calculus
integration
conics
parabola
analytic geometry
graphing lines
slope

1)$ l_1: y \equal{} \sqrt {3}x \plus{} a$ intersect $ C$ at exactly one point$ \implies x^2 \equal{} \sqrt {3}x \plus{} a$ has a double root$ \implies (x_3 \minus{} \frac {\sqrt {3}}{2})^2 \equal{} a \plus{} \frac {3}{4} \equal{} 0$
2)$ l_2: y \equal{} \frac { \minus{} 1}{\sqrt {3}}x \plus{} b$same thing$ \implies (x_1 \plus{} \frac {1}{2\sqrt {3}})^2 \equal{} b \plus{} \frac {1}{12} \equal{} 0$
3)$ l_1$ and $ l_2$ intersect at $ x_2 \equal{} \frac {1}{2\sqrt {3}}$
then$ A \equal{} \int_{x_1}^{x_2}[x^2 \plus{} \frac {1}{\sqrt {3}}x \plus{} \frac {1}{12}]dx \plus{} \int_{x_2}^{x_3}[x^2 \minus{} \sqrt {3}x \plus{} \frac {3}{4}]dx$

Hmm...

$ \phi\minus{}\frac{\pi}{4}\equal{}\theta\equal{}\arctan{\sqrt{3}}\equal{}\frac{\pi}{3}\Rightarrow \phi\equal{}\frac{7\pi}{12}$
$ m_2\equal{}\displaystyle\tan{\left(\frac{\pi}{4}\plus{}\frac{\pi}{3}\right)}\equal{}\frac{1\plus{}\sqrt{3}}{1\minus{}\sqrt{3}}\equal{}\minus{}(2\plus{}\sqrt{3})$

Solve $ C^{\prime}\equal{}\{m_1, m_2\}$ to find $ x_1, x_2$.
Then $ y\equal{}x_i^2$ for $ i\equal{}1,2$ and you get your points of tangency.
After that, solve for $ \ell_1\equal{}\ell_2$ to find $ x_i$.

$ A\equal{}\displaystyle\int_{\minus{}(\sin\theta\plus{}1)}^{\sin\theta\minus{}1} (x^2\plus{}(2\plus{}\sqrt{3})x\plus{}7/4\plus{}\sqrt{3})\,dx\plus{}\int_{sin\theta\minus{}1}^{\sin\theta} (x^2\minus{}x\sqrt{3}\plus{}3/4)\,dx$
[hide="I got"]$ \boxed{A\equal{}\sqrt{3}\plus{}\frac{1}{3}}$[/hide]
Did it by hand...  :stretcher:

AYAMANE, the slope of $ l_2$ is incorrect.

igiul, the intervals of Integration for finding the area are incorrect.

[[quote="kunny"]Consider the parabola $ C: y \equal{} x^2$ in the $ x \minus{} y$ plane. Let $ l_1$ be the tangent line of $ C$ with slope $ \sqrt {3}$ and $ l_2$ be the tangent line of $ C$ with negative slope which makes angle $ 45^\circ$ with $ l_1$. Find the area bounded by $ C,\ l_1$ and $ l_2$.[/quote]

Slope of $l_2=\tan \biggl(\frac {\pi}{3}+\frac {\pi}{4}\biggr)=\tan \biggl(\frac {7\pi}{12}\biggr)=-(2+\sqrt 3)$

Every tangent to $ C: y \equal{} x^2\ {}$ will be of the form $\ y=mx-\frac{m^2}{4}\ {}$ and point of contact being $\biggl(\frac m2,\frac {m^2}{4}\biggr)$

$\implies l_1: y=\sqrt {3}x-\frac{3}{4}\ {}$ and point of contact being $\biggl(\frac {\sqrt {3}}{2},\frac {3}{4}\biggr)$
and
$ l_2: y=-(2+\sqrt 3)x-\frac{7+4\sqrt 3}{4}\ {}$ and point of contact being $\biggl(-\frac {2+\sqrt 3}{2},\frac {7+4\sqrt 3}{4}\biggr)$

Point of intersection of $l_1$ and $l_2$ is $\biggl(-\frac {1}{2},-\frac {3+2\sqrt 3}{4}\biggr)$

Hence Required area $=\int_{-\frac {2+\sqrt 3}{2}}^{-\frac 12}\biggl( x^2+(2+\sqrt 3)x+\frac{7+4\sqrt 3}{4}\ {}\biggr)dx +\int_{-\frac 12}^{\frac {\sqrt {3}}{2}} \biggl(x^2-\sqrt 3x+\frac 34\ {}\biggr)dx=\boxed{\frac{5+3\sqrt3}{6}}$


articles
email
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MATHCOUNTS
AoPS Books
\/closed

This might be kind of stupid, but I thought it might be a good idea to have some sort of an e-mail newsletter or something, that posts "math news", helpful formulas, tips, etc.

It could be a monthly e-mail newsletter or something like that...this is probably a stupid idea, but what do you all think?  :?:

Or you can subscribe to the magazine M&IQ  :D . They usually contain written articles by mathematicians all over the world.

I guess, but the thing about my idea was that it would sort of be focused on improving problem solving skills and such.

AoPS books don't give you [b]up-to-date[/b] information though.

An email might be nice, but rrusczyk does have updates in his blog every now and then.

Well, anything they wrote in these articles would be useless to the majority of the people, if they were about problem solving. MathCounts stuff is useless to people beyond that, etc.

then again--not all the STUDENTS (the parents will) have e-mails.


algebra
difference of squares
special factorizations
number theory unsolved
number theory

For $ a,b \in \mathbb{Q}$, assume that the equation $ ax^2 \plus{} by^2 \equal{} 0$ has at least one nonzero rational solution. Prove that the equation $ ax^2 \plus{} by^2 \equal{} t$ has at least one rational solution for all $ t \in \mathbb{Q}$. :)

[quote="brunojoyal"]$ a(p/q)^2 \plus{} b(p'/q')^2 \equal{} 0$
$ a(pq')^2 \plus{} b(p'q)^2 \equal{} 0$

Hence the equation has a solution in coprime integers $ x,y$. Moreover

$ (x^2/y^2) \equal{} \minus{} b/a$

Hence $ \sqrt { \minus{} b/a}$ is rational. But then if

$ x \equal{} 1 \plus{} \frac {t}{4a}$

$ y \equal{} \frac {1 \minus{} \frac {t}{4a}}{\sqrt { \minus{} b/a}}$

we have

$ \left(1 \plus{} \frac {t}{4a}\right)^2 \plus{} b/a\left(\frac {1 \minus{} \frac {t}{4a}}{\sqrt { \minus{} b/a}}\right)^2 \equal{} t/a$

$ a \left (1 \plus{} \frac {t}{4a}\right)^2 \plus{} b\left(\frac {1 \minus{} \frac {t}{4a}}{\sqrt { \minus{} b/a}}\right)^2 \equal{} t$[/quote]

Great.How could you think of such a solution "$ x \equal{} 1 \plus{} \frac {t}{4a}$

$ y \equal{} \frac {1 \minus{} \frac {t}{4a}}{\sqrt { \minus{} b/a}}$" :o  :o  :o . Thanks for the solution.

You're welcome! A nice problem which I'm sure is related to some deep theorems.

I set out to solve

$ x^2\plus{}(b/a)y^2 \equal{} t/a$

$ x^2\minus{}(\sqrt{\minus{}b/a}y)^2 \equal{} t/a$

with $ y'\equal{}\sqrt{\minus{}b/a}y$,

$ x^2\minus{}y'^2\equal{}t/a$

which has the solution

$ x\equal{}1\plus{}t/4a$
$ y'\equal{}1\minus{}t/4a$

Now just solve for $ y$ and voila!

Always remember that $ (1\plus{}m/4)^2\minus{}(1\minus{}m/4)^2 \equal{} m$. That is the key identity. A difference of squares is always easier to manipulate than a sum of squares.


[quote="GameBot"]What is the remainder of $ 8^6 \plus{} 7^7 \plus{} 6^8$ divided by 5?[/quote]

$ 8^6 \equiv \minus{}1^3 \equiv 4 \pmod 5$
$ 7^7 \equiv 2^7 \equiv 3 \pmod 5$
$ 6^8 \equiv 1 \pmod 5$

Adding gives $ 3$.


LaTeX
inequalities
complex numbers
imaginary numbers

Hi, what is wrong with the following "proof":

1 = $\sqrt{(-1)(-1)} = \sqrt{-1}*\sqrt{-1}= i*i = -1$

hence:

$1 = -1$

Edit: $\LaTeX$ed by mod who likes $\LaTeX$

$\sqrt{AB}=\sqrt{A}\sqrt{B}$ works iff $A>0$ and $B>0$.

Therefore, $\sqrt{-1\cdot -1}\ne\sqrt{-1}\sqrt{-1}$ ;)

The proof is wrong because $\sqrt{-1*-1}\neq \sqrt{-1}\sqrt{-1}$ .

And welcome to the Art of Problem Solving forum!

ah.. ok..

is there any nice proof for that statement? or atleast a good explination exept that "that's how it is, check calc"

thx =)

thank you 4everwise.. sry I am still learnign math and I have to admit I dont really understand what you  just said ^^

can you explain it a bit? I have just begin with imaginary numbers and such..

EDIT:: NWM seems like it was a "view"-problem from my side..

yes that is true, but why? any proof or anything?

if this was a task of yours to give your teacher, what would you say? =)

thx

It means that the square root of the product of two negative numbers doesn't equal the product of the square roots of the negative numbers.

Does that help?

seems like it was a "view"-problem from my side..  (or you edited it =) )

yes that is true, but why? any proof or anything?

if this was a task of yours to give your teacher, what would you say? =)

thx

$\sqrt{-x*-x}=\sqrt{x^2}=x$

$\sqrt{-x}*\sqrt{-x}=i\sqrt{x}*i\sqrt{x}=-x$

Because $-x \neq x$, $\sqrt{-x*-x} \neq \sqrt{-x}*\sqrt{-x}$

That's what I would say.  ;)

[quote="CruS"]yes that is true, but why? any proof or anything?

if this was a task of yours to give your teacher, what would you say? =) [/quote]
Well, I would say you don't have to prove it. What I mean by that is, for real number $x$, $\sqrt{x}$ is defined only when $x$ is positive. However, for convenience, we wanted to [b]define[/b] a square root of a negative number. We created imaginary numbers, and there are some rules that must be followed in order to make it work. The statement is just a part of the rule of imaginary numbers.

(EDIT- I edited some explanations.)

[quote="CruS"]but where does the "i" come from and what is neqx? [/quote]

$i$ is simply a notation for $\sqrt{-1}$

neqx.  :blush: That was a typo. I meant to type neq, which displays $\neq$ in Latex.

[quote="CruS"]ok.. sorry for being so stupid =)

but where does the "i" come from and what is neqx?

thx in advance[/quote]

Hello. Well, first of all, in $\LaTeX$, \neq means "not equal to..."  About the $i$: we know that $i=\sqrt{-1}$, as a result $y=\sqrt{-x}=\sqrt{(-1)(x)}=\sqrt{-1}\sqrt{x}=i\sqrt{x}$.  What I want to know is that someone, involving this forum said that $\sqrt{ab}\neq\sqrt{a}\sqrt{b}$ if $a<0$ and/or $b<0$.  How would that statement explain my above work...if it is correct?

EDIT: I think the $\sqrt{ab}\neq\sqrt{a}\sqrt{b}$ inequality should only apply if both $a,b<0$.

Masoud Zargar

I think my original statement might be wrong...

How about this: $\sqrt{AB}=\sqrt{A}\sqrt{B}$ iff $A\ge 0$ [b][u]or[/u][/b] $B\ge 0$

(Note: At least $A$ or $B$ has to be positive; both $A$ and $B$ could be positive)

This probably makes more sense.
And that explains boxedexe's point as well.

Okay, now that works...I thought I have totally messed up in math. :)

Masoud Zargar

what about this:

sqrt(-x*-x) = -sqrt(x)*-sqrt(x) ?

In this case, both $a,b<0$, therefore,
$\sqrt{(-x)\cdot (-x)}\neq\sqrt{-x}\sqrt{-x}$. You can also say that this is incorrect by saying that$\sqrt{(-x)\cdot (-x)}=\sqrt{(-x)^2}=\sqrt{x^2}=x$.
Since
$\sqrt{-x}\sqrt{-x}=i\sqrt{x}i\sqrt{x}=x\cdot i^2=x\cdot (-1)=-x$, $-x\neq x$.
As a result, we can say that
$\sqrt{(-x)^2}\neq \sqrt{-x}\sqrt{-x}$.

EDIT: Sorry, I accidently thought that you wanted to see how that a and b worked in the squareroot rule, but now I saw that it was something else :blush:.

Masoud Zargar

Well, as far as I can see, my words imply that $i=\sqrt{-1}$.  By the way, anotehr thing is that your results are incorrect because the same way we say $i^2=\sqrt{-1}\sqrt{-1}=-1$,
we can say that $[i^2=\sqrt{-1}\sqrt{-1}=-1]\neq[\sqrt{(-1)^2}=\sqrt{1}=1]$.
This is due to the fact that $\sqrt{-1}\sqrt{-1}=\sqrt{-1}^2=[(-1)^{\dfrac{1}{2}}]^2=(-1)^1=-1\neq 1$.

Masoud Zargar

Can someone explain about the self-contradictory nature of defining positive and negative for square roots of imaginary numbers?

[quote="CruS"]but.. wouldnt that mean i = -1 ?
if you read what you said in a previous post[/quote] Well,
$\sqrt{-x*-x}=\sqrt{x^2}=x$
$-\sqrt{x}*-\sqrt{x}=-1*\sqrt{x}*-1*\sqrt{x}=1*x=x$

The second equation didn't have an $i$ term in it.  ;)

But does

$\sqrt{x} \cdot \sqrt{x} = \sqrt{x \cdot x}$

for all real x?

........................

How are positives and negatives for square roots of imaginary numbers self-contradictory? I would say that since the square root of an imaginary number is another imaginary number, $a + bi \text{ where } a \neq 0 \text{ and } b \neq 0$, and imaginary (not pure imaginary) numbers are neither positive nor negative, then that's the end of that, and there's no contradiction.

[quote="mathnerd314"]But does

$\sqrt{x} \cdot \sqrt{x} = \sqrt{x \cdot x}$

for all real x?

........................

[/quote]

Try x=-1. I dont' think that equality works for that. for positive integers, however, it does work. (Wait a second, my algebra book never said for positive integers, or did it? I'll have to check on that).

Because we're all assuming that it does. Since $i=\sqrt{-1}$ we know that $i^2=\sqrt{-1} \cdot \sqrt{-1}$ but that doesn't mean that $i^2=-1$, because of what you're getting at.

Well, when we have $\sqrt{A}\sqrt{B}=\sqrt{AB}$ iff $A\ge 0$ or $B\ge 0$ or both.  In this case, $i^2=\sqrt{-1}\sqrt{-1}\neq \sqrt{1}$ because, as we previously said, this only works iff $A\ge 0$ or $B\ge 0$ or both. According to this rule, which I came across while taking a look at this forum (I haven't seen this rule in any math book I have read :( ),
\[ i^2=\sqrt{-1}\sqrt{-1}=\sqrt{-1}^2=[(-1)^{\dfrac{1}{2}}]^2=(-1)^1=-1 \].
$\therefore$ our rule is correct.

Masoud Zargar


Let $G \subset \mathbb{R}^2$ be a subset of the plane such that
(i) every vertical line meets $G$ in precisely two points; and
(ii) every non-vertical line meets $G$ in a dense set of points.
Must $G$ have a subset $H$ such that every vertical line meets $H$ in precisely one point, while every non-vertical line still meets $H$ in a dense set of points?

I am very unsure of this proof.  I think it is right, though.

======================

yes.  Call the conjugate of the point $x \in G$ as the unique point $y \ne x$ on the vertical line containing $x$.   When I say $X$ is "red dense", I mean that the set of all red points in $X$ are dense, and similarily for blue.

--

color points in H as red, not in H but in G as blue.  A line from H hits a dense set of blue and red points.  We claim it hits also a dense set of red points.  Suppose not.  We will construct H' from H satisfying the required properties.

On a line $L$ for which the red points are not dense, let $B, R$ be the blue and red points on this line.  There exists some point $x$ in $\mathbb{R}^2$ for which points in $R$ are not arbitrarily close to.  This means there is an interval $I$ for which all points on $L$ with ordinates in $I$ are blue (call these points $L'$ points).  Let us color some fraction of these points red, and their conjugate blue, so that $L'$ is red dense and blue dense. It is possible because $B \cup R$ is dense (given as a property of G.)

If, for such a recoloring, we have no line $M$ such that there exists a new point $y$ (not there previously) for which red points in $M$ are not arbitrarily close to $y$, we are done.  Suppose that there is such an $M$ and $y$.  It implies that there exists an interval $J \subset I$ for which points in $M$ with ordinate in $J$ are all blue (and $y$ has ordinate in this interval).  [This is the same reasoning as in the above paragraph.]  Call these points $M'$ points.  Because originally, $M$ was red dense, a dense set of $M'$ must be recolored points.  So the conjugate of these points must be a dense set of red points, which is a subset of $L'$.  Finally, color $M'$ in the same way we colored $L'$.  The result is that $L'$ and $M'$ are both red dense and blue dense.  Because we only dealt with conjugates, changing $M'$ points did not change any other points.

--

Finally, because every problem can be fixed by recoloring using the above strategy, we can fix all problems and demonstrate such a subset H.

But, following this approach you will only deal with a countable number of non-red dense problems. Unfortunately, there could be a noncountable number of such problems. ;)

Pierre.

So you claim the converse, that there is no subset H with all of the (possibly non-countable) recommended recolorings?

I do not claim anything about the initial problem. Only that your answer is not complete....

Pierre.

Does anyone have an idea how should the set H look like? It's surely a graph of a function, that is not continous, but I wonder what's the type of this function.


geometry
geometric transformation
reflection
analytic geometry
angle bisector
congruent triangles
cyclic quadrilateral

Let a quadrangle  $ABCD$ with angle $\angle DAB > 180^{\circ}$   (where  $\angle DAB$  is an inner angle). We also know that:

 $AB \cdot CD = AD \cdot BC$ . If  point $P$ is symmetrical with point $A$ with respect to $BD$ ,
prove  that  $\angle PCB = \angle ACD$ .

 [u] Babis[/u]

I dont know if it's correct because I found it very easy  :?

[hide]$\frac{AB}{BC} = \frac{AD}{CD}$. Hence $\Delta ABC \sim \Delta ADC$.

But for ASA we have $\Delta ABC \equiv \Delta ADC$.
So $AC$ is the angle bisector of $\Delta BCD$ and also the altitude.
Hence $P \in line AC$.
So $\angle ACD = \angle ACB = \angle PCB$.[/hide]

I was just trying to prove that two triangles are similar , but I considered this easy possibility as imposimple !. In this way I was entered in a great adventure ! Any way, thank you.
 [u]Babis[/u]

The problem is from the Polish MO 2005, and the corresponding round has ended yesterday.

[quote="Andreas"]$\frac{AB}{BC} = \frac{AD}{CD}$. Hence $\Delta ABC \sim \Delta ADC$.[/quote]

Sorry, but this conclusion is wrong...

Here is one possible solution, though not too elementary (I use Apollonius circles and inversion):

[color=blue][b]Problem.[/b] Let ABCD be a quadrilateral such that $AB\cdot CD=BC\cdot DA$, and let P be the reflection of the point A in the line BD. Prove that < BCP = < ACD, where we are using directed angles modulo 180°.[/color]

[i]Solution.[/i] The equation $AB\cdot CD=BC\cdot DA$ rewrites as $\frac{AB}{DA}=\frac{BC}{CD}$, or, equivalently, $\frac{BA}{DA}=\frac{BC}{DC}$. Also, since P is the reflection of the point A in the line BD, we have BP = BA and DP = DA, so that $\frac{BP}{DP}=\frac{BA}{DA}=\frac{BC}{DC}$. Hence, the points P, A and C lie on an Apollonius circle with respect to the segment BD; the center M of this Apollonius circle must lie on the line BD, and the points B and D must be inverse to each other with respect to this Apollonius circle (in fact, the endpoints of a segment are always inverse to each other with respect to any Apollonius circle with respect to this segment). Since the radius of this Apollonius circle is MC, we thus have $MB\cdot MD=MC^2$, what rewrites as $\frac{MB}{MC}=\frac{MC}{MD}$. Together with < BMC = - < CMD, this yields that the triangles BMC and CMD are inversely similar, so that < DCM = - < CBM. Hence,

  < ACD = < ACM - < DCM = < ACM - (- < CBM) = < ACM + < CBM.

On the other hand, since the points P, A and C lie on our Apollonius circle, whose center is M, the central angle theorem for directed angles modulo 180° shows that < MCP = 90° - < PAC. But since $AP\perp BD$ (what is clear because the point P is the reflection of the point A in the line BD), we have < (AP; BD) = 90°, and thus

  < MCP = 90° - < PAC = < (AP; BD) - < (AP; AC) = < (AC; BD).

Thus,

  < ACD - < BCP = (< ACM + < CBM) - (< BCM + < MCP)
                         = (< ACM - < BCM) + < CBM - < MCP
                         = < ACB + < CBM - < MCP = < (AC; BC) + < (BC; BD) - < (AC; BD)
                         = < (AC; BD) - < (AC; BD) = 0°,

so that < ACD = < BCP, and the problem is solved.

  darij

Nice solution darij. Mr Stergiou do you have a more elementary solution ?thanks

Darij: I skim through your solution and it looks exactly the same as mine which I sent for competition  :) (Apollonius and inversion as well). I've heard that others proved it with some reflections instead of inversion (but Apollonius circle was still used). I like this one  :)

To silouan

 No I have not. I want to put coordinates, one point (0 , 0) and another(1,0). I do not know if computations take a long. I' ll try it , just to see wht will hapen.

Babis

[b]Thanks to ZIBI[/b]
 for doing the solution(analytical) , that I suggested to my last post. He saved me much time !Of course he could take one point as (  1 , 0) instead of (b , 0) , to simplify the problem.

  [u]Babis[/u]

Hi everyone,

My solution shares most of ideas of Darij's solution, but I did something rather different.

I did consider the same Apollonius circle $\Gamma$ and proved that $P$, $A$ and $C$ are on $\Gamma$. But I finished the problem in another way.

One useful fact is that if $X$ and $Y$ belong to the Apolonius circle of $M$ and $N$, then the bisectors of angles $\angle MXN$ and $\angle MYN$ meet on a point in the line $MN$. This follows almost immediately from the bisectors theorem, that is, if the bisector of $\angle MXN$ meets $MN$ on $T$ then ${MT\over NT} = {MX\over NX}$. Also, $T$ belongs to the Apolonius circle.

With this in mind, let the bisectors of $\angle BCD$ and $\angle BPD$ meet on $T$. Since $C$ and $P$ belong to $\Gamma$, $T$ is on $BD$. Now we have more chances of solving the problem, since bisectors induce some equal angles. But the most remarkable fact is that, since $A$, $C$, $P$ and $T$ belong to $\Gamma$, $CAPT$ is a cyclic quadrilateral. Recalling also that $DAT$ and $DPT$ are congruent triangles (by simmetry), we have $AT=PT$.

Some angle chasing finishes the job. Indeed, $\angle TCD = \angle TCB$ ($CT$ bisects $\angle BCD$). Note that $\angle PCB = \angle TCB \pm \angle TCA$ and $\angle ACD = \angle TCD \pm \angle TCP$, where the sign on the $\pm$ are the same in both equations and depends on the positions of $A$ and $P$. Hence it suffices to prove that $CT$ also bisects $\angle ACP$. But this follows from the cyclic quadrilateral $CATP$ with $AT=PT$: $\angle TCP = \angle TAP = \angle TPA = \angle TCA$.

the problem will be solved by a very well known theorem,which its name is Apolonius Circle.

theorem:let $ABC$ be a triangle ,and$AM,AN$  are the inner and outher bisectors

respectivly.

the locus of points like$P$  in which:$PB: PC$ =fixed=$AB: AC$,

is a circle,with the diameter $MN$.

i my self have a nice proof for it by harmonic...but i think it doesnt need proof... and u

know its proof. ;)

lets come back to our problem:

$AB.CD=AD.BC$

 then:

$AD: AB=CD: CB$

it means that $A$ is on the apolonius circle of triangle $CBD$ .

$P$is also on this apolonius circle,because:the angles are equal ,and:

$PD: PB=AD: AB=DC: BC$

the apolonius circle, intersects $DB$in point$K$ ,which $CK$  is the bisector

of$\angle A$.

and $CF$  is the outer bisector.then we know that:.

$\angle KCB=\angle KCD$

$KF$ is the diameter of circle.and $A,P$are are symmtrictoeach other with respecting

to $KF$.

it means that the arcs $AK,PK$are equal.and:

$\angle ACK=\angle KCP$

then:$=\angle KCB-\angle KCP=\angle KCD-\angle KCA$

and:$\angle PCB=\angle ACD$

..........................................

This problem, in a other (equivalent) enunciation, becomes a new ( or wellknown ?) property of the harmonical quadrilateral:
" Let $ABCD$ be a harmonical convex quadrilateral, i.e $AB\cdot CD=AD\cdot BC$. Denote the reflection $P$ of the point $A$ w.r.t. the diagonal $BD$ and the point $E\in BD$ for which $\frac{EB}{ED}=\frac{AB}{AD}\ \left (=\frac{CB}{CD}\right)$. Then the points $A$, $C$, $E$, $P$ are conciclically. "

[u]Remark.[/u] The solution offered by [i]Ashegh[/i] is "suddenly" and I like it.

Sorry but I didn't read carefully the previous solutions and maybe my solution is similar to other .
But here it goes :
From the fact that $ \frac {AB}{AD} \equal{} \frac {CB}{CD}$ we conclude that the points A,C lie on the Apollonius circle of B,D
and let this circle meet BD at F,G . (F is betwen B,D)  .It's easy to see that P lies on the Apollonius circle of B,D .
Now because CF is the bisector of $ \angle{BCD}$ it suffices to prove that $ \angle{FCP} \equal{} \angle{ACF}$ but the arcs
AF,FP are equal because P is the symmetric of A and we are done .

My solution using isogonal conjugate point
Let internal bisector of $ \angle CBA,\angle CDA$ meet at $ X$ and internal bisector of $ \angle CBP,\angle CDP$ meet at $ Y$ . From $ \frac{BA}{BC}\equal{}\frac{DA}{DC}$, $ X$ lie on $ CA$. Similarly, $ Y$ lie on $ CP$. It is easy to show $ XB,YB$ is isogonal line wrt $ \angle CBD$ using $ XB,DB,YB$ is internal bisector of $ \angle CBA,\angle ABP,\angle CBP$. Similarly, $ XD,YD$ is isogonal line wrt $ \angle CDB$. Hence $ X,Y$ are isgonal points wrt $ \triangle CBD$. Therefore, $ \angle PCB\equal{}\angle YCB\equal{}\angle XCD\equal{}\angle ACD$.

See an extension of this problem at http://www.mathlinks.ro/Forum/viewtopic.php?t=182355


For triangle angels A, B, C find the smallest real number D such that
$ sin^{2}A \plus{} sin^{2}B \minus{} cosC\leq D$ holds.

Calculus not allowed of course.

[hide="as far as I got..."]$ C\equal{}180\minus{}A\minus{}B$

$ \cos{C}\equal{}\cos{180\minus{}A\minus{}B}\equal{}\minus{}\cos{A\plus{}B}\equal{}\sin{A}\sin{B}\minus{}\cos{A}\cos{B}$

$ \sin^2{A}\plus{}\sin^2{B}\plus{}\sin{A}\sin{B}\minus{}\cos{A}\cos{B}$[/hide]

Am I missing something? :huh:

[quote="1=2"][hide="as far as I got..."]$ C \equal{} 180 \minus{} A \minus{} B$

$ \cos{C} \equal{} \cos{180 \minus{} A \minus{} B} \equal{} \minus{} \cos{A \plus{} B} \equal{} \sin{A}\sin{B} \minus{} \cos{A}\cos{B}$

$ \sin^2{A} \plus{} \sin^2{B} \plus{} \sin{A}\sin{B} \minus{} \cos{A}\cos{B}$[/hide]

Am I missing something? :huh:[/quote]
[hide="Yes  :lol: "]
Notice that by Cauchy, we have that
\[ (\cos^2 A \minus{} \sin^2 A)(\cos^2 B \minus{} \sin^2 B)\ge (\cos A\cos B \minus{} \sin A\sin B)^2
\]\[ \cos 2A\cos 2B\ge \cos^2 (A \plus{} B)
\]\[ 2\sqrt {\cos 2A\cos 2B}\ge 2\cos (A \plus{} B)
\]\[ \cos 2A \plus{} \cos 2B\ge 2\sqrt {\cos 2A\cos 2B}\ge 2\cos (A \plus{} B)
\]\[ \frac {\cos 2A \plus{} \cos 2B}{2}\ge \cos (A \plus{} B)
\]\[ \frac {\cos^2 A \plus{} \cos^2 B \minus{} \sin^2 A \minus{} \sin^2 B}{2}\ge \cos (A \plus{} B)
\]\[ \frac {\cos^2 A \plus{} \sin^2 A \plus{} \cos^2 B \plus{} \sin^2 B}{2}\ge \sin^2 A \plus{} \sin^2 B \plus{} \cos (A \plus{} B)
\]\[ 1\ge \sin^2 A \plus{} \sin^2 B \plus{} \cos (A \plus{} B) \equal{} \sin^2 A \plus{} \sin^2 B \minus{} \cos C
\]
[/hide]

EDIT: Sorry, Brut3Forc3, I will edit my solution (completely.)

Does this work now?

Quatto, I'm a bit confused by your solution. It seems that at the end, you were looking for a lower bound instead of an upper... Could just be me though.  :huh:

Argh... hang on... I messed up on my first step, I said that $ (\cos^2 A \minus{} \sin^2 A)(\cos^2 B \minus{} \sin^2 B)\ge \cos A\cos B \minus{} \sin A\sin B$, but the opposite is true.

I can't seem to prove my conjecture for the value of $ D$, but here is my idea:
[hide="Can anyone follow this up?"]
If someone can prove that $ \cos 2A \plus{} \cos 2B\ge 2\cos (A \plus{} B)$, then we are done.

After we can prove the lemma (if it even is true, but it does seem so), then we could do the following:
\[ \cos^2 A \minus{} \sin^2 A \plus{} \cos^2 B \minus{} \sin^2 B \equal{} \cos 2A \plus{} \cos 2B\ge 2(\cos A\cos B \minus{} \sin A\sin B)
\]\[ \cos^2 A \minus{} \sin^2 A \plus{} \cos^2 B \minus{} \sin^2 B \equal{} 2\cos (A \plus{} B)
\]\[ 2 \equal{} \cos^2 A \plus{} \sin^2 A \plus{} \cos^2 B \plus{} \sin^2 B\ge 2\cos (A \plus{} B) \plus{} 2\sin^2 A \plus{} 2\sin^2 B
\]\[ 1\ge \cos (A \plus{} B) \plus{} \sin^2 A \plus{} \sin^2 B \equal{} \sin^2 A \plus{} \sin^2 B \minus{} \cos C
\]\[ D \equal{} \boxed{1}
\]
[/hide]


Έστω τρίγωνο $MAB$  με σταθερή βάση  $AB$. Υποθέτουμε ότι το $M$ κινείται στο επίπεδο έτσι ώστε τα εξωτερικά κατασκευασμένα τετράγωνα $MAKC$,
$BMHZ$, $MBLI$ να ικανοποιούν τη σχέση  $(AMCK)+(BMZH)+(ABLI)+4(MAB)=2002$.
όπου με $(...)$ συμβολίζουμε το εμβαδό. Βρείτε το γεωμετρικό τόπο του $M$

[hide="Ωραία άσκηση..."]

Έστω $AB=2a$
Ορίζω σύστημα συντεταγμένων με αρχή το $O$, και τα σημεία  $A(-a,0),B(a,0),M(x,y)$ και $H(x,0)$ η προβολή του $M$ στον άξονα $x'x$
$(MAB) = \frac{1}{2}2ay \Rightarrow  4(MAB) = 4ay$

$AM^2+BM^2+AB^2+4(MAB) = 2002\Rightarrow$
$AM^2+BM^2+(2a)^2+4ay = 2002\Rightarrow$
$AM^2+BM^2+4ay = 2002-(2a)^2 = 2(1001-2a^2) = 2k^2$

$AH=|a+x|$
$BH=|a-x|$

Για κάθε περίπτωση ($H$ μεταξύ των $A,B$ ή εκτός) ισχύουν τα ίδια

$AM^2 = (a+x)^2+y^2 = a^2+x^2+2ax+y^2$
$BM^2 = (a-x)^2+y^2 = a^2+x^2-2ax+y^2$

$AM^2+BM^2+4ay =2(a^2+x^2+y^2+2ay) = 2k^2 \Rightarrow$
$x^2+(a^2+y^2+2ay) = k^2 \Rightarrow$
$x^2+(y+a)^2 = k^2 \Rightarrow$
Κύκλος με κέντρο $(0,-a)$ και ακτίνα $k = \sqrt{1001-2a^2}$

.. και ελπίζω να μην έχω κάνει λάθος[/hide]

[quote="pontios"][hide="Ωραία άσκηση..."]

[/hide][/quote]

Και ωραία και απλή λύση


geometry
3D geometry
algebra
polynomial
function
Putnam
complex numbers

Find the sum cube of real roots of x^3-2*x^2-x+1=0

You should first check that all the roots are real, no?  (Since it asks for the sum of the cubes of the real roots, you aren't necessarily summing the cubes of all the roots.)

Anyhow, that's not very hard -- it's negative at -infinity, positive at 0, negative at 1, and positive at infinity, so that's 3 roots.

I don't know what you mean by "elementary symmetric polynomials" -- do you mean the symmetric polynomials which we get as the coefficients of the expression $(x-a_1)(x-a_2)\cdots(x-a_n)$?  If so, the answer is yes.  For 2 variables, it's quite easy to do.  For larger numbers you can still do it, but it's decidedly trickier.  This is actually somewhat similar to a question (A4, I think) on the Putnam exam this year.  Also, the numbers don't have to be real -- complex numbers work just fine, as well.

For an explicit formula look here:
http://www.mathlinks.ro/Forum/viewtopic.php?t=20017&highlight=

[quote="jazzmaster"]I have a question:

Let [tex]a_1,...,a_n[/tex] be real numbers.
If we know all the [url=http://mathworld.wolfram.com/SymmetricPolynomial.html]elementary symmetric polynomials[/url] of this numbers. Can we write [tex] a_1^{m}+...+a_n^{m} [/tex] as a function of this polynomials?[/quote]

Yes, and the relevant functions follow from the Newton-Girard formulas:
[url=http://mathworld.wolfram.com/Newton-GirardFormulas.html]mathworld.wolfram.com/Newton-GirardFormulas.html[/url]


number theory
relatively prime
number theory unsolved

Suppose p and q be distinct primes.
Then p^(q-1) + q^(p-1) is congruent to 1 (mod pq)
Either prove this statement or give a counter-example.

By Fermat's little theorem p^(q-1) is congruent to 1 (mod q). q^(p-1) is congruent to 0 (mod q) then p^(q-1)+q^(p-1) is congruent to 1 (mod q). Similarly, q^(p-1) is congruent to 1 (mod p) and p^(q-1) is congruent to 0 (mod p) so p^(q-1)+q^p-1 is congruent to 1 (mod p). As p and q are relatively prime, then p^(q-1)+q^(p-1) is congruent to 1 (mod pq).


[quote="tkhtn"]For $a,b,c,d\in R$.
Prove that inequality:
$(a-b)(c-d)+(a-d)(b-c)+(d-a)(b-c)\ge 0$.[/quote]
$(a-d)(b-c)+(d-a)(b-c)=0\Rightarrow$ we must prove $(a-b)(c-d)\ge 0$
And this is certainly wrong

[quote="tkhtn"]For $a,b,c,d\in R;a+d=b+c$.
Prove that inequality:
$(a-b)(c-d)+(a-c)(b-d)+(d-a)(b-c)\ge 0$.[/quote]
If I am right:
\[ (a-b)(c-d)+(a-c)(b-d)+(d-a)(b-c)=2(a-b)(c-d) \ge 0 \]
(because $a-b=c-d$)


Let $ r$, $ s$ be two positive integers and $ P$ a 'chessboard' with $ r$ rows and $ s$ columns. Let $ M$ denote the maximum value of rooks placed on $ P$ such that no two of them attack each other.

(a) Determine $ M$.

(b) How many ways to place $ M$ rooks on $ P$ such that no two of them attack each other?

[Note: In chess, a rook moves and attacks in a straight line, horizontally or vertically.]

(a) Obviously the answer is $ \min\{r,s\}$,because if we have more than $ \min\{r,s\}$,then either in a row or in a column we have more than one rook.On the other hand if we have $ \min\{r,s\}$,then we obviously can arrange them in a chessboard,such that non of them attack another.
(b) WLOG assume than $ r\equal{}\min\{r,s\}$,so $ M\equal{}r$.There are $ \binom{rs}{r}$ ways to place rooks in a cheerboard.And there are exactly
$ \binom{rs\minus{}2}{r\minus{}2}(\binom{r}{2}s\plus{}\binom{s}{2}r)$ ways to arrange the rooks,when there exist two attacking each other rooks.
So the answer is:
$ \binom{rs}{r}\minus{}\binom{rs\minus{}2}{r\minus{}2}(\binom{r}{2}s\plus{}\binom{s}{2}r)$

The second part can be answered more simply: we need $ r$ columns for the $ r$ rooks, and we can choose which column and row pair up in $ r!$ ways for a total of $ \binom{s}{r} r! \equal{} \frac{s!}{(s \minus{} r)!}$ arrangements.

[quote="JBL"]The second part can be answered more simply: we need $ r$ columns for the $ r$ rooks, and we can choose which column and row pair up in $ r!$ ways for a total of $ \binom{s}{r} r! \equal{} \frac {s!}{(s \minus{} r)!}$ arrangements.[/quote]Or $ \frac {r!}{(r \minus{} s)!}$ if $ r>s$. :wink:


Tomorrow is Pi Day and I figured if anyone was doing anything fun it would be you guys!

So open question: Anyone doing anything interesting for Pi Day?

(Side note: In Japan tomorrow is [url=http://en.wikipedia.org/wiki/White_day]White Day[/url] which might interfere for people there.)

[color=green]Note from moderator: there is [i]never[/i] a good reason to block-quote the entire post preceding yours.[/color]

Arguing with the girls in my maths set who seem to think that its just about eating pie and nothing to do with recital competitions.

no yell pi at 1:59.

We're going to have a party and pie some of the teachers  :D

At least you all should be consistent. Do something on $ e$-, $ \gamma$-, $ \zeta(3)$-, and $ 1$-day !!! And on any other date that is contained in the reals.

That would mean celebrating everyday  :rotfl:   :D

I'm going to put the digits of Pi on a piece of paper taped to my locker. Also, the Pi symbol. Plus, the estimated fractions, etc...Shout Happy Pi Day when it's 1:59 pm,  the works. My teachers don't really care so we can't bring in an actual pie. I'm also going to wear my brother's Pi pin that goes up to 100 digits.

For Pi Day, we're having a giant online gathering time (see [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=188613]here[/url]) on AoPS.  :)

At my faculty, they are organizating a party. At least many people sends mails to congratulate everyone... so Happy Pi day to everyone :P

Wow, I don't think I've seen a single person with that signature since people started it a month ago.

Hmmm well the only thing I'm going to do is have a party in math. The only pi day that was of any interest to me was in 7th grade, when the teacher actually took it seriously. The other years, the teacher just let it pass.

In absense of proper pi day garb, I'll be wearing a shirt emblazoned with the distance formula and using electrical tape to put $ \pi$ and its digits on myself (as well as friends). Close enough, right?

The proper thing to do is shout pi when it has been exactly 14.159 days of March, which is around 3:49 AM (yes, I actually know someone devoted enough to wake up then).

I'm devoted enough to be awake at that time without thinking of $ \pi$  :wink:

well how many hrs exactly from now is it for the record breaking experience?
this is urgent!!!  :D

a random thought:
what do you think people did on 3/14/1592?

Neither AoPS nor accurate watches didn't exist at that time, so nothing.

Some kid at my school stuck a "Happy Pi Day" sign on the back of another kid and he didn't realize he had it on until the end of the day.  :D

We had pie in math class today  :) . Our teacher was nice enough to buy her entire class (periods 1-7) pie.

You gotta be kidding me.

I knew like 80 at the end of 8th grade but memory has left only 40 or so.

[quote="Nerd_of_the_Ages"]Hopkins had a Pi Contest. People who couldn't to 15 digits got whipped cream splattered on their face. It was disturbing. The most anyone could memorize was 20 something. 2 people could only do 3 digits. When the leadership teacher said that only people who could get at least 5 digits could try, half the people wanting to try left.[/quote]

something like that happened at my school, horner. we had a 1 dollar pie throwing thing where you could throw a pie at a teacher, the VP, or the principal. then, we had a pi-reciting contest, which was won by a guy who memorized 100 digits.  :|

ps: horner is the rival school of hopkins (really)

[quote="Nerd_of_the_Ages"]Hopkins had a Pi Contest. People who couldn't to 15 digits got whipped cream splattered on their face. It was disturbing. The most anyone could memorize was 20 something. 2 people could only do 3 digits. When the leadership teacher said that only people who could get at least 5 digits could try, half the people wanting to try left.[/quote]

Wow, I only know 15 digits of Pi. I barely get through the whipped-cream-in-face number of digits. The person in my school who knows the most amount of digits: 3.14159...That's all. 6 digits.

[quote="MustItAlwaysBeMe"]I memorized the first 50 in 6th grade, and I still remember them. :w00t:[/quote]

Nice. I don't see how people can forget digits. I've accumulated over 100 now.

in response to the first post, Koreans also celebrate White Day! :D


function
calculus
derivative
quadratics
integration
inequalities
analytic geometry

It would be more helpful to either put all your links into one message,
or use the Subject Title Header to tell people at least what the topic is,
so they don't have to do a repeated goose-chase.

Just edit the header.

Ultracold neutrons enter a long cavity of the length L between a horizontal mirror M at the height z = 0 and a horizontal absorber A at the height z = H through an opening W on the left with a constant horizontal velocity component $v_x = 10\ \text{m/sec}$ and a wide range of the vertical velocities $v_z$. Once in the cavity, they travel to the right in the Earth's gravitational field, elastically bouncing from the mirror, unless they reach the absorber, in which case they are absorbed. A detector D at the right end of the cavity measures the neutron transmission rate N(H) as a function of the cavity height H.

[color=blue][b](1)[/b][/color] Calculate the classical range of neutrons velocities $v_z(z)$ of the neutrons that can reach the detector D, assuming that the cavity length is large.

Assume that a neutron at rest is dropped from the height z = H. Its height z(t) is given by solving the differential equation z'' = -g with the initial conditions z(0) = H, z'(0) = 0, i.e.,

$z = -\frac{gt^2}{2} + H$

$v_z = z' = -gt,\ \ t = -\frac{v_z}{g},\ \ z = -\frac{v_z^2}{2g} + H$

Thus the velocity $v_z < 0$ is given by

$v_z = -\sqrt{2g(H - z)}$

When the neutron elastically bounces from the mirror M, its velocity becomes $v_z = +\sqrt{2g(H - z)} > 0$. Neutrons with the vertical velocity components $v_z$ outside of the interval $\left(-\sqrt{2g(H - z)}, +\sqrt{2g(H - z)}\right)$ have the maximum of their parabolic paths higher than H, the position of the absorber A. If the cavity length L is large, they will reach the absorber A at some time, regardless of the horizontal velocity component $v_x$, . Only the neutrons entering the cavity at the height z with the vertical velocity components $v_z \in \left(-\sqrt{2g(H - z)}, +\sqrt{2g(H - z)}\right)$ can reach the detector D.

[color=blue][b](2)[/b][/color] Calculate the minimum classical range $L_c$ necessary for all neutrons with the vertical velocity components of the above range to be absorbed, regardless of z. Use $v_x = 10\ \text{m/sec},\ H = 50\ \mu \text m$.

For a neutron entering the cavity at the height $z$ with the vertical velocity component $v_z$, its height $\zeta(t)$ is given by solving the differential equation $\zeta'' = -g$ with the initial conditions $\zeta(0) = z$ and $\zeta'(0) = v_z$, i.e.,

$\zeta(t) = -\frac{gt^2}{2} + v_z t + z$

Assuming $v_z > 0$, the time $t_{\text{max}}$ of reaching the maximum height is given by putting the derivative to zero:

$\zeta'(t) = 0,\ \ -gt + v_z = 0,\ \ t_{\text{max}} = \frac{v_z}{g}$

The time of bouncing from the mirror M at z = 0 is given by putting the function to zero, solving the resulting quadratic equation for t and taking the positive root:

$\zeta(t) = -\frac{gt^2}{2} + v_z t + z = 0,\ \ t_{\text{min}} = \frac{v_z + \sqrt{v_z^2 + 2gz}}{g}$

The time difference between the minimum and maximum heights is equal to $t_{\text{min}} - t_{\text{max}}$ and the full period between 2 maxima is twice this difference:

$T = 2(t_{\text{min}} - t_{\text{max}}) = 2 \left(\frac{v_z + \sqrt{v_z^2 + 2gz}}{g} - \frac{v_z}{g}\right) = \frac{2 \sqrt{v_z^2 + 2gz}}{g}$

The neutrons entering the cavity at the height z and with the initial vertical velocity component $v_z^2 = 2g(H - z)$ just touch the absorber at their maximum positions and they have to spend at least one full period in the cavity to reach their maximum positions with certainty:

$T_{\text{max}} = \frac{2 \sqrt{v_z^2 + 2gz}}{g} = \frac{2 \sqrt{2g(H - z) + 2gz}}{g} = 2 \sqrt{\frac{2H}{g}}$

The neutrons entering the cavity at the height z and with the initial vertical velocity component $v_z^2 > 2g(H - z)$ obviously reach the absorber and their time intervals between the positions $z = H \rightarrow z = 0 \rightarrow z = H$ are shorter than $T_{\text{max}}$, hence, they are all absorbed within this time interval. Thus the minimum classical length of the cavity is

$L_c = v_x T_{\text{max}} = 2v_x \sqrt{\frac{2H}{g}}$

Substituting the given values,

$L_c \doteq 2 \cdot 10 \cdot \sqrt{\frac{2 \cdot 50 \times 10^{-6}}{9.81}} \doteq 0.0633 \times 10^{-3}\ \text m = 63.3\ \text{mm}$

[color=blue][b](3)[/b][/color] Calculate the classical neutron transmission rate $N_c(H)$ in terms of $\rho$, the number of neutrons per unit time, per unit velocity, per unit height entering the cavity, with all heights z and vertical velocity components $v_z$ equally probable.

The neutrons with the vertical velocity components $v_z$ in the interval $\left(-\sqrt{2g(H - z)}, +\sqrt{2g(H - z)}\right)$ are transmitted. This interval depends on the initial neutron height z, which varies from 0 to H. Therefore, the classical transmission rate is

$N_c(H) = \rho \int_0^H 2\sqrt{2g(H - z)}\ \text dz = 2\rho \sqrt{2gH} \int_0^H \sqrt{1 - \frac z H}\ \text dz =$

$= 2\rho H \sqrt{2gH} \int_0^1 \sqrt{1 - \zeta}\ \text d\zeta = 2\rho \sqrt{2gH^3} \cdot \left[ -\frac{2 (\sqrt{1 - \zeta})^3}{3} \right]_0^1 = \frac{4\rho \sqrt{2gH^3}}{3}$

Instead of monotonical increase of the transmission rate N(H) with the cavity height H, an experiment shows abrupt increases at discrete heights $H_1, H_2, H_3, ...$. The vertical motion of the neutrons bouncing on the mirror is quantizied. The action of these neutrons along the vertical direction is an integer multiple of Plank's constant.

$S = \int p_z(z) dz = nh,\ \ h = 1, 2, 3, ...$

where $p_z$ is the vertical component of the classical momentum and the integral is taken over the whole bouncing cycle. Only these neutron states are allowed in the cavity.

[color=blue][b](4)[/b][/color] Calculate the discrete heights $H_n$ and the discrete energy levels $E_n$ associated with the vertical motion, using the above Bohr-Sommerfeld quantization rule. Give the numerical values for $H_1$ and $E_1$.

The action is defined as

$S = \int_{t_1}^{t_2} L(z, v_z)\ \text dt$

where L = K - V is the Lagrangian of the system and K, V its kinetic and potential energies. Let $t_{\text{min}} < t_{\text{max}}$ be the times a neutron reaches the successive minimum and maximum positions. The difference $\frac T 2 = t_{\text{max}} - t_{\text{min}}$ is equal to half period of the bouncing cycle. Considering just the vertical motion (1D problem), the Lagrangian and the action over the bouncing cycle with the period T are

$L = \frac{mv_z^2}{2} - mgz$

$S = \int_{t_{\text{min}}}^{t_{\text{min}} + T} \left( \frac{mv_z^2}{2} - mgz\right) \text dt = \int_{t_{\text{min}}}^{t_{\text{min}} + T} \frac{mv_z^2}{2}\ \text dt - \int_{t_{\text{min}}}^{t_{\text{min}} + T} mgz\ \text dt = $

$= 2 \int_{t_{\text{min}}}^{t_{\text{max}}} \frac{mv_z^2}{2}\ \text dt - \int_0^0 mg\ \text dz = \int_{t_{\text{min}}}^{t_{\text{max}}} mv_z^2\ \text dt = \int_0^{z_{\text{max}}} p_z\ \text dz = \frac 1 2 \oint p_z\ \text dz$

In the 1st integral, we used the fact that the kinetic energy $K = \frac{mv_z^2}{2}$ is an even function of time with respect to $t_{\text{max}}$, then substituted $p_z = mv_z,\ v_z\ \text dt = \frac{\text dz}{\text dt}\ \text dt = \text dz$ and changed the integration limits to $z(t_{\text{min}}) = 0,\ z(t_{\text{max}}) = z_{\text{max}}$. In the 2nd integral, we substituted $z\ \text dt = \text dz$ and changed the integration limits to $z(t_{\text{min}}) = z(t_{\text{min}} + T) = 0$. It follows that the 2nd integral is equal to zero. Thus the action S over the bouncing circle is half of what was given in the problem conditions :?. (I would appreciate someone checking this conclusion.) To calculate the action integral, we express the vertical momentum component $p_z$ as a function of the height z and the neutron energy $E_z$ corresponding to the vertical motion.

$E_z = mgz_{\text{max}},\ \ z_{\text{max}} = \frac{E_z}{mg}$

$z = -\frac{gt^2}{2} + z_{\text{max}},\ \ t = \pm \sqrt{\frac{2(z_{\text{max}} - z)}{g}} = \pm \sqrt{\frac{2(E_z - mgz)}{mg^2}}$

We have to use the minus sign for the square root, because we are setting up the action integral with the integration limits from 0 to $z_{\text{max}} = z(t_{\text{max}})$, but we are calculating the time backwards from $t_{\text{max}}$.

$p_z = mv_z = mz' = -mgt = mg  \sqrt{\frac{2(E_z - mgz)}{mg^2}} = \sqrt{2m(E_z - mgz)}$

$S = \int_0^{z_{\text{max}}} p_z\ \text dz = \int_0^{\frac{E_z}{mg}} \sqrt{2m(E_z - mgz)}\ \text dz = \sqrt{2mE_z} \int_0^{\frac{E_z}{mg}} \sqrt{1 - \frac{mgz}{E_z}}\ \text dz =$

$= \sqrt{2mE_z} \cdot \frac{E_z}{mg} \int_0^1 \sqrt{1 - \zeta}\ \text d\zeta = \frac{\sqrt{2mE_z^3}}{mg} \cdot \left[-\frac{2(\sqrt{1 - \zeta})^3}{3} \right]_0^1 = \frac{2\sqrt{2mE_z^3}}{3mg}$

If the given path integral is quantizied as

$2S = \oint p_z\ dz = \frac{4\sqrt{2mE_z^3}}{3mg} = nh,\ n = 1, 2, 3, ...$

then the energy levels $E_{zn}$ are

$\frac{4\sqrt{2mE_{zn}^3}}{3mg} = nh,\ \ E_{zn} = \frac{\sqrt[3]{18n^2h^2 mg^2}}{4},\ \ n = 1, 2, 3, ...$

and the discrete heights $H_n$ of the absorber corresponding to abrupt increases of the neutron transmission rate are

$H_n = \frac{E_{zn}}{mg} = \frac{\sqrt[3]{18n^2h^2 mg^2}}{4g} = \frac 1 4 \cdot \sqrt[3]{\frac{18n^2h^2}{m^2g}},\ \ n = 1, 2, 3, ...$

Substituting n = 1, Plank's constant $h \doteq 6.63 \times 10^{-34}\ \text{J sec}$, the neutron mass $m = 1.67 \times 10^{-27}\ \text{kg}$ and the gravitational acceleration $g \doteq 9.81\ \text{m/sec}^2$,

$E_{z1} = \frac{\sqrt[3]{18h^2 mg^2}}{4} \doteq \frac{\sqrt[3]{18\cdot 6.63^2 \times 10^{-68} \cdot 1.67 \times 10^{-27} \cdot 9.81^2}}{4} \doteq \frac{\sqrt[3]{1.272 \times 10^{-90}}}{4} \doteq$

$\doteq 0.271 \times 10^{-30}\ \text J \doteq 1.69 \times 10^{-12}\ \text{eV}$

$E_{zn} = n^{2/3} E_{z1},\ \ n = 2, 3, ...$

$E_{z2} \doteq 0.430 \times 10^{-30}\ \text J \doteq 2.68 \times 10^{-12}\ \text{eV}$

$E_{z3} \doteq 0.563 \times 10^{-30}\ \text J \doteq 3.52 \times 10^{-12}\ \text{eV}$
...

where $1\ \text{eV} \doteq 1.602 \times 10^{-19}\ \text J$ and $e \doteq 1.602 \times 10^{-19}\ \text C$ is the elementary charge. The 1st abrupt increase of the neutron transmission rate should be observed at the absorber height $H_1$ corresponding to the 1st energy level $E_{z1}$.

$H_1 = \frac{E_{z1}}{mg} \doteq \frac{0.271 \times 10^{-30}}{1.67 \times 10^{-27} \cdot 9.81} \doteq 26.2 \times 10^{-6}\ \text m = 16.5\ \mu \text m$

$H_n = n^{2/3} H_1,\ \ n = 2, 3, ...$

$H_2 \doteq 26.2\ \mu \text m,\ \ H_3 \doteq 34.4\ \mu \text m,\ ...$

Whew !!! I have never used the old quantum theory with the Bohr-Sommerfeld ad-hoc quantization rules in my life and it amazes me that this thing still being invoked in a context other than historical. This is like trying to start a fire with a flint. If I have time, I will try to post a normal quantum mechanical solution.

Yetti

Assume a cavity with the height $H < H_2$. Classically, all neutrons with energies in the range considered in the question (1) are allowed, while quantum mechanically, only neutrons at the energy level $E_1$ are permitted. According to the time-energy Heisenberg principle of uncertainty, this reshuffling requires a minimum time of flight. The uncertainty of the vertical motion energy will be significant, if the cavity length is small, giving rise to the widening of the energy levels.

[color=blue][b](5)[/b][/color] Estimate the minimum time of flight $t_q$ and the minimum cavity length $L_q$ necessary to observe the 1st sharp increase in the neutron transmission rate. Assume $v_x = 10\ \text{m/sec}$.

This question is difficult to answer, because the form the time-energy Heisenberg uncertainty principle is not given and the satisfactory sharp increase in the neutron transmission rate is not defined. The exact form of the time-energy Heisenberg uncertainty principle is

$\triangle E \cdot \triangle t \ge \frac{\hbar}{2},\ \ \hbar = \frac{h}{2 \pi}$

The meaning of this inequality is that the energy of a system that has been in existence only for a finite time $\triangle t$ has an energy spread at least $\triangle E$, where $\triangle E$ is square root of the variance of the Hamiltonian operator. But the old quantum theory gives the time-energy uncertainty only as

$\triangle E \cdot \triangle t \approx h$

where the energy spread is not even exactly defined. Since the answer to the question (4) called for the use of the old quantum theory, I will use this 2nd form of the time-energy uncertainty principle. As to the definition of a "sharp" increase in the neutron transmission rate, we can demand that $\triangle E_{z} < \frac{E_{z1}}{2}$. We can also demand that the energy spreads of the energy levels $E_{z1},\ E_{z2}$ do not overlap, i.e.,

$\triangle E_{z} < \frac{E_{z2} - E_{z1}}{2} = \frac{E_{z1} (2^{2/3} - 1)}{2} \approx 0.3\ E_{z1}$

which is slightly stronger. This is what I am going to do. The minimum time of flight is then

$t_c = \triangle t > \frac{h}{0.3\ E_{z1}} \doteq \frac{6.63 \times 10^{-34}}{0.3 \cdot 0.271 \times 10^{-30}} \approx 6.55 \times 10^{-3}\ \text{s} = 6.55\ \text{ms}$

and the minimum cavity length

$L_c = v_x t_c > 10 \cdot 6.55 \times 10^{-3} = 65.5 \times 10^{-3}\ \text m = 65.5\ \text{mm}$

The ordinary quantum mechanical solution of the problem consists of finding the energy levels of a neutron in the one dimensional potential well V(z) defined as

$V(z) = mgz$ [color=white].[/color] for [color=white].[/color] $z \ge 0$, [color=white].[/color] $V(z) = \infty$ [color=white].[/color] for [color=white].[/color] $z < 0$

We will skip the indices z of the neutron momentum $p_z$ and energy $E_z$. In the region $z \ge 0$, the Schrödinger equation for the neutron stationary state $\left|\psi\right>$ is

$\widehat H \left|\psi\right> = \left(\frac{\widehat p^2}{2m} + mg \widehat z\right) \left|\psi\right> = E \left|\psi\right>$

Solving the problem in the coordinate representation leads to a 2nd order differential equation for $\psi(x) = \left<x|\psi\right>$, while solving the problem in the momentum representation leads to a 1st order differential equation for $\psi(p) = \left<p|\psi\right>$. Hence, we go to the momentum representation.

$\left<p|\ \widehat p\ |\psi\right> = p \cdot \psi,\ \  \left<p|\ \widehat z\ |\psi\right> = i \hbar \frac{\text d \psi}{\text dp}$

$\frac{p^2}{2m}\psi + i \hbar mg \frac{\text d\psi}{\text dp} = E \psi$

$\frac{\text d\psi}{\text dp} + \frac i \hbar \left(\frac{E}{mg} - \frac{p^2}{2m^2 g}\right) \psi = 0$

$\frac{\text d\psi}{\psi} = \frac i \hbar \left(\frac{E}{mg} - \frac{p^2}{2m^2 g}\right) \text dp$

$\ln \psi = \pm \frac i \hbar \left(\frac{Ep}{mg} - \frac{p^3}{6m^2g}\right) + C$

$\psi(p) = A \exp \left[ \pm \frac i \hbar \left(\frac{Ep}{mg} - \frac{p^3}{6m^2g}\right) \right]$

Obviously, for every energy level E, we have 2 linearly independent orthogonal eigenstates with p = |p| or p = -|p|, because a neutron with the energy E can be moving up or down. By the superposition principle, the neutron is generally in the suprrposition of these 2 states, hence

$\psi\left(E = \frac{p^2}{2m}\right) = A \psi(|p|)  + B \psi(-|p|) =$

$= A \exp \left[ +\frac i \hbar \left(\frac{E|p|}{mg} - \frac{|p|^3}{6m^2g}\right) \right] + B \exp \left[ - \frac i \hbar \left(\frac{E|p|}{mg} - \frac{|p|^3}{6m^2g}\right) \right] =$

$= A' \cos \left[ \frac i \hbar \left(\frac{E|p|}{mg} - \frac{|p|^3}{6m^2g}\right) \right] + B' \sin \left[ \frac i \hbar \left(\frac{E|p|}{mg} - \frac{|p|^3}{6m^2g}\right) \right]$

because A, B, A', B' are arbitrary complex constants. The neutrons cannot penetrate the infinite potential barrier at z = 0, when $|p| = \sqrt{2mE}$ and since the the wave function is continuous, it has to vanish at this momentum. Consequently, the cosine constant A' = 0 and

$\frac 1 \hbar \left(\frac{E\sqrt{2mE}}{mg} - \frac{(\sqrt{2mE})^3}{6m^2g}\right) = n\pi$

This is the sought after condition for the discrete energy levels.

$\sqrt{\frac{2mE^3}{m^2g^2}} - \sqrt{\frac{8m^3E^3}{36m^4g^2}} = n\pi \hbar$

$\sqrt{\frac{2E^3}{mg^2}} - \sqrt{\frac{2E^3}{9mg^2}} = n\pi \cdot \frac{h}{2\pi}$

$\sqrt{\frac{2E^3}{mg^2}} \left(1 - \frac 1 3\right) = \frac{nh}{2}$

${\frac{2E^3}{mg^2}} \cdot \frac 4 9 = \frac{n^2 h^2}{4}$

$E^3 = \frac{9 n^2 h^2 mg^2}{32} = \frac{18 n^2 h^2 mg^2}{64}$

$E_n = \frac{\sqrt[3]{18 n^2 h^2 mg^2}}{4}$

which is the same result as the one obtained by the old quantum theory. I edited the previous post, because I dropped a factor of 2 and consequently, the previously calculated energy levels were off by a factor of $2^{2/3} \doteq 1.6$. The problem can certainly be solved in the coordinate representation, yielding the same energy levels, but the solution is rather ugly - it requires using the Bessel functions.


function
algebra
polynomial
number theory unsolved
number theory

Prove that there do not exist polynomials $ P$ and $ Q$ such that

$ \pi(x)\equal{}\frac{P(x)}{Q(x)}$ for $ x\equal{}1,2,3,...$


search
combinatorics unsolved
combinatorics

This is usually described as a "spreading infection", and is famous for having a one-word solution ("perimeter"). See here:

http://www.mathlinks.ro/viewtopic.php?search_id=199859112&t=28298

[quote="handrake"]places on some of the normal t
squares containing coins. (Two if they m
number of can place on[/quote]

post deleting FAIL

  dg


In triangle ABC with H- orthocenter, S - area of Triangle and p - half perimeter. Prove inequalities:
$\frac{BC}{AH}+\frac{AC}{BH}+\frac{AB}{CH}\geq \frac{p^{2}}{S}$

[quote="gemath"]$(\frac{a}{AH}+\frac{b}{BH}+\frac{c}{CH})(aHA+bHB+cHC)\ge (a+b+c)^{2}=4p^{2}$ note $aHA+bHB+cHC=4S$
we have this problem is the fact!  :lol:[/quote]

Good solutions!!! :wink:


function
number theory unsolved
number theory

I've read that it's relatively easy to prove it in $ P[x]$, $ P$ a prime subfield, instead of in $ \mathbb Z$. How "easy" is it?

I'm not sure what the Riemann Hypothesis in $ P[x]$ is supposed to be; do you mean Fermat's Last Theorem?

We have a tool in $ P[x]$ we don't have in $ \mathbb{Z}$, which is [url=http://mathworld.wolfram.com/MasonsTheorem.html]Mason's Theorem[/url].  See for yourself how easy it is :)

Would someone be kind enough to explain this problem in detail please? When I first seen it this problem it seemed to me your were talking about [/i]function fields[/i] but now it seems you are talking about something else. Are you?


The product of a two-digit number and 5 is equal to the product of the same two-digit number with its digits reversed and 6. What is the sum of the digits of the number?

Use $\text{\LaTeX}$. It helps readers understand and grasp the concept better. Thank you. For more info on $\text{\LaTeX}$, go to [url]http://www.artofproblemsolving.com/Wiki/index.php/LaTeX:About[/url].

The original number is divisible by six because $x = \text{something}\cdot6,$ where x is equal to the original number. This applies to the reverse number, too. The one's digit must either be 0 or 5. However, since the original number is a 2 digit one, the one's digit of the reverse number is 5. The ten's digit of the original number is then 5. The only number in to the 50's divisible by 6 is 54.

$5+4 = \boxed{9}.$


The summary of a survey of 100 students listed the following
totals:
\\
\\
59 students did math homework
\\
49 students did English homework
\\
42 students did science homework
\\
20 students did English and science homework
\\
29 students did science and math homework
\\
31 students did math and English homework
\\
12 students did math, science and English homework
\\
\\
How many students did no math, no English and no science
homework?

We use PIE:

$ 100\minus{}59\minus{}49\minus{}42\plus{}20\plus{}29\plus{}31\minus{}12\minus{}x\equal{}0 \implies x\equal{}\boxed{18}$.


Ring Theory
abstract algebra
invariant
superior algebra
superior algebra unsolved

Let $ A$ be a ring with $ n$ elements,$ n \geq 3$ , so that the sum and the product of its nonnzero elements are nonzero.Prove that  $ n \equal{} 4$ and there is an isomorphism from $ A$ to $ Z_4$ .

Do your rings necessarily have unity?  Your other thread implies yes.  If so...

To start with, consider the subring generated by $ 1$, isomorphic to some $ \mathbb{Z}/k$, $ k$ a factor of $ n$.  The product of its nonzero elements must be nonzero.  Through some pendantic counting of powers of 2, we can see that $ (k \minus{} 1)!$ is nonzero mod $ k$ only when $ k$ is odd, or $ k \equal{} 2$ or $ 4$.

Now, the decomposition theorem for abelian groups implies that the additive group of $ R$ cannot have only elements of odd order, so in particular $ k$ is even, hence $ k \equal{} 2$ or $ 4$.  But then $ 4a \equal{} 0$ for every $ a\in R$, implying that $ n$ is a power of $ 2$.  Looking again at the decomposition theorem, we see that the additive group of $ R$ cannot have more than one 2-component, so the additive group of $ R$ is cyclic, say $ R \equal{} \langle x\rangle$.

The product of the nonzero elements of $ R$ is then $ (n \minus{} 1)! x^{n \minus{} 1}$, and since $ n$ is even, the same argument as above shows that $ n$ is $ 2$ or $ 4$.

Finally, we have a ring $ \{0,x,2x,3x\}$.  Either $ R\equal{}\langle 1\rangle$, or $ 2x\equal{}1$.  But then $ 1\plus{}1\equal{}4x\equal{}0$, so $ 2x\equal{}0$, contradiction.

EDIT: Sorry for so many edits.  Should be good now.

Thank you for the answer.However,I have a small question:I think I missunderstanding the part with the decomposition theorem.Why it is involving the fact that R cannot have only elements of odd order?

We can sum the elements in any finite $ \mathbb{Z}$-module. The sum is invariant under the isomorphism $ x\mapsto -x$, so it is either the identity or an element of order 2. If there are multiple elements of order 2, which one? None of them. $ \sum_{x\in G\oplus H}x=|H|\sum_{x\in G}x+|G|\sum_{x\in H}x$, and both terms are zero if both $ G$ and $ H$ have even order.

[quote="gordon_c"].Why it is involving the fact that R cannot have only elements of odd order?[/quote]

If you look only at the group $ (A,\plus{})$ you can easily prove that $ \sum_{a \in A} a \equal{} \sum_{a \in B} a$ , where $ B\equal{}\{x \in A | ord(x) \equal{}2\}$, or in another way $ B$ is the set of all elemnts $ x \in A$ with the property that $ x\equal{} \minus{}x$.

If $ B$ doesn't have any element then $ \sum_{a \in A} a \equal{}0$.

Actually, it seems that a version of my argument works for rings without unity:

As stated, the characteristic $ m$ of our ring must be even.  Another glance at the decomposition theorem shows that if $ p$ is an odd prime factor of $ m$, we can find an element $ a$ of additive order $ 2p$.  Then $ (2p\minus{}1)!a^{2p\minus{}1}$ must be nonzero, so $ (2p\minus{}1)!$ is nonzero mod $ 2p$, giving $ 2p\equal{}2$ or $ 4$, contradiction.  Hence $ m$ is a power of $ 2$, so the additive group of $ R$ is cyclic and we proceed as before.


Let $ PB\cap DR \equal{} K$. Verify that $ QR\parallel AC\iff QR\perp BC\iff \angle QRB \equal{} \angle BKR\iff \angle QRD \equal{} \angle PBC$. So we'll be done if we can show that $ \angle QRD \equal{} \angle PBC$. But this follows from the fact that $ \angle B$ is the common angle of the right triangles $ PBC$ and $ RBK$. So we are done.

EDIT: Fixed typo.

[quote="nayel"]Let $ PB\cap DR \equal{} K$. Verify that $ QR\parallel AC\iff QR\perp BC\iff \angle QRB \equal{} \angle BDR\iff \angle QRD \equal{} \angle PBC$. So we'll be done if we can show that $ \angle QRD \equal{} \angle PBC$. But this follows from the fact that $ \angle B$ is the common angle of the right triangles $ PBC$ and $ RBK$. So we are done.[/quote]

If I'm not mistaken $ \angle QRB \equal{} 90^\circ \neq \angle BDR$ :huh:

Some hints for my proof:

[hide]Use nayel's point $ K$ and the Fact that if $ A_1A_2B_2B_1$ and $ B_1B_2C_2C_1$ are cyclic quadrilaterals and $ A_1A_2, B_1B_2, C_1C_2$ are concurrent, then $ A_1A_2C_2C_1$ is a cyclic quadrilateral.[/hide]

[quote="The QuattoMaster 6000"][quote="bertram"]$ \Delta ABC$ is a triangle with $ \angle C \equal{} 90^\circ .$ $ P$ is a point on $ AC$ and $ Q$ is chosen on $ AB$ such that $ PQ\perp AB.$ In addition, take point $ D$ on $ AB$ with $ CD\perp AB$ and choose $ R$ on $ BC$ so that $ DR\perp BP.$ Prove that $ QR\parallel AC.$[/quote]
Very nice problem!
[hide="Solution"]
Denote the intersection between $ DR$ and $ BP$ to be $ K$. Now, we know that $ \angle RKP \plus{} \angle RCP \equal{} 90 \plus{} 90 \equal{} 180$, so $ RKPC$ is cyclic, so $ BR(BC) \equal{} BK(BP)$.  Also, $ \angle DKP \equal{} 90 \equal{} \angle PQD$, so $ \angle DKP \plus{} \angle PQD \equal{} 180$, so $ PQDK$ is cyclic.  This implies that $ BK(BP) \equal{} BD(BQ)$.  So, $ BR(BC) \equal{} BD(BQ)$ and thus $ RDQC$ is cyclic.  So, $ \angle CRQ \equal{} \angle CDQ \equal{} 90\implies \angle BRQ \equal{} 90$.  So, $ \angle BRQ \equal{} 90 \equal{} \angle BCA\implies RQ\parallel AC$.  [/hide][/quote]

Yop, that was my solution too. The real problem with this problem is not to get carried away by the abundance of cyclic quadrilaterals. ;)

[quote="nayel"][quote="bertram"][quote="nayel"]Let $ PB\cap DR \equal{} K$. Verify that $ QR\parallel AC\iff QR\perp BC\iff \angle QRB \equal{} \angle BDR\iff \angle QRD \equal{} \angle PBC$. So we'll be done if we can show that $ \angle QRD \equal{} \angle PBC$. But this follows from the fact that $ \angle B$ is the common angle of the right triangles $ PBC$ and $ RBK$. So we are done.[/quote]

If I'm not mistaken $ \angle QRB \equal{} 90^\circ \neq \angle BDR$ :huh:[/quote]

Sorry I think it's correct now. :maybe:[/quote]

I can follow your solution till the proof of $ \angle QRD \equal{} \angle PBC.$ But how do you do that proof? I can see absolutely no relation between $ \angle CBP$ and $ \angle QRB$ that doesn't make use of $ \angle QRB \equal{} 90^\circ$ :(

[quote="bertram"][/quote][quote="nayel"][quote="bertram"][quote="nayel"]Let $ PB\cap DR = K$. Verify that $ QR\parallel AC\iff QR\perp BC\iff \angle QRB = \angle BDR\iff \angle QRD = \angle PBC$. So we'll be done if we can show that $ \angle QRD = \angle PBC$. But this follows from the fact that $ \angle B$ is the common angle of the right triangles $ PBC$ and $ RBK$. So we are done.[/quote]

If I'm not mistaken $ \angle QRB = 90^\circ \neq \angle BDR$ :huh:[/quote]

Sorry I think it's correct now. :maybe:[/quote][quote="bertram"]

I can follow your solution till the proof of $ \angle QRD = \angle PBC.$ But how do you do that proof? I can see absolutely no relation between $ \angle CBP$ and $ \angle QRB$ that doesn't make use of $ \angle QRB = 90^\circ$ :([/quote]
No, his solution works, bertram.
[hide="Further Explanation"]
Notice that $ \angle RBP + \angle DRB = 90 = \angle RBP + \angle CPB\implies \angle CPB = \angle DRB$.  Thus, $ \angle CRK = 180 - \angle DRB = 180 - \angle CPB$.  This implies that $ CRKP$ is cyclic and the solution follows.  Is this what you meant, nayel? [/hide]
EDIT: Oh, sorry bertram... I thought I saw something else... this post is useless...  :(

[quote="The QuattoMaster 6000"][/quote][quote="bertram"][/quote][quote="nayel"][quote="bertram"][quote="nayel"]Let $ PB\cap DR = K$. Verify that $ QR\parallel AC\iff QR\perp BC\iff \angle QRB = \angle BDR\iff \angle QRD = \angle PBC$. So we'll be done if we can show that $ \angle QRD = \angle PBC$. But this follows from the fact that $ \angle B$ is the common angle of the right triangles $ PBC$ and $ RBK$. So we are done.[/quote]

If I'm not mistaken $ \angle QRB = 90^\circ \neq \angle BDR$ :huh:[/quote]

Sorry I think it's correct now. :maybe:[/quote][quote="bertram"]

I can follow your solution till the proof of $ \angle QRD = \angle PBC.$ But how do you do that proof? I can see absolutely no relation between $ \angle CBP$ and $ \angle QRB$ that doesn't make use of $ \angle QRB = 90^\circ$ :([/quote][quote="The QuattoMaster 6000"]
No, his solution works, bertram.
[hide="Further Explanation"]
Notice that $ \angle RBP + \angle DRB = 90 = \angle RBP + \angle CPB\implies \angle CPB = \angle DRB$.  Thus, $ \angle CRK = 180 - \angle DRB = 180 - \angle CPB$.  This implies that $ CRKP$ is cyclic and the solution follows.  Is this what you meant, nayel? [/hide][/quote]

The cyclic quadrilateral also follows from thales's theorem (two right angles)  :oops: but I think I got the solution.

I'm very sorry. I don't think my solution was complete anyway. Sorry.... :(


Let $ a,b,c>0,a\plus{}b\plus{}c\equal{}1$. Prove that:
$ a^ab^bc^c\plus{}a^bb^cc^a\plus{}b^ac^ba^c \le 1$

Very nice inequality!

From weighted AM-GM, we have

$ 1)\quad \frac {a^2 \plus{} b^2 \plus{} c^2}{a \plus{} b \plus{} c}\ge (a^ab^bc^c)^{\frac {1}{a \plus{} b \plus{} c}}\implies a^2 \plus{} b^2 \plus{} c^2\ge a^ab^bc^c$

$ 2)\quad \frac {ab \plus{} bc \plus{} ca}{a \plus{} b \plus{} c}\ge (a^bb^cc^a)^{\frac {1}{a \plus{} b \plus{} c}}\implies ab \plus{} bc \plus{} ca\ge a^bb^cc^a$

$ 3)\quad \frac {ac \plus{} ba \plus{} cb}{a \plus{} b \plus{} c}\ge (a^cb^ac^b)^{\frac {1}{a \plus{} b \plus{} c}}\implies ab \plus{} bc \plus{} ca\ge a^cb^ac^b$

Adding the three inequalities we get $ (a \plus{} b \plus{} c)^2 \equal{} 1\ge a^ab^bc^c \plus{} a^bb^cc^a \plus{} a^cb^ac^b$


inequalities induction function calculus derivative

Prove Bernoulli's Inequality, use induction if you can.

You can't use induction to prove the inequality: it only proves it for when the exponent is an integer.

[hide="Ridiculous proof which only works for integers"]For $ n\equal{}2$, we have $ (1\plus{}x)^2 >1\plus{}2x$ by the trivial inequality.

For $ n>2$, we have $ (1\plus{}x)^{n\minus{}1} > 1\plus{}(n\minus{}1)x$ by the inductive hypothesis, so $ (1\plus{}x)(1\plus{}x)^{n\minus{}1} > (1\plus{}x)(1\plus{}(n\minus{}1)x)$. And expanding $ (1\plus{}x)(1\plus{}(n\minus{}1)x)$, we find that it is greater than or equal to $ 1\plus{}nx$, again by the trivial inequality.[/hide]

[hide="Ridiculous proof which actually works for all values"]Consider the function $ (1\plus{}x)^n \minus{}1\minus{}nx$. Its derivative is $ n((1\plus{}x)^{n\minus{}1}\minus{}1)$ which is negative over the interval from $ \minus{}1$ to $ 0$ exclusive and positive from $ 0$ to infinity exclusive. Since it is continuous, it obtains its minimum value for $ x\geq \minus{}1$ at $ 0$. Plugging $ 0$ in for $ x$, you get a value of $ 0$. Therefore, the value of this function is greater than $ 0$ for $ x\geq \minus{}1$ not equal to $ 0$.[/hide]

I don't see why you decided to post this even though you asked me for the solution earlier and I told it to you...

Darn. I knew a full induction solution was impossible... What I asked you to post on was the Germany problem, not the inequality one :wink:

You don't need calculus to prove it for non-integers.  Show that $ (1 \plus{} px)^q > (1 \plus{} qx)^p$ for integers $ q > p$ and argue by continuity.

We can also consider a function $ f$ defined by $ f(x)\equal{}(1\plus{}x)^{n} \ge 1\plus{}nx$ , and using its derivable . . . it will be simple

[i]$ \forall x\ge \minus{} 1$ and $ n\in N$, we have $ \implies (1 \plus{} x)^n\ge 1 \plus{} nx$

If $ 1 \plus{} nx\leq 0$ ,then the inequality is true , because $ (1 \plus{} x)^n > 0$

If $ 1 \plus{} nx > 0$,we have:

$ \sqrt [n]{(1 \plus{} nx)(1\cdot 1 \cdots 1)} \leq \frac {(1 \plus{} nx) \plus{} (n \minus{} 1)}{n} \equal{} x \plus{} 1 \implies (1 \plus{} nx)\leq (1 \plus{} x)^n$

$ \forall x > \minus{} 1$ and $ \forall r \in R , r > 1$,we have:
$ (1 \plus{} x)^r > 1 \plus{} rx$

$ \forall x > \minus{} 1$ and $ \forall r \in R ,0 < r < 1$,we have:
$ (1 \plus{} x)^r < 1 \plus{} rx$[/i]


modular arithmetic
number theory solved
number theory

Prove that there is no square number that end in 4 equal digits.
$x^2=\overline{a_na_{n-1}...a_4a_3a_2a_1}$
Prove that there is no natural $x$ for which $a_4=a_3=a_2=a_1$

Well, there is the trivial case of $a_4 = a_3 = a_2 = a_1 = 0$  :P

All perfect squares are $\equiv 0$, or $1 \pmod{8}$, so taking $\mod 1000$, we check for the nine cases $111, 222, \ldots, 999$, we find that the only possibilities are $777$ and $888$, yet we know that $x^2 \equiv 0, 1, 4, 9, 6, 5 \pmod{10}$, so contradiction.


$ x$ and $ y$ are positive integers such that $ x^{2y}\equiv 84\pmod{100}$, and $ x^{3y}\equiv 52\pmod{100}$. Find the value of $ x^y \pmod{100}$.

"Brute force" Solution

[hide]We first examine the exponents and realize the first must be a perfect square and the second must be a perfect cube.  The only possible units digits of the first expression are $ 2$ and $ 8$, but only $ 8$ works for the second expression.  Narrowing our guesses down significantly, we examine perfect squares and cubes to discover that $ x\equal{}28, y\equal{}1$.  Thus $ x^y\equiv 28 \pmod {100}$.[/hide]

This is a cool problem!!
Anyway, we observe that $ x^{2y}\cdot x^y\equal{}x^{3y}$ so we have

$ 84n\equiv 52\mod 100$

$ 21n\equiv 13\mod 25$

Then you can easily compute the answer by finding the inverse of $ 21$ mod $ 25$ but I'm too lazy to do it. Also azn, there are more solutions.

[hide="finish"]Let $ n\equal{}x^y$. Then we have $ 84n\equiv 52 \pmod {100} \iff 21n\equiv 13 \pmod {25}$. Solving gives $ n\equiv 3 \pmod {25}$, therefore $ n\equiv 3, 28, 53, 78 \pmod{100}$. Observe that $ n$ must be even, so $ n\equiv 28, 78\pmod{100}$. Checking shows both solutions satisfy the given congruences. [/hide]

This is a rather strange problem considering that $ x^y$ can be immediately replaced by another variable and this doesn't affect the integrity of the problem.  Perhaps we should ask for the minimum possible value of $ x \plus{} y$ or something interesting along those lines?


Points $A_{1}, B_{1}, C_{1}$ are given inside an equilateral triangle $ABC$ such that $\widehat{B_{1}AB}= \widehat{A1BA}= 15^{0}, \widehat{C_{1}BC}= \widehat{B_{1}CB}= 20^{0}, \widehat{A_{1}CA}= \widehat{C_{1}AC}= 25^{0}$.
Find the angles of triangle $A_{1}B_{1}C_{1}$.

I saw similar problem in IMO Compendium book.

it was already posted here: [url]http://www.mathlinks.ro/Forum/viewtopic.php?t=126187[/url]


a function satisfies that f(2+x)=f(2-x), this function has 4 real roots, find the sum of the roots

The function is semmetry with respect to line $x=2$, so the sum of the roots is $8$.

why is the sum 8 if the function is respect to line x=2?

$\dfrac{x_1+x_2}{2}=\dfrac{x_3+x_4}{2}=2$ because of semmetry.


Prove that if $ a\neq b\neq c\neq d$, there do not exist positive integers $ a,b,c,d$ satisfying equation $ \frac{1}{a}\plus{}\frac{1}{b}\equal{}\frac{1}{c}\plus{}\frac{1}{d}$

I think it is true, but I would like to know a "real mathematical" solution of that problem

Thanks in advance :p

I'm not sure I understood the problem exactly, but is the following a counterexample?
\[ \frac{1}{5} \plus{} \frac{1}{20} \equal{} \frac{1}{6} \plus{} \frac{1}{12} \, .\]

Uhm... thanks, I'll have to think once again about what I wanted to ask for :D

Take $ 2$ times the reciprocal of each side to get:
\[ \frac{2}{\frac{1}{a}\plus{}\frac{1}{b}}\equal{}\frac{2}{\frac{1}{c}\plus{}\frac{1}{d}}\]
Which simply states that the harmonic means of two pairs of numbers are equal.


We put 4 stars in a 5 x 5 board such that no 2 or more are in the same row or the same column and also the 4 stars are in squares which have the sama colour. In how many ways can we make this distribution?

[hide]Ok, I'm terrible at combinatorics, but is it 24?[/hide]

It's 96. But I don't know how. I got 1000 but then I noticed that I didn't consider they had to be in squares with the same colour. :D

First consider white squares.  All except one column will have exactly one star.  The column without a star cannot be a 3-white-square column (i.e columns 2 and 4).  Therefore, there are two 2-square and two 3-square columns with stars.  There are $2$ available spots in the first 2-white-square column, leaving $1$ in the other 2-white-square column.  The are $3$ spots in the first 3-white-square column and $2$ available ones in the next.  There are $3$ combinations of occupied columns, giving $3(2\times1\times3\times2)=3(12)=36$ possibilities for white squares.

Similar logic shows there to also be $3(2\times1\times3\times2)+2(2\times3\times2\times1)=3(12)+2(12)=60$ possibilities with the black squares (this is slightly more complicated since both 2- and 3-square columns can have no star), giving a grand total of $96$.


Prove that :

$ \mathrm{log}_2\sqrt[5]{2^{13}}<\mathrm{log}_3 18<e$

[quote="Pain rinnegan"]Prove that :

$ \mathrm{log}_2\sqrt [5]{2^{13}} < \mathrm{log}_3 18 < e$[/quote]

It's easy to see that this is equivalent to $ \frac{13}{5} < 2\plus{} \mathrm{log}_3 2 <e$

Left inequality is $ 3<5log_3 2$ $ \Rightarrow 3^3<2^5$, which is clearly true.

For the right inequality , we just need an estimation of $ e$, i.e. $ 2\plus{}\frac{2}{3}<e$

Then it's enough to prove $ log_3 2 <\frac{2}{3} \Leftrightarrow 2^3<3^2$


[img]http://img502.imageshack.us/img502/6073/sdfsdfyo5.png[/img]

Why this happens?

(Both square roots and boxes are compiled on the same PDF.)

I doubt that even stevem will be able to answer your question without seeing the code that produced this output. ;)

Exactly. Unfortunately my abilities don't extend to mind reading. I can't say I even understand what the problem is. And can I remind you that you said in another thread:
[quote="Field"]Now I get how to give the minimal example, sorry.[/quote]

Ohhh, sorry, I thought it wasn't a code problem.

For example, I use this code

[code]\shadowbox{$\dint {\Big[ {f_1 (x) \pm f_2 (x)} \Big]\,dx}  = \int {f_1 (x)\,dx}  \pm \int {f_2 (x)\,dx}$}[/code]

which generates the second box.

But this code

[code]\shadowbox{$\dint {\frac{1}
{{\sqrt {a^2  - u^2 } }}\,du}  = \operatorname{arcsen} \frac{u}
{a} + k,\,a > 0{\text{ y }}\int {\frac{1}
{{a^2  + u^2 }}\,du}  = \frac{1}
{a}\operatorname{arctg} \frac{u}
{a} + k,\,a > 0.$}[/code]

generates the first box and there're such lines which protrude. (Both boxes were centered with \begin{center}...)

The same with the square root,

[code]$$\int {\sqrt {4x - 1} \,dx}$$. (This generates the first square root.)

$\dint {\sqrt {4x - 1} \,dx}$. (This generates the second square root.)[/code]

(\dint it's command to substitute \displaystyle\int.)

Ah, you are just unhappy about how everything looks on the screen :) As far as I know, it is a fault of the viewer (at least, I'm getting such pictures time to time and, when I print them, they come out perfectly fine). The problem seems to be that the precision with which $ \text{\LaTeX}$ defines the graphic elements exceeds the viewer resolution and everything is rounded to the next pixel or something like that. Just try to zoom in and you'll see that your boxes will become fine.

1. Still no minimal document which is a short, [b]complete[/b] document showing the problem.
2. I still don't know what the problem is. This is what the following minimal example produces:
[code]\documentclass{article}
\usepackage{amsmath}
\usepackage{amssymb}
\usepackage{fancybox}
\begin{document}
\newcommand{\dint}{\displaystyle \int}
\shadowbox{$\dint {\Big[ {f_1 (x) \pm f_2 (x)} \Big]\,dx}  = \int {f_1 (x)\,dx}  \pm \int {f_2 (x)\,dx}$}

\shadowbox{$\dint {\frac{1}
{{\sqrt {a^2  - u^2 } }}\,du}  = \operatorname{arcsen} \frac{u}
{a} + k,\,a > 0{\text{ y }}\int {\frac{1}
{{a^2  + u^2 }}\,du}  = \frac{1}
{a}\operatorname{arctg} \frac{u}
{a} + k,\,a > 0.$}

$$\int {\sqrt {4x - 1} \,dx}$$. (This generates the first square root.)

$\dint {\sqrt {4x - 1} \,dx}$. (This generates the second square root
\end{document}[/code] and it seems fine to me.

There is no problem whatsoever. The complaint was about slightly protruding horizontal lines on the left in the first box. If you try to zoom in and out your picture, you may be lucky to reproduce the effect on your viewer at some scales but it'll disappear at most scales. (Note that your box sizes are clearly different from Field's ones).

Yes, that worked out.

I just changed it to 122,1% and square roots look fine, but not the box. (Well, I think that's not big deal.)

Some viewers have a "magnifying glass" tool that allows you to see how some small fragment of your output looks in detail, so you can just move it over suspicious places to check if they are fine or not. Otherwise, just zoom in until the lines become more than one pixel thick. If you see any problem at this level, it is a real one. If not, it is just a visual effect of insufficiently high viewer resolution.


if it's like that...

maybe you wrote it wrong

-jorian

btw, if it uses the term "that number" and you have infinite numbers, there's something wrong

[quote="kstan013"][quote="epatjn"]how can the answer to this be 1.1?

Mathcount warmup 10 2006:  A quarter of one percent of 40 is less than what number?  Express your answer as a deciment to the nearest tenth.[/quote]It probably said "one less than" what number. If you did that, it would be 1% of 40 is .4, and a qaurter of that is .1. .1 is 1 less than 1.1 :D  :!: This is why I like algebra :D[/quote]

yeah i remember this problem it does involve what number is one more than ...

[quote="epatjn"]how can the answer to this be 1.1?

Mathcount warmup 10 2006:  A quarter of one percent of 40 is less than what number?  Express your answer as a deciment to the nearest tenth.[/quote]

Maybe they're right, your question must be wrong...

0.1 < 0.1+x, where x = any number more than 0

:)

The question, as presented in the School Handbook is:

[quote]A quarter of one percent of 40 is one less than what number?  Express your answer as a decimal to the nearest tenth.[/quote]


On Line 12 of Page 42 of Allen Hatchers' "Algebraic Topology", it says "a free group is $ Z_2 * Z_2$". Can someone tell me what is the $ Z_2$ here and why $ a^2 \equal{} e \equal{} b^2$ in this case?

Now it's a lot clearer to me. One more stupid question on this free product: Can someone tell me why $ Z_2 * Z_2$ is not a free group?

[quote="jmerry"]It's a notation for the cyclic group of order 2, which I don't like since it conflicts with notation for another algebraic object.

The free product in this case is the dihedral group on $ \mathbb{Z}$[/quote]

I see. Thanks.
[quote="jmerry"]There are already relations involving the generators we started with: they each have order 2. Free groups have no elements of finite order.[/quote]

Please use the reply button rather than the quote button, or edit the quotes if you do use the quote button. Quoting the entire preceding post just clutters up the thread; you're assumed to be responding to it anyway.


The "one" and I have agreed that question one should be disqualified. However if you didn't answer it you are still out. BTW people who missed it were 1,5,7,8,9,13,14. The answer was C. However people who are eliminated for not answering the question are...2,10 and 11.

New first question! Remember Private Message me the answer!

How many teams have made an 18 game winning streak during the season in the NBA?

A)9
B)12
C)4

This ends on Thursday this week at 7:00 Centeral!

I meant more than 18 games straight so really 19+!

If I may ask: Does this mean for this season only? (O.o I'm not sure...)
Also: Is it at least 18, or exactly 18?


logarithms
calculus
calculus computations

studie the convergence of the serie $\sum_{n\ge1}({\frac{logn}{log(n+1)}})^{n^{a}}$ with $a$ is a real number :D

$\ln(n+1)=\ln n+\ln\left(1+\frac1n\right)=\ln n+\frac1n+O(n^{-2}).$

So $\frac{\ln n}{\ln (n+1)}=1-\frac1{n\ln n}+o(n^{-2}) n)^{-1}).$

$\ln\left(\frac{\ln n}{\ln (n+1)}\right) =-\frac1{n\ln n}+o(n^{-2})$

$n^{\alpha}\ln\left(\frac{\ln n}{\ln (n+1)}\right) =-\frac{n^{\alpha-1}}{\ln n}+o(n^{\alpha-2}).$

If $\alpha\le 1,$ then this tends to zero, which means that the terms of the series tend to 1 and the series diverges.

Now try applying the root test. To take the $n$th root of the term is to divide the logarithm of the term by $n.$ That gives us a leading term in the logarithm of $-\frac{n^{\alpha-2}}{\ln n}.$ For $\alpha>2,$ this tends to $-\infty,$ meaning the $n$th root tends to zero, meaning that the series converges by the root test. For $1<\alpha\le 2,$ the root test is inconclusive.

But fix some $p>1$ and consider comparing the terms of our series to $\frac1{n^{p}}.$

Can we say (assuming $\alpha>1$) that

$-\frac{n^{\alpha-1}}{\ln n}<-p\ln n?$

This is the same as saying that $n^{\alpha-1}>p\ln^{2}n.$

This is true for all sufficiently large $n.$ Since it is true, the original series converges for all $\alpha>1$ by comparison to some convergent "$p$-series." It diverges for $\alpha\le1$ because the terms don't go to zero.


Let $ ABC$ be a triangle with $ AB\equal{}AC$. The perpendicular from $ C$ to $ AB$ meets the circumcircle $ \omega$ of $ ABC$ again at $ K$. The line $ \ell$ through $ K$ which is parallel to $ BC$ when reflected over $ KC$ meets $ \omega$ again at $ P$. Show that $ AP\perp BC$.

[quote="nayel"]Let $ ABC$ be a triangle with $ AB \equal{} AC$. The perpendicular from $ C$ to $ AB$ meets the circumcircle $ \omega$ of $ ABC$ again at $ K$. The line $ \ell$ through $ K$ which is parallel to $ BC$ when reflected over $ KC$ meets $ \omega$ again at $ P$. Show that $ AP\perp BC$.[/quote]
[hide="Solution"]
Let $ \angle BAC \equal{} 2\alpha$.  Thus, $ \angle ABC \equal{} \angle ACB \equal{} 90 \minus{} \frac {\angle BAC}{2} \equal{} 90 \minus{} \alpha$.  It follows that $ \angle AKC \equal{} \angle ABC \equal{} 90 \minus{} \alpha$ and $ \angle KCB \equal{} 90 \minus{} \angle ABC \equal{} \alpha$.  Let $ L$ be the point with $ KL\parallel BC$ and $ L$ be on $ AC$.  Hence, $ \angle PKC \equal{} \angle CKL \equal{} \angle KCB \equal{} \alpha$.  This means that $ \angle AKP \equal{} \angle AKC \plus{} \angle CKP \equal{} 90 \minus{} \alpha \plus{} \alpha \equal{} 90$, so $ AP$ is a diameter of $ \omega$.  Hence, $ \angle ABP \equal{} 90\implies \angle PAC \equal{} \angle PBC \equal{} 90 \minus{} \angle ABC \equal{} \alpha$, so $ \angle PAB \equal{} \angle BAC \minus{} \alpha \equal{} \alpha \equal{} \angle PAC$.  This means that $ \angle PCB \equal{} \angle PAB \equal{} \alpha \equal{} \angle PAC \equal{} \angle PBC$, so $ PB \equal{} PC$ and $ AB \equal{} AC$, so $ ABPC$ is a kite.  This means that $ AP\perp BC$, as desired. [/hide]


Asymptote
conics
hyperbola
inequalities
complex numbers
algebra proposed
algebra

[quote="Dimitris X"]Let $ a,b,x,y \in \mathbb{R}$ such that $ (a,b) \not \equal{} (0,0)$ and $ (x,y) \not \equal{} (0,0)$ and is true that:

$ a(x^2 \minus{} y^2) \minus{} 2bxy \equal{} x(a^2 \minus{} b^2) \minus{} 2aby$

and

$ b(x^2 \minus{} y^2) \plus{} 2axy \equal{} y(a^2 \minus{} b^2) \plus{} 2abx,$

prove that $ x \equal{} a$ and $ y \equal{} b.$

P.s
I think it is a very nice problem,which i solved the last 20 minutes during the contest...... :)[/quote]
 :lol: i solve it too nice and clever problem it takes me 35 minutes to think the solution but ok ..HINT: [hide]:P try to solve it with complex numbers [/hide]

$ (a(x^2 \minus{} y^2) \minus{} 2bxy)^2 \plus{} (b(x^2 \minus{} y^2) \plus{} 2axy)^2$
$ \equal{} a^2(x^2\minus{}y^2)^2\plus{}(2bxy)^2\plus{}b^2(x^2\minus{}y^2)^2\plus{}(2axy)^2$
$ \equal{} a^2x^4 \minus{} 2a^2x^2y^2\plus{} a^2y^4 \plus{} 4b^2x^2y^2 \plus{} b^2x^2\minus{}2b^2x^2y^2 \plus{} b^2y^4 \plus{} 4a^2x^2y^2$
$ \equal{} (a^2\plus{}b^2)(x^2\plus{}y^2)^2$
And in the same way:
$ (x(a^2 \minus{} b^2) \minus{} 2aby)^2 \plus{} (y(a^2 \minus{} b^2) \plus{} 2abx)^2 \equal{} (a^2\plus{}b^2)^2(x^2\plus{}y^2)$.
Hence:
$ (a^2\plus{}b^2)(x^2\plus{}y^2)^2 \equal{} (a^2\plus{}b^2)^2(x^2\plus{}y^2) \iff$
$ a^2\plus{}b^2 \equal{} x^2\plus{}y^2 \iff$
$ x^2\minus{}a^2 \equal{} b^2\minus{}y^2$

$ a(x^2 \minus{} y^2) \minus{} 2bxy \equal{} x(a^2 \minus{} b^2) \minus{} 2aby \iff$
$ ax(x\minus{}a) \minus{} 2by(x\minus{}a) \minus{} ay^2 \plus{} xb^2 \equal{} 0$
$ xb^2 \minus{} ay^2 \equal{} (x\minus{}a)b^2 \plus{} a(b^2\minus{}y^2) \equal{} (x\minus{}a)b^2 \plus{} a(b^2\minus{}y^2) \equal{} (x\minus{}a)b^2 \plus{} a(x^2\minus{}a^2)$
$ \equal{} (x\minus{}a)(b^2 \plus{} ax\plus{}a^2)$
Hence:
$ (x\minus{}a)(ax\minus{}2by\plus{}b^2\plus{}ax\plus{}a^2) \equal{} 0 \iff$
$ (x\minus{}a)(2ax\minus{}2by\plus{}a^2\plus{}b^2) \equal{} 0$

In the same way:
$ b(x^2 \minus{} y^2) \plus{} 2axy \equal{} y(a^2 \minus{} b^2) \plus{} 2abx \iff$
$ (b\minus{}y) ( 2yb \minus{} 2ax \plus{} x^2\plus{}y^2) \equal{} 0$

Hence: If we have neither $ b\minus{}y\equal{}0$ nor $ x\minus{}a\equal{}0$ we must have:
$ 2yb \minus{} 2ax \plus{} x^2\plus{}y^2\equal{}0, 2ax\minus{}2by\plus{}a^2\plus{}b^2\equal{}0 \Rightarrow$
$ a^2\plus{}b^2\plus{}x^2\plus{}y^2\equal{}0$, which isn't possible.
So either $ x\equal{}a$ or $ b\equal{}y$.

Assume that $ x\equal{}a$ and $ b \neq y$. Then $ b^2\minus{}y^2 \equal{} a^2\minus{}x^2 \equal{} 0 \Rightarrow b \equal{} \minus{}y$. Hence: $ 2yb\minus{}2ax\plus{}x^2\plus{}y^2 \equal{} 0 \iff$
$ \minus{}2y^2\minus{}2x^2\plus{}x^2\plus{}y^2 \equal{} 0 \iff x^2\plus{}y^2 \equal{} 0$ which is impossible. So $ x\equal{}a \Rightarrow b\equal{}y$. In same way we find $ b\equal{}y \Rightarrow a\equal{}x$, and therefore $ a\equal{}x \iff b\equal{}y$.
Since at least one of them hold, we get $ a\equal{}x,b\equal{}y$.

$ [a(x^2\minus{}y^2)\minus{}2bxy]\plus{}i[b(x^2\minus{}y^2)\plus{}2axy]\equal{}[x(a^2\minus{}b^2)\minus{}2aby]\plus{}i[y(a^2\minus{}b^2)\plus{}2abx]$
so $ (a\plus{}bi)(x\plus{}iy)^2\equal{}(a\plus{}bi)^2(x\plus{}iy)$ so $ x\plus{}yi\equal{}a\plus{}bi$ so$ a\equal{}x ,b\equal{}y$

GR8 solutions....
There's another way ut - however mine is not as great as urs....to look at it like 2 hyperbolae...

[hide]$ (x^2\minus{}y^2)\minus{}\frac{2b}{a}xy\plus{}x\frac{b^2\minus{}a^2}{a}\plus{}2by\equal{}0$

& $ (x^2\minus{}y^2)\plus{}\frac{2a}{b}xy\minus{}y\frac{a^2\minus{}b^2}{b}\minus{}2ax\equal{}0$

1 of the Asymptotes turn out to be

For $ H_1$ :-  $ y\equal{}\frac{b}{a}x\minus{}b$ and for the 2nd hyperbola $ y\equal{}\frac{ax}{b}\plus{}\frac{(b^2\minus{}a^2)}{2b}$....Drawing the graphs and analysing a few (2 I think) inequalities, we get only 2 solutions viz $ (0,0)$ & $ (a,b)$.[/hide]

Thanx for the solutions!


Find all function $ f: R \mapsto R$ such that:
      $ f(x^{2}+yf(x))= (f(x))^{2}+xf(y)$     $ \forall x,y\in R$


linear algebra
matrix
IMO Shortlist
algebra
system of equations
arithmetic sequence
IMO Longlist

In what case does the system of equations

$\begin{matrix} x + y + mz = a \\ x + my + z = b \\ mx + y + z = c \end{matrix}$

have a solution? Find conditions under which the unique solution of the above system is an arithmetic progression.

[b]Please post your solutions.[/b] [url=http://www.mathlinks.ro/LaTeX/AoPS_L_About.php]Use $\LaTeX$ please![/url] Please omit jokes/smilies etc. Comments and generalizations are welcome. If you noticed that the comment has been discussed elsewhere, please provide a link. If you don't know the link(s) do NOT post and state that the problems has been discussed many times. If the provided solution is not complete, right or in $\LaTeX$-style I would be happy if you could (re-) post your/the solution in this thread again! At the end of your (re-)written solution post the links to those insufficient solutions as follows:

[hide="Remark"]
- first link, e.g. http://www.mathlinks.ro/Forum/viewtopic.php?p=346917#p346917
- second link
etc.
[/hide]

[b]Happy Problem-Solving![/b] :)

We consider the cases $m = -2$ and $m = 1$ first. If $m = 1$ then we obviously need $a = b = c$, and in this case, there are infinitely many solutions. If $m = -2$ then we need $a + b + c = 0$ (just add the inequalities to see this), and in this case there are infinitely many solutions as well (since we get two equations in three variables, and if $(x,y,z)$ is a solution, then so is $(x + k,y + k,z + k)$.) Now, assume that $m \neq -2$, $m \neq 1$.

Set $S = a + b + c$. Adding the given equations gives \[ x + y + z = \frac{S}{m + 2}. \] Hence \[ a = x + y + mz = \frac{S}{m + 2} + (m - 1)z \Leftrightarrow z = \frac{1}{m - 1} \cdot \left(a - \frac{S}{m + 2}\right). \] Similarly \[ y = \frac{1}{m - 1} \cdot \left(b - \frac{S}{m + 2}\right),\ x = \frac{1}{m - 1} \cdot \left(c - \frac{S}{m + 2}\right). \] Now, it becomes obvious that $x,y,z$ are in arithmetic progression if and only if $a,b,c$ are in arithmetic progression. So, that's it.


is this possible to partetion $ \mathbb{R}^2$ into two  $ L_5$ connected dense sets?

[quote="coquitao"]By the way,

What is an $ L_{5}$ set, esteemed [b]hamidreza[/b]?  :([/quote]

a subset $ C\subset R^n$ is L_{n} connected if each two distinct points can be joined by polygonal an having no more than

$ n$ segments lying in $ C$

for $ t$ a real positive put $ T_t\equal{}[it,t\minus{}it]\cup[t\minus{}it,\minus{}t\minus{}it]\cup[\minus{}t\minus{}it,it]$ and we define

$ A_1\equal{}\{$T_t | t is a rational and  $ t>0\}\minus{}\{[0,it]|$t is a rational and$ t>0 \}$

 and $ S_1\equal{}A_1\cup\{[0,it]| t\in R^\minus{}\}\cup(0,0)$

 $ A_2\equal{}\{T_t | t$is a irrational and$ t>0\}\minus{}\{[0,it]|t$ is a irrational and$ t<0\}$

and $ S_2\equal{}A_2\cup\{[0,it]| t\in R^\plus{} \}$

it is easily checked that $ S_1$ and $ S_2$ give desired partition. :wink:


This problem was on my math team tryout, more or less.

$ 4$ students are taking a test in a room with $ 15$ desks.
If all the desks are lined side by side and the teacher teacher cares about the order of the students and that at least one seat separates each student, how many ways are there to seat the $ 4$ students?

[hide="Solution"]Since atleast $ 1$ desk separates each of the $ 4$ students, take out $ 3$ seats, seat the students and then insert the $ 3$ desks back. There are $ \binom{12}{4}$ ways to seat the students. We also care about the order in which the students are seated, so we multiply by $ 4!$. Thus, the answer is $ 4!\binom{12}{4} \equal{} \boxed{11880}$ ways.[/hide]

But how many ways are there to take those 3 seats out?


[quote="chess64"]A given sequence $r_1, r_2, \ldots , r_n$ of distinct real numbers can be put in ascending order by means of one or more "bubble passes". A bubble pass through a given sequence consists of comparing the second term with the first term, and exchanging them if and only if the second term is smaller, then comparing the third term with the second term and exchanging them if and only if the third term is smaller, and so on in order, through comparing the last term, $r_n$, with its current predecessor and exchanging them if and only if the last term is smaller.

Suppose that $n=40$, and that the terms of the initial sequence are distinct from one another and are in random order. Let $p/q$, in lowest terms, be the probability that the number that begins as $r_{20}$ will end up, after one bubble pass, in the 30th place. Find $p+q$.[/quote][hide]
WLOG let the numbers be 1, 2, 3, 4, ... 40 in some order, since all that matters is that they are distinct

First of all, we can note that, given one complete bubble change on a string of numbers, the largest one will be at the end. Therefore if r20 ends up in the 30th place in the sequence r1, r2, r3, ..., r30, r31, then r20 is greater than all of r1-r19 and r-21-r30 but is less than r31. Thus r20 is greater than 29 of the numbers. and r31 is the greatest of the first 31 numbers. Since the probability that r31 is the greatest of the first 31 is 1/31, then the chance that r30 is the greatest of the first 30 is 1/30. So our answer is $031$[/hide]

This looks suspiciously like Kalva's solution... :huh:

Oh, btw, you forgot to multiply the probabilities to get [hide]1/930, so it's supposed to be $931$ not $031$ (maybe that was just a typo?)[/hide]


Hey guys,

Did you know that MOOD is DOOM spelled backwards?

Thanks.

WHOA THIS IS A PRETTY HELPFUL TOPIC

ON ANOTHER NOTE,

DOOM DOOM DOOM

And COOL is OLCO spelled from the middle.

WHY IS THERE A RANDOM LOWER CASE "K" IN THE TITLE OH IS IT BECAUSE THEY DON'T LET YOU HAVE ALL CAPS IN TITLES THANKS


geometry
geometric transformation
reflection

[quote="Valentin Vornicu"][quote="semy"]May the ML contest problems be freely shared?
I hope that I could use these problems for training my students and introduce them to Math Letter, a students' journal of math in Korea.[/quote]
You can use them as long as you specify the source of the problems, namely this site MathLinks.ro.

I will have to reflect about publishing the questions in a journal. So please do not do it for now.[/quote]

Okay.
If any decision is made, please inform me.
Only the Korean translated version would be introduced with your activities here in that students' journal.
The journal is similar to Canadian Crux with Mayhem, Hungarian K"omal, and Excalibur of Hong Kong or so, not a formal journal.


Are there any particles that carry "magnetic charge", and if so, why are they so difficult to detect?
 ;)  :rotfl:

One of the four equations (of electromagnetism) as developed by Maxwell states that no magnetic monopoles exist.  Apparently, it is given by this equation:

$\nabla \cdot \mathbf{B} = 0$

But some GUT's predict that they do.  So people have been looking about, and we haven't seen any yet

The way I read about monopoles (from an EM book) was that if we formulate our entire theory of EM and fields assuming that monopoles exist, there are no faults or contradictions and all the Maxwell equations become symmetric and 'beautiful' As Griffiths says, Maxwell's eqns 'ask' for monopoles! Since our mathematics becomes beautiful (very !) if monopoles exist, we hypothesise their existence! Actually, the Maxwell eqn $\vec{ \nabla} \cdot \vec{B} = 0$ becomes similar to Gauss's law for charges and all 4 Maxwell eqns become symmetric in $B$ and $E$ .... That's why monopoles exist! I Dont know where they 'officially come from. Though logically, it does seem my 'definition' is correct!


group theory
abstract algebra
superior algebra
superior algebra unsolved

[quote="rustam"]Let $ G$ be a finite group and $ P$ be a Sylow p-subgroup of $ G$. Let H be a subgroup of $ G$ such that $ N(P) \subseteq H$. Prove that $ N(H)=H$.[/quote]

[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=155364]Pretty Straightforward[/url]


It would be really cool for all the mathlinks users to have the posibility to talk to eah other when they are on-line.
PM is very annoing, it takes a lot of time.
I see somebody on-line , I want to ask him a question , PM is too slow for that.
Valentin?

well I have gave some thought about that, but a chat room would seriously dim the memory of the system, in result more php errors. at the moment not acceptable.

anyway, I'll make it a poll to check out most of the active users' opinions

Hi members of mathlinks,

why don't you just provide your messenger information as yahoo, msn, icq and so on. All people you want to contact can be included in your messenger program and others can add you to their programs.  :) So go and get your messenger program !!!   :D

I think it will not be any problem at all for Valentin to add a further field in the personal profile.

Anytime soon, I'll get ICQ, so maybe it will be our way of communicating on-line.

Or why don't you just use:

MSN messenger:

- http://messenger.msn.com/

or YAHOO messenger:

- http://messenger.yahoo.com/

So see you soon.  :)


Let $x;y;z;t \in N$ such that:$xyzt=1989^{1991}$
Prove that:$(x+y+z-t) \vdots 1992$

what do you mean by $\vdots$? if you mean $x+y+z-t$ [b]divides[/b]$1992$ then my reasoning is right i think.

It means that $x+y+z-t$ is divisible by $1992$.

[quote="Rust"]Obviosly it is wrong.[/quote]
Can you prove it is wrong?


What is the remainder when 3x.2^(n+3)-4x.2^(n+2)+5x.2^(n+1)-8 is divided by x+1(n is a non-negative integer)?
I applied the Remainder theorem by replacing x=-1 and I got the variable results depending on each specific values of n. But the book gives the answer: -20 :huh:

The book is wrong. I remember going over this problem for a really long time with my sister, considering we both knew how to do the problem. The answer should indeed be in terms of $ n$.


calculus
derivative
function
analytic geometry
graphing lines
slope
calculus computations

A mountain climber on one edge of a deep canyon 800 ft wide sees a large rock fall from the opposite edge at time=0. As he watches the rock plummet downward, his eyes first move slowly, then faster, and than more slowly again. Let a be the angle of depression of his line of sight below the horizontal. At what angle a would seem to be moving the most rapidly. That is, would da/dt be maximal? Remember the rocks acceleration would be -32 ft/sec^2

first I'm going to define the distance the rock falls in terms of time
$acceleration = 32ft/s^{2}$
integrated with respect to time:
$v=32f/s^{2}t$
again:
$d=16ft/s^{2}t^{2}$

the angle is given by $a=acrtan(\frac{16ft/s^{2}t^{2}}{800ft})$ or $a=acrtan(\frac{1ft/s^{2}t^{2}}{50ft})$
the change of angle with respect to time is:
$a'=\frac{1}{1+\frac{t^{4}}{2500}}\frac{t}{25}$ or $a'=\frac{25t}{25+\frac{t^{4}}{100}}$
the second derivitive, which we'll use to see where a' local maxs/mins are is:
$a''=-\frac{2500(3t^{4}-2500)}{(t^{4}+2500)^{2}}$
the denomonater is defined for all of t, so looking at the numerator we see that a''=0 when:
$2500*2500=7500t^{4}$ or $t=(\frac{2500}{3})^{1/4}$ or roughly 5.37
we could take another derivitive to see if its turning up (minimum) or turning down (maximum) but I think 1 quotent rule is enough for now, so I gonna see which is greater, a' at $(\frac{2500}{3})^{1/4}$ or a' at $(\frac{2500}{3})^{1/4}+0.01$

well actually I'm not, I dont have my calculator with me, i've been doing this on my computer calculator so I will look at the function a' (and a'').  As I look at the function a', I can see that even though t^4 is divided by 100 it will out grow t for t is very large even though t is multiplied by 25, meaning for all of t is very large a' is decreasing b/c the denomonator is growing faster than the denominator.  Now, looking at a'', there is only 1 spot where that function crosses the t axis (horizontal, equals zero) therefore because a' is decreasing for t is very large and there is only 1 change in direction of slope $t=(\frac{2500}{3})^{1/4}$  must be a maximum, which is the answer.