Hello every1,

Recently I saw this curious problem in CLRS which demands me to produce the second best minimum spanning tree of a connected graph.

A formal definition of second best minimum spanning tree is given below.

Take the set of weights of all the spanning trees of the graph. Sort it. Delete the minimum value from this set. The minimum value of the modified set is the weight of the second best minimum spanning tree.

[b]So, i request u to kindly consider this problem and post an efficient algorithm to determine the weight of the asecond best minimum spanning tree.[/b]

U can assume that all the edges of the input graph are of different weights.[b] Also, i request u to prove that though minimum spanning tree is unique, the second best minimum spanning tree need not be unique....[/b]

Thank You and have a nice day

well... for 2nd best MST, u have to find first MST then u have to delete the edges of MST one by one, and then u have to find the smallest edge each time that joins the different two sets.

[quote]Also, i request u to prove that though minimum spanning tree is unique, the second best minimum spanning tree need not be unique.... [/quote]

The minimum spanning tree may not necessarily be unique. Consider for example a triangle graph where two of its edges have equal weights.


geometry
circumcircle
parallelogram
trapezoid
geometry proposed

Let triangle ABC satisfying its circumcircle (O) and A-excircle (Ia) has the same radius. The A-excircle touches BC,CA,AB at A',B'C'. Prove that the circumcenter O of ABC is the orthocenter of the triangle A'B'C'.

This is Iran TST 2006, Day 2, Problem 5. See http://www.mathlinks.ro/viewtopic.php?p=487755

Nice and easy  :lol:
1.quadrilateral OLIaK is a parallelogram
2.△OCIa △NIaC are congruence   =>  OCIaN is a isosceles trapezoid
Done ！

Denote D the projection  of the vertex A on the opposite side BC,then I,O,D are conlinear
This problem appeared in Chinese National Mathematics Competition 1998


geometry
geometric transformation
rotation
geometry unsolved

Hello,

I just thought about this one because of a post that I saw that was posted long ago. Prove or disprove the following statement:

Given any triangle $\triangle ABC$, there exists a point $M$ in the interior of $\triangle ABC$, such that there are points $D$, $E$ and $F$ such that $A$, $M$ and $D$ are collinear, and so are $B$, $M$ and $E$, and $C$, $M$ and $F$, and also $\triangle DEF$ is equilateral. I think the answer is no, but Geometry have never been my stronger subject.

Best,

[quote="djimenez"]Given any triangle $\triangle ABC$, there exists a point $M$ in the interior of $\triangle ABC$, such that there are points $D$, $E$ and $F$ such that $A$, $M$ and $D$ are collinear, and so are $B$, $M$ and $E$, and $C$, $M$ and $F$, and also $\triangle DEF$ is equilateral.[/quote]

Well, maybe I don't precisely understand your question, but I think that there are such points D, E, F for [i]every[/i] point M in the plane of the triangle ABC. The proof of this is quite simple. I will restate the assertion:

[color=blue]For every triangle ABC and for every arbitrary point M in its plane, there exists an equilateral triangle DEF such that its vertices D, E, F lie on the lines AM, BM, CM, respectively.[/color]

[i]Proof.[/i] We assume, for the sake of simplicity, that the point M lies inside of the triangle ABC. In all the other cases, the proof is analogous.

Let XYZ be an arbitrary equilateral triangle lying somewhere in the plane. The locus of all points P inside this triangle XYZ satisfying < YPZ = < BMC is a circular arc through the points Y and Z (in fact, the angle < BMC is a fixed angle, so our locus is the locus of all points from which a fixed segment is seen under a fixed angle). Similarly, the locus of all points P inside the triangle XYZ satisfying < ZPX = < CMA is a circular arc through the points Z and X. It is clear that the point Z is one of the two common points of these two circular arcs; let T be the other common point. Then, we have < YTZ = < BMC and < ZTX = < CMA. Consequently,

  < XTY = 360° - < YTZ - < ZTX = 360° - < BMC - < CMA = < AMB.

The equations < YTZ = < BMC, < ZTX = < CMA and < XTY = < AMB signify that the lines XT, YT and ZT intersect under the same angles as the lines AM, BM, CM. Now, let D, E, F be three points on the extensions of the segments AM, BM, CM beyound the point M such that DM = XT, EM = YT and FM = ZT. Then, on the one hand, < EMF = < BMC, what, combined with < YTZ = < BMC, yields < EMF = < YTZ. On the other hand, we know that EM = YT and FM = ZT. Thus, the triangles EMF and YTZ are congruent. Consequently, EF = YZ. Similarly, we can show that FD = ZX and DE = XY. But since the triangle XYZ is equilateral, we have YZ = ZX = XY. Thus, EF = FD = DE, and it follows that the triangle DEF is equilateral. Since we also know that the points D, E, F lie on the lines AM, BM, CM, respectively, we have thus shown the existence of an equilateral triangle DEF such that its vertices D, E, F lie on the lines AM, BM, CM, respectively. Note that, once we have found [i]one[/i] such triangle DEF, we can subject it to an arbitrary dilatation with center M, and automatically get [i]infinitely many[/i] equilateral triangles DEF with vertices D, E, F lying on the lines AM, BM, CM, respectively.

  Darij

Another way of proving this more general assertion:

Let $D$ be any point on $MA$. At most one of the images of $MC$ through a rotation centered at $D$ of angle $\pm\frac \pi 3$ is parallel to $MB$, so assume the image of $MC$ through the rotation of angle $\frac \pi 3$ cuts $MB$ at $E$. The image of $E$ through a rotation of center $D$ and angle $-\frac\pi 3$ is $F$, it lies on $MC$, and $DEF$ is equilateral.

You know, I am kind of stupid lately.

I forgot to mention that $D$, $E$ and $F$ are points of the excircle of $\triangle ABC$


For $ x,y,z,m>0$ prove that the following inequality holds:

$ \frac{x^2}{(my\plus{}z)(mz\plus{}y)}\plus{}\frac{y^2}{(mz\plus{}x)(mx\plus{}z)}\plus{}\frac{z^2}{(mx\plus{}y)(my\plus{}x)}\geq\frac{3}{(m\plus{}1)^2}$

[b]Romanian Mathematical Journal GAZETA MATEMATICA (5-6/2002), problem no. 24699[/b]

We have $ (my\plus{}z)(mz\plus{}y)\leq\frac{1}{2}(m\plus{}1)^2(y^2\plus{}z^2)$ as it is equivalent to $ 2yz(m^2\plus{}1)\leq(y^2\plus{}z^2)(m^2\plus{}1)$, which follows from AM-GM.

Therefore
\[ \sum \frac{x^{2}}{(my\plus{}z)(mz\plus{}y)}\geq\frac{2}{(m\plus{}1)^2}\sum\frac{x^2}{y^2\plus{}z^2}\geq\frac{3}{(m\plus{}1)^2}\mbox{,}\]
where the last step follows from the Nesbit-Inequality.  :)

[b]Thank you, Martin N.[/b]!
I don't know the author's solution, but we have the same way of thinking! I'm sure that this is the shortest solution to this problem.

[b]REMARK[/b]
For $ x,y,z,m,n>0$, using $ 2yz(m^{2}\plus{}n^2)\leq(y^{2}\plus{}z^{2})(m^{2}\plus{}n^2) ,$ we get
$ \left.(my\plus{}nz)(mz\plus{}ny)\leq \frac{1}{2}(m\plus{}n)^2(y^2\plus{}z^2)\ \Leftrightarrow\ \frac{x^2}{(my\plus{}nz)(mz\plus{}ny)}\geq \frac{2x^2}{(m\plus{}n)^2(y^2\plus{}z^2)}\right.$
and another two similar relations.
Using them, we are able now to generalize the initial problem:

For $ x,y,z,m,n>0$ the following inequality holds
$ \left.\frac{x^{2}}{(my\plus{}nz)(mz\plus{}ny)}\plus{}\frac{y^{2}}{(mz\plus{}nx)(mx\plus{}nz)}\plus{}\frac{z^{2}}{(mx\plus{}ny)(my\plus{}nx)}\geq\frac{3}{(m\plus{}n)^{2}} \right.$


induction
inequality system
algebra
Sequence
IMO Shortlist

Given a sequence $\{a_n\}$ of real numbers such that $|a_{k+m} - a_k - a_m| \leq 1$ for all positive integers $k$ and $m$, prove that, for all positive integers $p$ and $q$, \[|\frac{a_p}{p} - \frac{a_q}{q}| < \frac{1}{p} + \frac{1}{q}.\]

We will prove by induction on p that $ |a_{pk}-p\cdot a_k|\leq p-1 $ for any $ p\geq 1$ and any k. Suppose it is true for p and prove it for p+1. But we have $ |a_{(p+1)k}-a_{pk}-a_k|\leq 1 $ and also $ |a_{pk}-p\cdot a_k|\leq p-1 $ so
 $ |a_{(p+1)k}-(p+1)a_k|\leq p$ and the induction is done. Now we apply this and write $ |a_{pq}-pa_q|\leq p-1<p$ and $|a_{pq}-qa_p|\leq q-1<q $., THus, $ p+q>| pa_q-qa_p| $ and we are done.


If a, b, c are positive reals, then prove

[tex]\frac{b+c-a}{a\left( b+c\right) }+\frac{c+a-b}{b\left( c+a\right) }+\frac{a+b-c}{c\left( a+b\right) }\geq \frac{9}{2\left( a+b+c\right) }[/tex].

  Darij

Actually it seems you meant something different:

[tex]\frac{2}{b+c}+\frac{2}{c+a}+\frac{2}{a+b}\leq \left( \frac{1}{b}+\frac{1}{c}\right) /2+\left( \frac{1}{c}+\frac{1}{a}\right) /2+\left( \frac{1}{a}+\frac{1}{b}\right) /2=\frac{1}{a}+\frac{1}{b}+\frac{1}{c}[/tex],

so that

[tex]\frac{b+c-a}{a\left( b+c\right) }+\frac{c+a-b}{b\left( c+a\right) }+\frac{a+b-c}{c\left( a+b\right) }[/tex]
[tex]=\frac{a+b+c}{a\left( b+c\right) }+\frac{a+b+c}{b\left( c+a\right) }+\frac{a+b+c}{c\left( a+b\right) }-\left( \frac{2}{b+c}+\frac{2}{c+a}+\frac{2}{a+b}\right)[/tex]
[tex]\geq \frac{a+b+c}{a\left( b+c\right) }+\frac{a+b+c}{b\left( c+a\right) }+\frac{a+b+c}{c\left( a+b\right) }-\left( \frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)[/tex]
[tex]=\left( \frac{a+b+c}{a\left( b+c\right) }-\frac{1}{a}\right) +\left( \frac{a+b+c}{b\left( c+a\right) }-\frac{1}{b}\right) +\left( \frac{a+b+c}{c\left( a+b\right) }-\frac{1}{c}\right)[/tex]
[tex]=\frac{1}{b+c}+\frac{1}{c+a}+\frac{1}{a+b}\geq \frac{9}{\left( b+c\right) +\left( c+a\right) +\left( a+b\right) }=\frac{9}{2\left( a+b+c\right) }[/tex].

But you're right, it's quite easy. I should try to think up something more interesting.

  Darij

No, Darij, ma_go meant exactly what he wrote.

We observe that the inequality can be written as $\frac{1}{a} - \frac{1}{b+c} + \frac{1}{b} - \frac{1}{c+a} + \frac{1}{c} - \frac{1}{a+b} \geq \frac{9}{2(a+b+c)}$, and the rest follows from what he wrote.

I guess in the end it boils down to something vaguely similar to yours, but without so much calculation.  So I suppose it is rather trivial after all. :(

Oh yes, Valiowk, it's indeed simple, I was so stupid...

  Darij


To promote her school's annual Kite Olympics, Genevieve
makes a small kite and a large kite for a bulletin board display.
The kites look like the one in the diagram.

[asy]for ( int x = 0; x <= 6; ++x )
{
    for ( int y = 0; y <= 7; ++y )
    {
        dot((x,y));
    }
}
draw((0,5)--(3,7)--(6,5)--(3,0)--cycle);[/asy]

For her small kite, Genevieve draws the kite on a one-inch grid.  For the large kite
she triples both the height and width of the entire grid.
\\ \\
The large kite is covered with gold foil.  The foil
is cut from a rectangular piece that just covers the entire grid.  How
many square inches of waste material are cut off from the four
corners?

if you look at the small sample, the area cut off is 21.  The larger should be 9*, so 189.


inequalities
LaTeX
inequalities proposed

$ a,b,c$ are three different positive real numbers. Prove that:\[ \left|\frac{a\plus{}b}{a\minus{}b}\plus{}\frac{b\plus{}c}{b\minus{}c}\plus{}\frac{c\plus{}a}{c\minus{}a}\right|>1\]

The enquality equivalent :
$ |\frac{\sum_{sym}a^{2}.b\minus{}6abc}{a^{2}.c\plus{}c^{2}.b\plus{}b^{2}.a\minus{}a^{2}.b\minus{}b^{2}.c\minus{}c^{2}.a}| > 1$.

$ \Leftrightarrow |\sum_{sym}a^{2}.b\minus{}6abc| > |a^{2}.c\plus{}c^{2}.b\plus{}b^{2}.a\minus{}a^{2}.b\minus{}b^{2}.c\minus{}c^{2}.a|$.(1)

By AM-GM , $ \sum_{sym}a^{2}.b\ge 6abc$.

So (1)$ \Leftrightarrow\sum_{sym}a^{2}.b\minus{}6abc>|a^{2}.c\plus{}c^{2}.b\plus{}b^{2}.a\minus{}a^{2}.b\minus{}b^{2}.c\minus{}c^{2}.a|$.(2)
We consider two cases

[u]Case 1[/u] $ a^{2}.c\plus{}c^{2}.b\plus{}b^{2}.a > a^{2}.b\plus{}b^{2}.c\plus{}c^{2}.a$.

Then (2)$ \Leftrightarrow\sum_{sym}a^{2}.b\minus{}6abc > a^{2}.c\plus{}c^{2}.b\plus{}b^{2}.a\minus{}a^{2}.b\minus{}b^{2}.c\minus{}c^{2}.a$

Or $ 2(a^{2}.b\plus{}b^{2}.c\plus{}c^{2}.a) > 6abc$
It's right ( by AM - GM ), and the equal sign not hold because $ a,b,c$ are diferent numbers .

[u]Case 2[/u] $ a^{2}.c\plus{}c^{2}.b\plus{}b^{2}.a < a^{2}.b\plus{}b^{2}.c\plus{}c^{2}.a$.
Similary.

Done :)

[quote="Phạm Thành Quang"][quote="Omid Hatami"]$ a,b,c$ are three different positive real numbers. Prove that:
\[ \left|\frac{a\plus{}b}{a\minus{}b}\plus{}\frac{b\plus{}c}{b\minus{}c}\plus{}\frac{c\plus{}a}{c\minus{}a}\right|>1\]
[/quote]

Easy...

$ \frac{a\plus{}b}{a\minus{}b}\equal{} x;\frac{b\plus{}c}{b\minus{}c}\equal{} y;\frac{a\plus{}c}{c\minus{}a}\equal{} z$

$ \rightarrow xy\plus{}yz\plus{}xz \equal{} 1$

$ (x\plus{}y\plus{}z)^{2}\geq 3(xy\plus{}yz\plus{}xz) \equal{} 3\rightarrow |x\plus{}y\plus{}z|\geq\sqrt{3}> 1$[/quote]
Very nice  :D .These problem is an old inequality . :)

Just some corrections.

Actually, $ xy\plus{}yz\plus{}zx \equal{}\minus{}1$. But this is not a big deal, since $ (x\plus{}y\plus{}z)^{2}\equal{} x^{2}\plus{}y^{2}\plus{}z^{2}\plus{}2(xy\plus{}yz\plus{}zx) \equal{} x^{2}\plus{}y^{2}\plus{}z^{2}\minus{}2 > 1$, because $ x^{2}$, $ y^{2}$ and $ z^{2}$ are all bigger than $ 1$.

Note also that the bound cannot be improved: if $ a \equal{} c^{3}$ and $ b \equal{} c^{2}$, then $ x \equal{} y \equal{}\frac{c\plus{}1}{c\minus{}1}$ and $ z \equal{}\frac{1\minus{}c^{2}}{1\plus{}c^{2}}$. Letting $ c\to\infty$, we have $ x,y\to 1$ and $ z\to\minus{}1$, and $ x\plus{}y\plus{}z$ goes to $ 1$.

There is also another reasoning:
Obviously the inequality is equivalent to proving $ \sum_{sym}a^{2}b\minus{}6abc\geq |(a\minus{}b)(b\minus{}c)(c\minus{}a)|$ now if we replace each
$ a,b,c$ with $ a\minus{}t,b\minus{}t,c\minus{}t$ where $ t\le\min{a,b,c}$ then the LHS decreases while RHS doesnt change thus we can safely suppose $ c\equal{}0$ then we are left with $ a\plus{}b\geq |a\minus{}b|$ which is easy... :lol:

there is a (somehow) similar problem:
determine the maximum value of:
$ f(a,b,c)\equal{}\left|\frac {a \minus{} b}{a \plus{} b} \plus{} \frac {b \minus{} c}{b \plus{} c} \plus{} \frac {c \minus{} a}{c \plus{} a}\right|$
where $ a,b,c$ are the side lengths of a nondegenerate triangle.

[quote="Zarif"]This is second my second solution.[/quote]

But it's already posted. Can't you read the above posts?

[quote="Pain rinnegan"]But it's already posted. Can't you read the above posts?[/quote]
It's not like anything bad happens if the same solution is posted twice... Furthermore it can be helpful to ones proof writing skills to write down the proof, and the site is a nice place to save the solutions.

@Zarif: $ xy\plus{}yz\plus{}zx\equal{}\minus{}1$ in your second solution, so it is wrong. Se cyshine correction further up for more details. It's more convenient to other users if your write it in LaTeX and post it directly on the site. See [url]http://www.mathlinks.ro/LaTeX/AoPS_L_About.php[/url].[/url]

Let $x = \frac{a+b}{a-b}$ , $y =\frac{b+c}{b-c}$ , $z=\frac{c+a}{c-a}$ . It is easy to see $xy+yz+zx=1$ .
We can see $(x+y+z)^2 \geq 3(xy+yz+zx) = 3$ .
Now we can see
$$|x+y+z| \geq \sqrt{3} > 1$$ . $\blacksquare$


Is there a formula for the number of non-congruent triangles that have the same vertices as another regular polygon? For example, the number of non-congruent triangles that have the same vertices as a regular pentagon is 2. Is there a general formula for more than just a triangle. I think I have gotten somewhere for this, but I can't seem to finish it.

[quote="PowerOfPi"] Is there a general formula for more than just a >> triangle. <<
[/quote]
I don't know what you mean by this sentence.

I thought you were referring to more than a pentagon (regular),
and not a triangle, where you are looking for a possible formula
for the number of noncongruent triangles that can be drawn inside
of regular polygons (with sides $ \ge 4$).

I am thinking about the general "number of non-congruent $ n$-sided polygons using vertices of an $ m$-sided regular polygon".

For example, the number of quadrilaterals that can be drawn in a regular pentagon is just one. There are two ways to draw a quadrilateral in a regular hexagon, and so on.

So far, I have gotten that this is the same as distributing $ m\minus{}n$ identical objects into $ m$ groups. I will post a full explanation later if I get further. I was just looking for a simpler way to do it.

What's the formula for "distrubuting m-n objects identical objects into m groups"?

That is partly what I am trying to figure out. See my other post [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=316623]here[/url]


In a special football game, a team scores 7 points for a touchdown and 3 points for a field goal. Determine the largest mathematically unreachable number of points scored by a team in an (infinitely long) game.

[hide]11[/hide]

chicken mcnugget formula.

You are supposed to use Bezout's Formula ;)

isn't bezout about intersecting functions?

i can't see a way to set this problem in functions. (it looks easy though :( )


what do you think of High school musical?

the movie?

well I guess it was somewhat different compared to other Disney movies. a little less childish?

And the singing was really just.... bizarre   :)

Can't wait til the pop up edition on friday!!!!!!!!!

lol
-jorian
ps: i might watch it b/c i dont have anything going on then...my sis + bro will want to watch it, so ill be nice :P

i thought it was okay...but some parts were just plan cheesy...but yah i agree with ignite..."were all in this togehter".....high school musical 2 wasn't that bad...the bad thing is that zack efron can't even sing....they remix his voice sooo much it's not funny...but yet he's still famous...


geometry
rectangle
trapezoid
geometry solved

Triangle $QAP$ has the right angle at $A.$ Points $B$ and $R$ are chosen on the segments $PA$ and $PQ$ respectively so that $BR$ is parallel to $AQ.$ Points $S$ and $T$ are on $AQ$ and $BR$ respectively and $AR$ is perpendicular to $BS$, and $AT$ is perpendicular to $BQ.$ The intersection of $AR$ and $BS$ is $U,$ The intersection of $AT$ and $BQ$ is $V.$ Prove that

(i) the points $P, S$ and $T$ are collinear;

(ii) the points $P, U$ and $V$ are collinear.

Can anybodyhelp?
thanks

[color=darkred][b]An equivalent enunciation.[/b]
Let $ABRQ$ be a rectangle trapezoid $(AQ\parallel BR\ ,\ AB\perp AQ)$.
Denote the points : $U\in AR\ ,$ $BU\perp AR\ ;$ $V\in BQ\ ,$ $AV\perp BQ\ ;$ $S\in AQ\cap BU\ ;$ $T\in AV\cap BR\ .$ Prove that the lines $AB\ ,\ RQ\ ,\ UV\ ,\ ST$ are concurrently.[/color]

[color=darkblue][u]Very nice problem ![/u]  I shall come back soon (with the proof, if it is possibly).[/color]

[[b]Moderator edit:[/b] This problem is also problem 4 of the Norway olympiad (Niels Henrik Abel Contest) 1997-98 finals .]

Indeed it is not hard.

(i) is trivial, because \[ \triangle TBA \sim \triangle BAQ \Rightarrow TB=\frac{AB^2}{AQ} \] and \[ \triangle SAB \sim \triangle ABR \Rightarrow SA= \frac{AB^2}{BR} \], which together imply $\frac{BT}{BR}=\frac{AS}{AQ}$.

For (ii), Let $H=BS \cap TA$ and $K = BQ \cap RA$. Then $HK//AQ$ as \[ \frac{TH}{HA}=\frac{BT}{AS}=\frac{BR}{AQ}=\frac{RK}{KA}. \]
Let $HK$ cuts $PA$ at $X$. Since \[ \frac{BX}{XA}=\frac{BK}{KQ}=\frac{BR}{AQ}=\frac{PB}{PA} \] and by Ceva we have \[ \frac{AX}{XB}=\frac{AU}{UK}\frac{KV}{VB}, \] the by Menelaus $P,V,U$ are collinear.

[quote="Virgil Nicula"][color=darkred][b]An equivalent enunciation.[/b]
Let $ABRQ$ be a rectangle trapezoid $(AQ\parallel BR\ ,\ AB\perp AQ)$.
Denote the points : $U\in AR\ ,$ $BU\perp AR\ ;$ $V\in BQ\ ,$ $AV\perp BQ\ ;$ $S\in AQ\cap BU\ ;$ $T\in AV\cap BR\ .$ Prove that the lines $AB\ ,\ RQ\ ,\ UV\ ,\ ST$ are concurrently.[/color]

[/quote][color=darkblue][i]I came back and I saw the nice Ypsoi's proof which  suggestes me interesting lemma :[/i][/color]

[color=red][b]Lemma.[/b] Let $ABCD$ be a trapezoid $(AD\parallel BC)\ .$ Prove that if two from the following three statements are truly then and the other statement is truly :
$1.\blacktriangleright\ AB^2=BC\cdot AD\ .$
$2.\blacktriangleright\ AC\perp BD\ .$
$3.\blacktriangleright\ AB\perp AD\ .$ [/color]

Here is a short proof using the Pascal theorem.

[quote="ice_age"]Triangle QAP has the right angle at A. Points B and R are chosen on the segments PA and PQ respectively so that BR is parallel to AQ. Points S and T are on AQ and BR respectively and AR is perpendicular to BS, and AT is perpendicular to BQ. The intersection of AR and BS is U, The intersection of AT and BQ is V. Prove that

(i) the points P, S and T are collinear;
(ii) the points P, U and V are collinear.[/quote]

We rewrite < QAP = 90° as $AQ\perp AB$. Also, from BR || AQ and $AQ\perp AB$, it follows that $BR\perp AB$. Since $AR\perp BS$, we have < AUB = 90°, and similarly < AVB = 90°. Thus, the points U and V lie on the circle with diameter AB. The tangent to this circle at the point A is the line AQ (because it is perpendicular to the diameter AB and passes through the endpoint A of this diameter), and the tangent to this circle at the point B is the line BR (because it is perpendicular to the diameter AB and passes through the endpoint B of this diameter). Now, applying the Pascal theorem to the cyclic hexagon AAUVBB inscribed into the circle with diameter AB, we see that the points $AA\cap VB$, $AU\cap BB$ and $UV\cap BA$ are collinear, where the lines AA and BB are the tangents to the circle with diameter AB at the points A and B, respectively, i. e. the lines AQ and BR. Now, $AA\cap VB=AQ\cap VB=Q$, $AU\cap BB=AU\cap BR=R$ and $UV\cap BA=AB\cap UV$. Thus, the points Q, R and $AB\cap UV$ are collinear. In other words, the point $AB\cap UV$ lies on the line QR. Applying the Pascal theorem to the cyclic hexagon AAVUBB inscribed into the circle with diameter AB, we see that the points $AA\cap UB$, $AV\cap BB$ and $VU\cap BA$ are collinear. Now, $AA\cap UB=AQ\cap UB=S$, $AV\cap BB=AV\cap BR=T$ and $VU\cap BA=AB\cap UV$. Thus, the points S, T and $AB\cap UV$ are collinear, i. e. the point $AB\cap UV$ lies on the line ST.

Altogether, we see that the point $AB\cap UV$ lies on the lines QR and ST. Thus, this point coincides with the point of intersection of the lines AB and QR, i. e. with the point P, and hence it follows that the point P lies on the lines UV and ST. In other words, the points P, U, V are collinear, and the points P, S, T are collinear. The problem is solved.

  Darij

[quote="ice_age"]Triangle QAP has the right angle at A. Points B and R are chosen on the segments PA and PQ respectively so that BR is parallel to AQ. Points S and T are on AQ and BR respectively and AR is perpendicular to BS, and AT is perpendicular to BQ. The intersection of AR and BS is U, The intersection of AT and BQ is V. Prove that

(i) the points P, S and T are collinear;
(ii) the points P, U and V are collinear.

Can anybodyhelp?
thanks[/quote]

Do you have any other problem from this IMO training?? And another question, I see people here solving problems from actual competitions. Do the solutions get transferred to the main page where the database of the problems is?? I've seen some of the problems with solution but for instance, problems from the turkey olypiad were solved recently here. Wouldn't it be nice to move the solutions people post there??

Daniel


This is harder than most of the questions on this website. Define time

Hm, this seems to belong to the Physics forums, don't you think?

However, I am not sure how many people on AoPS can supply a proper scientific definition of time (eg. the one used in Einstein's General Relativity) -- and even scientific definitions seem to have faults... As far as I know, there is no general definition of time.

For practical (ordinary, every-day) purposes, time can be defined simply as yet another coordinate (or dimension, if you will) which is used to define an exact position of an event in a certain system of reference. The nature of this coordinate, alas, seems to remain a mystery -- but is, actually, of no importance if you're not a physicist or a philosopher.

time is just something measured by regular cycles, though regular is defined by time and the whole thing get more and more complicated until the tortoise wins his race against achilles :rotfl:

This is indeed one of the more difficult questions to answer, and in fact, occupies a quarter of our philosophy course.  The central idea is that time can be viewed as either static or dynamic, sometimes called the "Parminadean conception of time" as opposed to the "Heraclitean conception of time".

The former, static, refers to a "block universe", where we view time as streaks through a fourth dimension.  This conception of time is the most generally accepted today in the sciences (largely as a result of Einstein).

The latter, dynamic, refers to a something in which there is neither past nor future, but always an (if i may so call it) eternal present.  This may sound crazy, but addresses the problem of the assymetry between past and future.  That is, we often believe the past is not open to change, whilst the future is... whereas the a dynamic theory claims that both the past and present are open.

Having said this, this is only a tiny portion of the debate in the philosophy of time, and I am sure any philosophers out there will both vehemently agree and disagree with what i have just said.

this should be in the college forum or something!

No, the physics forum is more appropriate, as djole suggested.

Because Joe is always right, off to the physics forum it goes.

This was moved off to the physics forum before I even got a say, but I don't know whether Time should really be in the physics forum.  Besides, it is philosophy, not physics, that deals with the conception of Time.  Really, in physics, only one model is proposed of time - the block universe - and it might be hard to get an overall view of what time actually is.


[quote="ayush_2008"]If a+b+c=0 and a,b,c are not equal then find the value a^2/ (2a^2+bc) +b^2/ (2b^2+ac) +c^2/ (2c^2+ab)[/quote]

$ \frac {a^2}{2a^2 \plus{} bc} \equal{} \frac 13\frac {3a^2\minus{}a(a\plus{}b\plus{}c)}{a^2 \plus{} bc \minus{} a(b \plus{} c)}$ $ \equal{} \frac 13\frac {a(a\minus{}b)\plus{}a(a\minus{}c)}{(a \minus{} b)(a \minus{} c)}$ $ \equal{} \frac 13(\frac {a}{a \minus{} b} \plus{} \frac {a}{a \minus{} c})$

so the required sum is $ \frac 13(\frac {a}{a \minus{} b} \plus{} \frac {a}{a \minus{} c}$ $ \plus{} \frac {b}{b \minus{} c} \plus{} \frac {b}{b \minus{} a}$ $ \plus{} \frac {c}{c \minus{} a} \plus{} \frac {c}{c \minus{} b}) \equal{} 1$


Hi all,
Reading this forum I have learned that I should use align, multiline and split for my numbered, multiple-line equations and inequalities, instead of eqnarray.
But I am still having trouble making my output look nice.
These two ugly examples use the (bad) eqnarray environment.
Please could anyone show me how to make them beautiful using the new environments ?
\begin{eqnarray}
\label{eq: new inductive step}
&				& \|(2V_{n_1} - I) \circ \cdots \circ (2V_{n_j} - I) \circ i y_j\|	\leq \\
& 			& \|(2V_{n_1} - I) \circ \cdots \circ (2V_{n_{j-1}} - I) \circ i x_{j-1}\| + \|(2V_{n_1} - I) \circ \cdots \circ (2V_{n_{j-1}} - I)((2V_{n_j} - I) \circ i y_j - i x_{j-1})\| \nonumber \\
& <			& \|(2V_{n_1} - I) \circ \cdots \circ (2V_{n_{j-1}} - I) \circ i y_{j-1}\|\|x_{j-1}\| + 2 (2K + 1)^{j-1} \sum_{k = M_{n_j - 1} + 1}^{M_{n_j}} s_k \alpha_k \nonumber \\
& =			&	\|(2V_{n_1} - I) \circ \cdots \circ (2V_{n_{j-1}} - I) \circ i y_{j-1}\|\|i x_{j-1}\| + 2 (2K + 1)^{j-1} \sum_{k = M_{n_j - 1} + 1}^{M_{n_j}} s_k \alpha_k \nonumber \\
& \leq	& \|(2V_{n_1} - I) \circ \cdots \circ (2V_{n_{j-1}} - I) \circ i y_{j-1}\| \left ( \|(2V_{n_j} - I) \circ i y_j\| + \right . \nonumber \\
&				& \left . \mbox{} + \|(2V_{n_j} - I) \circ i y_j - i x_{j-1}\| \right ) + 2 (2K + 1)^{j-1} \sum_{k = M_{n_j - 1} + 1}^{M_{n_j}} s_k \alpha_k \nonumber \\
& <			& \|(2V_{n_1} - I) \circ \cdots \circ (2V_{n_{j-1}} - I) \circ i y_{j-1}\| \left ( \|(2V_{n_j} - I) \circ i\| + 2 \sum_{k = M_{n_j - 1} + 1}^{M_{n_j}} s_k \alpha_k \right ) + \nonumber \\
&				& \mbox{} + 2 (2K + 1)^{j-1} \sum_{k = M_{n_j - 1} + 1}^{M_{n_j}} s_k \alpha_k \nonumber \\
& \leq	& \|(2V_{n_1} - I) \circ \cdots \circ (2V_{n_{j-1}} - I) \circ i y_{j-1}\| \left ( 1 + \sum_{k = M_{{n_j - 1}} + 1}^{M_{n_j}} s_k (\delta_k + 2 \alpha_k) +  \right . \nonumber \\
&				& \left . \vphantom{\sum_{k = M_{n_j - 1} + 1}^{M_{n_j}}} \mbox{} + 2 \beta_{n_j} \right ) + 2 (2K + 1)^{j-1} \sum_{k = M_{n_j - 1} + 1}^{M_{n_j}} s_k \alpha_k. \nonumber
\end{eqnarray}
and
\begin{eqnarray}
\label{eq: big important inequality}
&				& \|(2V_{n_1} - I) \circ \cdots \circ (2V_{n_P} - I) \circ i\| \leq \\
& 			& \left ( \prod_{j=2}^P \left ( 1 + \sum_{k = M_{n_j - 1} + 1}^{M_{n_j}} s_k (\delta_k + 2 \alpha_k) + 2 \beta_{n_j} \right ) \right ) \cdot \left ( 1 + \sum_{k = M_{n_1 - 1} + 1}^{M_{n_1}} s_k \delta_k + 2 \beta_{n_1} \right ) + \nonumber \\
&				& \mbox{} + \sum_{j=2}^{P-1} \left [ \left ( \prod_{i=j+1}^P \left ( 1 + \sum_{k = M_{n_i - 1} + 1}^{M_{n_i}} s_k (\delta_k + 2 \alpha_k) + 2 \beta_{n_i} \right ) \right ) 2 (2K + 1)^{j-1} \sum_{k = M_{n_j - 1} + 1}^{M_{n_j}} s_k \alpha_k \right ] + \nonumber \\
&				& \mbox{} + 2 (2K + 1)^{P-1} \sum_{k = M_{n_P - 1} + 1}^{M_{n_P}} s_k \alpha_k. \nonumber
\end{eqnarray}
Thank you very much !
First post - be gentle !  :)

You've done well with such complicated material. How about this for the second example?
[code]\begin{equation}
\label{eq: big important inequality}
\begin{split}
&\|(2V_{n_1} - I) \circ \cdots \circ (2V_{n_P} - I) \circ i\| \leq \\
&\qquad \left ( \prod_{j = 2}^P \left ( 1 + \sum_{k = M_{n_j - 1} + 1}^{M_{n_j}} s_k (\delta_k + 2 \alpha_k) + 2 \beta_{n_j} \right ) \right )  \cdot \left ( 1 + \sum_{k = M_{n_1 - 1} + 1}^{M_{n_1}} s_k \delta_k + 2 \beta_{n_1} \right )  \\
&\qquad + \sum_{j = 2}^{P - 1} \left [ \left ( \prod_{i = j + 1}^P \left ( 1 + \sum_{k = M_{n_i - 1} + 1}^{M_{n_i}} s_k (\delta_k + 2 \alpha_k) + 2 \beta_{n_i} \right ) \right )  2 (2K + 1)^{j - 1} \sum_{k = M_{n_j - 1} + 1}^{M_{n_j}} s_k \alpha_k \right ]  \\
&\qquad + 2 (2K + 1)^{P - 1} \sum_{k = M_{n_P - 1} + 1}^{M_{n_P}} s_k \alpha_k.
\end{split}
\end{equation}[/code]
If you'd prefer the equation number on the last line instead of the middle then replace
\usepackage{amsmath}
by
\usepackage[tbtags]{amsmath}.
I leave the first example as an exercise :)

stevem,
Thank you very much for your help and your kind comments !
That is fantastic !
Your version is much better looking than mine !
I will do the first example the same way.
Many thanks again !

A note: some of your expressions are very long. On this forum, they get split (with spacing that doesn't look very good). In a document, they might well run off the side of your page into unprintable territory. Watch out for this, and split further if necessary.

jmerry,
Thank you very much for your comment !
I will look out for that.
I really appreciate you guys taking the time to help. :D


If $ X > 3$ find $ Y$ such that $ Y \equal{} |X| \minus{} |3\minus{}X|$

$ X$ is positive here, so we know that $ |X| \equal{} X$.  Also, we know that $ |3 \minus{} X| \equal{} X \minus{} 3$.  We want to find $ X \minus{} (X \minus{} 3)$; all we have to do is make this equal to $ X \minus{} X \plus{} 3$, so our final answer is $ \boxed{3}$.


search
linear algebra
matrix
vector
linear algebra unsolved

Suppose $e_{1},e_{2},\dots,e_{m}\in\mathbb R^{n}$ that for each $i\neq j$ we know that $(e_{i},e_{j})<0$. Prove that $m\leq n+1$

This has been posted before, by alekk I believe (one of the threads at least; I think there were several).

Since I don't know what to search for to find one of those threads, here's a proof:

Let $M$ be the matrix with columns $e_{i}$, and let $A=M^{T}M$. $A$ is symmetric with positive entries on its main diagonal and strictly negative entries off the diagonal. $A$ is also positive semidefinite, so no eigenvalue is negative.
Choose $c$ greater than any $\langle e_{i},e_{i}\rangle$; $B=cI-A$ has all entries positive. Let $\lambda$ be the 2-norm of $B$; the maximum of $\frac{|Bv|}{|v|}$. This $\lambda$ is the square root of the largest eigenvalue of $B^{T}B$, and $v$ is an eigenvector. If we replace negative entries with positive entries, we can only increase this number, and any maximal $v$ is a vector with nonnegative entries. It's easy to show that the entries of $v$ are actually positive; replacing a zero entry with $\epsilon$ changes $|v|$ negligibly compared to $|Bv|$.
Since $B$ is symmetric, we can write $B=X^{T}DX$, with $X$ orthogonal and $D$ diagonal. Then $B^{T}B=X^{T}D^{2}X$, and the eigenvectors of $B$ are the eigenvectors of $B^{T}B$. These eigenvectors are orthogonal, so there's only one eigenvector $v$ with positive entries, and this is the unique eigenvector of $B$ with eigenvalue $\lambda$.

Now, since $\lambda$ is the largest eigenvalue of $B$, $c-\lambda$ is the smallest eigenvalue of $A$. If $c-\lambda>0$, the rank of $A$ is $m$ and $m\le n$. If $c-\lambda=0$, the rank of $A$ is $m-1$ and $m\le n+1$.

There are certainly many other ways to write the middle section; this is just the first way I thought of to put it together.

Here's the [url=http://www.mathlinks.ro/Forum/viewtopic.php?highlight=negative+vector&t=31714]topic[/url], jmerry.

I think that your proof is the same as Moubinool's (or some variation of it). Sorry if I'm saying stupid things right now, but I'm very tired  :(

Bye Radon's theorem if $m\geq n+2$ then there exist $\alpha_{1},\alpha_{2},\dots,\alpha_{m}\in\mathbb R$ that $\sum_{i=1}^{m}\alpha_{i}e_{i}=0$ and $\sum_{i=1}^{m}\alpha_{i}=0$. So all of $\alpha_{i}$ do not have the sam sign. WLOG you could write $\lambda_{1}e_{1}+\dots+\lambda_{k}e_{k}=\lambda_{k+1}'e_{k+1}+\dots+\lambda_{m}'e_{m}$  that $\lambda_{1},\dots,\lambda_{k},\lambda_{k+1}',\dots,\lambda_{m}'\geq 0$ and all of them are not zero. Suppose $\lambda_{1},\lambda_{k+1}'>0$. So \[0>\sum_{i,j}\lambda_{i}\lambda_{j}'(e_{i},e_{j})=(\sum_{i=1}^{k}\lambda_{i}e_{i},\sum_{i=k+1}^{m}\lambda_{i}'e_{i})=(\sum_{i=k+1}^{m}\lambda_{i}'e_{i},\sum_{i=k+1}^{m}\lambda_{i}'e_{i})\geq0\] Contradiction solves the probelm. :D


geometry
circumcircle
incenter
symmetry
geometry unsolved

[b]1.[/b] Let $ \{N\} \equal{} EU\cap AO; \ \{M\} \equal{} EU\cap AB$ and AD altitude of A in $ \triangle ABC$.
[b]2.[/b]$ \angle ANM \equal{} \angle ADC \equal{} 90^0$ and line AO and AD isogonal conjugate in $ \triangle ABC \Rightarrow$
$ \Rightarrow \triangle{AMN}\sim \triangle{ACD} \Rightarrow \angle AMN \equal{} \angle ACD \Rightarrow \\
\Rightarrow MBCE \minus{} cyclic \ quadrilateral \Rightarrow \angle AEU \equal{} \angle ABC$;    (1)
[b]3. [/b]$ \triangle{AEU}\equiv \triangle{AFU} \Rightarrow \angle AEU \equal{} \angle AFU;$         (2)
[b]4. [/b]   (1) and (2) $ \Rightarrow \angle AFU \equal{} \angle ABC \Rightarrow$  FU||BC.

[quote="jayme"]Dear Mathlinkers,
let ABC a triangle, O the center of the circumcircle, I the incenter, E, F the points of contact of the incircle with AC, AB, U the point of intersection of the perpendicular to AO through E with AI.
Prove  that FU is parallel to BC.
Sincerely
Jean-Louis[/quote]
[hide="Solution"]
Let $ AI$ meet $ BC$ at $ P$.  Now, $ \angle OAC\equal{}90\minus{}\angle ABC$.  Since $ (AO, UE)\equal{}90$, we see that $ \angle AEU\equal{}\angle ABP$.  Also, $ \angle BAP\equal{}\angle UAE$, so $ \triangle AEU\sim \triangle ABP$, so $ \frac{AU}{AP}\equal{}\frac{AE}{AB}\equal{}\frac{AF}{AB}$, so $ UF\parallel BC$, as desired. [/hide]

Dear Mathlinkers,
my problem can be associated with this reference :
http://www.mathlinks.ro/viewtopic.php?t=223729
Sincerely
Jean-Louis

[b]1. and 2. [i]Alternative proof for [/i](1):[/b]
$ EM\perp AO \Rightarrow \ line \ EM \ antiparallel \width \ respect \ to \ side \ BC \ in \triangle{ABC}\Rightarrow \\
\Rightarrow MBCE - cyclic \ quadrilateral\Rightarrow \angle AEU = \angle ABC.$    (1)

ÐOAC=90º-ÐB and, with EU and AO perpendicular , ÐAEU=ÐB; due to symmetry, ÐAFU=ÐAEU, thence ÐAFU=ÐABC, FU||BC
(Here Ð means 'angle').

Best regards,
sunken rock


function
limit
real analysis
real analysis solved

The function $f(x)$ which is defined at $x>0$ satisfies the following three conditions $(a),(b),(c)$

$(a)\ f''(x)<0$
$(b)\ f(x)>0$
$(c)\ \lim_{x\rightarrow \infty} f(x)=1$

Find $\lim_{x\rightarrow \infty} f'(x)$.

You have that $f'(x)$ is monotonically decreasing and it can't get less than $0$ (Otherwise the limit c) wouldn't exist).
This gives that $d=\lim_{x\rightarrow \infty} f'(x) $ exists.
But now you easy get a contradiction if $d$ is not zero.

Hmm... I think I did something wrong because I didn't use b) ...  :?

[quote="ZetaX"]This gives that $d=\lim_{x\rightarrow \infty} f'(x) $ exists.
But now you easy get a contradiction if $d$ is not zero.
[/quote]
Now we can apply Lagrange's Mean-value Theorem,
$f(x+1)-f(x)=f'(\xi)$
Taking $x\to\infty$ for both sides, we get $d=0$.


algorithm
algebra
polynomial
number theory
greatest common divisor
superior algebra
superior algebra unsolved

Show that the gcd of the following pairs of polynomials is 1 using the euclidean algorithm.    I may just be making computational errors but I'm stuck and I can't seem to find mine. Please help.

x^3 + x^2 -2x -2 and x^4 – 2x^3 + 3x^2 - 6x

for f,g polynomials if  degree(f)>degree(g) do gcd(f,g)=gcd(f-g,g)
repeat the step until one of the polynomials is of degree 1 i think

Not really... that's for numbers. Here we should use long division.


find all integer pairs $(x,\ y)$ that $x^{2}+x+11=y^{2}$

[hide="Tell-all (do not look if you want to solve it yourself)"]$(2x+1)^{2}+43=(2y)^{2}$[/hide]

Since squares are just the sum of consecutive odd integers, I just did $\frac{43-1}{2}$, then $\frac{21-1}{2}$.  One could similarly do that for the -11.


calculus
derivative
limit
calculus computations

The existence of a derivative at $ c$ is equivalent to saying that there exists a number $ f'(c)$ such that $ \Delta y \equal{} f'(c) \Delta x \plus{} \Delta x E(c, \Delta x)$, where $ E(c, \Delta x) \equal{} \frac {\Delta y}{\Delta x} \minus{} f'(c)$ and $ \lim\limits_{\Delta x \to 0} E(c, \Delta x) \equal{} 0$.

Why do we define it like this? What is the point? Why not define it as $ \Delta y \equal{} f'(c) \Delta x$? This is an $ o(n)$ term.

That last definition you propose doesn't work because $ \frac{\Delta y}{\Delta x}$ is not a constant: it just happens to approach a limit as $ \Delta x\rightarrow 0$.

(For example, if $ y\equal{}x^2$ and $ c\equal{}1$, then $ \frac{\Delta y}{\Delta x}\equal{}\frac{x^2\minus{}1}{x\minus{}1}\equal{}x\plus{}1\equal{}2\plus{}\Delta x$; we can't just say $ \frac{\Delta y}{\Delta x}\equal{}2$.)


This is where we post links to a bunch of poems.

Example:

[url]http://bau2.uibk.ac.at/sg/poe/Work.html[/url]

As the link suggests, that is a collection of Edgar Allen Poe's work. Try finding "The Raven" and "The Bells".

oh come on! Don't tell me you don't know where to find poems!!!


[quote="GameBot"]What is the units digit of $ 29^{57}$?[/quote]

Fixed (an error?! :o  :wink:  :P), and $ 29^{57}\equiv( \minus{} 1)^{57}\equiv \minus{} 1\equiv\boxed9\pmod{10}$.

When I saw this problem it came out as 29^5*7
the 5 was in subscript but not the 7 =[


geometry
symmetry
geometric transformation
homothety

$\triangle{DCG}$ is equilateral. $ABCD$, $CIHG$, and $GFED$ are squares. $M$, $N$, and $O$ are midpoints. Prove that $\triangle{MNO}$ is equilateral.

[hide="Er..."] If $\triangle DCG$ is equilateral then isn't this obvious by symmetry? [/hide]

ya i agree..are u looking for a geometric proof or something??

I guess a proof would be nice...using a homothety if possible.  :ninja:


Slove the system of the equations: $\begin{cases}6(x-\frac{1}{y})=3(y-\frac{1}{z})=2(z-\frac{1}{x}=xyz-\frac{1}{xyz}\end{cases}$

[hide]Call all those equalities equal to a.
Multiply the first three together.  $a^{3}=36(xyz-x-y-z+1/x+1/y+1/z-1/xyz)$=$a^{3}=36(a-a/6-a/3-a/2)=0.$  Therefore, $a=0$ and x,y,z=1 or -1 are the only solution.[/hide]
I didn't see that Farenhajt already posted it, sorry.

Mods, merge [b]dengmi[/b]'s post to [url=http://www.mathlinks.ro/Forum/viewtopic.php?t=123800]this topic[/url], and delete [url=http://www.mathlinks.ro/Forum/viewtopic.php?t=123796]this topic[/url]. Thanks :)


linear algebra
matrix
linear algebra unsolved

see:   http://www.artofproblemsolving.com/Forum/posting.php?t=106550

Let Un the (n,n) matrices whose entries are all either 0 or 1.
i) Find explicitly a matrix A in Un such that A^2,...,A^n are in Un and A^(n+1) is not in Un.
ii) Prove or disprove:
if A,..,A^(n+1) are in Un, then for all positive integers k,
A^k is in Un.

:blush: If n=2 then A, A^2 are in Un implie that for all k, A^k is in Un.
  Thus here we assume in i) and ii) that n>=3.
   Remark: ii) is true for n=3,4.


Outline a synthesis of 1,3-dimethylcyclopent-3-en-1-ol from 2-methyl-1,3-butadiene.

2-methyl-1,3 butadiene + heat(hv)-> 1 methyl cyclobutene
1 methyl cyclobutene +PCC/ch2cl2/n2 atmosphere-> methyl cyclo butenone(allylic oxidation).
tiffeneu demyanov rearrangement to give methyl cyclopentenone which on reaction with ch3mgbr gives
1,3-dimethylcyclopent-3-en-1-ol
is this okay  :?:

[quote="VARUNARASU"]1 methyl cyclobutene +PCC/ch2cl2/n2 atmosphere-> methyl cyclo butenone(allylic oxidation).[/quote]

There are three allylic positions available, two of them secondary. Which one will be oxidized?

[quote="VARUNARASU"]tiffeneu demyanov rearrangement to give methyl cyclopentenone[/quote]

With which reagents would you do that?

[quote="carcul"]
With which reagents would you do that?
[/quote]
ch3no2+base
foll by ni/h2+hno2


The unknowns $ a,b,c,d$ are in $ \mathcal{M}_2(\mathbb{C})$.
1) Solve the system:
$ \{a\plus{}b\plus{}c\equal{}0,a^2\plus{}b^2\plus{}c^2\equal{}0,a^3\plus{}b^3\plus{}c^3\equal{}0,a^2\not\equal{}{0}\}$.
1) Solve the system:
$ \{a\plus{}b\plus{}c\plus{}d\equal{}0,a^2\plus{}b^2\plus{}c^2\plus{}d^2\equal{}0,a^3\plus{}b^3\plus{}c^3\plus{}d^3\equal{}0,a^4\plus{}b^4\plus{}c^4\plus{}d^4\equal{}0,a^2\not\equal{}{0}\}$.

I give the solution for system 1) (with 3 equations):
I show that the system has no solutions.
$ a^2\equal{}a_0I\plus{}a_1a,b^2\equal{}b_0I\plus{}b_1b,c^2\equal{}c_0I\plus{}c_1I$, where $ a_i,b_i,c_i\in\mathbb{C}$.
$ a^3\equal{}a_0a_1I\plus{}(a_0\plus{}a_1^2)a,\cdots$.
If 2 among $ a,b,c$ commute then $ a,b,c$ pairwise commute and are simultaneously triangularizable. Thus considering the diagonals we obtain: $ spectrum(a)\equal{}spectrum(b)\equal{}spectrum(c)\equal{}\{0\}$ and $ a^2\equal{}0$, a contradiction.
We have the following linear system in the unknowns $ a,b,c$:
$ \{a\plus{}b\plus{}c\equal{}0,a_1a\plus{}b_1b\plus{}c_1c\equal{}\minus{}(a_0\plus{}b_0\plus{}c_0)I,(a_0\plus{}a_1^2)a\plus{}(b_0\plus{}b_1^2)b\plus{}(c_0\plus{}c_1^2)c\equal{}\minus{}(a_0a_1\plus{}b_0b_1\plus{}c_0c_1)I\}$; it's matrix is $ \begin{pmatrix}1&1&1\\a_1&b_1&c_1\\a_0\plus{}a_1^2&b_0\plus{}b_1^2&c_0\plus{}c_1^2\end{pmatrix}$ and it is of rank<3, else $ a,b,c$ would be scalar matrices. Moreover if it is of rank 2 then, for example, $ a\equal{}uc\plus{}vI$ and $ ac\equal{}ca$, a contradiction. Then it's rank is 0 or 1 and $ a_1\equal{}b_1\equal{}c_1$, therefore $ a_0\plus{}b_0\plus{}c_0\equal{}0$; $ a_0\plus{}a_1^2\equal{}b_0\plus{}b_1^2\equal{}c_0\plus{}c_1^2$ therefore $ a_0\equal{}b_0\equal{}c_0$ and the relation $ a_0a_1\plus{}b_0b_1\plus{}c_0c_1\equal{}0$ is satisfied.
Conclusion: $ a\plus{}b\plus{}c\equal{}0,a^2\equal{}a_1a,b^2\equal{}a_1b,c^2\equal{}a_1c$. In particular $ a_1\not\equal{}0$ and we may assume $ a_1\equal{}1$. $ a,b,c$ are projectors s.t. $ a\plus{}b\plus{}c\equal{}0$. $ a,b,c$ are not scalar matrices then $ trace(a)\equal{}trace(b)\equal{}trace(c)\equal{}1$ that is contradictory with $ trace(a\plus{}b\plus{}c)\equal{}0$.  QED.
PS: System 2) (with 4 equations) admits solutions.


Is it possible to find an algorithm for finding the number of odd and even entries in any nth row in pascals triangle.
I don't want hints its either you can solve it or you can't!

The number of odd entries in the $ n$th row of Pascal's Triangle is $ 2^a$ where $ a$ is the number of $ 1$'s in the binary representation of $ n$.


integration
limit
function
real analysis
real analysis unsolved

[b]Ascertain[/b] for $ a\ >\ 0$ :

$ \odot\ \ \int_0^\infty e^{\minus{}x}x^a \mathrm {dx}\ .$

$ \odot\ \ \lim_{n\to\infty}\ \frac {n!\cdot n^a}{a(a\plus{}1)(a\plus{}2)\ldots (a\plus{}n\minus{}1)(a\plus{}n)}\ .$

It's Gamma. :) From the definition.. The first one is the classic definition, and the second one in Euler's definition.

hello
[hide="1"]
By definition of Gamma function
$ \int_0^\infty e^{\minus{}x}x^a \mathrm {dx}\ .$
=Gamma[a+1,x]
check this ::http://reference.wolfram.com/mathematica/ref/Gamma.html
[/hide]
thank u


This forum was created at the request of Dr. George Thomas, founder of MathPath.

I believe Dr. Thomas intends this as a place where former and prospective students can discuss the program... to find out how things work at MathPath; to talk about the topics that are taught at MathPath; to talk about the non-math activities that occur at MathPath; to connect with others who have been to MathPath or who might be going to MathPath, and probably for lots of other good discussions.

I do not believe Dr. Thomas intends this as a place to discuss the Qualifying Quiz.  So, unless Dr. Thomas explicitly relaxes these rules, DO NOT DISCUSS any aspect of the Qualifying Quiz.

[list]Do not discuss the problems.

Do not discuss your solutions.

Do not ask for help with the problems.

Do not discuss difficulty of the questions.

Do not discuss strategies for taking the test.

Do not say anything that will give a student who reads this forum any advantage or appearance of any advantage over a student that does not have access to this Forum.

You may state whether or not you have been admitted to the camp.
[/list]

I will contact Dr. Thomas, to see if he wants to provide further clarification.  Meanwhile, feel free to discuss the [b]MathPath program[/b], but not the [b]MathPath Quiz[/b].

I agree absolutely with Mr. Verser.

In order that it does not confer an unfair advantage to prospective applicants, please do not discuss the problems in the current MathPath Qualifying Quiz. That is, discussion of the problems, their solutions, solution strategies, or asking for help with an aspect of a problem should be avoided.

You can ask for clarification of the statement of a problem.  You can also write to camp@mathpath.org for clarification.
Clarifications may also be found on the [url=http://www.mathpath.org/clarify.htm]Clarifications page[/url] of the [url=http://www.mathpath.org]MathPath website[/url].

--- Thomas / MathPath

This is a reminder that the Qualifying Quiz window is still open.  Do not discuss any aspect of the Qualifying Quiz.

[quote="daermon"]okay, when can we talk about our solutions?[/quote]
Exceptions to the deadline have sometimes been granted to National MATHCOUNTS Competitors, who can't meet the official deadline.  There have also been rare exceptions in which students may be admitted to MathPath at the last minute.

In the past, the staff has used student solutions to the qualifying quiz as a teaching tool during the first few days of camp.  [You might see good solutions, with exceptionally clear write-ups.  You might see solutions with exceptional insight, but with poor write-ups.  Part of what you will learn at MathPath will be how to write better solutions.]

So, unless Dr. Thomas say otherwise on this thread, assume you may not discuss your solutions until you arrive at MathPath.

[quote="funcia"][quote="rcv"]
In the past, the staff has used student solutions to the qualifying quiz as a teaching tool during the first few days of camp.  [You might see good solutions, with exceptionally clear write-ups.  You might see solutions with exceptional insight, but with poor write-ups.  Part of what you will learn at MathPath will be how to write better solutions.]
[/quote]

I hope they don't say who the students with the bad write ups are. Do they?[/quote]

I went to MathPath last year and they didn't attach names to the solutions they presented.

Yes, they don't say who wrote the solutions.

Last year one of my solutions evidently had the best idea of everyone but the actual solution was completely wrong (the one about the bees), and it was shown.

I don't think anyone solved the one with the bees...

There were no names attached to either good or bad write-ups. I think some of the really bad solutions were from people not accepted... there was one that looked more like someone's notes than a solution.

Since the Qualifying Test season is upon us, I remind everyone to read and comply with the rules posted in the first two posts of this thread.  [size=150][b]In summary, do not discuss the Qualifying Test.[/b][/size]


limit
advanced fields
advanced fields unsolved

Let $ f(n)$ denote the maximum possible number of right triangles determined by $ n$ coplanar points. Show that \[ \lim_{n\rightarrow \infty} \frac{f(n)}{n^2}\equal{}\infty \;\textrm{and}\  \lim_{n\rightarrow \infty}\frac{f(n)}{n^3}\equal{}0 .\]

[i]P. Erdos[/i]

I guess the maximum will occur in the following configuration: [embedded picture]


parameterization
geometry
3D geometry
number theory unsolved
number theory

Find all positive integers x,y,z such that
$x^3 + y^3 + 2 = z^3 $

seems to me that the only solutions are $(5,6,7)$ and $(6,5,7)$... can't prove it though  :?
looking forward to see someone solve this actually :)

Actually, for each $ t\in\mathbb{N}$ it holds:

\[ (6t^3\minus{}1)^3\plus{}(6t^2)^3\plus{}2\equal{}(6t^3\plus{}1)^3\]

The solutions mentioned above are precisely of this form (for $ t\equal{}1$ and $ t\equal{}2$).

However, there is [b]no[/b] reason to assume that all solutions are given by this parametrization, since similar equations (that is, involving three cubes and a constant) may have very large solutions out of any scheme. Moreover, if we allow integer solutions (instead of restricting to natural ones) to the given equation, there is an example of solution not of the given form (for $ t\in\mathbb{Z}$, of course):

\[ 3528875^3\minus{}1214928^3\plus{}2\equal{}3480205^3\]


Prove that if $x^2+y^2+1=xyz$, where x, y, z are positive integers, then z=3.

$x^2-(zy)x+y^2+1=0$
If $x>y>1$,then suppose $x_1=zy-x=\frac{y^2+1}{x}<y<x$
And east to see $(x_1,y,z)$satisfied the equation,too.
And if $y>x_1>1$,then we can continue to get $(x_1,y_1,z)$ and so on.
Since $x,y$ are positive integers,so at last we get $(a,1,z)$.
So $a^2-az+2=0$ Since $a$ is positive integer $a|2,a=1,2$
Easy to see $z=3$.

[quote="jin"]$x^2-(zy)x+y^2+1=0$
If $x>y>1$,then suppose $x_1=zy-x=\frac{y^2+1}{x}<y<x$
And east to see $(x_1,y,z)$satisfied the equation,too.
And if $y>x_1>1$,then we can continue to get $(x_1,y_1,z)$ and so on.
[/quote]

   Sorry for my stupid question. But how can we continue the step for $(x_1,y_1)$. SInce at the quadractic equation you prove that it has 2 root:$x,x_1$ and $x >y>x_1$. But how can you find $x_2$?
   Thanks you very much :)

[quote="jin"]I mean we can find $y_1<x_1<y$ and $(x_1,y_1,z)$ also a solution.[/quote]

  I am so dumb in number theory. But may you show me how to find it.
  As I mention before. You prove that there is exist $(x,y)$ such that $x>y$ and then you prove that the other solution is $x_1$ satisfy $x>y>x_1$. But this is a quadractic form and it should have only 2 roots. How can you find another root $x_2$ for this? I am not sure about this step since you just postulate it without proof

I mean we consider a solution such that $x>y$,we want to find a smaller one.
then consider the equation about $t$,$t^2-(yz)t+y^2+1=0$
One solution is $x$,the other is $x_1$,so we can know $x_1<y<x$.
And $(x_1,y,z)$ is a solution too.
So $(y,x_1,z)$ is also a solution such that $y>x_1$
Then we can find a smaller one and a smaller smaller one,since it can't continue infinitely.
So at last we get $x=y=1$


Is it okay to email twice if I'm not sure that my email went through?

thank you for the response

We'd prefer if you used a different method (mail or fax), and attached a note explaining what you think happened.  Getting two emails is somewhat confusing for us.

Let me add -- only do this if you have a legitimate reason for suspecting that your email didn't go through.


take a look on this web site !
http://mathworld.wolfram.com/MacBeathCircle.html
the geometry graphic is amazing, it is so beautyfull !!!
I need a soft to make the picture like that, i tried with Cabri, Geometer Sketch Pad, C.A.R ... but the result is not good, the " resolution " of my pictures is so poor ...
help me please
thank you alot
(i have mathematica 5.2, but i dont think that soft is used to make the graphic on mathworld.wolfram.com ?)

you can try Geometry Expressions

a god software

or the best of ever is AutoCad ;)

OMG, that s exactly what i need, but do you know where could i download that ?
the price of that soft is too high and the demo version limited user only 20 minutes,
if you have it, please send me that soft
my email address is nguyenhaduy1990@yahoo.com,
thank you alot :D

http://www.geometryexpressions.com/downloads.php

[quote]
Full Featured 30 day Trial Version

Get a first hand look at what Geometry Expressions can do by downloading our 30 day trial. The trial is fully featured but expires after 30 days. All constraints, constructions and measurements are available for you to use.
[/quote]

Is 30 days the trial...

After that You can just go to the clock in the windows and set a dat.

When I use it i set the date to 21 Feb 2006, the date I got it

;)

How about KSEG?

You could get Google Sketchup... :D

If any of you want the image as an avatar:


Some pictures from HMMT 2007 are up. I'll put up more over the next few days. Both the staff and the competitors did a great job. I know the staff works very hard  - many of the names are already posted here or listed in the program. I'll be happy to to credit to anyone in the pictures (I know a good portion of the staff already). Jut PM me to let me know it ok with you.

The gallery.

http://www.artofproblemsolving.com/Forum/album.php?t=136947


I've been hearing these people on the radio talking about the economy, and honestly I think that some/lots of them have no idea what they are talking about. The economy now has values of money in the millions, billions, and trillions. Some people have no idea what these numbers mean.

http://vimeo.com/4428480

[img]http://9gag.com/photo/4041_540.jpg[/img]

[hide="a trillion dollars in 100 dollar bills"][img]http://ttoes.files.wordpress.com/2009/04/image0066.jpg[/img][/hide]

What is your opinion on this?
[hide=" "]I know I'm gonna be called an idiot again.[/hide]

I'm not quite sure what you want us to debate.  I don't think the reporters are dishonest in what they're reporting.

I get angry when anyone whose not an expert acts like they are.

The sad thing is that people like her [url]http://en.wikipedia.org/wiki/Jenny_McCarthy#Activism_and_autism_controversy[/url] get credibility

I'm not sure how much credibility she actually has.  Most doctors will dispute her and I think most people in the public no better.

Well my dad studied economics and he says that the inflation rate will become 10-15 percent because Obama says the solution to the recession is to just print money...so in 20 years the median salary will almost be a million...eek...

[quote="fooblah"]eek...[/quote]

The people who spread these kinds of ideas are people like Peter Schiff, economic advisor to Ron Paul's presidential campaign.

While I'm not an expert on Obama's economic policies and plans, you can rest assured that it's not quite, "print, print, print!"

Also, I'll point out that your post was essentially off topic.

There is no topic here to be off of ... I'm basically just waiting for sufficient quantity of posts like fooblah's to pile up before I banish it to G&FF (or whatever it's called now).  Though I do think fooblah is invoking his or her father in exactly the way some earlier posters have complained.  "Studied economics," like, took a couple of classes in college?   :roll:  I'm also curious about what is [i]inherently[/i] troublesome about a median income of a million dollars?

Incidentally, we would need to have above 16% annual inflation for 20 consecutive years to reach the point that current household median income of about 50,000USD would be worth 1,000,000USD in 2030's money -- the suggestion that this would happen  "because of Obama" (who can only be in office for 8 years) is completely ridiculous.

On what I gather to be the intended subject of this thread: it's quite clear that economic decisions have not purely economic outcomes (i.e., they affect real people, not just toy models of the economy), and so it is fully appropriate for non-economists to take part in discussions of what the appropriate economic decisions are.  This is, after all, a nation that likes to call itself a democracy.  If people on the radio talking about the economy are idiots with nothing useful to say, the appropriate solution is to not listen to them (or, I suppose, to call in and tell them they're idiots).

[quote="JBL"]I'm also curious about what is inherently troublesome about a median income of a million dollars?[/quote]
I think the point he is making is that a million dollars will be meaningless due to inflation, although I doubt that will happen.
[quote="JBL"]If people on the radio talking about the economy are idiots with nothing useful to say, the appropriate solution is to not listen to them (or, I suppose, to call in and tell them they're idiots).[/quote]
For some reason, I don't think Rush Limbaugh or Glenn Beck will even give you the opportunity to call them idiots. :wink:

I agree, there's nothing wrong with that.  If everyone is making 1 million in 61 years, then the mega-rich now will be making a lot more.  Everything is relative.

Like many people who have commented already, there is no purpose of this thread.

I personally find it humorous that a student in high school shows discontent for non-economists for not being economically accurate. In what basis, do you find these "some/lots have no idea"? Also, who do you exactly mean by "some/lots"? I am not being bitter about the thread, but like JBL said, this could easily go in Games and Factory due to ambiguity.

And, as others have noted, let's stick to the thread, and stop saying "Because of _____."

I'm a middle schooler and I expected for this to happen. :(

[quote="JBL"][quote="JRav"][quote="JBL"]I'm also curious about what is inherently troublesome about a median income of a million dollars?[/quote] I think the point he is making is that a million dollars will be meaningless due to inflation, although I doubt that will happen.[/quote]  But again, so what?  There's nothing inherently special about a million dollars -- it's just a number that happens to have nice orthography in the number system we use.  Half of households will be making a million dollars eventually in any case, barring major disasters: annual inflation of 5% takes 50,000USD to 1,000,000USD in 61 years, just about my current life expectancy and certainly within the life expectancy of my students.  Inflation also has the positive side-effect that it may eventually kill off the cursed penny -- either that, or cause a re-valutation of the dollar that would make the penny worth non-trivially little.)[/quote]
The issue is that when the government prints money no value has been produced and the effect is that the government gets richer and everybody else gets poorer. It's not that different from demanding that everybody who owns US currency should give away a percentage of their money. Usually one talks about inflation causing uncertainty, but certainly the same uncertainty would be caused if government agents could sporadically come around and demand money from you for no reason.

I think state monopoly on currency is [i]extremely[/i] immoral, and view it no different than extortion from MC gangs. So it upsets me a lot that many discuss this issue so lightly.

[quote="Kalle"]The issue is that when the government prints money no value has been produced and the effect is that the government gets richer and everybody else gets poorer.[/quote]  This is something that also happens when, say, the government imposes any broad-based tax, which is something that all non-crazy people agree is worthwhile in order that public goods will actually be produced.

"The existence of universally-accepted currency" (not the currency itself, mind you) is a public good and is exactly the sort of thing that needs to be a government monopoly if it's going to work at all.

Not true. Disregarding arbitrary lines drawn in the sand, there are already many different currencies, and there are many agents who trade currencies with you for a fee. While I must admit I do not know US laws, there is no reason to forbid private currencies other than the state retaining its power. If one really thinks a private currency cannot work, then would be no need to ban it because it would not exist anyway.

And yes, I think taxes are not morally defendable either. It amazes me that people are so quick to accept that just because an entity of people calls themselves a country, and that most people within that country think it is OK to take someone's property, then that morally justifies taking people's property against their will. I just think that an instance of violence (including theft) can't be morally justified even if it is without a doubt "for the greater good". I also have faith in people and think the position that we need coercion in order to both survive and do good deeds is ridiculous and dishonest. People solve things.

Of course, extreme left academics and activists can justify this within their philosophical framework, but they admit and encourage that it is ok to use violence against the rich. Both violence in its usual sense and in a broader scope that includes theft.

So, just letting you know that I am "crazy" I guess. You probably already suspected it  :wink:. As for how this ties in with the topic, I guess this is an instance of a non-economist talking about the economy.

Normally, in a situtation like this, I would spend time explaining to you what I believe to be wrong about your arguments or claims (much of which has to do with the existence of public goods, of which "universally accepted currency" is surely one).  But when you attribute defenses of taxation (a system that has existed as long as, and is integral to, civilization) to "extreme left academics and activists" it makes me think that really you are far more interested in grinding some philosophical/political axe than you are in making statements that are accurate characterizations of the world we live in.  As a result, it basically strikes me as not worth my effort to respond to you with anything other than this out-of-hand dismissal.

It was not my intention to attribute taxation to the extreme left. Rather, my not-so-clear point is that you rarely see anyone else admit that taxation is an act of violence that somehow has to be justified, and that bothers me.

I do know what a public good is, and I think that many when specifying which goods are public or merit are exaggerating (but I DO believe in state monopoly of violence in form of police, military and judicial system). As you say, I doubt some kind of general discussion will lead us anywhere, but I am interested in hearing what currency has to do with public goods and why it leads to needing to ban private currencies. My position is not as offensive as you think.

In fact, lots of private currencies exist (though they often aren't called that and are usually denominated in some other currency) -- for example, in addition to my credit card (essentially a private form of currency), I have "frequent flier miles" that I can exchange for flights on certain airlies and I carry "rewards cards" and "gift certificates" for various commercial establishments that are exchangeable for goods or services of various sorts.  So, I suppose it's possible that after reading this you'll just agree that you are wrong about the non-existence of private currency (and so moot the rest of my argument), and in any case I will happily submit that this sort of currency is not a public good.

Otherwise, we have to have some discussion about the "universally accepted" part.  And here, the point is just that there are only two reasons for me to accept your currency in exchange for something: either I believe your currency to be inherently valuable (but now we're really just talking about barter, which I hope we can agree is a terrible way to try and run an economy) or because I believe that I'll be able to exchange it for things that I do value.  Now, if you and I happen to live in the same neighborhood and use the same neighborhood bakery and are trading "rewards points" for this neighborhood bakery, that works fine, provided that I believe that the bakery won't close down before I have a chance to redeem the points.  But if I'm trying to conduct business on an even remotely larger scale, this simply fails: if I don't trust the institution backing your currency or if the people selling things I want to buy don't trust that institution, your currency is worthless to me.  This means that institutions can only produce universally-accepted currency if they operate on a scale as large as any business that might be conducted.  So, the range of possible producers of a universally-accepted currency is really quite limited, and also any universally-accepted currency must be widely distributed.  But in this case, universally-accepted currencies trivially meet the definition of public goods: my holding onto some currency doesn't reduce its availability for others (because it's widely available), and no one can prevent me from using it.

The language is quite tricky here, because if I understand you right you don't mean that the currency itself is public, but rather the service of backing the currency up is a public good. However, I believe that this is not such a big problem. As you said, there are credit cards, and at least in sweden commercial agents (e.g. retail stores) do pay the credit card institutions to be able to accept payments from credit cards. And they pay quite a lot too - smaller stores usually charge a fee for using a credit card (although it is disputed if this is legal). Of course, they also have to pay to handle currency, for transport and change and such.

I certainly believe large financial institutes would be willing to back a currency up to make profits. I as a consumer would gladly pay to convert my currency to for example a currency set against the gold standard, given that the latter currency is in use. To introduce such a new general currency would be highly illegal or severely restricted in many countries, and I maintain that it is mainly for control purposes. Many people believe that government is needed to do well in a project as big as controlling a currency. However, when government feels free to print money clearly this means that a private agent could do better. To prevent this, government forbids it. I take the childish view that this is simply unfair.

[quote="JBL"]"The existence of universally-accepted currency" (not the currency itself, mind you) is a public good[/quote]   :wink:   (Sorry, am just heading off to bed ... will edit this for actual content tomorrow some time.)

This article/blog post may be interesting (I saw it on Cogito, many people have probably already seen it...):
[url]http://terrytao.wordpress.com/2009/05/04/the-federal-budget-rescaled/[/url]


induction
modular arithmetic
strong induction

Pls see the attached and explain to me with steps.. THZ A LOT!!

First, use induction to prove that all terms of the sequence are integers.

Then, look at the sequence $ \mod 2$:  $ 1,0,1,1,0,1,1,0,1,1,0,1\dots$.
it's easy to prove by induction that $ a_{n}\equiv 0\pmod 2\iff n=3k+2$.

Now assume that $ a_{n}=0$, for some $ n$. Since 0 is even, we must have $ a_{n-1}\equiv 1\pmod 2$ and $ a_{n-2}\equiv 1\pmod 2$. Thus $ a_{n-1}a_{n-2}$ is odd. Hence $ 0=a_{n}=a_{n-1}-a_{n-2}\Rightarrow a_{n-1}=a_{n-2}$.

So from $ a_{n-1}=5a_{n-2}-3a_{n-3}$ we get $ 3a_{n-3}=4a_{n-2}$. Thus $ 3\mid a_{n-2}=a_{n-1}$. We will now prove for all $ k<n$, that $ 3\mid a_{k}$. For the base case, we have just shown $ 3\mid a_{n-1}=a_{n-2}$. now assume that for some $ k$, $ 3\mid a_{n}, a_{n-1}, a_{n-2},\dots, a_{n-k}$.

if $ a_{n-k-1}a_{n-k}$ is even, then $ a_{n-k+1}=a_{n-k}-a_{n-k-1}\Rightarrow 3\mid a_{n-k-1}$.

else $ a_{n-k}=5a_{n-k-1}-3a_{n-k-2}\Rightarrow 3\mid a_{n-k-1}$. So by induction we have proven that for all $ 1\le k\le n, 3\mid a_{k}$. But $ a_{1}=1$ and $ a_{2}=2$, which aren't divisible by $ 3$. Thus our initial assumption was false and we conclude that $ a_{n}\neq 0$ for all $ n$.

Note:

1. Induction was used three times, so imagine how a powerful tool it can be!

2. In the last part, we used strong induction in the opposite direction: $ P_{n}\Rightarrow P_{n-1}$.

3. We also used that if $ a,b,c$ are integers such that $ a+b=c$ and $ 3$ divides any two of them, then $ 3$ divides the third one as well.

[quote="nayel"]if $ a_{n-k-1}a_{n-k}$ is even, then $ a_{n-k+1}=a_{n-k}-a_{n-k-1}\Rightarrow 3\mid a_{n-k-1}$.

else $ a_{n-k}=5a_{n-k-1}-3a_{n-k-2}\Rightarrow 3\mid a_{n-k-1}$.[/quote]

There are some mistakes in the above argument. I cannot edit my post so here's the correct version:

If $ a_{n-k}=a_{n-k-1}-a_{n-k-2}$, we get $ a_{n-k-1}-a_{n-k-2}\equiv 0\pmod 3\Longrightarrow a_{n-k-1}\equiv a_{n-k-2}\pmod 3$. Thus $ a_{n-k-1}= 5a_{n-k-2}-3a_{n-k-3}\Longrightarrow 4a_{n-k-1}\equiv 0\pmod 3\Longrightarrow 3\mid a_{n-k-1}$.

Else, $ a_{n-k}=5a_{n-k-1}-3a_{n-k-2}\Longrightarrow 3\mid a_{n-k-1}$.


geometry
3D geometry
LaTeX
trigonometry
complex numbers
imaginary numbers

What are the cube roots of 27$\i$?

Also, anyone know how to display LaTeX for a Mac? I already have installed TeXShop and its necessary components, I just can't get the thing to work! :blush:

You don't need to install anything to write LaTex on the forums.

Really? :blush:

I read some of the posts on the LaTeX tab in the forum, so I just enclose whatever I want in between dollar signs, following the rules for symbols? How come when I type $\sqrt{a^2+b^2}$, which is how you are supposed to type the square root of a^2 + b^2, nothing happens? Could you help me?

Uh...never mind! UBER :blush:

It just never showed up on the preview screen...I am sorry I wasted your time...but can you solve my problem?

So, the problem, in LaTeX, is $What are the cube roots of 27i?$

Wow...how do I fix the messed up spacing?  :o

$\text{What are the cube roots of 27i?}$

\text{ message }

To answer your question, lets whip out demoivre's theorem.

(this is all in degrees, btw)

Rewrite this equation as $27cis{90}$  ($\cos{x}+i\sin{x}$)
Raise to the 1/3rd power
${27cis{90}}^{\frac{1}{3}}=3cis{30 +120k}$

We add 120 because it is 360/3, because the roots of 27i are evenly spaced about a circle of radius 27 in the complex plane.

$3cis{30}= \frac{3\sqrt{3}}{2}+\frac{3i}{2}$
$3cis{150}=\frac{-3\sqrt{3}}{2}+\frac{3i}{2}$
$3cis{270}= -3i$ (our expected root)

[quote="sapphyre571"]just by looking, the cube root must be $-3i$
$3i^{3} = 27\times{\sqrt{-1}}^{3}$
$=-27i$
so in order to reach positive $27i$
we must multiply by $-1$
and since $\sqrt[3]{-1} = -1$
the cube root of $27i$ must be $-3i$
wow
a complex number that I solved without help
sweet[/quote]
Well, that's one of them, but there are 3 in total  :)

[quote="sapphyre571"]just by looking, the cube root must be $-3i$
$3i^{3} = 27\times{\sqrt{-1}}^{3}$
$=-27i$
so in order to reach positive $27i$
we must multiply by $-1$
and since $\sqrt[3]{-1} = -1$
the cube root of $27i$ must be $-3i$
wow
a complex number that I solved without help
sweet[/quote]

but its not that hard with demoivre's theorem... and you only found one of the three roots, as any real or imaginary number has $n^{}$ $n^{th}$ roots, as Pakman had already found.

And because $x^3-27i=0$ has 3 roots. Just for fun [hide="Cube roots of 1"] are $1cis0=\boxed{1}$, $1cis120=\boxed{-\frac{1}{2}+\frac{\sqrt3}{2}i}$, and $1cis240=\boxed{-\frac{1}{2}-\frac{\sqrt3}{2}i}$. :D [/hide]

ok nevermind
i am definitely not that far into complex numbers yet :blush:
ugh

***walks off to reread (again) the complex numbers section in AoPS
because he isn't very good at reading comprehension
or complex numbers...***

:D

I have read your guys's responses, and think it is cool you attempted to answer the question. As for the answer, many of you simply listed -3i, when in fact there are multiple answers. I won't ruin the excitement, but was wondering if this is the kind of stuff that is acceptable for the "Intermediate topics" of the "High school" forum, for I am still in middle school, just advanced in math.

Actually, I guess Pakman2012 answered right on the money, with the 3-answer idea, as others suggested, and since it wasn't hidden info, I guess it's okay to release the spoiler after all...

\frac{3\sqrt{3}}{2}+\frac{3i}{2},
\frac{-3\sqrt{3}}{2}+\frac{3i}{2},
AND, OF COURSE
-3i

[quote="Nickzdabomb"]What THE? Arg...my apologies, I still don't understand why the LaTeX code isn't working for me...just look at Pakman2012's post to see the 3 answers.[/quote]

... lol try not to post like 4 times in a row  :D  if you think the code may not be working, use the preview button right next to the submit button to look at your post before positng.

also, to get $\LaTeX$ code to work on the forum, you need to add a currency sign ( $\$$) before AND after the math code. Hope this helps! :D

(and also, for a spoiler you can write it in [hide]hidden text![/hide] by selecting the text you've written and clicking the "hide" button right under where it says the subject of the post.

[quote="Nickzdabomb"]:oops:

Sorry about that...and, yes, that advice is quite beneficial, thanks![/quote]

:D no prob


greatest common divisor
number theory
relatively prime

What is the least three-digit positive integer that has 2, 5 and 7 as factors?

[hide][color=green]2*5*7 = 70
70*2 = 140
140 is the least 3 digit number with 2, 7, and 5 as factors.[/color][/hide]

[hide]
$2cdot5\cdot7=70\\
70\cdot2=140$
140 is the smallest 3 digit number that has 2, 5 and 7 as factors.[/hide]

[hide]since the three numbers are relatively prime, their GCF is 2*5*7 or 70, but 70 isn't a three digit number, so check multiples
140 is the smallest multiple of 70 that is a three digit number.[/hide]

[hide]Multiply the 3 numbers together to get 70. However, this isn't a 3-digit number so multiply by 2 to get 140 as the answer.[/hide]


logarithms
geometry
3D geometry
ratio
complex numbers

According to my Algebra II textbook, if $x$ is a negative number and $m$ is not an integer, $x^m$ is undefined.

Could someone please explain why this might be so? Because when I typed in (-8)^(1/3) into my calculator, I got -2, seeming to indicate that it is defined.

Thanks.

The statement requires qualification.

(i)  If $x$ is a negative number and $m$ is a [i]rational number with a even denominator[/i], then $x^m$ is not [i]real[/i].  (This is because you would be taking the square root of a negative number.)

(ii)  If $x$ is a negative number and $m$ is an [i]irrational real number[/i], then $x^m = e^{ m \ln x}$, which is (always, I think) [i]not real[/i].  (This is more complicated.)

If $x$ is negative and $m$ is irrational, then
(1) if you only define these things in terms of real numbers, it will be undefined
(2) if you allow yourself the use of complex numbers, it will represent an infinite set of complex numbers, just as a rational power represents a finite set of complex numbers.  There is no convenient way to choose among these infinitely many roots, unlike the case of rational numbers in which we can take, for instance, a canonical square or cube root (even though every number has 2 square and 3 cube roots).

[quote="t0rajir0u"](i)  If $x$ is a negative number and $m$ is a [i]rational number with a even denominator[/i], then $x^m$ is not [i]real[/i].  (This is because you would be taking the square root of a negative number.)[/quote]

What about the case where $m=\frac 22$? After all, there's an even denominator. Would you have to say "...denominator, when written in lowest terms"? This could get messy.

Yes, he meant "when written in lowest terms."

are imaginary numbers considered irrational? Can they be?

we just went over this (topic) a few days ago, and this was one of the questions that also appeared.

That depends on what exactly your definition happens to be.  For example, a reasonable definition of irrational is "a real number which cannot be expressed as the ratio of two integers," in which case complex numbers cannot be irrational.  Frankly, though, the choice of definition is fairly arbitrary -- the numbers exist regardless of what names you call them, so as long as it's clear what you're talking about in any given instance, there is no problem.

[quote="pleurestique"]essentially, irrational numbers are more irrational than imaginary numbers because irrational numbers are in fact real[/quote]

That sounds like a pretty "rational" explanation. :rotfl:

But you still need to define "rational".

Some definitions.

$\mathbb{N}$ - the set of counting numbers.  Usually $1, 2, 3, ...$ and occasionally $0, 1, 2, 3, ...$

$\mathbb{Z}$ - the set of integers.  $..., -3, -2, -1, 0, 1, 2, 3, ...$

$\mathbb{Q}$ - the set of rational numbers.  This is defined as any number representible in the form $\frac{p}{q}$, where $p, q \in \mathbb{Z}, q \neq 0$.

$\mathbb{R}$ - the set of real numbers.  The reals are difficult to rigorously define (at least, for me), so suffice it to say that the reals constitute all possible locations of a point along a number line.

$\mathbb{R} / \mathbb{Q}$ - the set of real numbers, without the rationals.  Hence, the set of irrationals.  Note that the irrationals are real.

$\mathbb{C}$ - the set of complex numbers, representible in the form $a + bi$, where $i = \sqrt{-1}, a, b \in \mathbb{R}$.  This includes the imaginary numbers, where $a = 0$.  Note that none of the imaginary numbers belong to the irrationals, and vice versa.  However, the components of a complex number - $a, b$ - can be either irrational or rational.

In short:  No.


We are currently designing our Spring class schedule and would like input from students who plan to take classes during the Spring semester.  Keep in mind that nearly all of the classes we hold will end well before the end of the school year (before final exams).

[quote="NightFlarer"]So I suppose both Intermediate Counting/Probablity and Intermediate Algebra wont both be offered in the Spring like I had hoped?
Or is this topic just to see which class is more popular?[/quote]

It is unlikely, though not impossible that we will offer both.


anybody have any recommendations for national mathcounts preparatory books/resources?

Please use the search function on AoPS- there are already threads on this topic.


rotation
analytic geometry
probability
geometry
geometric transformation
complex numbers

1. Triangle PQR has points P(0,0), Q(3,12), R(6,0).  PQR is then rotated counterclockwise such that the median lies on the y-axis.  Find the new coordinates of point Q.

2.  Randomly pick integers a, b, c, d ranging from -3 to 3.  What is the probability that ad-bc=0?

3.  4 circles with radii 2r are centered at (r,r), (-r,r), (-r,-r), (r,-r).  Find the area of the region that all four overlap in.

#2:
[hide]
ad-bc = 0

We can make a plot of points. ad=bc.
Case 1:
For a, we have 7 picks. For d, we have 7 picks. Now, for b, we have 2 picks, and for c, only one pick.
Case 2:
0 times anything is zero. So, for a, we have 7 picks. d is 0. Then, b we have 7 picks and c is zero. And anycombination of those. Of course, minus the 0(a)=0(a) since those are included in case 1.

Total possibilities to satisfy: (7^2)(2)+4(7^2)-4(7)

Total number of picks possible: 7^4.

So, probability is:

(6(7^2)-4(7))/(7^4) = 38/343

I have the feeling I made a mistake somewhere. Not sure.
[/hide]

[hide="#1"]If $S$ is the midpoint of $QR$, then $S({9\over 2},6)$ and $|PS|={15\over 2}$.

Using complex numbers, the original $S$ corresponds to $s={9\over 2}+6i$ and the rotated image $S'$ corresponds to $s'=|PS|i={15\over 2}i$. The rotation factor is $f={s'\over s}={15i\over 9+12i}={4\over 5}+{3\over 5}i$.

The number corresponding to $Q$ is $q=3+12i$, hence the rotated image $Q'$ corresponds to $q'=qf=(3+12i)\left({4\over 5}+{3\over 5}i\right)=-{24\over 5}+{57\over 5}i$. Therefore $Q' \left(-{24\over 5},{57\over 5}\right)$.[/hide]


Over past month, I worked on old AMC's, AHSMEs, all the current level AMC-10 tests from various nations, and any other high school math contest I can touch to practice for my AMC. My score increased and usually it's at least 110 and at most 122.5. I didn't take any real ones (because I'm out of it) but I'm pretty sure I can do well on AMC. But now, AMC's right front of me.

What should I do now? Should I spend time on doing new problems or going back and do AMC-10 problems all again?

Any help is thankful.

i am probably in a very similar situation as you are, Silverfalcon. i am going over what i can in the Art of Problem Solving books. In addition, i am going over the mock AMC contests on the forum, especially the questions that i missed. Understanding concepts that you have missed on a AMC test or are a little fuzzy on is probably more important than trying to grasp a new concept now. Good Luck!!!   :D

Does anybody know if the 2003 and 2004 AMC 10's tests are avaliable? (I took the 2004 one officially but i accidently shredded it :P  )

if you haven't gone through basically all of aops 1 thats a very good idea. aops 2 is a bit harder, but some sections in there are really useful too. i find that doing the same problems again dont help as much because i remember the solutions, but some of the tougher ones (say the last 5 or so, last 10 on AHSME's) are really good practice. if possible, see if you can solve every amc10 problem and almost every AHSME problem (thats tough, though) that you dont remember the solutions to if you've already seen them.

overall, just do problems that you have not solved before and you should get good practice.

[quote="Silverfalcon"]

What should I do now? Should I spend time on doing new problems or going back and do AMC-10 problems all again?

[/quote]

AMCs are like next week so there isn't time to do new problems. You should go back to AMC-10s and review, review, and review! And good luck!


[quote="1/(ln x)"]The decimal representation of $\frac{m}{n}$, where m and n are relatively prime positive integers and m<n, contains digits 2, 5, and 1 consecutively, and in that order. Find the smallest value of n for which this is possible.[/quote]

[hide]I'm not sure i quite understand this question.. wouldn't it just be 51?[/hide]

My first post in this forum --

What!? This is AIME #14? Seems easy and it only took me ~10 seconds, but I'm probably wrong...

[hide][b] 10 [/b] 251/10. You have 2.51. The numbers are relatively prime. If it's not ten, then it's something lower.[/hide]

Edit: Bleh, n>m. Nvm I'm wrong.

JUST IGNORE ME -- you can read what I wrote in the next post anyways under the quote

[quote="Fanatic"]:blush:  :blush:

Possible answer:

[hide]500. Originates from 103/500, which originates from 2510/10000 [/hide][/quote]

Your only option is .251 because n>m, right...? unless i misunderstood the problem AGAIN....which is likely
wait, can you use 0 as a digit? so wouldnt it be 1000 beacuse it's 251/1000?

maybe the problem means that the decimel contains those digits, but not necessarily just those digits. so it could be .02939384251251251251 or .01251

UGH. I was taking a shower thinking about this problem when i realized how blatantly stupid I was to attach a zero at the end. Sorry folks. Well anyhow,

[hide] 2512/1000 = 1256/5000 = 314/1250 = 157 / 675.

My new answer: 675

Why this is possibly right:

1. You can't attach anything before the 251, because then you'd have a number that isn't divisible by 5 or 2
2. You can't attach anything before 251 and after the 251, because the lowest 100000 divides into by 2's is 3375, which is too high (PLEASE let my calculations be right). Under this logic, you can't put two digits after the 251 either.
3. You can't attach anything else after the 251 and make the eventual denominator smaller than 675. I tried miserably with 0's, the odds are self explanatory, 4 doesn't work as well, neither does 6 or 8.

[/hide]

OK I THINK THIS ONE MAY FINALLY BE RIGHT  :blush:

EDIT: NO !*** IT WAS WRONG AGAIN

[quote="1/(ln x)"]The decimal representation of $\frac{m}{n}$, where m and n are relatively prime positive integers and m<n, contains digits 2, 5, and 1 consecutively, and in that order. Find the smallest value of n for which this is possible.[/quote]

[hide="This question is cool"]
Lemma:  We can make a surjection between values of $\frac{m}{n}$ that begin with $0.251$ and values of $\frac{m}{n}$ that do not begin with $0.251$.

Proof:  Suppose $\frac{m}{n} = 0.d_1 d_2...d_k 251...$.  Then $\frac{10^k m - d_1 d_2 ... d_k}{n} = 0.251...$.  Hence we can assume WLOG that the decimal representation begins $0.251...$.  Q.E.D.  Note that this transformation does not affect the value of $n$, which is why it is valid for solving this problem.

Now:  $0.251 < \frac{m}{n} < 0.252$
$251n < 1000m < 252n$
$n < 250(4m - n) < 2n$

In all likelihood, the smallest solution comes when $4m - n = 1$, meaning $n < 250 < 2n$.  Then $4m = 1 + n \ge 126 \implies 1 + n = 128 \implies n = \boxed{127}$.  For verification purposes, $\frac{32}{127} = 0.2519685...$
[/hide]

Edit:  Fanatic, the decimal doesn't have to end.  :D

[quote="Fanatic"]*beats self on head for thinking 7th grader could possibly solve a seemingly easy [b] #14 [/b] AIME problem... UGH[/quote]

Ummm actually a 7th grader most certainly could. I solved a #14 (if not this one!) at some point in 8th grade with some luck and a lot of time. 7th grade is not far from 8th grade, and #14's can sometimes be easier than you would expect ;).

Nuh uh, it's easy, but it's not THAT easy.

Yeah, I'd say 4 or 5, maybe 6. But 13? Definitely not.

[quote="darktreb"]How about 13 from that year? Riiiiiiiiiiidiculous, should have been #1 or 2.[/quote]

Ahahahaha... at least they could have made it a tetrahedron.


percent
calculus
Columbia
geometry
geometric transformation
reflection

The world's greatest (Don't contradict me because the Barnes and Noble Editors said this) memoir is about to published. Clinton who is considered this generation most brilliant politician. It shows the hard work Clinton has done.
[quote]It is the fullest, most concretely detailed, most nuanced account of a presidency ever written, and a testament to the positive impact on America and on the world of his work and his ideals. [/quote]

Plus if you are a really really Republican. I suggest that you read Dear Americans by Ronald Reagan. He may be crappy on internal government, but had the coolness that charmed everyone.

I am not going to read it.  It would probably be interested but isn't it something like 900 pages?  That would take me forever.  I haven't read any memoir before so don't consider this as an attack against Clinton.

I'm going to read it. It sounds quite interesting, and I'm curious to find about Clinton's own perspective on his political career.

I might read it just because I need something to read but it will probaly only make me hate Clinton more.

I have enougth summer reading to do, and I am not that interested in Clinton's life.

Where is the 'No way!  Why would I waste my time like that?' option?

...although I agree with what tetrahedr0n said.

[quote="churchilljrhigh"]Holy Cow! Preorders are exceeding 2 miliion. I wonder how many Harry Potter had. Order your copy NOW!
Double Cow! Harry Potter sold 5 million in the first 24 hours. But we still have hope! THe book hasn't started selling![/quote]
So essentially what you're saying is that about seven tenths of one percent of the country wants to read this book badly enough that they've preordered it. To me, that's not particularly impressive. It's only one thirtieth of one percent of the world's population, after all.

i second that "why would i waste my time on this trash?" i don't give a crap why he did any of the trashy things he did.
you know all of the people who pre-ordered that book are going to flip to the part where he talks about monica. the american public is crazy like that.

[quote="tokenadult"]The reviews of the book seem mostly to agree that it's too long and not honest enough.[/quote]

not honest enough? you dont say! a guy that lied in federal court? NOT HONEST ENOUGH? *gasp*... [/sarcasm]

[quote="bleumoose"]not honest enough? you dont say! a guy that lied in federal court? NOT HONEST ENOUGH? *gasp*... [/sarcasm][/quote]

You could at least be factually correct yourself.  Federal court, it was not.

Confuted -- you're obviously missing the point.  2 million people may not be a lot, but compared to similar events it certainly is.  Actually, the thing that your line most reminds me of is Bush responding to hundreds of thousands of war protestors by saying that he doesn't listen to focus groups.  <sigh>

I will probably not read it, unless someone I know lends it to me.  Or at least I would wait for the soft-cover :)

[quote="JBL"]You could at least be factually correct yourself.  Federal court, it was not.[/quote]

I'm trying to remember in which of several cases the deposition in which "It all depends on what the definition of 'is' is" occurred. William Jefferson Clinton has been a party to a lot of litigation in the last several years. A party to a case has just as great an obligation to tell the truth in a pretrial deposition as in testimony in a trial, so "lying in federal court" may not be a wholly inaccurate characterization, if the case was in fact federal. (I can't remember whether the lying pertained to a state case or a federal case, but that's not a big deal either. Lying under oath IS a big deal, although every lawyer sadly knows that it happens every day.)

[quote="Rep123max"]If you say Clinton had the chance to get Bin Laden one more time without any facts except you saying it (no citing from an actual book-by actual, I mean not by Hannity, Limbaugh, etc.) I will go crazy.[/quote]
http://www.infowars.com/saved%20pages/Prior_Knowledge/Clinton_let_bin_laden.ht
http://www.msnbc.msn.com/id/4540958/
http://www.amazon.com/exec/obidos/tg/detail/-/0895260743/104-9857695-9484745?v=glance
http://www.cnn.com/2002/US/09/03/clinton.bin.laden/
http://www.freerepublic.com/focus/news/535793/posts

So perhaps I don't have a book to cite from, and I'm not going to go buy one, but there are a few of the myriad of links I found searching for "clinton bin laden" on Google. One link that I included was to the amazon.com listing of a book about the subject; I also got MSNBC and CNN in there. Hopefully that's reliable enough to convince you.

[quote="gauss202"]As I've said before, politics is more about people's passions and personal interests than it is about either logic or truth.[/quote]

which is why i attempt to stay out of it. but that is me, i hope that some of the logical people on this forum will make an attempt in politics, we could use you!

The U.S. gets hundreds of terrorist threats a year.  There was nothing credible to the 9/11 one.  THe U.S.S. Cole had been bombed not long before they got the warning.  Security was tightened more than usual.  In the following months though as always happenes we relaxed our stance on security.  That made it relitively simple for the highjackers to get the planes.  The security level was about average on 9/11.  If any good came out of the attack it has taught us that we are not immune to attacks.  Far from it in fact.  Bush had nothing to do with 9/11.  Clinton did not have anything to do with it.  It happened at a time when we had our guard down and that was our fault.  Bush did nothing to prevent it true, but he also didn't get much warning on it either.  He was unlucky to be the President during the term when the terrorists attacked.  Clinton had a relitively safe term where nothing happened.  That probally didn't have anything to do with  him.   He just had quiet terms while the highjackers were planning the attacks.  Think about it we know that Osama Bin Ladin wanted the attacks launched a year earlier.  That didn't happen so it fell in Bushs term.  If it had not then we would all be blaming Clinton for it.  THe point is it was no ones fault and it never will have been anyones fault except the terrorists.  Hopefully we have learned from it.

It is something about americans that none of us want to take the blame.  We always try to lay it on someone else.  Thats not right and something that needs to be over come.  Bush had his faults but so did Clinton they were president at two different times that can not really be compared.

[quote="gauss202"]As I've said before, politics is more about people's passions and personal interests than it is about either logic or truth.[/quote]

Passion is such an easy fruit to grow fat on.

Hmm...This is interesting.  Have any of you read Michael Moore's [u]Dude, Where's My Country[/u]  I think you should all read it before you Clinton's book.  Haha, some of the stuff makes me want to move to Canada!  Haha, yeah, once you go see Farenheit 9/11, go see Bowling For Columbine.  In case none of you knew this before, days after 9/11 Osama's family, who were probably the ONLY ones who knew of his whereabouts, were ESCORTED by OUR planes to Iraq (or somewhere, correct me if I'm wrong) without proper questioning.   All this occured when NO COMMERCIAL AIRLINES WERE ALLOWED TO TRAVEL.   So don't tell me that, "Oh, Clinton should've taken him when we had the chance."  Um, we had the CHANCE again, and we shouldn't have screwed it up.  Seriously, don't you LEARN from you mistakes.  Also, haven't we killed more Iraqis than what happened on 9/11?  It's like, "Girlfriend, you just did-n't mess wit my man!  Well, I did ALL yours so it even.  Aiight?"   Well...perhaps a better analogy, "You killed 3,000 people in my country!  Oh, we just did the same (or more).  Oh well, sorry for the spillover."  Haha, sorry if this isn't coherent.  I'm tired.

[quote="SingTheSorrow"]Osama's family, who were probably the ONLY ones who knew of his whereabouts[/quote]

As much as I love things with which to blame the Bush Administration, this is almost certainly untrue.  The bin Laden family is huge, and while it would have been appropriate to interview them before allowing them to leave the country, it is very unlikely that any of them had close ties to or knew the whereabouts of ben Laden.  Besides which, we (roughly) knew where he was -- with the Taliban in Afghanistan.

But why give THEM planes to get away, and then attack and imprison random Indian, Pakistani, Afghani people?   Seems kind of retarded.

I don't think is was a good idea -- I just didn't think your characterisation of it was accurate.

Could the moderator kindly splice the last few posts onto the Fahrenheit 9/11 thread?

As I have read about this in the mass media, the large Bin Laden family includes people with divergent opinions about politics and religion--just like most families I know in the United States and in Taiwan. Because Muslim men can take multiple wives, families can become HUGE in just a few generations, and Osama Bin Laden has a lot of relatives who are, according to what I read in the press, not at all in agreement with him about the Saudi monarchy or the role of Islam in government.

Blaming someone because of something a relative of his did, even a relative with the same family name, is just the same, old same-old of guilt by association that sophisticated thinkers know better than to engage in. It would be no more valid to criticize Michael Moore's movie if one of his relatives were found to be a child-molesting pervert or a drug dealer than it is to criticize peaceful members of the Bin Laden family because their most famous relative is an avowed terrorist.

Based on the clips I've seen, and I hope to see the full-length movie soon, the most slamming parts of the movie Fahrenheit 9/11 to George W. Bush are the parts that show George W. Bush being himself. By contrast, the speculation about the Bin Laden family is much less convincing to me.

[quote]Roger Moore's movie if one of his relatives were found to be a child-molesting pervert or a drug dealer[/quote]

Wait, do you mean Mike Moore, or do you mean Roger Moore (the person who played James Bond after Sean Connery)?  Did this really happen w/Roger Moore?

[quote="JS1527"]Wait, do you mean Mike Moore, or do you mean Roger Moore?[/quote]

I meant Michael Moore (and have edited my post to reflect that), the director of the film Fahrenheit 9/11 and earlier the director of the film Roger and Me, which is where the name Roger got into my mind from.

I know this is an old subject but I would like to share something with you all. Everytime you mention George Bush, Bill Clinton, or the name of any other government executive name, someone is watching. People do not realize that the government has to keep a close watch because of what has been happening in the world in the past couple years. So remember, everytime you mention a government official or a politicians name, someone knows. Someone is watching whether that person be good or bad. No I am not talking about normal people such as you and I, but rather ones we should perhaps... fear. They're watching. Beware of what you say.

No duh someone is watching. I'm watching this topic, and so are the several hundred other people on this forum. You wouldn't post something on a public forum if you wouldn't want anyone to read it. I really don't see your point.

No need to be paranoid, this isn't Communist Russia in the 1940s where if you insult Stalin, you'll be beheaded.  It's America.


ABC is a triangle.G is the barycenter of the ABC.G' is the conjugate point of G.O is the centre of the circumcirle of ABC.
BG' intersects AC at D.CG' intersects AB at E. l1 is perpendicular to OG' and it intersects DE at F and BC at J.
Prove JG'=2G'F

Let $ AM\parallel{}l1$，so we just need proof $ AL\equal{}ML$ .
It's well knows $ l1\parallel{}PQ\parallel{}AM$，and $ P,A,D,C$ is harmonious，so we can easy get $ AL\equal{}ML$ .
So $ JG'\equal{}2G'F$，$ JF\equal{}KF$，and we can see $ K,F,G',J$ is harmonious，In fact it's abaot a problem of harmonic pencil of lines.

Sorry I can't understand what you mean?WHY IS IT enough to prove that AL=ML?
And why is l1 is parrelel to PQ?

Sorry，I'm going to down，who can help me to explaine for $ gb2124$？
Or，next I will give a explaine for you.


Men, women and children make purchases at a counter in a shop according to independent Poisson processes with rates 10, 12 and 4 purchases per hour respectively. If the store opens at 9am, find an expression for the probability that exactly 30 purchases have been made by noon. Given that 30 purchases have been made by noon, find an expression for the probability that equal numbers of men, women and children made the purchases.

Note: the sum of independent Poisson RV's is a Poisson RV, with the expectation that it would have to have.


let a,b,c be positive numbers such that: abc=a+b+c+2
prove that :abc-2(1/a+1/b+1/c) ≥ 5

we need to prove that:
$ abc \minus{} \frac {2\sum ab}{abc} \geq 5$
$ (\sum a \plus{} 2)^2 \minus{} 2(\sum ab) \geq 5(\sum a \plus{} 2)$
$ \sum a^2 \geq \sum a \plus{} 6$
remember that:
abc=a+b+c+2 then $ \sum a \geq 6$
$ \sum (a^2 \plus{} 4)\geq \sum 4a$
$ 3\sum a \geq 18$
sum up and
Q.E.D :D