abstract algebra
group theory
superior algebra
superior algebra unsolved

if $ Aut(G)$ is cyclic then $ G$ is abelian .

Could we have an abelian not cyclic group $ G$ with $ Aut(G)$ is cyclic.

 :huh:  :huh:

this seems to be somewhat complicated ;-). I can show that $ G$ can't be finitely generated. but then ...

what about to assume $ G$ is finite for first steps?

let me see ur proof [b]-oo-[/b] :wink:

Note that $ Aut(A) \times Aut(B) \subset Aut(A \times B)$. Note that a cyclic group has at most one element of order $ 2$, so we already get a contradiction if each automorphism group of $ A$ and $ B$ has an element of order $ 2$.
Write $ G$ as sum (or product) of groups $ G_p$, $ p$ prime or $ 0$, where $ G_p$ is a $ p$-group (which for $ p \equal{} 0$ shall mean that it is torsion free). Every $ G_p$ has an automorphism $ x \mapsto \minus{} x$, which is nontrivial if $ 2 G_p \neq 0$ and has order $ 2$ then; also, if $ 2G_p\equal{}0$, then $ p\equal{}2$ and $ G_2$ is a $ \mathbb F_2$ vectorspace, giving that it must have dimension $ 1$ (otherwise the automorphisms aren't even abelian). In total, if $ |G_p|>2$, it has a nontrivial automorphism of order $ 2$. So if there is more than one such factor $ G_p$ bigger than $ 2$, the automorphisms aren't cyclic.
So either $ G \equal{} G_p$ for any $ p$ or $ G \equal{} (\mathbb Z / 2 \mathbb Z) \times G_p$ for $ p$ odd or infinity; in the latter case, an automorphism maps the generator of $ \mathbb Z / 2 \mathbb Z$ to itself (it is the only element of order $ 2$). We are thus left to check the property for $ p$-groups.
Until here, no finiteness assumption was needed, only that we can factor the group into torsion and torsionfree parts.
But if we use the classification of finitely generated abelian $ p$-groups, we get that $ G_p$ is isomorphic to some product of groups of type $ \mathbb Z / p^s \mathbb Z$ (note this is true for $ p \equal{} 0$, too!). Again, if there is more than one factor, the automorphism group isn't cyclic by the same reason as before.
So $ G_p \equal{} \mathbb Z / p^s \mathbb Z$, so $ G$ is cyclic.

The general question may be even independent from ZFC, as some other general properties about abelian groups are.

It is possible.  There exists a torsion-free abelian group, cardinality of the continuum, with Aut group cyclic of order 2.  I believe a detailed exposition of this is given in Fuchs' "Abelian Groups", but I don't have my copy handy for the next week or two.  I found some references to this group given, suggesting it is considered on page 180.

Also note that every abelian group with order $ > 2$ has an automorphism of order 2, and so $ Aut(G)$, if cyclic, always contains a copy of $ \mathbb{Z}/2\mathbb{Z}$.

Steve

EDIT - I found it, in Fuchs' "Infinite Abelian Groups", volume II, pages 268ff.  Of interest is the Hallett-Hirsch theorem on page 270 (describing all possible finite automorphism groups for torsion-free abelian groups), and Example 1 on page 271, which constructs a torsion-free abelian group with automorphism group cyclic of order 4.  The group I mentioned above is contained in exercises 1 and 2 on page 274 [though I could swear it was done in the text in the 1958 edition; oh well].

[quote="ZetaX"]
Until here, no finiteness assumption was needed, only that we can factor the group into torsion and torsionfree parts.[/quote]

It is not in general true that abelian groups have their torsion piece as direct summand.  What further conditions are you assuming? A sufficient condition is if $ tor(G)$ has finite exponent, since it is a pure subgroup (this is often called the Kulikov-Prufer theorem), but for example,
$ \displaystyle\prod_{p\in\mathbb{P}}\mathbb{Z}_p$ cannot be split over its torsion subgroup.

Steve


Sas grafw mia askisi me dianismata pou me provlimatise:

Estw ta mi sineutheiaka dianismata $\overrightarrow{a},\overrightarrow{b}\neq \overrightarrow{0}$. Na vrethei i lisi tis eksiswsis ws pros $x$. [color=red](EDIT: Evgala tin proipothesi $x<0$ giati linontas tin katalava oti pio poli mperdeuei para dieukolinei. Den xreiazetai loipon auti i proipothesi)[/color]

$\frac{\overrightarrow{a}^{2}+x\cdot\overrightarrow{a}\cdot\overrightarrow{b}}{|\overrightarrow{a}|\cdot |\overrightarrow{a}+x\cdot \overrightarrow{b}|}=\frac{\overrightarrow{b}^{2}+\overrightarrow{a}\cdot \overrightarrow{b}}{|\overrightarrow{b}|\cdot |\overrightarrow{a}+\overrightarrow{b}|}$

Alexandros

μια διόρθωση(πιστεύω να είναι σωστή :oops: )
Αφού είναι μη συγραμμικά είναι και μη μηδενικά. :roll:

To mideniko dianisma einai siggramiko me ola apo ton orismo. Alla ok, egw apla to tonisa gia na exei noima to klasma!

Asxolitheite me tin askisi! Einai poli omorfi kai sinama tha sas kanei na anazitisete diaforous tropous gia na tin lisete. Episis nomizw tha ksekatharisete kai polla pragmata sta dianismata pou den isxioun opws stous pragmatikous (p.x. h prosetairistikotita tou pollaplasiasmou ktl ktl...)

Alexandros

όντως ωραία , θυμάμε κάτι παλιές που για να τις λύσω πολλαπλασίαζω με κατάλληλο διάνυσμα ( εδώ μάλλον όχι).

[hide]Ανοίγω τον τύπο του εσωτερικού γινομένου διώχνω τα μέτρα α και β και μετά μπλέκω ... :huh: [/hide]

Eixa prospathisei ki egw stin arxi me ton tropo pou anafereis, alla distixws den mporousa meta na vgalw mia kaliteri sxesi apo tin sxesi pou sindeei ta sinimitona dio dianismatwn ki etsi prospathisa na to dw alliws. Pragmati edw den xreiazetai "magoulikia" kai pollaplasiasmoi me katallilo dianisma. Thelei apla kalo xeirismo sto ksekinima tis.

Giauto tin parethesa! EPeidi mou arese!

a... evgala tin proipothesi $x<0$ (koitakste kai stin ekfwnisi) dioti pio poli mperdeuei para dieukolinei.
Alexandros

Η άσκηση είχε μπει σε έναν διαγωνισμό της ΕΜΕ παλιά .
Βάζω τα βασικά στοιχεία της λύσης μου .
Από τη δοθείσα βγαίνει ότι η γωνία
μεταξύ των διανυσμάτων $\vec{a}$ kai $\vec{a}+x\vec{b}$
είναι ίση με τη γωνία μεταξύ των διανυσμάτων $\vec{a}$  και
$\vec{b}+\vec{a}$

Μετά δεν είναι δύσκολο να βρούμε ότι
αν $x>0$  τότε $\vec{x}=\frac{\left|\vec{a}\right|^{2}}{\left|\vec{b}\right|^{2}}$

Kai an $x<0$  tote
$x=\frac{-\vec{a}^{2}}{\vec{b}^{2}+2\vec{a}\cdot\vec{b}}$

[quote="silouan"]
Μετά δεν είναι δύσκολο να βρούμε ότι
αν $x>0$  τότε $\vec{x}=\frac{\left|\vec{a}\right|^{2}}{\left|\vec{b}\right|^{2}}$

Kai an $x<0$  tote
$x=\frac{-\vec{a}^{2}}{\vec{b}^{2}+2\vec{a}\cdot\vec{b}}$[/quote]

Den iksera oti eixe mpei se diagwnismo tis EME. Pote?

Pantws nomizw oti akrivws apo ekei kai meta einai i diskolia tis askisis. Apo tin isotita autwn twn gwniwn na pas se lisi tis eksiswsis. Ektos an einai kati poli poli profanes pou den vlepw.

Oi liseis einai akrivws autes pou grafeis pragmati. An mporeis, grapse tin skepsi sou apo ekei kai pera. Egw as poume gia na tin lisw den xrisimopoiisa katholou gwnies parolo pou ki emena auti itan i prwti skepsi.

Alexandros

Λοιπόν φτιάξε τα διανύσματα (όχι παράλληλα ) κατασκεύσε τα $\vec{a}+\vec{b}$  κτλ και βρες όμοια τρίγωνα που σχηματίζονται σε κάθε περίπτωση .
(Αν δεν βρεις κάτι πες μου να το γράψω αναλυτικά)

Grafw tin lisi mou gia to thema auto to opoio me paidepse einai i alitheia giati exw na dw kairo eksiswseis me dianismata.

Loipon, oli mou i idea (afou den perpatise auto pou skeftika stin arxi me tis gwnies) itan na kanw sintomes prakseis se auto ton tropo pou tha efarmoza. Nomizw oti kai auto akoma einai mia poli diskoli diadikasia kai anadeikniei omorfia, mesa ston xamo twn praksewn.

Etsi loipon thimithika tin idiotita twn analogiwn pou leei oti

$\frac{a}{b}=\frac{c}{d}\Longleftrightarrow \frac{a+b}{a-b}=\frac{c+d}{c-d}$.

Ypswnoume loipon kai ta dio meli eis to tetragwnoki exoume

$\frac{\vec{a}^{4}+2x(\vec{a}\vec{b})\vec{a}^{2}+x^{2}(\vec{a}\vec{b})^{2}}{\vec{a}^{2}(\vec{a}^{2}+2x\vec{a}\vec{b}+x^{2}\vec{b}^{2})}=\frac{\vec{b}^{4}+2(\vec{a}\vec{b})\vec{b}^{2}+(\vec{a}\vec{b})^{2}}{\vec{b}^{2}(\vec{a}^{2}+2\vec{a}\vec{b}+\vec{b}^{2})}\Longleftrightarrow$

(idiotita analogiwn pou egrapsa parapanw)

$\frac{2\vec{a}^{4}+4x(\vec{a}\vec{b})\vec{a}^{2}+x^{2}\vec{a}^{2}\vec{b}^{2}}{x^{2}[(\vec{a}\vec{b})^{2}-\vec{a}^{2}\vec{b}^{2}]}=\frac{2\vec{b}^{4}+4(\vec{a}\vec{b})\vec{b}^{2}+\vec{a}^{2}\vec{b}^{2}+(\vec{a}\vec{b})^{2}}{(\vec{a}\vec{b})^{2}-\vec{a}^{2}\vec{b}^{2}}\Longleftrightarrow$

( To $(\vec{a}\vec{b})^{2}-\vec{a}^{2}\vec{b}^{2}\neq 0$ dioti ta $\vec{a},\vec{b}$ einai mi sineutheiaka ara diagrafontai)

$x^{2}\vec{b}^{4}-\vec{a}^{4}+2\vec{b}^{2}(\vec{a}\vec{b})x^{2}-2\vec{a}^{2}(\vec{a}\vec{b})x=0\Longleftrightarrow$

$(x\vec{b}^{2}-\vec{a}^{2})[x\vec{b}^{2}+\vec{a}^{2}+2x(\vec{a}\vec{b})]=0\Longleftrightarrow$

$x\vec{b}^{2}=\vec{a}^{2}$ 'h $x\vec{b}^{2}+2x(\vec{a}\vec{b})=-\vec{a}^{2}$

ki etsi pairnoume tis dio liseis $x=\frac{\vec{a}^{2}}{\vec{b}^{2}}$ kai $x=-\frac{\vec{a}^{2}}{\vec{b}^{2}+2\vec{a}\vec{b}}$

Elpizw na sas arese o tropos proseggisis.

Tha epanelthw molis xalarwsw apo tis eksetaseis mou me askiseis pou fainetai i megali aksia twn analogiwn sto na mas vgazoun apo ti diskoli thesi se periptwseis praksewn. Einai texni na kaneis grigores prakseis. Fantasteite stin prwti isodinamia na min pairname idiotites analogiwn kai antitheta kaname apeutheias xiasti ti prakseis tha eixame na antimetwpisoume.

Silouan den mporw na katalavw ton tropo me ta omoia trigwna. An mporeis se parakalw parethese tin apodeiksi sou. Sigoura tha einai pio omorfi afou linetai gewmetrika.

Alexandros


Prove that, if $z^{5}=1$,

$\frac{z}{1+z^{2}}+\frac{z^{2}}{1+z^{4}}+\frac{z^{3}}{1+z}+\frac{z^{4}}{1+z^{3}}=2$

and

$\frac{z}{1-z^{2}}+\frac{z^{2}}{1-z^{4}}+\frac{z^{3}}{1-z}+\frac{z^{4}}{1-z^{3}}=0$

I know you could do these if you clear the denominators and keep reducing the powers of $z$ using $z^{5}=1$, but is there a nicer way than that? But if you have any solution, even the one I mentioned above, I'd be glad to see it. Thanks. :)

This may be your solution. I wonder what a nice solution is. A geometric one, maybe?

Anyways. Ahh, [b]M4RI0[/b], you beat me this time. My solution is slightly different from yours though the essential idea is the same!


[hide] 9 (C-6) = (F-6) and F+1=5(C+1). Solving for F gives F= 69?, so is Mr. Frost 69 yrs old?[/hide]

[hide]
Carl's age is $13$.
Mr.Frost's age is $69$.
[/hide]

[quote="ZhangPeijin"]Carl how old are you[/quote]

Well if you work through the algebra you'll figure out...

Ok I got it you're 13..
Mr. Frost doesnt look like hes 69 though

Except that my mother won't even be that old for another 14 years!!

Man Leet you couldn't think of someone else who was actually within say 35 years of that age???

Thanks Peijin, I'm 33

[quote="13375P34K43V312"]Six years ago, Mr. Frost was nine times as old as Carl. Next year, Mr. Frost will be five times as old as Carl. How old is Mr. Frost?[/quote]
[hide]
$f-6=9(c-6)$
$f+1=5(c+1)$
$-7=4c-59$
$52=4c$
$c=13$
$f=69$[/hide]

[quote="13375P34K43V312"]Six years ago, Mr. Frost was nine times as old as Carl. Next year, Mr. Frost will be five times as old as Carl. How old is Mr. Frost?[/quote]

[hide]
carl=l (for lowell on the redsox team)
frost=r (for ramirez on the redsox team)

$r-6=9(l-6)$

$r+1=5(l+1)$

$52=4l$

$l=13$

$r=69$

carl (whoever that is) is 13 and mr. frost is 69

GO REDSOX
[/hide]jorian

Wow, that's really cruel.  Maybe you should think about what you're posting before you post it, everyone who thought that Mr. Frost was 69.  That's really low 133... (w/e).

not really. do you know they're on the same team? that might help...

whoops.  i thought that they were on separate teams.  he was just joking then.  i thought that they didn't know each other at all. :blush:

dude

he's [b]69[/b] years old

i'm gonna open a store

instead of being called forever 21 or w/e the name of that store is

it'll be called forever 69

Wow, this topic is never ending.  The answer was said and it is still refuted. hahahaha  :)


Factor 8051 into a product of positive prime numbers.


geometry rectangle probability inequalities

Consider a square ABCD with side length 15. Four lines parallel to AB split the square into strips with widths 1, 2, 3, 4 and 5 (not necessarily in that order.) Then four more lines parallel to BC split the square in the same fashion, not necessarily in the same order as the first. Now we have a grid of 25 rectangles.

Five of these rectangles belong to the set R. They are arranged diagonally from A to C. (see pic)

1. What is the maximum area of R?
2. Minimum?
3. Most common? (If lines are selected randomly, which area has the highest probability?)

[hide]
This screams rearrangement inequality.  We have one set of integers, lets call it A, that is made up of the width of segments on side AB, and another set, B, that is made up of the width of segments on AD.  We want to find the maximal value of $R=a_1b_1+a_2b_2+a_3b_3+a_4b_4+a_5b_5$.  The rearrangemet inequality tells us that R is maximized when $a_1 \leq a_2 \leq a_3 \leq a_4 \leq a_5$ and $b_1 \leq b_2 \leq b_3 \leq b_4 \leq b_5$.
Thus we see that the maximum value of R is $1^2+2^2+3^2+4^2+5^2=55$.

The rearrangment inequality also tells us that the minimum value occurs when $a_1 \leq a_2 \leq a_3 \leq a_4 \leq a_5$ and $b_1 \geq b_2 \geq b_3 \geq b_4 \geq b_5$.  Thus the minimum value of R is $1 \cdot 5 + 2 \cdot 4 +3 \cdot 3 + 4 \cdot 3 + 5 \cdot 1 =  35$.
[/hide]

I have an idea to make the third part of the problem simpler, but I don't know any way to solve it without counting all the possibilities.

[hide]The area of $R$ is equal to $a_1b_1+a_2b_2+a_3b_3+a_4b_4+a_5b_5$ where each $a$ term is from the set ${1,2,3,4,5}$ so that each number in the set is used once and where each $b$ term is also from the set ${1,2,3,4,5}$ so that each number in the set is used once.  Right now, we have $10!$ possibilities to consider.  But, it does not matter which order the $a$ terms or the $b$ terms are in.  All that matters is which $b$ terms are matched with each $a$ term.  Therefore, we may assume $a_1=1$, $a_2=2$, and so on without loss of generality.  We now have only $5!$ possibilities to consider.[/hide]


calculus
function
trigonometry
calculus computations

-Rolling a Ball Bearing-
A ball bearing is placed on an incline plane and begins to roll. The angle of elevation of the plane is θ.
The distance (in meters) the ball bearing rolls in t seconds is

s(t) = 4.9(sin θ)t^2

a. Determine the speed of the ball bearing.

b. What value of θ will produce the maximum speed at a particular time.

Speed is given by $ \left|s'(t)\right|$.  That should clear up the first part.  For the second part, you want to maximize the spped as a function of angle.  Note that in reality, $ s\equal{}s(\theta,t)$ is a multivariable function.  So we have
$ \frac{\partial s}{\partial \theta}\equal{}4.9t^{2}\cos{\theta}\equal{}0\Rightarrow\theta\equal{}\frac{\pi}{2}$
which makes sense.


In triangle $ABC$, $\measuredangle CBA=72^\circ$, $E$ is the midpoint of side $AC$, and $D$ is a point on side $BC$ such that $2BD=DC$; $AD$ and $BE$ intersect at $F$. The ratio of the area of triangle $BDF$ to the area of quadrilateral $FDCE$ is

[asy]
size(200);
defaultpen(linewidth(0.7)+fontsize(10));
pair B=origin, C=(15,3), D=(5,1), A=7*dir(72)*dir(B--C), E=midpoint(A--C), F=intersectionpoint(A--D, B--E);
draw(E--B--A--C--B^^A--D);
label("$A$", A, dir(D--A));
label("$B$", B, dir(E--B));
label("$C$", C, dir(0));
label("$D$", D, SE);
label("$E$", E, N);
label("$F$", F, dir(80));[/asy]

$\text{(A)} \ \frac 15 \qquad \text{(B)} \ \frac 14 \qquad \text{(C)} \ \frac 13 \qquad \text{(D)} \ \frac 25 \qquad \text{(E)} \ \text{none of these}$

probably wrong solution using mass points...
[hide]
Set weights of 2 to B, 1 to C, thus 3 to D, 1 to A, 2 to E, making F 4.  So BF=FE, so if we extend a line parallel to AD from E, intersecting BC at G, EBG is similar to FBD by a scale of 2, so the height from F of FBD is half the height from E of EBC.  Denoting BD = b, height from F = h, we have BC = 3b, height from E = 2h, making EBC 6 times as large as FBC.  So the desired ratio is 1/5, or A
[/hide]

:D I love mass points.  I didn't even need to use the angle given.

that was the point of giving the angle...all it does is that somebody could not assume it was say an isosceles right triangle and brute force it, eventhough if you know these types of problems, it is obvious the angle really does nothing

[quote="4everwise"]...$AE$ and $BE$ intersect at $F$...

[img]http://www.artofproblemsolving.com/Forum/album_pic.php?pic_id=466[/img][/quote]

$AE$ and $BE$ intersect at $E$... :D

You could also solve this problem by adding a line through E parallel to AD.  Then BD/DC = 1/2 and DG/DC = 1/2, therefore, BD/DG = 1 (or BD = DG).  From there BF/BE = 1/2 and BD/BC = 1/3.  [BDF]/[BCE] = 1/6 and [BDF]/[CDFE] = 1/5.

Could anyone tell me what is [b]mass points[/b]?

Thanks ;)  :)

[quote="chess64"]http://mathcircle.berkeley.edu/mpgeo.pdf[/quote]

Thanks :)  ;)

Hasn't anybody tried using area ratios yet?  :D

[hide="Solution"]Answer: $\text{(A)}$

[img]http://www.artofproblemsolving.com/Forum/album_pic.php?pic_id=535[/img]

First draw line segment $EG$ such that $EG||ED$ and $G$ is the midpoint of $DC$.
We know that $FD$ is the midpoint of $BE$ because it passes the midpoint of $BG$ and is parallel to one side of the triangle.

Now, we see that $\frac{\text{area} \triangle BDF}{\text{area}\triangle BEG}=\frac 14$, and $\frac{\text{area}\triangle EBG}{\text{area}\triangle EBC}=\frac 23$.

Thus, $\frac{\text{area}\triangle BDF}{\text{area}\triangle BEC}=\frac 16$, and $\frac{\text{area}\triangle BDF}{\text{area} CEFG}=\frac{1}{5}$.[/hide]

[quote="4everwise"]Answer: $ \text{(A)}$

[img]http://www.artofproblemsolving.com/Forum/album_pic.php?pic_id=535[/img]

First draw line segment $ EG$ such that $ EG\parallel{}ED$ and $ G$ is the midpoint of $ DC$.
[/quote]
I'm don't see how it's possible to have $ EG\parallel{}ED$ from my drawing...
also, how come your picture isn't showing up?  :huh:

We know BD=x, DC=2x, by Menelaus, BF=FE=n. So [FBD]=(1/2)xnsinFBD and [EBC]=(1/2)2n3xsinFBD. So [FBD] is 1 part, [EBC] is 6 parts, so [FDCE] is 5 parts. Thus the answer is A.

Yet another way to solve this:
[hide]
1. Join F and C:
[asy]
defaultpen(linewidth(0.7)+fontsize(10));
pair B=origin, C=(15,3), D=(5,1), A=7*dir(72)*dir(B--C), E=midpoint(A--C), F=intersectionpoint(A--D, B--E);
draw(E--B--A--C--B^^A--D);
draw(F--C);
label("$A$", A, dir(D--A));
label("$B$", B, dir(E--B));
label("$C$", C, dir(0));
label("$D$", D, SE);
label("$E$", E, N);
label("$F$", F, dir(80));
[/asy]

2. Let the areas of $\triangle BFD$ and $\triangle AFE$ be $x$ and $y$, respectively:
[asy]
defaultpen(linewidth(0.7)+fontsize(10));
pair B=origin, C=(15,3), D=(5,1), A=7*dir(72)*dir(B--C), E=midpoint(A--C), F=intersectionpoint(A--D, B--E);
draw(E--B--A--C--B^^A--D);
draw(F--C);
label("$A$", A, dir(D--A));
label("$B$", B, dir(E--B));
label("$C$", C, dir(0));
label("$D$", D, SE);
label("$E$", E, N);
label("$F$", F, dir(80));
label("$x$",(3.2,1.2));
label("$y$",(4,4.7));
[/asy]

3. Then, the area of $\triangle CFE$ must be $y$ (since $AE=EC$) and the area of $\triangle FDC$ must be $2x$ (since $DC=2BD$):
[asy]
defaultpen(linewidth(0.7)+fontsize(10));
pair B=origin, C=(15,3), D=(5,1), A=7*dir(72)*dir(B--C), E=midpoint(A--C), F=intersectionpoint(A--D, B--E);
draw(E--B--A--C--B^^A--D);
draw(F--C);
label("$A$", A, dir(D--A));
label("$B$", B, dir(E--B));
label("$C$", C, dir(0));
label("$D$", D, SE);
label("$E$", E, N);
label("$F$", F, dir(80));
label("$x$",(3.2,1.2));
label("$y$",(4,4.7));
label("$2x$",(5.6,1.9));
label("$y$",(8.3,3.6));
[/asy]

4. The areas of $\triangle ABE$ and $\triangle CBE$ are equal (since $AE=CE$). The area of $\triangle CBE$ is $x+2x+y=3x+y$. Therefore, the area of $\triangle AFB$ must be $3x$:
[asy]
defaultpen(linewidth(0.7)+fontsize(10));
pair B=origin, C=(15,3), D=(5,1), A=7*dir(72)*dir(B--C), E=midpoint(A--C), F=intersectionpoint(A--D, B--E);
draw(E--B--A--C--B^^A--D);
draw(F--C);
label("$A$", A, dir(D--A));
label("$B$", B, dir(E--B));
label("$C$", C, dir(0));
label("$D$", D, SE);
label("$E$", E, N);
label("$F$", F, dir(80));
label("$x$",(3.2,1.2));
label("$y$",(4,4.7));
label("$2x$",(5.6,1.9));
label("$y$",(8.3,3.6));
label("$3x$",(1.9,3));
[/asy]

5. The area of $\triangle ACD$ is twice the area of $\triangle ABD$ (since $CD=2BD$):
$y+y+2x = 2(x+3x)$
Hence, $y=3x$.
Hence, the area of quadrilateral $FDCE$ is $y+2x = 3x+2x = 5x$.
Hence, the answer is $\frac 15$.

[/hide]

[quote="anthonyjang"][quote="chess64"]http://mathcircle.berkeley.edu/mpgeo.pdf[/quote]
link doesn't work. Any other good explanation of mass points? I have heard of them, but don't know what they are and their applications.[/quote]
link works for me.
google it or something

[color=darkred][b][u]PP[/u].[/b] Let $\triangle ABC\ ,\ \left\{\begin{array}{ccc}
D\in (BC) & ; & DB=m\cdot DC\\\
E\in (AC) & ; & EC=n\cdot EA\end{array}\right|$ and  $F\in AD\cap BE$ . Find the ratio $\frac {[BDF]}{[FDCE]}$ .[/color]

[b][u]Proof[/u].[/b] Observe that $\left\{\begin{array}{c}
\frac {DB}{m}=\frac {DC}{1}=\frac {BC}{m+1}\\\\
\frac {EC}{n}=\frac {EA}{1}=\frac {AC}{n+1}\end{array}\right|$ . Apply the [u][b]Menelaus[/b]' theorem[/u] to $\overline{AFD}/\triangle BCE\ :\ \frac {AE}{AC}\cdot\frac {DC}{DB}\cdot\frac {FB}{FE}=1$ $\iff$

$\frac {FB}{FE}=$ $\frac {AC}{AE}\cdot \frac {DB}{DC}=$ $m(n+1)$ , i.e. $\frac {FB}{m(n+1)}=$ $\frac {FE}{1}=$ $\frac {BE}{mn+m+1}$ . Therefore, $\frac {[BDF]}{[BCE]}=$ $\frac {BD}{BC}\cdot\frac {BF}{BE}=$

$\frac {m}{m+1}\cdot\frac {m(n+1)}{mn+m+1}$ $\iff$ $\frac {[BDF]}{m^2(n+1)}=$ $\frac {[BCE]}{(m+1)(mn+m+1)}=$ $\frac {[FDCE]}{mn+2m+1}$ $\iff$ $\boxed{\frac {[BDF]}{[FDCE]}=\frac {m^2(n+1)}{mn+2m+1}}$ .


a1 + a2 + a3 + ...  + an = c; [1 ,2, 3, ... are in subscript]
c is a given constant.

Find limiting values (max and min) of a1*a2 + a2*a3 + a3*a4 + ... + an*a1.

Please give the proof along with the answer.

Thank You

[hide=" HINT"]Cauchy Swarz and some algebra[/hide]

that was the best way u could have helped...
got my answer.. n benefited too...
i appreciate it..
thanx
 :)


real analysis
integration
limit
real analysis unsolved

Let $g_\varepsilon=m(B(0,\epsilon))^{-1}\chi_{B(0,\varepsilon)}$. Prove that if $f: \mathbb{R}^{n}\to \mathbb{R}$ is continuous then $g_\varepsilon *f(x)\to f(x)$ for all $x\in \mathbb{R}^{n}$ as $\varepsilon\to 0+$.

This can be rewritten as

$\frac1{m(B(x,\varepsilon))}\int_{B(x,\varepsilon)}f(y)\,dy\to f(x)$ as $\varepsilon\to0^{+}.$

This is a close relative of the Fundamental Theorem of Calculus. Note that

$f(x)=\frac1{m(B(x,\varepsilon))}\int_{B(x,\varepsilon)}f(x)\,dy$

Subtract that from the limit we want, and we want to show this:

$\lim_{\varepsilon\to0^{+}}\frac1{m(B(x,\varepsilon))}\int_{B(x,\varepsilon)}(f(y)-f(x))\,dy = 0.$

Now, what can you obtain from the continuity of $f$ at $x?$


Describe all real numbers x for which {x} + {1/x} = 1.

(Here {a} = a - [a] denotes the fractional part of a,
for example, {11.4} = 0.4, {-2.73} = 0.27, {5} = 0.)

Does there exist a rational number x for which {x} + {1/x} = 1 ?

WLOG 0<x<=1 since symmetrical.  (If x is a solution, so is 1/x, and f[x]=-f[-x].) So {x}+{1/x}=x-Floor[x]+(1/x)-Floor[1/x]=x+(1/x)-Floor[1/x]=1, x+(1/x)=1+Floor[1/x]=k. Solving we have x^2+1=kx, x^2-kx+1=0, discriminant k^2-4>=0, |k|=|1+Floor[1/x]|>=2 has solutions. So all x are given by the roots of x^2-kx+1=0, where |k|>=2, k integer.

Then in order for solution to be rational, note that k^2-4 must be square of integer. k^2-4=j^2, k^2-j^2=4, (k+j)(k-j)=4. (k+j=4, k-j=1 has no integer k solution), but (k+j=2, k-j=2 has solution k=2, j=0). So the only rational x that satisfies {x}+{1/x}=1 are x=+-1.

There is something wrong in your solution ... x = 1 is no solution !

Sorry, x=+-1 are degenerate cases. So I suppose there are no rational x. The problem was "x-Floor[x]+(1/x)-Floor[1/x]=x+(1/x)-Floor[1/x]=1" for x=1, which is not true. Quite a sneaky problem.

This is a British problem. Have a look at

http://www.kalva.demon.co.uk/bmo/bsoln/bsol902.html


calculus
derivative
search
calculus computations

y=$e^{-x}(cos2x+sin2x)$ show that y''+2y'+5y=0

Is there any short cuts to solve this problem?

As opposed to plugging it into the equation and taking the derivatives you can solve the ODE:
[hide]$y''+2y'+5y$ has the auxillary equation $m^{2}+2m+5$.  Assume that $e^{mx}$ is a solution .  Solving for $m$ we have $m =-1 \pm 2i \implies e^{(-1\pm2i)x}$ is a solution.  Use Euler's identity to find that $e^{(-1\pm2i)x}= e^{-x}(\cos2x+\sin2x)$.   [/hide]

[quote="Zakary"]illcrowflu, how often does that technique work?  Can you tell me more information about it or tell me enough so I can efficiently look it up?[/quote]
Any book has this method. Search for "Ordinary Differential Equations with Constant Coefficient".


Let $ f$ be a function from $ N^\plus{}\rightarrow N^\plus{}$,$ p$,$ k$ are two given positive number,and for any positive $ n$ we have $ f^{(p)}(n)\equal{}n\plus{}k$.Prove that: the necessary and sufficient condition that $ f$ exists is $ p|k$

[quote="math10"]Let $ f$ be a function from $ N^ \plus{} \rightarrow N^ \plus{}$,$ p$,$ k$ are two given positive number,and for any positive $ n$ we have $ f^{(p)}(n) \equal{} n \plus{} k$.Prove that: the necessary and sufficient condition that $ f$ exists is $ p|k$[/quote]

The condition is sufficient since $ k\equal{}mp$ $ \implies$ $ g^{(p)}(n)\equal{}n\plus{}k$ with $ g(x)\equal{}n\plus{}m$

In order to show that the condition is necessary :
1) obviously $ f(x)$ is injective
2) $ \forall i\in[0,p\minus{}1]$, let $ A_i\equal{}\{x\in[1,k]$ such that $ f^{(i)}(x)\in[1,k]$ and $ f^{(i\plus{}1)}(x)\notin[1,k]$
$ \forall i\in[1,p\minus{}1]$ : $ x\in A_i$ $ \implies$ $ f(x)\in A_{i\minus{}1}$ and so $ |A_i|\leq |A_{i\minus{}1}|$ (remember $ f(x)$ is injective)
$ \forall i\in [0,p\minus{}2]$ : $ x\in A_i$ $ \implies$ $ f^{(i\plus{}1)}(x)>k$ $ \implies$ $ f^{(p)}(f^{(i\plus{}1)}(x)\minus{}k)\equal{}f^{(i\plus{}1)}(x)$ $ \implies$ $ x\equal{}f(y)$ where $ y\equal{}f^{(p\minus{}i\minus{}1)}(f^{(i\plus{}1)}(x)\minus{}k)$ (remember $ f(x)$ is injective) $ \implies$ $ y\in A_{i\plus{}1}$ $ \implies$ $ |A_{i\plus{}1}| \geq |A_i|$

So all $ A_i$ have the same cardinal $ m$
And since $ A_i\cap A_j\equal{}\emptyset$ $ \forall i\neq j$ and $ \cup A_i\equal{}[1,k]$, we get $ k\equal{}pm$
Q.E.D.


I have a question:

Are infinitely many pairs of primes p, p+2? I mean has this been proved?
And if it has, do you think it can be used in competitions?

I think that the silence on this topic would mean that in the competitions it cannot be used
:)
Even if it were true
:)

Well, thanks anyway :D I looked it up on the internet. It's not proved, it's just a conjecture.


Εστω $ x,y,z$ ανηκουν στους θετικους ρητους και $ x^2 \plus{} y^2 \plus{} z^2 \equal{} xyz$ Ν.δ.ο.
$ xy \plus{} xz \plus{} yz\geq2(x \plus{} y \plus{} z) \plus{} 9$
Πισω στη λυση της γραφουν να προσπαθησετε και την δυσκολοτερη
$ xy \plus{} xz \plus{} yz\geq4(x \plus{} y \plus{} z) \minus{} 9$
Για την πρωτη εχω βρει μια αρκετα ομορφη λυση...Την δευτερη δεν την εχω προσπαθησει και πολυ.....

[quote="NickNafplio"]Ωραιο θεματακι

Για το 1ο εχουμε:

$ xyz \equal{} \sum{x^2} \geq \sum{xy} \Rightarrow xyz\sum{xy} \geq (\sum{xy})^2 \geq 3xyz(x \plus{} y \plus{} z) \Rightarrow \sum{xy} \geq 3(x \plus{} y \plus{} z)$.

Αρκει να δειξουμε τωρα οτι $ x \plus{} y \plus{} z \geq 9$.

Εχουμε ομως:
$ (\sum{x^2})(x \plus{} y \plus{} z) \equal{} xyz(x \plus{} y \plus{} z) \leq \frac {(\sum{xy})^2}{3} \leq \frac {(\sum{x^2})(x \plus{} y \plus{} z)^2}{9} \Rightarrow x \plus{} y \plus{} z \geq 9$
και η αποδειξη ολοκληρωθηκε.

Το 2ο θα το κοιταξω σε λιγο.[/quote]
Η δικια μου λυση ειναι η εξης για το πρωτο
Απο το $ xyz\equal{}x^2\plus{}y^2\plus{}x^2\leftrightarrow xy\equal{}\frac{x^2\plus{}y^2\plus{}z^2}{z}$
Αντιστοιχα και τα αλλα.
Αρα αρκει ν.δ.ο $ \frac{x^2\plus{}y^2\plus{}z^2}{x}\plus{}\frac{x^2\plus{}y^2\plus{}z^2}{y}\plus{}\frac{x^2\plus{}y^2\plus{}z^2}{z}\geq2(x\plus{}y\plus{}z)\plus{}9$
Δηλαδη μετα απο λιγες πραξεις$ (x^3y\plus{}y^3z\plus{}z^3x)\plus{}(x^3z\plus{}y^3x\plus{}z^3y) \geq x^2yz\plus{}xy^2z\plus{}xyz^2\plus{}9xyz$
Ειναι ομως $ x^3y\plus{}y^3z\plus{}z^3x\geq x^2yz\plus{}xy^2z\plus{}xyz^2$ πυ βγαινει ευκολα ειτε με B.C. ειτε με βαρη και $ x^3z\plus{}y^3x\plus{}z^3y\geq 9xyz$ η αλλιως αρκει (χρισημοποιωντας την am-gm για το $ RHS$) $ xyz\geq27$ πραγμα που προκυπτει απο τη δοσμενη με απλη am-gm.....

[quote="new_member"]mporei na mou eksigisei kapoios pou xrisimopoithike to gegonos oti oi arithmoi anikoun sto Q+.
Den isxyei ean x,y,z einai sto R+?[/quote]

Χωρίς να έχω κοιτάξει την άσκηση (sorry...), οι περιορισμοί και οι ανισότητές σου (που δεν είναι γνήσιες) αφορούν μόνο συνεχείς συναρτήσεις, οπότε αν κάτι ισχύει στο $ \mathbb{Q}$ μάλλον θα περνάει και στο $ \mathbb{R}$.

Ωστόσο, μπορεί να λέω χαζομάρες επειδή το κοίταξα μόνο για δευτερόλεπτα... :oops:

Cheerio,

Durandal 1707

[quote="new_member"]mporei na mou eksigisei kapoios pou xrisimopoithike to gegonos oti oi arithmoi anikoun sto Q+.
Den isxyei ean x,y,z einai sto R+?[/quote]
Η αληθεια ειναι οτι στρην εκφωνηση της ασκησης λεει θετικους πραγματικους αλλα απο κεκτιμενη ταχυτητα εγραψα θετικους ρητους..... :oops:


Find all functions $ f :\mathbb R\to\mathbb R$ satisfying
$ f(x - f(y)) = f(x)+xf(y)+f(f(y))$
for all $ x, y\in\mathbb R$.

$ f(x\minus{}f(y))\equal{}f(x)\plus{}xf(y)\plus{}f(f(y))$ posted before


find all polynomials $ P(x) \equal{} ax^3 \plus{} b x^2 \plus{} cx \plus{} d ; P(x \plus{} y)\geq P(x) \plus{} p(y)$

$ p(y)\le p(x \plus{} y) \minus{} p(x) \equal{} y[a(y^2 \plus{} 3yx \plus{} 3x^2) \plus{} b(y \plus{} 2x) \plus{} c]\to 0$, when $ y\to 0$, therefore $ p(0)\le 0$.
$ p(x \plus{} y) \minus{} p(x) \minus{} p(y) \equal{} xy[3a(x \plus{} y) \plus{} 2b ] \minus{} d\ge 0$. If $ x \equal{} t,y \equal{} \minus{} t, t\to \infty$, we get $ b\le 0$, $ x \equal{} t \plus{} 1,y \equal{} \minus{} t\to a\le0$.
$ x \equal{} t,y \equal{} t\to a\ge 0$. Therefore $ a \equal{} 0$ and $ b\ge 0$ or $ b \equal{} 0$.
It give $ P(x) \equal{} cx \plus{} d,d\le 0$.
It is true for any polinomial function.


inequalities
trigonometry
inequalities unsolved

[quote="e.lopes"]$\frac{\cos(B-C)}{\cos A}=\frac{\sin(B+C) \cos(B-C)}{\sin A \cos A}=\frac{\sin 2B+\sin 2C}{\sin 2A }$.
so, by AM-GM, the result follows.  :)[/quote]
[quote="shyong"]For $A,B,C>0$ and $A+B+C=\frac{\pi}{2}$
Prove that
$\cos (A-B)\cos (B-C)\cos (C-A)\geq 8\sin A\sin B\sin C$[/quote]
$A+B+C=\frac{\pi}{2}!$ ;)

[quote="e.lopes"]$\frac{\cos(B-C)}{\cos A}=\frac{\sin(B+C) \cos(B-C)}{\sin A \cos A}=\frac{\sin 2B+\sin 2C}{\sin 2A }$.
so, by AM-GM, the result follows.  :)[/quote]

abit mistake there but you almost got it anyway  :)


abstract algebra
function
group theory
induction
vector
number theory
greatest common divisor

Show that a finitely generated abelian group which is torsion-free is in fact free abelian.

i usually walk away from 'free group' problems
but if your group would be abelian, it certainly by the classificiation theorem would be isomorphic to a product of a bunch of copies of Z , with also a couple of finite cyclic groups
as soon as you actually would have such a nontrivial finite cyclic group, just pick a generator of that cyclic finite group and you have an element of finite order

but probably, you are trying to prove the classification theorem and you need this fact... :ninja:

I think I've finally got it. Let $(a_i)$ with $1\leq i\leq n$ a $Z$ generator of this group. Consider the function $f(k_1,...,k_n)=k_1 a_1+...+k_n a_n$ defined on $Z^n$. This function is surjective and also your group $G$ is isomorphic to $Z^n/ Ker f$. Thus $H=Ker f$ is a subgroup of $Z^n$ such that $Z^n/H$ is torsion free.Let's prove that $Z^n/H$ will be of the form $Z^k$ in this case. The proof is by induction on $n$. If $n=1$ it is obvious. Suppose now that $H$ is a subgroup in $Z^n$ such that $Z^n/H$ is torsion free. Take $v$ a non zero vector in $Z^n$ and $c$ its content (the gcd of its components). Then it is easy to see that $\frac{1}{c} v$ is in $H$ (because $Z^n/H$ is torsion free). Also, it is well known that a vector all of whose components are relatively prime can be extended to a $Z$ base of $Z^n$. Consider such a base $v_1,...,v_n$ with $v_1=v$. It is also obvious that $Z^n/H$ is isomorphic to $Z^{n-1}/ Ker g$ where $g$ is the natural morphism from $Z^{n-1}$ into $Z^n/H$. Now just apply the inductive hypothesis.
      Thus we have shown that $G$ is isomorphic to $Z^k$ and the conclusion follows.

[quote="harazi"] Take $v$ a non zero vector in $Z^n$ and $c$ its content (the gcd of its components). Then it is easy to see that $\frac{1}{c} v$ is in $H$ (because $Z^n/H$ is torsion free). [/quote]

I think $v$ must be a non-zero vector of $H$.
The proof is beautiful.

You can find it in any group theory book; this allows you to classify all finitely generated abelian groups. See for instance Algebra -Lang-, Th. 8.5 p.46.


$x$, $y$, and $z$ are the angles of a triangle. Find the maximum value of
\[\sin^{2}x+\sin y \sin z \cos x \]

[hide="Partial"]
$\sin^{2}x+\sin y \sin z \cos x$
$\Rightarrow 1-\cos^{2}x+\sin y \sin z\cos x$
$\Rightarrow 1-\cos x(\cos{x}-\sin{y}\sin{z})$

if $\cos{x}<0$ then
$\Rightarrow\cos{x}-\sin{y}\sin{z}<0$
$\Rightarrow\cos x(\cos{x}-\sin{y}\sin{z})>0$
$\Rightarrow\sin^{2}x+\sin y \sin z \cos x<1$

if $\cos{x}=0$ then
$\Rightarrow\sin^{2}x+\sin y \sin z \cos x=1$

if $\cos{x}>0$ then

$\Rightarrow\cos{x}<\sin{y}\sin{z}$, for the expression to be greater then one

$\Rightarrow\cos{x}=-\cos{(180-x)}=-\cos{(y+z)}=-(\cos{y}\cos{z}-\sin{y}\sin{z})=\sin{y}\sin{z}-\cos{y}\cos{z}$

so

$\rightarrow\sin{y}\sin{z}-\cos{y}\cos{z}<\sin{y}\sin{z}$
$\rightarrow-\cos{y}\cos{z}<0$
$\rightarrow \cos{y}\cos{z}>0$

STuck..??
[/hide]

EDIT: I made a dumb mistkae but i still cant wither away at it

[hide="Manipulative (incomplete)"] Let $z = 180-(x+y)$.  Then we wish to maximize

\begin{eqnarray*}\sin^{2}x+\sin y \cos x \sin (x+y) &=& \sin^{2}x+2 \sin^{2}y \cos^{2}x+2 \sin y \sin x \cos y \cos x \\ &=& \frac{1-\cos 2x+(1-\cos 2y)(1+\cos 2x)+\sin 2y \sin 2x}{2}\\ &=& \frac{2-\cos 2y-\cos 2x \cos 2y+\sin 2x \sin 2y}{2}\\ &=& \frac{2-\cos 2y-\cos (2x+2y)}{2}\\ \end{eqnarray*}

We want to minimize $\cos 2y+\cos (2x+2y)$.  Ideally, they are close to $-1$.  [/hide]

I had gotten to the same point where t0rajir0u posted, but couldn't find a nice answer. Any help?

Really?  I thought my expression was wrong because the limiting case I found gives the wrong answer when I plug it back into the original expression.

hmm I dont think thats been expanded correctly..

if we let $z=180-(x+y)$ then

\begin{eqnarray*}\sin^{2}+\sin{y}\cos{x}\sin{(x+y)}&=& \sin^{2}x+\sin^{2}y\cos^{2}x+\sin{x}\sin y\cos x\cos y\\&=& \frac{4\sin^{2}x+4\sin^{2}y\cos^{2}x+\sin{(2x)}\sin{(2y)}}{4}\\&=& \frac{4-4\cos^{2}x(1-\sin^{2}y)+\sin{(2x)}\sin{(2y)}}{4}\\&=& \frac{4(1-\cos x\cos y)(1+\cos x\cos y)+\sin{(2x)}\sin{(2y)}}{4}\end{eqnarray*}

...?!?

Oh.  For some reason I was using $\sin (x+y) = 2 \sin x \cos y+2 \sin y \cos x$, which would explain a [b]lot.[/b]    :oops:


Bacterial digestion is an economical method of sewage treatment.  The reaction: [url]http://www.msnusers.com/dlf62lv8o768ej84v84e04i2t4/Documents/chem.htm[/url] is an intermediate step in the conversion of the nitrogen in organic compounds into nitrate ions.  How much bacterial tissue (C5H7O2N) is produced in a treatment plant for every 1.0 x 10^4 kg of wastewater containing 3.0% NH4 ions by mass?  Assume that 95% of the ammonium ions are consumed by the bacteria.

I have no clue how to work this problem...Please help me.

i cant see the equation but:

for all stoichiometry problems, start with all your known facts, then build up to come to a step that you know how to do.

for this particular problem, you have to convert the NH4 into number of moles, then do the reverse percentage and using the equation, you will get the solution

I'm not sure how to convert the 3.0% of NH4 into a mass.  Please help.

Start with the 3.0% of NH4.  Multiply it by what is it 3% of and you have a mass of NH4.

If you want N, simply multiply this result of the first calculation by the molecular weight of Nitrogen (14 I believe) and divide by the molecular weight of NH4 (18 I believe).

Note I said the molecular weight of N, since I choose to use the general term, sometimes you may have the atom in interest showing up in more than one place, such as the H in the NH4.

You should be "off to the races."


Hello to all mathlinkers.please recommend me a good collection of function problems.
thank you very much.

There was a book by B.J.Venkatachala I forgot the name, you can try that!

FUNCTIONAL EQUATIONS BY B.J.VENKATACHALA...PRISM PUBLICATIONS


how do we convert pentanedioicacid  to 5-hydroxypentanoic acid . or reduce only one ofthe carboxy groups. I'm struck . both always react .tried heating and getting anhydride , then adding an alcohol to getester and then convert the acid to acid chloride and the n to amide and so on , but at some stage , I get caught and can't proceed further. help.

This is already being discussed [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=235476]here[/url], so why making a new post?

Theoretically, this forum has two moderators, but seems that neither one is interested in fulfilling their obligation.


ratio
geometry
perimeter
trigonometry
geometry unsolved

Let ABC be an equilateral triangle with the side length is 1. An changed line cut the edges AB, AC at M, N;respectively such that perimeter os triangle AMN is the same perimeter of quadrilateral BCNM. Let $S_{1},S_{2}$ be areas of AMN, BCNM, respectively. Prove that $\frac{S_{1}}{S_{2}}\geq 1$

See the attached diagram.

$AM+AN=MB+BC+CN=s={3\over 2}$. If $x : = BM$, then the other segments can be denoted as shown.

\begin{eqnarray*}{S_{1}\over S_{2}}\geqslant 1 &\iff& \frac{S_{1}}{{\sqrt{3}\over 4}-S_{1}}\geqslant 1\\ &\iff& S_{1}\geqslant{\sqrt{3}\over 8}\\ &\iff&{1\over 2}(1-x)({1\over 2}+x)\sin 60^\circ\geqslant{\sqrt{3}\over 8}\\ &\iff& (1-x)({1\over 2}+x)\geqslant{1\over 2}\\ &\iff&{x\over 2}-x^{2}\geqslant 0\\ &\iff& x({1\over 2}-x)\geqslant 0\\ &\iff& BM\cdot CN\geqslant 0\end{eqnarray*}

which is obviously always fulfilled, and the equality holds iff $BMNC$ is degenerate, i.e. iff $M\equiv B$ or $N\equiv C$.


search
real analysis
real analysis unsolved

From my lecture in class today:

Compute $ \sum_{n\equal{}0}^{\infty}\frac{1}{1\plus{}n^2}.$

[hide="And what class was that?"]The title of the class is "Fourier Analysis."[/hide]

It was discussed [url=http://www.mathlinks.ro/viewtopic.php?search_id=783191057&t=174442]here[/url].


AB = 80 ; BD = 20 ; AC = 60 ; EC = 20 & BC = 50

Find the area of shaded region in figure below.

[hide]
Let $\triangle ADE = a$

$\triangle DEC = b$

$\triangle DCB = c$

$a+b: c \implies 3: 1$ same height.

$a: b \implies 2: 1$

$a=2b$

$\frac{3b}{c} = 3 \implies b=c$

$a: b: c \implies 2: 1: 1$

$|\triangle ABC| = \sqrt{95\cdot15\cdot35\cdot45} = 1498.123827$

$\implies b = 374.531$

[/hide]

oopsies made a little mistake earlier. :blush:

[quote="sen"][hide]
Let $\triangle ADE = a$

$\triangle DEC = b$

$\triangle DCB = c$

$a+b: c \implies 3: 1$ same height.

$a: b \implies 2: 1$

$a=2b$

$\frac{3b}{c} = 3 \implies b=c$

$a: b: c \implies 1: 2: 2$

$|\triangle ABC| = \sqrt{95\cdot15\cdot35\cdot45} = 1498.123827$

$\implies b = 599.25$

[/hide][/quote]

thanks !

[hide]I got $18.75\sqrt{399}$, which is same like sen's answer.[/hide]

[hide]
By Heron's, we get the area of $ABC$ to be $25 \sqrt{19 \cdot 3 \cdot 7 \cdot 9} = 75 \sqrt{399}$
By area ratios, we find $BDC = \frac 1 4 ABC \implies DAC = \frac 3 4 ABC$
Also, $DEC = \frac 1 3 DAC \implies DEC = \frac 1 4 ABC = \frac{75\sqrt{399}}4$
[/hide]

[hide]The area of triangle CEP (the shaded region) is 1/3 that of triangle APC, since it's base is 1/3 that of triangle APC, and they share the same height.
The area of triangle APC is 3/4 that of triangle ABC, since it's base is 1/4 that of triangle, and they share the same height.
The area of triangle CEP is therefore (1/3)(3/4)=1/4 that of triangle ABC.
The area of CEP $=\frac{\sqrt{95(15)(35)(45)}}{4}=\frac{75\sqrt{399}}{4}$.[/hide]


calculus
integration
function
calculus computations

For real number $x$,find $cosx-\frac{1}{3}cos3x+\frac{1}{5}cos5x-...$

Not a simple problem; this belongs to Fourier series.

One possible formula for this function is $\text{Re}(\arctan(e^{ix})).$

Here's the [hide="real answer:"]
$f(x)=\begin{cases}\frac{\pi}2,&-\frac{\pi}2<x<\frac{\pi}2\\-\frac{\pi}2,&\frac{\pi}2<x<\frac{3\pi}2\\ 0,&x=\frac{\pi}2\text{ or }x=\frac{3\pi}2\end{cases},$

continued to be periodic of period $2\pi.$

To see that this is correct, compute the Fourier series of this function.[/hide]

Thank you very much.
I got the first expression,but I used to think it's always $\frac{\pi}2$. :lol:


Equilateral triangle ABC has AB=1. A circle K has the altitude from A to BC as diameter. The circle K intersecta AB and AC at P and Q, resp. What is PQ?


Let $a$ and $b$ be positive integers for which there exists a non-constant real function $f$
satisfying \[ f (\frac{ax+by}{a+b}) = af(y)+bf(x) \] Prove that $a = b$.

$a$ and $b$ can't be positive integers because pluggin $y=x$ we get:

$f(x)=(a+b)f(x)$ and as $f(x)$ is nonconstant we get that $a+b=1$ which is a contradiction

i am sorry  :blush:
\[ f (\frac{ax+by}{a+b}) = \frac{af(y)+bf(x)}{a+b} \]

The problem is absurd. Take a,b to be rational and f to be a cauchy function.
then f(ax+by /a+b) = f(ax/a+b) +f(by/a+b) = a/a+b f(x) + b/a+b f(y)

I don't see how you can prove a=b then!

[quote="ali"]... a/a+b f(x) + b/a+b f(y)
[/quote]

Careful! The problem states a/a+b [b]f(y)[/b] + b/a+b [b]f(x)[/b]


Find the area bounded by the region $ |x\plus{}y|\leq3$.

how do you graph this?

you graph it in two parts: one where x+y is positive, and one where its negative. so you get: x+y<=3 and x+y>=-3. don't think it has an area though, looks unbounded...

thats what i thought, but thats not one of the answer choices.

Choices: 3,6,9,12,18

Maybe they mean that it is bounded by the axes as well? In that case it would be $ 2\cdot\frac12\cdot3^2\equal{}\boxed9$.

Are you sure it's not $ |x|\plus{}|y| \leq 3$ ?

In which case, split the square down its diagonals, and put the four pieces together into two 3x3 squares to see it has area 18.


[quote="TALENTO"]Find four positive integers $ X, Y, Z, W$ all different from each other, that are between $ 200$ and $ 500$  and to check that $ X^2 \plus{} Y^3 \equal{} Z^2 \plus{} W^3$.

Greetings  :D[/quote]

$ 460^2\plus{}225^3\equal{}243^2\plus{}226^3$

Чтоб найти решения уравнения:

$X^3+Y^2=Z^3+W^2$

Можно воспользоватся формулами. Все числа которые фигурируют, целые и задаются нами.

$X=dbs(2bs+6as\pm2pb)$

$Y=d(p^2b^2+3a^2s^2+b^2s^2-36adb^3s^4\pm2psb^2)$

$Z=dbs(2bs-6as\pm2pb)$

$W=d(p^2b^2+3a^2s^2+b^2s^2+36adb^3s^4\pm2psb^2)$

Можно воспользоваться и другими:

$q=3a^2+3ab+b^2$

$j=(q+2d)b+2a$

 тут внимательно надо быть и подставить соответствующее $q$ .

Тогда:

$X=ja$

$Y=j(a+(q+d)b)$

$Z=j(a+b)$

$W=j(a+db)$

Ясно, что это выписаны формулы не совсем примитивных решений. Сокращать надо будет на соответствующие коэффициенты.


Hi  fanatics of Numb3rs !

What do you think about that:[url]http://www.weallusematheveryday.com/tools/waumed/what_is_waumed.htm[/url] ?
I think that's a very good idea ! :roll:

I actually (no sarcasm intended) think it's a great idea! I know if I learned math in the context of catching "bad guys" for the FBI I probably would have liked it much more when I was younger. My fear is that it may not be entirely comprehensive (I don't know how you'd teach fundamental mathematical concepts by relating it to all the combinatorics/computational mathematics he does in the TV show...) So I'm skeptical that it'll turn out well, I do however, like the [b]idea[/b].

It looks nice, I've taken a look at the topics and they are indeed interesting but I have no idea how they can actually be incorporated (I mean some of them are easy to, but some are really hard) into a regular math class.

(btw my signature has been a quote from Num3rs for a long while now :P).

I watch NUMB3RS every now and then. I don't really believe all the things Charlie says but hey, it's fiction.

I think it was April, or early May, there was this awards banquet for a local math competition, and the two guest speakers were writers for the We All Use Math Every Day program. They showed clips from NUMB3RS and talked about the corresponding activity for each. It was really interesting, unlike the guest speakers from previous years.

In Computer Science at school one day a while back the teacher taught us some basic encryption via matrices by a NUMB3RS segment. It was a stenography episode. I'm all for it, especially since it teaches some mundane topics in an exciting way.

That's awesome : ) And so is TI's idea =P I'm a big Numb3rs fan myself.

I got the impression that some of you expressed doubt in the accuracy of the mathematics on show, but I'll let you know that people at Wolfram Research are actually consultants for the math in the show, so it should very accurate.

As for the feasibility of the program succeeding, I should mention that in every show there's an "Analogy Scene" where Charlie explains the basic idea of his mathematical approach in a rather accessible analogy. I think they will be basing the curriculum off of these ideas. Furthermore, the show airs once a week every Friday, so I'd guess that they would spend every week on a different one of these analogies. This is a very uneducated guess though... I don't have the time right now to look into it >_<

I think it would, of course, be way more educational if the TI program made its own curriculum pertaining to WAUMED, but I guess the connections to the show is a way to get the kids enthusiastic about the curriculum... so it will be interesting to see how it works out.

Hi all and thanks for your answers ! I feel no more ALONE when I'm watching NUMB3RS !

Because the French part of MathLinks is not too visited...I announce here the site :[url]http://numb3rs.singularity.free.fr[/url]

[b]Et voilà ![/b] :roll:

I looked at the WAUMED curriculum materials about a year ago and found them pretty weak and useless.  Of course, I teach college students and strong high-school students, so perhaps the materials are better for "average" students.

[quote="ktm"]... but I'll let you know that people at Wolfram Research are actually consultants for the math in the show, so it should very accurate.[/quote]

And there are consultants from Caltech, etc... but that doesn't stop the creators/directors/producers of the show from messing with the math.  Check out the NUMB3RS blog at [url]http://www.atsweb.neu.edu/math/cp/blog/[/url] for some examples/details.

[quote="smbelcas"]I looked at the WAUMED curriculum materials about a year ago and found them pretty weak and useless.  Of course, I teach college students and strong high-school students, so perhaps the materials are better for "average" students.

[/quote]

I think that NUMB3RS is intended  [b]only for vulgarize[/b] maths for the great public ! NUMB3RS don't replace a math course !

Well, they are trying to make mathematics more popular among people who otherwise would have no natural inclination towards it and it is a good thing. Somehow for many of us, mathematicians, the value and the beauty of mathematics is so apparent that we do not think any effort is needed for promoting it among the general public, which is wrong. Love for mathematics is an "acquired taste" thing and it requires a considerable effort for an average person to get to the point where he/she can start understanding and appreciating it. In absence of true understanding, what rules is "public opinion", which (in the USA, at least) currently basically reduces to the idea that "socializing and sports are cool and good for everyone" and "science is boring and incomprehensible (though useful) and suitable only for nerds". So, in my humble opinion, any attempts to change this public opinion should be welcome. The fact that in the process of popularization, the subject may be severely distorted is not so important. Those who get interested enough to go into the nitty-gritty will always be able to find a lot of resources to straighten the matters out and for others just to get the general idea (or merely the idea that "math. may be an interesting thing after all") is good enough. Of course, I do not suggest that NUMB3RS or something similar should [b]replace[/b] the regular school curriculum but something like that should be added to it in reasonable quantities. Mathematics as it is usually taught reminds me a bit of teaching an English language course covering spelling, grammar, some bits of history, philology, etc. but not touching a single fiction book or poem written in the language. Whether it is possible to design a mathematical analogue of "Little Red Riding Hood" or "Harry Potter" for children is a good question. The only person I know of who seems to have tried to do it in honest is Lewis Carroll. But it seems that today's "education bosses" have never heard of him. Martin Gardner was another really great Anglo-Saxon math. educationist but you don't hear his name in math. education circles often either. So, in my opinion, NUMB3RS is good even if they sometimes don't really understand what they are talking about and make silly mistakes in their presentation. Just to have something devoted to mathematics that people may watch on TV as a football game is great. The perception of a "normal" person seems to be rather emotional than logical and what one usually is looking for in popular shows and books is intensity of feelings rather than intensity of thought. So, if one can make mathematics itself (not just the person's life story somehow related to it as in "Pie", "Proof", or even "Beautiful mind") an intense emotional adventure, that's already quite a feat. Anyway, this is just my humble opinion of the subject :).

Thank you fedja ! I'm totally agree with you !

And now a book about Numb3rs ! Et voilà !

http://www.amazon.com/Numbers-Behind-NUMB3RS-Solving-Mathematics/dp/0452288576

have a lot of fun !

I heard of Numb3rs, but I never watched it (I'm guessing you need cable, witch I don't have). Although my overall thoughts are:

Promoting math? I'm all for it  :D !

[quote="Nerd_of_the_Ages"]Promoting math? I'm all for it  :D ![/quote]

Yup  :D .

The new season of NUMB3RS starts Friday, September 28th on CBS at 10pm eastern time.

Whee... I don't really watch TV, so what channel is CBS on (and does it need cable)?

You do need cable, but CBS is on the basic package (in SD area it's about 10 bucks a month).

That depends on the market- CBS has broadcast channels in most large cities. Valentin only speaks for San Diego.

oh, ok. Thanks. I don't think I'm gonna watch it, because I dont wanna pay for cable, and I dont really have time.


My club has 25 members.  In how many ways can I choose members to form a 4-person executive committee?

I also didn't understand anything here what $ 4!$ mean

$ 4!\equal{}4 \cdot 3 \cdot 2 \cdot 1$.  To generalize, $ n!\equal{}n \cdot (n\minus{}1) \cdot (n\minus{}2) \cdot \ldots \cdot 2 \cdot 1$.

For this problem, it is simply $ \binom{25}{4}$ ($ 25$ choose $ 4$), or $ 12650$.

[hide=Solution]
$\binom{25}{4}=\frac{25\times24\times23\times 22}{4!}=\boxed{12{,}650}$ ways.
[/hide]


[quote="Morrison & Boyd"]
The behaviour of o groups is different than m or p groups.They always tend to increase the acidity whether they are electron donating or eletron releasing[/quote]
why??

i thik coz the bulky ortho groups disturb the planarity of the system...

one of the reasons for ortho effect is the steric crowding due to bulky substituents at ortho position. if you are talking abt acidity,(mainly for benzoic acids), it increases due to ortho effect as conjugate base is more stable(lesser steric effect). and decreases basicity(for amines), again due to the same reason.

@Varunarasu how can loss of just one $ \ce{H}$ make the steric hindrance so less that the conjugate acid becomes so stable....after all if there is an +M group then that will destabilize the negative charge
and i suppose this will be more than the reduction in steric factor

[quote="madness"]Yes i think its bcos of steric inhibition of resonance which reduces a negative factor for acidity thereby incr acidity  :)[/quote]
yes it doesn't reduce the acidity due to steric factor as varunarasu said but how does it increase the acidity :?:

i think usually resonance effect decreases acidity but here as bulky grps are there @ the ortho posn, the compound is made non planar thus reducing the acid weakening resonance effect (as planarity is a need for resonance) and hence the compd becomes more acidic  :)

no u don't get me madness....i know they can't sufficiently reduce the acidic nature but that in no way guarentees that they will increase it's acidity

consider the normall compd without subsn okay? Let it have some acidity. Now when the compd is substituted @ ortho posn ,the acidity will be more bcos of steric inhibition of resonance  :)
What dont u get there? We are talking only abt relative acidities i.e. btwn the unsubstituted and the ortho substituted compds  :)

[quote="madness"]consider the normall compd without subsn okay? Let it have some acidity. Now when the compd is substituted @ ortho posn ,the acidity will be more bcos of steric inhibition of [/quote]
how  :huh: ...due to the steric factor the resonance is incomplete or not as good as it would have been in a p-substituted compound thus doesn't sufficently 'destabilize' the negative charge ...
BUt how does it INCREASE the acidity :?:
it doesn't Decrease the acidity much ...but the question is HOW does it INCREASE the acidity :?:  :?:

I think the resonance in benzoic acid is primarily a -R and what happens is that the inflow of electrons due to resonance makes that carbonyl carbon less electrophilic and thus thirst of electrons is not there which inductively means that the $ \ce{O\minus{}H}$ bond need not be cleaved and so they are weaker acids than aliphatic acids.
Also $ \ce{\minus{}COOH}$ undergoes out of twisting due to steric effects and so resonance is not very effective.

Suppose a -R group like $ \ce{ \minus{} NO_2}$ is there in the ortho it makes the $ \ce{\minus{}COOH}$ to come back to the same plane and now the nitro group shows it's -R and -I effect and makes the carbonyl more electrophilic which is inductively felt by the $ \ce{\minus{}O\minus{}H}$ group and $ \ce{H^ \plus{} }$ has to go away and thereby the acidity increases.

chemrock my question was that even if there are +R groups still the acidity [b]increases[/b] why??

I think like VARUNARASU said the conjugate base is more stable

and my reply same as i gave for varunarasu...how can loss of 1 small hydrogen decrease the steric factor so brilliantly :)

[quote="pardesi"]
@Varunarasu how can loss of just one  make the steric hindrance so less that the conjugate acid becomes so stable....after all if there is an +M group then that will destabilize the negative charge
and i suppose this will be more than the reduction in steric factor
[/quote]
i said 'one of the reasons'

In Benzoic acids, resonance stabilization due to benzene is not very great as the COO- and the benzene ring are not always in the same plane. For eg, the pka of phenol is 10 and the pka of its aliphatic counterparts is around 16-18. But the pka of benzoic acid is 4.17 and the pka of acetic acid is around 4.8 if the benzoate was resonance stabilized by the benzene ring, surely its acidity wud have been much greater than its aliphatic counterpart( that does not mean +R and –R effects do not come into the picture while considering the acidities of benzoic acids). Also the ortho effect is more than just due to sheer bulkiness of the group, it is similar to torsional energy of the single bonds (although steric factor is also important) for eg. Boat conformation of cyclohexane is less stable than the chair conformation by about 11kcal/mole. Surely this is due to reasons other than just bulkiness or eclipsed conformations of the hydrogens.


AMC
USA(J)MO
USAMO
AIME
AMC 10
AMC 12
ARML

i think the USAMO selection process was not very good.  Sophomores have too much advantages.  It is harder for a junior or a freshman to get into USAMO than for sophomores to get into USAMO.  I think they should just leave 60 spots for senior, 60 for junior, 60 for sophomore, 60 for freshman, and 10 for everyone else.

Quotas are almost always a bad idea.  They wouldn't solve the supposed problems, and would probably make the process less fair, since in certain grades would be easier/harder to get into USAMO/

I personally think the USAMO selection process is fairly fair, and if it favors people who took certain tests, well, it's [i]supposed[/i] to favor those tests.

Except that 60 freshmen never qualify, and it's hardly likely that there are 60 freshmen on an equivalent level as many of the sophmores and juniors and seniors.

you would have liked the way they did it until 3 years ago. (numbers may not be exact, but here's the gist of it) They took the first 120 index scorers regardless of grade, then the next 20 in grades 11 and below, then the next 20 in grades 10 and below, the next 20 in grades 9 and below, then the state people. The system was changed when the AHSME exam was replaced by the AMC 10 and 12. Since then they've fine-tuned from year to year. I don't think there is a perfect system for USAMO selection, but I do think the AMC organizers are committed to trying to be as fair as possible given the constraints of the organization and the exam process.

[quote="billzhao"]I personally think that the grade level should not be taken into consideration when it comes to selection. It seems to me that in an elite group of competitors, this is very little correlation between grade and skill, or sometimes even experience.[/quote]

I agree. I'd say to just take the students with the top 200 indices. Simple as that.

You are all missing the point: they don't want 9th graders because they think 9th graders are smarter.  They want 9th graders because they think that, having been to MOP, they will become smarter.  (Smarter here is used only to mean that they will perform better.)  There's an obvious logic in taking 9th graders: they're good enough to get to the appropriate level (which very few younger students are), so you don't have to teach them the really basic stuff, and if you start with people that young, you have a good shot at getting 1 or 2 years out of them on the US team.  (Remember that the point of all of this isn't just for your own benefits as students -- they're also trying to win IMO :))  I personally think that the system should be simplified -- the number of times you've had to answer some questions is a good indication of this -- but that the basic goals of the current system are pretty good.  If I'd known I'd had a shot at being on the IMO team from 8th or 9th grade, I might have actually worked hard enough to be competitive for it, and I think the same is true of other people as well.

Okay, but what does AMC then do about the following?

Student A takes the 2004 AMC 10B, gets 120, qualifies for AIME, takes AIME I gets 8 correct, has a USAMO index of 200.

Student B takes the 2004 AMC 12A, gets 105, qualifies for AIME, takes AIME I, gets 9 correct, has a USAMO index of 195.

This is a simplified version (no grade/age considerations explicitly mentioned here) of the situation in 2001, for example.

Steve Dunbar
AMC Director

[quote]You are all missing the point: they don't want 9th graders because they think 9th graders are smarter. They want 9th graders because they think that, having been to MOP, they will become smarter.[/quote]

Joel, we were making a reference to USAMO selection, not MOP selection. If the AMC wanted to invite a group of 30 young and promising kids to MOP with my system, they would simply take the 30 students in the lowest grades.

[quote]Okay, but what does AMC then do about the following? [/quote]

Well, going along with my system, the AMC would have a couple of options here. Both would be drastically different from the current way things are run, but both make the selection process much simpler.

1) They could simply say hey, if a student wants to be in the elite group of USAMO qualifiers, he/she must take the AMC12. The purpose of the AMC10 would only be for younger students to practice and to spread the love of mathematics and problem-solving in this country, but no one taking the AMC10 would be able to qualify for the USAMO. If a student is serious about the AMC contests and is serious about making USAMO, MOP, etc..., he/she would have to take the harder AMC12, not the AMC10.

2) On the other hand, if the AMC still wants to give younger kids a leg up in the process, they could simply treat an AMC10+AIME index the same as an AMC12+AIME index. Mr. Dunbar, with this new selection process, kids in 10th grade and below would not consider to take the AMC12 instead of the AMC10 because both tests are weighted equally in this system. Therefore, your hypothetical situation with those two students would not arise.

Or you can do what SFBA ARML does with the AMC scores. We subtract some amount (usually about 25 points) from AMC10 score to estimate the AMC12 score.

[quote="AMCDirector"]Please describe what you you believe would be a good selection process for the USAMO, using the available reliable information that we have, namely the score on the max[ AMC {10|12} {A|B}] and the AIME {I|II}.

Steve Dunbar
AMC Director[/quote]

My suggestion: Let students take both AIME 1 and 2 to eliminate the heartbreaking factor of careless mistakes and leave all other rules stay as they are; and if practically possible, allow only those AIME 1 participants with at least a score of, lets say, 5 to take AIME 2

If one of AMCs major objectives is to start students early, then AMC8 and the splitting of AHSME into AMC10/12 have achieved that.

If another objective is to define talents early, the current USAMO Selection rules allow for that.

If some students start AMC too late and suffer the penalty of being juniors or seniors, it will be impractical to change the rules to accommodate them but in the process go against the above two objectives. After all, if a junior/senior is good enough to be in the top 160, he/she is still in!

The glaring problem with the current system of USAMO selection, both empathically and practically, is that it eliminates students who are USAMO/IMO capable due to bad day, careless mistakes, etc. The illogical thing is, careless mistakes do not reduce an USAMO/IMO score anywhere close to that of an AIME score.

Basically, we need AIME to help advance the more capable students, but we do not need AIME to also eliminate some of those with a wrong logic. I can think of at least one senior whose exclusion from USAMO this year is simply tragic. The general sentiment in the AoPS community suggests there may be a few more of such tragedies.

So if they score 5+ on AIME 1 and think they should have gotten 10+, let them prove it  for everybodys benefit.

[quote="MathChauffeur"]

The glaring problem with the current system of USAMO selection, both empathically and practically, is that it eliminates students who are USAMO/IMO capable due to bad day, careless mistakes, etc. The illogical thing is, careless mistakes do not reduce an USAMO/IMO score anywhere close to that of an AIME score.

Basically, we need AIME to help advance the more capable students, but we do not need AIME to also eliminate some of those with a wrong logic. I can think of at least one senior whose exclusion from USAMO this year is simply tragic. The general sentiment in the AoPS community suggests there may be a few more of such tragedies.

[/quote]

I think we are agreed that the USAMO serves 2 functions - first, it is a "prize" of sorts for doing well in the earlier stages of the competition. Second, it is the keystone of the selection process for further training and development of US teams for the IMO. This second function is arguably the most important. So we need 2 additional things - (a) to identify promising younger people who have time to learn to do this before they run out of eligibility, and (b) to include and rank the relatively small number of people who can already do this.

(a) I agree with the poster who suggested that if you want to identify the top x number of younger students for something like the 9th grade MOP or other special training, go ahead and do so. They do not have to be designated USAMO qualifiers for this purpose. I disagree with the suggestion that people be required to take the AMC-12 to advance. The Computing Olympiad has a system like that, but people have multiple contests spread over the year to figure out which division they belong in. With AMC, most students have only one shot per year, and may not know how strong they are compared to other students around the country. It's too easy for students (or their teachers) to choose wrong. Some of the people who take the 10 are taken by surprise at how well they do. Some teachers and schools do not give their brightest students in grades 10 and below the option of taking the 12.

(b) I totally agree with Mathchauffeur's quote above. I can even add to the roster of heart breaking stories of people who scored high on USAMO and had MOP/IMO potential, but who failed to qualify the following year. There were at least 2 of those last year. My community had a senior who had been an honorable mention the year before, and a sophomore who started qualifying for USAMO in 7th grade and was ranked around 50th at the end of MOP 2002. Both spent an entire year working on USAMO problems, and were poised for MOP/IMO in 2003. But they each missed the AIME cut-off by a point and didn't get to take the USAMO. Meanwhile, the 3 local kids who did qualify all made zero. I also know of 2 people in the past who eventually made IMO teams but who missed out on a potential year of MOP because they had a bad AIME day and didn't get to take the USAMO during one year after doing well on it the previous year.

Most people who qualify for the USAMO cannot actually do USAMO problems (the most common score is zero out of 42. Half the scores are six or less). There are few enough kids who can do USAMO problems that it seems like a resource we should be trying very hard to conserve. If there is room for 100 qualifiers every year who are going to make a zero, there should be room for an extra 5-6 people who have already proved that they can do USAMO problems but who fail to qualify for one reason or another. A second shot at AIME is one idea. Another thought is to allow the director to invite a few people to take the USAMO each year based on high scores the previous year. (This would need to be done in some objective way that would avoid having the director inundated with pleas begging for special consideration.)

One possibility could be reserving the next year's spots for the top x% of this year's USAMO contestants.

I know what you might say, that this could encourage them to slack off because of that reservation, but I personally know that it's not true.

Olympiad is very different from AMC or AIME. Once a student passes the "threshold" into the olympiad world, he/she shouldn't be forced to go back and deal with the petty little contests like AMC or AIME (no offence to the contests). It makes sense to allow them to concentrate on the olympiads, rather than the things you need to get into an olympiad.

Speaking from experience, once you've done enough of the pre-olymp material, there really isn't any incentive to linger around any longer. The olympiad problems are much more fun to do. Futhermore, every pre-olym problem look just like something you've done before, and their approaches can almost always be found on sight. So AMC and AIME eventually becomes nothing more than a race against time. And after a while, we just become sick of it. LoL, one of my friends, who was a Canadian bronze medalist at the IMO last year, he got a one on the AIME. Now that says something.

I agree with the posts above.  In particular (or in addition?), I feel that there should be greater MOP opportunities for younger students qualifying for the USAMO.  As an 8th grader, my views are biased, but it really is quite difficult to learn how to do USAMO-level problems on your own.  Speaking of this, since there don't seem to be 30 9th grade qualifiers this year (only 25), will the remaining 5 slots for MOP be given to other students (e.g., 8th graders)?

I would support past high-scorers on USAMO being protected against a potential bad day, but I think that using AMC scores to help determine USAMO qualification is a bad idea. For example, in MATHCOUNTS, chapter scores are not used in conjunction with state scores to determine who gets to go to nationals. It should be the same with the AMCs--when one qualifies for AIME, one should have a clean slate. If someone had a "bad day" on the AMC, it could hold them back from qualifying for USAMO in the same way that a low AIME score could.
However, to account for the "bad day" factor, perhaps one could decide beforehand whether to take both AIMEs and average them or just take one and go with that score. This would lessen the detrimental effects of stupid mistakes, but if someone were confident in their ability to avoid them, then they could just take one.

[quote="JSRosen3"]I would support past high-scorers on USAMO being protected against a potential bad day, but I think that using AMC scores to help determine USAMO qualification is a bad idea. For example, in MATHCOUNTS, chapter scores are not used in conjunction with state scores to determine who gets to go to nationals. It should be the same with the AMCs--when one qualifies for AIME, one should have a clean slate. If someone had a "bad day" on the AMC, it could hold them back from qualifying for USAMO in the same way that a low AIME score could.
[/quote]

But doesn't this just increase the "bad day" factor?  If you let USAMO qualification depend solely upon one exam, then it depends on how you do on that one day, rather than how you do on two of three days (if you take both AMCs).

It's arguable that MathCounts' system is not perfect either. It is also completely different in terms of test material and test format. So, really, we cannot use it to compare.

The USAMO selection process is fair in my opinion...what's not fair is how they change MOP selection every year.

I made the USAMO this year and last year (i'm a sophomore now).  If i took it a year earlier or a year later as a freshman (2002 or 2004) i would've gotten into MOP because in 2002 they took everyone and in 2004 they're going to take 30 freshmen or lower.

Oh well...

[quote="ComplexZeta"]Why do you insist that most people who make the USAMO can't do USAMO problems? Sure, most people probably can't get three questions, but why not one? I think the median score last year was 8 or something.[/quote]

I added up everyone in each score group for the 3 years 2000-2002 (I'm missing 2003) and got the following distribution:

score: 35-40   (People: 16)
           28-34  (People 37)
           21-27   (People: 50)
             7-13    (People: 201)
              0-6     (People: 425)
                                               TOTAL PEOPLE 825

I don't know the average for any of those years, but given how left-skewed the distribution curve is for the score groups, I'm willing to guess that the people in the 0-6 group are also skewed to the left, meaning most of them made near zero. That's over half the total number of USAMO-takers. That also goes along with the anecdotal evidence of local kids I've known of and their scores.

[quote="ComplexZeta"]Why do you insist that most people who make the USAMO can't do USAMO problems? Sure, most people probably can't get three questions, but why not one? I think the median score last year was 8 or something.[/quote]

The [url=http://www.unl.edu/amc/e-exams/e8-usamo/e8-1-usamoarchive/2003-ua/03usamoscoring.html]score distribution from 2003 on USAMO[/url] shows roughly half the participants getting scores of 8 (out of 42) or lower, with thirty-some students scoring 0. That shows the rarity of developed USAMO-taking ability even among the select group who are invited to take the USAMO.

[quote="tokenadult"]
The [url=http://www.unl.edu/amc/e-exams/e8-usamo/e8-1-usamoarchive/2003-ua/03usamoscoring.html]score distribution from 2003 on USAMO[/url] shows roughly half the participants getting scores of 8 (out of 42) or lower, with thirty-some students scoring 0. That shows the rarity of developed USAMO-taking ability even among the select group who are invited to take the USAMO.[/quote]

From that chart, 61 people got 2 or less. That's about a quarter of the test-takers. It looks like around half got at least one problem. That's slightly better than the previous 3 years, but still a pretty limited success rate.

OK, so people get 0's. Do you think that they aren't good at math? No. Obviously to get that far you are good at math. But you shouldn't say that "Oh, well so many people get 0's so we should have better selection" because I doubt many other people could get much either (the occasional exception is what we are arguing about here, so what does getting 0 on the USAMO have anything to do with it). No matter how great a selection is, at a level that high there will be 0's. I agree to letting people deemed "capable" from previous success to have an advantage, but there is no sense in showing the "limited success" of USAMO participants because it does not prove anything.

[quote="ComplexZeta"][quote="Eugevir"]I would like to point out that last year's USAMO had a ridiculously easy number one, usually not the case, which is why scores were higher at the low end than on most tests.[/quote]

But don't you think #1 on 2002 was just as easy as #1 on 2003?[/quote]

I'd say so.  2002 and 2003 both had an unusually easy low end (#2 on 2002 was also extremely easy for a #2), and their #1's were nothing near as hard as the 2000 and 2001 #1's... however it's hard to say if this is a blip or a trend.

[quote="tokenadult"]
The [url=http://www.unl.edu/amc/e-exams/e8-usamo/e8-1-usamoarchive/2003-ua/03usamoscoring.html]score distribution from 2003 on USAMO[/url] shows roughly half the participants getting scores of 8 (out of 42) or lower, with thirty-some students scoring 0. That shows the rarity of developed USAMO-taking ability even among the select group who are invited to take the USAMO.[/quote]

Give us a few years.  They'll have to write a harder test to get this distribution in 2006 :wink: .

[quote="rrusczyk"]Give us a few years.  They'll have to write a harder test to get this distribution in 2006 :wink: .[/quote]

LOL at your gung-ho spirit. I'm trying to make the big push this year and next to get my son acquainted with the fine art of writing essay proofs, so I hope that by the time he can get past the AIME, he can also do respectably on the USAMO. I'm looking forward to the new plans AoPS has for enhancing math education of young people.

justdudxit wrote:
Any one from AIME cares to comment why are we not allowed to appear for both AIME 1 and AIME 2. It should be the same logic as you could do both AMC A and B. If it is a question of money (or time) to hand grade the AIME papers, I am positive that the number of additional students would not be that high.

tokenadult replied:
I'm not from AMC, but one bit of logic I can see is that the examination is the American INVITATIONAL mathematics examination. If your friend invites you to dinner, and then invites another friend to dinner on a separate occasion, it's not standard practice for you to attend BOTH dinners. The preliminary tests (at this level, the AMC10/12) are tests offered to all eligible comers who pay the fee, and they can use those tests to qualify for an invitation (singular) to take the AIME, but AIME is meant to be a one-shot chance each year at the USAMO..

justdudxit replied:
According to the above  logic, if someone appears for AMC A and AMC B, and qualified for AIME in both, then she/he should be able to appear for both AIME1 and AIME 2. Some of us would love to have that chance.
It is more of a question of money and time.  Only couple of years before, someone had only "one-shot chance" each year to AIME.
After all, the goal is to promote interest in math while selecting a USAMO team.

[quote="justdudxit, helpfully moving over comments posted to another thread,"]According to the above  logic [in tokenadult's reply], if someone appears for AMC A and AMC B, and qualified for AIME in both, then she/he should be able to appear for both AIME1 and AIME 2.[/quote]

Well, there used to be exactly one shot at taking the AIME, until quite recently. (This is readily known to anyone who orders the old AIME tests from AMC, something that I recommend that everyone preparing for the AIME do.)

[quote="justdudxit"]Some of us would love to have that chance [to take the AIME twice in one year].
It is more of a question of money and time.  Only couple of years before, someone had only "one-shot chance" each year to AIME.
After all, the goal is to promote interest in math while selecting a USAMO team.[/quote]

Yes, and as one goes higher in the AMC test echelon, the selecting-the-team rationale gains prominence as against the promoting-interest-in-math rationale, because presumably everyone sitting for the AIME is already interested in math, but the IMO team can never be more than six people and one or two alternates. So there has to be some pretty serious winnowing of the candidates going on between 13,000 people taking the AIME and 300 or so taking the USAMO and a few dozen going to MOP and only 6 going on to IMO.

Earlier in this thread, I proposed that a person who gets a MOP-qualifying score on USAMO one year be allowed to take USAMO the next year simply by TAKING the AIME, regardless of the score the second year. This rewards people with proven ability to write essay-proof answers, which is a rare ability in teenagers. My intended meaning is that one could grandfather-in USAMO qualification for a select group of previous USAMO qualifiers, to help them overcome the difficulty and randomness of the AIME, while still leaving all other criteria for selecting people for MOP and for the IMO team as they would be without my suggestion being implemented.

By contrast, letting people take AIME twice could result in a LOT more AIME tests to score at AMC headquarters, with only minutely better selection of eventual IMO participants.

That's just my humble and entirely unofficial opinion.


I need help please. How do you solve this problem using extreme principle?

In the plane, n lines are given (n>=3), no two of them are parallel. Through every intersection of two lines, there passes at least an additional line. Prove that all lines pass through one point.

Thanks.

http://www.ams.org/featurecolumn/archive/sylvester1.html

http://en.wikipedia.org/wiki/Sylvester–Gallai_theorem

(the latter link isn't working correctly, just search wikipedia for 'Sylvester-Gallai')

it is no one's fault but your own if you are too lazy to check out the above links for yourself.


algebra
polynomial
real analysis
real analysis unsolved

Hi,
  Let p be a normalised polynomial of degree n+1:
p(x)=x^(n+1)+...+a
show that for -1<x<1,sup(abs(p(x))<=1/2^n
Good luck

In the title:   minoration of sup(absp(x))
In the post:  sup(abs(p(x))<=1/2^n wich is a majoration of sup(abs(p(x))

I don't understand!

I think the problem is the majorize the quantity q(p) = sup|p(x)| where -1  \leq  x  \leq 1, as p(x) ranges over the set of E(n+1) the polynomial with degree n+1 and leading coefficient equals to 1.

And, this is a well-known property of Tchebychev's polynomials T_n of the first kind that for any p from E(n+1), we have q(p)  \leq 1/2 n  and equality occurs if and only if p = (T_n)/2 n .

Pierre.

Thanks Pierre,
  It is a well known result, ok but I was looking for an elementary proof of it if it does exist.... I have already seen the one where we use the fact that Chebychev polynomials. If someone knows another proof would be kind to right it down.

What's the bound for the general case, <P_n,P_m> = a_n*\delta_{nm} (a_n not necessarily being 1)?


[quote="thaithuan_GC"]Let $ A\in N$ be a non-empty set satisfying the condition: if $ a\in A$ then $ 4a$ and $ [\sqrt {a}]$ are both in $ A$ (where [x] is the integer part of x). Prove that $ A \equal{} N$ (N is the set of all natural numbers)[/quote]

deadline is 15 sept 2008, as for the rest of your topics (which i have deleted for that reason).

anyway this problem has been given at a contest long ago, but i'm not telling more details. ;)

  dg


calculus
integration
trigonometry
calculus computations

$ \int \frac {1}{sin^5(x) \plus{} cos^5(x)}dx$

[quote="atomicwedgie"]What happened to $ \int \frac {\sin x \plus{} \cos x}{\sin^4 x \plus{} \cos^4 x} \, dx$?[/quote]

actually i got it...so deleted it :)


let $a_1,b_1,a_2,b_2>0$ show that if
$\frac{a_1 x+b_1 y}{a_2 x+b_2 y}+\frac{a_1 y+b_1 x}{a_2 y+b_2 x}\geq 2\frac{a_1+b_1}{a_2+b_2}$
holds for two unequal positive $x,y$  then it will hold for all $x,y>0$

Yes,It's very easy!
The inequality is equivalent to:
$\frac{(x-y)^2((a_{1}b_{2}-a_{2}b_{1})(b_{2}-a_{2})}{(a_{2}+b_{2})(a_{2}x+b_{2}y)(a_{2}y+b_{2}x)}$ $\geq$ $0$.
So if it holds for two unequal positive $x,y$ then $(a_{1}b_{2}-a_{2}b_{1})(b_{2}-a_{2})$ $\geq$ $0$ and the inequality will hold for all $x,y>0$.


linear algebra
matrix
algebra
polynomial
linear algebra unsolved

For example,if $A^2=O_n$ then the conclusion follows easily from $(I_n+\frac{1}{2}A)^2=I_n+A$.

 For $A$ a matrix of order $2$ the conclusion is obvious. ;)

Here is a more general problem:

 Let $A,B$ two matrices of order $n$ such that there is $k>1$ with $B^k=O_n$.Prove that if $AB=BA$ then
 $det(A+B)=detA$ ;)

 $Solution.$
   First,let's assume that $A$ is invertible.It is obvious that $BA^{-1}=A^{-1}B$ so $det(A+B)=detA\cdot det(BA^{-1}+I_n)=detA\cdot f_{BA^{-1}}(-1)$,where $f_BA^{-1}$ is the characteristic polynomial of $BA^{-1}$.But $(BA^{-1})^k=B^k A^{-k}=O_n$ so $f_{BA^{-1}}(x)=(-1)^n x^n$ so $f_{BA^{-1}}(-1)=1.$

 The other case is very easy.

[quote="EB"]Prove that if the matrix [b]A[/b] is nilpotent then [b]I[/b] + [b]A[/b] is invertible.[/quote]

Solution:

Let $p(x)=det(xA+I), \forall x \in {\mathbb C}$. Since $A$ is nilpotent $=> \exists k$ such that $A^k=0(*)$. Well now we have $x^kA^k-I=(xA-I)M(**)$,where $M$ is an ugly matrix that do not interest us. From $(*)$ and $(**)$ we get that $det(xA-I)detM \neq 0 => xA-I$ is invertible for any $x \in {\mathbb C}$. Let's put now $x \to -x$ and we get $det(xA+I) \neq 0, \forall x \in {\mathbb C}$, and for $x=1$ we get $det(A+I) \neq 0$.

q.e.d.

There's a very simple solution for the original problem. The inverse of $I+A$ is the sum $\sum_{k=0}^{\infty}(-1)^kA^k,$ provided that sum converges. If $A$ is nilpotent, the series becomes a finite sum and thus converges.

By a related argument, if all of the eigenvalues of $A$ have absolute values that are strictly less then 1, then $I+A$ is invertible. Nilpotent matrices (all eigenvalues zero) become a special case of this.

Even simplier:since $A$ nilpotent,all of it's eigenvalues are $0$.Therefore $det(A+I) \neq 0$ (otherwise it must have eigenvalue $-1$).


with $ a,b,c>0$.Prove that
  $ \frac{a}{b\plus{}c}\plus{} \frac{b}{c\plus{}a}\plus{} \frac{c}{a\plus{}b} \plus{}\frac{4abc}{(a\plus{}b)(b\plus{}c)(c\plus{}a)}\geq 2\plus{}\frac{4(a\minus{}b)^4(b\minus{}c)^4(c\minus{}a)^4}{(a\plus{}b)^4(b\plus{}c)^4(c\plus{}a)^4}$

[quote="2424"]with $ a,b,c > 0$.Prove that
  $ \frac {a}{b \plus{} c} \plus{} \frac {b}{c \plus{} a} \plus{} \frac {c}{a \plus{} b} \plus{} \frac {4abc}{(a \plus{} b)(b \plus{} c)(c \plus{} a)}\geq 2 \plus{} \frac {4(a \minus{} b)^4(b \minus{} c)^4(c \minus{} a)^4}{(a \plus{} b)^4(b \plus{} c)^4(c \plus{} a)^4}$[/quote]
Assume $ a \ge b \ge c$
We have: $ LHS\minus{}2 \ge \frac{(a\minus{}b)^2(a\plus{}b\minus{}c)}{(a\plus{}b)(b\plus{}c)(c\plus{}a)}$.
So,we only need to prove:
$ (a\plus{}b\minus{}c)(a\plus{}b)^3(b\plus{}c)^3(c\plus{}a)^3 \ge 4(a\minus{}b)^2(b\minus{}c)^4(a\minus{}c)^4$
on the other hand, $ (b\plus{}c)^3(c\plus{}a)^3>(b\minus{}c)^3(a\minus{}c)^3$
                            $ a\plus{}b\minus{}c >a\minus{}c$
                            $ (a\minus{}c)^2 \ge 4(a\minus{}b)(b\minus{}c)$
                             $ (a\plus{}b)^3 >(a\minus{}c)^2(a\minus{}b)$
So,we have Q.E.D

[quote="quykhtn-qa1"][quote="2424"]with $ a,b,c > 0$.Prove that
  $ \frac {a}{b \plus{} c} \plus{} \frac {b}{c \plus{} a} \plus{} \frac {c}{a \plus{} b} \plus{} \frac {4abc}{(a \plus{} b)(b \plus{} c)(c \plus{} a)}\geq 2 \plus{} \frac {4(a \minus{} b)^4(b \minus{} c)^4(c \minus{} a)^4}{(a \plus{} b)^4(b \plus{} c)^4(c \plus{} a)^4}$[/quote][/quote]
Assume $ a \ge b \ge c$
We have: $ LHS \minus{} 2 \ge \frac {(a \minus{} b)^2(a \plus{} b \minus{} c)}{(a \plus{} b)(b \plus{} c)(c \plus{} a)}$.
Why?
I think it's not true,my friend! :lol:
if true,can you prove it?

[quote="2424"][quote="quykhtn-qa1"][quote="2424"]with $ a,b,c > 0$.Prove that
  $ \frac {a}{b \plus{} c} \plus{} \frac {b}{c \plus{} a} \plus{} \frac {c}{a \plus{} b} \plus{} \frac {4abc}{(a \plus{} b)(b \plus{} c)(c \plus{} a)}\geq 2 \plus{} \frac {4(a \minus{} b)^4(b \minus{} c)^4(c \minus{} a)^4}{(a \plus{} b)^4(b \plus{} c)^4(c \plus{} a)^4}$[/quote]
Assume $ a \ge b \ge c$
We have: $ LHS \minus{} 2 \ge \frac {(a \minus{} b)^2(a \plus{} b \minus{} c)}{(a \plus{} b)(b \plus{} c)(c \plus{} a)}$.
Why?
I think it's not true,my friend! :lol:
if true,can you prove it?[/quote][/quote]
 $ LHS\minus{}2\equal{}\frac{a(a\minus{}b)(a\minus{}c)\plus{}b(b\minus{}c)(b\minus{}a)\plus{}c(c\minus{}a)(c\minus{}b)}{(a\plus{}b)(b\plus{}c)(c\plus{}a)} \ge \frac{a(a\minus{}b)(a\minus{}c)\plus{}b(b\minus{}c)(b\minus{}a)}{(a\plus{}b)(b\plus{}c)(c\plus{}a)}\equal{}\frac{(a\minus{}b)^2(a\plus{}b\minus{}c)}{(a\plus{}b)(b\plus{}c)(c\plus{}a)}$

[quote="quykhtn-qa1"][/quote][quote="2424"][quote="quykhtn-qa1"][quote="2424"]with $ a,b,c > 0$.Prove that
  $ \frac {a}{b + c} + \frac {b}{c + a} + \frac {c}{a + b} + \frac {4abc}{(a + b)(b + c)(c + a)}\geq 2 + \frac {4(a - b)^4(b - c)^4(c - a)^4}{(a + b)^4(b + c)^4(c + a)^4}$[/quote]
Assume $ a \ge b \ge c$
We have: $ LHS - 2 \ge \frac {(a - b)^2(a + b - c)}{(a + b)(b + c)(c + a)}$.
Why?
I think it's not true,my friend! :lol:
if true,can you prove it?[/quote]
 $ LHS - 2 = \frac {a(a - b)(a - c) + b(b - c)(b - a) + c(c - a)(c - b)}{(a + b)(b + c)(c + a)} \ge \frac {a(a - b)(a - c) + b(b - c)(b - a)}{(a + b)(b + c)(c + a)} = \frac {(a - b)^2(a + b - c)}{(a + b)(b + c)(c + a)}$[/quote][quote="quykhtn-qa1"][/quote]
yes,you true! :)
And We also have $ LHS\geq 2+\frac{4(a-b)^2(b-c)^2(c-a)^2}{(a+b)^2(c+a)^2(a+b)^2}$ :lol:

[quote="2424"][/quote][quote="quykhtn-qa1"][/quote][quote="2424"][quote="quykhtn-qa1"][quote="2424"]with $ a,b,c > 0$.Prove that
  $ \frac {a}{b + c} + \frac {b}{c + a} + \frac {c}{a + b} + \frac {4abc}{(a + b)(b + c)(c + a)}\geq 2 + \frac {4(a - b)^4(b - c)^4(c - a)^4}{(a + b)^4(b + c)^4(c + a)^4}$[/quote]
Assume $ a \ge b \ge c$
We have: $ LHS - 2 \ge \frac {(a - b)^2(a + b - c)}{(a + b)(b + c)(c + a)}$.
Why?
I think it's not true,my friend! :lol:
if true,can you prove it?[/quote]
 $ LHS - 2 = \frac {a(a - b)(a - c) + b(b - c)(b - a) + c(c - a)(c - b)}{(a + b)(b + c)(c + a)} \ge \frac {a(a - b)(a - c) + b(b - c)(b - a)}{(a + b)(b + c)(c + a)} = \frac {(a - b)^2(a + b - c)}{(a + b)(b + c)(c + a)}$[/quote][quote="quykhtn-qa1"]
yes,you true! :)
And We also have $ LHS\geq 2 + \frac {4(a - b)^2(b - c)^2(c - a)^2}{(a + b)^2(c + a)^2(a + b)^2}$ :lol:[/quote][quote="2424"][/quote]
Of course!
We have: $ LHS-2 \ge \frac{(a-b)^2(a+b-c)}{(a+b)(b+c)(c+a)}$ and $ (a-c)(b-c) <(a+c)(b+c)$
So,we only need to prove: $ (a+b-c)(a+b) > 4(a-c)(b-c)$
Which is obvious true because $ (a+b-c)(a+b)-4(a-c)(b-c)=(a-b)^2+3c(a+b)-4c^2 >0$

We have:
$ LHS \ge 2\plus{}\frac{4abc(a^2\minus{}b^2)^2}{(a\plus{}b\plus{}c)(a\plus{}b)^2(b\plus{}c)^2(c\plus{}a)^2}$


I think it's true!
we can write a/(ab+b^2)=a/b-a/(a+b)
We have a/b+b/c+c/a >=6 and a/(a+b)+b/(b+c)+c/(c+a)<=3/2
so ........

we can put $\frac{a}{b}\geq \frac{b}{c}\geq \frac{c}{a}$

then $\frac{a}{a+b}\geq \frac{b}{b+c}\geq \frac{c}{c+a}$

WLOG $abc=1$

so we have $\sum{\frac{a^{2}}{ab+b^{2}}}= \sum{\frac{a}{b}\frac{a}{a+b}}\geq \frac{1}{3}\sum{\frac{a}{b}}\sum{\frac{a}{a+b}}$ (chebyshev)

then because of $\frac{a}{b}+\frac{b}{c}+\frac{c}{a}\geq a+b+c$ for $abc=1$ we have
$LHS \geq \frac{1}{3}\sum{a}\sum{\frac{a}{a+b}}= \frac{1}{6}\sum({a+b})\sum{\frac{a}{a+b}}\geq \frac{1}{6}\sum{a^{\frac{1}{2}}}$ (CSB)
$\geq \frac{1}{6}3^{2}= \frac{3}{2}$ (AM-GM)

 :D

[quote="eagerkill"]I think it's true!
we can write $a/(ab+b^{2})=a/b-a/(a+b)$
We have $a/b+b/c+c/a \geq 6$ and $a/(a+b)+b/(b+c)+c/(c+a)\leq 3/2$
so ........[/quote]

you did a type mistake... $a/b+b/c+c/a \geq 3$

and how did you get $a/(a+b)+b/(b+c)+c/(c+a)\leq 3/2$

it's wrong try $a=0, b>c$

nice solution Adalbert

Put b+c=x
                    c+a  =y
                    a +b =z   Hence a=(y+z-x)/2

                           Apply A.M-G.M Inequality  ,
                                        To get the required answer, i.e                        [a/( b+c)]+[b/(c+a)]+[c/(a+b)] >=1.5  :P


Determine all pairs of natural numbers $ (a,b)$ such that

$ a^{5a} \equal{} b^b$

If $ b < a$ then $ b^b < a^a < a^{5a}$. On the other hand, if $ b \ge 5a$ then $ b^b \ge (5a)^{5a} > a^{5a}$. Therefore we conclude that $ 5a > b \ge a$.

Let $ g \equal{} (a,b)$ and let $ a \equal{} gA,b \equal{} gB$. Then $ (A,B) \equal{} 1$ and our equation becomes $ g^{5gA \minus{} gB}\cdot A^{5gA} \equal{} B^{gB}$. Taking the $ g$-th root on both sides we get $ g^{5A \minus{} B}\cdot A^{5A} \equal{} B^B$. Hence $ A\mid B^B$ which implies $ A \equal{} 1$. Thus $ g^{5 \minus{} B} \equal{} B^B$. Checking $ B \equal{} 1,2,3,4$ we find that only the pairs $ (B,g) \equal{} (1,1),(4,4^4)$ work. Hence all solutions are $ (a,b) \equal{} (1,1),(4^4,4^5)$.


I am curious what you guys consider to be the fundamental "purpose" of college.  Specifically, do you believe that the goal of college is to learn and prepare for a [u]career[/u], or to learn about your [u]interests[/u].  Personally, I am deeply interested in mathematics, and I would like to learn about it more in college.  Hence, I am considering majoring in (pure) mathematics.  However, a lot of people I talk to tell me that I should look for a major that uses math but is more practical (finance, economics, engineering, etc.); that is, I should pick a major with a career in mind.  I am having troubling determining if college should be used to prepare for a career (major in math related field like engineering) or used to learn and explore one's interests (major in pure math).  What are your thoughts?

Not too long ago, my thoughts on majors in pure vs.  applied mathematics were "Applied math is for mellow people with no true fire who can't handle the whole generality. Real men study theoretical math and render the applications semi-trivial."

I've now switched significantly on this. Oh sure, chumps do major in applied math, no question. But I now consider theoretical math  with no known potential applications a "just-for-fun" activity, and a major in it the smart person's version of a major in dance.

The purpose of college is to prepare you for and help you get an initial bearing on life. That [i]includes[/i] figuring out whether or not you want to be more geared towards your career or your interests. That said, there are two opposing forces to keep in mind:

[b]1.[/b] Exploring your interests will ultimately help you decide on a suitable career for yourself, and possibly open your eyes to career paths you did not previously consider.
[b]2.[/b] If you are well prepared for a fruitful career, you may have more resources later in life to pursue your interests.

Pure math is actually an incredibly flexible major from what I gather, both career- and grad school- wise. If it's what you're interested in, why not?

Pure math is only "incredibly flexible" if you actually use your electives strategically to explore and at least somewhat develop fields to which you might want to apply your mathematics. You could then double major between math and something you might want to apply it to, or major in math and minor in one or two other things.


inequalities
email
algebra
polynomial
Cauchy Inequality
inequalities proposed

Let $a,b,c$ be nonegative real numbers such that
\[a^{2}+b^{2}+c^{2}=1,\]
prove that
\[\frac a{\sqrt{1+bc}}+\frac b{\sqrt{1+ca}}+\frac c{\sqrt{1+ab}}\leq\frac32.\]
Have you got a nice solution for it?

[quote="pvthuan"]Let $a,b,c$ be nonegative real numbers such that
\[a^{2}+b^{2}+c^{2}=1, \]
prove that
\[\frac a{\sqrt{1+bc}}+\frac b{\sqrt{1+ca}}+\frac c{\sqrt{1+ab}}\leq\frac32. \]
Have you got a nice solution for it?[/quote]
Here is my solution, MrThuan!
Applying Cauchy Inequality, we have
\[\left(\sum\frac a{\sqrt{1+bc}}\right)^{2}\le \left(\sum a \right) \left( \sum\frac{a}{1+bc}\right) \]
We have the following result:
\[\sum\frac{a}{1+bc}\le \sum\frac{a}{(a+b)(a+c)}\]
Actually, we have
\[LHS-RHS = \sum\frac{a((a+b)(a+c)-a^{2}-b^{2}-c^{2}-bc)}{(1+bc)(a+b)(a+c)}=\sum\frac{ab(a-b)-ca(c-a)}{(1+bc)(a+b)(a+c)}=\sum\frac{ab(a-b)}{(1+bc)(a+b)(a+c)}-\sum\frac{ab(a-b)}{(1+ac)(b+a)(b+c)}=-\sum\frac{ab(a-b)^{2}(1-c^{2})}{(a+b)(b+c)(c+a)(1+ac)(1+bc)}\le 0 \]
Hence, it is sufficient to prove that
\[\left(\sum a\right)\left(\sum\frac{a}{(a+b)(a+c)}\right) \le \frac{9}{4}\]
\[\Leftrightarrow (a+b)(b+c)(c+a) \ge \frac{8}{9}(ab+bc+ca)(a+b+c) \]
By AM - GM Inequality, it is true. :)

Good, Can. Come on your contribution by email to me.

[quote="pvthuan"]Let $a,b,c$ be nonegative real numbers such that
\[a^{2}+b^{2}+c^{2}=1, \]
prove that
\[\frac a{\sqrt{1+bc}}+\frac b{\sqrt{1+ca}}+\frac c{\sqrt{1+ab}}\leq\frac32. \]
Have you got a nice solution for it?[/quote]

I wonder if the condition $a^{2}+b^{2}+c^{2}=1$ can be extended into $a^{n}+b^{n}+c^{n}=1$ ($n \geq 2$) and the expression can still have a maximum?

[quote="pvthuan"]Let $ a,b,c$ be [b]nonegative[/b] real numbers such that $ a^{2} \plus{} b^{2} \plus{} c^{2} \equal{} 1$, prove that

$ \frac a{\sqrt {1 \plus{} bc}} \plus{} \frac b{\sqrt {1 \plus{} ca}} \plus{} \frac c{\sqrt {1 \plus{} ab}}\leq\frac32.$

Have you got a nice solution for it?[/quote]A short proof see : http://www.mathlinks.ro/Forum/viewtopic.php?t=197878

Let $ a,b,c$ be [b]real[/b] numbers such that $ a^{2} \plus{} b^{2} \plus{} c^{2} \equal{} 1$, then the following inequality

$ \frac {a}{\sqrt {k \plus{} bc}} \plus{} \frac {b}{\sqrt {k \plus{} ca}} \plus{} \frac {c}{\sqrt {k \plus{} ab}}\leq\frac {3}{\sqrt {3k \plus{} 1}}$

holds if and only if $ k\geq 0.666952\cdots$, which is a root of the irreducible polynomial

$ 186624k^8 \plus{} 145152k^7 \plus{} 172800k^6 \minus{} 248832k^5 \minus{} 14028k^4 \plus{} 20644k^3 \minus{} 3891k^2$

$ \plus{} 288k \minus{} 9$.


geometry
geometric transformation
rotation
absolute value

Here's a problem I'm stuck with:

Two small bodies of masses $m_1$ and $m_2$ are connected by a non-elastic string of length $l$ and can move on a horizontal frictionless plane. At one moment, the velocity of one body is zero and the velocity of the other body is $v$ and is orthogonal to the string connecting the bodies. Find the tension of the string.

Move into the center of mass reference frame.
I think you can solve it from there.  :)

On a first looking I would say we need the acceleration of the particle with zero velocity.

Sorry, I didn't notice the [b]frictionless[/b], therefore CM is an inertial frame.
So, move your obsever ovethere and it will move with a velocity $\,V\,=\,\frac{m_1v}{m_1+m_2}\,\,\,$,  and " see" both particles twirling around it (we assume, wlg, that the instant center of rotation is the body whose mass is $m_2$).

[hide="Click to reveal hidden content"]
 The relative velocities of the bodies will be $\,v_1\,=\,\frac{m_2v}{m_1+m_2},\,\,\,\,\,\,v_2\,=\,-\,\frac{m_2v}{m_1+m_2}\,\,$ (the sign means that v_2 and V are in ooposite directions), and for the rotation radius we'll have $\,r_1\,=\,\frac{m_2l}{m_1+m_2},\,\,\,\,\,r_2\,=\,\frac{m_1l}{m_1+m_2}\,\,$
The tention, T, will be the centripetal force for either of the bodies (in absolute value): $T\,=\,\frac{m_1v_1^2}{r_1}\,=\, \frac{m_1v_2^2}{r_2}.\,\,$
[/hide]


analytic geometry
algebra
linear equation

Goldfish are sold at $ 15$ cents each. The rectangular coordinate graph showing the cost of $ 1$ to $ 12$ goldfish is:

$ \textbf{(A)}\ \text{a straight line segment} \qquad \\
\textbf{(B)}\ \text{a set of horizontal parallel line segments}\qquad \\
\textbf{(C)}\ \text{a set of vertical parallel line segments}\qquad \\
\textbf{(D)}\ \text{a finite set of distinct points}\qquad
\textbf{(E)}\ \text{a straight line}$

Unsure how to insert graphs but the equation for this graph would be:
$ y \equal{} 15g$ where $ g$ is the number of goldfish and $ y$ is the cost, with $ 1 \le g \le12$

As this is a graph of points of a linear equation (to indicate the 1 -12 goldfish), wouldnt the answer be $ D$?


I just downloaded TeXnicCenter and am going through the configuration wizard. It says "Enter the full path of the directory, where executables (latex, tex, etc.) of your TeX distribution are located:"
However, I have no idea what it wants. Could someone help? :(

Just use the Browse button to navigate to where latex.exe can be found. It should be in ....\MiKTeX\bin where .... depends on your machine

I'm really confused about this.... :(
It hasn't asked me at all if it wants to configure with MikTex, and I can't find latex.exe anywhere.....I think the LaTeX giude here is messed up or something...

OK. Have you installed MiKTeX first? This provides all the latex facilities and [i]will[/i] include the latex.exe file wherever you have chosen to install MiKTeX. Look for the directory in which you've installed MiKTeX (c:\texmf\miktex or c:\Program Files\miktex or whatever) and inside that you'll find \bin and inside that you'll find latex.exe and lots of other files such as tex.exe.

TexnicCenter doesn't mention MiKTeX because it doesn't care what LaTeX distro you've installed; there are a number of them but MiKTeX is the most common one for Windows which is why the AoPS guide suggests downloading it.

Thank you so much! It works now! Thank you, thank you!

i was having the same problems. but for some reason when i look in those folders i don't see \bin  :(

Try searching your drive(s) for latex.exe; if it's not there then you may need to re-install MiKTeX.