Σκαρωθηκε για test

Ν.δ.ο.αν η εξίσωση $ ax^3\plus{}3b^2x\plus{}c\equal{}3bx^2\plus{}x$ όπου $ a,b,c$ ακέραιοι, εχει τρεις ακέραιες ρίζες και άνισες αυτές θα είναι διαδοχικές.

Όμορφη !
Έστω $ x,y,z$ οι ρίζες του πολυωνύμου. Τότε από τους τύπους του Vieta αφού οι αριθμοί $ x \plus{} y \plus{} z,xy \plus{} yz \plus{} zx$ είναι ακέραιο)ι, θα πρέπει $ a|3b$ και $ a|3b^2 \minus{} 1$. 'Αρα $ a|3b\cdot b \minus{} 3b^2 \plus{} 1$ άρα $ a|1$ άρα $ a \equal{} 1$ ή $ a \equal{} \minus{} 1$ . Θα θεωρήσουμε την περίπτωση $ a \equal{} 1$ (η άλλη είναι παρόμοια). Τότε είναι $ x \plus{} y \plus{} z \equal{} 3b$ και $ xy \plus{} yz \plus{} xz \equal{} 3b^2 \minus{} 1$. Υποθέτουμε λόγω συμμετρίας ότι ο $ x\geq y\geq z$
Παρατηρούμε ότι $ (x \plus{} y \plus{} z)^2 \equal{} 3(xy \plus{} yz \plus{} zx) \plus{} 3$. 'Ομως η τελευταία είναι ισοδύναμη με
$ (x \minus{} y)^2 \plus{} (y \minus{} z)^2 \plus{} (x \minus{} z)^2 \equal{} 6$. Όμως αφού οι αριθμοί είναι διαφορετικοί, θα ισχύει $ x \minus{} y\geq 1$ , $ x \minus{} z\geq 2$ και $ y \minus{} z\geq 1$
Άρα $ (x \minus{} y)^2 \plus{} (y \minus{} z)^2 \plus{} (x \minus{} z)^2\geq 6$. Επομένως παντού παραπάνω ισχύει η ισότητα. Άρα $ x \minus{} y \equal{} 1$ , $ y \minus{} z \equal{} 1$ , $ x \minus{} z \equal{} 2$
και συνεπώς οι ρίζες είναι της μορφής $ (x,x \minus{} 1,x \minus{} 2)$ που είναι και το ζητούμενο.

Σαν τα χιόνια ....(σου είναι φανταζομαι γνωστή η τοπικη εκφραση)
Sil kαλο διάβασμα....

Είναι $ k(x\plus{}p)(x\plus{}q)(x\plus{}r)   \equiv ax^{3}\plus{}3b^{2}x\plus{}c\minus{}(3bx^{2}\plus{}x)$

όπως είπε ο Σιλουανός πάνω είναι $ a|1$ άρα $ a\equal{}k \in$ {-1,1}

για $ a\equal{}k\equal{}1$ έχουμε

$ x^3\plus{}x^2(p\plus{}q\plus{}r)\plus{}x(pq\plus{}qr\plus{}rp)\plus{}pqr \equiv  ax^{3}\plus{}3b^{2}x\plus{}c\minus{}(3bx^{2}\plus{}x)$

άρα $ p\plus{}q\plus{}r\equal{}\minus{}3b$ , $ pq\plus{}qr\plus{}rp\equal{}3b^2\minus{}1$, $ pqr\equal{}c$ (η τελευταία δν χρειάζεται στη λύση)

λύνοντας τώρα την 1η εξίσωση ως προς $ r$ και αντικαθιστώντας στη 2η είναι

$ pq\plus{}(p\plus{}q)(\minus{}p\minus{}q\minus{}3b)\equal{}3b^2\minus{}1 \Leftrightarrow pq\minus{}(p\plus{}q)^2\minus{}3b(p\plus{}q)\equal{}3b^2\minus{}1 \Leftrightarrow$

$ \minus{}p^2\minus{}q^2\minus{}pq\minus{}3bp\minus{}3bq\minus{}3b^2\plus{}1\equal{}0 \Leftrightarrow p^2\plus{}p(q\plus{}3b)\plus{}(q^2\plus{}3bq\plus{}3b^2\minus{}1)\equal{}0$

θεωρώντας τριώνυμο ως προς $ p$ είναι

$ D \equal{} q^2\plus{}6bq\plus{}9b^2\minus{}4q^2\minus{}12bq\minus{}12b^2\plus{}4 \equal{} \minus{}3q^2\minus{}6bq\minus{}3b^2\plus{}4 \equal{} \minus{}3(q\plus{}b)^2\plus{}4$

αν υπάρχουν τρεις ρίζες της αρχικής (δηλ. όλες) και αφού p=-ρίζα της αρχικής,
είναι $ D>\equal{}0 \Leftrightarrow 3(q\minus{}b)^2<\equal{}4 \Rightarrow \minus{}1<\equal{}q\minus{}b<\equal{}1$ αφού έχουμε ακεραίους

άρα $ q \in A\equal{}$ {b-1,b,b+1} αφού είναι είναι ακέραιος

ομοίως, $ p,q,r \in A$ άρα, αφού είναι τα αντίθετα των ριζών, θα είναι

άρα οι ρίζες της αρχικής είναι $ x_1,x_2,x_3 \in$ {-(b-1),-b,-(b+1)} και αφου είναι διαφορετικοί αριθμοί, θα είναι και διαδοχικοί,
αφού ανήκου σε αυτό το σύνολο

ομοίως για $ a\equal{}k\equal{}\minus{}1$

P.S. Έχει πολύ πλάκα που βγαίνει χωρίς εξεζητημένη γνώση...  :P


quadratics
algebra
Diophantine equation
Additive Number Theory

Find all integers $m>1$ such that $m^3$ is a sum of $m$ squares of consecutive integers.

Suppose that $ m^3$ is the sum of $ m$ squares $ (n\plus{}1)^2,\ (n\plus{}2)^2,\ \dots,\ (n\plus{}m)^2$. Then
\[ m^3 \equal{} \sum_{i\equal{}1}^m (n\plus{}i)^2 \equal{} \sum_{i\equal{}1}^{m\plus{}n} i^2 \minus{} \sum_{i\equal{}1}^n i^2 \equal{} \frac{(m\plus{}n)(m\plus{}n\plus{}1)(2m\plus{}2n\plus{}1)}{6} \minus{} \frac{n(n\plus{}1)(2n\plus{}1)}{6} \equal{} \frac{2m^3 \plus{} (6n \plus{} 3) m^2 \plus{} (6n^2 \plus{} 6n \plus{} 1) m}{6}.\]
Hence, we need to solve
\[ n^2 \plus{} (m\plus{}1)n \minus{} \frac{4m^2 \minus{} 3m \minus{} 1}{6} \equal{} 0\]
in positive integers. The discriminant of this quadratic equation w.r.t. $ n$ is $ \frac{11 m^2 \plus{} 1}{3}$ and it must be a square of some integer $ q$. Positive solutions to this generalized Pell equation $ 11 m^2 \plus{} 1\equal{}3q^2$ is given by the following recurrent sequence:
$ m_0 \equal{} 1,\ m_1 \equal{} 47,\ m_{k\plus{}1} \equal{} 46 m_k \minus{} m_{k\minus{}1}$
which can be alternatively described as
\[ m_k \equal{} \frac{11\plus{}2\sqrt{33}}{22}(23\plus{}4\sqrt{33})^k \plus{} \frac{11\minus{}2\sqrt{33}}{22}(23\minus{}4\sqrt{33})^k.\]

Corresponding values of $ n$ are given by
$ n_0 \equal{} 0,\ n_1 \equal{} 21,\ n_{k\plus{}1} \equal{} 46 n_k \minus{} n_{k\minus{}1} \plus{} 22$.


geometry
inequalities
triangle inequality

O is any point inside triangle ABC.prove that
  (AB+BC+AC)>(AO+BO+CO) :wink:

Here is a picture:

 [geogebra]dd37958872bfe94ac401f4ab1e3e1f7e3164fbe7[/geogebra]

If you just think about it, you will find that it is always true.

@powerofpi, that's not a proof...

See, geogebra doesn't work on my computer so just draw this out...

Draw any point inside triangle ABC. Then draw a line from the vertex that intersects the point "O". Then draw another  line from another vertex(let's say B), that intersects our original line AO. Notice that no matter what kind of triangle it is that AB is always the hypotenuse as long as it's INSIDE the triangle. AO is just a leg.(A hypotenuse is obviously longer than a leg because of triangle inequality) Now just apply this to the other sides and we're done. Notice how any segment(AO, BO, CO), is always shorter than a median, and a median is always shorter than an actual side of triangle ABC.

Ray $ BO$ cuts $ AC$ at $ X.$ By triangle inequality in $ \triangle ABX$ and $ \triangle COX,$ we get

$ OX\plus{}XC > OC \ , \ AB\plus{}AX > OB\plus{}OX.$

$ OX\plus{}XC\plus{}AB\plus{}AX > OB\plus{}OC\plus{}OX  \Longrightarrow AB\plus{}AC > OB\plus{} OC.$

Similarly we get $ BA\plus{}BC > OA\plus{}OC \ , \ CA\plus{}CB > OA\plus{}OB.$

Adding the three previous inequalities gives $AB\plus{}BC\plus{}CA > OA\plus{}OB\plus{}OC.$


Given:
$(x^2-3x-2)^2-3(x^2-3x-2)-2-x=0$,

Prove roots of x^2-4x-2=0 are also roots of given equation(not sure why that's even necessary)

find all roots to the given equation.

[quote="Misha123"]The solutions are:  $1\pm\sqrt{5}$ and $2\pm\sqrt{6}$. Am I right?

Misha[/quote]
ya....., proof?

Denote $x^2-4x-2$ by $T$ and you get: $(T+x)^2-3(T+x)-2-x=0$ and

$T(T+2x-2)=0$.

Now you solve both quadratic equations. Sorry if it's too short proof but I have no time at the moment :( .

Misha

ya.. that was the solution in the book...... but I don't think the first part was necessary at all

here's what I did:

Let $x^2-3x-2=K$, add $x^2-2x$ to both side of given equation we have

$K^2-2K=x^2-2x$
$K^2-2K-(x^2-2x)=0$

quadratic formula:

$K=\frac{2 \pm \sqrt{4-8x+4x^2}}{2}$

$=1 \pm (x-1)$
solve each quadratics yields the solutions

Denote $f(x) = x^2 - 3x -2$, then the equation is equivalent to $f(f(x)) = x$. Now, any fixed points of $f$ is a root of the equation. So we first find the two roots of $f(x) = x$. After knowing what the two roots are, we can then factor and reduce the ortiginal quartic into a quadratic.


[quote="236factorial"][quote="Spencer"]I have a question. When does your name like New Member, Pointcare Conjecture, PvsNP, ect. change? Is it after a few posts?[/quote]

so many times this has been posted...

anyway,

20 posts: P Vs. NP
50 posts: Hodge Conjecture
100 posts: Poincare Conjecture
250 posts: Riemann Hypothesis
500 posts: Yang-Mills theory
1000 posts: Navier Stokes equations
2500 posts: Birch and Swinnerton dyer

I've been through every one of them, recently, I've been promoted in rank as birch and swinnerton dyer.[/quote]

Sorry. I didnt want to waste time looking for it.

[quote="Spencer"]Yay! Im almost at Riemann Hypothesis.[/quote]

yes you are, just don't spam to get your ranking up.  ;)

[quote="236factorial"]I've been through every one of them, recently, I've been promoted in rank as birch and swinnerton dyer.[/quote]

Didnt you have only a 1000 posts..like 20 days ago?

yeah . thats right. 236 factorial just got like about 1000 posts in about a month. how come. i've been working for 2 mths and only got 500.  ;)  :(

I didnt get to 5 posts until like Febuary.

20 posts: P Vs. NP
50 posts: Hodge Conjecture
100 posts: Poincare Conjecture
250 posts: Riemann Hypothesis
500 posts: Yang-Mills theory
1000 posts: Navier Stokes equations
2500 posts: Birch and Swinnerton dyer
thats the order.

You have to have a member of the AoPS staff, i.e. employed by AoPS or ML somehow, to be an administrator.

[quote="MysticTerminator"]whoa, is mildorf an admin? that would be the coolness[/quote]

he's not even a moderator  ;)

[quote="MysticTerminator"]whoa, is mildorf an admin? that would be the coolness[/quote]

He's employed by AoPS? O.O

Anyhow, I didn't say that all people somehow working for AoPS would be admin. *mutters*

i am a Navier stokes equation.  How long would it take you to solve that?

Or you could develop your own math forum and community, and wait for a merge offer from AoPS-Mathlinks. Of course, it'd take some time to get a couple thousand members, and thats only if the admin here want to merge together.

[quote="paladin8"][quote="236factorial"][quote="h78reg"]just wondering, how do you become an administrator? :D moderators have already been discussed.[/quote]

To be an admin, you have to teach on the online classes  ;)

I've posted 1461 in the past month.[/quote]

 :o  :o  :o  :o  :o  :o  :o  :o  :o  :o  :o  :o  :o  :o  :o  :o  1461!!!!?!?!?!?!?!?!?!?!?!?!!?!?!?!?!?! That's almost 50 posts a DAY! How in the world do you do that? There's no way you can have 50 posts a day without spamming.[/quote]

NO WAY?!!?!??!!
Maybe I did spam, but I hope not much. Helps when you're 7 hours online daily.  ;)

[quote="paladin8"][quote="h78reg"]yea, 7 hours a day leaves you with 7 posts per hour, which is possible.[/quote]

That's still more than a post every 10 mintues CONTINUOUSLY for 7 hours a day. And in another thread 236factorial mentions that he posts 80-120 posts a day. Now if that's not ridiculous I don't know what is.[/quote]

Can you guys stop talking about me or I'll lock this :dry: because it is possible to post one every 3 minutes

whoops, I'm not a moderator  :rotfl:

Heh I was wondering about this too. Maybe we should get a sticky about it?

Please excuse my opinion on this matter, but it's difficult not to spam when you spend most of your time in the fun and games forum.  ;)

[quote="MithsApprentice"]This discussion is becoming ridiculous. Please get back on track or at least discuss something of merit. -.-

And I'd listen to some of paladin's advice about spam, 236factorial.[/quote]

..... the topic question has been answered.............

  In my opinion, the reason a few of our members have considerably-huge post counts is because they are solving problems below there problem solving capacity.  While this is all fine and dandy, solving problems below your capabilities does not strengthen one's mathematical skill, one should be focusing on problems within and slightly above there grasp.  I use to average 10-15 posts per day, until someone kindly pointed out that it was rather ridiculous for me to post in a forum I was obviously advanced for.  Another contributing factor: AoPs is not always the best means for students to asorb and learn new material.  Although AoPS is an excellent resource for students who are in need of other peers to discuss mathematical ideas, some students find AoPS is just another means of socialization, and utilize it for this purpose.  One again, this is coming from a first-hand view.

I would say that it is quite sad for some of our members to be spending time doing problems below their ability or posting mostly on G&F only, because 80-120 posts a day means They pretty much have nothing outside of AOPS in their life. I pity thee. (not just factorial236, but other so-called "spammers" ive seen in the fora)

Haha, imagine an IMOer posting about 200 times a day in the Middle Scool forums...

it seems most IMOers on AOPS aren't above Riemann Hypothesis. they probably post in olympiad section and college playground most of the time though. ;)

by the way, is your signature really true?

yes it is. try it on a calculator and you'll see. :)

[quote]and what is the point of this thread?[/quote]

well, initially, the point was to discuss the ranks in AOPS, though now the thread is seriously off-topic. ;)

[quote="G-UNIT"] In my opinion, the reason a few of our members have considerably-huge post counts is because they are solving problems below there problem solving capacity.  While this is all fine and dandy, solving problems below your capabilities does not strengthen one's mathematical skill, one should be focusing on problems within and slightly above there grasp.  I use to average 10-15 posts per day, until someone kindly pointed out that it was rather ridiculous for me to post in a forum I was obviously advanced for.  Another contributing factor: AoPs is not always the best means for students to asorb and learn new material. [/quote]

I don't think anyone has told me to move out yet. Everyone seems to not notice I'm any older than they are (because unlike some say, I'm not any smarter than the basics level  :roll: ). Maybe i should try reducing the amount of posts I post in the clasroom math level (because I'm obviously way over that, not to brag or anything), and see how it goes. I don't think forcing a post stop is necessary, but it might just happen if this thing happens that doesn't involve me at all.

So, I just won't go on classroom that often anymore. See a big drop in my post rate hopefully.

- Lock up the Never ending story (unless something is done to change it to make it a more worthwhile. I've been trying to post more than just 5 words but taht doesn't last long).

[quote]and what is the point of this thread?[/quote]

What is the point of posting that?

G-UNIT, it obviously depends on the person - not everyone appreciates the benefits of the forums and not everybody uses it wisely...  :roll:

[quote]and what is the point of this thread?[/quote]

[quote]What is the point of posting that?[/quote]

lock this thread please

Yeah, because a 7 month old thread is taking up WAY too much space on these forums.


Let $ ABCD$ is a quadrilateral circumscribed about the circle circle $ (O)$, let $ AB, BC, CD, DA$ tangen with circle $ (O)$ at $ M,N,P,Q$ respectively. Let $ AN$ meet $ (O)$ at point $ E$, let $ AP$ meet $ (O)$ at point $ F$. Prove that $ ME$, $ QF$, $ AC$ are concurrent.

[quote="darij grinberg"][quote="thanhnam2902"]Let $ ABCD$ is a quadrilateral inscribed in circle $ (O)$, with $ AB, BC, CD, DA$ tangen with circle $ (O)$ at $ M,N,P,Q$ respectively. Let $ AN$ meet $ (O)$ at point $ E$, let $ AP$ meet $ (O)$ at point $ F$. Prove that $ ME$, $ QF$, $ AC$ are concurrent.[/quote]

I think you want to say that the quadrilateral ABCD is circumscribed about the circle (O), not inscribed in the circle (O).
The problem itself has been solved at  http://www.mathlinks.ro/viewtopic.php?t=343 .

  darij[/quote]

Yes, thank you very much. I will repair it now.


geometry
invariant
circumcircle
geometric transformation
reflection
trigonometry
function

Let  $\triangle EFG$ be the orthic triangle of $\triangle ABC$, and let $H$ be the orthocenter of  triangle $\triangle ABC$. Let $A^\prime$ the midpoint of $[AH]$, $B^\prime$ the midpoint of $[BH]$ and $C^\prime$ the midpoint of $[CH]$. Let $A^{\prime\prime}$ the symetric of $A^\prime$ with regard to $(FG)$, $B^{\prime\prime}$ the symetric of $B^\prime$ with regard to $(GE)$ and $C^{\prime\prime}$ the symetric of $C^\prime$ with regard to $(EF)$.
Show that exists an indirect similarity, with $H$ as invariant point, that transforms $\triangle A^\prime B^\prime C^\prime$ in $\triangle A^{\prime\prime}B^{\prime\prime}C^{\prime\prime}$.

By orthic triangle I mean the triangle formed by bases of altitudes that is the triangle who's vertices are the triangle's altitude's feet.

By orthic triangle do you mean the triangle formed by bases of altitudes? :maybe:

Yeah, that's what he means. (The triangle who's vertices are the triangle's altitude's feet)

The 9-point circle of the triangle $\triangle ABC$ is the circumcircle of both the orthic triangle $\triangle EFG$ and the triangle $\triangle A'B'C'.$ Moreover, the triangles $\triangle A'B'C' \sim \triangle ABC$ are centrally similar with the similarity center H and $\triangle EFG$ is the circumcevian triangle of the triangle $\triangle A'B'C'$ with respect to the orthocenter H. Thus the problem is equivalent to a slightly simpler one:

Let $\triangle EFG$ be the circumcevian triangle of the triangle $\triangle ABC$ with respect to its orthocenter H and let $A', B', C'$ be the reflections of the vertices $A, B, C$ in the sides $FG, GE, EF.$ Then the triangles $\triangle A'B'C' \sim \triangle ABC$ are oppositely similar with the common orthocenter H.

The triangle $\triangle HEC$ is isosceles with $\angle CHF = \angle HCF = \angle A,$ hence $CF = CH.$ The triangle $\triangle C'CF$ is isosceles with $C'F = CF,$ hence $C'F = CF = CH = 2R |\cos C|.$ Similarly, $A'F = AF = AH = 2R |\cos A|,$ where $R$ is the circumradius of $\triangle ABC.$ From the circumcircle of $\triangle ABC,$ $\angle CFA = 180^\circ-\angle B,$ $\angle CFE = \angle CAE = 90^\circ-\angle C$ and $\angle AFG = \angle ACG = 90^\circ-\angle A,$ so that

$\angle C'FA' = |\angle CFA-2(\angle CFE+\angle AFG)| = |180^\circ-3\angle B|$

By the cosine theorem for the triangle $\triangle C'FA',$

$C'A'^{2}= C'F^{2}+A'F^{2}+2C'F \cdot A'F \cos (3 B) =$

$= 4R^{2}(\cos^{2}C+\cos^{2}A+8 \cos C \cos A \cos^{3}B-6 \cos C \cos A \cos B) =$

$= 4R^{2}(\cos^{2}C+\cos^{2}A+2 \cos C \cos A \cos B-8 \cos C \cos A \cos B \sin^{2}B) =$

$= 4R^{2}(\cos^{2}C+\cos^{2}A+2 \cos C \cos A \cos B-8 \cos C \cos A \cos B \sin^{2}B) =$

$= 4R^{2}\sin^{2}B (1-8 \cos C \cos A \cos B) = CA^{2}\eta^{2}$

and by cyclic exchange, $A'B'^{2}= AB^{2}\eta^{2},\ B'C'^{2}= BC^{2}\eta^{2},$ where $\eta = \pm \sqrt{1-8 \cos A \cos B \cos C}$ is the similarity coefficient of the triangles $\triangle A'B'C' \sim \triangle ABC.$ The following trigonometric identity was used:

$\sin^{2}B = \sin^{2}(C+A) =$

$= \sin^{2}C \cos^{2}A+\cos^{2}C \sin^{2}A+2 \sin C \sin A \cos C \cos A =$

$= \cos^{2}C+\cos^{2}A-2 \cos^{2}C \cos^{2}A+2 \sin C \sin A \cos C \cos A =$

$= \cos^{2}C+\cos^{2}A-2 \cos C \cos A \cos (C+A) =$

$= \cos^{2}C+\cos^{2}A+2 \cos C \cos A \cos B$

The base $C'C$ of the isosceles triangle $\triangle C'CF$ with $C'F = CF = 2R |\cos C|$ and the angle $\angle C'FC = |180^\circ-2 \angle C|,$ i.e., with the base angles $\angle C'CF = \angle CC'F = \angle C$ is equal to

$C'C = 2 CF \cos C = 4R \cos^{2}C$

$CC'$ is one side of the triangle $\triangle C'CH$ and $CH = 2R |\cos C|$ is another side of this triangle. As the line $CC'$ passes through the circumcenter $O$ of $\triangle ABC,$ by a simple angle chase, $\angle C'CH = |\angle A-\angle B|.$ Using the cosine theorem for the triangle $\triangle C'CH,$

$C'H^{2}= CC'^{2}+CH^{2}-2 CC' \cdot CH \cos (A-B) =$

$= 4R^{2}[4 \cos^{4}C+\cos^{2}C-4 \cos^{3}C \cos (A-B)] =$

$= 4R^{2}\cos^{2}C \{1-4 \cos C [\cos (A+B)+\cos (A-B)]\}=$

$= 4R^{2}\cos^{2}C (1-8 \cos C \cos A \cos B) = CH^{2}\eta^{2}$

and by cyclic exchange, $A'H^{2}= AH^{2}\eta^{2},\ B'H^{2}= BH^{2}\eta^{2}.$ It follows that the similar triangles $\triangle A'B'C' \sim \triangle ABC$ have the common orthocenter $H.$

Clearly, the similarity coefficient $\eta = 0$ iff the triangle $\triangle ABC$ is equilateral, $\angle A = \angle B = \angle C = 60^\circ$ (when the triangle $\triangle A'B'C'$ degenerates to the point $H$). Since $\eta$ is a continuous function of the angles $\angle A, \angle B, \angle C,$ the similarity $\mathcal S: \triangle ABC \rightarrow \triangle A'B'C'$ is a continuous function of these angles. A reflection in a line (parity change) is a non-continuous transformation, hence the similarity $\mathcal S$ either contains a reflection for all non-equilateral triangles or it does not contain it for all non-equilateral triangles. Therefore, it is sufficient to decide this for one arbitrary non-equilateral triangle $\triangle ABC,$ for example, an isosceles triangle with $\angle A \rightarrow 90^\circ,$ which trivially gives an oppositely similar triangle $\triangle A'B'C',$ i.e., the similarity $\mathcal S$ always contains a reflection in a line and always results in oppositely similar triangles.

[color=darkblue][b]Very nice metrical proof, Yetti ![/b] I'll try to simplifiy the calculus in the your proof (in the its first part).

The following relations (in a triangle $ABC$) $\boxed{\cos^{2}A+\cos^{2}B+\cos^{2}C=1-2\cos A\cos B\cos C}\ \ (1)$
and (from the trigonomety) $\cos 3x=\cos x\left(4\cos^{2}x-1\right)$ are well-known. Thus, $\boxed{\cos 3A=\cos A\left(1-4\sin^{2}A\right)}\ \ (2)\ .$
Therefore, $\left(\frac{B'C'}{2R}\right)^{2}=\left(\cos^{2}B+\cos^{2}C\right)+2\cos B\cos C\cos 3A=$ $\left(1-\cos^{2}A-2\cos A\cos B\cos C\right)+2\cos A\cos B\cos C\left(1-4\sin^{2}A\right)=$ $\sin^{2}A(1-8\cos A\cos B\cos C)=\left(\frac{OH\sin A}{R}\right)^{2}\Longrightarrow \boxed{\ \frac{B'C'}{BC}=\frac{OH}{R}\ }\ \ (3)$. Analogously is proved $\frac{HA'}{HA}=\frac{OH}{R}\ \ (4)$,  i.e. $\triangle A'B'C'\sim\triangle ABC\ .$

[b]Question.[/b] Try to prove the third and fourth relations syntetically.
[b]The Yetti's proof is great because for the time being uniquely and a syntetical proof is very difficulty and yet it doesn't exist ![/b]

[b]Remark.[/b] The lines $AA'$, $BB'$, $CC'$ are concurrently in the circumcenter of the triangle $ABC\ .$[/color]

[hide="only for Nazarian."][color=darkblue]The power $p(H)$ of the orthocenter $H$ of the acute triangle $ABC$  with respect to the its circumcircle $C(O,R)$ is $p(H)=-HA\cdot 2HD$, where $D\in BC$ and $AD\perp BC$. But $p(H)=OH^{2}-R^{2}$, $HA=2R\cos A$ and $HD=2R\cos B\cos C$. Therefore, $OH^{2}=R^{2}+p(H)=R^{2}-8R^{2}\cos A\cos B\cos C$, i.e. $\boxed{OH^{2}=R^{2}(1-8\cos A\cos B\cos C)}$. [b]Nazarian[/b], understand you why ... ?.

[b]Remarks.[/b] $\boxed{\mathrm{H\ldots\ldots G\ldots O}}$, i.e. $HO=3\cdot GO\ .$ The power of the centroid $G$ is $p(G)=-GA\cdot GA'$, where the point $A'$ is the second intersection of the $AG$ with the circumcircle of the triangle $ABC$. Thus, $p(G)=GO^{2}-R^{2}=-\frac{2}{3}\cdot AM\left(\frac{1}{3}\cdot AM+MA'\right)$, where the point $M$ is the middlepoint of the side $[BC]$. Therefore, $GO^{2}=R^{2}-\frac{2m^{2}_{a}}{9}-\frac{2}{3}\cdot MA\cdot MA'=$ $R^{2}-\frac{2m^{2}_{a}}{9}-\frac{2}{3}\cdot MB\cdot MC=$ $R^{2}-\frac{2m^{2}_{a}}{9}-\frac{a^{2}}{6}=$ $R^{2}-\frac{4m^{2}_{a}+3a^{2}}{18}=$ $R^{2}-\frac{a^{2}+b^{2}+c^{2}}{9}$ $\Longrightarrow$ $\boxed{GO^{2}=R^{2}-\frac{a^{2}+b^{2}+c^{2}}{9}}$ and $\boxed{p(G)=-\frac{a^{2}+b^{2}+c^{2}}{9}}$. From the relation $HO=3\cdot GO$ results $HO^{2}=9R^{2}-\left(a^{2}+b^{2}+c^{2}\right)$. But $\sum a^{2}=2R^{2}\sum 2\sin^{2}A=$ $2R^{2}\sum (1-\cos 2A)$ and $\sum \cos 2A=-1-4\prod\cos A$. Therefore, $HO^{2}=9R^{2}-2R^{2}\left[3-\left(-1-4\prod\cos A\right)\right]=$ $R^{2}(1-8\cos A\cos B\cos C)$ $\Longrightarrow$ $\boxed{HO^{2}=R^{2}(1-\cos A\cos B\cos C)}\ .$[/color][/hide]

Slightly different notation (D, E, F instead of E, F, G) of the simplified problem:

Let $\triangle DEF$ be the circumcevian triangle of the triangle $\triangle ABC$ with respect to its orthocenter H and let A', B', C' be reflections of the vertices A, B, C in the sides EF, FD, DE. Then the triangles $\triangle A'B'C' \sim \triangle ABC$ are oppositely similar with the common orthocenter H.

DA, EB, FC are bisectors of the angles $\angle D, \angle E, \angle F$ of the circumcevian triangle $\triangle DEF,$ hence H is its incenter and A, B, C are midpoints of the arcs EF, FD. DE opposite to D, E, F, respectively. A', B', C' are reflections of these midpoints in the sides EF, FD, DE, hence by definition, the circumcircle (O') of the triangle $\triangle A'B'C'$ is the Fuhrmann circle of the triangle $\triangle DEF.$ Let G, M be the orthocenter and Nagel point of $\triangle DEF.$ GM is a diameter of the Fuhrman circle (O'). Indeed, according to Honsberger's Episodes, let the D-altitude of $\triangle DEF$ meet its circumcircle (O) again at U and let $AX \parallel DU$ be the circumcircle diameter. The orthocenter G and the point A' are reflections of U, A in EF, therefore DUAX and GA'AX are both isosceles trapezoids with the same parallel bases and equal angles and DGA'X a parallelogram, $GA' \parallel DX.$ As AX is a diameter of (O), $DX \perp DA,$ hence $GA' \perp DA.$ Let $\triangle D''E''F''$ be a triangle having $\triangle DEF$ as its medial triangle. DED''F is a parallelogram, hence DD'', EF cut each other in half. AA', EF also cut each other in half, hence DD'', AA' cut each other in half, so that DAD''A' is also a parallelogram, $D''A' \parallel DA.$  Since DA is a bisector of $\angle D,$ D''A' is a bisector of $\angle D''.$ The bisectors of $\triangle D''E''F''$ meet at its incenter, identical with the Nagel point M of $\triangle DEF.$ Since $MA' \equiv D''A' \parallel DA \perp GA',$ $\angle GA'M = 90^\circ$ is right and A' lies on the circle with diameter GM. Similarly, B', C' lie on this circle. Let N be the 9-point circle center of $\triangle DEF.$ According to the  [url=http://www.mathlinks.ro/Forum/topic-14728.html]easy problem (nine point center N as midpoint of IF)[/url], N is the midpoint of HO' (the incenter and Fuhrman circle center of $\triangle DEF$). Since N is also the midpoint of GO (the orthocenter and circumcenter), OHGO' is a parallelogram and OH = O'G. Since $G \in (O')$, OH is equal to the radius of Fuhrmann circle (O').

Earlier, we showed that $GA' \perp DA.$ Since $DA \perp BC$ is the A-altitude of $\triangle ABC,$ we get $GA' \parallel BC$ and similarly, $GB' \parallel CA,\ GC' \parallel AB.$ It follows that $\angle C'A'B' = \angle C'GB' = \angle CAB$ and similarly, $\angle A'B'C' = \angle ABC,$ $\angle B'C'A' = \angle BCA,$ so that the triangles $\triangle A'B'C' \sim \triangle ABC$ are similar, having equal angles. Since $\triangle ABC$ is inscribed in (O) with circumradius R and $\triangle A'B'C'$ in (O') with circumradius OH, their similarity coefficient is $\eta = \frac{OH}{R}.$ Let the altitudes DG, EG, CG of $\triangle DEF$ meet (O') at D', E', F'. As $G \in (O'),$ by a trivial angle chase, the triangles $\triangle D'E'F' \sim \triangle DEF$ are similar, having equal angles. Since $GA' \parallel BC$ and $F'G \equiv FG \parallel CO \equiv CC'$ (both perpendicular to DE),

$\angle A'GF' = \angle BCO = 90^\circ-\angle A = \frac{\angle D}{2}= \frac{\angle E'GF'}{2}.$

Thus GA' bisects the angle $\angle E'GF' \equiv \angle EGF = \angle D,$ so that A' is the midpoint of the arc E'F' opposite to D', which means that D'A' bisects the angle $\angle D' \equiv \angle F'D'E'.$ Similarly, E'B', F'C' bisect the angles $\angle E' \equiv \angle D'E'F',\ \angle F' \equiv \angle E'F'D'.$ The angle bisectors D'A', E'B', F'C' of $\triangle D'E'F'$ meeting at its incenter are identical with the altitudes A'D', B'E', C'F' of its circumcevian triangle $\triangle A'B'C'$ with respect to this incenter, identical with the orthocenter of this circumcevian triangle. (By a simple angle chase, we can also show $\angle A'B'D' = \angle ABD,$ so that A'D' is the A'-altitude of $\triangle A'B'C'$ and similarly, B'E', C'F' are its B'-, C'-altitudes.) As GM is a diameter of (O'), $D'G \perp D'M.$ From $DD' \equiv D'G \equiv DG \perp EF \parallel E''F''$ we then get $D'M \parallel E''F''.$ M is the incenter of $\triangle D''E''F''$ and $D \in E''F'',$ hence DD' is equal to the inradius of this triangle, i.e., twice the inradius r of its medial triangle $\triangle DEF,$ DD' = 2r (and similarly, EE' = FF' = 2r). Triangle $\triangle AEH$ with $\angle AEH = \angle AHE = \angle C$ is isosceles and AH = AE. AEA'F is a rhombus and $\angle EAF = 180^\circ-\angle D.$ Let K be the intersection of the rhombus diagonals $AA' \perp EF,$ yielding $\angle AEK = \frac{\angle D}{2}.$ Let T be the tangency point of the incircle (H) of $\triangle DEF$ with DE, HT = r. The right angle triangles $\triangle AEK \sim \triangle HDT$ are similar, having equal angles $\angle AEK = \angle HDT = \frac{\angle D}{2},$ therefore

$\frac{AA'}{AH}= \frac{2 AK}{AE}= \frac{2 HT}{HD}= \frac{DD'}{DH}$

As $DD' \equiv DG \parallel AO \equiv AA',$ the triangles $\triangle AA'H \sim \triangle DD'H$ are centrally similar with similarity center H, hence $H \in A'D'.$ Similarly, $H \in B'E',\ H \in C'F',$ so that the altitudes of $\triangle A'B'C'$ concur at H, which is therefore its orthocenter.

I say it was far easier, thus preferable, with nothing else but the cosine theorem. For example, see [url=http://www.mathlinks.ro/Forum/viewtopic.php?t=42587]Orthocentre of Fuhrmann triangle[/url] in Geometry Unsolved Problems, with no solution posted.

Dear Mathlinkers,
you can see for example
http://perso.orange.fr/jl.ayme/   vol. 5  Le P-cercle de Hagge p. 11.
Sincerely
Jean-Louis

[quote="yetti"]
Let $ \triangle DEF$ be the circumcevian triangle of the triangle $ \triangle ABC$ with respect to its orthocenter H and let A', B', C' be reflections of the vertices A, B, C in the sides EF, FD, DE. Then the triangles $ \triangle A'B'C' \sim \triangle ABC$ are oppositely similar with the common orthocenter H.
[/quote]
We can use $ (x,y)$ coordinates to prove that $ A',B',C',H$ are concyclic and there will be just a few computations. Then
we use the following lemma to prove the part 'with the common orthocenter $ H$'  :
if $ ABCD$ is a quadrilateral and $ P$ is a point inside $ ABCD$ such that $ \angle PAD \equal{} \angle PBC$ and $ \angle PDA \equal{} \angle PCB$ then :

1- $ P$ exists .
2-if $ K$ is the intersection of the perpendicular bisectors of $ AB,CD$ then $ \angle PAD \equal{} \frac {1}{2} \angle CKD$ .

EDIT :
this method also works for the generalization :
[quote="treegoner"]Let $ ABC$ be the triangle and $ P$ be a point. Let $ A_1B_1C_1$ be the circumcevian triangle of $ P$ wrt $ \Delta ABC$. Call $ A',B',C'$ the reflections of $ A_1,B_1,C_1$ across $ BC,CA,AB$ respectively. Then $ A'B'C'$ and $ A_1B_1C_1$ are similar. Furthermore, the barycentric coordinates of $ P$ wrt $ A'B'C'$ is the same of $ P$ wrt $ A_1B_1C_1$. [/quote]


Fie $ a,b,c\in [0;1]$.Atunci are loc inegalitatea:$ \displaystyle\sum_{cyc}\frac{1}{5\minus{}ab}\geq \frac{\sqrt{a}\plus{}\sqrt{b}\plus{}\sqrt{c}}{4}$

Pt inceput avem $ \sum_{cyc} \frac 1{5 \minus{} ab}\ge\frac {(1 \plus{} 1 \plus{} 1)^2}{15 \minus{} (ab \plus{} bc \plus{} ca)} \equal{} \frac 9{15 \minus{} (ab \plus{} bc \plus{} ac)}$

Ramane de demonstrat $ \frac 9{15 \minus{} (ab \plus{} bc \plus{} ac)}\ge\frac {\sqrt {a} \plus{} \sqrt {b} \plus{} \sqrt {c}}4\Leftrightarrow$
$ 36\ge[15 \minus{} (ab \plus{} bc \plus{} ac)](\sqrt {a} \plus{} \sqrt {b} \plus{} \sqrt {c})$

Mai trebuie gasit maximul produsului.

Partea cu [color=blue]Inegalitatea Cauchy-Schwartz[/color] e gresita, deoarece:

Cum \[ a,b,c\in[0;1]\Rightarrow \sqrt{a}\plus{}\sqrt{b}\plus{}\sqrt{c}\geq a\plus{}b\plus{}c\geq a^2\plus{}b^2\plus{}c^2\geq ab\plus{}bc\plus{}ca.\]

Ahh..corect am gresit in cauchy:D inapoi la lucru.Poti sa postezi rezolvarea intre timp?

$ (a\minus{}1)(b\minus{}1) \ge 0 \Rightarrow ab\plus{}1 \ge a\plus{}b \Rightarrow 1\minus{}a\minus{}b \ge \minus{}ab \Rightarrow 6\minus{}a\minus{}b \ge 5\minus{}ab \Rightarrow$

$ \Rightarrow  \frac {1} {6\minus{}a\minus{}b} \le \frac{1} {5\minus{}ab}$

$ \Rightarrow \displaystyle\sum\frac{1}{5\minus{}ab}\ge \displaystyle\sum\frac{1}{6\minus{}a\minus{}b}\ge \frac{9}{18\minus{}2(a\plus{}b\plus{}c)}$

$ \frac {\sqrt{a}\plus{}\sqrt{b}\plus{}\sqrt{c}}{4}\le\frac{\frac{a\plus{}1}{2}\plus{}\frac{b\plus{}1}{2}\plus{}\frac{c\plus{}1}{2}}{4}\equal{}\frac{a\plus{}b\plus{}c\plus{}3}{8}$

Este destul sa aratam ca daca $ a\plus{}b\plus{}c\equal{}s$ atunci:

$ \frac{s\plus{}3}{8}\le\frac{9}{18\minus{}2s} \Leftrightarrow 0 \le (s\minus{}3)^2$

De fapt inegalitatea este ceva mai generala: [i]Fie $ x_{1},x_{2}\dots x_{n}\in [0;1].$ Atunci are loc inegalitatea:

$ \sum^{n}_{i \equal{} 1} \frac {1}{2k \plus{} 1 \minus{} x_i x_{i \plus{} 1}} \ge \frac {\sum^{n}_{i \equal{} 1} \sqrt [k]{x_i}}{2k}$, unde $ k\geq\frac{1}{2}.$[/i]

Pentru demonstratie:  :arrow: [url=http://www.mathlinks.ro/viewtopic.php?t=222610/]vezi aici[/url]


geometry
projective geometry
geometry proposed

Let $ ABCD$ be a quadrilateral, and $ E$ be intersection points of  $ AB,CD$ and $ AD,BC$ respectively. External bisectors of $ DAB$ and $ DCB$ intersect at $ P$, external bisectors of $ ABC$ and $ ADC$ intersect at $ Q$ and external bisectors of $ AED$ and $ AFB$ intersect at $ R$. Prove that $ P,Q,R$ are collinear.

Should be one of the results in http://www.mathlinks.ro/viewtopic.php?t=30921 .

  darij

[quote=sumita]$ AD \cap BC \equal{} F,AB \cap DC \equal{} E$
external bisector $ DAB \equal{} L_{1}$
external bisector $ ABC \equal{} L_{2}$
external bisector $ BCD \equal{} L_{3}$
external bisector $ CDA \equal{} L_{4}$
external bisector $ AFB \equal{} L_{5}$
external bisector $ AED \equal{} L_{6}$
$ L_{3} \cap L_{4} \equal{} A_{1}$
$ L_{3} \cap L_{5} \equal{} C_{1}$
$ L_{5} \cap L_{4} \equal{} B_{1}$
$ L_{1} \cap L_{2} \equal{} A_{2}$
$ L_{2} \cap L_{6} \equal{} B_{2}$
$ L_{1} \cap L_{6} \equal{} C_{2}$
but:
$ A_{1}B_{1} \cap A_{2}B_{2} \equal{} Q$
$ A_{1}C_{1} \cap A_{2}C_{2} \equal{} P$
 $ C_{1}B_{1} \cap C_{2}B_{2} \equal{} R$
but$ A_{1}A_{2},B_{1}B_{2},C_{1}C_{2}$ go through incircle of $ DFC$ now from desargue theorem we coclude $ Q,P,R$ are collinear.[/quote]

go through excircle od $ DFC $


what is the mechnaism of $ \ce{R \sbond CH2 \sbond NH2} \plus{} \ce{HNO2 \minus{} > ROH}$

Atually it is R-CH2-OH

1.O=N-OH + H+ --> O=N-OH2+ -- >O=N+ + H2O
2.R-CH2-NH2 + +N=O --> R-CH2-NH2+-N=O (Lewis acid base pair reaction)
3.R-CH2-NH2+-N=O -->R-CH2-NH-N=OH+ (Transfer of proton) <-> R-CH2-NH+=N-OH
4.R-CH2-NH+=N-OH -->R-CH2-N=N-OH2+ (Transfer of proton) --> R-CH2-N=N+ +H2O
5.R-CH2-N=N+ -->R-CH2+ + N2
6.R-CH2+ + H2O -->R-CH2-OH

This topic has been moved out of the Organic Chemistry Metathread.  Perhaps the title is not clear enough, but the metathreads are only for links to other threads.  Please post a link to this topic in the metathread; I would do it but am not sure what description is best.

While one can write the mechanism above for that reaction, keep in mind that there are other products formed, and because of that the reaction mentioned have no utility in synthesis.

whats the mech of this reaction n-propylamine +hno2-> cyclopropane

Again, that reaction is ridiculous, because there are several other products formed. However, the mechanism begins as usual with diazotization, followed by nitrogen lost and formation of a carbocation: this carbocation can then cyclize to cyclopropane with hydride lost to another carbocation, or it can eliminate H+ to form propene, which than can isomerize to cyclopropane.


modular arithmetic
number theory solved
number theory

Prove that there exist infinitely many integers $n$ such that $n^{2}+1$ divides $n!$.

Oh,  this is very easy. Let n=2m^2 then
n^2+1=4m^4+1=(2m^2-2m+1)(2m^2+2m+1). If m\equiv 1 \pmod 5
then 2m^2+2m+1\equiv 0 \pmod 5. Since 2m^2-2m+1 and 2m^2+2m+1
are coprime, and 2m^2-2m+2, (2m^2+2m+1)/5 <2m^2=n we can found that n! devided by n^2+1.

In this way you can prove that for each a>0 there exist infinitely many
n such that n^2+1 devides [an]! (where [] -- integer part)

We can consider the negatice Pell ecuation: x^5-5y^2=-1 which has (x,y)=(2,1) as minimal solution, so it is solvable and it has an infinity of solutions (x,y) in N.
Now we consider those solutions for which y>5 so we get that 5<y<2y<=x and 4y^2<=5y^2-1=x^2. So 2(x^2+1)=5y*2y divides x!.

cheers! :D :D


Which of the following are true?

I. $\sin{x}\le{x}$ for $0\le{x}\le{\frac{\pi}{2}}$
II. $\sin{x}\ge{\frac{2}{\pi}x}$ for $0\le{x}\le{\frac{\pi}{2}}$
III. $\sin{x}\ge{\frac{2\sqrt2}{\pi}x}$ for $0\le{x}\le{\frac{\pi}{4}}$
IV. $\sin{x}\ge{\frac{3}{\pi}x}$ for $0\le{x}\le{\frac{\pi}{6}}$
V. $\sin{x}\ge{\frac{3\sqrt3}{2\pi}x}$ for $0\le{x}\le{\frac{\pi}{3}}$

I tried some values for each and got lucky for the answer, but I know there's a systematic way to do this.

[hide]For the first one, draw a sector of central angle $x$ and radius 1. $\sin x$ will be the length of the straight-line distance from the top of the sector (where it turns from a segment into an arc) to the $x$-axis, while $x$ is the length of the arc. Clearly $\sin x \leq x$ because the straight-line distance is shortest.

For the other four, note that the sine curve is concave down on the interval $0\leq x\leq \frac{\pi}{2}$ (and, in fact, $0\leq x\leq \pi$). As long as any curve is concave down between two values of $x$ (say, $a$ and $b$), the curve between $x=a$ and $x=b$ is always higher than the line connecting the two points on the curve at $x=a$ and $x=b$.

This means that if the line $y=mx$ intersects the graph of $y=\sin x$ at $x=c$ where $0<c\leq \frac{\pi}{2}$, then as long as $0\leq x\leq c$, $\sin x\geq mx$. This will work for ALL of the other four choices ($c$ is given in each case).

It follows that all of the choices are true.[/hide]


In a triangle  $ ABC$, prove that

$ \cos A \plus{} \cos B \plus{} \cos C \equal{} 1\plus{} 4\sin \frac{A}{2} \sin \frac{B}{2} \sin \frac{C}{2}$.

[hide="my solution"]$ \cos A \plus{} \cos B \plus{} \cos C$

$ \equal{} 2cos\frac {A \plus{} B}{2}.cos\frac {A \minus{} B}{2} \plus{} 1 \minus{} 2sin^2\frac {C}{2}$

$ \equal{} 1 \plus{} 2sin\frac {C}{2}(cos\frac {A \minus{} B}{2} \minus{} sin\frac {C}{2})$

$ \equal{} 1 \plus{} 2sin\frac {C}{2}(cos\frac {A \minus{} B}{2} \minus{} cos\frac {A \plus{} B}{2})$

$ \equal{} 1 \plus{} 4sin\frac {A}{2}sin\frac {B}{2}sin\frac {C}{2}$[/hide]

To generalize $ \cos a \plus{}\cos b \plus{} \cos c\plus{}\cos(a\plus{}b\plus{}c)\equal{}4\cos \frac{a\plus{}b}{2}\cos \frac{b\plus{}c}{2}\cos \frac{a\plus{}c}{2}$

$ (\cos a\plus{}\cos b)\plus{}(\cos c\plus{}\cos(a\plus{}b\plus{}c))\equal{}2\cos \frac{a\plus{}b}{2}\cos \frac{a\minus{}b}{2}\plus{}2\cos  \frac {a\plus{}b\plus{}2c}{2}$
$ \cos \frac{a\plus{}b}{2}\equal{}2\cos \frac{a\plus{}b}{2}(\cos \frac{a\minus{}b}{2}\plus{}\cos \frac{a\plus{}b\plus{}2c}{2})\equal{}4\cos \frac{a\plus{}b}{2} \cos \frac{a\plus{}c}{2} \cos \frac{b\plus{}c}{2}$
q.e.d  $ a\plus{}b\plus{}c\equal{}\pi$ $ a\equal{}A,b\equal{}B,c\equal{}C$.Hence result
 :P


Find the number of quadratic polynomials [b]ax² + bx + c [/b]such that:

a) a, b, c are distinct.

b) a, b, c ε {1, 2, 3, …2008}

c) x + 1 divides ax² + bx + c

[b]a) 2013018            b) 2013021            c) 2014024             d) 2018040       e) none of these[/b]

[hide="Solution"]Since $ x\plus{}1$ divides $ ax^2\plus{}bx\plus{}c$, we have $ a\minus{}b\plus{}c\equal{}0\implies a\plus{}c\equal{}b$ by the remainder theorem.

Doing casework for values of $ b\equal{}1, 2, \cdots 2008$, there are $ \frac{2007\cdot 2008}{2}\equal{}2015028$ pairs $ (a,c)$ satisfying given conditions.

However, we must subtract the number of pairs such that $ a\equal{}c$ (There cannot be occasions where $ a\equal{}b$ or $ b\equal{}c$, since then we will have $ c\equal{}0$ and $ a\equal{}0$ respectively).

There are 1004 pairs such that $ a\equal{}c$, since if $ a\equal{}c>1004$, the value of $ b$ will be more than 2008.

Hence, the number of quadratic polynomial satisfying above conditions is $ 2015028\minus{}1004\equal{}2014024$.

Answer: $ (c) 2014024$[/hide]


What is the work done to uniformly charge a sphere with a radius R to a charge Q? Consider 2 cases, conductor sphere and insulator sphere. assume the sphere is solid.

dude, although it doesn't matter for conducting sphere,but are the spheres solid or not???

For the conductive sphere, the work should be $ W = \frac{1}{2}CV^{2}$.

So, say we have a conducting solid sphere of radius $ R$. If we put some charge on let's say $ Q$ , then it will uniformly distribute itself over the surface of the sphere. The energy of this electrostatic configuration can be easily calculated by the energy of the field they create:
$ E_{cond}=\oint_{all space}{\frac{\epsilon_{0}E^{2}}{2}dV}=\int_{R}^\infty \frac{1}{2}\epsilon_{0}(\frac{Q}{4 \pi \epsilon_{0}r^{2}})^{2}4\pi r^{2}dr= \int_{R}^\infty \frac{Q^{2}}{8 \pi \epsilon_{0}r^{2}}dr =$
$ =\frac{Q^{2}}{8 \pi \epsilon_{0}R}$
For an insulating solid sphere I'll assume that we distribute the charge uniformly over the volume. Then the energy is equal
$ E_{ins}=\oint_{all space}{\frac{\epsilon_{0}E^{2}}{2}dV}= \int_{0}^{R}\frac{1}{2}\epsilon_{0}(\frac{Qr}{4\pi \epsilon_{0}R^{3}})^{2}4\pi r^{2}dr+\int_{R}^\infty \frac{1}{2}\epsilon_{0}(\frac{Q}{4 \pi \epsilon_{0}r^{2}})^{2}4\pi r^{2}dr =$
$ = \frac{Q^{2}}{40\pi \epsilon_{0}R}+\frac{Q^{2}}{8 \pi \epsilon_{0}R}= \frac{3Q^{2}}{20\pi \epsilon_{0}R}$

Don't forget that, as soon as there is an infinitesimal charge $ dq$ on the sphere, it will repell other charges that we want to tranfer to the sphere.


OK, everyone and the WOOT page (under My Classes) says that I will be able to drop WOOT only before the 3rd class begins.

However, I distinctly remember an announcement posted here that said I can drop the class before some day in October. Yet, the 3rd class starts September 28th.

Can someone clear up my confusion? I was initially planning on taking the Polynomials B class and seeing how tough that was before I made my final decision as to whether this class is too hard for me.

The third class refers to Polynomials C. It does not count the double offerings. So you can drop it before or on Oct. 2 which is the day after the Thursday class of Polynomials B.


[quote="kunny"]$ \cos 4x \equal{} \cos 2(2x) \equal{} ....$[/quote]

That's not very helpful ...

Just use $ \cos 2\theta \equal{}2\cos ^ 2 \theta \minus{}1$ :wink:
$ \cos 4x \equal{} \cos 2(2x) \equal{} 2\cos ^2 (2x) \minus{} 1 \equal{} 2(2\cos ^ 2 x \minus{} 1)^2\minus{}1$


Yes,kind of.I skipped some details which were obvious. ;)

solution very good from both of you  ;)  ;)  :D


Prove that a plane can not be covered by the interior parts of a finite number of parabolas.

http://www.artofproblemsolving.com/Forum/viewtopic.php?search_id=1260093072&t=139345
Edit:  See also http://www.artofproblemsolving.com/Forum/viewtopic.php?search_id=758949085&t=140462 .


when a 1000g of 1m aqueous solution of glucose is cooled down and maintained at a temp. -3.534 degree celcius how much ice is seperated out if $ K_{f}$ for water is 1.86 K Kg $ mol^{-1}$

The freezing point of an 1m glucose solution is -0.186ºC, so at that temperature (-3.534ºC) all the solution is expected to be solid. What am I missing?

no that's what i got but according tto the book(this seems logical now) that  whenit starts freezing the molality increases hence the freezing point of the solution increases


inequalities
trigonometry
function
inequalities proposed

Prove that in an [b]acute[/b] triangle $ ABC$ there is the relation $ 2\cdot\sum \cos\frac {B\plus{}C}{4}\ge \sum\cos\frac A2\plus{}\frac {3\sqrt 3}{2}$ .

Can you post you solution? Thank you very much.

Hi [b]Virgil Nicula[/b]. your problem is nice. Thank.

[quote="Virgil Nicula"]Prove that in an [b]acute[/b] triangle $ ABC$ there is the relation $ 2\cdot\sum \cos\frac {B \plus{} C}{4}\ge \sum\cos\frac A2 \plus{} \frac {3\sqrt 3}{2}$ .[/quote]
This inequality holds for all triangle $ ABC$ indeed. here is my solution for it
Setting $ a \equal{} \frac {B \plus{} C}{2},b \equal{} \frac {C \plus{} A}{2},c \equal{} \frac {A \plus{} B}{2}$ then $ 0 < a,b,c \le \frac {\pi}{2}$ and $ a \plus{} b \plus{} c \equal{} \pi$, the inequality becomes
$ \sum \left( 2\cos \frac {a}{2} \minus{} \sin a \minus{} \frac {\sqrt {3}}{2}\right) \ge 0$
Consider the function
$ f(x) \equal{} 2\cos \frac {x}{2} \minus{} \sin x \minus{} \frac {\sqrt {3}}{2} \plus{} x \minus{} \frac {\pi}{3}$ with $ 0 < x \le \frac {\pi}{3}$
We have
$ f'(x) \equal{} \minus{} \sin \frac {x}{2} \plus{} 1 \minus{} \cos x \equal{} \sin \frac {x}{2}\left( 2\sin \frac {x}{2} \minus{} 1\right)$
Since $ 0 < x \le \frac {\pi}{2}$, we have
$ f'(x) \equal{} 0 \Leftrightarrow x \equal{} \frac {\pi}{3}$
Moreover, for $ x > \frac {\pi}{3}$, we have $ f'(x) > 0$ and for $ x < \frac {\pi}{3}$, we have $ f'(x) < 0$. From now, we can obtain
$ f(x) \ge f\left(\frac {\pi}{3}\right) \equal{} 0$
$ \Rightarrow 2\cos \frac {x}{2} \minus{} \sin x \minus{} \frac {\sqrt {3}}{2} \ge \frac {\pi}{3} \minus{} x$
$ \Rightarrow \sum\left( 2\cos \frac {a}{2} \minus{} \sin a \minus{} \frac {\sqrt {3}}{2}\right) \ge \sum \left( \frac {\pi}{3} \minus{} a\right) \equal{} 0$
We have done. :)

[quote="Virgil Nicula"]Prove that in an [b]any[/b] triangle $ ABC$ there is the relation : $ 2\cdot\sum \cos\frac {B+C}{4}\ge \sum\cos\frac A2+\frac {3\sqrt 3}{2}$ .[/quote]
[color=darkblue]This inequality is true in [u]any triangle[/u] . Apply [u][b]Popoviciu[/b]'s inequality[/u] to the [u]concave[/u] function $f:(0,\pi)\to(-1,1)$ , $f(x)=\cos\frac x2$:

$\implies\ \begin{array}{cccc}\cos\frac A2+\cos\frac B2+\cos\frac C2+3\cdot\cos(\frac {\frac A2+\frac B2+\frac C2}3)\ \le\ 2\cdot[\cos(\frac {A+B}4)+\cos(\frac {B+C}4)+\cos(\frac {C+A}4)]\end{array}$

$\implies\ \frac {3\sqrt{3}}2+\sum\ \cos\frac A2\ \le\ 2\cdot\sum\ \cos(\frac {B+C}4)$ , as desired .[/color]

[quote="kuing"][quote="Mateescu Constantin"][quote="Virgil Nicula"]Prove that in an [b]any[/b] triangle $ ABC$ there is the relation : $ 2\cdot\sum \cos\frac {B\plus{}C}{4}\ge \sum\cos\frac A2\plus{}\frac {3\sqrt 3}{2}$ .[/quote]
[color=darkblue]This inequality is true in [u]any triangle[/u] . Apply [u][b]Popoviciu[/b]'s inequality[/u] to the [u]concave[/u] function $f(x)=\cos\frac x2$ on the interval $\left(\ 0\ ,\ \frac {\pi}2\ \right)\ \ :$

$\implies\ \begin{array}{cccc}\cos\frac A2+\cos\frac B2+\cos\frac C2+3\cdot\cos\left(\frac {\frac A2+\frac B2+\frac C2}3\right)\ \le\ 2\cdot\left[\cos\left(\frac {A+B}4\right)+\cos\left(\frac {B+C}4\right)+\cos\left(\frac {C+A}4\right)\right]\ \implies\end{array}$

$\implies\ \frac {3\sqrt{3}}2+\sum\ \cos\frac A2\ \le\ 2\cdot\sum\ \cos\left(\frac {B+C}4\right)$ , as desired .[/color][/quote]
$f(x)=\cos\frac{x}{2}$ or $f(x)=\cos x ?$[/quote]

What is it unclear ? The function is $\boxed{\ f(x)=\cos\frac x2\ }$ with $x\in \left(0,\pi\right)$ .


Everyone in my class is required to find an information on the internet about a math related topic, write a 2 page essay about it, and give a 10 minute presentantion in class. Topics suggested but not limited to are: platonic solids, roman numerals, tower of hanoi, kognensberg bridge, graph theory, data encryption, etc. Any topic that involves some type of math is acceptable such as su doku and other things.

I just need help on a few topics AoPS knows that are quite famous and can be explained to a class that is not so great in math in 10 minutes. Any ideas?

Walk in, say some things about 4d hypercubes, then sit down.

I think data encryption would be a good topic to talk about.  It can be explained relatively well in 10 minutes depending on how in-depth you plan to go.

Fibonacci sequence is also a good one, I think.

Tower of Hanoi is cool. I like the name  :D . I am still wonderring why they name it after the capital of Vietnam  :lol: .


AMC AIME geometry USA(J)MO USAMO MATHCOUNTS quadratics

I think i'm the only one, but I think it's easier.  Most likely because I have to brute force most of the problems i get right, and these were mostly counting problems.

[quote="bthings_2000"]Perhaps I'm just looking for self-validation because I did worse this year than last year... but in reading the forums, it seems like a lot of people had a tougher time with this one. What do you think?[/quote]

i agree that scores seem lower than last years'. but notice that your sample is not representative of everyone taking the test.

I thought it was much easier, I was averaging 5s on AIMEs earlier and I got an 8

This was a good year for careless mistakes..
and wtf is up with all the number theory problems at the beginning-there was no geometry until number 9...

very true, any voluntary response sample is not sufficiently randomized to draw valid conclusions over the population.

But it does offer some indication; unlike some of the threads (eg. give your AMC profile), there is really no reason for anyone to feel better or worse about answering this poll, so it must be somewhat telling.

lol.. got a 6 this year.. ya definately it's significantly harder, because last year I know how to do 1-12 in the AIME I...

no matter the difficulty level, my 3 won't land me in Usamo

interesting... so far 2/3 put it in one of the two "harder" categories.

I thought it was about the same. I actually thought the first 7 were super easy, but #4 was kind of lame so that brings me down from "easier" to "about the same." Then #10 was evil, but that's about it (no comment on #13 or #15, though).

yeah, what ccy said (although i got #4 right). Counting is definitely not my strength, i did horrible. Hopefully next year there are less counting problems.

In terms of how easy it was to get a good score, this seems to be on or around par with the AIMEs of the mid-90s. However, in terms of quality of problems, it's pretty similar to recent years' AIMEs, just more computation.

yes

more computation

a lot more computation

also the problems were more tricky than difficult...you look at them and it is easy to miss the trick...i guess it is fair, but i prefer the problems that are straightforward what they are, but finding the answer is tough

Too. Much. Geo.

Got that out of my chest. Because I can usually do geo when it's before question #10...

The geo problems were much more algebraic this year.  And they're giving you more computations so that you can be prepared for the no-calculator amc next year ;)  And it was much easier if ur careful, but harder if you are not.  #2 was just stupid, though I got it right.

if by 'difficult' you mean sucky, then yes i would say it was difficult

[quote="indianamath"][quote="ccy"]It's not harder, just more Mathcounts-like omg.  #4 was an old Mathcounts problem, if I remember correctly.

Die, MC.

Let's petition Mathcounts to sue CAMC for infringing on its patent on computation stupidity. :mad:  :mad:  :mad:  :mad:[/quote

no #4 was NOT an old MATHCOUNTS problem. that problem made it so that all the planets had to be on the same side. (i've done all MATHCOUNTS problems)[/quote][/quote]

Actually, that MC problem did not. I missed it the first time I took it, and never forgot. Although before I had to guess that the planets would lie on the same lie as they started, and on the AIME I figured out a really good way to do it.

I'm not qualified to comment on the geometry, because I suck at geometry -- and, BTW, WTF was with all of it in 9 - 15? That really screwed me over in terms of the number of questions that I could answer, because I can usually answer up to about number five or six for geometry problems and up to 12 or 13 for others.

So, anyway, I answered 1 - 8, 10, and 11 — that is, all the non-geo problems up to 12 — and I honestly didn't think any of them were all that hard conceptually.

On the other hand, I thought that they invited a lot careless errors (I ended up with a 7).

Overall, I'd say that it was easier in terms of conceptual difficulty than most of the past AIMEs, but harder to get a good score.

The geometry was fine...the last problem had a nasty quadratic to solve...which I solved [incorrectly] in the last 10 mins of the test. I just think that people did not search for good solutions...they were like HEY I CAN BRUTE FORCE--->>>>LETS DO THAT.

Since I don't have my test, I think #9 is the circle with the tangent one? I think a lot of people made mistakes when they thought that the leg was $\frac{5}{3}$ of the radius, instead of $\frac{3}{5}$. However, if you looked at the diagram closely, it is counter intuitive, there is no way that leg can be larger than the radius. I think it's just that people plug in formulas so much, they just forget to check the really obvious things.

[quote="bpms"][quote="probability1.01"]Uhh... what? #12 was bad no matter what, #9 was also somewhat bad no matter what. #13 was fine, but #15 involved a certain amount of necessary computation. I think it wasn't that bad overall other than #12, but certainly nothing really nice.[/quote]
Is there a pure geo solution for 9, I used trig (not that messy)[/quote]

Well I used a sort of scaling of the given triangle with hypotenuse the line segment between the two circles. Then I set up an easy system of equations to solve for $r$.

[quote="probability1.01"]Uhh... what? #12 was bad no matter what, #9 was also somewhat bad no matter what. #13 was fine, but #15 involved a certain amount of necessary computation. I think it wasn't that bad overall other than #12, but certainly nothing really nice.[/quote]

9 was just like c/(2+ab/(a+c)+ab/(b+c))

not amazing obviously...but that is not that bad...although i managed to reduce incorrectly

#12...i ono...i think some of those cyclic quadrilaterals...i did not finish that solution...but it looks pretty easy

i should have been less lazy and searched for the elegant solutions...man...like i see them after i have time to just sit and think...but i ono...oh well

It seemed a lot easier to me, but with all the time I've spent on here getting better, I'm only voting for a "just a bit easier," because some of that is probably due to my improvement, rather than difference in difficulty...

Hmm, I found a almost pure geometry solution to number 9.

Used only similar triangles and pythagorean theorem.

I totally blanked on a better method for #12 so I did arguably the most brute-force in my life.  Find all coordinates, line equations, and intersections, then shoestring it.

UGHH...that was about 30 minutes because I messed up a few times and the door in our room was shaking heavily.

I thought it was much easier, because I solved the first twelve, while I usually get killed by AIME's.

[quote="1/(ln x)"]I'd have to say that this test was much harder than the last couple of years (unless if you are good at geo and counting, which I am not). Too many of the problems were brute forceable which gave an advantage of people who are good at brute forcing over those who actually try to look for a different route. This is just my opinion, but this test seemed to be loaded with geometry and counting, there didn't seem to be that many number theory/pure-algebra problems (#7 is the only one that comes to my mind, but there are probably more).[/quote]
#14 was algebra, but yeah, a lot of geo this year.

I gave this aime a cleverness score of 1.5/15=10%, which is a massive failure.
#2: wtf, read "besides" but took a while debating whether a guy would be moving beside it if no path beside the walkway was introduced
#3: pretty dumb problem
#4: somewhat tricky... i even checked 105 then decided that it didnt work(even drawing out the diagram and the positions of the three points at 105...
#8: lot of steps... wtf
#9: typical
#10: not actually that hard? it's only doing casework on two columns since you have one column symmetric and the last column predetermined by the other three, so there are like 4 calculations total involving small numbers
#12: massive ugliness, still have not determined why my answer is wrong
#13: not bad, but still a tad too many computations, but typical for later geo problems
#14: intuition alone is enough here
#15: wtf, was looking for some cool translation solution

15 was perfectly good

13...not really...it was easy to find the coordinates of the intersection of the plane of the midpoints and the lateral edges using similarity

dot product gives you the dihedral angle with practically no computation

trivial base:base=area:area computations give the area of the projection of the intersection onto ABCD

but we can all agree that #12 was bad

There really wasn't a terrible load of thinking or creativity involved.

For someone who can easily plow through a whole bunch of brute-force work without making mistakes (aka "math machine"), then this test was very easy.

For people who prefer problem solving and looking for nice, clean solutions, this test had the tendency to be strategically and/or psychologically difficult.

[quote="probability1.01"][quote="DanK"][quote="probability1.01"]#9 was also somewhat bad no matter what.[/quote]

What makes you say that?  It was really easy if you used half-angle identity for tangent, which gave you a direct relation of the length of the radius to the sides of the triangle.  Now if I could just learn to add, I would have had that one right...[/quote]

Well I dunno I guess I just didn't like the numbers. It only took me about 4 minutes to solve and check, but the ratio of time spent doing routine stuff vs. time spent solving the problem conceptually was probably well over 40.[/quote]

Wow, dude, you did all the conceptual stuff for that problem in 6 seconds!  You really are a beast.

Well, yea considering he's an IMO silver medalist  :wink:

This is kinda weird... I just saw the problem right now, and found the exact answer assuming that both circles were excircles. Is this a coincidence or am I being stupid and making careless errors?  :|


calculus
integration
function
floor function
calculus computations

Let $ \lfloor x \rfloor$ be defined as $ \lfloor x \rfloor \equal{} k \in \mathbb{Z}$ such that $ x \in [k, \, k\plus{}1)$. The floor function.

Let $ \{x \}$ be defined as $ \{ x \} : \equal{} x \minus{} \lfloor x \rfloor$. The fractional part function. (convention for $ x < 0$ does not concern us here)

Determine $ \int\limits_0^1 \, \int\limits_0^1 \left\{ \frac{x}y \right\} \left\{ \frac{y}x \right\} \, dx \, dy$.

Sending this back up to bring it under your attention again.
Think about the symmetries in the function that is to be integrated.

Perhaps it should be in Unsolved and Proposed Problems instead?

See here

http://www.math.leidenuniv.nl/~naw/home/ps/pdf/uitwerking-2008-1.pdf

Yes well the solution in the link you posted is of course correct, but not complete (this is also where the problem came from).

Can someone complete the solution?


My school has only a few spots for lots of people trying. I am probably one of the best at "hands on" projects, but there may be some kids who guess at the multiple choice questions better than me.  I don't even know what all the topics of the test are but I think there are a bunch of areas.  Is there any sample questions or past tests available - I googled for 2 hrs with no luck, so I thought maybe one of my fellow AoPS'ers could help! How 'bout it? Did anyone here ever take a qualifying test? Even if I don't make it this yr, I'd like to start preparing for next.

yes, i do it. hey guyinPA, might i know your name. roygbiv-rgbiv-rg ring a bell? i also PMed you.

im on my schools team. and i hope you realize there is a MUCH better place for this. [url]scioly.org[/url]


Prove the following two theorems:

1. Let f(z) = u(x,y) + iv(x,y).  The function u(x,y) and v(x,y) are continuous at any point where f(z) is continuous.  Conversely, at any point where u and v are continuous, f(z) is also.

2. If f(z) is continuous in some region R, then |f(z)| is also continuous in R.  If R is bounded and closed there exists a positive real number, say M, such that |f(z)| <= M for all z in R.  The equality holds for at least one value of z in R.

yeah, this isn't complex analysis

these are basic point-set topological facts about continuity with respect to the product topology and continuous mappings of compact sets. perhaps try thinking about them in that framework to see if that helps (well not necessarily in that generality; for example, replacing "general topological space" with "$ \mathbb{R}^n$" would suffice for your purposes here) and if you don't get anywhere, repost the topological statements in the topology forum


Why are there 3 irrational numbers among six irrational number each whose pairwise sums are all irrational. Is it also true for five irrational numbers each time ?

for 5 it's not: $1 + \sqrt 2 , 1 - \sqrt 2, 1 + \sqrt 3, 1 - \sqrt 3, 1 + \sqrt 5, 1 - \sqrt 5$, look not even for 4.

I think the proof for 3 has something to do with these. Although I won't know how to make it rigorous :)

[quote="orl"]Why are there 3 irrational numbers among six irrational number each whose pairwise sums are all irrational.[/quote]

Cuz R(3, 3) = 6, isn't it?

  Darij

Ah well, now it seems you have forced me to write up the complete solution. But it's trivial, believe me.

It's a well-known (and easy) fact that if six people are in a room, and if any two of these people can either be friends of each other or enemies of each other, then you will always find either three people who are pairwisely friends of each other, or three people who are pairwisely enemies of each other.

Now assign to each of our six irrational numbers a guy, and two guys will be called friends of each other if and only if the sum of the corresponding irrational numbers is rational. Then, by the above fact, you will either find three people who are pairwisely friends of each other, or three people who are pairwisely enemies of each other. This means that among your six irrational numbers, you can either find three ones whose pairwise sums are all rational, or three ones whose pairwise sums are all irrational. The first of these two cases is impossible, since, if x, y, z were three irrational numbers whose pairwise sums are all rational, then, since $\displaystyle x = \frac{\left(z+x\right)+\left(x+y\right)-\left(y+z\right)}{2}$, the number x would be rational, too, contradicting our assumption that all three numbers x, y, z are irrational. Hence, the latter case must occur, i. e. among our six numbers we can find three whose pairwise sums are all irrational. Qed..

R(a, b) is a [url=http://mathworld.wolfram.com/RamseyNumber.html]Ramsey number[/url]. The concept of Ramsey numbers generalizes the above fact about the six people.

  Darij

I knew it was trivial, but I had no clue how to express it.


trigonometry
inequalities
inequalities unsolved

ABC IS AN ACUTE TRIANGLE;show that:
cyclic sum[SQROOT(1+8cos^2A)/sinA] IS greater than 6

There was a mistake ,the right one is:
ABC IS AN ACUTE TRIANGLE;show that:
cyclic sum[SQROOT(1+8cos^2A)/sinB] IS greater than or equal 6


I ordered the Art of Problem Solving Volume 1 with Solutions text! I'm excited to get it and just felt like sharing.  Just wondering, I am goign to be a senior in high school, and although when I was in middle school, I was very good at contest math and problem solving and thinking, I have become quite rusty.  I was just wondering if there is any sites that feature problems from state math leagues and the AMC [b]FOR FREE[/b].  I browse this forum but most of the problems that are posted are solved before I find the solution so it does little good to post the solution, but do any people have any sites? I plan to post a lot more so I look foward to meeting all of you.

Yep.  A good place to start would be to click on the "Resources" button on the left hand bar, then select "problems."  Scroll down to find problems.  I recommend Elias Saab's "Math Drills/Math Counts" practice page at [url]http://mathcounts.saab.org/[/url].  If these are too easy, then you could always check out Flanders Mathematical Olympiad problems (don't be fooled by the name...it's not olympiad (proofs), it's AMC-level problems).  It doesn't have solutions, but does that matter?  You have us.  Which is sort of like an interactive solutions manual.

Happy Solving.

P.S. There are plenty of problems in the AoPS books.  Do all of those first.

Ok, I plan to go through it and do everything I can, then read the other books in the resources section and move onto my zeitz book and the AoPS Vol 2, thanks for the help!

You have the Zeitz book?  Lucky ducky.

Oops, sometimes I sound like I'm four years old.  But I can't help it - you're just lucky.

Oh, and do AoPS2 before the Zeitz book (which has been called "AoPS3").

I guess I did it all wrong then. I did Zeitz then AoPS2 and never AoPS1. Perhaps I could have done well on contests in 9th grade if I'd had a better order. Oh wait. I read Zeitz summer after 9th grade, so that's not true anyway.

Yea, I bought Zeitz because a couple of students who were really good (but now in college) in math recommended it to me.  I will follow your plan fanatic, along with the plan this site recommends.  Hehe, I'm going to be a senior in high school so I don't have all the time in the world but I plan to have fun with it all.

Yes indeed.  Have lots of fun.  Meanwhile I'll sit here and wish I had that book <sarcasm> Nah, there's plenty of stuff to do here...lotsa math.  Well, have fun!

It came!! but I can't even read it yet due to all the homework I have from computer classes and I cant even get my c++ program to work!


superior algebra
superior algebra solved

$R$ is a ring. $\mathfrak a$ is an ideal of $R$ and it has identity (there exists $e\in\mathfrak a$, such that $e\mathfrak a=\mathfrak a e=\mathfrak a$), $A$ is an ideal of $\mathfrak a$. Show that $A$ is an ideal of $R$.

Let $ c\in \mathfrak a$ with $ce=e$. Then for $d \in \mathfrak a$ we find $f \in \mathfrak a$ with $ef=d$. Multiplying by $c$ to the left we gef $ef=cef=d$ so $cd=d$. Now let $ue=c$. If $ef=d$ then multiplyiing by $u$ we get $f=du$. Hence if $d\in A$ then any $f$ with $ef=d$ we get $f \in A$.

Now let any $x \in R, y \in A$. Let $ez=y$ then $ z \in A$ by the observation above.
Then $xy=xez=(xe)z$. But since $\mathfrak a$ is an ideal in $R$ we get $xe \in \mathfrak a$ and since $A$ is an ideal in $\mathfrak a$ we get $(xe)z \in A$ hence $xy \in A$.
By a similar 'reversed' argument we get $yx \in A$ and hence $A$ is an ideal in $R$


Hi,

What's the difference between WinShell and WinEdt, except that WinEdt cost something ?...

I dunnot but I tend to like WinEdt over most LaTeX shells.  :?

Well,

I do like a lot winedt, but it cost something. Many people install it and reinstall it each, I don't know, fourty days or so. I strongly recomend TeXNic. It is free (well, the project accept donations, but that's up to you) and is as efficient as WinEdt, and almost as friendly.

Of course, if you leave the windows addiction, you can find excelent editors in basically any distribution of Linux, a lot better than WinEdt and FREE.

Best,

David

The biggest problem with TeXnicCenter is that it doesn't automatically match up braces.  Why an editor specifically designed for TeX doesn't have this basic feature is beyond me.

I use Crimson Editor.

[quote="DPatrick"]The biggest problem with TeXnicCenter is that it doesn't automatically match up braces.  Why an editor specifically designed for TeX doesn't have this basic feature is beyond me.[/quote]
[b]TeXnicCenter 1 Beta 6.30 released[/b]

[i]The brand new version 1 Beta 6.30 (codename "Firenze") of TeXnicCenter has been released today and is available for download.[/i]

Here is what Pavel Vacha and Tino Weinkauf have done for you:
 :arrow: Highlighting of Parenthesis: If the cursor is near an opening/closing parenthesis, it will be highlighted green/red depending on whether it has a matching closing/opening brace or not and the counterpart will also be highlighted.
 :arrow: Brace Block Selection: Press Ctrl-M and the whole text enclosed by the next matching parenthesis will be selected. Press Ctrl-M again to grow the selection to the upper level of matching parenthesis.
 :arrow: Auto-Preview after Building Output: Compile and View your document in one step by simply pressing Ctrl-F5.
 :arrow: Several small features and bug fixes.

Have a lot of fun and happy TeXing...

Wow, the new version of TeXnicCenter is so much nicer!

Thanks Myth for posting this and letting us all know about the new version.

[quote="djimenez"]Well,Of course, if you leave the windows addiction, you can find excelent editors in basically any distribution of Linux, a lot better than WinEdt and FREE.[/quote]

And there are a plethora of great TeX editors for the Macintosh.

[quote="Myth"]WinEdit must die![/quote] :lol:  :lol:  Mikhail I think that you mean WinEdt, not WinEdit (which is another windows editor, but not for $\LaTeX$) :P

[quote="Valentin Vornicu"]Seems better but still doesn't quite equalize WinEdt (this is ok since it's a free product). It only matches up simple braces, not the dollar signs, or accolades ...  :?  :([/quote]
You are not right. TxC highlights inline equations with special colors. It is much more better than only match up dollar signs.
I even can't imagine what accolades mean here...

Well I installed this 6.30 version  and it does not highlight anything :(
Does it not do it by default? I mean do I have to change some settings to do that?

By accolades I ment \[ ... \] or \{ ... \} or [ ... ] :)

TxC hates people who don't like it.

I didn't perform any additional changes, but I installed it over 6.20. Really, for unknown reason TxC highlights only {}, (), [] and doesn't highlights \{\}, \[ \]  :?

accolade... chocolate... No one other associations. My dictionary agrees with me.

[quote="Myth"]TxC hates people who don't like it.[/quote]I've got nothing against it  :) I like them both, it's just that I like WinEdt better :D

Probably you wouldn't want it to highlight \{...\}.

E.g.:

\[|x| = \left\{ \begin{array}{ll}
x \text{ if } x \geq 0 \\
-x \text{ if } x < 0
\end{array}\right.
\]

\{ with no \}.  Now if it automatically matched up \left's and \right's, that would be cool.

[quote="DPatrick"]Probably you wouldn't want it to highlight \{...\}.  [/quote]WinEdt matches them up and there are no problems  :?  :?

I agree that you wouldn't want it to match \{ with \}, but I'm not sure why you'd want to use that construction when the cases environment already exists.

[quote="DPatrick"]Now if it automatically matched up \left's and \right's, that would be cool.[/quote]

Yeah, that would.  Whenever I'm going to be using the floor function (or similar), I just define a new command, \floor{stuff}, as \left\lfloor stuff \right\rfloor just so I don't have to type \left and \right.

Every time I'm trying to run dvi. I have some strange error:
,,One or more error occured!
Please check the options!"
But options are not the problem because I haven't changed anything (program calls only). Same error on Winshell 2.6 and some older ones.
Any suggestions?

Dvi worked correctly at the begining. I didn't used it for some time and when I tried it I had this error :/


ratio
inequalities
algebra
polynomial
Rational Root Theorem

Prove that the x^(1/x) will never be rational for integer x when x >1

I don't know how to prove things but I'll give it a go:

[hide]

Assume temporarily that x^(1/x) is rational.  Then it can be expressed as a ratio of integers.  Let [tex]x^{\frac{1}{x}}=\frac{a}{b}[/tex], where a and b are integers that are relativley prime.  Raise everything to the xth power to get [tex]x=(\frac{a}{b})^x=\frac{a^x}{b^x}[/tex].  But that does not make any sense.  Remember that a and b are relativley prime, so in the numerator you will have x a's and in the denominator you will have x b's, but all of those are relativley prime (as assumed before), so they will never simplify out to be one integer x.  This is a contradiction and it follows that [tex]x^{\frac{1}{x}}[/tex] is irrational for all x>1. [/hide]

[hide]If x^1/x = p/q with (p,q)=1 then x *q^x = p^x. Then q | p which only possible if q=1. Now 2^x > x. So there are no solution. [/hide]

Julie, your proof almost works. What you want to say is that [hide]it is either an integer or irrational because of what you did. But you can show that it's always between 1 and 2, so it can't be an integer.[/hide]

I am confused.  How can it be an integer?  I said they were already in simplest form, and a != b, so it can't be an integer...

[hide]

Okay, so I need to prove that:

[tex]1 < x^{\frac{1}{x}} < 2[/tex]

Raise everything to the xth power:

[tex]1^x < x < 2^x[/tex]

[tex]1 < x < 2^x[/tex]

It's given that x>1, and you can prove that for x>1, x < 2^x (using mathematical induction).

This shows that the inequality is true when x>1.

[/hide]

That might be really off.

[quote]I am confused. How can it be an integer? I said they were already in simplest form, and a != b, so it can't be an integer... [/quote]

a/b could be an integer when b=1.

For example, why doesn't your proof prove that [tex]4^{1/2}[/tex] is irrational? You can use exactly the same argument. In general, your proof shows that [tex]a^{1/b}[/tex] is either irrational or an integer for positive integers a and b.

Idea:

[hide]

We know that if x^(1/x) = a, then

1 < a < 2

x^(1/x) = a

x = a^x

a^x - x = 0

Now we solve for rational a.  By the Rational Root theorem, the only possible rational a is an integer factor of x.  But if x is not one, then a is not an integer.  Thus there are no rational numbers a that make the statement a^x - x = 0 true, and x^(1/x) cannot be rational.[/hide]

Tokenadult doesn't like having problems with the word ``prove'' in Getting Started. It should be in Intermediate. I'll move it over there. (This post was in response to Scrambled's post, which Scrambled deleted while I was posting.)

In my opinion problems without the word 'prove' in it shouldn't be on this website. But lets not get into that debate again.

Even if "prove" is not in a question, you still have to "prove" your answer is right. So I don't see how the word "prove" should not be in getting started problems.

scrambled can just ask Is x^(1/x) a rational, but still you can't just say yes/no.


probability counting derangement MATHCOUNTS

4 letters and 4 envelopes are addressed to 4 different people. If the letters are randomly inserted into the envelopes, what is the probability that a. exactly 1 will be inserted into the correct envelope.and b. at least one will be inserted into the correct envelope.

[hide]
The probability that one letter goes into the right envelope is $.25$, but there is more than one way that this can happen.  Using the binomial distribution we have:
$\binom{4}{1}(.25)^{1}(.75)^{3}\approx .42$.
The probability of at least one right is the probability that one, two, three, or four are right.  Alternatively, it is 1 minus the probability that none are right. The probability that none are right is $.75^{4}$ so then $1-(.75)^{4}\approx .68$ [/hide]

i don't think that thats th right answer. In the back of my book it gives a different answer but i don't know how to get it.

Is the answer for the first one [hide].667?[/hide]

[hide="b"]
This would be 24-derangements(ways none are right)/24
So $\frac{15}{24}=5/8$
Is this right?
[/hide]

yeah. those are the correct answers. How did you find them?

[hide=" Part a"]
Label the envelopes and corresponding letters A,B,C,and D. Any of the four can be the one that is correct.  Suppose letter A goes into envelope A.  Then the rest of the letters must be in the wrong envelopes and this can happen two ways.  Letter-Envelope:
A-A, B-C,C-D,D-B  or A-A,B-D,C-B,D-C.  The probabilty of one of these situations occuring is:
$\left(\frac{1}{4}\right)\cdot\left(\frac{2}{3}\right)\cdot\left(\frac{1}{2}\right)\cdot\left(1\right)$.  Multiply that by 2 since there are two ways this can happen.  Then multiply that by 4 since any of the four can be the one in the correct position. [/hide]

[quote="illcrowflu"]
Label the envelopes and corresponding letters A,B,C,and D. Any of the four can be the one that is correct.  Suppose letter A goes into envelope A.  Then the rest of the letters must be in the wrong envelopes and this can happen two ways.  Letter-Envelope:
A-A, B-C,C-D,D-B  or A-A,B-D,C-B,D-C.  The probabilty of one of these situations occuring is:
$\left(\frac{1}{4}\right)\left(\frac{2}{3}\right)\left(\frac{1}{2}\right)\left(1\right)$.  Multiply that by 2 since there are two ways this can happen.  Then multiply that by 4 since any of the four can be the one in the correct position. [/quote]
yes that way works only when there are less than or equal to 4 envelopes.  If you try that way with 5 envelopes or more, it wont be right (because they will be dependent on each other)  To get exactly one right, it is basically multiply the derangement of 3 envelpes by 4, the derangements of 3 envoples (3 choices, none of the 3 are right) is 2, so 4*2 (beucase there are 4 spots taht can be right)=8, so 8/(4!)=8/24=1/3 for exactly 1 right

How would I find the dearrangements for larger numbers? like for part b it is like the dearrangemnt of 24. How do i solve that?

[quote="illcrowflu"].667?[/quote]

[quote="Art of Owna"]so 8/(4!)=8/24=1/3 for exactly 1 right[/quote]

Wait, so is the answer 1/3, or 2/3? Im confused...

Using partial derangements, I too got 1/3. But if i read turtlecloud correctly, 2/3 is the right answer?

edit: turtlecloud, partial derrangements can be explained here:
http://mathworld.wolfram.com/PartialDerangement.html
Hope that link helps.

opps. I meant 1/3. I somehow thought that .667 is 1/3.
i still dont understand the derangement. the link to mathworld was confusing. Can someone explain it?

There was a big discussion about it in the Mathcounts Forum. I calculated it by
$\frac{4!}{e}$ to the nearest whole number


function
trigonometry
calculus
calculus computations

How many zeros does the function y = sin(lnx) have for 0<x<1 ?

Not sure if I'm doing this right...

$ 0 \equal{} \sin(\ln(x))$

$ \sin^{ \minus{} 1}(0) \equal{} \ln(x)$

There are infinitely many solutions for $ \sin^{ \minus{} 1}(0)$, but I'm only interested in $ \minus{} \pi$ for the moment.

$ \minus{} \pi \equal{} \ln(x)$

$ e^{ \minus{} \pi} \equal{} x$

Now, since $ \minus{} 2\pi \equal{} \sin^{ \minus{} 1}(0)$, we have

$ e^{ \minus{} 2\pi} \equal{} x$ as another solution. Indeed, in general;

$ e^{ \minus{} \pi n} \equal{} x$ will work for $ n \in \mathbb{N}$


LaTeX
ratio
trigonometry
geometric sequence

There aren't any real roots as far as I know.

Maybe you can use the fact that it's $\frac{x^{5}-1}{x-1}$

[quote="SplashD"]Maybe you can use the fact that it's $\frac{x^{5}-1}{x-1}$[/quote] Oh, thanks. :D

[quote="lotrgreengrapes7926"]Is there any way to solve $x^{4}+x^{3}+x^{2}+x+1=0$?[/quote]

[hide]You could also divide by $x^{2}$ throughout, because you know x won't equal 0, and you get a suspiciously symmetric equation, namely $x^{2}+x+1+\frac{1}{x}+\frac{1}{x^{2}}$.  See if you can make a nice substitution anywhere in there, and the rest is algebra.

[hide]Try $y=x+\frac{1}{x}$.  :)
[/hide]
[/hide]

This is in ACoPS.

divide the equation thru by $x^{2}$

$x^{2}+x+1+\frac{1}{x}+\frac{1}{x^{2}}=0$

regroup to make $x^{2}+\frac{1}{x^{2}}+x+\frac{1}{x}+1=0$

subsitute in $u=x+\frac{1}{x}$

so $u^{2}=x^{2}+x+\frac{1}{x^{2}}$

now equation is $u^{2}+u-1=0$ and $u=\frac{-1+-\sqrt{5}}{2}$ and plugging back in $x=\frac{u+-\sqrt{u^{2}-4}}{2}$

please dont make me type the answer.

P.S. how do you latex plus-minus sign?

\pm with dollar signs around it gives you plus/minus.

You all seem to have entirely ignored SplashD's insightful comment on this problem  :wink:

Can you find a nice form for all four roots?

Fifth roots of unity except for 1. This follows from SplashD's comment.

This is another way you can get that SplashD got, you have a geometric sequence with first term 1 and common ration x, then you use the forumla to get what he got.

[quote="jhcreinhardt"][quote="t0rajir0u"]I might as well mention that there is a way to solve quartics generally, but it's entirely unnecesary here.[/quote]

surely you're not talking about the quartic formula...[/quote]

Hahah, surely he is. I don't think there is any other general way to solve it, so I feel obliged to post the following:

http://en.wikipedia.org/wiki/Quartic_formula#The_general_case.2C_along_Ferrari.27s_lines

For those who wish to tackle the problem this way, I wish you good luck, lol.  :)

[quote="jhcreinhardt"]http://planetmath.org/encyclopedia/QuarticFormula.html

for the actual formula...scary stuff[/quote]

LOL that stuff spans over 2 pages wide! If you try to actually use it to solve, you probabtly can't even fit it on your paper lol.  :rotfl:

[quote="lotrgreengrapes7926"][quote="jhcreinhardt"]ok thanks

anyway heres the answer

$x=\frac{-1\pm\sqrt{5}\pm i\sqrt{2\sqrt{5}\pm10}}{4}$[/quote]
Is this technically correct? I mean, this implies there are $6$ solutions.  :maybe:[/quote]

Actually, it implies there are 8 -- the problem is that two of those $\pm$ signs operate in parallel.  If you go and look at jhcreinhardt's solution, about fifth post in the thread, you'll see that $u$ has a $\pm$ in it -- both appearances of $u$ must have this symbol going the same direction, but that fact disappears when the substitution is made..

yes that is what it boils down to.  it is very hard to show the solutions w/o using cis.

You can use roots of unity via $x^{5}-1=0$.  Since the equation is a reciprocal equation, it can take the the form $f\left(x+\frac{1}{x}\right)$.  Both of these will give you results.  Something is hidden within these two solutions.  Find it. :)

Masoud Zargar

What do you mean by reciprocal equation?

[quote="boxedexe"] Something is hidden within these two solutions.  [/quote]

[hide="Hint"] $\cos x = \frac{e^{ix}+e^{-ix}}{2}$ [/hide]

R u just saying that it could be put into a general cis form?

i think the ideas he was implying had already been said...


[quote="math10"]Let $ x,y,z > 0$.Prove that:   $ \sum_{cyc}\frac {3y^2 \minus{} 2xz \plus{} 2z^2}{x^2 \plus{} 2xz}\geq 3$
I found a solution using the Cauchy-Schwarz.But I want every to find nice solutions  :D  (for a problem don't  nice)[/quote]
By CS and Vasc's ineq, easily to prove the genaral problem:
$ \sum\ \frac {x^2 \plus{} my^2 \plus{} nz^2}{x^2 \plus{} 2yz} \geq\ 1 \plus{} m \plus{} n$  :wink:

purposes of this topic is to find solutions of this problem , not general( because the work you do is not difficult)

Are you sure?  :wink:
[quote="math10"]Let $ x,y,z > 0$.Prove that:   $ \sum_{cyc}\frac {3y^2 \minus{} 2xz \plus{} 2z^2}{x^2 \plus{} 2xz}\geq 3$
I found a solution using the Cauchy-Schwarz.But I want every to find nice solutions  :D  (for a problem don't  nice)[/quote]
This problem can be rewrite to: $ \sum\ \frac{x^2\plus{}3y^2\plus{}2z^2}{x^2\plus{}2yz}\geq\ 6$
It's the cases with $ m\equal{}3; n\equal{}2$

[quote="math10"]Let $ x,y,z > 0$.Prove that:   $ \sum_{cyc}\frac {3y^2 \minus{} 2xz \plus{} 2z^2}{x^2 \plus{} 2xz}\geq 3$
I found a solution using the Cauchy-Schwarz.But I want every to find nice solutions  :D  (for a problem don't  nice)[/quote]
I think my proof is not true :
Firstly,
\[ \frac {3y^2 \minus{} 2xz \plus{} 2z^2}{x^2 \plus{} 2xz} \ge \frac {2y^2 \plus{} x^2}{x^2 \plus{} 2xz}
\]
and
\[ \frac {2y^2 \plus{} x^2}{x^2 \plus{} 2xz} \ge \frac {y^2 \plus{} 2yz}{x^2 \plus{} 2xz}
\]
So
\[ LHS \ge \sum_{cyc} \frac {y^2 \plus{} 2yz}{x^2 \plus{} 2xz} \ge 3
\]
By $ AM \minus{} GM$ inequality:
I think you should detail your symbol $ \sum_{cyc}$ as a clear formula.


Techincally, it's still winter, at least in the Northern hemisphere.
My spring break isn't for another month. I'll probably spend it studying for
APs, so I'm not exactly looking forward to it.

I also have no school right now for a week
YAAAAAAAAAY!!!!!!!!!!!!   :D

[quote="plokoon51"][quote="mcalderbank"]YAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAY. No school! :D[/quote]

your spring break is starting??? wow, thats early. my spring break starts in a week...[/quote]

Mine too, but when you're a grad student, spring break or not is not a huge difference  :( .

Best,

Spring brake: study for precalc and chem and prays to god i can skip those classes

[quote="Bictor717"]I'll probably spend it studying for APs, so I'm not exactly looking forward to it.[/quote]
Same here.
With that in mind, it's really painful hearing about all these other people going to all these other places over the world...
But I almost never go anywhere during Spring Break because of the tremendous probability of getting flattened..
Home at the city will be nice and empty :)

[quote="Bictor717"]I'll probably spend it studying for APs, so I'm not exactly looking forward to it.[/quote]

ah...it's lucky to be a freshman sometimes...

Bah, I've been on Spring Break since the beginning of March and have been spending a ton of time just prepping for the AIME. Bleh...


How would you do this problem?

Given a regular hexagon, a second hexagon is created by joining the midpoints of the sides of the first hexagon, then a third hexagon is created by joining the midpoints of the sides of the second hexagon. What is the ratio of the area of the third hexagon to the area of the first hexagon?

[hide]$\frac{19-8\sqrt{3}}{16}$?

The third hexagon would be similar to the first hexagon. Let's say the first hexagon has side 1. Draw the segment from the center of the hexagon to one of the vertices. We know this is 1. I then found the distance from the center of the hexagon to a vertex of the third hexagon as $\frac{4-\sqrt{3}}{4}$ and squared this. I don't think this really makes sense, sorry. [/hide]

[hide]Let's say the first hexagon initially had a sidelength of 1. The  area of the second hexagon is the first, minus, the area of the six triangles it leaves.

Area second : Area first
$\frac{6\cdot\frac{1}{2}\cdot\sin{60}-6\cdot\frac{1}{8}\cdot\sin{60}}{6\cdot\frac{1}{2}\cdot\sin{60}}= \frac{3}{4}$

So we have $(\frac{3}{4})^{2}= \frac{9}{16}$[/hide]

[hide]aw, crap. I planned out a solution for this on scrap paper and my little brother destroyed it......

take 2.

Draw a diagram. Call the side length (of hexagon 1)  x. Using 30-60-90 we see that the distance from the perimeter of  Hexagon #1 to the perimeter of Hexagon #3 is 1/2 of half the length of x. So the side length of Hexagon #3 is 3/4x.

Since the area of a regular hexagon includes squaring the side length, the ratio of hexagon #3 to hexagon #1 is $\left(\frac{3}{4}\right)^{2}=\frac{9}{16}$ [/hide]


geometry
3D geometry
number theory
greatest common divisor
least common multiple

How many positive integers less than 1000 are both perfect squares and perfect cubes?

[quote="ernie"]A number that is a perfect square can be expressed in form $ x^2$.  A perfect cube can be expressed in form $ x^3$.  Thus, a number that is both a square and a cube can be expressed in form $ x^{\text{gcd}(2,3)}$, or $ x^6$.  Thus, there are $ \boxed{3}$ numbers $ (1,64,729)$ that are both perfect squares and cubes.[/quote]

I think you mean lcm, not gcf.

There are 3 perfect sixth powers less than 1000.

[quote="mathproblem66"]How did you get 3 from x^6?[/quote]
Since we know we are looking for numbers of the form $x^6$, we can experiment with different values of $x$.

$1^6=1$
$2^6=64$
$3^6=729$

We see that three such values of $x$ satisfy the condition that $x^6$ is less than $1000$. Therefore, there are $\boxed{3}$ positive integers less than $1000$ that are both perfect squares and perfect cubes, namely $1$, $64$, and $729$.


Arround a circle are sitted n man and n-1 woman. prove there are two man such that between them there are n persons.

We must have $n$ pairs of the form $(m,w)$, where $m$ is a man and $w$ is a woman situated s.t. between the two there are $n$ people in a clockwise direction. Clearly, all such women are distinct, which means that there are at least $n$ women (obviously false, since there are only $n-1$).


algebra
polynomial
superior algebra
superior algebra solved

Let A \in Mn(Q) such that det(A+2^(1/n).I_n) = 0
Prove that det(A-I_n) = detA + (det(A+In))^n

Mihail Bencze, Brasov
GM7-8/1999

I might be wrong, but I think x n - 2 is the minimal polynomial of 2 1/n in Q[x].

If that's right, then P(x)=det(A+xI_n)=x n -2 and it's easy to draw the conclusion P(-1)=P(0)+(P(1)) n .


search
arithmetic sequence
algebra unsolved
algebra

Are there such natural numbers $ a_1, a_2, a_3, a_4, a_5$ :
$ 1)$ the sequence $ a_1, a_2, a_3, a_4, a_5$ is an arithmetic progression
$ 2)$ $ a_1a_2a_3a_4a_5 \equal{} b^{2008}$ where b is some natural number

This is a problem from [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?search_id=2012039856&t=231231]Tournament of Towns 2008[/url].


function
calculus
integration
geometry
probability
absolute value
expected value

Let X be a continuous random variable with density function

f(x)= |x|/10  for -2 less than or equal to x  which is less than or equal to 4

and 0 otherwise.

find expected value.

My attempt:

I multiply the above function by x and conclude that the integral of the function  x^2/ 10 evaluated from -2 to 4 gives the expected value.

Problem:

I split the integral in two and integrate from -2 to 0 and then again from 0 to 4.

For some reason, I believe the integral from -2 to 0 will give me a negative number but I should take the absolute value and add it to the integral from 0 to 4.

so I get 12/5.

but the solution is 28/15, which suggests a negative value from -2 to 0.

I reason that because I'm taking the expected value of something, this integral is allowed to be negative. But if they had asked me to find the area under a curve, I would have to take the absolute value of the integral from -2 to 0. Is that reasoning correct?

You're not being asked to take the area under a curve. You're asked for an expected value.

A [i]probability[/i] must be nonnegative, and following on this, a [i]probability density[/i] must be nonnegative. But an [i]expected value[/i] may be positive, negative, or zero.

Let's stay away from continuous distributions for a moment to remove a distraction - but it still makes no difference as far as the concept is concerned.

Suppose $ X$ takes on the value $ 1$ with probability $ \frac23$ and the value $ -3$ with probability $ \frac13.$ That's a perfectly legitimate probability distribution. Then
\[ E(X)=(-3)\cdot\frac13+(1)\cdot\frac23=-1+\frac23=-\frac13\]
On average, this distribution is negative, which is a matter of what the values are and what the probabilities are - the only way to be sure was to add them all up to see where we stood.

Here's a continuous example: $ X$ has probability density $ \frac{|x|}4$ for $ -2\le x\le 2$ and zero otherwise. Graph this. One thing you notice from the graph is that this is all perfectly symmetrical and perfectly balanced around $ x=0.$ And hence, we [i]must[/i] have that $ E(X)=0.$ I'm so sure of that that it's a waste of time to write the integral, but just so you can see it:
\begin{align*}E(X)&=\int_{-2}^2x\cdot\frac{|x|}4\,dx\\
&=\int_{-2}^0-\frac{x^2}4\,dx+\int_0^2\frac{x^2}4\,dx\\
&=-\frac{8}{12}+\frac{8}{12}=0\end{align*}


What volume would 32.0 grams of N0[size=59]2[/size] gas occupyat 309 kPa and 18.0 degrees celcius?

Again, as the molar mass of nitrogen dioxide is 46.0 g/mol, the number of moles is

$n(NO_{2}) = \frac{32.0g}{46.0g/mol}= 0.70 mol$.

Now use the perfect gas law:

$V = \frac{nRT}{P}= \frac{0.70mol\cdot0.0820578LatmK^{-1}mol^{-1}\cdot(273.15+18.0)K}{\frac{309kPa}{101.325kPa/atm}}= 5.48 L$.


geometry
3D geometry
number theory unsolved
number theory

[quote="skywalkerJ.L."]Prove that for any two distinct positive perfect cubes there is a perfect square between them.[/quote]

Assume that the statement is wrong. Then
$ \forall x \in N \exists k \in \{0,1,2,3,... \}$ so that $ x^3 \ge k^2$ and $ (x \plus{} 1)^3 \le (k \plus{} 1)^2$

$ \Rightarrow x^3 \plus{} 3x^2 \plus{} 3x \le k^2 \plus{} 2k \le x^3 \plus{} 2x^2$ , contradiction.

So the statement is true.

(Edited)

[quote="stephencheng"][quote="skywalkerJ.L."]Prove that for any two distinct positive perfect cubes there is a perfect square between them.[/quote]

Assume that the statement is wrong. Then
$ \forall x \in N \exists k \in \{0,1,2,3,... \}$ so that $ x^3 \ge k^2$ and $ (x \plus{} 1)^3 \le (k \plus{} 1)^2$

$ \Rightarrow x^3 \plus{} 3x^2 \plus{} 3x \le k^2 \plus{} k \le x^3 \plus{} x^2$ , contradiction.

So the statement is true.[/quote]

Excuse me,how do you get $ x^3 \plus{} 3x^2 \plus{} 3x \le k^2 \plus{} k \le x^3 \plus{} x^2$?

Isn't $ x^3\plus{}3x^2\plus{}3x\le k^2\plus{}2k$?


calculus
integration
factorial
number theory
prime factorization
prime numbers

Bertrand's postulate has definitely been proven, lots of times (although not by Bertrand).

[hide]The problem, so everyone remembers, is $m!=n^2$.

In the prime factorization of m, there are a number of prime factors.

For there to be a number $n^2=m!$, $m!$ must have a multiple of 2 for each exponent of each prime factor. Else there is no number such that $m!=n^2$. Try to find an integral  square root of $2^3\times 3^2$.

We must have 2 multiples of every prime in m! for the exponent to be a multiple of 2. Unless (1,1), which works.

So at what point do the primes become so spread out that the next prime is more than that prime's value?edit> By Bertrands, never.<edit  So, (1,1) is the only solution.

So that's why Bertrands was mentioned! [/hide]

Billy

MysticTerminator wrote:find all pairs of integers (m,n) such that m!=n:^2:. For the purposes of this problem, you may assume Bertrand's postulate to be true (I think it's been proved, but not sure). Bertrand's postulate says [hide]for any natural number n, there exists a prime between n and 2n[/hide] (i think)

Bertrands postulate actually says, if n > 1, there exists a prime number between n and 2n.

[hide="Solution"] I think the only thing that works is (1,1), because negative numbers can immediately be eliminated (negative numbers factorial are impossible, and the same goes to non-integers) 0! is 1, while 0 2 is 0. So now you have to deal with positive integers. With that postulate, 1,1 works, but with that postulate, you keep getting prime numbers that need to be squared, so the only working combination is (1,1)[/hide]

[quote="jli"][hide="MysticTerminater"]for any natural number n, there exists a prime between n and 2n[/hide]

what if n is 1?[/quote]

then m=1

Don Favor, jli was asking about Bertrand's Postulate... not about the actual problem, because Mystic Terminator forgot to say that n is greater than 1 for Bertrand's postulate.


integration
limit
logarithms
calculus
real analysis
real analysis unsolved

$\int_0^{+\infty}f(x)dx$ is limit,$xf(x)$ is descending in $[0,+\infty)$,
prove $\lim_{x \to +\infty} xf(x)\ln x=0$
I can only prove if  $xf(x)\ln x$ is descending.

Okay...
This solution is similar to the integral problem in http://www.mathlinks.ro/Forum/viewtopic.php?t=35662..

Using Cauchy's Convergence Principle, consider the integral
$\int_{\sqrt x}^x f(t)dt = \int_{\sqrt x}^x tf(t) \cdot \frac{dt}{t}$

thank you,
but I still don't get you meaning.
but I come out a solution with your hint.
let $t=e^{y^2}$

It is clear that $f(x)\geq 0$

$\int_{\sqrt x}^x f(t)dt = \int_{\sqrt x}^x tf(t) \cdot \frac{dt}{t} \geq \int_{\sqrt x}^x xf(x) \frac{dt}{t} = xf(x)\ln \sqrt{x} = \frac{1}{2}xf(x)\ln x$

Since $\int_0^\infty f(t)dt$ exists, then by Cauchy's Convergence Theorem, for $\epsilon > 0$, there exists $G>0$ s.t.
$\left|\int_{\sqrt x}^x f(x)dx\right|<\frac{\epsilon}{2}$ for $x > G^2$

Thus $|xf(x)\ln x| \leq \left|2\int_{\sqrt x}^x f(t)dt\right| < \epsilon$


There is already a topic for this.  Please see [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=200756]this[/url] topic.

Thanks!


You have a regular 52-card deck. How many 3-card hands are there, so that no two cards have the same suit or same number?

[quote="PenguinIntegral"]720*52?[/quote]
Can you explain?

I got:
[hide]$(13\cdot 4)(12\cdot 3)(11\cdot 2)=41184$, hope I multiplied them correctly...

For the first card, you can pick any of 52.
For the second card, you can only pick from 3 suites & 12 numbers.
For the last card, you can only pick from 2 suites & 11 numbers.
[/hide]

(EDIT- I completed my solution.)

My answer and yours doesn't not take into account that you could have one hand, say,
king jack queen and another king queesn jack. These are the same according to the problem but out solutions count them as the same.

Devide your answer by 6.

Oops :oops: Now I see I made a very simple mistake. So is the answer $6864$?

But why is it that your answer is still different? Can you explain where $720$ comes from? :?

[hide="Answer"]

  There are 52 choices for the first card.  The second card can be anything except the 12 other cards in the original  cards suit, the 3 other cards in the original cards number and finally the original card itself.  Therefore, it has $52-12-3-1=36$ choices.  The third card can be anything except the the 12 other cards in the second cards suit, the 3 other cards in the second cards numbers, the second card itself, any everything that applied for card two.  Therefore, it has $52-(16*2)=20$ choices.  This comes to a total of:

$ 52 \cdot 36 \cdot 20 = \boxed{37440}$ combinations.  [/hide]

i dunno if u understood this before because i didnt indicate it, but the order doesnt mean anything. for example 3♠4♣5♥ and 5♥3♠4♣ and 4♣5♥3♠ etc. all count for one kind of hand.

That doesn't make sense p.c, and a problem with those kinds of parameters needs to be announced in the topic. I'm not trying to put you down, I just thought that was a weird way to figure it out, since when you are playing cards a 5(heart)4(club)3(spades) is different than 3(heart)4(club)5(spades). And you must assume that they mean it the normal way unless they say so.

Like I said, this is not an attack on you, just a comment :)

Yeah, we're looking for combinations, not permutations.

i understand what ur talking about Windslicer. However i was not looking for the number of combinations u can put 3 cards, but how many hands u can make.  :) In a hand it doesnt really matter what order they are in. i think PenguinIntegral got it.    :P

Oops :blush:   I forgot about the two cards that are already taken out after the second draw.  It should be $52 \cdot 36 \cdot 22=\boxed{41184}$

I got 41184 too. Same method as frt. What's wrong with that method?


[quote="nhat"]Find the best constant $k \in R$ for which the inequality hold for all non-negative reals $x,y,z,t$ :
                 $(\frac{x^{m}+y^{m}+z^{m}+t^{m}}{4})^{\frac{3}{m}}\geq \frac{xyz+yzt+ztx+txy}{4}$
I guess the answer is $k=\frac{ln64-ln27}{ln4}$ but i only can prove for the case $k \geq \frac{ln9-ln4}{ln3}$.The case $k \in [\frac{ln64-ln27}{ln4},\frac{ln9-ln4}{ln3})$ is mysterious to me.So I'm very pleased and grateful for someone who can help me solve completely this problem. :(  :([/quote]
We see that $\frac{ln64-ln27}{ln4}<\frac 2{3}<\frac{ln9-ln4}{ln3}$. For $m=\frac 2{3}$, the inequality is problem 4 in Algebraic Inequalities (page 385):
If $a,b,c,d$ are positive numbers such that $a^{2}+b^{2}+c^{2}+d^{2}=4$, then
$(abc)^{3}+(bcd)^{3}+(cda)^{3}+(dab)^{3}\leq 4$.
We can completly solve the problem, for $k \geq \frac{ln64-ln27}{ln4}$, using Equal Variable Theorem. According to this Theorem, it suffices to consider $x\leq 1$ and $y=z=t=1$.