle that fits in the center with radius 1. So, make a right triangle with 3 of the circle's centers as vertexes.
so it has sides 2x, 2x, and 2x+2
so, 4x^2+4x^2=4x^2+8x+4
8x^2=4x^2+8x+4
4x^2-8x-4=0

(8±sqrt((-8)^2-4(4)(-4)))/2(4)

8±sqrt(64-(-64))/8
8±sqrt128/8
8±8sqrt2/8
1±sqrt2

must be plus since distance can't be negative
1+sqrt2

[quote="mathmath1"]Buzzer 11, The four circles surrounding the middle circle is much larger than the middle circle or they couldn't surround it![/quote]

That is what I did, i think.

Same with that swimming pool problem... This is not just a homework help site! We can help explain, but you should be able to think it through yourself once we've finished.

Why would i have a homework assignment on a Saturday? Come on get a life.

Think about it.
We spend our time solving this problem for you, simply because you either don't know how to solve it, or are too lazy to, or are simply taking advantage of us.
And then you aren't satisfied and start screaming "EXPLAIN YOUR ANSWERS!"
And then you say we don't have a life because we accuse you of this being your homework.
Is that how you ask for help?

[quote="ZhangPeijin"][quote="mathmath1"][quote="ZhangPeijin"]YOU SHOULD BE ABLE TO UNDERSTAND MY ANSWER IF YOU TRIED!
Anyways, I suspect this is a homework assignment you don't wish to do.[/quote]

and how or what do you do to go from r=1/[(sqrt2) -1] and then r=[(sqrt2)+1]/2[/quote]
It's called simplifying.
And tell me, why should I help you if you don't appreciate my help?[/quote]

I appreciate your help, but can you explain how that is simplified?

[quote="mathmath1"][/quote][quote="ZhangPeijin"][quote="mathmath1"][quote="ZhangPeijin"]YOU SHOULD BE ABLE TO UNDERSTAND MY ANSWER IF YOU TRIED!
Anyways, I suspect this is a homework assignment you don't wish to do.[/quote]

and how or what do you do to go from r=1/[(sqrt2) -1] and then r=[(sqrt2)+1]/2[/quote]
It's called simplifying.
And tell me, why should I help you if you don't appreciate my help?[/quote][quote="mathmath1"]

I appreciate your help, but can you explain how that is simplified?[/quote]
That was easy.
$ \frac{1}{\sqrt{2}-1}=\frac{\sqrt{2}+1}{(\sqrt{2}-1)(\sqrt{2}+1)}=\frac{\sqrt{2}+1}{1}$

[quote="ZhangPeijin"][/quote][quote="mathmath1"][/quote][quote="ZhangPeijin"][quote="mathmath1"][quote="ZhangPeijin"]YOU SHOULD BE ABLE TO UNDERSTAND MY ANSWER IF YOU TRIED!
Anyways, I suspect this is a homework assignment you don't wish to do.[/quote]

and how or what do you do to go from r=1/[(sqrt2) -1] and then r=[(sqrt2)+1]/2[/quote]
It's called simplifying.
And tell me, why should I help you if you don't appreciate my help?[/quote][quote="mathmath1"]

I appreciate your help, but can you explain how that is simplified?[/quote][quote="ZhangPeijin"]
That was easy.
$ \frac {1}{\sqrt {2} - 1} = \frac {\sqrt {2} + 1}{(\sqrt {2} - 1)(\sqrt {2} + 1)} = \frac {\sqrt {2} + 1}{1}$[/quote]

Tyfor explaining

Just one question; how do you get from r*(sqrt2)-r=1 to r= 1/[(sqrt2)-1]?

[quote="mathmath1"][/quote][quote="ZhangPeijin"][/quote][quote="mathmath1"][/quote][quote="ZhangPeijin"][quote="mathmath1"][quote="ZhangPeijin"]YOU SHOULD BE ABLE TO UNDERSTAND MY ANSWER IF YOU TRIED!
Anyways, I suspect this is a homework assignment you don't wish to do.[/quote]

and how or what do you do to go from r=1/[(sqrt2) -1] and then r=[(sqrt2)+1]/2[/quote]
It's called simplifying.
And tell me, why should I help you if you don't appreciate my help?[/quote][quote="mathmath1"]

I appreciate your help, but can you explain how that is simplified?[/quote][quote="ZhangPeijin"]
That was easy.
$ \frac {1}{\sqrt {2} - 1} = \frac {\sqrt {2} + 1}{(\sqrt {2} - 1)(\sqrt {2} + 1)} = \frac {\sqrt {2} + 1}{1}$[/quote][quote="mathmath1"]

Tyfor explaining

Just one question; how do you get from r*(sqrt2)-r=1 to r= 1/[(sqrt2)-1]?[/quote]
You can write $ r\sqrt{2}-r$ as $ r(\sqrt{2}-1)$
So it's just basically.
$ r\sqrt{2}-r=1$
$ r(\sqrt{2}-1)=1$
$ r=\frac{1}{\sqrt{2}-1}$
$ r=\frac{\sqrt{2}+1}{1}$
$ \boxed{r=\sqrt{2}+1}$

[color=indigo]Moderator's Notice: Please do not quote the entire immediate post above yours.[/color]

Why do change r to 1?

Because $ r\cdot \minus{}1 \equal{} \minus{}r$

[quote="ZhangPeijin"]Because $ r\cdot \minus{} 1 \equal{} \minus{} r$[/quote]

You know r=1 doesn't equal r*sqrt2 :arrow:  :!:

I realized tht yesterday!

I believe he was using the property of Distribution  :wink: .


calculus
integration
function
ratio
logarithms
topology
real analysis

Let $f$ be non-negative continious function on $[0,+\infty]$, such that $\int_{0}^{\infty}f(x)dx=\infty$. Prove that there exists $a>0$ such that $\sum_{n=1}^{\infty}f(na)=\infty$.

Nice problem :).

We want to find an $a>0$ such that for every $n\ge 1$ we have some $p\ge 0$ with $f(na)+f((n+1)a)+\ldots+f((n+p)a)>1\ (*)$. Of course, we can find such an $a$ for $n=n_1=1$. The set of $a$ satisfying $(*)$ for a fixed $p$ is an open set, so we can actually find an entire compact interval $I_1$ consisting only of such $a$. Now, within the interval $I_1$, we look for $a$ which satisfy $(*)$ for a pair $(n_2,p)$ with $n_2\ge 2$. Suppose we find such $a$ again; then we'll have a non-degenerate compact interval $I_2\subseteq I_1$. Now we repeat the process. If at each step we're able to find our $a$'s, the non-empty intersection of the intervals $I_j$ will contain an $a>0$ like the one we're looking for. This means that it suffices to prove the following (in our hypotheses, namely that $f: [0,\infty)\to[0\infty)$ is continuous, and its integral over the non-negative reals is $\infty$):

For every non-degenerate interval $I=[a,b]\subset(0,\infty)$ and any $n\in\mathbb N^*$ we can find $x\in I$ such that there is a $p\ge 0$ with $f(nx)+\ldots+f((n+p)x)>1$.

Suppose the contrary. This means that there is some $M>0$ such that for all $x\in I$ we have $f(x)+f(2x)+\ldots<M$. Integrating this over $I$, we get $\sum_{m\ge 1}\frac 1m\int_If\ \mbox dt<M(b-a)\ (**)$. Now let's see, for an arbitrary $n$, what we can say about the number of times the interval $[n,n+1]$ is covered by intervals $mI$. In order for this to happen, we must have $ma\le n<n+1\le mb$, which is equivalent to $\frac{n+1}b\le m\le\frac na$. The smallest $m$ for which this happens (call it $m_n$) is at most $\frac{n+1+b}b$, and the largest one is at least $\frac{n-a}a$. Their ratio is thus at least $\frac ba\cdot\frac{n-a}{n+1+b}$. For any $r\in\left(1,\frac ba\right)$, when $n$ is large enough this ratio is at least $r$and, the coefficient in front of $\int_n^{n+1}f\ \mbox dt$ that appears in $(**)$ will be at least $\frac 1{m_n}+\ldots+\frac 1{\lceil ing rm_n\rceil ing}$. As $n\to\infty$ this goes to $\ln r>0$. This means that we can find a positive integer $N$ such that the sum $(**)$ is at least $C\int_N^\infty f\ \mbox dt=\infty$ (where $C>0$ is a constant), and we have a contradiction.


[quote="ummagumma"]Find all the integer solution of the equation:

p^x - y^p = 1

where p is a prime and x,y are integer[/quote]
If you use dollar signs... [code]$p^x - y^p = 1$[/code] then you get $ p^{x}\minus{}y^{p}\equal{} 1$.

See http://www.artofproblemsolving.com/Forum/viewtopic.php?t=161505

In fact, this can be nicely generalized asking for any triple $ (x,y,z)$ of [i]integers[/i] such that $ x^{p}\plus{}y^{p}\equal{} p^{z}$, where $ p\in\mathbb{N}$ is a given prime. So solve it! :wink:


geometry
incenter
angle bisector
geometry unsolved

The incircle of the triangle $\triangle{ABC}$ has center at $O$ and it is tangent to the sides $BC$, $AC$ and $AB$ at the points $X$, $Y$ and $Z$, respectively. The lines $BO$ and $CO$ intersect the line $YZ$ at the points $P$ and $Q$, respectively.

Show that if the segments $XP$ and $XQ$ has the same length, then the triangle $\triangle ABC$ is isosceles.

A simple angle computation shows that $\angle BPZ=\frac{\angle ACB}2$. This means that $\angle OPY+\angle OCY=\pi$, i.e. $COPY$ is cyclic. Since $OY\perp CY$, we get $BP=OP\perp CP$. In the same way we get $CQ\perp BQ$, and this means that if $D=BQ\cap CP$, then $O$ is the orthocenter of $BCD$, whereas $XPQ$ is the orthic triangle of $BCD$, i.e. $O$ is the incenter of $XPQ$.

Assume now that $XP=XQ$. Then $XO$, which is the angle bisector of $\angle PXQ$, is also the $X$-altitude of the isosceles triangle $XPQ$, i.e. $XO\perp PQ=YZ$. But $YZ\perp AO\Rightarrow AO\|XO\Rightarrow AO\perp BC$, i.e. the $A$-bisector of $ABC$ is also its $A$-altitude. We thus get $AB=AC$.

http://www.cut-the-knot.org/wiki-math/index.php?n=MathematicalOlympiads.IbAMO2001Problem2

Vo Duc Dien

If $M$ si the midpoint of $BC$ then $MP=MQ$ then $X=M$ and we are done.

We claim that $\triangle QOX \cong \triangle QOY$. To see this, we observe that $OX \cong OY$ and $\angle QOX = 180° - \angle XOC = 180° - \angle COY = \angle YOZ$. From this it follows that $\angle OXQ = \angle QYO =\frac{\angle A}{2}$. Analogously we have $\angle OXP = \frac{\angle A}{2}$. Next we observe that $\angle QXB = 90°-\angle OXQ = 90°-\angle OXP = \angle CXP$ so $2\angle QXB=180°-\angle PXQ$ which is equal to $\angle XQP$ if we assume $XQ \cong XP$. This result implies that $ZY \| BC$, so $\triangle ABC \sim \triangle AZY$, which is known to be an isosceles triangle. $\square$


I have a while whit this problem and i haven´t  solve it:
There is a square zoo of side n. You put vegetarian and predators on it. To avoid stress on animals, no animal can border mostly whit predators (For example, animals at corners can´t border whit more than 1 predator, while animals at the center can´t whit more than two) What is the highest number of predators that can be at the zoo?
(Two animals border if they have a common segment in their cages)

i don't get it.. i mean how big are the cages, or how do you construct cages

I think he meant :

You have a $n \times n$ squares. You want to colour black some of the unit squares (the predators), but any square must have at most half of its neighbours coloured black.
What is the maximum number of black unit squares?

Pierre.

Yeah, I´ll try to redact it better for the next time. That was what i meant.
Anyway, has someone sove it? (And why did they move the topic, i tought that it was on Open Questions)

I moved it because it does not really looks like an open question...but more like an unsolved problem.
Please read the aim of the forum before posting something somewhere.

Pierre.

Just some thoughts on how to find bounds: look at the edges and put to the edges the colors of the faces it belongs to. Now every square has at most $2$ white edges on its boundary for all squares not the ones in the corners. and the ones in the corners have at least $1$ white square. Now every black square adds $4$,$3$ or $2$ balck edges, so here we can bound the number of black squares from above.


show that the equation (x^2)+(y^2)+(z^2)=n has no solution if n=(4^k)(8m+7) for any k,m in Natural number + {0}

thank you

You may want to rethink that part where you say that $ k,m$ are real numbers  :wink:
And (in the correct version) posted before.


geometry
incenter
circumcircle
parallelogram
rhombus
geometry proposed
U sviti matematyky

Let $ I$ be the incenter of triangle $ ABC$. The circumcircles of triangles $ AIC,AIB$ intersect sides $ AB,AC$ at $ K,N$ respectively. Let $ M$ be an arbitrary point on segment $ KN$. Prove that the sum of the distances from $ M$ to the sides of triangle $ ABC$ does not depend on the choice of $ M$.

Are you sure that this is the statement from the journal? I see two values for $ M \equiv K$ and $ M \equiv N$.

Most likely, $ \triangle ABC$ is not general, but such that circles $ \odot(AIC), \odot(AIB)$ intersect side segments $ AB, AC$ (not just sidelines) again at $ K, N.$ However, the proposition holds for a general $ \triangle ABC$ and for the line $ KN$ (not just segment), if the distances of $ M$ from $ BC, CA, AB$ are properly directed: Let $ D, E, F$ be feet of perpendiculars from $ M \in KN$ to $ BC, CA, AB.$ Let $ \overline{MD} > 0$ when $ M$ is on the same side of $ BC$ as $ A$ and $ \overline{MD} < 0,$ when $ M$ is on the opposite side of $ BC$ than $ A.$ $ \overline{ME}, \overline{MF}$ are directed similarly.
_____________________________________

Since $ \odot(AIC), \odot(AIB)$ are centered on the internal bisectors of the angles $ \angle B, \angle C,$ $ \overline {BK} \equal{} \overline{BC}$ and $ \overline {CN} \equal{} \overline{CB}.$ Then $ \overline{AK} \equal{} c \minus{} a$ and $ \overline {AN} \equal{} b \minus{} a.$ ($ K, N$ are on the segments $ AB, AC$ if $ a \equal{} BC \equal{} BK < BA \equal{} c$ and $ a \equal{} CB \equal{} CN < CA \equal{} b,$ $ a$ is the shortest triangle side.) Let $ X \in BC, Y \in CA, Z \in AB$ be feet of the external bisectors of the $ \angle A, \angle B, \angle C.$ From $ \overline{AY} \equal{} \frac {bc}{c \minus{} a}, \overline {AZ} \equal{} \frac {bc}{b \minus{} a},$ we get $ \frac {\overline {AN}}{\overline {AK}} \equal{} \frac {b \minus{} a}{c \minus{} a} \equal{} \frac {\overline {AY}}{\overline {AZ}},$ which means that the lines $ KN \parallel XYZ$ are parallel.
_____________________________________
_____________________________________

Let $ P \in XYZ$ be arbitrary and $ D, E, F$ feet of perpendiculars from $ P$ to $ BC, CA, AB.$ WLOG, assume $ P$ is on the finite segment $ YZ,$ so that $ \overline {PE}, \overline {PF} > 0$ and $ \overline {PD} < 0.$ (Proof for a segment of $ XYZ$ containing the infinite point of this line is the same, except 2 directed distances are negative and one positive.) Let perpendicular to the internal bisector $ AI$ of the angle $ \angle A$ through $ P$ cut $ AC, AB$ at $ P_b, P_c.$ $ P_bP_c$ is locus of points $ P',$ such that $ \overline{P'E'} \plus{} \overline {P'F'} \equal{} \text{const}$ (where $ E', F'$ are feet of perpendiculars from $ P'$ to $ AC, AB$). Let $ F_b$ be foot of the perpendicular from $ P_b$ to $ AB,$ so that $ \overline {PE} \plus{} \overline {PF} \equal{} \overline {P_bF_b}.$ The $ \triangle P_bPY \sim \triangle AXY$ are centrally similar, having parallel sides, with similarity center $ Y.$ Let a parallel to $ BC$ through $ P$ cut $ AC, AB$ at $ V, W$ and let a parallel to $ AB$ through $ P_b$ cut $ BC$ at $ U.$ Let the lines $ VPW, P_bU$ intersect at $ Q.$ The $ \triangle P_bQV \sim \triangle ABC$ are again centrally similar with the same similarity center $ Y,$ $ Y \in BQ.$ Thus the external bisector $ BYQ$ of the angle $ \angle B$ is the diagonal of the parallelogram $ UBWQ,$ it is a rhombus and

$ \overline {PD} \plus{} \overline {PE} \plus{} \overline {PF} \equal{} \overline {PD} \plus{} \overline {P_bF_b} \equal{} 0.$

Let $ I_a, I_b, I_c$ be the excenters of the original $ \triangle ABC.$ Let an arbitrary parallel to $ XYZ$ cut $ BC, CA, AB$ at $ X', Y', Z'.$ Let parallels to $ AX, BY, CZ$ through $ X', Y', Z'$ pairwise intersect at $ I_a^*, I_b^*, I_c^*.$ The $ \triangle I_a^*I_b^*I_c^* \sim \triangle I_aI_bI_c$ are centrally similar, having parallel sides. The $ \triangle AY'Z' \sim \triangle AYZ$ are centrally similar with similarity center $ A,$ hence $ I_a^* \in AI_a$ and likewise, $ I_b^* \in BI_b, I_c^* \in CI_c.$ Therefore, the incenter $ I$ is the similarity center of the $ \triangle I_a^*I_b^*I_c^* \sim \triangle I_aI_bI_c.$ Their orthic triangles $ \triangle A^*B^*C^* \sim \triangle ABC$ are centrally similar with the same similarity center. The line $ X'Y'Z'$ therefore cuts the lines $ B^*C^* \parallel BC,$ $ C^*A^* \parallel CA,$ $ A^*B^* \parallel AB$ at the feet $ X^*, Y^*, Z^*$ of the external bisectors of the $ \angle A^*, \angle B^*, \angle C^*$ of the $ \triangle A^*B^*C^*.$  Let $ M \in X'Y'Z' \equiv X^*Y^*Z^*$ be arbitrary, let $ D', E', F'$ be the feet of perpendiculars from $ M$ to $ BC, CA, AB$ and $ D^*, E^*, F^*$ the feet of the same perpendiculars in $ B^*C^*, C^*A^*, A^*B^*.$ Then

$ \overline{MD'} \plus{} \overline{ME'} \plus{} \overline{MF'} \equal{}$

$ \equal{} (\overline{MD^*} \plus{} \overline{D^*D'}) \plus{} (\overline{ME^*} \plus{} \overline{E^*E'}) \plus{} (\overline{MF^*} \plus{} \overline{F^*F'}) \equal{}$

$ \equal{} \overline{D^*D'} \plus{} \overline{E^*E'} \plus{} \overline{F^*F'} \equal{} \text{const}$

which is a constant independent of $ M \in X'Y'Z' \parallel XYZ.$
_____________________________________

The above result can be used in solving the following problems:
[url]http://www.artofproblemsolving.com/viewtopic.php?t=209599[/url]
[url]http://www.artofproblemsolving.com/viewtopic.php?t=209657[/url]


calculus
integration
quadratics
inequalities
real analysis
geometry
3D geometry

What would have been your favouite CHoice of Pattern for IITJEE..

During the 70's ,in the maths paper u would have two choices: New Maths or OLD maths
New Maths was CALCULUS
Old was ALGEBRA ORIENTED

20 sums , 3 hours , 11-12 cutoff.. THIS IS MY FAVOURITE

id go for old maths anyday :D

Well that's surely better with no need to study :)

but in phy i would have been bowled over so no need to brag too much :D

well--there shud be 150 ques.--3 hrs
And cut off at about 100-110

questions- infinity
time - 2*infinity
cutoff-0000

I heard that old MAths was very tough... So tough that the good students who prepared for the old, changed to the new pattern after seeing the paper

calculus depends on students--few r good at it -some r not--but i think old maths cud have been improved by practice

can you post example questions of the old math thingy? I dont think it can be sooo tough n all n i bet it would dearly suit chappli, soumya, vihang n pardesi n other olympiad guys :)

[quote="Nitishexpress"]@rituYOU r right... no one can be like MADNESS[/quote]

What does that mean?  :huh:

i secoond amdness. can neone post questions or atleast gimme a link for the 70's papers

This was an info i got from my father who took the iitjee in the 70's

and there ar others also who r good at both--like rohit flashed the names just before

but i think these olympiad guys will have a greater edge in the old pattern n theyll wipe through the paper :D :D

[quote]Nitishexpress wrote:
@rituYOU r right... no one can be like MADNESS

What does that mean?

[/quote]

Only u could go without studying

DONOT UNDERESTIMATE THE BABY,,, he would not even be happy with a AIR 1

my personal feeling is that calculus based papers involve application of formulae, thats all, atleast at jee level, but you wud be require talent ofr algebra, especially combinatorics and prob.

I strongly disagree with that, every portion requires soome basic formulas, even algebra has..,, there some great manupilations u can do in calculus sums,

no algebra doesnt require any formulae at all :) thnx for agreeing wid me on that varun, even i too feel as i have said a million times that calculus is very mechanical :)

Even though there are some good sums in calculus, how did we ever learn integration?? linear root quadratic n all, there's already a well defined procedure for that while for algebra, there's no such thing :)

i wud also agree there aren't any formulas in algebra--only pre-requisite understanding is req.--but prob. has some basic formulae--like
P=fav. outcomes/total outcomes!!!!! :D  :D

aw sheddup, one of the most logical chapters is combinatorics :D :D

i would obviously prefer AIEEE pattern for IIT too with only single choice questions!!!!! :D
easier to guess u see!!!! :rotfl:

@ae--checkout readnow's post on this topic :D  :rotfl:

Any other Nice  patterns......
I have always taught having exams based on ur stream would be the best...
Why would i have to study something which wont be useful for me.
This would be apllicable to many in AOPS

[quote="Euclidean Geometer"]@ Madness and Varun:
Indeed it is quite silly to say dat calculus is formula based. <No offence meant  :) > Please try solving the calculus problems i ve been posting on the jee thread, which are indeed within  the bounds of our jee syllabus. i don mean to say algebra is bad. Algebra rocks  :D  Especially my fav topic being theory of equations and inequalities :coolspeak:  But Calculus isnt as mechanical as u seem to be thinking.[/quote]

but u ought to agree that most of the integration sums have a std technique :D :)

and the 1st prin. proofs as well(though it is not be concerned with jee)

Yeah may be at this level we ve not really innovated with calculus. Algebra as of now has offered to us a lot more imagination. And needless to say geometry. But do you use apostol for calculus  :?: Its awesome.

yes i have it but Rudin is far better but apostol is better from jee point of view.. :)

I enjoy all spheres of mathematics except geometry. (pun intended). ;) Though combinatorics is my personal favorite, I give equal weightage to algebra and calculus.

Oh geometry in all its manifestations is the love of my life  :lol:
and if u ask ppl who know me really well (like alter ego or pradeep mathias), u ll probably learn dat i am kinda wierd  :!:  i like stuff which is generally disliked by others  :D

no surely not....ask soumya...i like geometry and inequalities to the core(in fact the only thing i like olympiads for) and till very recently disliked combi :rotfl:

calculus is the most wonderful subject..(esp integral calculus)... and i am dissapointed that our sir having taught calculus very well and still his students dont like that subject...

i don mean to say i hate algebra though... i am just saying i love calulus better...

[quote="Euclidean Geometer"]Books :rotfl:[/quote]

no, authors of books :rotfl:  :rotfl:  :rotfl:  :P

I like the pattern to be something like a mixture of OLD MATHS and NEW MATHS with both the objective and subjective parts like the papers in the early 2000's. :lol:  :D  I like those kinds of mixtures which test your knowledge and aptitude. :wink:

Yes i also think they were the best--REAL TESTERS


If cos(x)+cos(y)+cos(z)=sin(x)+sin(y)+sin(z)=0, the find the exact value of

cos(x-y)+cos(y-z)+cos(z-x)

i have an answer but not a solution.

[hide]
let x = 0 degrees
let y = 120 degrees
let z = 240 degrees

verify that the sums are equal

cos(x-y) + cos(y-z) + cos(z-x) = cos(-120) + cos(-120) +cos(240)
= -.5 + -.5 + -.5
=-1.5
[/hide]

Square the equation and add both sides still equal zero.  Then add both sides, which equates to zero.
$ cos^2(x)\plus{}cos^2(y)\plus{}cos^2(z)\plus{}2(cos^2(x)cos^2(y)\plus{}cos^2(y)cos^2(z)\plus{}cos^2(z)cos^2(y))\plus{}sin^2(x)\plus{}sin^2(y)\plus{}sin^2(z)\plus{}2(sin^2(x)sin^2(y)\plus{}sin^2(y)sin^2(z)\plus{}sin^2(z)sin^2(x))\equal{}0$
Then combine terms using $ cos^2(x)\plus{}sin^2(x)\equal{}1$ and $ cos(x\minus{}y)\equal{}cos^2(x)cos^2(y)\plus{}sin^2(x)sin^2(y)$.  If our desired sum is $ S$, then
$ 3\plus{}S\equal{}0\Longrightarrow S\equal{}\minus{}\frac{3}{2}$

@ZzZzZzZzZzZz: You made a tiny mistake. It should be $ 3\plus{}2S\equal{}0$, not $ 3\plus{}S\equal{}0$.

You've written your cosine angle identity incorrectly.  There's a lot going on there; you should check your work even if your final answer is correct because a lot of the expressions involved evaluate to $ 0$.

[hide="Alternately..."] We could use vectors!  We have three unit vectors that sum to zero; they clearly form an equilateral triangle.  We want to find the sum of their pairwise dot products, which is therefore $ 3 \cos 120^{\circ} \equal{} \minus{} \frac {3}{2}$.

A similar solution is possible with complex numbers.  :)[/hide]

[quote="t0rajir0u"]they clearly form an equilateral triangle[/quote]

I don't immediately see this, could you elaborate?

If three unit vectors $ a, b, c$ sum to zero, then the triangle $ 0, a, a \plus{} b$ is a triangle with side lengths $ |a| \equal{} 1, |b| \equal{} 1, |c| \equal{} 1$; equilateral.


algebra
binomial theorem
combinatorics unsolved
combinatorics

We know :
 $\sum_{n=1}^r S(r,n) =$ the number of partition of {1,2,...,r}
 Put $B_r=\sum_{n=1}^r S(r,n)$
 Show and prove the explicit formula of $B_r$ ?

plz give a more explanation about your problem. As i know stirling have two different kind numbers. The first kind is defined as the number of ways to arrange $r$ distinct objects around $n$ identical circles such that each circle has at least one object and the second kind is defined as the number of ways of distributing $r$ distinct objects into $n$ identical boxes such that no box is empty. If your meant was the first kind i think the answer is $r!$ and if your meant was the second kind i don't know what is the answer of it. :?

Do you mean this: http://www.mathlinks.ro/Forum/viewtopic.php?t=63592 ¿

with some analysis you can show that $B_{n + 1} = 1 + \sum\limits_{k = 1}^n {\left( {\begin{array}{*{20}c} n \\ k \\ \end{array} } \right)} B_k$ but i wonder if an explicit formula or a simple elementary proof exists !!

See the link in my post ;)
(I didn't know it's name there, so used $s_n$ instead of $B_n$)

I know with no proof :
 $S(r,k)=\frac{1}{k!}\sum_{j=0}^k(-1)^{k-j}{k\choose j}.j^r$
 and $B_r =e^{-1}\sum_{j=0}^{\infty}\frac{j^r}{j!}$
 Do you have a proof for it , Zetax?

For the second one:
Since $B_0=B_1=1$ holds also in this formula, it's enough to prove that it fulfills the recursion formula.
To prove: $\sum_{j=0}^{\infty} \frac{j^{r+1}}{j!} = \sum_{k=0}^{r} \binom{r}{k} \sum_{j=0}^{\infty}\frac{j^k}{j!}$.
Using binomial theorem gives that $\sum_{k=0}^{r} \binom{r}{k} \frac{j^k}{j!} = \frac{(j+1)^r}{j!}$, so we have
$\sum_{j=0}^{\infty} \sum_{k=0}^{r} \binom{r}{k} \frac{j^k}{j!} = \sum_{j=0}^{\infty} \frac{(j+1)^r}{j!} = \sum_{j=1}^{\infty} \frac{j^r}{(j-1)!} = \sum_{j=0}^{\infty} \frac{j^{r+1}}{j!}$
which was to be proven.

In your link , why computing $B_n$ , we just consider the set $\{1,2,...,n-1\}$ . Why we just choose $k$ numbers from $n-1$ numbers not $n$ numbers . Can you explain it for me , ZetaX ?

We want to calculate $s_n=B_n$ there (in contrary to the version of this formula maskman gave, where $B_{n+1}$ is calculated; but they are the same after shifting $n$ by one).
So we assume to have already calculated $s_0,s_1,s_2,...,s_{n-1}$ and want now to find the number of partitions of $\{1,2,3,...,n\}$. The idea is to choose any set $A$ of $k$ numbers from the first $n-1$ numbers and to partition the rest of $\{1,2,3,...,n-1\}$ into sets. But we neglegted to take $n$ into any set, so we put it into $A$. So indeed it can be said we choosed $k+1$ numbers with the condition to contain $n$. We do this because of not knowing enough about the partitions of $\{1,2,3,...,n\}$, but knowing already all we want about the smaller sets.


The set of numbers e^n, n = 0,±1, ±2, .. is countable.

True, because $ \mathbb Z$ is countable and $ x\to e^{x}$ is a bijection from that set to $ \mathbb Z$.


geometry
geometric transformation
reflection
articles
probability

Hello!
What do you think would be the best way to prevent terrorism on public transportation system? What would be the possible ways to increase safety on trains, buses, etc.?

Thank you for your input!

Well, assuming that Muslims were doing the bombings, you could fill bladders with pig blood and hang them in the bus/train/etc. Am I doing your homework for you?

[quote="sponge008"]Am I doing your homework for you?[/quote]

off topic, but you gotta wonder how many of these topics have been started because of homework.

Solution: Stop all public transportation systems.

You didn't say it had to be practical.

[quote="rookworm"]
Answer: They have a system of beliefs (Islam) that supresses free thought, art, science and fun-- and one that leads believers towards hatred, violence, racism, deception and resentment. Therefore they secretly resent other cultures, and wish to destroy them or convert them.
[/quote]
This is the standard official answer, I don't see where is your free thought. How much do you know about Islam? And where did you get the information? A joke says "I abominate the racists and the colors"
Coal ton after ton, and screaming "Dumn boiled water, my finger hurts", I can't find behavior stupider than this.

Randomly seeing this topic as the latest in the round table, I just had to post here.  Some of the above comments were quite offensive to me, and I'm not a Muslim myself, but at least I realize that terrorist attacks and such are not in any way a direct result of Islam.  In fact, some might argue that the west sparked this whole conflict, and with western religion playing a role here.  And besides, there are plenty of Christian fundamentalists who commit atrocious acts.  We don't say that Christianity teaches hatred, violence, and anger, now do we?  Keep in mind that we want to keep this place friendly for all reasonable users.

I completely second stephen's post.  I'm appalled by the racism in this thread.  Joel or Daniel, please delete this.

[quote="rookworm"][quote="shobber"]First you must know why they are doing terrorism attackings. Then you can come up with the solutions.[/quote]

Answer: They have a system of beliefs (Islam) that supresses free thought, art, science and fun-- and one that leads believers towards hatred, violence, racism, deception and resentment. Therefore they secretly resent other cultures, and wish to destroy them or convert them.

Solution: Step 1: Educate people about the nature of those beliefs; Step 2: Disengage from all involvement (trade, foreign aid, immigration, etc.) from the Muslim world, except possibly military intervention for security reasons.[/quote]

Wow, that's a new level on the ignorance chart. Jeez, this is so blatant that it's not even funny.

[quote="Buffalo"]How much do you know about Islam?[/quote]

Quite a bit. I know a lot of what is written in the Koran, Sura, and authentic Hadiths, as well as the arguments of Islamic apologists and their critics. If you follow some of the links on [url=http://www.jihadwatch.org]www.jihadwatch.org[/url] you might learn something yourself.

[quote="Buffalo"] And where did you get the information? A joke says "I abominate the racists and the colors"
Coal ton after ton, and screaming "Dumn boiled water, my finger hurts", I can't find behavior stupider than this.[/quote]

I'm not sure what your point is here (I gather you think I am a racist who likes to immerse himself in boiling water-- I assure you neither is the case), but if you want to see something stupider, I suggest you refocus your eyes so you may see your reflection in the computer monitor.

Let me remark that Islam is not a race, it is an idealogy, and anyone is free (in the personal sense) to follow it or not. And as a set of ideas and beliefs, it is fair game for unrestricted criticism.

[quote="Probaility"]And besides, there are plenty of Christian fundamentalists who commit atrocious acts. We don't say that Christianity teaches hatred, violence, and anger, now do we? [/quote]

1) name five (bet you can't)
2) we don't say that Christianity teaches hatred because it does not teach hatred. Nor would this necessarily be the case IF there was a Christian terrorism problem. But the fact is, these two religions are not comparable. There is ample documentation that Islam, as is practiced today, and as is written in their holy texts, encourges jihad, expansionism, subservience of non-muslims, etc. etc.

Sometimes the truth offends, unfourtunately....

I suggest everyone look at this site, read the headlines for a couple of days, and follow some of the links on the left column, in addition to reading the FAQ. you will find it most enlightening.

[url=http://www.jihadwatch.org]www.jihadwatch.org[/url]

tighten security to security like airport...

or

...eliminate public transport

no need to discriminate based on religious beliefs... that's just plain unfair

Post 2 is very offensive in my opinion. A moderator of an administrator should delete it.

[quote="rookworm"]
[quote="Probaility"]And besides, there are plenty of Christian fundamentalists who commit atrocious acts. We don't say that Christianity teaches hatred, violence, and anger, now do we? [/quote]

1) name five (bet you can't)
2) we don't say that Christianity teaches hatred because it does not teach hatred. Nor would this necessarily be the case IF there was a Christian terrorism problem. But the fact is, these two religions are not comparable. There is ample documentation that Islam, as is practiced today, and as is written in their holy texts, encourges jihad, expansionism, subservience of non-muslims, etc. etc.

Sometimes the truth offends, unfourtunately....

I suggest everyone look at this site, read the headlines for a couple of days, and follow some of the links on the left column, in addition to reading the FAQ. you will find it most enlightening.

[url=http://www.jihadwatch.org]www.jihadwatch.org[/url][/quote]

OK, I can't name five... I can't name five Muslim extremists either.  Just because I can't name them doesn't mean they don't exist.

For your second point, if you interpret everything in that way, then yes... But Christianity never did anything like that.  No missionaries ever forcibly converted Native Americans, and Christians never ever oppressed pagans--wait that word is demeaning and comes from Christianity that means it doesn't exist--I meant non-Christians.  The Crusades?  No, that was caused by Muslims who attacked Europeans... oh wait the crusades were fought in the Middle East well let's forget that for now.

Anyways, my point is that the circumstances can warp people's views of Christianity or Islam.  I admit I never read a Koran, but if I can trust common sense and the vast majority of authoritative sources plus a few quotes, then there is nothing in there that can only be interpreted in the way you describe above.

um, yeah. i agree with probability. what rookworm said might be true for some people, but really there's no difference between islam and christianity. you can't judge or predict the actions of today's muslims based on a text thousands of years old. a lot of what's in the christian bible isn't true today; same for the koran.

I agree with "probability" too.  Once again, we are letting a small minority cloud our perception of a larger group of people.  [i]The moderator needs to step in now to stop the ridiculous racism that occurred in the earlier posts.  [/i]

Christianity and Islam ARE in fact inherently linked, with some of the biblical characters appearing in both religions!

And you can't say that the Christian bible has no hatred involved, just as you can't say that there's no hatred involved in the Koran.  What about the smiting of all those who don't believe the Christian doctrine, as described in "Revelations"? (just look at it and tell me there's no violence involved).  The point i'm trying to make is that they can be compared, and that they both have their moments.

The most important point though, is that whatever the Christian Bible says, or the Koran says, it is, in the end, not the responsibility of the religion but rather the person involved.  A terrorist simply uses religion as an excuse to commit whatever crime they want to commit.

Now, back to the topic.  How to deal with increasing security of public transport?  Well for starters, we need to step up airport security (this is a topical issue in australia, although im not so sure for other places).  The local community is also helpful in identifying unusual activity, but once again, that runs the risk of misinformed people (like the racist pigs earlier) just continually calling a hotline.  Increasing security in the city (it is unlikely that a terrorist attack should occur outside a major population centre) will also help prevent terrorism.

Finally, we need to step up the emergency procedures should an attack happen.  A network between the community, intelligence offices, police, firestations, hospitals etc. must be quick, efficient and ready.  Evacuation procedures should be set in place, and a system of security cameras around greater population density areas (e.g. entrances to some train stations, around the city, etc.)  need to be established and easily reviewed.

[quote="shobber"]First you must know why they are doing terrorism attackings. Then you can come up with the solutions.[/quote]

Well, that presupposes that terrorism is a rational response to some perceived or real injustices, which I don't think is always (or even usually) the case.

I am really depressed, I didn't mean to spark any racism.

However, there are a few good ideas that I haven't thought of... so THANK YOU. Hmm... I wonder how to do the Work Cited pages for these ideas, do you have any suggestions?

"First you must know why they are doing terrorism attackings. Then you can come up with the solutions"

I have thought about that, but by the time you learn why, will it be too late? Different attacks have different perceptions behind them but I agree that those perceptions are meant to be destructive.

2. Thanks bleumoose, it is a relief to know that asking for help with homework on here is appropriate.

3. Stop all transporting system... hmm..my fault... I haven't state my questions clearly.

4. zeus_three, the last two paragraphs in your post is really helpful! Thank you!

okay, since I didn't think that there will be any response, I made my question general, but here's the specific one...

[b]How can we increase security and decrease threat of error on the mass transit system in such as way that the majority of people consent, affordable budge, respect certain privacy, took less time for everday security checks and avoid inconveniency?[/b]

Thank you for your unracist input!

i [i]wasnt[/i] condoning that, actually. i was making a comment about it. honestly, it seems rather like cheating to me, but whatever. now, if the question said "what do people think about ___?" then thats different. but if the question is simply "what should be done about ___?" then, in my very humble opinion, asking others to tell you what YOU think is wrong and counter-productive.

for those of you that would like to "step up airline security,":
[quote="benjamin franklin"]They who would give up an essential liberty for temporary security, deserve neither liberty or security.[/quote]

and for those of you that are butchering rookworm: yes! you are right! christianity has some REALLY BAD things happen in the past! yes! you are right! there are people today that do some REALLY BAD things in the name of "christianity"! however, i honestly don't believe that the muslim culture is doing enough to rid themselves of their own extremists. i know a lot of this is because muslims do not have a central leader, and thus it is extremely difficult for "the religion" to speak out as a whole (unlike with christianity where, like it or not, the Vatican still pretty much represents "christianity" in general). However, I still would like to see muslims speak out more against the "muslim" terrorists, because it is the extremists which are making their entire religion look bad by doing horrible things in the name of their religion.

with all of that said, i dont think its a good idea to have racial-profiling on airlines, because it goes against what america stands for. i would rather every person on the plane be checked individually than someone just checking "middle-eastern looking people," simply because i believe in the american ideology.

and please stop asking the mods to "delete the racist comments." they are not hateful slurs against the culture but merely someone's opinion. just because they go against the grain doesn't mean they should be gotten rid of. it's free speech. argue against and disagree all you want, i fully endorse that; but please don't censor opinions of that nature.

well, i hardly call filling bladders with pig blood and hang them in the bus/train/etc" a difference of opinion.

The Vatican hardly represents Christianity in general, especially in the United States (where evangelicals seem to hold more sway currently).  If, however, you mean that various Christian denominations are doctrinally more similar than their Muslim counterparts, then that is probably an accurate statement.

[quote="bleumoose"]with all of that said, i dont think its a good idea to have racial-profiling on airlines, because it goes against what america stands for. i would rather every person on the plane be checked individually than someone just checking "middle-eastern looking people," simply because i believe in the american ideology.

and please stop asking the mods to "delete the racist comments." they are not hateful slurs against the culture but merely someone's opinion. just because they go against the grain doesn't mean they should be gotten rid of. it's free speech. argue against and disagree all you want, i fully endorse that; but please don't censor opinions of that nature.[/quote]

Yes, I agree with that. We should all keep in mind that the majority of arabs are actually non-Muslim, and the majority of Muslims are non-arab (by far). So anyone who thinks that to criticise Islam is to be racist, is being racist him/herself by implicitly making that equivocation (i.e. Muslim=arab)!  :P

As for racial profiling, if we observe that terrorists are of a certain ethnicity (assuming that such a thing as "ethnicity" exists, which is really something subjective), in a greater proportion than the general population, Bayes' theorem tells us that a member of such a group will be more likely to be a terrorist (ignoring all other factors). When lives are at stake, I'm not so sure that it would be wise to ignore that information completely. Note, however, that you can't tell a terrorist just by ethnicity, and many non-mid-easterners have been would-be terrorists (Richard Reid, John Lindh, etc., etc.)

I agree that sponge's comments were inflammatory (especially considering his avatar!), not to mention unsanitary.  Besides, it wouldn't work (haven't you seen the movie [i]Carrie[/i]?). Much better would be to cut terrorism off at the root (over the long term), and in the mean time, have better intelligence/ surveillance/ mathematical models to catch current terrorists.

One last note: the Vatican only speaks for Catholics, who comprise about half of the world's Christians (about 1.1 billion out of 2.1 billion). Although Islam has no similar leadership (although al-Quaeda, et. al. wish to re-establish the Caliphate, the rough equivalent), there is a kind of orthodoxy in the way the religion is practiced. The Quran is taken to be the immutable word of god, and it is considered to be the literal truth. Moreover, Islamic law (sharia) is very specific about how to behave. So while there is no official leadership, there is a certain consensus about how things should be (notwithstanding minor differences between Shia and Sunni).

Anyway, enough for now.  :)

Well that seems impossible, you should have to check every bag every time someone gets of a bus

The mean thing about terrorist is how easy it is for  them to ruin our fun
They do one attack with airplanes : our airport security is tightened firmly, and me and my father have to take of our belts, shoes, and are checked manually (fouilleren, how to say in english?) i mean spread my legs and stuff you know (I had to do it twice, probably because I am a twenty year old not perfectly shaven male)
They do one attack on buses : even in another continent cops are watching you enter, there are plaques advising you to report any weird activity...

every time they do something, everyone gets more scared and we lose a facet of our daily life, the only thing the terrorists have to do is look at the next thing that is now the easiest target...

However, changing american policy and international policy WILL reduce the risk

it seems americans find not giving in very important (this is just something i think is true), i remember a speech by bush when the afghanistan war began about a girl whose daddy would go and could even die, but that it was worth it :|
here in europe, many people feel that peace is important , i think that has to do with us not having any major conflict on their soil since 1865
we see monuments of the war, damaged buildings, stories by my grandmother about my wounded grandfather
a few days ago, I passed an office of the national defence force in Ghent, it said in big letters "Priority to peace!"

It is true that if you give in you open the door for more, so I cannot say which is the best, but face it, when people are willing to spend years of their life, in order to end it by taking as many others with them, there has got to be reason
and that same reason will also cause injustice to those that have never had anything to do with terrorism, like for instance palestines, or poor iraqi people

I still think it's kind of funny when people apply the word "racist" to talking about religion. I guess it's just the harsh connotation that that word brings up, in this day and age. Is there even a word for what they're talking about (like religionist or something)?

Anyway, I recently read an excellent article on terrorism in general. It's 36 pages long, so only read it if you have the time to spend on the issue. You can find that article [url=http://www.au.af.mil/au/awc/awcgate/usmc/ceto/when_devils_walk_the_earth.pdf]here[/url].

Racism as applied to islam is somewhat accurate given that most muslims are from a certain region(the mideast). Although, it doesn't really matter, because the criticism is still right on target regardless of what word is used to describe the actions. Rookworm's initial posts essentialize muslims as terrorists and suggests that the U.S. treat all muslims like crap because a few have been bad.

Also, while it's clearly untrue that all arabic speaking people are muslim -- the majority actually are according to CIA world fact book. So I am not sure what you're talking about there Rookworm.

Anyways, you suggest cutting off terrorism at it's roots. While I agree terrorism should be dealt with in its beginnings -- it doesn't begin with people flying planes into buildings; it begins with poverty and oppression. In a Foreign Affairs article I read a few years ago, it  was reported that many palestinians (according to the U.N.) had only one meal a day  and great difficulties finding jobs. It's conditions like that which breed hopelessness and lead to terrorism rather than specifically islam as a religion.( this is an example, I am not saying all muslims are palestinians or anything)

I agree a lot with what fredbel6 says. By your bayesian analysis Rookworm -- the government could essentially oppress any group which is statistically different because they pose a slightly greater risk. For example, by the same reasoning,  it would be ok to preemptively jail all poor people because there's a slightly higher probability of robbery with in that particular population.

[quote="probability1.01"]For your second point, if you interpret everything in that way, then yes... But Christianity never did anything like that.  No missionaries ever forcibly converted Native Americans, and Christians never ever oppressed pagans--wait that word is demeaning and comes from Christianity that means it doesn't exist--I meant non-Christians.[/quote]

Umm, actually didn't the Catholic church burn heretics and such? I recall a certain Galileo had trouble with the church. Also Christian Spanish crusaders AND Christopher Columbus were known to force natives into renouncing their religions, converting to Christianity then enslaving them until they died. You can read the book [i]Lies My Teacher Told Me[/i] by James Lowen for more information.

My point is Muslims can do these things which we consider wrong in the name of their faith, but that isn't something we weren't also guilty of.

I think probability1.01 was being sarcastic in that instance. He's known for doing that. And Loewen has two e's. ;)

And BTW, agolsme, I think Rookworm's point was that the majority of [i]Muslims[/i] are non-Arabs, not the other way around (there's a difference).

There were two parts to his statement. I was contesting the first part.

[quote] [b]We should all keep in mind that the majority of arabs are actually non-Muslim[/b], and the majority of Muslims are non-arab (by far). [/quote]
bolded part is what i responded to.

Actually, I'm not even sure about that second part, but that's possible (but I doubt by far) considering there are significant Muslim populations in SE Asia. Of course, if by "Arab" you mean a very specific people confined to the Arabian peninsula...  :roll:

In any case, I'm not sure whether your statements can be considered "racist." We should not get hung up on this semantic issue. If you truly believe what you say, then I would like you to provide some specific examples of how mainstream Islam promotes anger, violence, etc.

Okay, the muslim=arab or not argument doesn't really matter, so I'll stop wasting your time :P

Has anyone read at least part of the 36-page article I linked to? I think it actually rather directly answers this question.

[quote="agolsme"]Racism as applied to islam is somewhat accurate given that most muslims are from a certain region(the mideast). Although, it doesn't really matter, because the criticism is still right on target regardless of what word is used to describe the actions. Rookworm's initial posts essentialize muslims as terrorists and suggests that the U.S. treat all muslims like crap because a few have been bad. [/quote]

Every statement in the above quotation is false. I never said muslims were essentially terrorists, nor did I say the US should treat them all like crap.

Most members of the Flat Earth Society are from the US. I do not stigmatize Americans (even "somewhat") when I criticize the belief that the Earth is flat. Similarly, one can criticize any system of beliefs without demeaning the inhabitants of any entire geographic region.

I do not consider "He's racist, delete his posts" to be either valid criticism or on target. This is simply a smear, having nothing whatsoever to do with good reasoning or argumentation. Even to consider this as an open question is fundamentally unreasonable. This is not a semantic issue, this is just verbal (written) abuse.

[quote="agolsme"]Anyways, you suggest cutting off terrorism at it's roots. While I agree terrorism should be dealt with in its beginnings -- it doesn't begin with people flying planes into buildings; it begins with poverty and oppression. In a Foreign Affairs article I read a few years ago, it  was reported that many palestinians (according to the U.N.) had only one meal a day  and great difficulties finding jobs. It's conditions like that which breed hopelessness and lead to terrorism rather than specifically islam as a religion.( this is an example, I am not saying all muslims are palestinians or anything)[/quote]

This is a standard platitude, and is also absolute nonsense. Do you think someone who eats "only one meal a day  and [has] great difficulties finding jobs" is capable of "flying planes into buildings"?

[quote="agolsme"]I agree a lot with what fredbel6 says. By your bayesian analysis Rookworm -- the government could essentially oppress any group which is statistically different because they pose a slightly greater risk. For example, by the same reasoning,  it would be ok to preemptively jail all poor people because there's a slightly higher probability of robbery with in that particular population.[/quote]

My conclusion was only that we should not discard good information in the name of political correctness. No mention of jail or preemptive punishment. Just clues as to how to find threats more efficiently.

As for the question at hand, I will say no more about it, because it is such a sensitve issue. If anyone wants to see any evidence of my previous claims, there is ample documatation if you care to look for it.

I agree. Let's drop the issue and talk about other ideas in protecting public transportation. BTW, if anyone wants evidence that the earth is flat, there's ample documentation if you look for it.

1. I'd call cutting off all aid, immigration,and trade "treating  muslims like crap". I am sorry if that phrasing was offensive to you.

[quote]

Answer: They have a system of beliefs (Islam) that supresses free thought, art, science and fun-- and one that leads believers towards hatred, violence, racism, deception and resentment. Therefore they secretly resent other cultures, and wish to destroy them or convert them.

Solution: Step 1: Educate people about the nature of those beliefs; Step 2: Disengage from all involvement (trade, foreign aid, immigration, etc.) from the Muslim world, except possibly military intervention for security reasons.[/quote]

2. I wasn't saying you should be censored. I was saying it a) doesn't really make a difference whether your statements were called racist or not b) your statements were essentialist.

3. Saying my argument is a "standard platitude" doesn't mean it's not true. Clearly starving people can't fly planes into building, but they can strap on a bomb and walk into a crowded area.

Anyways, I just wanted to clarify my arguments since you're misinterpretinig what I was saying.


Does anyone have a quick way to find the sum of the digits for every number from 0 to 1,000,000?  The way I think I might attempt it is the same way I would do it for 0 to 100:

[hide="0 to 100"]Each units digit occurs 10 times, so the sum of the units digits is $S_{1}=\left(45\right)\left(10\right)=450$.

Each tens digit occurs 10 times as well, giving a sum of $S_{2}=\left(45\right)\left(10\right)=450$.

So the sum of all of the digits from 0 to 100 is 900.[/hide]
[hide="My Attempt at 0 to 1,000,000"]Following the pattern from above, I might say that the sum of the digits is $S=\left(6\right)\left(45\right)\left(100,000\right)=27,000,000$.  Can anyone confirm?[/hide]

well, that would be the sums from 0 to 99 and from 0 to 999999. The solution works because

[hide]i'll show the case for 0 to 999999. This is all combinations of abcdef, where a,b,c,d,e,f are digits which may be 0.

Let x be a digit. Out of all of the 1000000 numbers, one tenth of them have a=x because for each combination of b,c,d,e,f, one number has a=x and nine don't

Similarly, the cases for b=x, c=x, d=x, e=x, and f=x are all one tenth, or 100000 numbers. Therefore each digit appears in each place 100000 times.[/hide]

Well, you're almost right.  You're missing the last case in both - $100$ and $1000000$ respectively.

[hide="Why your answers cant possibly be right"]
Both of them are divisible by $9$.  This would imply that the sum of the integers from $0$ to $100$ (equivalently to $1000000$) is divisible by $9$, which it's not: $\frac{100(101)}{2} = 5050$.
[/hide]

[quote="Philip_Leszczynski"]He's not talking about the sum of the integers, he's talking about the sum of the [i]digits[/i].[/quote]

Call $s(n)$ the sum of the digits of $n$.

Lemma:  $s(n) \equiv n \bmod 9$.

Proof:  $n = 10^k d_k + 10^{k-1} d_{k-1} + ... + 10 d_1 + d_0 \equiv d_k + d_{k-1} + ... + d_1 + d_0 \bmod 9 = s(n)$.

Now, for the body of the proof.  We wish to evaluate $\sum_{k=1}^{100} s(k)$.  Because we know $s(n) \equiv n \bmod 9$ we know $\sum_{k=1}^{100} s(k) \equiv \sum_{k=1}^{100} k = 5050 \equiv 1 \bmod 9$, but he gave his answer as $450 \equiv 0 \bmod 9$ - contradiction.

Hence his answer couldn't have been correct (and I didn't even have to evaluate $\sum_{k=1}^{100} s(k)$ :D )

My answer turns out to be 1 off.  The answer is actually 2,700,001 according to the person who posed the problem (I forgot to add that "1" in the 1,000,000).


Prove that $ 1111.....1$($ 91$  times) is a composite number

[quote="earldbest"]$ \sum_{i \equal{} 0}^{90}10^i$ $ \equal{} \sum_{i \equal{} 0}^{12}10^{7i} (1111111)$[/quote]

Brilliant.

What led you to do this?


search
modular arithmetic
inequalities
IMO Shortlist
number theory
prime numbers
number theory unsolved

Find every pair of positive integers $ (a;n)$ such that every prime divisor(s) of $ a^n\plus{}1$ is also prime divisor(s) of $ a\plus{}1$

Zetax, could you please give the direct link to that theorem?

Erken, I found it in the IMO Compedium, Problem N4 of IMO Shortlist 2000, but I do not understand the solution performed in it.

I will give that solution here (just to the point that I no more understand) so that everybody can discuss the problem:

Trivially, $ (a,1)$ and $ (1,n)$ satisfy our condition. Assume that $ a,n > 1$, we can easily prove that even $ n$(s) are not solution.

Consider odd $ n$, let $ p$ be a prime divisor of $ n$ and put $ n \equal{} pn_1$ where $ p,n_1$ are odd, by the statement "every prime divisor of $ a^n \plus{} 1$ is also prime divisor of $ a \plus{} 1$", there exists $ m$ such that $ a^n \plus{} 1|(a \plus{} 1)^m$.

Let $ b \equal{} a^{n_1}$ then since $ n_1$ is odd, $ a \plus{} 1|a^{n_1} \plus{} 1$, which leads to:

$ a^n \plus{} 1 \equal{} b^p \plus{} 1|(a \plus{} 1)^m|(a^{n_1} \plus{} 1)^m \equal{} (b \plus{} 1)^m$

Thus: $ P \equal{} \frac {b^p \plus{} 1}{b \plus{} 1} \equal{} b^{p \minus{} 1} \minus{} b^{p \minus{} 2} \plus{} ... \minus{} b \plus{} 1|(b \plus{} 1)^m$.

At here, the solution of the IMO compedium stated that "$ P\equiv p \pmod{b \plus{} 1}$ and so $ P$ has no primed divisor other than $ p$", how can we got that??????? :o

Dear Marshimaru , I have an other approach . It is easy to show that n must be an odd number . Because $ n>1$ so it has an odd prime divisor $ p$. Let consider prime divisor of $ \frac{a^p\plus{}1}{a\plus{}1}$ . Let $ q$ is a prime divisor of this number , use the fact that $ q|a\plus{}1$ so we must have $ q|p$ , both of two numbers are prime numbers so $ p\equal{}q$ . Therefore exist an positive integer $ k$ such that
$ \frac{a^p\plus{}1}{a\plus{}1}\equal{}p^k$ . Use the lifting exponent we have $ p\parallel{}\frac{a^p\plus{}1}{a\plus{}1}$ , it shows that $ k\equal{}1$ . Then by some inequalities , we obtain  $ a\equal{}2,p\equal{}3$ . There are some other steps , but I think you can solve them by self .  :)

Thank you TTsphn, I have succeeded understanding the solution finally :oops:, there are some left steps(!).

Actually yours comes from the same idea as in the IMO Compedium, after proving that $ \frac{a^p\plus{}1}{a\plus{}1}$ has no prime divisors other than $ p$, we can also prove $ n_1\equal{}1$ and $ p\equal{}3$ where $ n\equal{}pn_1$, from here we got $ (a,n)\equal{}(2,3)$.

Could anybody give the link directed to the Zsigmondy's theorem about prime divisors which ZetaX mentioned several posts above (with proof)?


This is posted before

http://www.mathlinks.ro/Forum/viewtopic.php?t=143656

Hope that my solution there is not too bad.

@bach. Are you Vietnamese IMOer this year?

thanks a lot. I think your solution is the best solution.I tried,but I couldn't find any nice solution.
Sorry,I'm not Do Xuan Bach.


Let $ f(x),\ g(x)$ be non zero polynomials with real coefficients such that $ f(x^3)\plus{}g(x)\equal{}f(x)\plus{}x^5g(x)$.
Find a possible $ f(x)$ that has the least degree.

We are to find non-constant $ f(x)$ of minimal degree such that $ (x^5 \minus{} 1)$ divides $ f(x^3) \minus{} f(x)$. I guess $ f(x) \equal{} x^5$.

$ f(x)\equal{}a(x^4\plus{}x^3\plus{}x^2\plus{}x)\plus{}b$ with $ a\ne0$.

[quote="xxp2000"]$ f(x) \equal{} a(x^4 \plus{} x^3 \plus{} x^2 \plus{} x) \plus{} b$ with $ a\ne0$.[/quote]

How could you get the solution?


Given the sequence: 0, 1/3,3/6,7/11,15/18,A,B,C,...
Find the sum of the numerator and denominator of C?
PS: Is it nice for you? :)

You're right, nayel! :) .I solved this problem by realizing that the increase of each numerator forms an geometric sequence(with r=2) and the denominator forms an odd arithmetic sequence(d=2).


I don't know this section fits or not..though..
Could you prove

sinx=x-x^3/3!+x^5/5!-x^7/7!....

And who first found this equation..:) I really like this equation..lol

Wow this is too much for me..lol
Though thanks a lot.


Is there a formula for regular polygons when you know only the radius?

$l_n=2R\sin \frac{\pi}{n}\ ;\ a_n=R\cos \frac{\pi}{n}\Longrightarrow$$P_n=nl_n=2nR\sin \frac{\pi}{n}\ ;$
$S_n=\frac 12 P_na_n=\frac 12nR^2\sin \frac{2\pi}{n}\Longrightarrow\boxed {\ P_n=2nR\sin \frac{\pi}{n}\ ;\ S_n=\frac n2\cdot R^2\cdot \sin \frac{2\pi}{n}}\ .$
[b]Remark.[/b] $\lim_{n\to\infty} P_n=2\pi R\cdot \lim_{n\to\infty} \frac{\sin \frac{\pi}{n}}{\frac{\pi}{n}}=2\pi R\ ;\ \lim_{n\to\infty} S_n=\pi R^2\cdot\lim_{n\to\infty}\frac{\sin \frac{2\pi}{n}}{\frac{2\pi}{n}}=\pi R^2\ .$
I used the remarkable limit $\lim_{x\to 0} \frac{\sin x}{x}=1\ \left(\lim_{n\to\infty}\frac{\pi}{n}=\lim_{n\to\infty}\frac{2\pi}{n}=0\right)\ .$
We obtained the perimeter and the area of a circle : $\boxed {P_{\infty}=2\pi R\ ;\ S_{\infty}=\pi R^2}\ .$

[hide="Derivation"]
Consider the regular $n$-gon with vertices $A_1, A_2, ... A_n$ inscribed in a circle of radius $r$.  We wish to find its area.  We accomplish this by drawing all of the radii to each vertex, dividng the $n$-gon into $n$ isosceles triangles.  What area do these isosceles triangles have?

Draw an altitude from the center to any side, that is, to the third side of any of these isosceles triangles.  This forms two right triangles with hypotenuse $r$ and an angle of $\frac{\pi}{n}$ (we divided the circle into $n$ triangles, then cut each angle in half by drawingn the altitude), so the altitude has length $r \cos \frac{\pi}{n}$ and the other leg (of the right triangles) has length $r \sin \frac{\pi}{n}$.  The base of the isosceles triangle is therefore $2r \sin \frac{\pi}{n}$, the area of each isosceles triangle given by $r^2 \sin \frac{\pi}{n} \cos \frac{\pi}{n} = \frac{1}{2} r^2 \sin \frac{2\pi}{n}$, and the area of the entire polygon

$\boxed{S_n = \frac{n}{2} r^2 \sin \frac{2\pi}{n}}$, as stated.
[/hide]

[hide="Remark"]
We expect that, as the number of sides increases, the area gets closer to the area of the circle, that is,

$\lim_{n \to \infty} \frac{nr^2}{2} \sin \frac{2\pi}{n} = \pi r^2$, and in fact we have

$\lim_{n \to \infty} \frac{n}{2} \sin \frac{2\pi}{n} = \pi$ (various methods, the most rigorous being the Sandwich Theorem, as proof), as expected.
[/hide]


Este no es un problema como tal, sino que reduje un problema que tengo a demostrar esto.
Dado el triangulo $\triangle ABC$, sea $I$ su incentro y sean $D,E,F$ las reflexiones de $I$ a travez de $BC,CA,AB$ respectivamente. Demuestre que $AD, BE, CF$ concurren.

This is a particular case of the following:

[url]http://www.xtec.es/~qcastell/ttw/ttweng/definicions/d_Kariya_p.html[/url]

Ya, que chevere esta eso, pero aun no logro demostrar que el caso particular sigue sirviendo.

Hola Leo, cómo andás tantos meses?, te cuento algo de tu problema. Hace un tiempo Carlos me mandó una generalización, supongo que debe ser la que grobber comenta (no me fijé). Pero a su vez ese problema que pienso es un caso particular de éste (mmm ahora que reflexiono debe ser éste el problema que grobber dice): Dado un punto P, sean D, E y F los pies de las perpendiculares a BC, CA y AB, respectivamente. Sea X un punto en PD, la perpendicular de PC por X intersecta a PE en Y; la perpendicular a AP por Y intersecta a PF en Z. Bueno, no cuesta nada ver que XZ es también perpendicular a BP. Demostrar que XYZ es perspectivo con ABC.
Bueno, creo que era así, de última después lo arreglo, pero casi seguro que era así. Tengo la solución pasada en .jpg, ¿puede mandar acá?, un gran saludo amigo (seguís en ee.uu.?)

Como ya han mencionado previamente este es un caso particular del llamado Teorema de Kariya

[b]Teorema:[/b] En un $ \triangle ABC,$ $ (I,r)$ es su incírculo y $ \triangle X_aX_bX_c$ su triángulo de contacto interno. La circunferencia $ (I,kr)$ corta a los rayos $ IX_a,$ $IX_b,$ $IX_c$ en $ D,E,F.$ Entonces $ \triangle ABC$ y $ \triangle DEF$ son perspectivos y ortológicos.

La demostración es relativamente simple usando el Teorema de Jacobi, que de hecho es un caso mas general que este teorema, o bien utilizando las ecuaciones en coordenadas baricentricas de la homotecia con centro $ I$ y coeficiente $ k$

La expresión general del perspector de Kariya es:

$ J  \left(\frac {1}{bc \plus{} kS_A}: \frac {1}{ac \plus{} kS_B}: \frac {1}{ab \plus{} kS_C}\right)$

En este caso en particular como $ D,E,F$ son las reflecciones de $ I$ en los lados del $ \triangle ABC,$ entonces $ k \equal{} 2,$ resultando que las rectas $ AD,BE,CF$ concurren en el conocido punto de Gray $X_{79}$ de $\triangle ABC.$

$ X_{79} \left(\frac {1}{bc \plus{} 2S_A}: \frac {1}{ac \plus{} 2S_B}: \frac {1}{ab \plus{} 2S_c}\right)$