Καλησπερίζω .Λοιπόν βάζω τα θέματα π μπήκαν σήμερα στον Μεσογειακό διαγωνισμό.
Θα παρακαλούσα όλους αυτούς που συμμετείχαν στον διαγωνισμό να βάλλουν τις Δικες τους λύσεις στα θέματα (γιατί καθεμιά είναι σίγουρα ξεχωριστή) τα οποία μου φάνηκαν πολύ ωραία ως πρώτη ματιά .

[color=brown]1[/color] Κάθε σημείο του επιπέδου χρωματίζεται με δύο χρώματα μπλε και κόκκινο. Υπάρχει τουλάχιστον ένα κόκκινο και τουλάχιστον ένα μπλε σημείο .
Είναι δυνατόν κάθε κύκλος ακτίνας 1cm να περιέχει επί της περιφέρειάς του ακριβώς:
ι)ένα μπλε σημείο
ιι)δύο μπλε σημεία

[color=green]2[/color]  Έστω $P$ ένα εσωτερικό σημείο ενός τριγώνου ΑΒC και έστω
$A_{1}B_{2}$,  $B_{1}C_{2}$,  $C_{1}A_{2}$  οι παράλληλες που άγονται από το σημείο P προς τις πλευρές AB,BC,CA αντίστοιχα με $A_{1},A_{2}\in BC$ , $B_{1},B_{2}\in AC$ και $C_{1},C_{2}\in AB$ .Νδο $(A_{1}A_{2}B_{1}B_{2}C_{1}C_{2})\geq\frac{2}{3}(ABC)$ όπου με (...) συμβολίζουμε το εμβαδόν

  [color=darkblue]3[/color] Θεωρούμε ABC τρίγωνο στο οποίο τα μήκη των πλευρών  a,b,c  είναι θετικοί ακέραιοι αριθμοί τέτοιοι ώστε ΜΚΔ(a,b,c)=1 .H εσωτερική διχοτόμος της γωνίας Α τέμνει την ΒC στο D .Να αποδείξτε ότι
To  ΑΒD είναι όμοιο προς το ΑΒC αν και μόνο αν ο αριθμός  c  είναι τέλειο τετράγωνο

[color=orange] 4[/color]  Έστω $m,n$ θετικοί ακέραιοι και έστω $x_{i,j}\in\left[ 0,1\right]$  for i=1,2,....,m και
j=1,2,....,n

Νδο  $\prod_{j=1}^{n}(1-\prod_{i=1}^{m}x_{i,j})+\prod_{i=1}^{m}(1-\prod_{j=1}^{n}(1-x_{i,j}))\geq 1$

Η τελευταία μοιάζει λίγο με Weirstrass II αλλά δεν ξέρω αν είναι εφαρμογή της. Θα το δω.

Σιλουανε ευχαριστω για την παραθεση των θεματων !!

Μπορεις να βάλεις και των μικρών??

 :)

Ta themata den itan asxima kai mallon arketa prosvasima. Elpizo na pigate kala osi simeteixate.
Sas grafo tis liseis mou :
1)
(i) Iparxei toulaxiston ena mple simeio, me kentro afto grapste kiklo aktinas 1, ean iparxei pano ston kiklo ena mple simeio tote apexei apo to kentro akrivos 1, ara iparxei (giati ?) kiklos aktinas 1 pou dierxete kai apo ta dio antifasi.
(ii) As bapsoume mple omokentrous kiklous, ton opioon i aktines diaferoun kata 2 monades. To ipoloipo epipedo kokkino. Tote profanos kathe kiklos aktinas 1 , eite tha temnei enan apo autous se dio simeia, eite tha efaptetai dio apo aftous.

2)
As apodeiksoume prota to parakato limma.
Esto trigono ABC , metablito simeio D stin BC , kai fero paralliles DK, DL pros tis AB, AC antistoixa (me K stin AC kai L stin AB). Tote to embadon (LDK) einai megiso otan to D ginei to meso tis BC.
Apodeiksi :
$\frac{LD}{AC} \cdot \frac{DK}{AB} = \frac{ BD}{BC} \cdot \frac{DC}{BC} \leq \frac{(BD+DC)^2}{4 \cdot BC^2} =\frac{1}{4}$ Me isotita otan D einai meso.

Ara (piso sto problima) an to P kinitai pano stin $C_2 B_1$ tote $(C_2 B A_1 P) +(B_1 C A_2 P)$ menei analoioto kai simfona me to parapano limma to $(A C_1 P B_2)$ tha ginei megisto otan to P brisketai sto meso tou tmimatos pou kinitai. Ara to P brisketai pano stin diameso AM (M meso tou BC) omoios gia tis alles pleures...

3)
Ta dio trigona einai omoia an kai mono an A =2C (gonies) ara isxuei oti $a^2-c^2=bc$ (Gnosti tautotita , tin afino os askisi). Ara $a^2=c \cdot (b+c)$. Tote prepei (b,c)=1, eidalos o koinos diaretis ton b,c tha diairouse to a , alla (a,b,c)=1. Tote omos (c,b+c)=1, ara to c tha prepei na einai teleio tetragono (giati ?)
To antistrofo den isxuei p.x parte to pithagorio trigono me pleures 3,4,5. (Silouane, mipos itheles na grapseis kati allo ?)

4)
Exo dio liseis giauto, tha grapso tin pio amesi kai tha sas doso mia mikri ipodeiksi gia tin alli.
I parasasti sta aristera einai grammiki se kathe tis metabliti ara pernei ta akrotata tis sta akra ton diastimaton pou orizete. (p.x an statheropoisete oles tis metablites ektos tis x11, tote einai gramiki sinartisi os pros aftin ..k.t.l )
Ara ola ta x_ij einai eite 0 eite 1 otan exoume elaxisto. An ola ta X_i1,X_i1,...X_i1 einai isa me 1 gia kapoio i, tote o protos oras tou athroismatos einai isos me 0 kai o deuteros isos me 1. An gia kathe i den isxuei to apopano tote o protos oros tou athroismatos einai panta toulaxtiston 1 kai o allos thetikos. Ara to elaxisto einai to 1  ;)  (mporeite na breite to megisto ?)

Mia alli proseggisi tha itan na ergastoume me pithanotites .... Ti mporei na antiprosopeoun ta x_ij ?

Stergios

ΦΙΛΟΙ ΠΡΕΠΕΙ ΝΑ ΔΕΙΤΕ ΤΙΣ ΠΡΟΤΕΙΝΟΜΕΝΕΣ ΛΥΣΕΙΣ ΑΠΟ ΤΗΝ ΕΤΑΙΡΕΙΑ ΓΙΑ ΤΟ ΠΡΟΒΛΗΜΑ 3 ΠΡΕΠΕΙ ΝΑ ΙΣΧΥΕΙ ΚΙ ΤΟ ΑΝΤΙΣΤΡΟΦΟ ΟΚ?
ΧΑΙΡΕΤΑΙ

Το πρώτο θέμα, το δεύτερο και από το τρίτο το πρώτο σκέλος τα έλυσα την ώρα του διαγωνισμού.Για το τρίτο,το αντίστροφο δεν κατάφερα να το αποδείξω και μάλλον ο Anto έχει δίκιο για το τρίγωνο με πλευρές 3,4,5 …  :o (άλλοστε οι πρώτες επίσημες λύσεις που βγήκαν δεν έχουν ολοκληρωθεί ακόμα).Βρήκα επίσης μια ωραία λύση στο τέταρτο που (όπως και πέρυσι) το έλυσα στο τρένο του γυρισμού,αλλά λόγω περιορισμένων γνώσεων Latex δεν μπορώ να τη γράψω.Η λογική παντως είναι να θεωρήσουμε ότι όλα τα x_ij παίρνουν τις ακραίες τιμές 0 ή1 .

1)ii)Εκτός από τη λύση  του Anto με τους ομόκεντρους κύκλους μπορούμε να θεωρήσουμε ότι όλα τα μπλε σημεία και μόνον αυτά  ανήκουν σε παράλληλες ευθείες με απόσταση 2cm.

3)Επίσης αποδεικνύεται ότι $c=d^2$ ,όπου $d=(a,c)$ .

Οι άλλες λύσεις μου είναι όμοιες με αυτές του Anto.

Στέλιος...  :)
________________________________

Λοιπόν αγαπητοί φίλοι...
Τα θέματατου διαγωνισμού ήταν πάρα πολύ όμορφα αν εξαιρέσουμε το τελευταίο που κατα τη γνώμη μου δεν ενδεικνυται για τέτοιου είδους διαγωνισμό!Αλλά:
1)ο Silouan μετέφερε με ακρίβεια τις διατυπώσεις (Αντο εκπλήσσεσαι???)
2)στο πρώτο θέμα η σωστή διατύπωση είναι η εξής:
[b]Υπάρχει χρωματισμός τέτοιος ώστε...[/b]  Για το β ερώτημα μπορούμε να κάνουμε κι άλλη μία -πιο αφηρημένη κατασκευή..(πώς??)
3)Στο δεύτερο θέμα πολύ απλά θέτουμε: ΒΑ1=χ, Α1Α2=ψ και Α2Γ=ζ και μετά χρησιμοποιούμε το εξής:
3(x^2+ψ^2+ζ^2)>=(χ+ψ+ζ)^2
4)[b][u]Τώα θα γίνω καυστικός!Δεν ξέρω αν φταίνε οι Ισπανοί ή η μετάφραση της ΕΜΕ αλλά είναι προφανές ότι κάποιος εξ' αυτών είναι ΑΣΧΕΤΟΣ![/u][/b].Για το α ωραίος και ο Αντο και ο Στέλιος (είχα την ίδια λύση)!Για το β είναι [u][b]προφανες ότι  ο Αντο έχει δίκιο!!!Ηταν απαράδεκτο!![/b][/u]
Κρίμα γιάτι μέρικοί θα προσπαθούσαν να αποδείξουν κάτι που δεν ισχύει!!
5)για το τελευταίο τι να πω??Όπως είπα μόνο αυτό δεν μου άρεσε ... αλλά αυτά συμβαίνουν και στις καλύτερες οικογένειες!!

(Αντο υπονοείς τίποτα για κέρματα) ???????? :roll:

Εις το επανιδείν!

Για το πρώτο ,τρίτο και τέταρτο θέμα οι λύσεις μου είναι σχεδόν όμοιες με αυτές του Αnto για τη δεύτερη έχω μια διαφορετική λύση .

Έστω $M_1=(C_2PA_1)$, $M_2=(B_2PC_1)$ , $M_3=(B_1A_2P)$ , $T_1=(A_1A_2P)$,
$T_2=(B_1B_2P)$, $T_3=(C_1C_2P)$ ,$T=(ABC)$.

Εύκολα βρίσκουμε ότι $(C_2A_1B)=M_1$ , $M_3=(A_2B_1C)$ kai $M_2=(AC_1B_2)$.

Παρατηρούμε ότι $2(M_1+M_2+M_3)+T_1+T_2+T_3=T$  (1)

άρα παίρνοντας και γράφοντας ισοδύναμα (με τη βοήθεια της (1)) αυτή που θέλουμε με  να αποδείξουμε καταλήγουμε ότι θνδο $3(T_1+T_2+T_3)\geq T$ όμως αυτό ισχύει αν χρησιμοποίησουμε Cauchy στο γνωστό θεώρημα που λέει ότι
$\sqrt{T}=\sqrt{T_1}+\sqrt{T_2}+\sqrt{T_3}$   :lol:  ;)

BTW ,το δεύτερο (ουσιαστικά ) ερώτημα από την άσκηση 3 είναι λάθος  ?????

ΦΙΛΟΙ ΑΠΟ ΟΤΙ ΕΧΩ ΚΙ ΤΙΣ ΕΠΙΣΗΜΕΣ ΛΥΣΕΙΣ ΕΙΝΑΙ ΔΥΝΑΤΟΝ ΤΟ ΑΝΤΙΣΤΡΟΦΟ ΑΝΑΚΟΙΝΩΣΕ Η ΕΜΕ ΟΤΙ ΤΟ ΑΝΤΙΣΤΡΟΦΟ ΔΕΝ ΓΙΝΕΤΑΙ? ΓΙΑΤΙ ΑΝ ΟΧΙ ΘΑ ΕΙΝΑΙ ΕΓΚΛΗΜΑ ΣΕ ΒΑΡΟΣ ΑΥΤΩΝ ΠΟΥ ΔΕΝ ΘΑ ΠΕΡΑΣΟΥΝ ΚΙ ΠΡΕΠΕΙ ΝΑ ΠΟΦΕΥΧΘΕΙ ΠΑΡΑΚΑΛΩ ΝΑ ΜΕ ΕΝΗΜΕΡΩΣΕΤΕ ΣΥΝΤΟΜΑ ΓΙΑ ΤΟ ΑΝ ΓΙΝΕΤΑΙ Η ΌΧΙ ΤΟ ΑΝΤΙΣΤΡΟΦΟ
ΧΑΙΡΕΤΑΙ


n points are given on a plane. For every integer k  (0<k<=n) ,can we draw a line that divides the set of points into two sets, one of them has k elements and the other is (n-k)? (there will be no point on the line)

[hide="Solution"]It's true. Let's consider all the lines througn our n points. Since that number is finite, there exists a line which is not parrell to any of theese lines. Lines parrell to it solve the problem.[/hide]


Just a LaTeX hint: the American standard is $\tan$ for tangent and $\cot$ for cotangent, and latex has built-in commands (\tan and \cot) for these.

What about this identity:

$a^{ \log_{a}x}=x$.

okay where are the $g$'s coming from?

is it supposed to be \[\frac{2}{3}\cdot 3^{\log_{6}{\tan^{2}{x}}}+\frac{3}{2}\cdot 2^{\log_{6}{\tan^{2}{x}}}= 6\left(3-\cot^{2}{x}\right)\]?

[quote="Teki-Teki"]okay where are the $g$'s coming from?[/quote]

$\text{tg}, \text{ctg}$ are used in some places for $\tan, \cot$.

Any body kill it ...???
SOrry for spamming. I only want to ...


pisqrt314
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#1
 FTW clan!!
about ftw clan
I think their should be FTW clans making it more challenging.

Should their be FTW CLANS?
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^ beat that ^

Posted: Sun Feb 01, 2009 5:53 pm

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dragon96
Navier-Stokes Equations

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#2
Describe how it would work? I'm guessing you mean something like the team round?
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Posted: Sun Feb 01, 2009 6:08 pm

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r15s11z55y89w21
Poincare Conjecture

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#3
It could be a good idea, but seeing how restless we have made levans with the lobby chat business, I doubt the poor guy has any leftover time to implement more stuff into the new FTW.

So, what I would suggest is, as levans says, remind others violating the terms of service to try to help keep lobby chat safe, productive, and clean, and he could use his time on the new FTW; he might work out a pleasing surprise in the new FTW in terms of what us users might like.

Seriously, to those spamming, flooding, swearing, being mean, advertising inappropriate content, etc.: I really don't see why you guys think it's so worth it to do all of this; even if you weren't muted for that, would it even help you in any way?
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Posted: Sun Feb 01, 2009 6:28 pm

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jjx1
Riemann Hypothesis

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#4
If we all improve our chat, levans might spend time on the new ftw. Maybe then will any of these ideas ever happen. I suggest everyone just not say stuff in chat gets you any risk of muted.

Posted: Sun Feb 01, 2009 6:39 pm

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pisqrt314
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#5
dragon96 wrote:
Describe how it would work? I'm guessing you mean something like the team round?

Anyway, a person becomes a clan leader and he schedules games. The clan trains the players and when they get better, then they play other clans.

Each Clan plays every other clan twice. Then the clans with the most wins goes to playoffs....
every clan has to consists of different level players.
Every week, one clan plays another one.
The clan leader schedules games and play with people in his clan. The clanmates teaches each other in the game or some other chat room.

_________________
 Math Problems I do each DAY: 196
^ beat that ^

Posted: Yesterday, at 4:09 pm

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dragon96
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#6
That would be more of a tourney than a game style. You can try this in the FTW other forum.
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Posted: Yesterday, at 6:15 pm

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pisqrt314
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#7
dragon96 wrote:
That would be more of a tourney than a game style. You can try this in the FTW other forum.

so how do I move my post their?

Posted: Today, at 12:41 pm

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r15s11z55y89w21
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#8
Two choices:

1) Copy+paste everything.
------OR------
2) Ask a mod/admin to do it for you.
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Posted: Today, at 3:18 pm

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You should have asked an admin...

And at least LOOK at what the heck you're posting!  Honestly, that's unreadable.

Mod/admin, please delete!!

This is clearly spam.

Locked.

ALso, whoever had this idea, forgot username, please don't worry about this. This would never happen unless we had a new ftw whih doesn't look too good.


Somebody help me,solving this equation :
$ \sqrt{1\plus{}\dfrac{\sqrt{x^2\plus{}28^2}}{x}}\minus{}\sqrt{{x}\sqrt{x^2\plus{}28^2}\minus{}x^2}\equal{}3$

thanks!

I think your equation is equivalent to:
$ x^6 \plus{} 786x^4 \minus{} 2529x^2 \plus{} 784 \equal{} 0$
which is a cubic in $ x^2$.  I don't think there are any nice roots.

[b]EDIT:[/b] Ignore this post....

The only root  is:
 $ x\equal{} \dfrac{\sqrt{47*4^2}}{47}$ , but how to get till here ? (an easier way),thanks !

An easier way compared to...?

Well, my naive answer is to substitute u = x^2 pound it through THIS:
http://www.math.vanderbilt.edu/~schectex/courses/cubic/

Assuming the root you have is correct wouldn't it be easier to write it as

$ x \equal{} \frac{4}{\sqrt{47}}$?

Of course it would be easier but I've rationalized it. Thanks 4 cubic formula !

Sorry my mistake, the first equation I posted is wrong.  It is actually equivalent to:
$ 658x^4 \minus{} 2527x^2 \plus{} 784 \equal{} 0$
Which is a biquadratic, easily solvable.  One of its roots is the one you want: $ \frac {4}{\sqrt {47}}$

[quote="Farenhajt"][hide="Solution"]Factor out $ |x|$ from the second term:

$ \sqrt {{\sqrt {x^2 + 28^2}\over x} + 1} - |x|\sqrt {{\sqrt {x^2 + 28^2}\over x} - 1} = 3$

Now put

$ u = \sqrt {{\sqrt {x^2 + 28^2}\over x} + 1}, v = |x|\sqrt {{\sqrt {x^2 + 28^2}\over x} - 1}$

We get

$ u - v = 3$

$ uv = |x|\sqrt {{x^2 + 28^2\over x^2} - 1} = |x|\cdot{28\over|x|} = 28$

Hence $ u(u - 3) = 28\iff u = 7\lor u = - 4$

We take only the positive solution because of the root.

\begin{eqnarray*}{\sqrt {x^2 + 28^2}\over x} + 1 = 49\iff {x^2 + 28^2\over x^2} = 2304 \iff {28^2\over x^2} = 2303\iff x = \pm\sqrt {28^2\over 49\cdot 47} = \pm{4\over\sqrt {47}}\end{eqnarray*}

However, the negative root doesn't work because of the second term in the initial equation, hence $ x = {4\over\sqrt {47}}$[/hide][/quote]

thank u Farenhajt. I've learnt a lot from this solution !


Given A = (–3,0), B = (4,6) and C = (–2,k),

Find k if
i) ABC is right angled at C.

ii) ABC is isosceles with AB = BC

I've got this so far, but dunno what to do next

$ AB_m\equal{}\frac{6\minus{}0}{4\minus{}(\minus{}3)}\equal{}\frac{6}{7}$

$ BC_m\equal{}\frac{6\minus{}k}{6}$

$ AC_m\equal{}\minus{}\frac{k}{5}$

$ AB\equal{}\sqrt{49\plus{}36}\equal{}\sqrt{85}$
$ AC\equal{}\sqrt{1\plus{}k^2}$
$ BC\equal{}\sqrt{36\plus{}(k\minus{}6)^2}$

i) $ AC^2\plus{}BC^2\equal{}AB^2\Rightarrow 1\plus{}k^2\plus{}36\plus{}(k\minus{}6)^2\equal{}85$ and solve for k.

ii) $ AB\equal{}BC\Rightarrow\sqrt{36\plus{}(k\minus{}6)^2}\equal{}\sqrt{85}$ and solve for k.


geometry
3D geometry
induction
AMC
AIME
modular arithmetic
algebra

What is the smallest number that when cubed ends in 888?

NOTE: My method is not recommended unless you can multiply somewhat quickly like I can, as this problem took me about 20 minutes. I hope someone can put a less time-consuming method. Also there's an explanation I requested in there, could someone please post/PM that?
[hide]
We know already that the number has to end with a $2$ for its cube to end with an 8. We can eliminate $2$ and $12$ since as an AoPSer you should probably know their cubes. Basically keep going up with numbers ending with $2$ until $92$ where you will observe something. $42^{3}$ is $74088$ and $92^{3}$ is $7768688$. I used induction and thought that every $50$ numbers will end in the same $2$ digits for their cube (which I had previously noticed was true with other numbers, could someone explain this) and so tried $142^{3}$ to get $2863288$ which was wrong and kept trying until I got the cube to end with $888$, the cube being $7077888$, which had a cube root of $192$
[b]ANSWER[/b]: $192$
[/hide]
Also bpms, do you happen to know which problem number on that AIME this particular one was? Thanks.

[hide="Some stuff you might need to know to understand the solutions"]
In case you aren't familiar, taking something mod x means you are putting every number into some "residue class" based on its remainder when divided by x. For example, $9 \equiv 4 \equiv-1 \pmod 5$.

In addition, if $p$ and $q$ are relatively prime, then $a \equiv b \pmod p$ and $a \equiv b \pmod q \iff a \equiv b \pmod pq$.

The CRT, or Chinese Remainder Theorem, basically says this is always possible and gives a method to find the right numbers (for example if you only know the residue of $a$ mod $p$ and $q$, it tells you how to find it mod $pq$).

Most of the time I go from one equation to another, I am expanding with the binomial theorem and killing any multiples of the mod I am taking things in.
[/hide]

[hide="Solution with less theory but more computation"]
$n^{3}\equiv 888 \pmod{1000}$

Obviously, only the first 3 digits matter. So let

$n = (100a+10b+c)$, where a,b,c are each digits.

Then $c^{3}\equiv 8 \pmod 10$, which means that $c = 2$.

$30bc^{2}\equiv 80 \pmod{100}\implies 12b \equiv 8 \pmod{10}\implies 2b \equiv 8 \pmod{10}$, so $b = 4$ or $b = 9$.

Now we have to be a bit careful since there are some carries involved. In the first case, $b = 4$, we have

$(100a+42)^{3}= 10^{6}a^{3}+3 * 10^{4}* 42 a^{2}+3 * 10^{2}* 42^{2}* a+42^{3}\equiv 3 * 10^{2}* 42^{2}* a+42^{3}\equiv 300 * 2^{2}* a+42^{3}\equiv 200a+(40+2)^{3}\equiv 200a+40^{3}+3 * 40^{2}* 2+3 * 40 * 2^{2}+2^{3}\equiv 200a+6 * 1600+480+8 \equiv 200a+600+480+8 \equiv 200a+88 \equiv 888$, so the smallest possible value of $a$ here is $4$, and we have one possible smallest solution is $442$.

On the other hand, if $b = 9$, then

$(100a+92)^{3}= 10^{6}a^{3}+3 * 10^{4}* 92 a^{2}+3 * 10^{2}* 92^{2}* a+92^{3}\equiv 3 * 10^{2}* 92^{2}* a+92^{3}\equiv 300 * 2^{2}* a+92^{3}\equiv 200a+(90+2)^{3}\equiv 200a+90^{3}+3 * 90^{2}* 2+3 * 90 * 2^{2}+2^{3}\equiv 200a+6 * 8100+1080+8 \equiv 200a+600+80+8 \equiv 200a+688 \equiv 888$, so the smallest possible value of $a$ here is $1$, yielding $192$, which is smaller than $442$, so the answer is 192.
[/hide]

[hide="Solution with more theory but less computation"]
$n^{3}\equiv 888 \pmod{1000}$

This is biconditional with the following:

$n^{3}\equiv 888 \pmod 8 \iff n^{3}\equiv 0 \pmod 8 \iff n \equiv 0 \pmod 2$

$n^{3}\equiv 888 \pmod{125}\iff n^{3}\equiv 13 \pmod{125}$

This second condition is a bit messier, but if this is to be true, then certainly, $n^{3}\equiv 3 \pmod 5$, so that $n \equiv 2 \pmod 5$. Then let $n = 5k+2$, $(5k+2)^{3}\equiv 13 \pmod{25}$, so $60k+8 \equiv 13 \iff 10k \equiv 5 \iff k \equiv 3 \pmod 5$, so, finally, let $n = 25k+17$, then $(25k+17)^{3}\equiv 13 \pmod{125}\iff 3*25*17^{2}*k+17^{3}\equiv 13 \pmod{125}\iff 3*25*2^{2}*k+(15+2)^{3}\equiv 13 \pmod{125}\iff 50k+3*15^{2}*2+3*15*2^{2}+2^{3}\equiv 13 \pmod{125}\iff 50k+25*3^{3}*2+5 * 6^{2}+8 \equiv 13 \pmod{125}\iff 50k+25*3^{2}+5 * (25+11) \equiv 5 \pmod{125}\iff 50k-25+55 \equiv 5 \pmod{125}\iff 25(2k+1) \equiv 0 \pmod{125}\iff 2k+1 \equiv 0 \pmod 5 \iff k = 2$.

So $n \equiv 2*25+3*5+2 \equiv 67 \pmod{125}$, and $n \equiv 0 \pmod 2$. CRT says that this means that $n \equiv 192 \pmod{250}$, so the smallest value of $n$ is 192.
[/hide]

[quote="Karth"]Where can we learn these concepts about Number Theory?[/quote]

aops1,2

A lot of it also becomes fairly intuitive if you think about it geometrically (a number is a line segment, and its divisors are line segments that can be repeatedly copied to come up with the original number).

Okay I explained that really poorly, but it helps (for me, at least).

[quote]
Also bpms, do you happen to know which problem number on that AIME this particular one was? Thanks.[/quote]
9

Yeah that sort of makes sense that its easier to do math mod 10 than mod 5, since we work in a base 10 system. I just sort of instinctively went down to the smallest mod possible, which apparently, as you point out, isn't the optimal way.

[quote="Karth"]Where can we learn these concepts about Number Theory?[/quote]

Right here!  :D

[b]Definition:[/b]  Given integers $a, b, m$, we say that $a \equiv b \bmod m \Leftrightarrow m | (a-b)$.

[b]Lemma 1:[/b]  $a | c, b | c, (a, b) = 1 \implies ab | c$ (elementary and left as an exercise)

[b]Lemma 2:[/b]  $a \equiv b \bmod p, a \equiv b \bmod q, (p, q) = 1 \implies a \equiv b \bmod (pq)$ (follows from lemma 1)

[b]Lemma 3:[/b]  $a \equiv b \bmod (pq) \implies a \equiv b \bmod p, a \equiv b \bmod q$ (obvious from definition)

[b]Lemma 4:[/b]  $a \equiv b \bmod (pq), (p, q) = 1 \Leftrightarrow a \equiv b \bmod p, a \equiv b \bmod q$ (lemma 2 + lemma 3 in a convenient package)

[b]Theorem 1:[/b]  Given remainders of some number $n$ when taken $\bmod m_{1}, m_{2}, ... m_{k}$ where all $m_{i}$ are pairwise relatively prime, we can uniquely identify $n \bmod (m_{1}m_{2}m_{3}... m_{k})$ (follows from lemma 4)

This is known as the Chinese Remainder Theorem.

JSteinhardt's theory-intensive solution is simply repeated usage of lemma 4 and theorem 1, with the exception of this possibly confusing step:

[quote="JSteinhardt"]$n^{3}\equiv 3 \pmod 5$, so that $n \equiv 2 \pmod 5$. [/quote]

This is easily confirmed just by cubing integers $\bmod 5$.  Alternately,

[b]Lemma:[/b]  Let $k$ be an integer and $p$ a prime such that $(k, \varphi(p)) = 1$.  Then $a^{k}\equiv b \bmod p$ has either zero or one solutions $\bmod p$.  (This is not so easy to prove, and is left as an exercise.)

In fact, we can generalize.

[b]Lemma:[/b]  Let $k$ be an integer and $p$ a prime.  Then the number of distinct $k^{th}$ powers $\bmod p$ is the smallest positive integer $d$ such that $\varphi(p) | (dk)$.

The final computation in the theory-intensive proof is generalized by the following, which I will not prove.  It is very interesting to think over.

[b]Theorem:[/b]  Let $f(x) \in \mathbb{Z}_{p}[x]$ (a polynomial with integer coefficients, for simplicity) for $p$ a prime be such that $f(x) \equiv 0 \bmod p$ has a solution $a$ and such that $f'(a) \not \equiv 0 \bmod p$.  Then

$f(x) \equiv 0 \bmod p^{n}$

Has a solution for arbitrary $n$.

[quote="Phelpedo"]A slightly simpler and more intuitive approach by kyyuanmathcounts during number theory class:

[quote="kyyuanmathcounts"]First, we consider the units digit.  After testing some multiples of 2, we find that the units digit MUST be 2.

Now, we must consider the 10's digit.  The last two digit of the number before cubing will be $10x+2$.  Cubing that yields $1000x^{3}+600x^{2}+120x+8$.  We can throw away the $1000x^{3}$ term and the $600x^{2}$ term, and the $8$ term because they will not affect the 10's digit.  Thus, we have $120x$ and we want it's 10's digit to equal 8.  Thus, $x=4$ or $x=9$.

Testing the 100's digit in the same manner, we get our smallest number as ${192}$ [/quote][/quote]
That is an extremely elegant solution.


modular arithmetic
induction
AMC
USAMO
number theory

Find all pairs of positive integers $\left(n;\;k\right)$ such that $n!=\left( n+1\right)^{k}-1$.

n+1 is a prime or else n! and n+1 are not coprime and the equality cannot take place.

So what needs solving is (p-1)!=pk-1. for p-prime.

Remark: This was also discussed at http://www.mathlinks.ro/viewtopic.php?t=4591 .

  darij

So, we have

\[ n,k \in \mathbb{Z^+} \qquad (n+1)^k=n!+1 \]

[b]Fact 1.[/b]If $l$ is composite and $l>4$ then we have $(l-1)!\equiv0 \pmod{l}$

[b]Proof[/b] Let be $l=p_1^{a_1}p_2^{a_2}...p_j^{a_j}$. If $j>1$ then all the $p_1^{a_i}$ figure in the product $(l-1)!$ cause they're less than $l$ and so we're done. Let instead be $l=p^a\mbox{ with }a>1$. It suffice that $p$ figures al least $a$ times in $(l-1)!$. That is

\[ ap<l=p^a \Longrightarrow a<p^{a-1} \]
Induction on $a$.
[list]
$a=2 \Longrightarrow p>2$ that is true cause we supposed $l>4$

$a<p^{a-1} \Longrightarrow a+1<p^a$ this is obvioulsy true.
[/list]

[b]Solution.[/b] Back to the problem. Let's test cases $n=1,2,3,4$ and we find the solutions $(1;1),(2;1),(4;2)$. For $n>1$ the $RHS$ is odd. So $n$ must be even, $n=2m$. We have

\[ (n+1)^k=n!+1 \Longrightarrow n^k+\binom{k}{1}n^{k-1}...+\binom{k}{k-1}n+1=n!+1 \]
\[ (n-1)!=n^{k-1}+kn^{k-2}...+k \Longrightarrow (2m-1)!=(2m)^{k-1}+k(2m)^{k-2}...+k \]

Let's consider the last equation. By [b]Fact 1[/b] we deduce $k\equiv0 \pmod{2m}$ and so $2m<k$ cause it can't be $k=0$. So, back to the text

\[ (2m+1)^k=(2m)!+1\mbox{ with } 2m<k \]

But with this condition it's easy to see that $RHS>LHS$. The only solutions are $(1;1),(2;1),(4;2)$.

why don't use AM-GM for an upper bound for k and v2(n!) to get some cases

The answers are $(n,k)=(1,1),(2,1),(4,2)$.
For sufficiently large $n$, the number of ending zeroes in $n!$ is simply the number of factors of 5's, or $\sum_{i=1}^\infty \left(\lfloor \frac{n}{5^i}\rfloor\right)$. The maximum number of ending zeroes on RS is simply the number of digits of $n-1$ or $\lfloor (logn) \rfloor$, due to the Binomial Theorem of $k$, and then subtracting $1$. Clearly, the number of zeroes on LS is bigger than this maximum on RS for $n\ge5$, and testing for $n=1,2,3,4$ yields the only cases.

[quote=harazi]Suppose n>5 and k>1.. Use Newton formula to find that n!=n^k+kn^(k-1)+...+k(k-1)/2n^2+kn. So, (n-1)!=n^(k-1)+...+k(k-1)/2n+k. Suppose n is not a prime. Then n|(n-1)!. So, n|k and so k>=n. So, n!+1>=(n+1)^n, false. So n is a prime. Since Mihai proved n+1 is a prime, n=2. So n<=4 or k=1 and now it's easy.[/quote]

I understand the logic, but why does it only work if $n>5$ and $k>1?$ Isn't $n!+1 < (n+1)^n$ true for all positive integer $n>1?$


A suitcase lock has 3 dials with the digits $ 0, 1, 2,..., 9$ on each.
How many different settings are possible if all three digits have
to be different?

Given the previous dials, there are $ 10 \cdot 9 \cdot 8 \equal{} 720$ settings where all the digits are different.


calculus
integration
logarithms
trigonometry
calculus computations


The best advice is simply do lots and lots of problems. There're no magic tricks. At the very least, go through every problem on the past ten years of CMO, perhaps even more.

[quote="eddyliu"]Hey anyone got any kind of idea of how to become super genious in math by CMO? i got 65 in COMC if anyone is wondering. and CMOR was a peice of cake after you sleep on the questions for a few days. suggestion from pros would be nice  :lol: by the way how many do u have to solve to make division 1.[/quote]

Nostalgic. :) I suggest you read the thread below:
[url]http://www.mathlinks.ro/viewtopic.php?t=80147[/url]

[quote="eddyliu"][quote="Mathgod"]
Nostalgic. :) I suggest you read the thread below:
[url]http://www.mathlinks.ro/viewtopic.php?t=80147[/url][/quote]

ya nice...if i was still in grade 10 when i asked this. grade 12 last shot :([/quote]

It's ok. Don't give up.

hey u guys got anymore of those crash course lectures like the one on inequality that Yufei posted?


geometry
circumcircle
incenter
geometry proposed

[quote="N.N.Trung"]Let a triangle $ ABC$. A point $ M$ lies on the interior of the triangle. Denote $ d_a,d_b,d_c$ are the distances from $ M$ to the sides of the triangle. Prove that $ d_ad_bd_c\ge r^3 \iff OM\le OI$ where $ O,I$ be the circumcenter and incenter, $ r$ is the inradius.[/quote]

 [geogebra]3219c3811ab86c14bb941da9d21d3ab9d846ec4d[/geogebra]

I think this problem is wrong... See picture. $ d_ad_bd_c \equal{} 0.5433 < r^3 \equal{} 2.4061$ but $ OM < OI$


floor function
pigeonhole principle
number theory unsolved
number theory

Prove there exists infinitely many positive integers n such that
$\lfloor x^{3/2}\rfloor+\lfloor y^{3/2}\rfloor = n $
has at least 1980 solutions in positive integers.

Consider all pairs $(x,y)$ s.t. $x,y\in \{1,2,3,...,m\}$, where m is a large integer.
For these $m^2$ pairs we have $2 \leq [x^{3/2}]+[y^{3/2}]\leq 2[m^{3/2}]$. Using pigeonhole principle we obtain that there is $n\in \{2,3,...,2[m^{3/2}]\}$ s.t. equation $[x^{3/2}]+[y^{3/2}]=n$ has at least $\sqrt{m}/2$ solutions. On the other hand, equation $[x^{3/2}]+[y^{3/2}]=n$ has at most $n$ solutions.
It implies result.


Put $x=\frac{1}{a}$ and the inequality is equivalent to
$\frac{x}{y+z}+\frac{y}{z+x}+\frac{z}{x+y}\geq \frac{3}{2}$ which is Nesbitt one and can be proved by Cauchy in Engel form.

[quote="parinaz.sa"]we can prove Nesbitt inequality by Algebraic substitution too.$a=y+z ,b=x+z,c=x+y$[/quote]
Or by Jensen, or by $\mathrm{AM-HM}$, etc., etc., etc., ...

[quote="Virgil Nicula"][color=darkblue]Often and I say in such cases a bit otherwise : [/color][color=darkred]It's like killing a [b][u]butterfly[/u][/b] with the atomic bomb ![/color][/quote]
Good point! :lol:
[quote="pohoatza"][quote="Rzeszut"][quote="pohoatza"]also we can apply muirhead after we do some long paperwork :)[/quote]
It's like killing a fly with a tank.  :wink:[/quote]

 :rotfl:[/quote]
Muirhead can be a good theorem as just a piece of art. It can be good to play with generalisations. But not to kill beautiful inequalities. Do you really think that solving by Muirhead an inequality which becomes the trivial Nesbitt after substitution $a=\frac1x$ etc. is a good idea? Similar to proving $x^{2}+y^{2}\geqslant \frac12$ for $x+y=1$ by Lagrange multiplier. Some others' (mainly moderators') opinions about Muirhead:
[url]http://www.mathlinks.ro/Forum/viewtopic.php?p=373993&highlight=hate+muirhead#373993[/url]
[url]http://www.mathlinks.ro/Forum/viewtopic.php?p=521106&highlight=hate+muirhead#521106[/url]
[url]http://www.mathlinks.ro/Forum/viewtopic.php?p=299450&highlight=hate+muirhead#299450[/url]
[url]http://www.mathlinks.ro/Forum/viewtopic.php?p=18894&highlight=hate+muirhead#18894[/url]
[url]http://www.mathlinks.ro/Forum/viewtopic.php?t=18937&highlight=hate+muirhead[/url]
And so on.


function
induction
calculus
derivative
Pascal\u0027s Triangle
AMC

To find out how many possible subsets a set with n elements has, is it 2^n? For example if the set is {1,2,3,4} does that mean there are 2^4 subsets? Can someone prove that? Thanks.

Yeah it's $2^n$.  Here's a proof.

The number of subsets are the ways to take 0 elements, 1 element, 2 elements, ..., n elements.  This can be represented
$\binom n 0 + \binom n 1 + \binom n 2 + \cdots + \binom n {n-1} + \binom n n$
This is the sum of the elements of the nth row of pascal's triangle, which we know is $2^n$.  You can prove that by doing binomial expansion on $(1+1)^n$

edit: Hmm, your proof is intriguing.  I've never seen that before, I like it.

[quote="deej21"]

edit: Hmm, your proof is intriguing.  I've never seen that before, I like it.[/quote]

It's a very easy subsets proof of the "sum-of-the-nth-row-of-Pascal's-triangle" combinatorics identity  :)  (Speaking of which, I have two other proofs - one based on generating Pascal's triangle through addition, and another by generating functions)

Just realize that $2^n$ includes the empty set, so if a problem asks for non-empty subsets you'd have to subtract $1$.

[quote="Calculus"]Just thought about it again... is this correct?

Each element has two choices; it is either in or not in the subset. Thus there are 2^n subsets?[/quote]
That's the simplest and most straightforward proof of this identity.  Any other proofs using Pascal's triangle seem to be to be a little more contrived.  You can also use a binomial expansion and prove the identity by induction, but that's much less fun.

[quote="Pakman2012"]I just like to think of it as

$(1+1)^n$

For the sum of the terms in Pascals triangle.

Perhaps its a little "cheaper"[/quote]
First you have to prove that the binomial expansion is indeed the proper expansion of $(x + y)^n$, which usually proceeds by induction, though can also be done using a combinatoric proof.

Oh. Lame.

That makes sense though haha.

[quote="Calculus"]Do you think you could show me your two other proofs? Thanks.[/quote]

Well, the first proof requires showing that adding consecutive entries in Pascal's triangle generates the correct binomial expansion, which can be done inductively or by a subsets proof - then you reason that, because each entry in a row is used twice to construct the entries in the next row, the sum of the entries in the next row is twice the sum of the entries in the previous row.  :)

(There is also an induction-based subsets proof.  Namely, if there are $2^n$ subsets for an $n$-size set, adding an element to the set doubles the number of possible subsets because for each subset of the $n$-size set you can either add the new element or keep it out. )

The generating functions proof is identical to the expansion of $(1 + 1)^n$, but the great thing about it is it can also be used to show ${n \choose 0} - {n \choose 1} + {n \choose 2} - {n \choose 3} +-... = 0$, the expansion of $(1 - 1)^n$.

[quote="Calculus"]
The generating functions proof is identical to the expansion of $(1 + 1)^n$, but the great thing about it is it can also be used to show ${n \choose 0} - {n \choose 1} + {n \choose 2} - {n \choose 3} +-... = 0$, the expansion of $(1 - 1)^n$.[/quote]
Another great consequence of this is if you take the derivative of $(x + 1)^n$ to get

$n(x + 1)^{n-1} = n \binom{n}{0}x^{n-1} + (n - 1) \binom{n}{1} x^{n-2} + ... + \binom{n}{n-1}$

which leads to the cool identity

$n2^{n-1} = n \binom{n}{0} + (n - 1) \binom{n}{1} + ... + \binom{n}{n-1}$, or, equivalently,

$n2^{n-1} = \binom{n}{1} + 2 \binom{n}{2} + ... + (n - 1) \binom{n}{n - 1} + n \binom{n}{n} = \sum_{k=1}^n k \binom{n}{k}$.

This is basically the elevator problem:
How many ways can a person starting on the zeroth floor get to the nth floor
He can stop at any number of floors but must always go up.
Then its just 2^n, because he can either stop at a each floor or not, which is the same as the set.
Each set can either include each number or not.


Find all functions $f: \mathbb R\to \mathbb R$ such that
\[\left | \sum_{k=1}^{n}2^{k}(f(x+ky)-f(x-ky)) \right |\leq 1 ,\]
 for all integers $n \geq 0$ and all $x,y \in \mathbb R$.

Hello spix !
[quote="spix"]Find all $f: R\to R$ such that:
 $\left | \sum_{k=1}^{n}2^{k}(f(x+ky)-f(x-ky)) \right |\leq 1 , \forall n \in N^{\star}, \forall x,y \in R$[/quote]

Let $A= \sum_{k=1}^{n}2^{k}(f(x+ky)-f(x-ky))$

We have $\left | A \right |\leq 1$

Let $B= \sum_{k=1}^{n+1}2^{k}(f(x+ky)-f(x-ky)) = A+2^{n+1}(f(x+(n+1)y)-f(x-(n+1)y))$

We have $\left | B \right |\leq 1$, so $\left | A+2^{n+1}(f(x+(n+1)y)-f(x-(n+1)y)) \right |\leq 1$

But : $\left | A+2^{n+1}(f(x+(n+1)y)-f(x-(n+1)y)) \right |$ $\geq$ $\left | 2^{n+1}(f(x+(n+1)y)-f(x-(n+1)y)) \right |-\left | A\right |$

So : $1 \geq$ $\left | 2^{n+1}(f(x+(n+1)y)-f(x-(n+1)y)) \right |-\left | A\right |$

So : $1+\left | A\right |\geq$ $\left | 2^{n+1}(f(x+(n+1)y)-f(x-(n+1)y)) \right |$

So : $\left | 2^{n+1}(f(x+(n+1)y)-f(x-(n+1)y)) \right | \leq 2$

And : $\left | f(x+(n+1)y)-f(x-(n+1)y) \right | \leq 2^{-n}, \forall n \in N^{\star}, \forall x,y \in R$

Let a and b two reals.

Let $x=\frac{a+b}{2}$ and $y=\frac{b-a}{2(n+1)}$

Then $\left | f(x+(n+1)y)-f(x-(n+1)y) \right | \leq 2^{-n}$ becomes $\left | f(b)-f(a) \right | \leq 2^{-n}, \forall n \in N^{\star}, \forall a,b \in R$

And f is any constant function.

--
Patrick


vector
algebra
polynomial
linear algebra
linear algebra unsolved

If you have the following basis vectors for a polynomial vector space
c_1*[-1,1,0]
c_2*[-1,0,1]

then how do you get to the form

(x-1)
(x^2 - 1)

as representing the same basis?

Those "vector" forms only make sense with respect to a standard basis for the polynomials. What do you think this standard basis is?


number theory
greatest common divisor
induction

[hide]
Without loss of generality, let our four numbers be $ a_1,a_2,a_3,a_4$ with $ |a_1| > |a_2|$. Then we have \begin{align*}
a_2^2 &= a_1a_3 + 1 \\
a_2a_4 + 1 &= a_3^2
\end{align*} Summing yields $ a_2(a_2+a_4) = a_3(a_1+a_3)$. But $ \text{gcd}\,(a_2,a_3) = 1$, and as all of the quantities are integers, it follows that $ a_3 | a_2+a_4$ and $ a_2 | a_1 + a_3$. Hence, \[ k = \frac{a_1+a_3}{a_2} = \frac{a_2+a_4}{a_3}\] where $ k \in \mathbb{Z}$, and further for any $ n$, $ \frac{a_{n-1}+a_{n+1}}{a_n} = k$. Assuming that $ a_{n-1},a_n$ are integers, then $ a_{n+1} = ka_n - a_{n-1}$ must also be an integer, so by induction all the elements are integers. Also, assuming that $ |a_{n-1}| > |a_n|$, then $ |a_{n+1}| = \frac{a_{n}^2-1}{|a_{n-1}|} < \frac{a_{n}^2-1}{|a_{n}|} < |a_{n}|$ (unless $ a_n = 0$, but then we're finished). It follows that the sequence $ |a_i|$ is integer and strictly decreasing, so $ 0$ must appear in the sequence.[/hide]

[quote="outback"]There is a row of real numbers, infinite in both directions. The square of each number is $ 1$ more than the product of the numberson either side. Prove that if four adjacent numbers are integers, then at least one number in the row is $ 0$.[/quote]
This was on the 2008 ELMO at MOP.  Here is my solution (quite similar to [b]azjps[/b]'s but a little bit different in the beginning.)
[hide="Solution"]
Assume that $ 0$ does not appear.  Let $ k$, $ a$, $ b$, and $ l$ be the integers that are consecutive.  We then have that $ \frac {a^2 \minus{} 1}{k} \equal{} b$, so\[ b^2 \minus{} 1 \equal{} \frac {a^4 \minus{} 2a^2 \plus{} 1}{k^2} \minus{} 1 \equal{} \frac {a^4 \minus{} 2a^2 \minus{} (k^2 \minus{} 1)}{k^2}\]Yet, $ al \equal{} b^2 \minus{} 1$, so we have that $ alk^2 \equal{} a^2 \minus{} 2a^2 \minus{} (k^2 \minus{} 1)$.  Since $ a|(a^4 \minus{} 2a^2)$, we have that $ a|(k^2 \minus{} 1)$.  This means that the number to the left of $ k$, say $ r$ is $ \frac {k^2 \minus{} 1}{a}$, which is an integer.  Similarly (inductively) we can show that all of the numbers in the sequence are integers.  Now, we  consider a number (integer) in the sequence, say $ m$.  Let the numbers to the left and right of $ m$ be $ x$ and $ y$ respectively.  Then, $ xy \equal{} m^2 \minus{} 1 < m^2$.  Without loss of generality, let $ |x| < |y|$ (if $ |x| \equal{} |y|$, then either $ \minus{} x^2 \equal{} m^2 \minus{} 1$, so $ x^2 \plus{} m^2 \equal{} 1$, so either $ x$ or $ m$ is $ 0$, which is a contradiction or $ x^2\equal{}m^2\minus{}1$, so $ x\equal{}0$, which is a contradiction as well.)  Then, $ |x| < m$ since we otherwise have that $ |x\parallel{}y| > m^2$, so $ xy$ is nonpositive, so $ m^2 \minus{} 1\le 0$, so $ m \equal{} \pm 1$ or $ m \equal{} 0$.  The second case is a contradiction, so we have that $ m \equal{} \pm 1$, so $ xy \equal{} m^2 \minus{} 1 \equal{} 0$, which means that $ x$ or $ y$ is $ 0$, which is a contradiction.  Hence, $ |x| < m$.  Similarly, if the number to the left of $ x$ is $ w$, then $ |w| < |x|$, and we keep going.  Eventually, we must reach $ 0$, so we have a contradiction.  Thus, $ 0$ shows up somewhere. [/hide]


I have posted a question on mathcounts, and I knew that some people answered it. But I can't find it now. May anyone please tell me the reason?

Most likely, it was inappropriate (e.g. too hard) for the forum it was placed in and a moderator moved it.

I have viewed my posts and searched for it, but I cannot find it.

[quote="N.T.TUAN"]Step 1: Login.
Step 2: click to ''View your posts''
If you not find it then It is deleted :P[/quote]It think this explanation is very clear.


$ \frac{108}{\sqrt{x^2\minus{}2916}}\equal{}\frac{378\minus{}x\minus{}\sqrt{x^2\minus{}2916}}{x\plus{}4}$

Is this what you mean? Because the "/" at 108 makes me think that it is also a fraction...

Could the answers be "58,5" and "62,5" by any chance?

hello, i think he meant
$ \frac{108}{\sqrt{x^2\minus{}2916}}\equal{}\frac{378\minus{}x\minus{}\sqrt{x^2\minus{}2916}}{x\plus{}4}$.
Sonnhard.

I would like a "non-calculator" solution more... I just done $ 54 < x \le 378$ as Graubner did by expanding with $ \sqrt {x^2 - 2916}$ and $ (x + 4)$ and picked some good numbers like those "58,5" and "62,5" with some trial-error... but apparently, my solutions were wrong !


for the answer to one of my problems in the "introduction to number theory" book (on page 12 of the answer book) it simply says this;

9409 = 1000 - 2(3)(100) + 3^2 = (100 - 3)^2 = 97^2

how did it get from the first step to the second step?!

This is simply about recognizing patterns; note that $ 10000$ and $ 9$ are both perfect squares, so it makes sense to try something with it: we end up with $ 10000\plus{}9\minus{}600$, or $ 10000\minus{}2(3)(100)\plus{}3^2\equal{}(100\minus{}3)^2\equal{}97^2$.

It helps to know that...

1. $ (a \plus{} b)^2 \equal{} a^2 \plus{} 2ab \plus{} b^2$
2. $ (a \minus{} b)^2 \equal{} a^2 \minus{} 2ab \plus{} b^2$
3. $ (a \plus{} b)(a \minus{} b) \equal{} a^2 \minus{} b^2$

... and the other way around.

(HINT: The 2nd formula-like thing is the one you're looking for!)

BTW I think this is mentioned in one of the AoPS books, but I can't remember which

It's from the Intro to Number Theory book.

EDIT: its probably in both as well as intro to algebra...

It is? I could have sworn it was from AoPS Vol 1 :huh:

Shouldn't it be $ 10000$ and not $ 1000$?

oh, I get it if it is 10,000! I like aops, but have they never heard of commas?!

(thx for all your help guys)

[quote="Waffle"]I generally don't put commas - I find they waste time.  :roll:

Just my opinion.[/quote]

Not to sound rude, arrogant, "goody two-shoes" or anything, but that post was a waste of time for you to type and everyone to read. Please don't do that again; mods or admins may punish you.

Sincerely

---EDIT---
Oops! I'm sorry; I meant to PM waffle, but I accidentally placed this post in the forum :blush: ! Sorry!


[quote="math154"]NO MORE ORCHESTRA PRACTICE RECORDS!!! (well, for a while, at least)[/quote]

Is there one due tomorrow?

Yeah, but I finished it while waiting for the extremely late bus. BTW, I got home at 5:00 and I was the FIRST stop. Waiting for the bus almost took the entire time you were in Anatomy. :wink:  :roll:


topology
real analysis
real analysis unsolved

I was fiddling with this proof given (in more general form) by jmerry a long time ago but for some reason this application is leading to a false conclusion.

The complete metric space $ [1,\infty [$ (in R, under the standard metric) has the representation $ [1,\infty [ \equal{} \bigcup\limits_{n \equal{} 1} {[n,n \plus{} 1]}$ and the Baire Category theorem (overkill here, but bare with me) shows that $ Y \equal{} \bigcup\limits_{n \equal{} 1} {]n,n \plus{} 1[}$ is a nonempty proper open subset of  $ [1,\infty [$. $ A \equal{} [1,\infty [\backslash Y$ is closed and nowhere dense in the space but $ A \ne \bigcup\limits_{n \equal{} 1} {([n,n \plus{} 1] \cap A)}$ because A is of second category in itself.

But this cannot be true.

It should be a simple error in reasoning somewhere but I cannot detect it. :blush:

EDIT: I found it. I was thinking nowhere dense means in every open set there is an open subset which does not intersect the set but (I'm guessing) the empty set doesn't count as an 'open subset'. If any set were to be nowhere dense you'd think it'd be N...

An empty set is open. Therefore, the condition should read "in every open set there is a [b]nonempty[/b] open subset..."

I prefer a different wording: a set is nowhere dense if its closure has empty interior.


When you approach a math contest problem, what do you think in your head while you're reading?
Do you reread the question to scan for small details and important words?

Do you immediately try to classify what type of problem it is such as logic or algebra?

What clues and signals do you see in the problem to use certain tools? For example, if I see a bunch of large equations, I might think to manipulate them all or some of them to get a prettier equation. If I see lots of shared or congruent angles, I might think of similar triangles. If I see logical statements in the forms if A, then B, then I think of the contrapositive and other stuff...

What are some other common hints in problems use certain tools or pursue certain tactics?

After you've found an answer to the problem, do you reread the problem to make sure that you've found what it's asking for? Do you tend to spend more time on understanding the problem than actually going through the steps to solve the problem?

When you check over your work on a math competition that you're doing under a timed situation, how do you determine which questions to check? Do you just check them all, and when do you decide to go back and check your work? [/youtube]

[quote="xpmath"]I don't really know what I think...I just try many things that could be useful for that type of problem.
Also, I don't really double check :P[/quote]

Yeah, double checking is something I've never got in the habit of doing.

I normally look for the easy marks first, so I'll read through as far as possible and then answer in the order of difficult which I consider for each question. I do categorise them, but my categories are "Geometry and Combinatorics" and "Everything else". If its in everything else, I'll take a swing at it, otherwise I will leave it until the very end

Well I'm not as good as a lot of people here on AoPS, but posting here helps me learn A LOT.

How do I solve problems...

When reading I rewrite any givens and underline what seems to be the real 'problem' in each question.  I've found that understanding what the problem is asking [i]is[/i] realizing the solution.  So be sure to be able to identify what sort of math you're dealing with. And to that end, try to be well rounded in non-discrete and discrete mathematics.

I do my best to improve my weaknesses, but I play to my strengths when the chips are down.  On any test where all problems have equivalent point value I solve the easiest first.  To me 'the easiest' doesn't mean the lowest level of math, it means the math I'm most comfortable with.  I'm not saying that you shouldn't try the questions that seem ridiculous, I'm just saying work on them later.

I know this sounds trivial, but be able to do things like square 37 without a calculator. Another good idea is learning divisibility rules. I believe calculators have been banned on the AMC....

Now as for checking questions, I normally 'come back' if I realize there's a problem.  Generally this happens when I look at the possible answers (I only work backwards if I have to).

I visit AoPS every day.   :lol:

[quote="aufha"]igiul said:[quote]Now as for checking questions, I normally 'come back' if I realize there's a problem. Generally this happens when I look at the possible answers (I only work backwards if I have to).
[/quote]

What are some signals that there is a problem? How do you realize that there is a problem without going back to check answers?[/quote]

Sometimes I wont "feel" right about an answer. It may seem too . $ x$, where $ x \in {}$obvious, small, low, ... generic quanitifier${ }$

Other times, I will realize that I made a computational mistake later on for no apparent reason.

[quote="aufha"]What are some signals that there is a problem? How do you realize that there is a problem without going back to check answers?[/quote]

Well occasionally I too feel uneasy about a problem, but as I continue with the test I often relate and apply apparently irrelevant details back to those problems.  Maybe the details come from the other problems themselves when you are looking at the same subject in a different manner. I always try to 'justify' my answer.

Most often the this is due to contradiction, so I may not know which answer is right, but I do know that mine is wrong.  If this is the case, its back to base one on that problem if I feel I can accommodate it.  But always if this happens, change your answer on the answer sheet!

Hope that helps!


function
geometry
geometric transformation
rotation
integration
calculus
real analysis

Given $ k \equal{} 1,\ldots,n$ (we are working in $ \mathbf R^n$), find Fourier transform of the function
\[ f(x) \equal{} \frac {x_k}{|x|}.
\]
Added: in the sense of tempered distributions.

Note that $ f$ is not integrable, and not in $ L^2.$ What you're asking for only makes sense if you're asking for it in the sense of tempered distributions.

Since $ f$ has mean value zero over any ball centered at the origin, $ \widehat{f}$ shouldn't contain any point masses. Since $ f$ is positively homogeneous of degree $ 0,$ we should have that $ \widehat{f}$ is positively homogeneous of degree $ \minus{} n.$ (The Dirac delta is positively homogeneous of degree $ \minus{} n$ but we have already eliminated it.) Investigating how this behaves under rotations leads us to this:
\[ \widehat{f}(\xi) \equal{} \text{p.v. }\frac {C\xi_k}{|\xi|^{n \plus{} 1}}
\]
for some (pure imaginary) constant $ C$ which depends on $ n.$

The only thing left to do will be to compute $ C.$ (I leave that out in part because I don't know your Fourier transform conventions.) The notation "p.v." means that this is to be interpreted as the distribution given by a principal value integral.

For more, look up the words "Riesz transform." (That's one generalization of the $ n\equal{}1$ object called the Hilbert transform.)

Here's a more constructive argument.

To establish the conventions, define

$ \widehat{f}(\xi)\equal{}\int_{\mathbb{R}^n}e^{\minus{}2\pi i\xi\cdot x}f(x)\,dx$

A specific computation is that the Fourier transform of $ e^{\minus{}\pi|x|^2}$ is $ e^{\minus{}\pi|\xi|^2}.$ And if we put that in the definition above and take $ \frac{\partial}{\partial \xi_k},$ we get that

if $ f(x)\equal{}x_ke^{\minus{}\pi|x|^2},$ then $ \widehat{f}(\xi)\equal{}\minus{}i\xi_ke^{\minus{}\pi|\xi|^2}.$

Then there is the scaling principle: if $ g(x)\equal{}f(ax)$ for some $ a>0,$ then $ \widehat{g}(\xi)\equal{}\frac1{a^n}\widehat{f}\left(\frac{\xi}{a}\right).$ In particular,

the transform of $ \sqrt{t}x_ke^{\minus{}\pi t|x|^2}$ is $ \minus{}i\,\frac{1}{t^{n/2}}\cdot\frac{\xi_k}{\sqrt{t}}e^{\minus{}\pi|\xi|^2/t}.$

Write $ \frac{x_k}{|x|}\equal{}\int_0^{\infty}t^{\minus{}1/2}x_ke^{\minus{}\pi t|x|^2}\,dx\equal{}\int_0^{\infty}\frac1t\left(\sqrt{t}x_k\right)e^{\minus{}\pi t|x|^2}\,dt$

Take the Fourier transform of this expression. (Consider it to converge weakly, in the distributional sense, and use that to justify what we're doing.) We get that the Fourier transform of $ \frac{x_k}{|x|}$ is

$ \int_0^{\infty}\frac1t\cdot\frac{1}{t^{n/2}} \cdot\frac{\xi_k}{\sqrt{t}}e^{\minus{}\pi|\xi|^2/t}\,dt$

Collect together the powers of $ t,$ make the appropriate change of variables in this integral, and if I didn't make any mistakes there, we get

$ \minus{}i\,\frac{\Gamma\left(\frac{n\plus{}1}{2}\right)}{\pi^{\frac{n\plus{}1}2}} \cdot\frac{\xi_k}{|\xi|^{n\plus{}1}}$ as the transform we seek.


how many squares are in 8 * 8 chess board?

[hide="hint"]not 64[/hide]

[hide][quote]how many squares are in 8 * 8 chess board? [/quote]

I wasn't sure whether the question was only for the squares on the grid lines, or diagonal as well.  I did only those on the grid lines.

size of squares :arrow: number of them

1x1  :arrow: 64
2x2  :arrow: 49
3x3  :arrow: 36
4x4  :arrow: 25
5x5  :arrow: 16
6x6  :arrow: 9
7x7  :arrow: 4
8x8  :arrow: 1

Its important to count the overlapping ones.

Since they are all squares, does any one know a formula for their sum?  I added them all up.

The answer is [b]204 squares[/b], not counting diagonals.[/hide]

Follow up question: how many squares in an $n$-by-$n$ chessboard?

[hide="hmmmmm..."]maybe $n^2+(n-1)^2+(n-2)^2+(n-3)^3 \cdots$
you would do this until you get n-x=1[/hide]

yeah, the sum comes out to be

S = n*(n+1)*(2n+1)/6

comes out to 204 for n=8 as other people showed.

Follow up question:

How many rectangles are in a 8 x 8 chess board?

response to follow-up question of how many RECTANGLES:

(9c2)^2 = 1296.

that's the slickest approach I know.  now can someone tell me where I got that from?  :D

Youre choosing the nine points and choosing to of them and then nine points on the side and choosing 2, making 4 points all with right angles

I think you got it.  the way I thought of it was that in an 8x8 chessboard, there's really a grid of 9x9 lines.  you have to choose any 4 of those (2 in horizontal, 2 in vertical) to define a rectangle.

What question you're answer, kishiman1??? Be more specific next time...

[quote="kishiman1"]sorry about that. I didnt read that hint at the top.[/quote]

Ok, for the next time.....


as you said $ \{x\} \equal{} x \minus{} [x]$,hence $ \{\minus{}1,3\} \equal{} \minus{} 1,3 \minus{} ( \minus{} 2) \equal{} 0,7$.


Let $ ABC$ be an isosceles triangle inscribed in a circle and $ M$ a point in the line $ BC$ ($ M$ out of the base $ BC$). Prove that $ \overline{MA}^2 \equal{} \overline{AB}^2 \plus{} \overline{MB}\cdot\overline{MC}$.

[quote="Virgil Nicula"]$ \boxed {\Longrightarrow}$ [color=darkred]Let $ ABC$ be a $ A$-isosceles  triangle. Prove that for any point $ X$

of the sideline $ BC$ exists the identity $ AX^2 \equal{} AB^2 \plus{} \overline{BX}\cdot\overline{CX}$.

$ \boxed {\Longleftarrow}$ Let $ ABC$ be a triangle for which exists a point $ X$ of the sideline $ BC$

so that $ AX^2 \equal{} bc \plus{} \overline{BX}\cdot\overline{CX}$. Prove that the triangle $ ABC$ is $ A$-isosceles

or the ray $ [AX$ is the bisector of the angle $ \widehat {BAC}\ .$

See  http://www.mathlinks.ro/viewtopic.php?t=177757 . [/color][/quote]
[color=darkblue][b][u]Proof.[/u][/b]

$ \boxed {\Longrightarrow}$ If $ AB \equal{} AC$, then the [b]Stewart's relation[/b]

$ AB^2\cdot\overline {CX} \plus{} AC^2\cdot \overline {XB} \plus{} AX^2\cdot \overline {BC} \plus{} \overline {CX}\cdot\overline {XB}\cdot\overline {BC} \equal{} 0\ \ (*)$

becomes $ AB^2\cdot \overline {CB} \minus{} AX^2\cdot\overline {CB} \minus{} \overline {CX}\cdot\overline {XB}\cdot\overline {CB} \equal{} 0\Longleftrightarrow$

$ AB^2 \minus{} AX^2 \plus{} \overline {BX}\cdot\overline {CX} \equal{} 0\Longleftrightarrow AX^2 \equal{} AB^2 \plus{} \overline {BX}\cdot\overline {CX}\ .$

$ \boxed {\Longleftarrow}$ If exists a point $ X$ of the sideline $ BC$ so that  $ AX^2 \equal{} bc \plus{} \overline{BX}\cdot\overline{CX}$,

then from the [b]Stewart's relation[/b] $ (*)$ obtain :

$ c^2\cdot\overline {CX} \minus{} b^2\cdot\overline {BX} \plus{} \left(bc \plus{} \overline {BX}\cdot\overline {CX}\right)\cdot \overline {BC} \minus{} \overline {CX}\cdot\overline {BX}\cdot\overline {BC} \equal{} 0\Longleftrightarrow$

$ c^2\cdot\overline {CX} \minus{} b^2\cdot\overline {BX} \plus{} bc\cdot\overline {BC} \equal{} 0\Longleftrightarrow$

$ c^2\cdot\overline {CX} \minus{} b^2\cdot\overline {BX} \plus{} bc\cdot\left(\overline {BX} \minus{} \overline {CX}\right) \equal{} 0\Longleftrightarrow$

$ b(c \minus{} b)\cdot \overline {BX} \plus{} c(c \minus{} b)\cdot\overline {CX} \equal{} 0\Longleftrightarrow$ $ b \equal{} c$ or $ b\cdot\overline {BX} \plus{} c\cdot\overline {CX} \equal{} 0$, i.e.

$ \triangle\ ABC$ is $ A$-isosceles or $ \frac {\overline {XB}}{\overline {XC}} \equal{} \minus{} \frac cb$  (the ray $ [AX$ is the bisector of the angle $ \widehat {BAC}$).[/color]

[quote="Virgil Nicula"] [color=darkred][b][u]A similar problem.[/u][/b] Prove that for the triangle $ ABC$ exists a point $ X$ which belongs

to the sideline $ BC$ so that $ AX^2 \equal{} \overline {BX}\cdot\overline {CX} \minus{} bc$ if and only if $ b\ne c$ and in this case
the point $ X$ is uniquelly (the ray $ [AX$ is the exterior bisector of the angle $ \widehat {BAC}$). [/color][/quote]

[quote="Virgil Nicula"][color=darkred][b][u]An easy extension.[/u][/b] Let $ ABC$ be a triangle.  Acertain for any $ m\in\mathcal R$ the set  $ D(m)$

of the points $ X$ which belongs to the sideline $ BC$  and $ AX^2 \equal{} mbc \plus{} \overline{BX}\cdot\overline{CX}$.[/color][/quote]

[color=darkblue][b][u]Notation.[/u][/b] The [b]Chasles' relation[/b] : for the points $ X(x)$ and $ Y(y)$

the algebraic measure of the orientated segment $ \overline {XY}$ is $ y \minus{} x$. [/color]

Thanks Virgil. I edited the sign of minus in my first post. When i finished the proof i saw your solution. Got one more mistake in the book where i found it.

Let AM intersect the circumcircle in D. $ \angle BMD \equal{} \frac {\widehat{AB}\minus{}\widehat{CD}}{2}\equal{}\frac{\widehat{AC}\minus{}\widehat{CD}}{2}\equal{}\angle AMC$.
Or $ \Delta ACD \sim \Delta AMC \Rightarrow \frac{AC}{AM}\equal{}\frac {AD}{AC}$, so $ AC^2\equal{}AM.AD\equal{}AM(AM\minus{}DM)\equal{}AM^2\minus{}AM.DM\equal{}AM^2\minus{}MC.MB$


What is the area enclosed by the graph of $ |3x| \plus{} |4y| \equal{} 12$?

$ \textbf{(A)}\ 6\qquad
\textbf{(B)}\ 12\qquad
\textbf{(C)}\ 16\qquad
\textbf{(D)}\ 24\qquad
\textbf{(E)}\ 25$

[hide="Answer"]The equation goes through the axes at points $(0,\pm 3)$ and $(\pm 4,3)$. Therefore, we have four congruent right triangles with legs $3$ and $4$, so the answer is $4\cdot \frac{1}{2}\cdot 3\cdot 4=24\Rightarrow \boxed{D}$.[/hide]

actually, you made a typo :)

I think you meant $ (\pm 4, 0)$ instead of $ (\pm4, 3)$

I know I am reviving a thread that is so old that I have not seen a thread that is older than this one, but I thought I have a valid justification: I have an alternative solution.

[hide="My Solution"]
As [b]JesusFreak197[/b] mentioned, the graph passes through the vertices $(0, \pm3)$ and $(\pm4, 0)$. This is the same as a rhombus with diagonals being the segment between $(0,3)$ and $(0,-3)$ and the segment between $(-4,0)$ and $(4,0)$. Thus, the diagonals have lengths $6$ and $8$. The area of a rhombus with diagonals $a$ and $b$ is $\tfrac{1}{2}ab$. Thus, the desired area here is: \[ \dfrac {1}{2} \cdot 6 \cdot 8 = \boxed {\textbf {(D) } 24} \]
[/hide]

l3xl + l4yl = 12
=> 3x + 4y =12;
=> x/(12/3) + y/(12/4)=1;
Thus, it cuts x-axis at 4 units, and it cuts y-axis at 3 units.
Again evaluating the equation by removing modulus, we get 4 straight lines which intercept at x and y axes forming 4 congruent triangles. thus, the total area enclosed is:-
 4*[(1/2)*4*3]
=4*6
=24 units.
Thus; answer is (D)


A sequence of positive integers begins 2009, 2008, 1, 2007, etc. Beginning with the
third term, each term is the absolute value of the difference of the previous two
terms. What is the 640th term of this sequence?

I think its 1583, I kinda made the problem up so.... ya

Here is another problem:

1,2,4,7,8,10,11,13,14,....,17291729 excludes all integers that are either multiples of 3 or 5. What is the 1337th term in this sequence?

[hide]
2506?
[/hide]


probability geometry trigonometry circumcircle calculus integration inequalities

[hide]The probability of one randomly chosen point being inside the circle is $ \frac{\text{area of circle}}{\text{area of square}}\equal{}\frac {9\pi}{144} \equal{} \frac {\pi}{16}$.

For three randomly chosen points to be inside the circle, the probability is $ \left(\frac {\pi}{16}\right)^3 \equal{} \frac {\pi^3}{4096}$.[/hide]

I think you misinterpreted -- the circle is not a particular, predefined circle. Here's a rephrase:

[i]What is the probability of choosing 3 points for which there exists at least one circle of radius 3 that contains all three chosen points?[/i]

And this is comparatively quite nontrivial, I'd think  :wink:

[quote="blackbelt14253"]Three points are selected inside square ABCD with side length 12. What is the probability that all three lie inside a circle of radius 3?

Anyone have a solution?[/quote]

Well this is probably (definitely) not right... whatever
[hide="augh"]Call the points complex numbers $ z_1, z_2, z_3$. The points lie inside a circle of radius 3 iff there exists a point $ z$ such that
\[ |z_1 \minus{} z|\le 3,\]
\[ |z_2 \minus{} z|\le3,\]
\[ |z_3 \minus{} z|\le 3.\]
If they do lie inside a circle of radius 3, then $ z \equal{} \frac {z_1 \plus{} z_2 \plus{} z_3}{3}$ should work as a possible point (center of gravity? iono).

But yeah so plug $ z$ in and get
\[ |2z_1 \minus{} z_2 \minus{} z_3|\le 9,\]
\[ | \minus{} z_1 \plus{} 2z_2 \minus{} z_3|\le 9,\]
\[ | \minus{} z_1 \minus{} z_2 \plus{} 2z_3|\le 9\]
but then let $ |z_1 \minus{} z_2| \equal{} a$, $ |z_2 \minus{} z_3| \equal{} b$, $ |z_3 \minus{} z_1| \equal{} c$ be the side lengths of the triangle formed by the three points, then
\[ |a \plus{} b|\le 9,\]
\[ |b \plus{} c|\le 9,\]
\[ |c \plus{} a|\le 9.\]
Assume $ a\ge b\ge c$ then $ b \plus{} c > a$ and then we just need $ |a \plus{} b|\le 9$ and $ |a \plus{} c|\le 9$.

And then I finished off my amazing reasoning with a bogus argument, used some geometric probability, and got an answer of $ \frac {2}{3}$ which seems absurdly large. >_>" [/hide]

edit: oh yeah I also tried other stuff but couldn't get anything out of it:

[hide]Law of sines: $ \frac {a}{\sin A} \equal{} \frac {b}{\sin B} \equal{} \frac {c}{\sin C} \equal{} 2R$
But if there exists a point which is the center of a circle with radius 3 that meets the requirements, then the circumcenter of the triangle formed by the three points should work again: $ \frac {a}{\sin A} \equal{} \frac {b}{\sin B} \equal{} \frac {c}{\sin C}\le 6$

Also, using $ [ABC] \equal{} \frac {abc}{4R}$, get $ [ABC] \equal{} \frac {abc}{4R}\ge \frac {abc}{12}$ or $ 12[ABC]\ge abc$ and then I stopped there[/hide]

I'd be shocked if there was a nice solution. If you take one particular point, then the circle of radius 3 containing that point can be cut off by the side of the square...this makes for a nasty integral.

[quote="Flame"]and it nowhere says circle has to be inside square so that complicates things further[/quote]

The center of the circle might not be in the square, but some part of the circle must be in the square (since all three points are in the square)

[quote]The center of the circle might not be in the square, but some part of the circle must be in the square (since all three points are in the square)[/quote]

I don't think it matters whether the center is restricted to being in the square.

Yeah it does. Say your three points form a right triangle with sidelengths $ 3/\sqrt{2}, \; 3/\sqrt{2}, \; 3$, and the hypotenuse of the triangle is part of an edge  :wink:

In any case, not more than a half of the circle has to be outside the square, so the probability should be $ P(x) \geq \frac {\frac {9\pi}{2}}{144} \equal{} \frac {9\pi}{288} \equal{} \frac {\pi}{32}$ (shouldn't it?).

Or perhaps we should focus more on the distances of the points, since the points have to be within a distance of 6 from each other (unless someone can prove that you can choose two points inside a circle with a distance greater than the diameter of said circle)?

A rather lax upper bound comes about if we fix one point in the center and find the probability that two other points are both within 3 units of this point.

This gives a very lax upper bound of $ \left( \frac {9 \pi}{144} \right)^2 \approx 0.04$. This, of course, is [b]much [/b]higher than the actual probability (though it knocks out the answer above :) ).

Here's an approach that's probably nicer than using a six-way integral, but doesn't work when you get close to the edge.

Consider two points that are 3 units or less apart. Draw the two circles of radius 3 that pass through these two points, and their union is the area of the plane in which the third can reside. We can now vary the first point, bound and vary the second point, and then bound, and then bound and vary the two-circle union area for the third point. Of course, it'll be mega-piecewise (at least 9 pieces) because of the square edges.

Would the problem be neater if the three points were chosen from a circle? There would still be a boundary, but it would be more consistent. And we could assume that the first point lies on a particular radius.

Well, now that I think it a bit more, the probability should actually be $ P(x) \geq \left(\frac {\pi}{32}\right)^{2} \equal{} \frac {\pi^{2}}{1024} \approx 0.01$, since we're supposed to choose [i]three[/i] points, so perhaps we could set the range of $ P(x)$ to be something similar to $ 0.01 \leq P(x) \leq 0.04$?

well if anyone can figure it out (WITHOUT CALCULUS, i havent learned that yet) can you generalize it to a square with side m and circle with size n? (and maybe even k dots)

Suggestion: Consider the triangle connecting the three points.  If its circumcircle has radius greater than 3, then no circle of radius 3 contains them all.  If its circumcircle has radius less than or equal to 3, then take the circle of radius 3 centered at the circumcenter, and you've got a circle of radius 3 containing them all.

Using this, you can (at the least) write an inequality that holds if and only if the coordinates of the three points have the required circle.  But I'm guessing that this idea can lead to a better approach than that.

Maybe this as a start: You have two points, $ x$ and $ y$ that are $ a$ units apart.  Then for what points $ z$ will the circumradius of tri. $ XYZ$ be less than 6?  Well, $ a/\sin{\alpha} \leq 6$ iff $ \sin{\alpha} \geq a/6$.  So we're looking for points that will make certain angles with the first two we've chosen.  I'm no geometry expert, so maybe someone else can pick up on this.

Actually... that won't work.  Having circumradius greater than 3 does not imply they won't fit into some circle of radius 3.  Oh well, back to  6-integral methods.


Find all numbers whose prime divisors are $2$ and/or $5$ such that $25$ added to that number gives a prefect square.

[hide="Solution"] We'll work backwards.  We want $n$ such that $n^{2}-25$ only has prime divisors of $2$ and $5$.

Well, that factors to $(n+5)(n-5)$.  So $n-5$ must have only prime divisors $2$ and $5$.  Let $n-5 = 2^{a}5^{b}$.

[b]Case:[/b]  $b = 0$.  Then $n+5 = 2^{a}+10 = 2(2^{a-1}+5)$.  The second factor is odd, but it cannot be a power of $5$, so there are no solutions in this case.

[b]Case:[/b]  $b = 1$.  Then $n+5 = 10(2^{a-1}+1)$.  The second factor is odd, and is a power of $2$ when $a = 1$ and a power of $5$ when $a = 1, 3$, which gives $n+5 = 20, n^{2}-25 = \boxed{ 200 }$ and $n+5 = 50, n^{2}-25 = \boxed{ 2000 }$.  For $a > 3$ the expression is odd so it must e some power of $5$; call it $5^{b}$.  We require that $5^{b}-1 \equiv 0 \bmod 8 \implies b \equiv 0 \bmod 2$.  But then $3 | 5^{b}-1$, so there are no more solutions in this case.

[b]Case:[/b]  $b > 1$.  Then $n+5 = 10(2^{a-1}5^{b-1}+1)$.  If $a > 1$ the second factor isn't divisible by $2$ or by $5$, so $a = 1$.  But $5^{b-1}+1 \equiv 2 \bmod 4$, and it is larger than $2$, so it cannot be a power of $2$ and there are no more solutions in this case. [/hide]


integration
logarithms
real analysis
real analysis unsolved

I have little knowledge in differential,
So how to solved this differential equation:
$ f'(x)\equal{}1\minus{}(f(x))^2$,such that $ f(0)\equal{}0$.
Where $ f(x)$ is a differential in $ \mathbb{R}$.

I know the answer may should by $ f(x)\equal{}\frac{e^x\minus{}e^{\minus{}x}}{e^x\plus{}e^{\minus{}x}}$

You could use the notation:

$ f'(x)\equal{}\frac{df}{dx}$

Then you can write:

$ \frac{df}{dx}\equal{}1\minus{}f^2$

or

$ \frac{df}{1\minus{}f^2}\equal{}dx$

Now, integrate:

$ \int\frac{df}{1\minus{}f^2}\equal{}x\plus{}C$

You could use the fact that:

$ \frac{1}{1\minus{}f^2}\equal{}\frac{1}{2}\left(\frac{1}{1\plus{}f}\plus{}\frac{1}{1\minus{}f}\right)$

and then fact that:

$ \int \frac{df}{1\plus{}f}\equal{}\ln(1\plus{}f)$

$ \int \frac{df}{1\minus{}f}\equal{}\minus{}\ln(1\minus{}f)$

So the we have:

$ \int\frac{df}{1\minus{}f^2}\equal{}\frac{1}{2}\ln(1\plus{}f)\minus{}\frac{1}{2}\ln(1\minus{}f)\equal{}\frac{1}{2}\ln{\frac{1\plus{}f}{1\minus{}f}}\equal{}x\plus{}C$

$ \frac{1\plus{}f}{1\minus{}f}\equal{}Ce^{2 x}$

For the ease of writing, denote $ Ce^{2 x}\equal{}A$

$ \frac{1\plus{}f}{1\minus{}f}\equal{}A$

$ 1\plus{}f\equal{}A(1\minus{}f)$

$ 1\plus{}f\equal{}A\minus{}A f$

$ f(1\plus{}A)\equal{}A\minus{}1$

$ f(x)\equal{}\frac{A\minus{}1}{A\plus{}1}\equal{}\frac{A\minus{}1}{A\plus{}1}\equal{}\frac{Ce^{2 x}\minus{}1}{Ce^{2 x}\plus{}1}\equal{}\frac{Ce^{x}\minus{}e^{\minus{}x}}{Ce^{x}\plus{}e^{\minus{}x}}$

Since:

$ f(0)\equal{}\frac{C\minus{}1}{C\plus{}1}\equal{}0$ and so $ C\equal{}1$.

Therefore:

$ f(x)\equal{}\frac{e^{x}\minus{}e^{\minus{}x}}{e^{x}\plus{}e^{\minus{}x}}$

Hi,I believe you are right this time,Thank you very much.
But you have use the fact $ (f(x))^2<1$ in the solution,
I think it also need a explain although it is not hard :)

I have? It's just a way of solving these types of differential equations I've learned in college. I do not see where have I used that fact.. Could you point it out for me, please!

[quote="milin"]

$ \int \frac {df}{1 \plus{} f} \equal{} \ln(1 \plus{} f)$

$ \int \frac {df}{1 \minus{} f} \equal{} \minus{} \ln(1 \minus{} f)$

[/quote]

You have used here.

he actually didn't use that fact.
he just forgot to put absolute values ;)

But how can we do when there is some point x such that $ f(x)\equal{}1$???

For $ f'(x) \equal{} 1 \minus{} \{f(x)\}^2$, by $ f(0) \equal{} 0$, we have that $ f(x)$ isn't always 0. Thus we have

$ \frac {f'(x)}{1 \minus{} \{f(x)\}^2} \equal{} 1\Longleftrightarrow \frac {f'(x)}{1 \plus{} f(x)} \plus{} \frac {f'(x)}{1 \minus{} f(x)} \equal{} 2$

$ \therefore \frac {d}{dx} \ln \left|\frac {1 \plus{} f(x)}{1 \minus{} f(x)}\right| \equal{} 2$

$ \therefore \ln \left|\frac {1 \plus{} f(x)}{1 \minus{} f(x)}\right| \equal{} 2x \plus{} A$

$ \therefore \frac {1 \plus{} f(x)}{1 \minus{} f(x)} \equal{} Ce^{2x}$ where $ C \equal{} \pm e^A$.

$ f(0) \equal{} 0\Longleftrightarrow C \equal{} 1$, yielding $ f(x) \equal{} \frac {e^{2x} \minus{} 1}{e^{2x} \plus{} 1}$

[quote="bookworm_vn"][quote="zhaobin"]But how can we do when there is some point x such that $ f(x) \equal{} 1$???[/quote]

In fact, $ f(x) \ne 1$ for all $ x$.[/quote]
But when you haven't show that $ f(x) \equal{} \frac {e^x \minus{} e^{ \minus{} x}}{e^x \plus{} e^{ \minus{} x}}$,You can't assert that,although there is some way to show that,But we when you write solution you should give a reason.Ok,maybe there is something explain in the differential equation course,but I don't know,I am sorry.
My way to show $ f(x) \ne 1$,is that,if there is some point $ f(x_1)\equal{}1$,then $ f'(x_1)\equal{}0$,Then try to show $ f(x)\equal{}1$ for any $ x$.

Thank you kunny.But in your solution you also assume $ \frac{1\plus{}f(x)}{1\minus{}f(x)}>0$ without any reason.

[quote="zhaobin"]
Thank you kunny.But in your solution you also assume $ \frac {1 \plus{} f(x)}{1 \minus{} f(x)} > 0$ without any reason.[/quote]

It isn't so, $ \ln\left|\frac{1\plus{}f(x)}{1\minus{}f(x)}\right|\equal{}2x\plus{}A\Longleftrightarrow \left|\frac{1\plus{}f(x)}{1\minus{}f(x)}\right|\equal{}e^{2x\plus{}A}$

$ \Longleftrightarrow \frac{1\plus{}f(x)}{1\minus{}f(x)}\equal{}\pm e^{2x\plus{}A}\equal{}\pm e^A\cdot e^{2x}\equal{}Ce^{2x}$ where $ C\equal{}\pm e^{A}$.

Ok,Thank you,I am wrong in reading this.

Once we have a solution, we can invoke the existence-uniqueness theorem. $ \tanh$ is the only solution satisfying the initial conditions.

What do the general solutions (over $ \mathbb{R}$) look like? There's $ y \equal{} \tanh(x \minus{} a)$, $ y \equal{} 1$, $ y \equal{} \minus{} 1$, and $ y \equal{} \coth(x \minus{} a)$.

The latter doesn't cover all of $ \mathbb{R}$; the formula is a positive solution for $ x > a$ and a negative solution for $ x < a$.

[quote="zhaobin"][quote="bookworm_vn"][quote="zhaobin"]But how can we do when there is some point x such that $ f(x) \equal{} 1$???[/quote]

In fact, $ f(x) \ne 1$ for all $ x$.[/quote]
But when you haven't show that $ f(x) \equal{} \frac {e^x \minus{} e^{ \minus{} x}}{e^x \plus{} e^{ \minus{} x}}$,You can't assert that,although there is some way to show that,But we when you write solution you should give a reason.Ok,maybe there is something explain in the differential equation course,but I don't know,I am sorry.
My way to show $ f(x) \ne 1$,is that,if there is some point $ f(x_1) \equal{} 1$,then $ f'(x_1) \equal{} 0$,Then try to show $ f(x) \equal{} 1$ for any $ x$.[/quote]

OK, I have a question for you. Under what Cauchy condition there exists $ x$ such that $ f(x) \equal{} 1$. I mean $ f(0)$ equals to what?

[quote="zhaobin"][quote="bookworm_vn"][quote="zhaobin"]But how can we do when there is some point x such that $ f(x) \equal{} 1$???[/quote]

In fact, $ f(x) \ne 1$ for all $ x$.[/quote]
But when you haven't show that $ f(x) \equal{} \frac {e^x \minus{} e^{ \minus{} x}}{e^x \plus{} e^{ \minus{} x}}$,You can't assert that,although there is some way to show that,But we when you write solution you should give a reason.Ok,maybe there is something explain in the differential equation course,but I don't know,I am sorry.
My way to show $ f(x) \ne 1$,is that,if there is some point $ f(x_1) \equal{} 1$,then $ f'(x_1) \equal{} 0$,Then try to show $ f(x) \equal{} 1$ for any $ x$.

Thank you kunny.But in your solution you also assume $ \frac {1 \plus{} f(x)}{1 \minus{} f(x)} > 0$ without any reason.[/quote]
if a solution starts at a non-critical value, it can not reach a critical point in a finite amount of time, as Jmerry said this would contradict the existence-uniqueness theorem.  :)

[quote="zhaobin"]Hi,guys,Thank you very much.But I have said that I am unfamiliar with the differential equaitons.And thanks to jmerry's comment.[/quote]

oh, that's why you should improve your ODEs right now  :P

[quote="bookworm_vn"]
oh, that's why you should improve your ODEs right now  :P[/quote]
Thanks for your advice. :)


Find the smallest positive integer $ x$ that satisfies the congruence $ 723x \equiv 117 \pmod {720}.$

$ 723x\equiv 720x \plus{} 3x\equiv 3x\pmod{720}$, so we have $ 3x\equiv 117\pmod{720}$

Dividing by 3, we have $ x\equiv 39\pmod {240}$, so $ x \equal{} 240k \plus{} 39,\;k\in\mathbb{Z}$.

To minimize $ x$ we set $ k \equal{} 0$ and have $ \boxed{x \equal{} 39}$.

Checking, $ 723\cdot 39\equiv 28197\equiv 28080 \plus{} 117\equiv 117\pmod{720}$.

you can't divide in a congruence.... well you can but with restrictions.

Um...the restriction is simple...

If we have $ a\equiv b\pmod m$, then $ \frac ab\equiv1\pmod{\frac m{\gcd(b,m)}}$.


Hmm, i found this from Gabriel Carol.
God this creature is a real genuis,just look at these:
http://www.people.fas.harvard.edu/~gcarroll/archives/puzzles/panalogies.htm

Oops.I think i should take this to Round table ,sorry

Those anologies are sure different from the one's on the SAT's. They have weird variations. Atomic number? Are you kidding me?  :D

they are hard.  that's what i mean by nice :)


Let $a,b,x,y>0$. Prove that we have:

$(a \frac{x}{y}+b)^2+(a \frac{y}{x}+b)^2 \geq 2(a+b)^2$.

cheers! :D :D :cool:

With a straight forward application of AM-QM, we get

(ax/y+b) 2 +(ay/x+b) 2  \geq 2 [ a(x/y+y/x)/2 + b ] 2  \geq 2 (a+b) 2

(the last step follows from a well known fact that r + 1/r  \geq 2)

Equality occurs iff x=y.


[quote="duc 95"]I think it is right[/quote]

this is wrong.

[quote="hedeng123"][quote="duc 95"]Can you prove it without SOS ?[/quote]

try  $ a \equal{} \frac {1}{4},b \equal{} \frac {1}{2},c \equal{} 1.$[/quote]


Let $ a,b,c\geq0$.Prove that:
$ a^3\plus{}b^3\plus{}c^3\plus{}3abc \geq bc\sqrt{2(b^2\plus{}c^2)}\plus{}ab\sqrt{2(a^2\plus{}b^2)}\plus{}ca\sqrt{2(c^2\plus{}a^2)}$

See here : http://www.mathlinks.ro/viewtopic.php?t=6317

If $ x,y,z>0,$ then

$ x^3 \plus{} y^3 \plus{} z^3 \plus{} 3xyz\geq\frac {2(yz \plus{} zx \plus{} xy)(x^2 \plus{} y^2 \plus{} z^2)}{x \plus{} y \plus{} z}$

$ \geq yz\sqrt {2y^2 \plus{} 2z^2} \plus{} zx\sqrt {2z^2 \plus{} 2x^2} \plus{} xy\sqrt {2x^2 \plus{} 2y^2}.$

See also : http://www.mathlinks.ro/viewtopic.php?t=216948

Sorry, dear Ji Chen, can you please explain why this inequality is true?

Thank you very much. :)

$ \frac {2(yz + zx + xy)(x^2 + y^2 + z^2)}{x + y + z}
\geq yz\sqrt {2y^2 + 2z^2} + zx\sqrt {2z^2 + 2x^2} + xy\sqrt {2x^2 + 2y^2}.$

[quote="Ji Chen"]Dear manlio,
$ \sum{yz\sqrt {2y^2 \plus{} 2z^2}}\leq\sum{\frac {yz(3y^2 \plus{} 2yz \plus{} 3z^2)}{2(y \plus{} z)}}$
[/quote]
Beautiful hint, Jichen  :lol:
I tried to use: $ \sum{yz\sqrt {2y^2 \plus{} 2z^2}}\leq \sum\ \frac{2yz(y^2\plus{}z^2)}{y\plus{}z}$, and don't obtain a true result  :oops:

[quote="Ji Chen"]Dear manlio,

$ \sum{yz\sqrt {2y^2 + 2z^2}}\leq\sum{\frac {yz(3y^2 + 2yz + 3z^2)}{2(y + z)}}$

$ = \frac {2(yz + zx + xy)(x^2 + y^2 + z^2)}{x + y + z} - \sum{\frac {yz(y - z)^2[x(y + z - x)^2 + yz(x + y + z)]}{2(y + z)(z + x)(x + y)(x + y + z)}}.$

By the way,

$ (x + y + z)(x^3 + y^3 + z^3 + 3xyz) - 2(yz + zx + xy)(x^2 + y^2 + z^2)$

$ = \sum{x^2(x - y)(x - z)} = \frac {1}{2}\sum{(y - z)^2(y + z - x)^2\geq0}$

holds for all real numbers $ x,y,z.$[/quote]

Wonderful, Ji Chen. :)

Thank you very much :)


This isn't exactly related to $\LaTeX$, but how do you scan several files onto your computer and then merge them into one PDF file?

You can merge PDF's with Adobe's Acrobat "Standard" or "Professional" product.  It's not particularly cheap though.  Adobe gives away the "Reader" hoping folks like you will decide at some point that merging and "editing" PDF's will be worth a few hundred dollars (educational and other discounts can make the price reasonable though.)

The "editing" feature allows you to delete pages from a document.  The software is capable of creating a new document using subsets of the pages from several documents. I've used this software and it works reliably.

There are other sources of "free" software that perform some of the same tasks but probably won't be as friendly to use.  Try Google on "Merge PDF" or similar terms to find the "freebies."  I wouldn't put one on my computer unless someone I trusted had good experience with the product.

- Greg


calculus integration logarithms function geometry limit trigonometry

Thanks sos440. Later I will post sulution in more different way by complex variable

[quote="sos440"]Generally, for $n>1$, we have

$\int_{0}^{\infty}\frac{\ln{x}\,dx}{x^{n}+1}=-\frac{\pi^{2}}{n^{2}}\cot\left(\frac{\pi}{n}\right)\csc\left(\frac{\pi}{n}\right)$

Examples include

$\int_{0}^{\infty}\frac{\ln{x}\,dx}{x^{2}+1}= 0$
$\int_{0}^{\infty}\frac{\ln{x}\,dx}{x^{3}+1}=-\frac{2\pi^{2}}{27}$
$\int_{0}^{\infty}\frac{\ln{x}\,dx}{x^{4}+1}=-\frac{\pi^{2}}{8\sqrt{2}}$[/quote]
can you post your sulution?

I integrated the function $f(z) = \frac{\ln z}{1+z^{n}}$ along the path $C$, the boundary of pizza-shaped area, travelled counter-clockwise:

[img]http://myhome.shinbiro.com/~sosQED/nsol088.gif[/img]

where $\theta = \frac{2\pi}{n}$. For $n = 2, 3, ...$, this area contains exactly one root of $z^{n}+1 = 0$, namely $\omega = e^{i\theta/2}$. And then compare two sides of the equation

$2\pi i \underset{z=\omega}{\text{Res}}\, \frac{\ln z}{1+z^{n}}= \int_{C}\frac{\ln z}{1+z^{n}}\, dz$ $\quad \Longrightarrow \quad$ $\lim_{R\to\infty}\int_{C}\frac{\ln z}{1+z^{n}}\, dz = 2\pi i \frac{\ln \omega}{n \omega^{n-1}}= \frac{2\pi^{2}\omega}{n^{2}}$

If we use the fact that the integral of $f(z)$ over the arc part of the path $C$ becomes zero, then

\begin{eqnarray*}\lim_{R\to\infty}\int_{C}\frac{\ln z}{1+z^{n}}\, dz & = & \int_{0}^{\infty}\frac{\ln z}{1+z^{n}}\, dz+\int_{\omega^{2}\infty}^{0}\frac{\ln z}{1+z^{n}}\, dz\\ & = & \int_{0}^{\infty}\frac{\ln x}{1+x^{n}}\, dx-\int_{0}^{\infty}\frac{\ln (\omega^{2}x)}{1+(\omega^{2}x)^{n}}\, \omega^{2}dx\\ & = & (1-\omega^{2})\int_{0}^{\infty}\frac{\ln x}{1+x^{n}}\, dx-\frac{2\pi i \omega^{2}}{n}\int_{0}^{\infty}\frac{1}{1+x^{n}}\, dx \end{eqnarray*}

Finally, use the equation $\int_{0}^{\infty}\frac{dx}{1+x^{n}}= \frac{\pi}{n}\cot \frac{\pi}{n}$ then we have desired result.

Or:

Let $I(\alpha)=\int_{0}^{\infty}\frac{x^{\alpha}}{x^{n}+1}dx$.

A calculation shows that:

${I(\alpha)=\frac{\pi}{n}\frac{1}{\sin \frac{(\alpha+1)\pi}{n}}}$.

Then,

$I'(\alpha)=\int_{0}^{\infty}\frac{x^{\alpha}\ln x}{x^{n}+1}dx=-\frac{\pi^{2}}{n^{2}}\frac{\cos \frac{(\alpha+1)\pi}{n}}{\sin^{2}\frac{(\alpha+1)}{n}}$.
Now, let $\alpha \rightarrow 0$, and you are done.

I proved next:
$\int_{0}^{oo}\frac{\ln{x}dx}{x^{n}+1}=-1+2\sum_{k=1}^{oo}\frac{1+k^{2}n^{2}}{(1-k^{2}n^{2})^{2}}(-1)^{k+1}$
and how can I find this sum?

sos440: (I don't know some defination and theorem in complex analsisy, therefore I have some simpl question).
How ?
1)  $2\pi i \frac{\ln \omega}{n \omega^{n-1}}= \frac{2\pi^{2}\omega}{n^{2}}$
2)$\begin{eqnarray*}\lim_{R\to\infty}\int_{C}\frac{\ln z}{1+z^{n}}\, dz & = & \int_{0}^{\infty}\frac{\ln z}{1+z^{n}}\, dz+\int_{\omega^{2}\infty}^{0}\frac{\ln z}{1+z^{n}}\, dz\\ &$
What is it: $\omega^{2}\infty$ ?
3) How? $\& \int_{0}^{\infty}\frac{\ln x}{1+x^{n}}\, dx-\int_{0}^{\infty}\frac{\ln (\omega^{2}x)}{1+(\omega^{2}x)^{n}}\, \omega^{2}dx\\ = (1-\omega^{2})\int_{0}^{\infty}\frac{\ln x}{1+x^{n}}\, dx-\frac{2\pi i \omega^{2}}{n}\int_{0}^{\infty}\frac{1}{1+x^{n}}\, dx \end{eqnarray*}$
4) Hoe do you find it? ${I(\alpha)=\frac{\pi}{n}\frac{1}{\sin \frac{(\alpha+1)\pi}{n}}}$.
$\alpha$ is real?
5)how? $I'(\alpha)=\int_{0}^{\infty}\frac{x^{\alpha}\ln x}{x^{n}+1}dx$

1) $\omega^{n}= (e^{i\pi/n})^{n}= e^{i\pi}=-1$, so $\omega^{1-n}=-\omega$. And also, $\ln \omega = \ln (e^{i\pi/n}) = \frac{i\pi}{n}$.

2) $\int_{\omega^{2}\infty}^{0}f(z) \, dz$ is the conventional notation for the integral of $f(z)$ along the path $C = \{\omega^{2}x : x \in [0, \infty)\}$, travelled centripetal direction.


i hope i have not asked this question before... :| can you show :D

$ \sum_{k\equal{}1}^{\infty}\;\;\frac{(\minus{}1)^{k\plus{}1}}{k^{3}\cdot\dbinom{2k}{k}}\;\;\equal{}\;\;\boxed{\frac{2}{5}\,\zeta{(3)}}$

see : http://pagesperso-orange.fr/gery.huvent/articlespdf/Autour_primitive.pdf
page 11 formula (7)

Apery formula : http://pagesperso-orange.fr/gery.huvent/articlespdf/article3.pdf


Assume that a_i>0 for i=1,2,....n  with a_n+1=a_n. Prove that

 \sum (i=1,..,n) (a_i/(a_i+1))^n  \geq  \sum (i=1,...,n)(a_i+1/a_i)

Could you please write it more explicitely? The problem that I understand cannot be true. Thank you very much.

Sorry Harazi, I have forgotten brackets and also made a terrible mistake.

The correct inequality is:

Assume that a_i>0 for i=1,2,....n with a_(n+1)=a_1. Prove that

 sum(i=1,..,n) (a_i/(a_(i+1))^n  \geq sum  (i=1,...,n)(a_(i+1)/a_i)

Now everything is normal and -fortunately-I can solve it. It obviously reduces to proving that if a_1*a_2*...*a_n=1, then sum a^n_i>=sum 1/a_i. But this comes down to AM-GM and Cebasev, since 1/a_1=a_2*...*a_n<= 1/(n-1)(a^(n-1)_2+...+a^(n-1)_n). So sum 1/a_i<= sum a^(n-1)_i<=sum a^n_i. The last one follows from the fact that sum a^n_i>=(sum a^(n-1)_i)*(sum a_i)/n and of course sum a_i>=n.


Sam, John, and Sarah bought six books costing $ 5, 6, 7, 8, 10,$ and $ 15$ dollars. Sam bought two of these books, while John and Sarah bought the remaining books. If Sam paid four fifths of the amount paid by John, and Sarah did not spend more than $ 14$ dollars, then the amount paid by Sarah is:

$ \textbf{(A)}\ 5 \qquad \textbf{(B)}\ 6 \qquad \textbf{(C)}\ 8 \qquad \textbf{(D)}\ 11 \qquad \textbf{(E)}\ 13$

[hide="Solution"]
Let $ S_m,J,S_w$ be the amounts that Sam, John and Sarah used, respectively.

$ S_m\plus{}J\plus{}S_w\equal{}51$

We know that $ S_m\cdot \dfrac{5}{4}\equal{}J$ and that $ 12\le S_m\le 20\implies 15\le J\le 25$.

Since $ S_m$ and $ J$ are integers, $ S_m\equal{}\{12,16,20\}$ and $ J\equal{}\{15,20,25\}$, the only pair that makes $ S_w\le 14$ is when $ S_m\equal{}20$ and $ J\equal{}25$ and $ S_w\equal{}6$.

So the answer is $ \mathbf{(B)}$

Note: The books that Sam buys are the ones that cost \$5 and \$15. The books that John buys are the \$7, \$8 and \$10 ones.
[/hide]


analytic geometry
combinatorial geometry
combinatorics unsolved
combinatorics

There are $n$ distinct points on the plane. We color the middle of each line $A_iA_j, i \neq j$ red. Find the minimum number od red points

I also remember something about $2n-3$, but I don't know if it can always be reached. To show that we have at least $2n-3$ midpoints, proceed by induction. Assume that the last time we added a point, we added it on the boundary of or outside the convex hull of the other $n-1$ points. It's easy to show that we add at least two more midpoints, between our new point and two of the vertices of the convex hull.

Edit: this answer was meant to address [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=26300]this[/url].

Hmm;.. I splitted the topic. To which problem are you answering? :D

Pierre.

I have edited my post to answer your question, Pierre :).

That's what I suspected while reading your answer. :P

Pierre.

[quote="grobber"]I also remember something about $2n-3$, but I don't know if it can always be reached. [/quote]
???
Take $n$ points on line on the same distance from each other.

Oh, cool! :) I had a tendency to disregard degenerate cases :).

As for showing that there must be at least $2n-3$ distinct midpoints, we can set up coordinate axes such that the points have pairwise distinct $x$-coordinates.  WLOG, we can number the points $A_1, A_2, \ldots, A_n$ in order of increasing $x$-coordinate.  Then it's easy to observe that the midpoints of the segments $A_1A_2, A_1A_3, \ldots, A_1A_n, A_2A_n, A_3A_n, \ldots, A_{n-1}A_n$ are all distinct.

or you can project the points on a line, not perpendicular to a line formed by two of them, and clearly the number of midpoints is less in the line.