Λοιπόν αφού κανείς άλλος δεν ενδιαφέρθηκε κάνω την αρχή.
Θα ξεκινήσω με ανισότητες (το αγαπημένο μου) και με την στοιχειώδη ανισότητα των
Buniakowski-Cauchy-Swartrz .Αν σας αρέσει θα ακολουθήσουν και άλλα πολύ δυσκολότερα όπως το SOS, η Vornicu -Shur , η Karamata, η Muirhead, και τα ακόμη πιο δύσκολα  Half-Right Conceive theorem , Ηαlf-Left _Conceive theorem, N-1Equal variable του Vasc και ότι ποθεί η ψυχή σας

[color=red]Θεώρημα[/color]

Αν $a_1,....,a_n,b_1,....,b_n$  είναι πραγματικοί αριθμοί τότε ισχύει
$(a_1^2+a_2^2+.....+a_n^2)(b_1^2+b_2^2+......+b_n^2)\geq (a_1b_1+a_2b_2.....+a_nb_n)^2$

[color=blue]Απόδειξη[/color]
Υπάρχουν αρκετές. Μία από αυτές είναι η :
Από την ταυτότητα του Lagrange έχουμε ότι $\sum_{i=1}^na_i^2+\sum_{i=1}^nb_i^2-(\sum_{i=1}^na_ib_i)^2=\sum_{1\leq i<j\leq n} (a_ib_j-a_jb_i)^2$

από όπου αφού το δεύτερο μέλος είναι $\geq 0$ προκύπτει η ανισότητα Cauchy-Swartrz

[color=brown]Άμεσες συνέπειες του θεωρήματος[/color]

Το θεώρημα ισχύει για μιγαδικούς αριθμούς όπως και για διανύσματα αλλά και για ολοκληρώματα

Δεύτερη συνέπεια η ανισότητα S. Wagner που υποστηρίζει
 Αν $a_1,....,a_n,b_1,....,b_n$  είναι πραγματικοί αριθμοί και $x\in [0,1]$ τότε ισχύει
$(\sum_{i=1}^na_i^2+2x\sum_{i<j}a_ia_j)(\sum_{i=1}^nb_i^2+2x\sum_{i<j} b_ib_j)\geq (\sum_{i=1}^na_ib_i+x\sum_{i<j}a_ib_j)^2$

Και μια τρίτη συνέπεια η γνωστή και ως ανισότητα Andreescu που υποστηρίζει
Αν  $a_1,....,a_n,b_1,....,b_n$  είναι πραγματικοί αριθμοί τότε ισχύει και οι $b_i>0$
τότε ισχύει
$\frac{a_1^2}{b_1}+......+\frac{a_n^2}{b_n}\geq\frac{(a_1+....+a_n)^2}{b_1+...b_n}$

 [color=yellow]Ασκήσεις που έχει εφαρμογή η Cauchy-Swartrz[/color]
Άσκηση 1 .Αν $a,b,c,d,e$ είναι πραγματικοί αριθμοί και
$a+b+c+d+e=8$  και  $a^2+b^2+c^2+d^2+e^2=16$  να βρείτε τη μέγιστη τιμή του e.

Άσκηση 2.
Αν $x,y,z$ πραγματικοί αριθμοί και $x^2+y^2+z^2=2$  τότε νδο
$x+y+z\leq xyz+2$

Άσκηση 3.Αν $a,b,c>0$  νδο
$\frac{a^2}{b}+\frac{b^2}{c}+\frac{c^2}{a}\geq a+b+c+\frac{4(a-b)^2}{a+b+c}$

Άσκηση 4. Αν  $x,y,z>0$  και $xyz\geq 1$  νδο
$\frac{x^5-x^2}{x^5+y^2+z^2}+\frac{y^5-y^2}{y^5+x^2+z^2}+\frac{z^5-z^2}{z^5+y^2+x^2}\geq 0$

Άσκηση  5.
Για κάθε $x,y,z,a,b,c>0$  νδο
$\frac{a(y+z)}{b+c}+\frac{b(z+x)}{a+c}+\frac{c(x+y)}{a+b}\geq\sum {\sqrt{(x+y)(y+z)}}-(x+y+z)$

Ελπίζω να μην σας φανούν πολύ πολύ εύκολες , οι περισσότερες είναι πολύ γνωστές αλλά και ωραίες.
Θα βάλλω και λύσεις σε κάτι που θα μήνει άλυτο και αν λυθούν πολύ γρήγορα θα βάλλω και κάτι πιο hard...........
Πείτε μου τη γνώμη σας
Χαιρετώ

Μπράβο Σιλουανέ

πολύ καλή παρουσίαση . Οι ανισότητες όντως είναι γνωστές . Λόγω της δυσκολίας που έχω με το Latex δεν μπορώ να σας στείλω τις λύσεις μου .

Επιτέλους κάποιος ασχολήθηκε και με το δικό μου post , ευχαριστώ Θανάση .Όπως προανέφερα έπεται συνέχεια .
Λοιπόν η τελευταία είναι μια πολύ όμορφη άσκηση και κατασκευαστές είναι οι
Gabriel Dospinescu (harazi) και Titu Andreescu  :D .Νομίζω ότι αυτό τα λέει όλα.
Να η λυση της τελευταίας (δίνω λύση γιατί μάλλον αύριο θα γράψω κάτι καινούργιο)

$\frac{a(y+z)}{b+c}+\frac{b(z+x)}{a+c}+\frac{c(x+y)}{a+b}=(a+b+c)(\frac{y+z}{b+c}+\frac{z+x}{c+a}+\frac{x+y}{a+b})-2(x+y+z)\geq\frac{1}{2}(\sqrt{y+z}+\sqrt{z+x}+\sqrt{x+y})^2-2(x+y+z)=\sum\sqrt{(x+y)(x+z)}-(x+y+z)$

Σιλουανε ευχαριστώ.
Actually, την έλυσα χθες στον ύπνο μου (υπνοπαιδεία αν έχεις ακούσει  :D ).
ομοια λύση με την δικιά σου!

Τι προτίθεσε να βάλεις?

Πράγματι στον ύπνο σου μπορείς να λύσεις και δύσκολες ασκήσεις .
Κάτι στις ανισότητες θα βάλλω .Εσύ τι θα προτιμούσες ?? :D

Nαι καλη ιδέα.
κατι περι Cedasev, αναδιαταξη, σουρ εικάζω..... :D

Σε επόμενη φάση μπορεις  να βάλλεις στοιχειώδη γεωμετρία....... ;)

[quote="silouan"]Άσκηση 2.
Αν $x,y,z$ πραγματικοί αριθμοί και $x^2+y^2+z^2=2$  τότε νδο $x+y+z\leq xyz+2$[/quote][url=http://www.artofproblemsolving.com/Forum/viewtopic.php?f=371&t=263608]εδώ[/url]


Enuntul acestei probleme este urmatorul:

Fie O centrul cercului circumscris triunghiului oarecare ABC(neisoscel).
Demonstraţi că punctele în care tangentele în vârfurile A, B şi C intersectează
paralelele duse prin punctul O la dreptele BC, CA şi respectiv AB,
sunt trei puncte coliniare.

Solutiile promise le gasiti in attachment!

Foarte frumoasa solutia cu inversiune.

O alta metoda:

Fie, utilizand aceleasi notatii ca in desenul dumneavoastra, $ M$ punctul de intersectie dintre paralela prin $ O$ la $ BC$ cu tangenta in $ A$ la circumscris. Cum $ A$ apartine polarei $ d$ a lui $ M$, iar $ MO\| BC\perp d$, rezulta deci ca polara lui $ M$ fata de circumscris este chiar inaltimea din $ A$ a triunghiului. Cum insa inaltimile sunt concurente, rezulta deci ca punctele în care tangentele in varfurile $ A$, $ B$ şi $ C$ intersecteaza paralelele duse prin punctul $ O$ la dreptele $ BC$, $ CA$ si respectiv $ AB$ sunt coliniare.


What is the number of square inches in the area of a circle that has a circumference of 12$ \pi$ inches? Express your answer in terms of $ \pi$.

$ 2 \pi r \equal{} 12 \pi \implies r \equal{} 6$

$ \therefore \pi r^2 \equal{} \pi 6^2 \equal{} \boxed{36 \pi}$


reason why in the presence of trace of peroxide or in the presence of UV light $1-Octene$ reacts with
a.$CHCl_{3}$ to give $1,1,1-tricholorononane$
b.$CHBr_{3}$ to give $1,1,3-tricholorononane$

also predict the product for it's reaction with
$CBrCl_{3}$

[hide="ans"]$1,1,1-trichloro-3-bromononane$[/hide]

Presence of peroxides or irradiation with UV light are conditions favourable to the formation of radicals. Peroxides have the general structure ROOR', and because the oxygen-oxygen single bond is weak, it breaks in an homolytic fashion quite easily to produce radicals:

$RO-OR \longrightarrow 2RO \cdot$.

Now, the key to understand the products from the radical addition of halomethanes to alkenes are the homolytic bond dissociation energies (HBDE) of the relevant bonds. Tipically, one have the following order for the HBDEs:

$C-H > C-Cl > C-Br$.

This explains why in bromoform (CHBr3) the bond that is preferantly broken is C-Br and not C-H, because the first requires less energy to cleave than the second. For the chloroform (CHCl3) one should expect on this basis alone that it would be a C-Cl bond to break, and not C-H. But in this particular case the formation of the trichloromethyl radical is favourable because the unpaired electron can be delocalised to the chlorine atoms, renderind it more stable (a similar process would be much more difficult for bromoform, because, since the bromine atoms are much larger, orbital superposition would be much less efficient). So in the case of CHCl3 the reaction that occurs is:

$Cl_{3}CH+RO \cdot \longrightarrow Cl_{3}C \cdot$
$Cl_{3}C \cdot+CH_{2}=CH-C_{6}H_{13}\longrightarrow Cl_{3}C-CH_{2}-CH(C_{6}H_{13}) \cdot$
$Cl_{3}C-CH_{2}-CH(C_{6}H_{13}) \cdot+Cl_{3}CH \longrightarrow Cl_{3}C-CH_{2}CH_{2}-C_{6}H_{13}+Cl_{3}C \cdot$.

For CHBr3 the reaction is:

$Br_{3}CH+RO \cdot \longrightarrow Br_{2}HC \cdot$
$Br_{2}HC \cdot+CH_{2}=CH-C_{6}H_{13}\longrightarrow Br_{2}CH-CH_{2}-CH(C_{6}H_{13}) \cdot$
$Br_{2}CH-CH_{2}-CH(C_{6}H_{13}) \cdot+Br_{3}CH \longrightarrow Br_{2}CH-CH_{2}-CHBr-C_{6}H_{13}+Br_{2}HC \cdot$.

In the reaction with CBrCl3, the HBDEs between C-Cl and C-Br bonds comes into play:

$Cl_{3}CBr+RO \cdot \longrightarrow Cl_{3}C \cdot$
$Cl_{3}C \cdot+CH_{2}=CH-C_{6}H_{13}\longrightarrow Cl_{3}C-CH_{2}-CH(C_{6}H_{13}) \cdot$
$Cl_{3}C-CH_{2}-CH(C_{6}H_{13}) \cdot+Cl_{3}CBr \longrightarrow Cl_{3}C-CH_{2}-CHBr-C_{6}H_{13}+Cl_{3}C \cdot$.


Euler
geometry
circumcircle
incenter
geometry proposed

Geom 2

The tangents at B and C to the circumcircle of ABC intersect at A1. We obtain B1 and C1 by the same way. A?, B?, C? are the feet of the altitudes of ABC. Prove that the lines A1A?, B1B?, C1C? and Euler?s line are all concurrent.

It is very easy to see, that A?B?||A_1B_1,B?C?||B_1C_1,
A?C?||A_1C_1. So there exist a homotetion H: H(A?)=A_1,H(B?)=B_1,
H(C?)=C_1. We just need to prove, that the center of H lies on Euler's
line of ABC (denote it l). The circumcenter of A?B?C? is the center of
EUler's circle of ABC and lies on l. So we need to prove, that the the
circumcenter O of A_1B_1C_1 lies on l. Let I -- circumcenter of ABC
(and incenter of A_1B_1C_1), F-- orthocenter of ABC.
A_1I, B_1I, C_1I meet circumcircle of
A_1B_1C_1 in A_2, B_2, C_2. Then there is a homotetion H_1 such that
H_1(A_2)=A, H_1(B_2)=B, H_1(C_2)=C. Then H_1(O)=I, H_1(I)=F, so O,I,F lies on
one line. But I,F lies on l, so O\in l.


You may say $\sqrt{1}=1$, but in this case, $\sqrt{-1.-1}\neq i.i$. You cant just use i to replace $\sqrt{-1}$ since there is another root -i. $\sqrt{x}$ of positive real is defined as only the positive root, but $\sqrt{x}$ of negative real is not defined to be only one root, but two of them.


[quote="risinglegend"]which of the following expression s are perfect squares?
Write the expressions that are perfect squares in the form(x+k)^2.
a)x^2+6x+9
b)x^2+10x+5
c)x^2-x+1/4
d)x^2-3x=9/4
e)x^2-5x+25/2
f)x^2+4x+4.

Tats all^^, thanxx[/quote]

well first let's all put these in LaTeX so

$x^{2}+6x+9$
$x^{2}+10x+5$
$x^{2}-x+\frac{1}{4}$
$x^{2}-3x=\frac{9}{4}$
$x^{2}-5x+\frac{25}{2}$
$x^{2}+4x+4$

now let's tackle a)

$x^{2}+6x+9$

now this equation is in the form

$ax^{2}+bx+c$ (do you see what a, b, and c equal?)

now b=6. to see if they are perfect squares

take

$\left(\frac{b}{2}\right)^{2}$

if this equals c then it is a perfect square

to write the equation in the form of $(x+y)^{2}$ take half of b and that is your y

so for letter a

$x^{2}+6x+9 = (x+3)^{2}$

I usually factor those types of problems. Your method is sort of like factoring right?

$x^{2}+10x+5$
$x^{2}-x+1/4$
$x^{2}-3x=9/4$
$x^{2}-5x+25/2$
$x^{2}+4x+4$

let's find an example that is not a perfect square
correct! $x^{2}+10x+5$ is one of them.

just take half of $b$ as he mentioned.
so it would be

$(x+5)^{2}$
but that gives us $x^{2}+10x+25$
so just subtract 20.

$(x+5)^{2}-20$ is our desirable answer

[quote="risinglegend"]hmm, guys i dun reali understand the explanation. sry. but i dun noe wad to do with the question and to arrive at the answer. can u guys explain clearer? ty lots[/quote]

You need to determine whether the expression is a perfect square, then you have to express that as a perfect square if possible.

In short, this is the pattern for the perfect square with coefficient of $x^{2}$ $1$:

$x^{2}+2px+p^{2}=(x+p)(x+p)=(x+p)^{2}$

And you should notice whenever $p$ is substituted with some variable (it is simple factoring).

For example,

$x^{2}+6x+9$
$=x^{2}+2 \cdot 3x+3^{2}=(x+3)(x+3)=(x+3)^{2}$.

For longer explanation:
[hide]The expressions all have the form $x^{2}+ax+b$.

To arrive at the answer, you first notice the square root of $b$; that is, the constant term without $x$.

Here is one example:

$x^{2}-x+\frac{1}{4}$

Since you know that $\frac{1}{4}=\left (\frac12 \right)^{2}$, you can say that

$x^{2}-x+\frac14=\left( x-\frac12 \right)^{2}$

And then you can check by actually expanding the right side, which should be easier to understand.

$\left( x-\frac12 \right) \left(x-\frac12 \right)=x^{2}-x+\frac14$ indeed.[/hide]

If you expand it about some hundred times for practice  :P , you will immediately be able to notice what the coefficient $a$ should be if it is a perfect square: it is $\pm 2\sqrt{b}$. If not, then it is not a perfect square.

[hide="solutions for other problems"]
b) $x^{2}+10x+5=x^{2}+2 \cdot 5 \cdot x+5$
But the last constant term should have been $25$, since $5$ is in the middle - it was just to test the students. It is not a perfect square. Instead, the perfect square will be $(x+5)^{2}=x^{2}+10x+25$.

d) $x^{2}-3x=\frac94$
Umm, it should not be an equation. You should have made a typo. Isn't $=$ supposed to be a $+$?

e) $x^{2}-5x+\frac{25}{2}=x^{2}-2 \cdot \frac52 \cdot x+\frac{25}{2}$.
But the last constant should have been $\left( \frac{5}{2}\right)^{2}=\frac{25}{4}$. So it is not a perfect square. Instead, it would have been $x^{2}-5x+\frac{25}{4}$.

f) $x^{2}+4x+4=x^{2}+2 \cdot 2 \cdot x+2^{2}$
Nice! This simplifies to $(x+2)^{2}$.
Or you can factor: $=(x+2)(x+2)=(x+2)^{2}$.[/hide]

Basically the quadratic equation they're talking about is:
[b]To solve a quadratic equation, put it in the "ax^2 + bx + c = 0". In your case, it's the reverse. Basically the formula is
(a + b)^2 = a^2 + 2ab + b^2.[/b]

To put it very, very simply: In a perfect square, the number that's added to the x's must be the sqaure of half of the number that x is multiplied by.

Or you can use Viète theorem to check. If you want to know whether $ax^{2}+bx+c$  is a perfect square or not, let's check the value of $b^{2}-4ac$. If $b^{2}-4ac=0$ , so $ax^{2}+bx+c$ is a perfect square. (Remember $a;b;c;x \in Z$)


integration
calculus
calculus computations

If f'(x) = -3+4x and f(1) = 3, then f(-1) = ?
I am completely lost on this one, so any help would be nice :)

do you integrate the f'(x)?
after integrating that, i get f(x) = 2x^2-3x+C
is that right?

$ f(1)\minus{}f(\minus{}1)\equal{}[f(x)]_{\minus{}1}^1 \equal{}\int_{\minus{}1}^1 f'(x)dx$

$ \equal{}\int_{\minus{}1}^1 (\minus{}3\plus{}4x)dx\equal{}2\int_0^1 (\minus{}3)dx\equal{}\minus{}6$.

$ \therefore f(\minus{}1)\equal{}f(1)\minus{}(\minus{}6)\equal{}9.$

[quote="kunny"]$ f(1) \minus{} f( \minus{} 1) \equal{} [f(x)]_{ \minus{} 1}^1 \equal{} \int_{ \minus{} 1}^1 f'(x)dx$[/quote]

That is a very good way-you need not find the value of $ c$!


Prove that :$ tan\frac{3\pi}{11}\plus{}4sin\cdot\frac{2\pi}{11}\equal{}\sqrt{11}$

Two proofs of the identity above are presented [url=http://arxiv.org/PS_cache/arxiv/pdf/0709/0709.3755v1.pdf]here[/url].


algebra
polynomial
quadratics
Galois Theory
superior algebra
superior algebra unsolved

Does there exist a complex number $ a$ s.t $ a$ and $ a \plus{} 1$ have the same minimal polynomail over the rational numbers?

My answer is yes.  But can someone give me an example, like what a satisfies this condition?  Or any hint at a proof, if it's impossible to find any.

Why do you already have an answer if you have no proof?  :D

It's impossible.  I can't see an "elementary" way to see it, but it's immediate if you consider the fact that there is some automorphism of the splitting field that takes $ a\to a\plus{}1$.

Ok, I was thinking maybe we can take $ f(x) \equal{} x^2 \minus{} x \minus{} a$ for some $ a$, and then use Frobenius endomorphism to show that if u is a root of f, then u+1 is also a root of f

There is no such thing as the Frobenius endomorphism for fields of characteristic $ 0$. :)  Anyway, it is very easy to show that a quadratic would not work, since the sum of the roots must be rational.

ok, so does anyone know whether it exists and how to prove it

I proved that it doesn't exist in my first response. :)

To be completely clear: the automorphism exists because the Galois group is transitive. (this is a basic fact in Galois theory)  Hence, iterating the automorphism, we find that $ a \plus{} 2$, $ a \plus{} 3$, etc., are roots of the same polynomial.  Since $ \mathbb{C}$ has characteristic $ 0$, this is impossible.

oh, i see.  it was so much more basic than i was expecting.  gee, thanks alot!  I was kinda thinking too hard.

this does not have anything to do with galois theory and works in every field extension of char. 0.

if $ f$ is the minimal polynomial of $ a$ and $ a\plus{}1$, then $ f(x)\equal{}f(x\minus{}1)$, and $ f$ has infinitely many zeroes $ a,a\plus{}1,a\plus{}2,...$, a contradiction.


function
advanced fields
advanced fields unsolved

Assume $f: X\to Y$ is a function which sends connected sets to connected sets and compact sets to compact sets. Does this follow that $f$ is continuous? Clearly not in general! But what conditions should be put on $X$ or $Y$ to make it right?

There is a proof [url=http://www.mathlinks.ro/Forum/viewtopic.php?t=33685]here[/url] for the case when $X,Y$ are Euclidean (i.e. $X=\mathbb R^{n},Y=\mathbb R^{m}$ for some $n,m\ge 1$).

Thanks for the link grobber. Do you think that the same method works in metric spaces?(It of course needs some modification.) Or it fails?


Of the following expressions the one equal to $ \frac{a^{\minus{}1}b^{\minus{}1}}{a^{\minus{}3} \minus{} b^{\minus{}3}}$ is:

$ \textbf{(A)}\ \frac{a^2b^2}{b^2 \minus{} a^2}\qquad
\textbf{(B)}\ \frac{a^2b^2}{b^3 \minus{} a^3}\qquad
\textbf{(C)}\ \frac{ab}{b^3 \minus{} a^3}\qquad
\textbf{(D)}\ \frac{a^3 \minus{} b^3}{ab}\qquad
\textbf{(E)}\ \frac{a^2b^2}{a \minus{} b}$

[hide="Solution"]
We know that $ \dfrac{a^{ \minus{} 1}b^{ \minus{} 1}}{a^{ \minus{} 3} \minus{} b^{ \minus{} 3}} \equal{} \dfrac{1}{\dfrac{1}{a^3} \minus{} \dfrac{1}{b^3}} \cdot \left( \dfrac{1}{ab}\right)$

And $ \dfrac{1}{\dfrac{1}{a^3} \minus{} \dfrac{1}{b^3}} \equal{} \dfrac{1}{\dfrac{b^3 \minus{} a^3}{a^3b^3}} \equal{} \dfrac{a^3b^3}{b^3 \minus{} a^3}$

So $ \dfrac{1}{\dfrac{1}{a^3} \minus{} \dfrac{1}{b^3}} \cdot \left( \dfrac{1}{ab}\right) \equal{} \dfrac{a^3b^3}{b^3 \minus{} a^3}\cdot \left(\dfrac{1}{ab}\right) \equal{} \dfrac{a^2b^2}{b^3 \minus{} a^3}$

Hence the answer is $ \boxed{\mathbf{(B)}}$
[/hide]


[quote="Michael Niland"]a,b,c and d are real numbers satisfying a+b+c+d =0

Prove or disprove that abc +bcd +cda +dab <0[/quote]
This is not true: $a=b=c=d=0$ or $a=2,$ $b=c=-1,$ $d=0.$ ;)


I was flipping through my notebook of old contests today and I came upon this problem that seems that it is right but I got it wrong.

You'll have to look at the problem yourself because I can't put the graph in the post.

I said '97. Isn't this right?

it would help if you had the question, even if you don't have the graph

I believe that he means #5 in the .PDF file attached below.
The answer to the question [i]is[/i] '98 (as stated in the answer booklet), because the numbers are referring to the [b]end of year[/b] sales totals. The greatest difference was between the [i]end[/i] of 1997 and the [i]end[/i] of 1998, which means that 1998 had the largest increase.

EDIT: OK, somehow, the attachment is back now.

Oh yeah...I put '97 for that... I got #4, #5, and #28, wrong on that School Sprint Round...

Yeah, as Nebula said, it is just pretty much because it is the end of the year, so it is '98.

[quote="bpms"]What .pdf file?[/quote]
It happens almost every time... It disappeared.  :ninja:

[quote="anirudh"]i saw it today though...about 15 minutes ago...[/quote]
Yeah, it works for a while (I downloaded), but then disappears.

EDIT: nebula42's is back up. :rotfl:

Yes, it has disappeared. I've made a topic about it in the "New Server Bugs" forum [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=111577]here[/url], in case you're interested.

But seriously, let's get back onto the actual topic of the thread.  :wink:

And by the way, thanks for putting it back up.

EDIT: Wow, my old one is back now...  :huh:  :what?:


Find all positive integers n such that n multiplicated twice by itself(don't know how to superscript 3) + 27 is a perfect square.I ve just determined what vales the last digit of n can take but i dont think this is useful.So i dont have any idea.What about you??

Source:4th Junior Balkan,second problem

You can factorize the equation to a^2=(n+3)(n^2-3n+9). These both factors are coprime, because gcd(n+3,n^2-3n+9)|gcd(n^2+3n,n^-3n+9) and so gcd(n+3,n^2-3n+9)|gcd(n+3,6n-9)|gcd(27,n+3)=1 (n cannot be divisible by 3, if 3|n n=3k, 27k^3+27 and also 3k^2+3 are squres but 3k^2+3 is divisible by 3 and not by 9 ). So (n^2-3n+9) must be a square. For n>8 you have: (n-1)^2>n^2-3n+9>(n-2)^2. You must now only check the cases n<9.

[quote="Iris Aliaj"]Find all positive integers n such that n multiplicated twice by itself(don't know how to superscript 3)[/quote]

powers can be written in 2 ways:
- for 2 and 3 there is a button for that (not recommended): like n and n
- for just any power the international notation is n^3. or a^b. When you use latex you can make that even more attractive to the eye by putting dollar signs around it. $n^3$ and $a^b$.

[quote="EvaristeG"][quote="Misha123"] (n cannot be divisible by 3, if 3|n n=3k, 27k^3+27 and also 3k^2+3 are squres but 3k^2+3 is divisible by 3 and not by 9 )[/quote]

I was browsing the forum when I saw this topic and, as I came into this problem some days ago, I was surprised for this brief solution...I don't know whether I don't understand something or there's really something wrong with it :
$n=3k \Rightarrow 27k^3+27 $ is a square $\Rightarrow 3k^2 + 3$ is a square, but $3k^2+3$ cannot be divisible by 9, so it cannot be a square.
If the $3k^2+3$ is a typo and should be $3k^3+3$, I can't see why it cannot be divisible by 9; if it's not a typo, I don't see why $27k^3+27$ is a square should imply that $3k^2+3$ is a square.
Can anyone explain me ?

Thanks[/quote]

Because if 9k is a perfect square, then k is necessarily a perfect square.

Well, but if $27k^3+27$ is a square, why on earth should also $3k^2+3$ be a square. Pay attention to the exponents!!
Following your answer, I can say that
$27k^3 + 27$ is a square implies $3k^3+3$ is a square
But there's nothing contraddictory in this conclusion : 3k^3+3 can be divisible by 9, depending on the choice of k. If we pick $k=3h+2$, for example...

So?
Am I wrong or is the answer given to this problem that's wrong ?
I'll repeat : look at the exponents of $k$.


You want to see a chain reaction? Teach your school The Game and then one day just start screaming "I JUST LOST" at the top of your lungs. it's quite fun to see the results. (but of course some people will think you are an idiot)

[quote="ashpotter"]You want to see a chain reaction? Teach your school The Game and then one day just start screaming "I JUST LOST" at the top of your lungs. it's quite fun to see the results. (but of course [b]everybody[/b] will think you are an idiot)[/quote]

Fixed, and it gets old very very fast.

LOL, I was on this forum, and this guy said, "Testing new sig." And his sig said, "You just lost the game. Then, a bunch of people posted, "I just lost the game."

lol, we did this at mathpath after history of math.

but then some people came up with other random responses and it wasn't funny anymore.

[size=200]I just lost The Game again.[/size]

I lost.

[url=http://www.losethegame.com/]And for those who have no clue what we're talking about[/url]

I play with a 4th rule that few people know:

4.  If you say the word "Moo" without thinking about The Game, you win The Game.

I've won a couple times.  However, the problem with The Game is that so many connections are made that pretty soon, you're always thinking anbout The Game.  For example:

1=2's avatar
V
The Simpsons
V
The Simpsons Movie
V
Spiderpig
V
Pigs
V
Cows
V
The Game

I lost,

The point of the game is not to think about it.

1. You are always playing,

2. When you think of the game you lose,

3. Losing must be announced like "I lost"

I lost.

[quote="randomdragoon"]You can also win if you successfully die without thinking about the game in the past 5 minutes.[/quote]

die????

Yeah,  I don't think anyone is that ready too win the game.

I losted.

Of course if you lose on your deathbed and you die within five minutes, you lose for all eternity.

I was trying to annoy someone by siming the zoopals song. And one person right next to me said that i just won the game. It is possible to win. But there was this one idiot who right after i taught him the game (probably about 20 second interval) said moo right out of context. Then he said "look, I just won the game!!!"

I told him that that's not possible and he was accually thinking of the game when he said moo. But, he insisted that he was not thinking of the game.

Anyways, I just lost

Also, it just has to be verbal, you don't have to type it out every single time you visit this thread.

no, you win if you die while eating an orange, without thinking about the game.

I lost the game! I've lost the game infitite times.

This thread always makes me lose.  It's annoying.

Dood i just wawn the game man.... pr0 l337 pwnage yo.

[size=200] I won the freaking game dog![/hide]

Now I gotta go smoke a joint with my friend Ralph cuz I loike nookies.

On a MOP test, I realized, with little time left, that my proof did not work at all.  So instead of writing $ \mathbb{QED}$, I wrote $ \mathbb{I LOST}$.

[quote="The Zuton Force"]This is why local contests own MOP... you can BS in the last minute and get points  :D (if you BS tastefully)[/quote]

More like you can get all the right answers and get points unlike MOP

how does hangout have 0 posts. cant a mod just delete this topic to refrain the entire AoPS population from losing

yeah, that would be nice.

HangOut has 0 posts cuz he/she only posts in the games and fun factory, where post count doesnt count (otherwise my post count would be about 300 already)

ya cuz i dont have ant classes so i wont know wat to post u know?


algebra
polynomial
calculus
integration
trigonometry
Galois Theory
algebra unsolved

for b=(a_1)^(1/k1)+(a_2)^(1/k2)+??+(a_n)^(1/k_n),ai,ki are natural numbers
prove that there exist a integer coefficient polynomial whose first coefficient is 1,
and b is a root of this polynomial.
I think it is a hard and useful problem. :)
who can help me?

It is well known that the algebraic inetgers (i.e. numbers which are the roots of a monic polynomial with integral coefficients) form a ring. Yopur number $b$ is the sum of numbers which are clearly algebraic integers, so it is an algebraic integer too.

I have a proof not using rings inspired by [url=http://www.mathlinks.ro/Forum/viewtopic.php?t=4496]www.mathlinks.ro/Forum/viewtopic.php?t=4496[/url]

Let $\omega_m$ be $\sin(\frac{2\pi}{m})i+\cos(\frac{2\pi}{m})$ or the first root of unity for $x^m=1$.
Then your polynomial is:
$\prod (x+\omega_{k_1}^{s_1}\sqrt[k_1]{a_1}+\omega_{k_2}^{s_2}\sqrt[k_2]{a_2}+\omega_{k_3}^{s_3}\sqrt[k_1]{a_3}+\ldots+\omega_{k_n}^{s_n}\sqrt[k_n]{a_n})$
where $s_i$ runs over $1,2,\ldots,k_i$, so this product has $k_1k_2k_3\ldots k_n$ terms.

But don't know how exactly to prove that it has integer coefs. I don't "see" to it :)

Grobber. You mean that if you have polynomials P(x),Q(x) with integer coeficients and a,b numbers such that P(a)=0, Q(b)=0, then there's a polynomial R(x) with integer coefficients so that R(a+b)=0 ? Can you show how to construct such polynomial R?

Yes, that is what I mean, but it's not worth actually going thrugh the trouble of constructing the polynomial. A standard proof of this uses some facts about field extensions (stuff like $[K(a):K],[K(b):K]$ finite extensions implies that $K[a,b]$ is also a finite extension, so $a+b,ab$ are also algebraic over $K$). Try googling "field extension", "Galois theory", "algebraic numbers", "algebraic integers". I'm sure you'll find more info than I could ever provide :).

What about these constructions:

Let $P(x)$ and $Q(x)$ be polynomials with integer coefficients. Then we can write them as:

$P(x)=c\prod_{i=1}^n(x-a_i)\\
Q(x)=d\prod_{i=1}^m(x-b_i)$

Now we construct polynomials $S^+(x)$ and $S^{\times}(x)$ this way:

$S^+(x)=\prod_{i=1}^n\prod_{j=1}^m(x-a_i-b_j)\\
S^{\times}(x)=\prod_{i=1}^n\prod_{j=1}^m(x-a_ib_j)$

They have rational coefficients ($a_i$'s and $b_j$'s are symetrical so we can write it using symetric polynomials and they are rational) and they have roots that are sums or products of roots of $P(x),Q(x)$. Isn't this what we wanted? Is this idea ok? Can it be rewritten to acceptable proof?


How do you enter the online class when it starts???

You click on the the button that says Classroom on the left of your screen.

yah...why?...does it not work for you? :huh:

Well, you need [url=http://www.java.com/en/download/windows_xpi.jsp?locale=en&host=www.java.com]Java[/url]

If it doesn't work, maybe you're in a different time zone. Just a thought.

[quote="ZhangPeijin"]10minutes before your class starts, click the classroom tab on the left side of the screen abd you will go into the online classroom if you have Java.
BTW, multiple exclamation points are a sign of a diseased mind. :P[/quote]
WAIT, REALLY!!!!!!!!!!!!!!!!!

[quote="ZhangPeijin"]10minutes before your class starts, click the classroom tab on the left side of the screen abd you will go into the online classroom if you have Java.
BTW, multiple exclamation points are a sign of a diseased mind. :P[/quote]

Ten minutes is good, but you can get in any time during the class.

Don't go in at 8:59 PM and expect to find out what's going on.

!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!

Is it me or is every post in this thread rated "Spam"?
People abuse that thing way too much...  :ninja:

[quote="ZhangPeijin"]10minutes before your class starts, click the classroom tab on the left side of the screen abd you will go into the online classroom if you have Java.
BTW, multiple exclamation points are a sign of a diseased mind. :P[/quote]
Aaah indeed - in the words of Terry Pratchett: "...all those exclamation marks, you notice? Five? A sure sign of someone who wears his underpants on his head.'" I'm afraid you guys go well past that limit. :P. Maybe it decreases blood flow or something...


What's the diffeence between the various programs for Latex in Windows like MikTex, Web2C , teTeX , etc... Which one is best for windows???


geometry
circumcircle
geometric transformation
homothety
geometry proposed

[quote="77ant"]Three circles touch one another at $ A$, $ B$, $ C$ respectively.
A point $ D$ is on one of them (like below figure). $ DA$ meets one of other two circles at $ E$, and $ EB$ meets the other at $ F$, and $ FC$ meets the first circle at $ G$ (please look at below figure).
Prove that $ D$, $ O$, $ G$ are collinear. ( $ O$ is the circumcenter of the first circle )

Is it well-known ?  :([/quote]
Let $ O_1,O_2$ are center of $ (EAB)$ and $ (CBF)$
$ f \equal{} V_{C}^{\frac { \minus{} R}{R_2}}.V_{B}^{\frac { \minus{} R_2}{R_1}}.V_{A}^{\frac { \minus{} R_1}{R}}$
$ f: D \minus{} \minus{} > G \\
: O \minus{} \minus{} > O$

But $ \frac { \minus{} R}{R_2}.\frac { \minus{} R_2}{R_1}.\frac { \minus{} R_1}{R} \equal{} \minus{} 1$
so $ f$ is half turn with $ O$ is center
So $ D,O,G$ are concurrent
:D

Dear 77ant and Mathlinkers,
your problem is a generalization coming from Yaglom.
See my website : http://perso.orange.fr/jl.ayme    vol.1  "A propos du théorème de Boutin".
Sincerely
Jean-Louis

(For easier writing) let's call O, N and P the centers of the circles including D and A, A and E and B and F. Due to homothety around A, EN || OD, then FP || EN and, finally GO || FP, hence
OD || GO and GD is a diameter of the circle O.

Best regards,
sunken rock


Express $ \minus{}3.5\plus{}(2.25\plus{}4.75) \div \minus{}3.5$ in simplest form.

Remember the order of operations!

$ \minus{}3.5\plus{}(2.25\plus{}4.75)\div\minus{}3.5$
$ \equal{}\minus{}3.5\plus{}7\div\minus{}3.5$
$ \equal{}\minus{}3.5\minus{}2$
$ \equal{}\fbox{\minus{}5.5}$


inequalities
function
inequalities unsolved

Let $a;b;c>0$.Prove that:
$\sqrt{\frac{a}{4a+4b+c}}+\sqrt{\frac{b}{4b+4c+a}}+\sqrt{\frac{c}{4c+4a+b}}\leq 1$

[quote="VANCHANH123"][quote="Svejk"]By Jensen :$\frac{1}{3}\sum \sqrt{\frac{a}{4a+4b+c}}\leq \sqrt{\frac{1}{3}\sum \frac{a}{4a+4b+c}}\leq \frac{1}{3}$
\[\leftrightarrow \]
$\sum \frac{a}{4a+4b+c}\leq \frac{1}{3}$
But   $\sum \frac{a}{4a+4b+c}=\sum \frac{a^{2}}{4a^{2}+4ab+ac}\leq \frac{(a+b+c)^{2}}{4\sum a^{2}+5 \sum ab}\leq \frac{1}{3}$
\[\leftrightarrow \]
$\sum a^{2}\geq \sum ab$[/quote]
[b]False[/b][/quote]
Not all! :wink:

Why you said this problem is easy? I think it is hard. I read this problem last week from an articles of Hungkhtn sent to Mathematics and Youth Magazine when he tell me to check the errors. There are two solutions to this one, one by Convex function  (og Hung), but the other of mine is better because it is easier and shorter

(1) Prove that (symmetric inequality)
\[\sum \frac{a(a+b+c)}{(4a+4b+c)(4a+4c+b)}\le \frac{1}{9}.\]

(2) (Mine) Directly expanding to prove that
\[\sum \frac{a}{4a+4b+c}\le \frac{1}{3}.\]

To chienthan: Where do you find this problem?

@VANCHANH123 :Can you point out my mistake  :maybe: ?

You cant aply Cauchy inequality your  usage of the inequality is wrong
see line3 of your proof :(

[quote="GacKiem"]To chienthan: Where do you find this problem?[/quote]
I find this problem in a VietNam math forum!


For each equation of the ellipse find
-the length of the major and minor axis
-the coordinates of the foci

(1)       $4x^2+9y^2-32x+36y+64=0$
(2)       $4x^2+y^2+16x-6y-39=0$

[quote="Jhonny Ruiz"][hide="hint"]Write both equations in...$\frac{x^2}{a^2}+\frac{y^2}{b^2}$
I think.....[/hide][/quote]
yeah, thats the easiest way to do it

[quote="Jhonny Ruiz"]oh but wait I forgot the constant... it should be something like...
$\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$[/quote]
yeah

[/hide]

[quote="math92"]$4x^2+9y^2-32x+36y+64=0$
[/quote]

[hide]Rearrange first:

$4x^2-32x+64+9y^2+36x+36=36$
$4(x-4)^2+9(y+2)^2=36$

Now divide both sides by 36 to get:

$\frac{(x-4)^2}{9}+\frac{(y+2)^2}{4}=1$

From this, the major axis is 6 and the minor axis is 4 and the foci are $(\pm\sqrt5+4, -2)$ (Note: I'm not very sure)

I'm too lazy to do the other one...[/hide]


calculus
integration
trigonometry
calculus computations

In the $ xyz$ space with the origin $ O$, given a cuboid $ K: |x|\leq \sqrt {3},\ |y|\leq \sqrt {3},\ 0\leq z\leq 2$ and the plane $ \alpha : z \equal{} 2$. Draw the perpendicular $ PH$ from $ P$ to the plane. Find the volume of the solid formed by all points of $ P$ which are included in $ K$ such that $ \overline{OP}\leq \overline{PH}$.

The region of $ P$ is the intersection of $ K$ and inside of the paraboloid $ z\equal{}\minus{}\frac{1}{4}(x^2\plus{}y^2\minus{}4)$. So, the volume we want to find is
$ \int_{\minus{}\sqrt{3}}^{\sqrt{3}}\int_{\minus{}\sqrt{3}}^{\sqrt{3}}\int_{0}^{\minus{}\frac{1}{4}(x^2\plus{}y^2\minus{}4)}dzdydx$
$ \equal{}4\int_{0}^{\sqrt{3}}\int_{0}^{\sqrt{3}}\minus{}\frac{1}{4}(x^2\plus{}y^2\minus{}4)dydx\equal{}\boxed 6$.

What's wrong in my approach? There's no miss in calculating...

Although the stated locus is correct, your approach is incorrect because you need to take into consideration the intersection of the volume $ K$ with the paraboloid.  The limits of integration you provided are incorrect.

As mentioned, the locus is the intersection of $ K$ with the paraboloid $ z \le (4\minus{}x^2\minus{}y^2)/4$.  When $ z \equal{} 0$, we obtain the circle $ x^2 \plus{} y^2 \equal{} 4$ which has radius $ 2 > \sqrt{3}$, so there is a portion of the paraboloid above the $ xy$-plane that is not contained in $ K$.  We must first compute the volume of this region.

To this end, we observe that this aforementioned region is composed of four congruent pieces, each of which has volume

$ T \equal{} 2\int_{x\equal{}\sqrt{3}}^2 \int_{y\equal{}0}^{\sqrt{4\minus{}x^2}} \frac{1}{4}(4 \minus{} x^2 \minus{} y^2) \, dy \, dx$

$ \equal{} \frac{1}{2} \int_{x\equal{}\sqrt{3}}^2 \left[(4\minus{}x^2)y \minus{} \frac{y^3}{3}\right]_{y\equal{}0}^{\sqrt{4\minus{}x^2}} \, dx$

$ \equal{} \frac{1}{3} \int_{x\equal{}\sqrt{3}}^2 (4\minus{}x^2)^{3/2} \, dx$

$ \equal{} \frac{1}{3} \left[\frac{1}{4}x(10\minus{}x^2)\sqrt{4\minus{}x^2} \plus{} 6 \sin^{\minus{}1} \frac{x}{2} \right]_{x\equal{}\sqrt{3}}^2$

$ \equal{} \frac{1}{3}\left(\pi \minus{} \frac{7 \sqrt{3}}{4}\right)$.

We must subtract $ 4T$ from the volume of the paraboloid bounded above the $ xy$-plane

$ \int_{r\equal{}0}^2 2\pi r (1 \minus{} r^2/4) \, dr \equal{} 2\pi$,

giving the result

$ \frac{2\pi}{3} \plus{} \frac{7}{\sqrt{3}}$.

I had a mistake in comparing size of the bottom of $ K$$ (\left|x\right|\leq\sqrt {3}, \left|y\right|\leq\sqrt {3})$, and the bottom of the paraboloid $ (x^2 \plus{} y^2 \leq 4)$. The right answer is $ \frac {2\pi}{3} \plus{} \frac {7}{\sqrt {3}}$,atomicwedgie's.


calculus
integration
function
trigonometry
calculus computations

$ f(x)$ is  a continuous function with the piriodicity of $ 2\pi$ and $ c$ is a positive constant number.
Find $ f(x)$ and $ c$ such that $ \int_0^{2\pi} f(t \minus{} x)\sin tdt \equal{} cf(x)$ with $ f(0) \equal{} 1$ for all real numbers $ x$.

Functions $ f(x)$ and $ \sin(x)$ are periodic with period $ 2\pi$, so we can express them as Fourier series, ie,
\[ f(x) \equal{} \sum_{n \equal{} \minus{} \infty}^{ \plus{} \infty}d_ne^{inx},
\]
where $ i$ is as always the imaginary unit, and obviously, $ \sin(t) \equal{} \frac {e^{it} \minus{} e^{ \minus{} it}}{2i}$, and since for any integer $ m$,
\[ \int_0^{2\pi}e^{imx}dx \equal{} \delta_m,
\]
where $ \delta_m$ is the Kronecker delta symbol with value 1 if $ m \equal{} 0$ and value 0 otherwise, we may find trivially that the integral on the RHS of the given equality is
\[ \frac {d_{ \minus{} 1}e^{ix} \minus{} d_1e^{ \minus{} ix}}{2i}.
\]
Therefore, we may find that $ d_n \equal{} 0$ for all $ |n|\neq1$, and that $ d_{ \minus{} 1} \equal{} 2icd_1$, $ d_1 \equal{} \minus{} 2icd_{ \minus{} 1}$, or $ d_1 \equal{} 4c^2d_1$, $ d_{ \minus{} 1} \equal{} 4c^2d_{ \minus{} 1}$.  Then, either $ c \equal{} \frac {1}{2}$, or $ f(x) \equal{} 0$, the second case being the trivial solution.  In the first case, we find that $ d_{ \minus{} 1} \equal{} id_1$, and for any complex constant $ d_1$,
\[ f(x) \equal{} d_1\left(e^{ix} \plus{} ie^{ \minus{} ix}\right) \equal{} d_1\left(1 \plus{} i\right)\left(\sin(x) \plus{} \cos(x)\right) \equal{} d_1'\left(\sin(x) \plus{} \cos(x)\right).
\]
If $ f$ is real, then $ d_1'$ may take any real values, included $ d_1' \equal{} 0$ which leads again to the trivial solution.  So, the general non-trivial solution is that $ f(x) \equal{} k\left(\sin(x) \plus{} \cos(x)\right)$ for any constant $ k$, where we will choose $ k$ real if $ f$ is real, or we may take, in the most general case, $ k$ complex.  It is trivial to show that this function satisfies the given equality.


probability
projective geometry
combinatorics open
combinatorics

(i) Find the cardinality of the largest family of subsets of $ [n]\equal{}\{1,2,\dots ,n\}$ so that intersection of any two subsets of the family is unique. Enumerate them and the maximal sets with same restriction.

Experiments with smaller n reveals that the maximum cardinality of such a family must be $ n\plus{}1$. But I am not sure beyond $ n\equal{}6$.

(ii) Generalizing (i), intersection of any $ k$ subsets of the family is unique.
Enumerate them and the maximal sets with same restriction.

(i) Don't waste your time with trying to prove the $ n \plus{} 1$ guess, as it fails for $ n \equal{} 7$: Consider the Fano plane as a set of $ 7$ points, and its following subsets: every line, the whole plane, and the set of the vertices of the triangle together with its center. All pairwise intersections of these $ 7\plus{}1\plus{}1\equal{}9$ sets are easily seen to be distinct.

Other than that, I fear I can't help much here.

  darij

It seems like you can choose $ 2^{cn}$ subsets (with say $ c \equal{} 1/100$).  Let's choose $ 2^{cn}$ random subsets (say with replacement).  First, the chance that any two will be equal is exponentially small.  Second, if $ A$, $ B$, $ C$, and $ D$ are four random subsets, the chance that $ A \cap B \equal{} C \cap D$ is exponentially small.  Similarly, the probability that $ A \cap B \equal{} A \cap C$ is exponentially small.  Even if we sum over all quadruples and triples, the total probability of being bad is exponentially small.  So there is at least one choice that works.

Sorry, the argument above is without the details, but I think it works.


search
number theory unsolved
number theory

given the sequence
$ a_1\equal{}4, a_2\equal{}22$...
$ a_{n\plus{}1}\equal{}6a_n \minus{} a_{n\minus{}1}$
prove that these existing y such that $ a_n$ is a divisor of $ y^2$+7

[quote="kihe_freety5"]given the sequence
$ a_1 \equal{} 4, a_2 \equal{} 22$...
$ a_{n \plus{} 1} \equal{} 6a_n \minus{} a_{n \minus{} 1}$
prove that these existing y such that $ a_n$ is a divisor of $ y^2$+7[/quote]
Hmm ,for each n there exist an integer y such that $ a_n|y^2\plus{}7$ ? It is not very hard . :wink:
We have $ a_n(a_n\plus{}a_{n\plus{}2})\equal{}a_{n\plus{}1}(a_{n\plus{}1}\plus{}a_{n\minus{}1})\equal{}6a_n.a_{n\plus{}1}$
Therefore $ a_{n\plus{}1}.a_{n\minus{}1}\minus{}a_n^2\equal{}a_{n\plus{}2}a_n\minus{}a_{n\plus{}1}^2,\forall n\geq 2$. Then it is easy to check that $ a_n.a_{n\plus{}2}\minus{}a_{n\plus{}1}^2\equal{}a_1.a_3\minus{}a_2^2\equal{}28$ .
Let $ a_n\equal{}2^s.t$ where t is an odd integer . A old problem shows that the equation $ y^2\equiv \minus{}7 (\mod 2^n)$ always has solution  (you can search on our forum ) .And from equality $ a_{n\plus{}1}.a_{n\minus{}1}\equal{}a_n^2\plus{}28$ we can find an integer y such that  $ t|y^2\plus{}7$ . This problem claim from chinese remainder theorem .

Define the sequence $ (y_n)$ by $ y_1 \equal{} 1$, $ y_2 \equal{} 9$, and $ y_n \equal{} 6y_{n \minus{} 1} \minus{} y_{n \minus{} 2}$ for all $ n \ge 3$.  Then it can be shown that $ y_n^2 \plus{} 7 \equal{} a_n a_{n \minus{} 1}$ for all $ n \ge 1$, so
\[ \frac{y_n^2 \plus{} 7}{a_n} \equal{} a_{n \minus{} 1}.\]


Let $f_{n}: \left[0,1\right]\rightarrow\mathbb{R}$ be defined by $f_{n}\left(x\right)=n^{2}xe^{-nx}$.  Show that $f_{n}\rightarrow 0$ pointwise on $\left[0,1\right]$.


linear algebra
matrix
vector
linear algebra unsolved

whats the number of linear transformations from a linear space of dimension $m$ into one of dimension $n$?

$mn$.  Take the $m\times n$ matrices with one one and the rest zeroes.  There are $mn$ of these, and they form a basis for the space of all linear transformations from one to the other.  Since every linear transformation from a finite dimensional space into a finite dimensional space can be represented as a matrix, we're done.

The number depends on the field. The linear transformations form a vector space of dimension $mn$ (over the same base field).


The larger of two consecutive odd integers is three times the smaller. What is their sum?

$ \textbf{(A)}\ 4 \qquad \textbf{(B)}\ 8 \qquad \textbf{(C)}\ 12 \qquad \textbf{(D)}\ 16 \qquad \textbf{(E)}\ 20$

[hide="hint"]
Let the smaller integer be x and the larger integer be x+2. Then set it up as an equation.
[/hide]

$x+2 = 3x$
$2x = 2$
$x = 1$

So the smaller number is 1, the larger number is thus 3, and the sum is 4. $\boxed{A}$

I liked this problem.  It took me about two seconds.
My work: (in its entirety)
[hide]$1+3=4$[/hide]

This was also AMC12A #3
Since I didn't really have work:
[hide="my thought process"]
$1$ and $3$ work. $\boxed{4}$
[/hide]

Took me exactly 8 seconds on the test. :D

I instantly say 1 and 3 worked.

But I still did the algebra anyway just to make sure  :rotfl:

Just think of numbers 1 2 3 4 5 ......
you quickly see that 3 x 1 = 3
giving the result 3 + 1 =4
Answer is [b](A.) 4[/b]

[hide=Solution]
If we let $2a$ be the sum of the 2 odd numbers, then the problem is equivalent to $a+1=3(a-1)$ which gives $a=2$ so the sum is $2\cdot 2=\boxed{\textbf{(A)}4}$
[/hide]

>15 year bump :o
anyway
[hide=sol]set the smaller number as n. because they are consecutive odd integers, the larger number is n+2.
n+2=3n, so n=1, n+2=3, and the sum of the two numbers is 4[/hide]

oh and maybe it was like the first few problems

We have $a+2=3a\implies 2a=2\implies a=1$.

So the two integers are $1$ and $3$. Their sum is $\boxed{\textbf{(A)}\ 4}$.


when a problem ask for the number of proper factors of a number, lets say 12, would we include 1 or 12?

NO.

A proper factor is a factor other than 1 or the number.

YES and NO.

A proper factor is a factor that is less than the number. Therefore, 1 would be a proper factor, but 12 would not be (in your example).

Yongyi's right (as if you needed to be told that). In definition of perfect numbers.

Usually they tell you "including 1 and the number itself" or "including 1 but not the number" or "including watermelons but not 1 or the number itself".


AMC AIME USA(J)MO USAMO FTW blogs MATHCOUNTS

States is Saturday, and I'm really nervous. Does anyone know what has been the 1st place written score (the highest) in IL MC history?

Also, I have so many nervous things in mind:

1. I'm afraid that I'll make stupid mistakes.

2. I have to really get a 40+ or perhaps a perfect.

3. I really hope I don't get paired with Ravi Jagadeesan before countdown semifinals.

4. I have to make Nats, and if I don't, I'll quit AoPS.

5. I hope suma_milli doesn't get on my nerves during countdown (NO OFFENSE, but I trust her she won't) :rotfl:

6. I'm afraid suma_milli would kill me if I don't make nats, and i'd scream to her if she doesn't make nats (Or maybe not)

7. I really want Ravi's autograph.

8. I'm afraid I won't get enough sleep.

9. It's likely I would cram.

10. It's likely I'd focus too much on AIME or USAMO or USAMTS.

11. It's likely that I'd waste a lotta time on AoPS.

12. It's likely I won't make the semifinals for countdown...

13. It's likely I would have too much blood sugar b/c of rofling with suma_milli (no offense, maybe not)

14. I'm afraid people would make fun of me if I don't make nats.

15. Ravi would probably not find me on FTW if I don't make nats.

16. People would spam my blog while I'm gone.

17. Pythag011 is pressuring me a lot...
nerrrgh... :maybe:  :ninja:  :o

Haha maybe by some fluke I'll beat you in countdown.  Then again maybe not. :rotfl: :P

:rotfl:  :rotfl:  :rotfl:

gahh you're using too many razzes again =O

What's the next highest?

And wasn't that Gregory Gauthier's record?

Btw aryth i wonder what your record was too!

46?? I'm hoping for a really hard State test then...!

lol

aryth what was ur highest?

I wonder if there were any perfect 46's in one of the hardest state tests...

you know what's fun? going to MOP and taking the USEMO (Useless Math Olympiad).

"Now you will be taking a MathCounts State Sprint Round. You probably all know the rules already, except we're tweaking one of the rules a bit. For these 30 problems, you will have...

...five minutes."

USEMO is awesome yay

inequality bashing especially

I heard about that and that Carl owned it completely. is this true =P...
What a party. MC state sprint round with 5 minutes xD.

good luck on states, guys

Haha, I'm posting on the Illinois Forum.

How about if we just let this topic die? That way, reduce the spam in here, particularly when it's from foreigners to IL.

[quote="Aryth"]carl "cheated" with someone and somebody complained so the graders halved carl's score  :roll:[/quote]

this is a lie - it just made him score a 0 on #3 and sent him to 2nd place.

anyway, good luck on states to you middle schoolers - hopefully you will see the light beyond mathcounts soon (!)

[quote="suma_milli"]guess what?  My score on ftw < 1100

WOW.  I try not to go on so I don't have to look at my score.  It's all because when I started at the beginning of the summer I kept facing really good people.  >_<  and that was back when I couldn't get greater than 80 on amc 10s...[/quote]

I'm on the same page suma_milli
what was the score for last year? (highest and 4th place?)


What command do you use for making letters "curly"? (to see what I'm talking about, look at problem 3 in the current round of ML contest, at that "F")

I believe you're talking about the \mathcal font: $\mathcal{ABCDEFG}$. It comes in the amsmath package, so just put \usepackage{amsmath} in your preamble if you want to use this font.


A finite set of unit circular disks is given in a plane, such that the area of their union is $S$.
Prove that there exists a subset of mutually disjoint disks whose union has area $> \frac{2S}{9}$.

We will show that if we can choose at most $ n$ disjoint discs then S is at most $ \frac92n\pi$.
At first, let's take these n discs and define n sets $ A_1,A_2,...,A_n$ and put the n discs to the corresponding sets .
Then let's take one-by-one the other discs. For each discs, examine which of the n chosen disc intersects it, and put this disc into one of their corresponding sets in which set this disc intersects all the other discs. There has to be at least one such set, because if there wasn't, we could choose $ n \plus{} 1$ disjoint discs. (This examined + one from each set of the intersected $ A_i$s' + the not-intersected $ A_i$s) Thus we can put every disc into a set, and in each set every two discs intersect.

But then these sets are independent, we should only show that in each set, the union of its element has an area at most $ \frac92\pi$. (Then choosing one element from each set would complete the constraints.)

So now we have a couple of discs, each two of them intersect. That means, that the distance of each two centers of the discs is at most 2. Now because of Jung's theorem, these centers can be covered with a circle of radius $ \frac2{\sqrt3}$. Now translate this covering circle so that it has a center on its border but still covers all the centers. Then the intersection of this circle and one of radius 2 around this center (lying on the border) covers all the centers. But then a shell-like shape with the area of $ (\frac23\frac {(2 \plus{} \sqrt3)^2}3 \plus{} \frac163^2 \plus{} \frac16 \minus{} \frac23\sqrt3\frac1\pi)\pi$ covers the union of the circles, that is it's greater than the union's area. But it's less than $ \frac92\pi$ so we're done.

[quote="kdano"]Thus we can put every disc into a set, and in each set every two discs intersect.[/quote]
This is false: if $O_1O_2O_3O_4O_5$ is a regular pentagon (where the $O_i$ are disk centers) with side length $2-\epsilon$, then we can choose at most $n=2$ disjoint disks, but we cannot find a working partition $A_1\cup A_2$ since there is no $K_3$.

The only solution I have seen (which gets a bound of $\frac{\pi}{8\sqrt{3}}S>\frac{2}{9}S$) uses the probabilistic method with hexagon tilings of the plane.  For a similar problem, see this [url=http://adeadhamster.blogspot.com/2010/04/fun-little-math-problem-of-day-3.html]blog post[/url].

As noted by [b]kevinatcausa[/b] [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?f=47&t=366962]here[/url], it would be interesting to see how much the size restrictions on the disks matter.

[quote=phoenixfire][hide=Solution]Partition the plane into regular hexagons of area $8 \sqrt{3}$ i.e. in-radius $2$. Thus if two circles $\mathcal{C}_{1}$ and $\mathcal{C}_{2}$ have two interior points $P_{1}$ and $P_{2}$ respectively, such that $P_{1}$ and $P_{2}$ are in the same position of two distinct hexagons, then, as $P_{1} P_{2} \geq 2$, the two circles cannot have any interior point.

Choose one central hexagon, and for each circle, translate the intersection of its interior with hexagons to the corresponding place in the central hexagon. Given that the total area of the circles is $S$, by pigeonhole principle in the plane there is a point in the central hexagon covered at least $\frac{S}{8 \sqrt{3}}$ times. Hence, the circles whose image cover it are distinct and their area is at least $\frac{\pi}{8 \sqrt{3}} S > \frac{2}{9} S$.[/hide][/quote]
Very beautiful and clever solution!


This ebook continues the project of Vietnam Inequality Forum : batdangthuc.net (but it’s been  temporarily closed). I’ve collected almost inequalities from 2008 Mathematical Competition over the world. Almost solutions were posted by Mathlinksers and I only added some my own solutions and some useful comments.
I want to give it as a small present to all members of Mathlinks. Hope you enjoy it!

As an update to this old topic, I've edited the first post and attached the PDF file.

Enjoy! :)


geometry
circumcircle
incenter
geometric transformation
reflection
geometry proposed

Let triangle ABC, $ \angle A\equal{}60^o, AB>AC$. O is the circumcenter, I is the incenter of triangle ABC. Prove that $ 2\angle AOI\equal{}3\angle ABC$.

[quote="mr.danh"]Let triangle ABC, $ \angle A \equal{} 60^o, AB > AC$. O is the circumcenter, I is the incenter of triangle ABC. Prove that $ 2\angle AOI \equal{} 3\angle ABC$.[/quote]
Just reflect $ AO$ W.R.T $ \overrightarrow{AI}$. Now, $ O\to H$. As, $ O$ and $ H$ are isogonal conjugates.
Now, we need to prove that, $ 2\angle AHI \equal{} 3\angle ABC$.
Which is just [url=http://www.mathlinks.ro/viewtopic.php?p=799589#799589]APMO 2007[/url]  :D

As the solution is much more straight forward than the original APMO problem I am giving solution.(and surprizingly the main problem can be proved from here just after a reftection)
$ \angle CIB\equal{}90^\circ\plus{}\frac A2\equal{}120^\circ$ and $ \angle COB\equal{}2\angle A\equal{}120^\circ$
So, $ CIOH$ cyclic.
and $ \angle AHI\equal{}\angle AOI\equal{}2\angle B \minus{} \angle COI\equal{}\frac{3}{2}\angle B$
So,$ 2\angle AHI\equal{}2\angle AOI\equal{}3 \angle ABC$
Done!  :)


geometry MATHCOUNTS probability trigonometry

Hello, this is my first post on the boards. Tommorow I will be going to a 6th grade regonal mathbee, and I need to cram. I have to say, even though I am not very good at problem solving, Im the best in my school. I have all sorts of problems with the mathcounts drills im doing,
here:http://mathcounts.saab.org/mc.cgi
Its free, and the classes on this site are like $500 apiece, so....
Mostly, I can get 1/4 of the problems right. Probablility is hard, but im having trouble with geometry, mostly circles. I have Vol 1: the basics, and I am having trouble solving the problem in the circle section, where you have three different sized circles, each tangent to each other, and a (non-right)triangle formed the centers. The problem is to solve for the radius of the small circle(I think, I don't have my book with me) in terms of the sides of the
triangle. I am completely stumped.  I draw more triangles and parralell lines, but that doesn't work. Ive tried setting one side the 1, and solving with trig, but I can't get the three measurments required to do that. I would be great if someone could look up the problem and give a detail explanation. Thanks!!!

Hey, I'm the one who posted this question. I'll wait to see if other people get it and if they don't I'll post the solution.  ;)

Really? Its in the book, so I thought.....nevermind.
Well, I leave in 45 minutes, I hope to be a part of this community, you all seem so good at problem solving... :D

I still can't find the answer. I don't want to beg for it, but i have really tried hard and im stuck.

Im getting really frusterated now. Im trying tangent lines and right angles. Not working.
Does anyone have the answer?

If i did not misunderstand the problem, the answer is $\frac{a+b-c}{2}$, where $a, b, $ and $c$ are the three sides of the triangle and $a<b<c$.

I will soon give you a complete proof. :)

Oh, nevermind, I derrived it from what you gave me. PM me if you want it.


How many people here have made it through so far? Managed to pick up 12th place in the first online round which I'm rather happy with.. need a few more good rounds yet though :)

Yea, I did really badly on the first subround, so I didn't make it then, even I though I should have.  And then, I kind of overestmiated my abilities in getting up at 4 in the morning for the third subround, so that didn't really work out either.

advanced with a 400, which is a good indication that i'll probably not pass the next one
good luck all the way through moutain view, MMM and bus!

I just realised I'm in the same region as China.

Aw man.

Then again, the majority of Chinese qualifiers were in the same round as me, so I must have beaten a lot of them.

Then again, they came 2nd, 4th, and 6th in that round, and also made up most of the top 50.

Oh well, I'll just do my best :D

h_s_potter2002 - think of me, all rounds from now on are 4am my time ;)

Interning in Japan for the summer, so the scheduling isn't much better for me either =P

Oh man I did so bad today, at least I won't have to travel anywhere for the next round.

I thought I did badly, but it seems one hard (and two easy) was enough for me to comfortably progress.. I was annoyed, since I knew I could solve D hard, just misread the constraints and couldn't fix in time..

Oh well, I'm saving my best for the last rounds.

I was very disappointed to see my D-large fail... and I thought I knew how to do C but later I found out that it won't work... and then I didn't have enough time to get event the small one. So I guess I'm not as good as I used to be :D.

I made it through round 3, getting all of A along with C-small and D-small. I think I knew how to do B but it looked too painful to code. A took me _way_ too long to debug, but I'm happy that I solved it. Unfortunately, I'm under 18, so I won't be able to go to the later rounds. The online rounds were fun, though.

I know it's kind of late, but congratulations to TripleM for making it to the onsite finals. :)
Is there anyone else here who made it? If so, good luck!
(As for me, I managed to place 10th in round 3 but unfortunately wasn't eligible for the local onsites. At least I can get a Google t-shirt. :) )

EDIT: I wasn't eligible because I'm not 18.

Thanks.. it was kind of crazy, having solved absolutely nothing at the halfway point in the contest and getting all the way from last to 8th, but I just hope I can have a good round the whole way through in America!


Two tanks pump water out of a tower. Pump A can empty the tower in 5 hours. When pump A and B work together the tower can be emptied in 3 hours. how long, in hours, does it take for pump B to empty the tank?

[hide="Solution"]
Let the amount of water in the tower be $ x$.

Pump A can empty the tower in 5 hours, which means it can empty $ \frac{3x}{15}$ in an hour.

If Pump A and Pump B work together, they can empty the tower in 3 hours, which means they can empty $ \frac{5x}{15}$ in an hour.

Hence, Pump B will empty $ \frac{2x}{15}$ in an hour, which means it will take 7.5 hours for Pump B to empty the whole tower.

Answer: 7.5[/hide]

i remember that there was another way to do these, that was slightly easier...

it was something like

 \[ \frac{x}{5} \plus{} (something that goes here)\equal{}1\]

I wrote out that descriptive solution because this is very basic problem, and if you do not understand how to solve this problem, you would not be able to solve more advanced problems.

It probably 'looked' complicated because it was quite long...

As a matter of fact, I can make one-line solution:

$ \frac{x}{3}\minus{}\frac{x}{5}\equal{}\frac{2x}{15}$, so it takes $ \frac{15}{2}\equal{}7.5$ hours.

yes, that one line solution was what i was looking for. we did this stuff in 7th grade and i vaguely remembered that there was a one line step to do it, however i did not remember how to acquire that solution.

tyty :P

Well, if you understand the concept, then the shortcut equation would look like
$ \frac{1}{5} \plus{} \frac{1}{B} \equal{} \frac{1}{3}$
And solving for $ B$, you get $ B \equal{} 15/2$

Remember that $ \frac {1}{x} \plus{} \frac {1}{y} \equal{} \frac {x\plus{}y}{xy}$
This might also be the one line equation you meant


Find all functions $ f: \mathbb{R^\plus{}} \to \mathbb{R^\plus{}}$ satisfying the equation
\[f(a^3\plus{}b^3)\equal{}(f(a)\plus{}b)(f(a^2)\minus{}af(b)\plus{}b^2),\]
for all $a,b \in \mathbb{R}^\plus{}$.

i can't solve it and need help..pco? :wink:

$ f(0) \equal{} f(0)^2$ we take $ k \equal{} f(0)$
if $ k \equal{} 0$ then $ f(b^3) \equal{} (k \plus{} b)(k \plus{} b^2) \equal{} b^3$
then $ \forall x\in R^ \plus{} : \ f(x) \equal{} x$
if $ k \equal{} 1$ then $ f(b^3) \equal{} (1 \plus{} b)(1 \plus{} b^2) \equal{} 1 \plus{} b \plus{} b^2 \plus{} b^3$
so $ 4\equal{}f(1\plus{}0^3)\equal{}(f(1)\plus{}0)(f(1)\minus{}f(0)\plus{}0^2)\equal{}(4\plus{}0)(4\minus{}1)$ gives contradiction
so $ f(x)\equiv x$

[quote="behemont"][quote="aviateurpilot"]$ f(0) \equal{} f(0)^2$ we take $ k \equal{} f(0)$
if $ k \equal{} 0$ then $ f(b^3) \equal{} (k \plus{} b)(k \plus{} b^2) \equal{} b^3$
then $ \forall x\in R^ \plus{} : \ f(x) \equal{} x$
if $ k \equal{} 1$ then $ f(b^3) \equal{} (1 \plus{} b)(1 \plus{} b^2) \equal{} 1 \plus{} b \plus{} b^2 \plus{} b^3$
so $ 4 \equal{} f(1 \plus{} 0^3) \equal{} (f(1) \plus{} 0)(f(1) \minus{} f(0) \plus{} 0^2) \equal{} (4 \plus{} 0)(4 \minus{} 1)$ gives contradiction
so $ f(x)\equiv x$[/quote]

sorry aviateurpilot, but we don't know what is $ f(0)$.. :P[/quote]
hhhh but we know that $ f(0)^2 \equal{} f(0)$
so if $ f(0) \equal{} 0$ i have found that $ f(x) \equal{} 0$
if $ f(0) \equal{} 0$ i get contradiction,
plz read ma solution again,  :D

 :arrow: me i konw that $ \mathbb{R}^\plus{}\equal{}\{x\in\mathbb{R}|\ x\ge 0\}$

i woudn't agree on that one..$ \mathbb{R}^\plus{}$ means only positive numbers, and not $ 0$... :!:

[quote="behemont"]Find all functions $ f: \mathbb{R^ \plus{} } \to \mathbb{R^ \plus{} }$ satisfying the equation

$ f(a^3 \plus{} b^3) \equal{} (f(a) \plus{} b)(f(a^2) \minus{} af(b) \plus{} b^2)$ $ ;$$ \forall a,b \in \mathbb{R}^ \plus{}$[/quote]

The difficulty here is the impossibility to use $ x\equal{}0$ or $ y\equal{}0$ since neither the function, neither the assertion are available for these values.

Here is a rather ugly demo (I would be interested in anything better).

Let $ P(x,y)$ be the assertion $ f(x^3\plus{}y^3)\equal{}(f(x)\plus{}y)(f(x^2)\minus{}xf(y)\plus{}y^2)$

$ P(x,y)$ $ \implies$ $ f(x^3\plus{}y^3)\equal{}(f(x)\plus{}y)(f(x^2)\minus{}xf(y)\plus{}y^2)$
$ P(y,x)$ $ \implies$ $ f(x^3\plus{}y^3)\equal{}(f(y)\plus{}x)(f(y^2)\minus{}yf(x)\plus{}x^2)$

From these two lines, we get $ f(x^2)\equal{}\frac{(f(y)\plus{}x)(f(y^2)\minus{}yf(x)\plus{}x^2)}{f(x)\plus{}y}\plus{}xf(y)\minus{}y^2$

So $ Q(x,y)$ : $ f(x^2)\equal{}\frac{f(x)((x\minus{}y)f(y)\minus{}y(x\plus{}y))\plus{}f(y)f(y^2)\plus{}f(y)x(x\plus{}y)\plus{}xf(y^2)\plus{}x^3\minus{}y^3)}{f(x)\plus{}y}$

Let $ f(1)\equal{}a$
$ P(1,1)$ $ \implies$ $ f(2)\equal{}a\plus{}1$
$ Q(2,1)$ $ \implies$ $ f(4)\equal{}2(a\plus{}1)$

$ Q(x,1)$ $ \implies$ $ f(x^2)\equal{}\frac{f(x)((a\minus{}1)x\minus{}(a\plus{}1))\plus{}x^3\plus{}ax^2\plus{}2ax\plus{}a^2\minus{}1}{f(x)\plus{}1}$
$ Q(x,2)$ $ \implies$ $ f(x^2)\equal{}\frac{f(x)((a\minus{}1)x\minus{}2(a\plus{}3))\plus{}x^3\plus{}(a\plus{}1)x^2\plus{}4(a\plus{}1)x\plus{}2a^2\plus{}4a\minus{}6}{f(x)\plus{}2}$

Equating the two RHS gives :
$ \frac{f(x)((a\minus{}1)x\minus{}(a\plus{}1))\plus{}x^3\plus{}ax^2\plus{}2ax\plus{}a^2\minus{}1}{f(x)\plus{}1}$ $ \equal{}\frac{f(x)((a\minus{}1)x\minus{}2(a\plus{}3))\plus{}x^3\plus{}(a\plus{}1)x^2\plus{}4(a\plus{}1)x\plus{}2a^2\plus{}4a\minus{}6}{f(x)\plus{}2}$

$ f(x)(f(x)\plus{}2)((a\minus{}1)x\minus{}(a\plus{}1))$ $ \plus{}(f(x)\plus{}2)(x^3\plus{}ax^2\plus{}2ax\plus{}a^2\minus{}1)$ $ \equal{}f(x)(f(x)\plus{}1)((a\minus{}1)x\minus{}2(a\plus{}3))$ $ \plus{}(f(x)\plus{}1)(x^3\plus{}(a\plus{}1)x^2\plus{}4(a\plus{}1)x\plus{}2a^2\plus{}4a\minus{}6)$

$ f(x)^2((a\minus{}1)x\minus{}(a\plus{}1))\plus{}$ $ f(x)(x^3\plus{}ax^2\plus{}2(2a\minus{}1)x\plus{}a^2\minus{}2a\minus{}3)\plus{}2x^3\plus{}2ax^2\plus{}4ax\plus{}2a^2\minus{}2$ $ \equal{}f(x)^2((a\minus{}1)x\minus{}2(a\plus{}3))\plus{}$ $ f(x)(x^3\plus{}(a\plus{}1)x^2\plus{}(5a\plus{}3)x\plus{}2a^2\plus{}2a\minus{}12)\plus{}x^3\plus{}(a\plus{}1)x^2\plus{}4(a\plus{}1)x\plus{}2a^2\plus{}4a\minus{}6$

$ f(x)^2(a\plus{}5)\minus{}f(x)(x^2\plus{}(a\plus{}5)x\plus{}a^2\plus{}4a\minus{}9)\plus{}x^3\plus{}(a\minus{}1)x^2\minus{}4x\minus{}4a\plus{}4\equal{}0$

$ (f(x)\minus{}x\minus{}a\plus{}1)(f(x)(a\plus{}5)\minus{}x^2\plus{}4)\equal{}0$

And so : $ \forall x$, either $ f(x)\equal{}x\plus{}a\minus{}1$, either $ f(x)\equal{}\frac{x^2\minus{}4}{a\plus{}5}$

The second possibility is impossible for $ x\leq 2$ since then we would have $ f(x)<0$ and so :

$ \forall x\leq 2$, $ f(x)\equal{}x\plus{}a\minus{}1$

Consider then $ x\leq 1$ : $ x^2\leq 2$ and $ x^3\plus{}1\leq 2$ so we can use $ f(x)\equal{}x\plus{}a\minus{}1$ in $ P(x,1)$ which gives $ (a\minus{}a^2)(x\minus{}1)\equal{}0$
And so $ a\equal{}1$ ($ a\equal{}0$ is impossible since $ f(x)>0$ $ \forall x$)

So, $ \forall x\leq 2$ $ f(x)\equal{}x$

Using then $ Q(x,1)$ for any $ x\leq 2$, we get $ f(x^2)\equal{}\frac{\minus{}2f(x)\plus{}x^3\plus{}x^2\plus{}2x}{f(x)\plus{}1}$ $ \equal{}\frac{\minus{}2x\plus{}x^3\plus{}x^2\plus{}2x}{x\plus{}1}\equal{}x^2$

And so $ f(x)\equal{}x$ $ \forall x\leq 4$ ...

And so $ f(x)\equal{}x$ $ \forall x$
And it's easy to verify that this function actually fits the initial requirements.


$ f: \mathbb{R}\rightarrow \mathbb{R}: \ \bigwedge\limits_{x,y\in R} f(x^2 \plus{} y) \plus{} f(f(x) \minus{} y) \equal{} 2f(f(x)) \plus{} 2y^2$

[hide="What I have by now"]$ f(0) \equal{} 0 \\
f(y) \plus{} f( \minus{} y) \equal{} 2y^2 \\
f((f(x))^2) \equal{} f(f(x^2)) \\
f(f(x)) \equal{} f(f( \minus{} x))$
I suppose, that the only solution is $ f(x)\equal{}x^2$, but I can't prove it.[/hide]

Taking $ y : \equal{} 0$, one has
\[ f\left(x^2\right) \equal{} f\left(f(x)\right)\,.
\]
Let $ c : \equal{} f(0)$.  Hence, $ c \equal{} f\left(0^2\right) \equal{} f\left(f(0)\right) \equal{} f(c)$.  Taking $ x: \equal{} 0$ and $ y: \equal{} c$, one gets
\[ 2c \equal{} f(c) \plus{} f(0) \equal{} 2f(c) \plus{} 2c^2 \equal{} 2c \plus{} 2c^2\,.
\]
That is, $ f(0) \equal{} c \equal{} 0$.

Now, if one replaces $ y$ by $ f(x)$, one obtains
\[ f\left(x^2 \plus{} f(x)\right) \equal{} 2f\left(f(x)\right) \plus{} 2\left(f(x)\right)^2\,.
\]
If one substitutes $ y$ by $ \minus{} x^2$, one has
\[ f\left(f(x) \plus{} x^2\right) \equal{} 2f\left(f(x)\right) \plus{} 2x^4\,.
\]
This shows that $ f(x) \equal{} \pm x^2$.

Note that $ f( \plus{} y) \plus{} f( \minus{} y) \equal{} 2y^2$ for all $ y$.  Thus, if there exits $ a \neq 0$ such that $ f(a) \equal{} \minus{} a^2$, then
\[ f( \minus{} a) \equal{} 2a^2 \minus{} f( \plus{} a) \equal{} 3a^2\,,
\]
contradicting the fact that $ f( \minus{} a) \equal{} \pm( \minus{} a)^2$.  Consequently, $ f(x) \equal{} x^2$ for all $ x$.

[quote="Batominovski"]Now, if one replaces $ y$ by $ f(x)$, one obtains
\[ f\left(x^2 \plus{} f(x)\right) \equal{} 2f\left(f(x)\right) \plus{} 2\left(f(x)\right)^2\,.
\][/quote]It should be: $ f(x^2\plus{}f(x))\equal{}2f(f(f(x)))\plus{}2(f(x))^2\equal{}2f(f(x^2))\plus{}2(f(x))^2$ (it's explained in the first line of your message)

[quote="Batominovski"]If one substitutes $ y$ by $ \minus{}x^2$, one has
\[ f\left(f(x) \plus{} x^2\right) \equal{} 2f\left(f(x)\right) \plus{} 2x^4\,.
\][/quote]It should be: $ f(f(x)\plus{}x^2)\equal{}2f(f(\minus{}x^2))\plus{}2x^4$ and we must notice, that $ f(f(x))\equal{}f(x^2)\equal{}f((\minus{}x)^2)\equal{}f(f(\minus{}x))$, so: $ 2f(f(\minus{}x^2))\plus{}2x^4\equal{}2f(f(x^2))\plus{}2x^4$, later it's obvious ;) : $ (f(x))^2\equal{}x^4 \iff (f(x)\minus{}x^2)(f(x)\plus{}x^2)\equal{}0$ and we have to only check these two options.

Thanks  :lol:

Sylwek Could You explain why in Your opinion threr sholuld be:

$ f(f(x) \plus{} x^2) \equal{} 2f(f( \minus{} x^2)) \plus{} 2x^4$

maybe I'm blind but J can't see it.

Sylwek, Batominovski made $ y : \equal{} f(x)$, but you are substituting $ x$ by $ f(x)$ too!


analytic geometry
graphing lines
slope
trigonometry
ratio

The vertices of Triangle ABC are A(-2, 1), B(4, 8) and C(14, -3).

(a) If point M(h, o) lies on line AC, find h.

(b) Show by means of slope that line segment BM is perpendicular to line segment AC.

(c) Find, to the nearest degree, the measure of angle A.

[hide]
First of all find an equation for the line AC, doing this we find it to be
$y =-\frac{1}{4}x+\frac{1}{2}$
I am assuming point M is (h, 0) so plugging in 0 for y in this equation yields x as $2$, so the value for h is $2$.

The slope of BM is $\frac{4}{1}$ and the slope of AC is $-\frac{1}{4}$. They are negative reciprocals and are thus perpendicular.

To find the measure of angle a, find the lengths of AB and AM, and then use the cosine ratios to approximate the angle

$\cos A = \frac{\sqrt{(2-(-2))^{2}+(0-1)^{2}}}{\sqrt{(8-0)^{2}+(4-2)^{2}}}= \frac{\sqrt{17}}{\sqrt{68}}= \frac{\sqrt{17}}{2\sqrt{17}}= \frac{1}{2}$

Therefore angle A has a measure of $60^{\circ}$ or $\frac{\pi}{3}$ radians.
[/hide]


It's a great strategy game. I'm on level 17. its crazy

[url=http://www.ebaumsworld.com/games/platformmaze.html]Platform Maze[/url]

This game is soooo hard. :wacko:
Congratulations for getting to level 17. How long did it take you to get there?

omg i love this game  :D  :P  im on lvl 18 and it took me in total about an hour and half of play time.. spent 30 minutes on lvl 17 XD

edit: yay lvl 20 now  :D

lol well that was fun, the game only has 20 lvls  :D  weren't that hard :roll:

I'm at level 9 now, I started just 10 minutes ago.

I love this game! :10: Lvl [b]15![/b] so far

That game is so annoying...

[quote="Ignite168"]eBaum's World sucks. It's run by egotistical people who have nothing to do than steal from other artists and pass it off as their own while making tons of money from it and lying from it and being immature.[/quote]

How so?

[quote="thatin"]hey, when I click the link a message appears that says that the visit of this site is prohibited. I know what it is but does this site deserve that characterization?[/quote]

Are you on a school network or something like that?

[quote="mathnerd314"][quote="thatin"]hey, when I click the link a message appears that says that the visit of this site is prohibited. I know what it is but does this site deserve that characterization?[/quote]

Are you on a school network or something like that?[/quote]

yes.

Bingo. The blocking software won't let you play games :(

At least that's how it is at my school.

miniclip isn't a well known site? Anyway I cannot play the game :noo:

Yay I understood how to play the game. Level 19.
I really love this game! :)

I finished it. :) . For some reason I got level 20 faster than level 19.


if $ a$ and $ b$ are integers such that $ ab\equal{}10$, then what is the probability that
$ a^{1\minus{}b}b^{\minus{}1}\equal{}10^{\minus{}1}$?

i_like_pie, what about negative integers?

[hide="Revised Solution (thanks to arcimedes)"]There are $ 8$ possibilities:

$ (10, 1)$
$ (5, 2)$
$ (2, 5)$
$ (1, 10)$
$ (\minus{}10,\minus{}1)$
$ (\minus{}5,\minus{}2)$
$ (\minus{}2,\minus{}5)$
$ (\minus{}1,\minus{}10)$

Of those, $ (10,1)$, $ (5,2)$, $ (1,10)$, and $ (\minus{}1\minus{}10)$ suffice. Therefore, the probability is $ \boxed{\frac{1}{2}}$.[/hide]

ab=10 so $ a^{\minus{}1}b^{\minus{}1}\equal{} 10^{\minus{}1}\equal{} a^{1\minus{}b}b^{\minus{}1}$ so $ a^{\minus{}1}\equal{} a^{1\minus{}b}$, since -1 is odd either -1=1-b => (5,2), a=1 => (1,10) or a=-1 => (-1,-10)

3/8

i_like_pie (10,1) doesn't work, 10^0*1^-1=1

this problem would have been hella good if it was ab=2400 or something instead


$(x+y+z)^{3}=x^{3}+y^{3}+z^{3}+3(x+y)(y+z)(z+x)\implies (x+y)(y+z)(z+x)=0$. The solutions are $(a,-a,b),(a,b,-a),(b,a,-a)$ where $a,b$ are arbitrary numbers.


geometry
trapezoid
trigonometry
geometry unsolved

Trapezoid $ ABCD$,$ DA$ is parallel to $ BC$ , $ AD \equal{} DB \equal{} DC$ and $ \angle BCD \equal{} 72$.point $ K$ is on $ BD$ such that $ AD \equal{} AK$.$ M$ is the midpoint of $ CD$.$ AM$ intersect $ BD$ at $ N$.prove that $ BK \equal{} ND$

nice problem :) .i change the names because i want to use some lenghs
in triangle $ ABC$ such that $ AB \equal{} AC$ and $ A \equal{} 36$ let $ AE$ parallel to $ BC$ and $ AE \equal{} AQ$ suvh that $ Q$ lies on $ AC$.if $ M$ be the midpoint of $ AB$ and $ EM$ intersect $ AC$ at $ P$ prove that $ AP \equal{} CQ$
$ AP \equal{} x$ then $ \frac {x}{b \minus{} x} \equal{} \frac {b}{a \plus{} b}$ ($ b \equal{} c$) .so $ x \equal{} \frac {b^2}{a \plus{} 2b}$. but we have $ CQ\equal{}b\minus{}a$ so we must prve that $ b^2\equal{}(a\plus{}2b)(b\minus{}a)\equal{}a(a\plus{}b)$ and the last quality is trivial.for proving it in $ ABC$ let $ R$be a point on $ BC$ such that $ CAM\equal{}36$ ($ M$is near$ C$) now use this $ \frac{BC}{CR}\equal{}\frac{AB}{AR}$

easier proof of last quality: by similirity of $ RBA$ and $ ABC$ we have $ b^2\equal{}(BC)(BR)\equal{}(a)(a\plus{}b)$ :)

[quote="behdad.math.math"][color=darkred]Let $ ABCD$ be a trapezoid so that $ AD\parallel BC$ , $ DA \equal{} DB \equal{} DC$ and $ m(\angle BCD) \equal{} 72^{\circ}$ . Denote the

midpoint $ M$ of $ [CD]$ , $ K\in (BD)$ such that $ KA \equal{} DA$ and $ N\in AM\cap BD$ . Prove that $ BK \equal{} ND$ .[/color] [/quote]

[color=darkred][b][u]Lemma[/u].[/b] In any triangle $ ABC$ exists the equivalence $ \boxed {\ A \equal{} 2B\ \Longleftrightarrow\ a^2 \equal{} b(b \plus{} c)\ \mathrm{or}\ b \equal{} c\ }$ .[/color]

[color=darkblue]Indeed, $ A \equal{} 2B\Longleftrightarrow$ $ a \equal{} 2b\cos B$ $ \Longleftrightarrow$ $ a^2c \equal{} b\left(a^2 \plus{} c^2 \minus{} b^2\right)$ $ \Longleftrightarrow$ $ b \equal{} c\ \ \mathrm{or}\ \ a^2 \equal{} b(b \plus{} c)$ .

[b][u]Particular case[/u].[/b] Let $ ABC$ be an  $ A$-isosceles triangle. Prove that $ A \equal{} 36^{\circ}\ \Longleftrightarrow\ b^2 \equal{} a(a \plus{} b)$ .

Otherwise, consider the point $ D\in (AC)$ for which $ DA \equal{} DB$ . Thus $ AD \equal{} DB \equal{} BC$ ,

$ DC \equal{} b \minus{} a$ and $ ABC\sim BDC$ $ \implies \frac ba \equal{} \frac {a}{b \minus{} a}\implies b^2 \equal{} a(a \plus{} b)$ .[/color]

[color=darkblue][b][u]Proof of the proposed problem[/u].[/b]  Denote $ KA \equal{} AD \equal{} DB \equal{} DC \equal{} a$ , $ BC \equal{} b$ and $ P\in AM\cap BC$ .

Therefore,  $ ADK\equiv DBC$ , i.e. $ \boxed {\ DK \equal{} b\ }$ . From above lemma applied to $ \triangle BDC$ obtain $ a^2 \equal{} b(a \plus{} b)\ \ (*)$ .

$ CP \equal{} DA \equal{} CD \equal{} a\implies BP \equal{} DP \equal{} a \plus{} b$ and $ \frac {BN}{a} \equal{} \frac {a \plus{} b}{2a \plus{} b}\stackrel {(*)}{\implies}$ $ BN \equal{} \frac {a(a \plus{} b)}{a \plus{} \frac {a^2}{b}}\implies$ $ \boxed {\ BN \equal{} b}$ .

If denote $ R\in (DP)$ so that $ DR \equal{} b$ , $ PR \equal{} a$ , then (isosceles) $ ADK\equiv DBC\equiv PCR\sim PBD$ and $ CR\parallel BD$ .[/color]


Here's a reaction sequence I made some years ago:

Propose structures for compounds A, B, C, and explain their formation.

What is the first reagent used and can you tell something about its use/function ?  :maybe:

Cu(acac)2 is a copper complex that has the same use as silver, copper, or light is the Wolff rearrangement.

No, it's not a ketene. Ok, explain to me what happens in the Wolff rearrangement.

Release of nitrogen gas and migration of alkyl group attached to carbonyl to the resulting electron deficient carbon (which was attached to diazo group) and formation of ketene.

In other words, we can say that there is the formation of a carbene and next there is alkyl migration to carbene carbon. What can happen if, in the same molecule where the carbene forms (by elimination of nitrogen gas), there is also a double bond?

Ok, i got your point but i wonder why would A rearrange to B. My (new) cyclic A is very stable..

Diels alder reaction . after wolff rearrangemetn
is A this?

Completely wrong. I have said already that there is no Wolff rearrangement here. Don't you students know other reactions of diazo compounds?

[quote="Carcul"]What can happen if, in the same molecule where the carbene forms (by elimination of nitrogen gas), there is also a double bond?[/quote]

I confess I have never heard of the Wolff Rearrangement but the first thing that popped to mind after seeing that is carbene addition to a double bond.

Very good BanishedTraitor. Can you then propose a structure for A?

Very good. Now what happens by heating at 580ºC?

Not even close. What reactions of vinylcyclopropanes do we know?

I don't know any , On searching on wiki , I got vinylcyclopropane rearrangement . is this what you want?

We have a vinylcyclopropane who is going to be heated at 580ºC. What do you think?

I don't know , it says the reaction proceeds at low temperatures easily due to high reactivity of the cyclopropane ring and the pi bond. I don't know any other [hide]( nor am I finding any on Wiki ) :blush:  my knowledge of organic chem is very less. please guide me to become better. [/hide]

What the Wikipedia article says is that the vinyl-cyclopropane rearrangement occurs at lower temperatures than most other sygmatropic rearrangements, which does not mean low temperatures. A typical vinyl-cyclopropane rearrangement occurs at 480ºC (with FVP) or 580ºC (heating in a suitable solvent). So, in this case, we definitly have a vinyl-cyclopropane rearrangement.


integration
logarithms
calculus
calculus computations

If we evaluate $\int_0^1 \frac{x-1}{\ln x} \ dx$ as follows:

$I(t) = \int_0^1 \frac{x^t-1}{\ln x} \ dx$

$I'(t) = \int_0^1 x^t \ dx = \frac{1}{t+1} \Rightarrow I(t) = \ln \left(t+1\right) \Rightarrow I(1) = \ln 2$

where does the $-\frac{1}{\ln x}$ come in?

What I mean is why $-1$ and not any other constant?

[quote="fedja"]Because you want $I(0)=0$ (otherwise the last implication in your post #1 is invalid)[/quote]

Thanks, I can't believe I forgot about initial conditions. :)


For the perimeter and the area of a square to be the same, the side of the square must be x. What is x?

$P=4s$
$A=s^2$
$s^2=4s$
divide by $s$
$s=4$
(A side can't be 0)

[hide]
For side length of square $s$ we proceed by setting the perimeter and area functions equal.

$P(s) = A(s)$
$4s = s^2 \Leftrightarrow s^2 - 4s = 0$
$s(s-4) = 0$
$s = 0$ or $s = 4$.

The former root is practically meaningless, but perhaps it is true that a square of side length zero (i.e. a point) has a perimeter that is numerically equal to its area.
[/hide]

[hide]$4s=s^2$

solve for s and you get

$\frac{s^2}{s}=4$

simplify and you get

$s=4$[/hide]

[hide]x=4 because there are four sides in a square.[/hide]

[hide]
This is a good MC Chapter problem

$4x=x^2\\4=x$
Done
[/hide]


email
analytic geometry
videos
integration
calculus
geometry
Mafia

Greece was awsome! Especially girls, they were so hot. We had hot guides too...like 5 of the guides were very beautiful...
we had a nice time, we went out to nice places and at night we hanged around with everybody..
post here what you liked and disliked of the imo 2004 @ greece

I must admit that compared with other years the guides were pretty hot :) also the nightlife in the city was a little better, however, it was not as good as in japan in total (thinking of organization, coordination, problem choosing, siteseeing, etc.).

PS I really loved how mexicans trashed their room in the last days :)

[quote="matty"]Well...then on behalf of the female guides I should say a big [b]Thaaaaank you!!![/b]
 :P[/quote]hey ... can you post your avatar so that people can see who you are? :P :)

hey matty, i don't remember you. i think you were in the same bus with me. i am from puerto rico...

[quote="matty"]Nah!!
I don't think so!
But I'm the guide of Romania!
 :P[/quote] aaa ... it's you :) I didn't recognize that after your username :) well nice to have you here also ..

[quote="manuel"]valentin, i think puerto rico was in front of your room.. do your remember?[/quote] I stayed in 1416 .... and no I don't remember any puerto rico guys :)

[quote="manuel"]oh no well it wasn't you... are you in the same team with gabriel emanuelle .. i guy named like that? was he from romania?[/quote]yes he is ... thye stayed at 923 :) maybe you are talking about that room :)

[quote="Valentin Vornicu"] aaa ... it's you :) I didn't recognize that after your username :) well nice to have you here also ..[/quote]

Well,thank you!
Nice to be here!!!
 :)

[quote="manuel"]ohh.. now i remember you matoula.. you had a friend from puerto rico, his name was carlos. i am the guy that was with him when you took us to Craft. the beer place.[/quote]

I remember taking Carlos to Craft with two other guys,I think,but I can't remember their faces.So I guess you are one of those guys (and you left immediately right?).
Well...I guess I don't remember you,then!
 :(

wow... italy in the bigs of basketball... cool  :D
in fact, after having defeated the not-so-big dream team...
anyway, i was the guest, ITA4 :)

hm.. what about the tunisian guide? i haven't seen her round for so long, just in the acropolis, in fact...

manuel, did you remember all guides i mentioned? :)

Hello, everybody! I watched the game and i should say that i was very happy for the final result... It is true that the US team lost many shots and especially the free shots and they were trapped in the Puerto Rico's zone defence... Congratulations Manuel... I think that Edmaries will be happy too... Greek team also won the Australian team... I think that Greece and Puerto Rico will fight for the qualification... C ya!

[quote="manuel"][quote="ma_go"]
hm.. what about the tunisian guide? i haven't seen her round for so long, just in the acropolis, in fact...

manuel, did you remember all guides i mentioned? :)[/quote]

I don't remember the tunisian guide.. i remember the one from ireland or iceland (not sure) that one was really hott.. and the one from your country , italy was hot too..[/quote]I wonder what the girls would say if they would read this topic? :)

Puerto Rico lost to Lithuania..  :( .. lets see if they can qualify.  :maybe:

ok ok.. the girls in the IMO were very hot, but speaking about basketball i have to say that we are the best!!!  did you watch the match against Serbia?? it was great the way we won!!

argentina is one of the best teams.... after spain..

argentina es una mierda ramiro! jaja mentira son muy buenos..
como andas che!

[quote="Valentin Vornicu"]
I wonder what the girls would say if they would read this topic? :)[/quote]

Hello, I was the guide of Finland and I can tell you that I am very flattered that someone mentioned me here..

well Nickolas I am Pascual from Colombia and of course i remember that night at Enzos, specially because i was very drunk (all of us in fact) and we danced a lot.
greece has been the best olympiad ever in the social aspect: Hot Girls, Party people, Beer! Also i remember that night they threw us away from the hotel, (Do you remember me Peter singing in Croatian or you were to drunk) and of course I am never going to forget craft, drinking a Jar of beer in one shot and of course chasing the girls!  Gretings to everybody and hope to party a lot with all of you in Cancun. Viva Amstel! and Viva Greece 2004!

Hi Pascual! Of course i remember! These nights were great! A lot of fun and drinking... I also remember a guy that he was solving Rubics Cube in less that 30 seconds at 4.00 in the morning!    :? At Enzo, the Mexicans layed down at the clubs floor, after drinking a lot! I also remember hotels lobby after the night with the beers... there were more than 100 empty beer cans! Cool!!!

hehe Nikos.... here U r?
I also remember some good mometns of IMO everything was fun:))))))))))))


Find all integers $a,b,n$ greater than $1$ such that $(a^{3}+b^{3})^{n}=4(ab)^{1995}$.

If $p\mid a$, then $p\mid ab$ and therefore $p\mid b$. Let $p^{k}\parallel a$ and $p^{l}\parallel b$. If $k>l$ then $p^{3ln}\parallel (a^{3}+b^{3})^{n}$ and $p^{(k+l)1995}\parallel 4(ab)^{1995}$ (if $p\neq 2$). So $3ln=(k+l)1995>3990l$, i.e. $n>1330$. On the other hand $(a^{3}+b^{3})^{n}\geq 2^{n}a^{3n/2}b^{3n/2}$, so $\frac{3n}{2}\leq 1995$, which is a contradiction. Simularly we get contradiction for $p=2$. So $a=b$, and $2^{n-2}=a^{3990-3n}$. Therefore $a=2^{m}$ and $m(3990-3n)=n-2$. So we only need to find all positive integers $n$ such that $3990-3n\mid n-2$.


The product of the digits of 3214 is 24. How many distinct
four-digit positive integers are such that the product of their
digits equals 12?

[hide]12's prime factorization is $ 2 \times 2 \times 3$. Thus the highest digit possible is six. We list the cases by brute force:

Highest digit 6:
$ 1126$ and permutations $ (12 \text{ total})$

Highest digit 4:
$ 1134$ and permutations $ (12 \text{ total})$

Highest digit 3:
$ 1223$ and permutations $ (12 \text{ total})$

Answer $ \fbox{36}$.[/hide]

If there are 0 ones, it is not possible, because abcd>12 if a, b, c, d>1 and are integers.  If there is 1 one, abc=12, a,b,c>1, and it has to be factored 2,2,3.  1,2,2,3 can be arranged 12 ways (4!/2! because of the two twos).  If there are 2 ones, ab=12, a, b>1, so it can be factored 2,6 and 3,4.  1,1,2,6 and 1,1,3,4 can each be written 12 ways for the same reason (except it's now repeating ones), so that makes 36.  With three ones, we get a digit of 12, which is impossible (from all of my sources, the problem was not written in base 13).  So 36.

We first have to figure out the different groups of 4 one-digit numbers whose product is 12. We obviously can't use 12 as one of the numbers, nor can we use 9, 8, or 7 (none divides 12). We can use 6, in which case one of the other numbers is 2 and the other two are 1s. So, we can have the number 6211, or any number we can form by reordering these digits. There are 4! ways to order these four numbers, but we have to divide by 2! because the two 1s are the same, so 4! counts each possible number twice. That gives us 4!/2! numbers that consist of 6, 2, and two 1s.
Next, we note that we can't have a 5, so we think about 4. If we have a 4, then the other three numbers are 3, 1, 1. Just as there are 12 ways to order the digits in 6211, there are 12 ways to order the digits in 4311. Finally, we check if there are any ways to get a product of 12 with digits that are 3 or less. There's only one such group, the digits in 3221. As with 6211 and 4311, there are 12 distinct ways to order the digits in 3221.
Combining our three cases, we have $ 12\plus{}12\plus{}12\equal{}\boxed{36}$  possible integers.


Im not sure if this is MATHCOUNTS level, but is there a way to solve this problem without a calculator?

  PURPLE
 -COMET
_________
    TEAM

Find the value of TEAM where every letter represents a different digit.

Dude, I spent like 20 minutes thinking M had to be somewhere from 3 to 8 inclusive, finally I tried M=2 and it worked and I'm like "wth?"

Thus being said, I don't have any suggestions

$ p\equal{}1$, otherwise TEAM will be a five digit integer.  Also, Purple cannot exceed Comet by more than 9999, which means that $ c\equal{}9$.  Also, $ U\equal{}0$, because $ 110,000\minus{}99,999\equal{}10,001$.

Thus, we can subtract 90,000 from both values to get:

 1RPLE
-OMET
_______
  TEAM
This should be easier.

Just brute-force and patience :roll: And follow mathcrazed's tips to make it less frustrating. I left my other team members to figure out #15 while I did #1 to #14.   It worked  :)  . edit:my team got a 14.

eh, you did miss #14, but that one was the one I used Stewart's on

Usually for these types of problems, it's easier to form it into a addition problem instead. I was so close to getting it, but then our team ran out of time. :mad:


articles
MIT
college
summer program
MathPath
email
algebra

A website that I found recently, [url]http://www.ratemyteachers.com[/url], was created for students to log on and rate their teachers, as well as posting comments about them. Apparently, very few teachers know about this.

I find it quite amusing to look at the comments that people post, past people's opinions on teachers that I now have and had. My question is, do you think that this site is bad? I mean, if a teacher randomly came across this site, and found that everyone hated them, well, wouldn't that make them feel bad?

A fair amount of schools have banned the use of the site for their teachers.

Your thoughts please.

I always thought that teachers generally know how well liked/disliked they are. In my school I think there is one clueless teacher who doesn't know that they are largely disliked by the student body, but I think that for the most part teachers know.

I just looked at it, and I found it kind of interesting, being able to see what your teachers are like, and knowing which teachers to transfer away from. The problem is, there are some total jerks that post things there, and some people that are just plain stupid. There aren't any teachers that don't have at least one or two mean comments, and I know that if I was a teacher, I would be offended by the site. The thing is, most teachers know how people feel about them already. I think that it is useful, and nothing is really said that isn't whispered behind the backs of teachers all around the school. This is just a way to make it viewable by everyone that wants to see.

someone told me about it in 6th grade, and i have posted a few things...we have some funny articles

i will go back and post some if i have time to do so

-jorian

I saw this website in 7th grade when a friend told me that someone was making fun of a ridiculous social studies teacher and asked me to join in and say something.

Since then, I haven't looked at it.

i didn't not know about this website until 10 seconds ago.
I don't think this website is bad because if a teacher feels bad they he/she might actually start teaching properly.

I used to visit this website a little bit in high school.  I see nothing wrong with it.  A lot of people will be immature and try to post wrong things, but most of the really bad things people will try and say will get deleted by moderators (in fact, all ratings [i]have[/i] to be checked by a moderator before they are posted on the site).

Also, in every class at MIT we have to fill out a survey about the class, the teacher, and our TA (teaching assistant).  It's helpful for the teachers, TAs, administration, and future students since the results of the surveys are posted online at the course catalog site.

yes that was a site i used to use (I am still a mod there). I have to approve of posts befor ethey get accpeted

Yes they have banned some schools...

As to teacher/TA surveys, we filled one out at MathPath, I felt bad giving someone a zero out of ten, but hey, she was a terrible, terrible teacher.

i believe they are on the wall of shame list if they are blocked

I don't like how "easiness" adds towards "total rating" because don't we want our teachers to be challenging?

yeah, that is not a good category to count toward your teacher's rating.

I find many of these comments contradictory to what I and many others think about some of my teachers. Maybe I'm being too hard on my teachers. Maybe these comments are old. I don't feel like posting my comments.

interesting site.

very interesting site

its cool

I saw this site a few months ago. It is quite interesting to see what other people have said about some of the teachers I've had.

I found about about it in 7th grade, when my science teacher found it during class and showed us all. He liked it. I have a few comments on there.

My school isn't there. I don't think i want to add it though....All the teachers in school find socially acceptable yet unfair reasons to cut ur marks and bug ur life if they know u hate them.Also,they access the computers frequently and may feel bad if they r genuinely good people. The best part is that we get to tell the principal whatever we want about the teachers and she joins in!!
We had a Kids Choice of Teachers Awards program last year..where a group of friends on email sent each other nominations like "Most boring", "Coolest","Best Dressed" and about 20 others. Then we passed it on to all our schoolmates till 150 voted for a teacher in each category. Then we had a virtual award ceremony! It was so much fun and the teachers had no clue! I think there should be something like that,a win-win thing. In the end, some teachers ARE quite nice!!(ONLY SOME)

I think it is potentially a good thing... we have a decent population of bad teachers at my school, and nobody seems to do anything about it.  I can see it being pretty devestating to a teacher, though, especially since kids would be likely to vent pent-up frustration on an anonymous website and say overly negative and generally damaging things.  Also, a lot of people don't like some of the best teachers at our school simply on teh basis of their class being dificult.

EDIT:
I figured I'd add/mention that I don't like their criteria at all.

[quote="k8reindeer"]I've discovered that by reading that website and some of the negative opinions on it I've actually begun thinking about what those people said and the negative aspects of my teachers that I'd never thought of before, and I have to be careful to make sure whatever opinion I have is my own opinion. A word of warning about that site: don't let what other people say influence how you think of a teacher.[/quote]

Why not? If those things are true about your teacher, but you just never realized them before, or they never applied to you, then why shouldn't it influence your opinion of a teacher?

Also, if you notice, easiness ISN'T in the criteria for the rating of a teacher. You can rate the teacher on easiness, but it is displayed separately from the rating of the teacher, which is only based on the other two factors.

I found what most people said about my teachers fair and accurate and I found that 5/1/1s were a lot more common than 1/1/1s meaning people at my school aren't grading a teacher on how easy their course is, but rather the opposite if there's a correlation at all.

I'm not sure everyone has the same rubric for easiness. Is a 5 very easy or very hard? I know people who've rated both ways.

[quote="Ignite168"]

Why not? If those things are true about your teacher, but you just never realized them before, or they never applied to you, then why shouldn't it influence your opinion of a teacher?

[/quote]

Most times the things people say are true, but on ratemyteachers they are really exaggerated. that extreme point of view kind of shifts my point of view closer to extreme, simply because it wouldn't have occurred to me to think of them that way. Trouble is, the points people make on ratemyteachers may be true but the exaggerations are not

Someone in my class asked my teacher if she checked hers and she said "I would never go on a site like ratemyteachers cuz the only people who would post there are the ones that either really like you or really hate you."  Which for the most part is probably true.  I don't know, I try to guess when I go on I only comment on those teachers I think are good and those I absolutely do not.

I find that false considering you have 6 teachers at the most in a semester and that's plenty of time to comment on all of them.

i think they mean it's easier to choose ratings :D

-jorian

[quote="mustafa"][quote="djshowdown2's teacher"]"I would never go on a site like ratemyteachers cuz the only people who would post there are the ones that either really like you or really hate you."[/quote]

That's a good point, I only rated my favorite teacher ever and my least favorite teacher ever. It was an extremely easy choice.[/quote]

yeah me too! Except it's my favorite few and least favorite few

Apparently my biology honors teacher has a turtle dance and a mitochondria dance, and there's an auction involving "Science Bucks" for a date to go out with him.

This probably means I should exchange my "Science Bucks" for real money.

Lol, currently I am on a school computer (on a temp account) and clicked the link to see if it was blocked (with no intention to post), and alas, it was.
This can really be paralleled to the whole Myspace.com debate (personal information found by teachers).
Really, what's the point of rating a teacher on public domain? So that people see how whiny and disputatious you are? There's another name for this-gossip. And we know we have all been hurt with gossip in school.
Seriously, to the people arguing for this site, what is the point of it? Does it make you HAPPY to insult someone behind your back? Even if it's good commentary, it's better to let a teacher know by working hard or being nice in class than by posting an anonymous comment.
~PoTW

My german teacher's son owns a website called  http://www.ratemyprofessor.com

My teacher told me he plans to sell it to Yahoo! for a lot of money.

RateMyProfessor is the poor man's PickaProf.  :D RMP derives grade distribution info from user reports -- you can imagine how representative such a sample can be. And what exactly is the purpose of uploading pictures of professors?

(Full disclosure: I have a chili pepper on RMP.  :roll:  )

[quote="mlok"](Full disclosure: I have a chili pepper on RMP.  :roll:  )[/quote]
And I don't. It's a small sample either way.

At my current university pickaprof.com is all the rage (I got 24 ratings there as of now.) Still a small sample, of course.

I rated ALL my teachers [yesterday was a bad day in terms of focus]. Fairly, I think. A lot of the teachers I really liked didn't get blank checks for 5's. I would recommend them to any student, but compared to my absolute favorites, they get fours.

Newsflash: Beavis and Butt-head switch from evaluating music videos to evaluating professors. ([url=http://www.prnewswire.com/cgi-bin/stories.pl?ACCT=104&STORY=/www/story/01-17-2007/0004507644&EDATE=]RateMyProfessors has been bought by MTV[/url].)


Team test scores for Mandelbrot [url=http://www.mandelbrot.org/scores/tmtestscrsrnd3.html]round three[/url] are up.

I'm excited.  Our team got perfect.

Here's a curiosity -- James Madison Memorial, which was not listed on the round 2 mandelbrot national team leaders (http://www.mandelbrot.org/scores/Rnd2NTm(Lead).html) scored 5 points fewer that Stuyvesant on round 3 ... but is now only trailing Stuy by 4 points (http://www.mandelbrot.org/scores/Rnd3NTm(Lead).html).  I wonder what happened there ...

But hey, we're still in 4th place.  Go Stuy!

Hmm ... what happened to Stuy's Division B team?  A strong 52 on the individuals ... and a 0 on the team round?  (Making them by 23 points the lowest scoring team on the past round that still makes it onto the high scoring teams list.  Although 10-15 places lower down than they used to be.)

I think that every now and then the scoring list is not complete with information for one reason or another.  When I was in high school my individual score and team were not listed after one round for some reason, though the scores were updated by the next round.  Sam does a good job, but you never know when the mail is slow or some other problem occurs.

Also, JBL, the San Diego Math Circle is gunning for Stuy!

My former high school rarely seems to be able to report scores on time. I really don't know why. I keep bugging them about it, but they don't do anything to change, sadly.


geometry
geometric transformation
reflection
AwesomeMath
summer program

The group of people producing the AwesomeMath summer camp are also producing an online journal.  Here's the first edition: [url]http://reflections.awesomemath.org[/url].  It's intended for a very experienced audience, including olympiad and collegiate mathematics.


Thirty-one books are arranged from left to right in order of increasing prices. The price of each book differes by $\$2 from that of each adjacent book. For the price of the book at the extreme right a customer can by the middle book and an adjacent one. Then:
(A) The adjacent book referred to is at the left of the middle book.
(B) The middle book sells for $\$36.
(C) The cheapest book sells for $\$4.
(D) The most expensive book sells for $\$64.
(E) None of these is correct.

$A$.

Say the first book is $2$, then the 31st book is $62$ and the 16th book is $32$ which is the middle book, $32+30(which is at the left)=62$.

[hide]
Let $x$ be the price of the leftmost book. The book adjacent to the middle book cannot be on its right side, as that would result in a negative value for $x$.
$x+60=x+30+x+32$
$x+60=2x+64$
$-x=4$, $x=-4$
Therefore, the book adjacent to the middle book must be on its left side.
$x+60=x+30+x+28$
$x+60=2x+58$
$x=2$
Thus, The adjacent book referred to is at the left of the middle book.[/hide]

Also, you can easily show that $B$, $C$, $D$ must all be false, because the price of any one book is not specified. Technically, the books could be priced at $\$ $126$ to $\$ $186$.