calculus
geometry
articles
probability
expected value

Greetings Matheletes!
Many of you have ventured through difficult arithmetic/geometric series, Counting/Probability, Calculus, Geometry, and of course dont forget addition/subtraction. :lol:  But this challenge you are about to experience is the toughest of the toughest. Me and a group of 20 others will be competing against the Brainy Dark Lord, and his Death Eaters. We will be going challenged to solve tough riddles, challenging math problems, dangerous expeditions, and much much more! My mentor Mr. BumbleBore recently passed away, and I am filled with contrition. But before Mr.BumbleBore died he told me a horrifying secret: The Dark Lord split his sole into 14 parts called MathCruxes. We must find all Math Cruxes and destroy them all!  First, I need people to join me in my quest to defeat The Dark Lord..

I am splitting us into 4 groups:
I will be the Leader  :)
[hide="Mathyndor"](Join if you are VERY good at math)[/hide]
[hide="RiddlePuff"]Join if you are a VERY cunning person[/hide]
[hide="ValorClaw"]Join if you are VERY brave[/hide]
[hide="SlytherBurn"]Join if you are a little arrogant and bossy, but dont worry this is a nice group :) ALso people in this group will be able to do a little of everything [/hide]

Dont spam please, and Have Fun! In this Game I will be posting Math Problems/Riddles/Adventures/Other Things!

Oh yeah, and when you join tell me what group you want to be in and maybe add a small reason why you picked that group...or you can just say you dont care about which group you get put into :D

cgyao15 does a little test and joins Mathyndor.

Admissions so far:

[hide="Mathyndor"](Join if you are VERY good at math)
cgyao15
[/hide]

[hide="RiddlePuff"]
Join if you are a VERY cunning person[/hide]

[hide="ValorClaw"]
Join if you are VERY brave[/hide]

[hide="SlytherBurn"]
Join if you are a little arrogant and bossy, but dont worry this is a nice group  ALso people in this group will be able to do a little of everything [/hide]

spaceguy..idk which group you want to be in... :maybe:

May I join? RiddlePuff? I'd join Mathyndor... but I'm really... i mean REALLY cunning  :whistling:

Admissions so far:

[hide="Mathyndor"](Join if you are VERY good at math)
[/hide]
cgyao15
[hide="RiddlePuff"]
Join if you are a VERY cunning person[/hide]
meewhee009

[hide="ValorClaw"]
Join if you are VERY brave[/hide]

[hide="SlytherBurn"]
Join if you are a little arrogant and bossy, but dont worry this is a nice group  ALso people in this group will be able to do a little of everything [/hide]

spaceguy..idk which group you want to be in... :maybe:[/

JOIN!!!!!

i would like to be in [size=200][u][i][b]SlytherBurn[/b][/i][/u][/size]

Admissions so far:

[hide="Mathyndor"](Join if you are VERY good at math)
[/hide]
cgyao15

[hide="RiddlePuff"]
Join if you are a VERY cunning person[/hide]
meewhee09
[hide="ValorClaw"]
Join if you are VERY brave[/hide]
Chompy
[hide="SlytherBurn"]
Join if you are a little arrogant and bossy, but dont worry this is a nice group  ALso people in this group will be able to do a little of everything [/hide]
hjoon0125

spaceguy..idk which group you want to be in... :maybe:[/

Admissions so far:

[hide="Mathyndor"](Join if you are VERY good at math)
[/hide]
cgyao15

[hide="RiddlePuff"]
Join if you are a VERY cunning person[/hide]
meewhee09
yaofan
louiethechewy123
[hide="ValorClaw"]
Join if you are VERY brave[/hide]
Chompy
[hide="SlytherBurn"]
Join if you are a little arrogant and bossy, but dont worry this is a nice group  ALso people in this group will be able to do a little of everything [/hide]
hjoon0125

spaceguy..idk which group you want to be in... :maybe:
7 people joined....13 more slots are open :P
Come on ppl get more active :P Dont give the Dark Lord free time :ninja:
Oh yeah...lets name The Dark Lord: Tom Math Riddle

lolz!!!!!

But I'm not anything...... Isn't there like the average people for like Huffelpuff?

Ok, I'll be Riddlepuff because it's like Hufflepuff.

WHERE ARE THE REST OF THE SLYTHERBURNS??  I have checked the Quicktopic multiple times, and nobody besides me has posted any other message besides "Confirm."   :(

magical bump.

Where are the rest of the SlytherBurns?  Is this game dead/mod-abandoned?

Both.

I say we should let this die. Other more interesting games are up now.

I'm just rlly busy. I'll revive this thread soon.

...umm....
this game has been going on quite a while?
Also, if you can join, as im not the mod, please choose a section to be in

No kidding.  There are like three active SlytherBurns right now.

ValorClaw I join because I am brave not good at math not cunning and not bossey (well except to my brother)

@pryopower: weren't you in my class earlier?

(sorry if i spell your name wrong)


I believe almost all Indians here (or atleast the older, wiser ones - not me!!!) know about the famous story of Arjuna in Mahabharata, when his master Dronacharya asks everybody what they see before hitting a pigeon's eye with a bow. People give different responses but Arjuna, the master archer, says he sees just the pigoen's eyes and succesfully shoots it.

Is there any story etc. in US, UK which is exactly same as this one and is well known so that if you write it in a letter, the other person eill understand??? Please help!!!!! :help:

This was one of the favorite story which   S. Chandrasekhar (Chicago Astorphysics  Professor , and nobel laureate)    told his students ..

Does it have a english equivalent!!!

My basic question is that is there a quote in US that expresses the same idea as the story and is wellknown so that everyone can understand???

No suggestions still??? Please help!!! :help:

The often used quote is "Be focused"  or  "keep the goal in focus"   ..

Is there some short American Story whcih conveys the same meaning as the Indian Story above???


geometry
3D geometry
probability
MATHCOUNTS

ok... so the  AoPS counting and probability book says something like "43,252,003,256,000" configurations, and something like 8!x10!x2^12x3^10......
but how do they come up with that stuff?
anyone have an answer?

If you read more carefully, you would know that "it is a [i]very[/i] difficult number to count." Usually, if someone gets recognition to be the first person to count it, [i]very recently[/i], it's probably pretty hard, and you can hopefully infer that it's not at this level. If not, you would notice that this was not included as an exercise or problem in the book, nor did it even challenge you to try to prove it (like the geometry book challenges you to prove the simson line, 9-point circle, morley's theorem, etc., even though they're tough to do with the experience the book gives you), and infer that it is not at this level.

You don't seem to have infered it from all of these clues, though.  :wink:

well, im pretty sure you cant just do $ 27!$. a start-each move you can move 6 different faces 2 different directions. so thats 12.

If the answer was 27!, it wouldnt be called very hard.

The methods used to calculate the number of positions of a rubik's cube are [b]far[/b] beyond mathcounts level (and beyond normal high school level too). Wikipedia outlines it at [url]http://en.wikipedia.org/wiki/Rubik%27s_Cube_group[/url].


inequalities
trigonometry
inequalities unsolved

[quote="giannis"][i]Let a triangle ABC , Proove that[/i]

$cosA+cosB+cosC\leq 3/2$[/quote]

Just write $\frac{a^{2}+b^{2}-c^{2}}{2ab}$ instead of $cosA$ etc.
Then gets Schur :)

[quote="giannis"][i]Let a triangle ABC , Proove that[/i]

$cosA+cosB+cosC\leq 3/2$[/quote]

Let  $S=cos(A)+cos(B)+cos(C)$

$a=\frac{(cos(B)+cos(C))}{2}$
 $b=\frac{(cos(A)+cos(C))}{2}$
 $c=\frac{(cos(A)+cos(B))}{2}$

we have

$S=a+b+c=cos(\frac{B-C}{2})sin(\frac{A}{2})+cos(\frac{A-C}{2})sin(\frac{B}{2})+cos(\frac{A-B}{2})sin(\frac{C}{2})$
then

$S \leq sin(\frac{A}{2})+sin(\frac{B}{2})+sin(\frac{C}{2}) \leq  3sin(\frac{A+B+C}{6}) = \frac{3}{2}$

[quote="giannis"][i]Let a triangle ABC , Proove that[/i]
\[\cos A+\cos B+\cos C\leq\frac{3}{2}\]
[/quote]
It's easy and well-known problem  :D
We have:
\[\cos A+\cos B+\cos C\leq\frac{3}{2}\]

\[\Longleftrightarrow 2\cos\frac{A+B}{2}\cos\frac{A-B}{2}+1-2\sin^{2}\frac{C}{2}\le\frac{3}{2}\]

\[\Longleftrightarrow 4\sin^{2}\frac{C}{2}-4\sin\frac{C}{2}\cos\frac{A-B}{2}+1\ge 0 \]

\[\Longleftrightarrow \left(2\sin\frac{C}{2}-\cos\frac{A-B}{2}\right)^{2}+\sin^{2}\frac{A-B}{2}\ge 0 \]

[quote="giannis"][i]Let a triangle ABC , Proove that[/i]

$cosA+cosB+cosC\leq 3/2$[/quote]

Use $(\sum \frac{\vec{AB}}{AB})^{2}\geq 0$.

by the inflection point theorem it is enough to consider $A=B$
so we need to prove
$3+2\cos 2A\geq 4\cos A\longleftrightarrow (\cos A-\frac12)^{2}\geq 0$
 :D

[quote="Albanian Eagle"] inflection point theorem [/quote]
What is that?  :maybe:

refering to
http://www.mathlinks.ro/Forum/viewtopic.php?t=64933
 :lol:

[quote="Till"][quote="andy_tok"][quote="arkhammedos"]$cosx$ is concave so
$\frac{cosA+cosB+cosC}{3}\leq cos(\frac{A+B+C}{3})$ :maybe:[/quote]
cosX is concave if the triangle is acute[/quote]
cos X is concave in $]0,\pi [$
 :D[/quote]
No, that will be a sine curve.
cosX is only concave in $[(2k-\frac{1}{2})\pi,(2k+\frac{1}{2})\pi]$ :wink:

Using Jensen's inequality , we can proove it easy when $0<A,B,C<90$ , but what if $90<A,B,C<180$?

[quote="arkhammedos"][quote="giannis"] but what if $90<A,B,C<180$?[/quote]
[b]what?[/b] :mad:[/quote]
yes rem it is convex but i am talking about angels theire sum will not be 180 (triangle).

OK, yes that's true.
Maybe he meant A, B, or C is in the range from 90 to 180.


inequalities
combinatorics unsolved
combinatorics

I have come across the following problem:
[quote]Prove that $ K_5$ (a complete graph with 5 nodes) is not planar. (Note that if $ K_5$ were planar then any region of the graph would have at least 3 boundary arcs and therefore $ 3R\le2A$)[/quote]
I don't quite understand how the inequality comes into play,  :?:  can someone please help.

Thank you very much.  :P

The graph has 10 edges. The above inequality implies that the number of regions into which it divides the plane is at most $ [20/3]\equal{}6$. On the other hand, the graph has 5 vertices, and at each vertex 4 angles are formed. This creates 20 angles. Since $ 6\cdot 3<20$, at least one of the regions has more than 3 angles, which leads to a contradiction because all the vertices of this region must be connected to each other.

[quote="mlok"]The graph has 10 edges. The above inequality implies that the number of regions into which it divides the plane is at most $ [20/3] \equal{} 6$. On the other hand, the graph has 5 vertices, and at each vertex 4 angles are formed. This creates 20 angles. Since $ 6\cdot 3 < 20$, at least one of the regions has more than 3 angles, which leads to a contradiction because all the vertices of this region must be connected to each other.[/quote]

Oh I see, thank you very much mlok.
I just wonder where is the inequality comes from?  :?:

Thank you very much.  :P

If you add together the numbers of sides in each region, the result is (a) at least 3 times the number of regions; (b) exactly twice the number of edges.


Για  να  γυρίσουμε  λίγο  και στην  ανάλυση   :D  :D

      Να  υπολογιστεί  η  σειρά  : $\sum \limits_{n=1}^{\infty} \frac{1}{(2n-1)(3n-1)}$

Μετά χρησιμοποιούμε
$\frac{1}{2}-\frac{1}{5}+\frac{1}{8}-\frac{1}{11}+......=\frac{\pi\sqrt{3}}{9}-\frac{1}{3}ln\ 2$

και  $1-\frac{1}{3}+\frac{1}{5}-\frac{1}{7}+\frac{1}{9}-......=\frac{\pi}{4}$

Μετά ένα κολπάκι  ;)   και τελικό αποτέλεσμα αν δεν κάνω λάθος κάπου είναι
$ln\frac{4}{\sqrt{27}} + \frac{\pi}{2\sqrt3}$

Σωστά ??

[quote="silouan"]
Μετά ένα κολπάκι  ;)   και τελικό αποτέλεσμα αν δεν κάνω λάθος κάπου είναι
$ln\frac{4}{\sqrt{27}} + \frac{\pi}{2\sqrt3}$

Σωστά ??[/quote]

Σωστός !  :)

Οταν ειδα το προβλημα του Γιωργου δεν εδειξα ιδιαιτερο ενδιαφερον. Πιστευω οτι δεν ειναι τοσο διδακτικη(ιδιαιτερα για διαγωνισμους).Εννοω ρε παιδια οτι δεν ειναι αυτη η φιλοσοφια των διαγωνισμων δηλαδη το να βαζεις ασκησεις που απαιτουν μονο γνωσεις.Σιγουρα χρειαζονται και τετοιες γνωσεις.Παντως το σιγουρο ειναι οτι ολοι μας πρεπει να ασχολουμαστε με ολους τους τομεις των μαθηματικων και με αυτα που δεν μας αρεσουν. :huh:

Θα μου επιτρέψεις να διαφωνήσω μαζί σου .Μαθηματικά δεν  είναι μόνο οι διαγωνισμοί .
Η άκσηση είναι διδακτικότατη όπως θα δεις και από τη λύση του Γιώργου αν τη βάλλει.
Μην κοιτάς που εγώ την έλυσα με κάπιους τύπους που σε εμένα ήταν πολύ οικείοι τουλάχιστον .Αν προσέξεις το σπάσιμο στην αρχή ,χρησιμοποιείται κατα κόρον στους διαγωνισμούς και πάρα πάρα πολύ στην ελλάδα. ;)
Από ότι ξέρω (μου το είπε ο Γίωργος σήμερα) η λύση του στην άσκηση είναι πολύ διαφορετική από τη δική μου και χρησιμοποιεί ένα δικό  του τρόπο υπολογισμού αθροίσματος.......

Συμφωνω φιλε Σιλουαν οτι μαθηματικα δεν ειναι μονο διαγωνισμοι εξαλλου αυτο ειπα(οτι καλο ειναι να ασχολουμαστε με οσο το δυνατον περισσοτερους τομεις των μαθηματικων).θα ηθελα να δω και την λυση του Γιωργου.Εγω ηθελα να πω οτι απο την μικρη εμπειρια που εχω οτι σημαντικο ρολο σε ενα προβλημα παιζει και η παρουσιαση του.(θα μου αρεσε περισοτερο αν το ειχε παρουσιασει πιο προσεγμενα.Επισης θα ηταν πιο ομορφο αν στην λυση της ασκησης υπηρχε και ενα "διδαγμα"...ενα ωραιο τεχνασμα.

Λοιπόν  θα  γράψω  την  περιβόητη  λύση  για  να  την  κρίνετε  και  εσείς  καλύτερα

           Εχουμε  λοιπόν  ότι
 $S= \sum_{n=1}^{\infty}\frac{1}{(2n-1)(3n-1)} =\sum_{n=1}^{\infty} (\frac{2}{(2n-1)} - \frac{3}{(3n-1)} )$
 $S= \sum_{n=1}^{\infty} (\frac{2}{(2n-1)}-\frac{2}{2n} +\frac{3}{3n} - \frac{3}{(3n-1)})$
 $S= \sum_{n=1}^{\infty} (\frac{2}{(2n-1)}-\frac{2}{2n}) -\sum_{n=1}^{\infty}(\frac{3}{3n-1} - \frac{3}{3n})$

   και  τώρα  το  κόλπο .......  θεωρούμε  τις ακολουθίες   συνάρτησεων $f_n(x) = x^{2n-2}-x^{2n-1}$  και  την  $g_n(x) = x^{3n-2}-x^{3n-1}$  οι  συναρτήσεις   αυτές  είναι  συνεχείς  και  παραγωγίσιμες  στο  (0,1)   έτσι  έχουμε  ότι

          $\int_{0}^{1} f_n(x) dx =\int_{0}^{1}x^{2n-2}-x^{2n-1} dx = \frac{1}{2n-1}-\frac{1}{2n}$    όμοια  ισχύει  ότι

   $\int_{0}^{1} g_n(x) dx = \frac{1}{3n-1}-\frac{1}{3n}$

     συνεπώς  $S = 2\int_{0}^{1}\sum_{n=1}^{\infty}f_n(x) dx -3\int_{0}^{1}\sum_{n=1}^{\infty}g_n(x) dx$

             άρα  $S=2ln2 - 3\int_{0}^{1}( x - x^2 + x^4 -x^5...+ x^{3n-2}-x^{3n-1}+...)dx$

   παρατηρούμε  λοιπόν  ότι  οι  ποσότητες  που  βρίσκονται  μέσα  στο  ολοκλήρωμα  είναι  γεωμετρικές  πρόοδοι με  άπειρους  όρους  και   | x|<1  μιας  και  χΕ(0,1)

  έτσι  έχουμε
                        $S = 2ln2 - 3\int_{0}^{1}x(1-x)(1 + x^3 + x^6 + ...)dx$

                        $S= 2ln2 - 3 \int_{0}^{1} \frac{x(1-x) }{ 1-x^3}dx$

                        $S= 2ln2 - 3 \int_{0}^{1} \frac{x}{ x^2 + x +1}dx$

  Το  τελευταίο  ολοκλήρωμα  ανάγεται  σε  τοξο  εφαπτομένης  και  βγαίνει  ρε  παιδιά  όμως  βαριέμε  να  γράφω  άλλο :D  :D
  οπότε ......................... $S= 2ln2 -\frac{3}{2}ln3 + \frac{\sqrt{3}\pi}{6}$

απλά εκπληκτικό .Τώρα βρήκα χρόνο να την κοιτάξω αναλυτικά ;)  .Μάλλον τώρα θα έλεγα ότι δεν είναι απλά διδακτική αλλά κάτι παραπάνω..... Συνέχισε ρε Μέγιστε να βάζεις τέτοιες ασκήσεις (όχι μόνο ανάλυσης)  για να ας φτιάχνεις .
Οι θεώρηση <<των ακουλουθιών deskos>> (έτσι θα τις λέω από δω και πέρα) είναι εκπληκτική......
Μετά απλά πρέπει να ξέρεις  να ολοκληρώνεις σωστά  ;)

Α με την ευκαιρία ....
Το τελευταίο ολοκλήρωμα $\frac{\pi}{3\sqrt 3}$  δεν κάνει??

Λοιπον , τα παρακατω μπορουσα να  τα γραψω σε π.μ αλλα καλο ειναι να τα ακουσουν ολοι.οι ανωνυμοι τελειωσαν.δεν υπαρχει κανενας ανωνυμος οπως βλεπετε. νομιζω οτι πειραχτηκες  λιγο απο το ποστ του Αρχιμηδη.αναφερομαι στο δευτερο ποστ σου Σιλουαν. ο Αρχιμηδης 6(  το πιο νεο  ενεργο μελος αν δεν κανω λαθος του φορουμ) λεει την αποψη του για μια ασκηση.ουτε βρισιες ουτε ειρωνιες.΄'δεν εδειξα ιδιαιτερο ενδιαφερον','δεν ειναι ΤΟΣΟ διδακτικη', πως αλλιως δηλαδη να το πει? .Σκεφτεται κατι, το λεει χωρις να κακολογισει.Καθενας ειναι ελευθερος να μιλαει ελευθερα ανεξαρτητως αν ειναι νεος ή παλιος.το λεω γιατι δεν νομιζω να εχει διαθεσει να προσβαλει τον Γιωργο. τα ειπα  ολα αυτα με το φοβο να παρεξηγηθω. Με εκτιμηση....

Λοιπόν Κώστα μάλλον με παρεξήγησες λίγο . ;)  .Δεν ήθελα να ειρωνευτώ κανέναν και ούτε κάτι τέτοιο .Εκτιμώ κάθε μέλος και δεν θέλω να αρχίσουν πάλι τα ίδια .Ίσως να μη σου άρεσε ο τρόπος που το έγραψα (μπορεί να υπερέβαλα λίγο ) αλλά εντάξει αυτά γίνονται .Εγω πάντως ζητώ συγγνώμη και αυτό ήταν όλο  ;)  Να είστε καλά και να συνεχίσουμε έτσι με μαθηματικά :lol:

Και εγω παιδια ζητω συγνωμη αν απο την πλευρα μου εαν εθειξα καποιον.Πραγματικα δεν ειχα τετοιο σκοπο.Ελπιζω το φορουμ να συνεχισει ετσι με τις ωραιες λυσεις που προτεινονται και χρηστες οι οποιοι τελευταια δεν εμφανιζονται στο φορουμ να στελνουν πιο συχνα τις λυσεις τους.Ετσι το φορουμ θα παρει μια νεα πνοη. ;)

λοιπον ισως ημουν και εγω λιγο υπεβολικος. δεν αμφιβαλλω οτι μπορει να σε παρεξειγησα.δεν τρεχει τιποτα οκ? αντε καλες ασκησεις και συγχαρητηρια στον Γιωργο Δεσκο για την  πολυ εξυπνη λυση του. :)  :)  :)

Εννοείτε ότι δεν τρέχει τίποτα μια παρέα είμαστε.... :D
Καλή συνέχεια με τις ασκήσεις .


This tournament will consist of a bracket style in which each round you will go head to head with another individual.  The entry limit will be 32 people.  In each round you will be paired with another person who you will compete with.  There will be two parts to every round, a no calculator part and a calculator part.  In the Final Four the rules will change.  Each round will consist of 10 problems and you will have 20 minutes to solve them.  You will have a week for each round.  You can them PM me the answers.  When you sign up, please tell me your FTW rating, so I can seed you accordingly.  Please join in this thread.  Only the first 32 will be allowed in so hurry!

I also need help
If you would like to
1. Grade
2. Review the test Before hand to check it
Please PM me.

I'll join. I can grade other peoples papers after I submit if you want.

me join.  i don't know my ftw rating, but you can check the leaderboard.

Sign-ups
1.Bogtro 1859
2. Ernie 1830
3. AIME15 2034
4. Issabella2296 1796

Also:

Using ftw ratings to seed isn't the best way... maybe use mc/amc/aime scores?

[quote="isabella2296"][quote="pythag011"]maybe use mc/amc/aime scores?[/quote]

What about people who didn't do MC/AMC/AIME this year? Or people who live in another country?[/quote]

Yes, but ftw ratings isn't that good of a way in my opinion...

Alternatively, create a "first round" thats only used to seed people?

Sign-ups
1.Bogtro 1859
2. Ernie 1830
3. AIME15 2034
4. Issabella2296 1748
5. Yaofan 1319

@pythag
Yes you have a point, but it is just easier if we use FTW rankings.  Anyway are you joining?

Dude, FTW ratings are 100% accurate as to picking people in descending level of intelligence :D

Mieh, pythag has a pretty good point.

I'll join. Rating of 1406 due to epic failage in the last few days  :(

i'll join (1635)

EDIT:my rating is now 1674

If you didn't begin now and wait to have $ 32$ participant you will see that they will reduce to $ 10$ participants without previous notice.

We will start on Monday.
I will put up Round 1 then.

These are the questions for the first part of Round 1.  You have 20 minutes to answer these ten questions without a calculator.  Please PM me the answers after you are done.

About a Week

Special thanks to Twin Prime Conjecture for help with this rounds problems.


floor function
combinatorics proposed
combinatorics

find the number of 0-1 sequences of length $ n$ such that before $ k$'th 1 there are at most $ k$ 0's (it holds for $ k\equal{}1,\dots,n$)

Notice that a 0-1 sequence of lenth $ q$ that satisfies the required condition can always be obtained from a sequence of length $ p$ (where $ p < q$) (which also satisfies the required condition) by attaching $ q \minus{} p$ zeroes.

ex. 0111 and 011100000000 are both valid. So the 0-1 sequence can have as many 0's at the right most end as possible.

So, now we need to determine the number of 0-1 valid sequences $ Lm$, of length $ m$ (for each $ m\in {1,2,..n}$) [u]that end with [b]1[/b][/u] and then add $ n \minus{} m$ 0's to make its lenght equal to n.

I am guessing that to determine the number of valid 0-1 sequences that end with 1, we need some kind of a recursive relation for $ Lm$.

I just figured out that there is no simple recursive relationship for number of 0-1 valid sequences of length m, Lm [u]that end with 1[/u].  :mad:  :mad:  :mad:

[b]primenumber[/b] can u please post your solution to this problem if you know it. Does any one else in here have the solution?

unfortunatelly no, it is an excercise and I can't solve it :(

Seems related to a [url=http://mathworld.wolfram.com/DyckPath.html]Dyck path[/url], where you start at $ (1,0)$, $ 0$ is going up, and $ 1$ is going right. Not exactly the same though.

If you write the $ 0$s as $ \minus{}1$s, the partial sums are paths that never go more than one unit below the $ x$-axis, which is almost the definition of a Dyck path except that in addition they're allowed to end anywhere above one unit below the $ x$-axis (and not just on the axis itself).

The terms of the series are: 1, 2, 3, 6, 10, 20, 35, 70, 126, 252, 462, 924, 1716 ....

The nth term is given by [b]C(n,floor(n/2))[/b]

This series can be found in OEIS at: [url]http://www.research.att.com/~njas/sequences/A001405[/url]

However, I think that it is very difficult to prove that the number of 0-1 sequences with the given condition is given by the above term. Now that we know the answer let's try to prove it.  :wink:

It's unclear to me which of the following two questions was intended:

1) Compute the number of sequences of 0s and 1s of length $ n$ such that for each $ i$, the number of 0s among the first $ i$ entries is no more than one larger than the number of 1s among the first $ i$ entries.

2) Compute the number of sequences of 0s and 1s of length $ n$ such that for each 1 in the sequence, the number of 0s that precede it is no more than one larger than the number of 1s that precede it.

In the first case, we have the following sequences: $ \varnothing; 0, 1; 01, 10, 11; 010, 011, 100, 101, 110, 111; \ldots$, and the number of these sequences of length $ n$ is $ \binom{n}{\lfloor n/2 \rfloor}$, as KRIS claimed.  We can show this as follows: first, note that there is a bijection between these sequences and sequences of length $ n \plus{} 1$ in which the number of 0s among the first $ i$ entries is no larger than the number of 1s among the first $ i$ entries.  (Just add a 1 at the beginning.)  Let the number of such sequences of length $ n$ be $ a_n$.  Then by keeping track of the first time we reach a point where we've seen the same number of 1s and 0s, we get $ a_{n \plus{} 1} \equal{} a_n \plus{} \sum_{i \equal{} 1}^{\lfloor n/2\rfloor} C_{i \minus{} 1} a_{n \plus{} 1 \minus{} 2i}$ where $ C_k$ is the $ k$th Catalan number.  Now we can finish it off easily with generating-functional methods, for example.

In the second case (which is what the words of the problem appear to actually describe), we have more sequences: $ \varnothing; 0, 1; 00, 01, 10, 11; 000, 010, 011, 100, 101, 110, 111; \ldots$.  In this case, as KRIS noted in an earlier post, we're just counting the objects from case 1, possibly with a bunch of 0s appended.  Thus the answer is $ b_n \equal{} \sum_{i \equal{} 0}^n a_i$ and is given by [url=http://www.research.att.com/~njas/sequences/A036256]A036256[/url] in the [url=http://www.research.att.com/~njas/sequences]OEIS[/url].


induction probability calculus derivative

9.	 You are in charge of a project to design a card-shuffling machine for a deck of n cards, initially ordered from 1 to n. The machine will repeatedly take the top card and re-insert it randomly into the deck, so that all positions for the inserted card are equally likely (including the top and bottom). The machine can keep track of where each insertion happens, so that at every point the current order of the cards is known. Your job is to tell the machine in advance when it should stop shuffling. In your instructions, you can refer to the current state of the deck (e.g. "Go until card 2 is on the bottom, then go until card n-1 is third from the top, then shuffle 5 more times, then stop"). You want to program the machine so that when it stops, all permutations of the deck will be equally likely. What instructions should you give to the machine? (Be sure to prove that these instructions really do lead to all permutations being equally likely.) With these instructions, how many insertions will it take, on average, until the machine stops?

My idea for this one was to have it shuffle the deck till card n reaches the top and then shuffle it in. But I couldn't find out how to [i]prove[/i] that all permutations are equally likely and I somehow ended up with n*(1+ 1/2 + 1/3 +1/4...+1/n, which I know is wrong. What were everyone else's solutions?

By the way, I'm curious to find how how to simplify the above equation.

I really liked this problem.
I believe that both your strategy and number of permutations are correct - the harmonic series cannot be simplified - I simply left it as $ n\sum_{i = 1}^{n}{\frac {1}{i}}$.
[hide="Part one solution outline"] I proved that the strategy worked by induction. Basically, right after a card is shuffled into the deck, it is equally likely to be anywhere (by definition). You can prove that if any given card is equally likely to be anywhere before any given shuffle, it will be equally likely to be anywhere after the shuffle no matter what. By induction, it follows that after a card is shuffled into the deck, it will always be equally likely to be anywhere; thus, once each card is shuffled into the deck at least once, each card will be equally likely to be in any position. Once the last card is shuffled into the deck, each card has been shuffled into the deck at least once as desired.[/hide]

For both parts of the question it might be worthwhile to induct directly on $ n$.

I realized that card 52 cannot be shuffled into the deck until it reaches the top. Thus, I got a harmonic series where the nth term was the average amount of turns it takes to place then nth card at any point below card 52, which I'm pretty sure is the same as what you guys are saying.

My final answer was ~236.

You misread the problem: there were not $ 52$ cards; there were $ n$ cards.

My solution to part (b) involved used calculus. Is there a way without calc?

Relevant text:

blahblah....let k be the number of cards below card n so that there are k places card n could be inserted below card n....blahblah

The weighted average of the number of steps needed until another card is inserted below card n will be the summation of the probability that s steps are needed multiplied by s, which will be the infinite series k/n+2((n-k)/n)k/n+3(k/n)((n-k)/n)^2k/n+ .....i(k/n)((n-k)/n)^(i-1)  The MacLaurin series for 1/(1-x)=1+x+x^2+...   for abs(x)<1  . Taking the derivative of both sides of the equation, multiplying by k/n  , and setting x to (n-k)/n  , we see that  k/n+2((n-k)/n)k/n+3(k/n)((n-k)/n)^2k/n+ .....i(k/n)((n-k)/n)^(i-1)=(k/n)/(1-(n-k)/n)^2=n/k.

Here are two other ways to evaluate such series':

[hide="Method One"]
Let $ S=x+2x^2+3x^3+4x^4+...$
Then $ xS=x^2+2x^3+3x^4+...$
Subtracting the first equation from the second yields:
$ S(x-1)=x+x^2+x^3+x^4+...$
Either repeat the process again (multiply the equation directly above by $ x$ and subtract the new equation from the old one), or recall the right-hand side is equal to $ \frac{x}{x-1}$ for $ \abs{x}<1$.[/hide]

[hide="Method Two"]
Let $ S=x+2x^2+3x^3+4x^4+...$
Let $ A=x+x^2+x^3+x^4+...$
Then $ A^2=(x+x^2+x^3+x^4+...)^2$
Expanding the right-hand side, we find that it is actually equal to $ xS$
Dividing both sides by $ x$, we find that $ S=\frac{A^2}{x}$, and, recalling that $ A$ is equal to $ \frac{x}{x-1}$ for $ \abs{x}<1$, we find our desired closed form.[/hide]


Let $ a,b,c$ be non-negative real numbers such that $ a\plus{}b\plus{}c\equal{}3$ and $ k>0$. Prove that

$ a^kb\plus{}b^kc\plus{}c^ka\le 3^{k\plus{}1}.\frac{k^k}{(k\plus{}1)^{k\plus{}1}}$

[quote="dduclam"]Let $ a,b,c$ be non-negative real numbers such that $ a \plus{} b \plus{} c \equal{} 3$ and $ k > 0$. Prove that

$ a^kb \plus{} b^kc \plus{} c^ka\le 3^{k \plus{} 1}.\frac {k^k}{(k \plus{} 1)^{k \plus{} 1}}$[/quote]

I think it must be:

$ a^kb \plus{} b^kc \plus{} c^ka\le\ max(3; 3^{k \plus{} 1}.\frac {k^k}{(k \plus{} 1)^{k \plus{} 1}})$

And It 's very hard at least for me.

[quote="dduclam"]Let $ a,b,c$ be non-negative real numbers such that $ a \plus{} b \plus{} c \equal{} 3$ and $ k > 0$. Prove that

$ a^kb \plus{} b^kc \plus{} c^ka\le 3^{k \plus{} 1}.\frac {k^k}{(k \plus{} 1)^{k \plus{} 1}}$[/quote]
I proved that for all $ k>0$, then
\[ a^kb\plus{}b^kc\plus{}c^ka \le \max\left\{ 3,\frac{3^{k\plus{}1}k^k}{(k\plus{}1)^{k\plus{}1}}\right\}\]
And my hint to a nice solution is
\[ a^mb\plus{}b^mc\plus{}c^ma \ge ab^m\plus{}bc^m\plus{}ca^m\]
for all $ a \ge b \ge c \ge 0, m \ge 1$
:)

[quote="can_hang2007"][quote="can_hang2007"][quote="dduclam"]Let $ a,b,c$ be non-negative real numbers such that $ a \plus{} b \plus{} c \equal{} 3$ and $ k > 0$. Prove that

$ a^kb \plus{} b^kc \plus{} c^ka\le 3^{k \plus{} 1}.\frac {k^k}{(k \plus{} 1)^{k \plus{} 1}}$[/quote]
I proved that for all $ k > 0$, then
\[ a^kb \plus{} b^kc \plus{} c^ka \le \max\left\{ 3,\frac {3^{k \plus{} 1}k^k}{(k \plus{} 1)^{k \plus{} 1}}\right\}
\]
And my hint to a nice solution is
\[ a^mb \plus{} b^mc \plus{} c^ma \ge ab^m \plus{} bc^m \plus{} ca^m
\]
for all $ a \ge b \ge c \ge 0, m \ge 1$
:)[/quote]
@Dduclam: My friend, as the title of this topic, I think you have a nice solution for it. So I would like to see your proof before I post my full proof. :)[/quote]

Ok, my friend :) I will post my solution tomorrow. Now I too busy  :(

[quote="can_hang2007"][quote="can_hang2007"][quote="dduclam"]Let $ a,b,c$ be non-negative real numbers such that $ a \plus{} b \plus{} c \equal{} 3$ and $ k > 0$. Prove that

$ a^kb \plus{} b^kc \plus{} c^ka\le 3^{k \plus{} 1}.\frac {k^k}{(k \plus{} 1)^{k \plus{} 1}}$[/quote]
I proved that for all $ k > 0$, then
\[ a^kb \plus{} b^kc \plus{} c^ka \le \max\left\{ 3,\frac {3^{k \plus{} 1}k^k}{(k \plus{} 1)^{k \plus{} 1}}\right\}
\]
And my hint to a nice solution is
\[ a^mb \plus{} b^mc \plus{} c^ma \ge ab^m \plus{} bc^m \plus{} ca^m
\]
for all $ a \ge b \ge c \ge 0, m \ge 1$
:)[/quote]
@Dduclam: My friend, as the title of this topic, I think you have a nice solution for it. So I would like to see your proof before I post my full proof. :)[/quote]

My proof.

[quote="dduclam"][/quote][quote="can_hang2007"][quote="can_hang2007"][quote="dduclam"]Let $ a,b,c$ be non-negative real numbers such that $ a + b + c = 3$ and $ k > 0$. Prove that

$ a^kb + b^kc + c^ka\le 3^{k + 1}.\frac {k^k}{(k + 1)^{k + 1}}$[/quote]
I proved that for all $ k > 0$, then
\[ a^kb + b^kc + c^ka \le \max\left\{ 3,\frac {3^{k + 1}k^k}{(k + 1)^{k + 1}}\right\}
\]
And my hint to a nice solution is
\[ a^mb + b^mc + c^ma \ge ab^m + bc^m + ca^m
\]
for all $ a \ge b \ge c \ge 0, m \ge 1$
:)[/quote]
@Dduclam: My friend, as the title of this topic, I think you have a nice solution for it. So I would like to see your proof before I post my full proof. :)[/quote][quote="dduclam"]

My proof.[/quote]
But it is not for $ k>0$, my friend. :(

[quote="can_hang2007"][/quote][quote="dduclam"][/quote][quote="can_hang2007"][quote="can_hang2007"][quote="dduclam"]Let $ a,b,c$ be non-negative real numbers such that $ a + b + c = 3$ and $ k > 0$. Prove that

$ a^kb + b^kc + c^ka\le 3^{k + 1}.\frac {k^k}{(k + 1)^{k + 1}}$[/quote]
I proved that for all $ k > 0$, then
\[ a^kb + b^kc + c^ka \le \max\left\{ 3,\frac {3^{k + 1}k^k}{(k + 1)^{k + 1}}\right\}
\]
And my hint to a nice solution is
\[ a^mb + b^mc + c^ma \ge ab^m + bc^m + ca^m
\]
for all $ a \ge b \ge c \ge 0, m \ge 1$
:)[/quote]
@Dduclam: My friend, as the title of this topic, I think you have a nice solution for it. So I would like to see your proof before I post my full proof. :)[/quote][quote="dduclam"]

My proof.[/quote][quote="can_hang2007"]
But it is not for $ k > 0$, my friend. :([/quote]

Yes, my friend. I will try to complete it, but now I have not times :(  I think some body will help me :)  Can you ?

[quote="dduclam"][/quote][quote="can_hang2007"][/quote][quote="dduclam"][/quote][quote="can_hang2007"][quote="can_hang2007"][quote="dduclam"]Let $ a,b,c$ be non-negative real numbers such that $ a + b + c = 3$ and $ k > 0$. Prove that

$ a^kb + b^kc + c^ka\le 3^{k + 1}.\frac {k^k}{(k + 1)^{k + 1}}$[/quote]
I proved that for all $ k > 0$, then
\[ a^kb + b^kc + c^ka \le \max\left\{ 3,\frac {3^{k + 1}k^k}{(k + 1)^{k + 1}}\right\}
\]
And my hint to a nice solution is
\[ a^mb + b^mc + c^ma \ge ab^m + bc^m + ca^m
\]
for all $ a \ge b \ge c \ge 0, m \ge 1$
:)[/quote]
@Dduclam: My friend, as the title of this topic, I think you have a nice solution for it. So I would like to see your proof before I post my full proof. :)[/quote][quote="dduclam"]

My proof.[/quote][quote="can_hang2007"]
But it is not for $ k > 0$, my friend. :([/quote][quote="dduclam"]

Yes, my friend. I will try to complete it, but now I have not times :(  I think some body will help me :)  Can you ?[/quote]
I have a nice solution for it, my friend. But I can't post it here, sorry. Wait for my new books, you will see the my solution there. :)


function
complex analysis
complex numbers
real analysis
real analysis unsolved

Every complex number is a zero of some entire function all of whose coefficients (of the power series) are rational.

Let $a$ be an arbitrary non-zero complex number. First, we choose $a_{1}$ so as to make the coefficient of $z$ in $\left(1-\frac{z}{a}\right)\left(1-\frac z{a_{1}}\right)$ rational. Then we choose $a_{2}$ that makes the coefficient of $z^{2}$ in $\left(1-\frac{z}{a}\right)\left(1-\frac z{a_{1}}\right)\left(1-\frac{z^{2}}{a_{2}}\right)$ rational, and so on. Our entire function will be $\left(1-\frac{z}{a}\right)\prod_{n\ge 1}\left(1-\frac{z^{n}}{a_{n}}\right)$. If the $a_{n}$ are chosen such that $(|a_{n}|)_{n}$ increases fast enough, this will indeed be an entire function.

More generally:

(i) every discrete sequence of complex numbers occurs among the zeros of some nonzero entire function with coefficients in $\mathbb{Q}$;

(ii) for every discrete sequence complex numbers there is an entire function with coefficients in $\mathbb{Q}(i)$ whose zeros are precisely the members of the corresponding sequence (and with corresponding multiplicities counted).


1. I've got about 12 files I'd like to email in.  Is it OK if I put them into a .tar archive, or should I just attach all 12 the one email?

2. THe directions say to make sure your username and USAMTS ID are on each page.  is it good enough for me to put that info in the filenames, or do I have to actually put it in my latex source code? (i.e. so that it appears on the pdf)

1. Attach 'em all or ZIP 'em, your call.

2. Definitely get it on every page of the output - that's what the graders look at; they won't see your filenames.  You can use the header or footer in LaTeX to put that on every page.

I have 10 files, and I think someone already said (I think it was DPatrick..?) that I could email it in two emails since yahoo only allows 5 per email, and just label the emails 1 of 2 and 2 of 2.


hello!
don't want to be impatient, but when are we supposed to recieve the results of the 4th round?
alex


calculus
integration
logarithms
real analysis
real analysis unsolved

$\sum_{n=1}^{\infty}\frac{1}{(n)^{n}}$.
What will be equivalent of this sum?
 :wink:

Extremal, what do you mean by "equivalent of this sum"?

I mean find this:
$\sum_{n=1}^{\infty}\frac{1}{(n)^{n}}\approx ??$.
and else : I note that above integral $x^{x}$ and $x^{-x}$ was posted here and solved by sos440
[quote="sos440"]Expand

\begin{eqnarray*}\int_{0}^{1}x^{-x}\, dx & = & \int_{0}^{1}e^{-x \ln x}\, dx\\ & = & \int_{0}^{1}\sum_{n=0}^{\infty}\frac{(-x \ln x)^{n}}{n!}\, dx\\ & = & \sum_{n=0}^{\infty}\int_{0}^{1}\frac{(-x \ln x)^{n}}{n!}\, dx \end{eqnarray*}

Then use the substitution $t =-(n+1)\ln x$ for $n \geq 0$. Then $dx =-\frac{e^{-t/(n+1)}}{n+1}dt$. Then

\begin{eqnarray*}\int_{0}^{1}\frac{(-x \ln x)^{n}}{n!}\, dx & = & \int_{0}^{\infty}\frac{t^{n}e^{-t}}{n!(n+1)^{n+1}}\, dt\\ & = & \frac{\Gamma(n+1)}{n!(n+1)^{n+1}}\\ & = & \frac{1}{(n+1)^{n+1}}\end{eqnarray*}

We have

\begin{eqnarray*}\int_{0}^{1}x^{-x}\, dx & = & \sum_{n=0}^{\infty}\frac{1}{(n+1)^{n+1}}\\ & = & \sum_{n=1}^{\infty}\frac{1}{n^{n}}\end{eqnarray*}[/quote]


function
geometry
geometric transformation
real analysis
real analysis solved

hi all...
how to show the sup norm of fn(x)=xne^(-nx) (for all x>=0 and n>=1) is greater than zero?????

Let me see whether I can translate your question into something undestandable. Do you want to show that the $L^\infty(0,+\infty)$-norms of the functions $f_n(x)=nxe^{-nx}$ do not tend to $0$, i.e., that $f_n$ do not converge to $0$ uniformly on $(0,+\infty)$? (I tried to extrapolate from your post title and the content of your post, which is, mildly speaking, somewhat perplexing. I apologize if my translation is incorrect).

Then just note that $f_n(1/n)=e^{-1}>0$ does not depend on $n$  :) .


abstract algebra
superior algebra
superior algebra unsolved

Let $ G$ be a finite group and $ T$ be an automorphism which $ T(x)\equal{}x^{\minus{}1}$ for at least $ \frac 34$ of elements of $ G$,Prove $ \forall x \in G\ : T(x)\equal{}x^{\minus{}1}$ and $ G$ as an abelian group.

That's not true, you need more than $ \frac 34$: take $ D_4$, the symmetries of a square; the identity is an automorphism of the required type.
But see http://www.mathlinks.ro/viewtopic.php?t=27046 .


function
geometry
3D geometry
calculus
calculus computations

Describe the set of points whose distance from the x-axis equals the distance from the yz-plane:
A cylinder opening along the z-axis.
A cone opening along the y-axis.
A cone opening along the x-axis.
A cylinder opening along the x-axis.
A cylinder opening along the y-axis.
A line parallel to the z-axis.
A line parallel to the y-axis.
A line parallel to the x-axis.
A cone opening along the z-axis.

Which one is true? What is the equation?

It's a cone opening along the X-axis but I have no clue as far as its equation is concerned.


Galois Theory
number theory
superior algebra
superior algebra unsolved

Let $G$ be a finite group. Because $S_{n}$ is realizable over $\mathbb{Q}$, there are number fields $L,K$ such that $L/K$ has Galois group isomorphic to $G$.

Is it possible to have $L/K$ unramified?

It is true, as can be seen by making a suitable base change: We are given $L,K, G$ with  $\textrm{Gal}(L/K)=G.$ One can then find a finite extension $K_{1}/K$ which is ramified enough to absorb all the decomposition groups of $G$ at primes where the extension is ramified, and such that $L,K_{1}$ are linearly disjoint over $K.$ The extension $LK_{1}/K$ has the properties you want.

It should be possible to construct the extension $K_{1}/K$ with basic knowledge of algebraic number theory; such constructions are quite well known to (algebraic) number theorists. The ones I know of use some form of class field theory, and with some looking around I might even be able to produce a theorem/proposition from some book/article that's relevant.


[quote="quykhtn-qa1"][quote="duythuc_lqd"]Let $ a,b,c\in [ \minus{} 1,1],ab \plus{} bc \plus{} ca \equal{} 1$.Prove that:$ \sum_{cyc}\frac {a^{2} \plus{} b^{2}}{(1 \minus{} a^{2})(1 \minus{} b^{2})}\geq \frac {9}{2}$[/quote]
We can assume $ a,b,c > 0$[/quote]
why??I don't understand.You can explain for me.

[quote="quykhtn-qa1"][quote="duythuc_lqd"]Let $ a,b,c\in [ - 1,1],ab + bc + ca = 1$.Prove that:$ \sum_{sym}\frac {a^{2} + b^{2}}{(1 - a^{2})(1 - b^{2})}\geq \frac {9}{2}$[/quote]
We can assume $ a,b,c > 0$
Then,we have : $ (1 - a^2)(1 - b^2) \le (1 - ab)^2 = c^2(a + b)^2 \le 2c^2(a^2 + b^2)$
So $ LHS \ge \frac {1}{2}(\frac {1}{a^2} + \frac {1}{b^2} + \frac {1}{c^2}) \ge \frac {1}{2}({\frac {1}{ab} + \frac {1}{bc} + \frac {1}{ca}) \ge RHS}$
Then,we have Q.E.D :)[/quote]
Wrong!! $ a,b,c$ are not positive yet. You cannot use the [b]AM-GM[/b] here:
[quote="duythuc_lqd"]$ \frac {1}{2}({\frac {1}{ab} + \frac {1}{bc} + \frac {1}{ca}) \ge RHS}$
Then,we have Q.E.D :)[/quote].
Check it again,please.

[quote="duythuc_lqd"]Let $ a,b,c\in [ \minus{} 1,1],ab \plus{} bc \plus{} ca \equal{} 1$.Prove that:$ \sum_{cyc}\frac {a^{2} \plus{} b^{2}}{(1 \minus{} a^{2})(1 \minus{} b^{2})}\geq \frac {9}{2}$[/quote]
We can solve this problem by trigonometric method.Let x=tan(A/2),y=tan(B/2),z=tan(C/2)

[quote="SUPERMAN2"][quote="duythuc_lqd"]Let $ a,b,c\in [ \minus{} 1,1],ab \plus{} bc \plus{} ca \equal{} 1$.Prove that:$ \sum_{cyc}\frac {a^{2} \plus{} b^{2}}{(1 \minus{} a^{2})(1 \minus{} b^{2})}\geq \frac {9}{2}$[/quote]
We can solve this problem by trigonometric method.Let $ x \equal{} tan\frac {A}{2}$, $ y \equal{} tan\frac {B}{2}$, $ z \equal{} tan\frac {C}{2}$[/quote]
Be cause of the fact that $ a,b,c\in [ \minus{} 1,1]$. Therefore, $ A,B,C$ are not 3 angles of a triangle. Then here come the difficulty (to me :D ). However, can you post your solution [b]Superman[/b]? I am curious.


analytic geometry
graphing lines
slope
probability and stats

How do you show that the simple regression line of y against x coincides with the simple regression line of x against y if and only if r^2 = 1.

Here r is the sample correlation coefficient between x and y.

If you normalize to standard deviation coordinates, the slopes of the two lines are $ r$ and $ \frac1r$, and both pass through the origin.

An expanded-out version of what jmerry just said is in a long post I made [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=299116]here[/url].

Thanks for the suggestion.

I am able to explain why the lines coincide with each other when r^2 = 1 and the intuition behind it, but understanding a derivation for this is very different.

I looked at the equations posted and understand where and how to place them.

Could you walk me through the derivation?

Thanks for your help.

Can anyone elaborate on the steps. I am having a hard time starting. I don't really understand how you are supposed to use the equations and the r^2 to show that they coincide.

Thanks.

the regression equation of y on x is
Y = mean(y) + r s(y)/s(x) (x-mean(x))
where s(x) is the sd of x.
the regression equation of x on y is
X=mean(x) +r s(x)/s(y)(y-mean(y))
which can be rearranged as
y= mean(y) +  s(y)/r s(x) (X- mean(x))

so the two regression lines coincide iff
their slopes are equal which leads to the result
directly...

think this will help you :)

Thanks for the insight

I will definitely work through your suggested steps.


I have a doubt on this....
If there are 2 children in a family,what is the probabilty of having a girl?

[hide="Solution"]
There are four possibilities:  boy-boy, boy-girl, girl-boy, and girl-girl.  Of these possibilities, 3 of them have a girl, and 1 of them does not, so the answer is $\frac{3}{4}$ chance of having at least one girl.[/hide]

Yeah, I agree with t0rajir0u, unless of course you already know something about them (such as they have a boy), or the probability of getting a girl is different from that of a boy.

[hide]There are 4 possible outcomes:
BB
BG
GB
GG
Of those 4, three include a girl, making the probability of picking a girl $\frac34$.[/hide]

[hide]
The probability of getting 2 boys is $\left(\frac{1}{2}\right)^2=\frac{1}{4}$, so the probability of getting at least one girl is $1-\frac{1}{4}=\frac{3}{4}$
[/hide]

[quote="shadysaysurspammed"]I was wondering whether the boy-girl and girl-boy case was the same...it might be the same,won't it?[/quote]

It isn't the same

[quote="b-flat"]For BG and GB, they are different because look at it as the first person is younger than the second person.  Then, they are different, but BB and GG won't matter if the first person is younger than the second person.[/quote]

Isn't it [b]possible[/b] for them to be of the same age :?

shadysaysurspammed's argument is actually one of the confusing points of the basic of probability. It [i]seems[/i] that B-G and G-B are the same, but they aren't the same because we consider each baby as distinguishable from the other.

Regardless of whether they're "the same" or not, the fact remains that having a boy and a girl occurs with probability $\frac{1}{2}$.  So if you really wanted to, you could consider that one case - it just doesn't have the same probability as boy-boy and girl-girl (which have probability $\frac{1}{4}$ each).

Look at it this way:  Since the two babies couldn't possibly come out of the mother at the same time, "the first baby" is distinguishable from "the second baby," so boy-girl (first boy, second girl) is a different case from girl-boy (first girl, second boy)  :D


calculus
ratio
geometry
similar triangles
calculus computations

Here is  two problems from my calculus textbook which I can't figure it out.  Can anyone help me out and show me how to solve these two problem?  Thanks!

1. A street light is at the top of a 20 ft tall pole. A woman 6 ft tall walks away from the pole with a speed of 8 ft/sec along a straight path. How fast is the tip of her shadow moving when she is 30 ft from the base of the pole?

2. Water is leaking out of an inverted conical tank at a rate of 10600.0  cubic centimeters per min at the same time that water is being pumped into the tank at a constant rate. The tank has height 6.0 meters and the diameter at the top is 7.0 meters. If the water level is rising at a rate of 27.0 centimeters per minute when the height of the water is 2.0 meters, find the rate at which water is being pumped into the tank in cubic centimeters per minute.

[hide="1"]
Let $a$ be the distance between the woman and the base of the pole, and let $b$ be the distance between the tip of the shadow and the base of the pole.

Since we have a pair of similar triangles, we have $\frac{a-b}{6}=\frac{a}{20} \Longleftrightarrow 14a-20b=0$

Then we use implicit differentiation.
$14\frac{da}{dt}-20\frac{db}{dt}=0$

We know that $\frac{db}{dt}=8$, so $\frac{da}{dt}=\frac{80}{7} (ft/sec)$

One thing I'm worried about is that I didn't use the number $30$ (the current distance between the woman and the base of the pole).
But I guess it doesn't matter anyway because we are dealing with ratios, not lengths.
[/hide]

Thanks, frt. That is correct by the way.  But what about #2,  Anyone?

I did this very quickly, so I could be wrong:

[hide="Problem 2"]$\frac{dV}{dt}=\frac{49\pi h^{2}}{144}\frac{dh}{dt}=k-10600$

...using the values you were given:

$\frac{dV}{dt}=\frac{49}{144}\left(\pi\right)\left(2^2\right)\left(27\right)=k-10600$

$k=10600+\frac{147\pi}{4}\approx 10715.45\,\frac{\textrm{cc}}{\textrm{min}}$[/hide]


Let $ I$ be an interval and let $ f: I \to \mathbb{R}$ be a continuous function. Prove that $ f$ is one to one iff $ f$ is strictly monotonic.

What happens if $ f$ is not continuous? What happens if $ I$ is not an interval. It would be false, but what are some counterexamples?

You probably have to invoke $ \epsilon\minus{}\delta$ definition of continuity?

[quote="ragingbullrocky"]What happens if $ f$ is not continuous?[/quote]

Let $ I\equal{}(0,2).$ Let $ f(x)\equal{}\begin{cases}x\plus{}1&0< x\le 1\\ x\minus{}1&1<x<2\end{cases}$

[quote]What happens if $ I$ is not an interval?[/quote]

Let $ I\equal{}(0,1)\cup(1,2).$ Let $ f(x)\equal{}\begin{cases}x\plus{}1&0< x < 1\\ x\minus{}1&1<x<2\end{cases}$

Note in this second case that $ f$ [b]is[/b] continuous on $ I.$ (Just not uniformly continuous.)

Many, many different examples are possible, based on a variety of ideas.

So if $ f$ is one to one, then $ f(x) \equal{} f(y) \implies x \equal{} y$ for $ x,y \in X$. Now since $ f$ is continuous, $ |x \minus{} y| < \delta \implies |f(x) \minus{} f(y)| < \epsilon$ for some $ y \in I$ (this holds for any $ \epsilon > 0$ and for some $ \delta > 0$). How would we deduce that $ f(x) < f(y)$ or $ f(x) > f(y)$ if $ x < y$? Conversely, if $ f(x) < f(y)$ and $ f(y) < f(x)$ for $ x < y$ and $ x,y \in X$, then $ f(x) \equal{} f(y) \implies x \equal{} y$?

Is this correct?

[quote="ragingbullrocky"]Is this correct?[/quote]

No. Certainly you don't have a proof of one-to-one implying monotone.

A correct proof needs to invoke the Intermediate Value Theorem somewhere.

Thanks. But the converse is correct? Well the IVT is for a closed interval $ [a,b]$. In this problem, $ I$ can be any interval. More specifically the IVT in my book is: Suppose we have a continuous function $ f: [a,b] \to \mathbb{R}$. Suppose that $ \gamma$ is between $ f(a)$ and $ f(b)$. Let $ c \equal{} \sup \{x \in [a,b]: f(x) \leq \gamma \}$. Then $ f(c) \equal{} \gamma$.

Suppose $ I$ is an interval in $ \mathbb{R}$ and let $ f: I \to \mathbb{R}$ be a continuous function. Prove that $ f$ is injective $ \Longleftrightarrow f$ is strictly monotonic.

$ \Rightarrow$ direction: Let $ a,b \in I$. Suppose $ a \neq b$. Then $ f(a) \neq f(b)$. So $ a<b$ or $ a > b$. And $ f(a) < f(b)$ or $ f(a) > f(b)$. From here, would you use the intermediate value theorem?

$ \Leftarrow$ direction: Suppose $ f$ is strictly monotonic. Then $ x \neq y \implies f(x) \neq f(y)$.

Is this correct?

$ (\Rightarrow)$ direction: Let us suppose $ f$ is not strictly monotonic. We are going to show that then $ f$ is also not one to one mapping.

If $ f$ is not strictly monotonic and there does not exist a pair $ x_{1}\neq x_{2}$, such that $ f(x_{1})=f(x_{2})$ (because that would instantly exclude $ f$ being an injection), then there exists a triple $ x_{1}<x_{2}<x_{3}$, for which either $ f(x_{1})< f(x_{2})$ and $ f(x_{2})>f(x_{3})$ or $ f(x_{1})>f(x_{2})$ and $ f(x_{2})<f(x_{3})$.

WLOG we can study the first option $ (x_{1}<x_{2}<x_{3}$, $ f(x_{1})< f(x_{2}) $ and $  f(x_{2})>f(x_{3}) )$. Since $ f$ is continuous $ f([x_{1},x_{2}]) \subset [f(x_{1}),f(x_{2})]$ and $ f([x_{2},x_{3}]) \subset [f(x_{3}),f(x_{2})]$.

Now since $ [ f(x_{1}),f(x_{2}) ) \cap [f(x_{3}),f(x_{2}) ) \neq \emptyset$, there exist $ x^{\ast} \in [x_{1},x_{2})$ and $ x^{\circ} \in (x_{2},x_{3}]$, for which $ f(x^{\ast})=f(x^{\circ})$ and $ x^{\ast}\neq x^{\circ}$.

Contradiction with the assumption, that there does not exist such a pair. Therefore $ f$ is not an injection.

[b]ragingbullrocky wrote:[/b]
[quote]More specifically the IVT in my book is: Suppose we have a continuous function $ f$. Suppose that $ \gamma$ is between $ f(a)$ and $ f(b)$. Let $ c \equal{} \sup \{x \in [a,b]: f(x) \leq \gamma \}$. Then $ f(c)\equal{}\gamma$.[/quote]

If $ I$ is a closed interval, we can directly make the procedure, which will be described, but if $ I$ is an open interval, then we first choose a closed interval within $ I$.

Let us have $ [a,b]\subset I$. Let $ f$ be continuous and injective function on $ I$.
We will show that $ f$ is strictly monotonic on $ I$. WLOG we can say $ f(a)<f(b)$.

If $ a<b$ and $ f(a)<f(b)$ and $ f$ is strictly monotonic, then for any pair $ c_{1},c_{2} \in [a,b]$, $ c_{1}<c_{2}$ we must have $ f(c_{1})<f(c_{2})$.

Now we choose $ c_{1}, c_{2} \in [a,b], c_{1}<c_{2}.$ According to IVT and injectivity of $ f$ we have $ c_{1}\equal{} \sup \{x \in [a,b]: f(x) \leq f(c_{1}) \}, c_{2}\equal{} \sup \{x \in [a,b]: f(x) \leq f(c_{2}) \}$.

We want to show that $ f(c_{1})<f(c_{2})$. We can assume contrary - $ f(c_{2})<f(c_{1})$. But then $ c_{2} \in \{x \in [a,b]: f(x) \leq f(c_{1}) \}$, hence $ c_{1} \neq \sup \{x \in [a,b]: f(x) \leq f(c_{1}) \}$. Contradiction. Hence $ f$ is strictly monotonic on $ [a,b]$.

Now if $ I$ is an open interval, we can come as close as we want to $ \inf I$ with $ a$ and to $ \sup I$ with $ b$. Therefore $ f$ is strictly monotonic on $ I$.

How do the above functions fail to satisfy the theorem?

You probably mean
[quote]
Let $ I\equal{}(0,2)$. $ f(x)\equal{}\begin{cases}x\plus{}1&0< x\le 1\\ x\minus{}1&1<x<2\end{cases}$
[/quote]
and
[quote]Let $ I\equal{}(0,1)\cup(1,2)$. $ f(x)\equal{}\begin{cases}x\plus{}1&0< x < 1\\ x\minus{}1&1<x<2\end{cases}$[/quote]

The first one is not continuous, hence you cannot apply IVT for it. But on the other you have the union of intervals, hence the function is strictly monotonic on each of the interval, but not on their union.

Let we have $ I\equal{}(0,1)\cup(1,2)$ and $ f(x)\equal{}x$. If we apply IVT for $ [0,2]$ and $ \gamma\equal{}1$, then we get
$ 1\equal{} \sup \{x \in [0,2]: f(x) \leq 1 \}$. But $ f(1)$ is not even defined. Therefore the function must be defined on all $ x\in[a,b]$ for IVT to hold.

Another example:

$ I\equal{}\mathbb{Z}.$ $ f(x)\equal{}\begin{cases}2k\plus{}1& x \equal{}2k \\ 2k & x\equal{}2k\plus{}1\end{cases}$

The function is continuous and injective, but not strictly monotonic.


Vieta
quadratics
function
geometry
3D geometry
algebra proposed
algebra

Solve in $\mathbb{R}$ the following equation:

                                \[ x^{3}-45x+110=0.  \]

This is a classic cubic equation, first solved by Cardano.

Let $x = t-u, \ -45 = t^3 - u^3, \ and \ 110 = -3tu$.

Use those relations to find t and u, and divide the equation by the resulting x to obtain the other two roots. Choose the real roots.

Can you please write a full solution? ;)

[quote="silouan"]It has not real solutions.[/quote]
That would be very curious ;)

[quote="ZetaX'']
That would be very curious ;)[/quote]

What do you mean ?

ANY cubic has at least one real solution.

[quote="silouan"]It has not real solutions.[/quote]

You are wrong. There are three distinct real solutions, which are one negative root and two positive roots.

I see nobody's actually posted a full solution yet.   :)

[hide="Solution"]
Make Vieta's substitution, $x = w + \frac{45}{3w}$, which reduces to the quadratic $(w^3)^2 + 110(w^3) + 15^3 = 0$, which has solutions $w^3 = \frac{-110 \pm \sqrt{110^2 - 4(15^3)} }{2} = -55 \pm 5 \sqrt{11^2 - 135} = -55 \pm 5i \sqrt{14}$.  (I think this is what the person who said it didn't have real solutions meant, but I think he did not realize that $x = w + \frac{15}{w}$ could be real even if $w^3$ was not :D )

So, we have $x = \sqrt[3]{-55 \pm 5i \sqrt{14} } + \frac{15}{ \sqrt[3]{-55 \pm 5i \sqrt{14} } }$, which gives six solutions, but in equal triplets, so there are three solutions (all of them are real, but actually calculating them explicitly with Euler's will only bring us back to our original cubic  :D )

By calculator, the solutions are $x \approx -7.69975, 4.58156, 3.11819$.
[/hide]

[quote="silouan"]It has not real solutions.[/quote]

   By using the property of cubic function that's continuos in R and also using limitation u can prove that all the cubic functions always have real zeros in R...
For example: $f(x)=x^3+ax^2+bx+c$ for x tend to negative infinite f(x) also tend to negative infinite...and similarly for x to possitive inf..But f(x) is continuos so the graph of f(x) will cut Ox at at least 1 zeros..
  There can be 2 possible case for cubic equations in R: They are:(1) 1 zeros and (2) 3 zeros..can not be 0 or 2...

You are right Arne i forget that any cube has at least a real root but i get in the trap of this.

[quote="t0rajir0u"]  (I think this is what the person who said it didn't have real solutions meant, but I think he did not realize that $x = w + \frac{15}{w}$ could be real even if $w^3$ was not :D )

[/quote]


geometry
geometric transformation
rotation
counting
distinguishability
function

You have 4 colors, red white blue and green. How many ways can you color the sides of a 4 sided-square?

Sorry I was unclear. red and white are different colors, you can't combine them.

Oops. I forgot a space. Maybe this will be easier to understand.

Wait. Is white-blue a color?

no, red white and blue are all different. YOu cant combine any of them.

Sorry, my English is bad. Is green a color too? Or does the comma imply that blue and green are the same color?

They are all different. Sorry, I am bad at answering questions.

Does distinguishable mean "able to tell apart"?

oh. My english is bad too. I am from China so yeah...

Rotations are undistinguishable.

oh yea. :blush:

What is the difference between the prefixes IN and UN?

Is it some weird English grammar rule?

That's a good question. I bet it's sarcastic AND rhetorical (I just learned TWO American words!)

well the language is english. and that was neither.

Oh. :wallbash:  :wallbash_red:  :stretcher:

Thanks for the enragement. I hope I used the right term. :maybe:

I believe you mean to use encouragement. Enragement is making you angry, like this post might.

What;s to be sorry about, you are learning.


A parallel-plate capacitor with plates of area A and plate separation d has the region
between the plates filled with two dielectric materials (e1,e2) as shown in Figure.Determine the capacitance.

[hide="Hint"]Cut the capacitor into small parts of length $ dx$ perpendicular to $ L$ and treat each small part as composite of two capacitors.[/hide]
[hide="Answer"]$ C\equal{}\frac{\epsilon_0 A}{d}\frac{e_1 e_2 \ln\frac{e_1}{e_2}}{e_1\minus{}e_2}$[/hide]

Can I see,your solution?I need to know it until tomorrow...

Consider small length $ \delta x$ at distance $ x$ from one end. At distance $ x$, thickness of $ e_1$ part is $ d\frac{x}{l}$ and that of $ e_2$ part is $ d(1\minus{}\frac{x}{l})$.
Capacitance of each part is $ C_1\equal{}\epsilon_0 e_1 w\delta x\frac{l}{xd}$ and $ C_2\equal{}\epsilon_0 e_2 w\delta x\frac{l}{(l\minus{}x)d}$
Equivalent Capacitance $ C\equal{}\left( \frac{1}{C_1}\plus{}\frac{1}{C_2}\right) ^{\minus{}1}\equal{}\epsilon_0 \frac{we_1e_2\delta x}{e_2\plus{}\frac{xd}{l}(e_2\minus{}e_1)}$
Because all the small capacitors $ \delta x$ are in parellel, we can integrate them.
$ \int_0^l \epsilon_0 \frac{we_1e_2\delta x}{e_2\plus{}\frac{xd}{l}(e_2\minus{}e_1)}\equal{}\epsilon_0 we_1e_2\frac{l}{d(e_2\minus{}e_1)}\ln (e_2\plus{}\frac{xd}{l}(e_2\minus{}e_1))|_0^l\equal{}\epsilon_0 we_1e_2\frac{l}{d(e_2\minus{}e_1)}\ln\frac{e_1}{e_2}$


geometry
probability
Alcumus
AMC
AIME
number theory

[quote="Romeo"]My state competition is soon, which AoPS book should I get?

Algebra is one of my best math topics, so I don't need that.
I can do "Intermediate" Geometry, I assume that'll be good enough for the test
So, should I get the Counting/Probability one or Number Theory

And what exactly is in the Number Theory book?

Thanks in advance for your help  :lol:[/quote]

For MC, try Intro to C & P, although Intro to Number Theory would be helpful too.  But don't try to cram it, you'll explode during the competition.

Counting and Probability is your best bet.  Don't try to do every problem in the book in 1 or 2 weeks, however.  Read over each chapter, and try a few review and challenge problems.  Then, move on.  Spend the most time on the last chapter, as it contains harder problems.

When you say "Intermediate" geometry, what do you mean?

Also. for a more well-rounded book, try AoPS Vol. 1.

I really have to disagree on this one. You can use Alcumus, the online version of the Counting and Probability book, AND Get the Number Theory book and use that.

Although im being a hypocrite here, try not to cram.

 but if u have to either cram it all or not at all.

I would reccommend C&P and alcumus tends to distract me as im on the computer so...

idk besides the fact that intermediate is a little more than mc needs.....

[quote="ReseesPieces"][b]Although im being a hypocrite here, try not to cram[/b].

 but if u have to either cram it all or not at all.

I would reccommend C&P and alcumus tends to distract me as im on the computer so...

idk besides the fact that intermediate is a little more than mc needs.....[/quote]

i'm pretty sure every1's done that once...lol :)

*frantically goes over AIME notes*

:D

I suppose Alcumus can HELP replace Intro to C&P.  However, if people start using Alcumus more and buying the book less, then AoPS is bound to make Alcumus cost money.

If you really want to master the Intro to C&P topics, use [b]both the book AND Alcumus[/b].


If you placed all the people in a line, you can make groups by placing dividers between two people.  There are n-1 gaps between two people, so we can put in up to n-1 dividers.  Between two people we can either place a divider between them or not , so we have [tex] 2^{n-1}[/tex] possible groups.

That doesn't quite work - how could you place dividers to get the AC B option? (where A and C aren't standing next to each other). As you can see from the example, there are 5 ways for n=3, not 4.

Yeah - That's was the main problem I faced.  I thought of a nice Pascal's combinatorics like solution before that realization.

Can you give a hint please, because I can't seem to find a pattern.

Does the problem have to do with partitions of n?

Actually, as far as I know, there is no closed form for the answer. There is a recurrence which isn't too hard to find - is that what you were looking for?

I really don't know what I'm looking for.  It was an open-ended question.

Ah.
Well in that case, take a look at (as always) [url=http://mathworld.wolfram.com/BellNumber.html]Wolframs site[/url].

Heres a hint about how to come up with the recurrence:
[hide="Hint"]Consider person A. How many different groups can he/she be in? How many ways are there to arrange the leftover people?[/hide]

Ok I can do the recurrence relation:
$B_{n+1}=\displaystyle\sum_{k=0}^{n}B_k\binom{n}{k}$
(that took a while to code)

This works by considering a group of n-k objects (there are $\binom{n}{n-k}$ such groups).  Then you have $B_k$ ways to arrange the leftover k objects into groups.  Since $\binom{n}{n-k}=\binom{n}{k}$ they wrote it like that.

[quote="Wumbate"]There are n different people standing around.  What is the maximum number of groups they can be seperated into?  For example, with 3 people (ABC), they can be like this: A B C; AB C; AC B; BC A; ABC.[/quote]
1+nc1+nc2+...nc[n/2]+1 i think
didnt look at the other posts

Is this sort of like the pigeonhole principle?


algebra
system of equations
number theory unsolved
number theory

If a triple of distinct primes $(p,q,r)$ has these given properties, then there exist natural numbers $a$, $b$ and $c$ such that

$(1) \;\; q+r = ap,$

$(2) \;\; r+2p = bq,$

$(3) \;\; p+3q = cr.$

$\bullet \max\{p,q,r\}= p \;\; \Rightarrow \;\; a = 1 \mbox{ by (1)}\;\; \Rightarrow \;\; p = q+r \;\; \Rightarrow \;\; q|r+2p = 2q+3r$
$\Rightarrow \;\; q|3r \;\; \Rightarrow \;\; q|3 \;\; \Rightarrow \;\; q = 3 \;\; \Rightarrow \;\; r | p+3q = r+12 \;\; \Rightarrow \;\; r|12$ $\;\; \Rightarrow \;\; r = 2$
$\Rightarrow \;\; p = q+r = 3+2 = 5.$

$\bullet \max\{p,q,r\}= q \;\; \Rightarrow \;\; b \leq 2 \mbox{ by (2).}$

$\star \: b = 2 \;\; \Rightarrow \;\; r+2p = 2q \;\; \Rightarrow \;\; 2|r \;\; \Rightarrow \;\; r = 2 \;\; \Rightarrow \;\; q = p+1$
$\Rightarrow \;\; q = 3 \;\; \& \;\; p = 2 = r \;\; \Rightarrow \;\; \mbox{CONTRADICTION!}$

$\star \; b = 1 \;\; \Rightarrow \;\; r+2p = q \;\; \Rightarrow \;\; r|p+3q = 7p+3r \;\; \Rightarrow \;\; r|7p \;\; \Rightarrow \;\; r|7$
$\Rightarrow \;\; r = 7 \;\; \Rightarrow \;\; p|q+r = 2r+2p \;\; \Rightarrow \;\; p|2r \;\; \Rightarrow \;\; p|2 \;\; \Rightarrow \;\; p = 2$
$\Rightarrow \;\; q = r+2p = 7+2 \cdot 2 = 11.$

$\bullet \max\{p,q,r\}= r \;\; \Rightarrow \;\; c \leq 3 \mbox{ by (3).}$

$\star \: c = 3 \;\; \Rightarrow \;\; p+3r = 3q \;\; \Rightarrow \;\; 3|p \;\; \Rightarrow \;\; p = 3 \;\; \Rightarrow \;\; q = r+1$
$\Rightarrow \;\; q = 3 \;\; \& \;\; r = 2 = p \;\; \Rightarrow \;\; \mbox{CONTRADICTION!}$

$\star \: c = 2 \;\; \Rightarrow \;\; p+3r = 2q \;\; \Rightarrow \;\; q|r+2p = 4q-5r \;\; \Rightarrow \;\; q|5r \;\; \Rightarrow \;\; q|5$
$\Rightarrow \;\; q = 5 \;\; \Rightarrow \;\; p = 10-3r \;\; \Rightarrow \;\; (p,r) \in\{(4,2), (1,3)\}$
$\Rightarrow \;\; \mbox{CONTRADICTION!}$

$\star \: c = 1 \;\; \Rightarrow \;\; p+3r = q \;\; \Rightarrow \;\; q|r+2p = 2q-5r \;\; \Rightarrow \;\; q|5r \;\; \Rightarrow \;\; q|5$
$\Rightarrow \;\; q = 5 \;\; \Rightarrow \;\; p+3r = 5 \;\; \Rightarrow \;\; \mbox{CONTRADICTION!}$

[b]Conclusion:[/b] There are only two solution of the Diophantine system of equations (1)-(3), namely $(p,q,r) = (5,3,2)$ and $(p,q,r) = (2,11,7).$

[quote="marko avila"]Find all triples of disinct primes (p,q,r) such that :

 p|q+r  q|r+2p and r|p+3q[/quote]
Maybe we can generalize the problem:
Find all triples of disinct primes (p,q,r) such that :

 $p|aq+br ,q|cr+dp ,r|ep+fq$where $a,b,c,d,e,f$ are positive integers.


[hide]
First one, 247 * 5 = 1235

Second one:
There are 12 ways to have 3 diff pos integers less than 5.
the ones in the order a<b<c are:
123, 124, 134, 234.
So 4/12 = 1/3

Not sure for 3rd[/hide]

[quote="kidwithshirt"]1. Use the digits, 2,4,5,7 each once to create two prime numbers. What is the smallest possible product of two prime numbers created from these digits.

(my coach said this was worded badly =\  :blush: )

2. The elements of the ordered triple (a,b,c) are different positive integers less than 5. What is the probability that a<b<c? Express answer as a common fraction.

3. If a mortorist increases her speed by 8 miles per hour, she will cover a distance of 120 miles in 30 fewer minutes. What is her original rate of speed in miles per hour?[/quote]

[hide]1. The smallest product is 457*2=954 (457 is prime)
$954$

2. Total number of possibilities is 4*3*2=24
The triples that work are (1,2,3), (1,2,4), (1,3,4), and (2,3,4)
So the probability is 4/24=1/6
$1/6$

3. There is a way of doing this using algebra, but I will use guess and check.
120=40*3 or 48*2.5
These satisfy the conditions in the problem.
So the motorist's speed originally is 40 mph
$40$[/hide]

[quote="236factorial"][hide="little more algebra for 3"]
From the formula d=rt, where r=rate in mph, d=distance in miles, and t=time in hours

$(r+8)(t-\frac{1}{2})=120$.

$rt+8t-\frac{r}{2}-4=120$.

Since rt=120,

$8t-\frac{r}{2}=4$, or multiplying it by 2, $16t-r=8$.

Substituting $t=\frac{120}{r}$ and rearranging,

$r^2+8r-1920=0$.

And r=-48 or 40. since r>0, r=40. [/hide][/quote]

Good one! I m still bit unclear on ur 120 -> rt part

Are you sure you can do that?

[hide="awesome solution for 2"]
There are $5*4*3=60$ choices.
For each that increases, it is like choosing combinations. (you can choose 1,2,3 1,2,4...) basically you are choosing 3 numbers out of 5 and arranging them so that they are in increasing order
$\binom 53=10$
Thus the answer is $\frac 16$[/hide]

[hide="1"] 914. this one is sort of tricky because you think the prime numbers are two-digit.  i used 2*457 to get my answer.[/hide]

[hide="2"] 1/6. there are four combinations: 1,2,3; 1,2,4; 1,3,4; 2,3,4. there are 6 total combination for each first number (6 for 1,_,_; 6 for 2,_,_) so 4/24 = 1/6[/hide]


geometry
inequalities unsolved
inequalities

[quote="nickolas"]In any triangle ABC  show that
[tex] m_a\cdot m_b\cdot m_c \geq r(m_a^2+m_b^2+m_c^2) [/tex]----------- (1)
[/quote]

Very nice problem!

$\displaystyle m_a+m_b+m_c \geq \frac{a^2+b^2+c^2}{2R} = (2S)\cdot\frac{a^2+b^2+c^2}{abc}$ ----------- (2)

(2) is true since $\displaystyle m_a \geq \frac{b^2+c^2}{4R}$, etc.

We will show (2) ==> (1).

Let $a'=m_a$ etc, be the sides of a new triangle $\Delta{A'B'C'}$, then it's medians are $m_{a'} = \frac{3}{4}a$, etc. And the area $S'=\frac{3}{4}S$. So (2) applies to triangle $\Delta{A'B'C'}$ implies that:

$\displaystyle \frac{3}{4}(a+b+c) \geq \left(2\cdot \frac{3}{4}S\right)\cdot \frac{m_a^2+m_b^2+m_c^2}{m_am_bm_c}$ which implies (1).


If $f: \mathbf{R}\rightarrow\mathbf{R}$,the following conditions are equivalent:$f(x)+f(y)=f(x+y)$ and $f(x)+f(y)+f(z)+f(xyz)=f(x+y+z+xyz)$?

The first obviously implies the second and the second implies (for x=y=z=0)
4f(0)=f(0), thus f(0)=0, so for z=0 the second implies the first.

Sorry,i wanted this $f(x)+f(y)=f(x+y)\leftrightarrow f(x)+f(y)+f(z)+f(xy)+f(xz)+f(yz)+f(xyz)=f(x+y+z+xy+yz+zx+xyz)$


With a,b,c,d>=0: a+b+c+d=1.prove that
3(ab+ac+ad+bc+bd+cd)<=32abcd+1 :P

But the condition is $a + b + c + d = 1$...

I have a generalization:
$0 \le a_1,a_2...a_n: \sum a_i =1 we have E_k =\sum a{}_i{}_1 a{}_i{}_2 ...a{}_i{}_k then F(a_1,a_2,...a_n)=E_k -f(k,n)E_n \le F(1/n,...,1/n) with f(k,n)max=n^n(nCk/n^k -(n-1)Ck/(n-1)^k) and1 \le k \le n-1$


I just started teaching grade 2 and have noticed many studnets have simply memorized 0-20. My question is about introducing math language.
I will have to teach studnets 0-20 and what the number look like and understand them, just not memorize 0-20. I was wondering if anyone had experience teaching this. My question is,  am I better off teaching 0-10 first or do you think 0-20 is too much to begin with..

Thanks,

LB

I think starting with 0-20 is good enough to begin with.

Hmm... I don't have much advice. I guess as they reach higher numbers they might recognize a pattern.

PS Welcome to Art of Problem Solving

I think it depends on how you think they will do...

And I think if you give them treats after each correct answer, they would certainly fall for it. (Or not..)

Thanks for the advice. I guess the best way is to take it step by step. I just wanted some
reassurance before I got started.
Thanks
LB

[quote="DonkeyKong"]
And I think if you give them treats after each correct answer, they would certainly fall for it. (Or not..)[/quote]

Personally, I'm not that sure that's a good idea, but you can give it a shot if you want. Maybe not candy direct on the spot. If I remember correctly, in second grade there was a reward system where you earn tokens, and tokens are picked out of a jar every once in a while, and if it's yours, you get to choose a prize.

I think this is better suited for the Round Table.

[quote="www.primarymath.ca"]I know students are motivated with candy, and with small toys maybe from the dollar store, but from experience what have you used to motivate studnets with a reward ? Free homework pass, etc... ?[/quote]

Well, it depends for every student, I guess... usually I always went for the HW passes. But candy and other toys do really attract some people.

[quote="lawrence"]I know students are motivated with candy, and with small toys maybe from the dollar store, but from experience what have you used to motivate studnets with a reward ? Free homework pass, etc... ?[/quote]

I don't know where you live, but the US is suffering from an obesity epidemic...  So candy isn't always the best reward.

[quote]Schools that run bake sales and let teachers reward students with candy risk having more overweight pupils, a study in Minneapolis suggests.[/quote]
Check out http://findarticles.com/p/articles/mi_qn4188/is_20051207/ai_n15903445 for the whole article.

Personally, I always thought that little kids were kinda like dogs - they go for affection more than concrete rewards.  When I learned my multiplication tables, my teacher took us out to lunch with her after we had reached a satisfactory level of knowledge...  So some one-on-one time works well.  A class reward is also kinda nice - if the entire class learns something then the class gets to go to the iMax or something like that...  Although it is rough if one kid doesn't make the cut.  I guess that I don't really know...


Let $ s_{1}$ be any positive integer. Define $ s_{n}$ to be so that $ \sum_{k \equal{} 1}^{n}s_{k}\equiv 0(\bmod k)$ with $ 0\leq s_{n}\leq n\minus{}1$. Show $ \exists N\in\mathbb{N}$ so that $ \forall p,q\geq N$ we have $ s_{p}\equal{} s_{q}$.

Note: $ p,q$ arent primes here.

[quote="sylow_theory"]Let $ s_{1}$ be any positive integer. Define $ s_{n}$ to be so that $ \sum_{k \equal{} 1}^{n}s_{k}\equiv 0(\bmod k)$ with $ 0\leq s_{n}\leq n\minus{}1$. Show $ \exists N\in\mathbb{N}$ so that $ \forall p,q\geq N$ we have $ s_{p}\equal{} s_{q}$. [/quote]

Very nice problem !
I think there is a typo error : "Define $ s_{n}$ to be so that $ \sum_{k \equal{} 1}^{n}s_{k}\equiv 0(\bmod n)$" instead of "$ \sum_{k \equal{} 1}^{n}s_{k}\equiv 0(\bmod k)$"

[hide="A solution"]We have, for $ n\geq 2$,  $ \sum_{k \equal{} 1}^{n}s_{k}\leq$s_{1}+$ \sum_{k \equal{} 2}^{n}(k\minus{}1)\equal{}s_{1}\plus{}\frac{n(n\minus{}1)}{2}$
So, whatever could be $ s_{1}$, it exists $ m\in\mathbb{N}$ such that  $ s_{1}\plus{}\frac{m(m\minus{}1)}{2}<m^{2}$ and then $ \sum_{k \equal{} 1}^{m}s_{k}\leq m^{2}$
Let then $ u\equal{}\frac{1}{m}\sum_{k \equal{} 1}^{m}s_{k}$. $ u$ is an integer (since $ \sum_{k \equal{} 1}^{m}s_{k}\equal{}0\pmod{m}$) and $ u<m$

Then $ s_{m\plus{}1}\equal{}\minus{}\sum_{k \equal{} 1}^{m}s_{k}\equal{}_{m}u\equal{}u\pmod{m\plus{}1}$ and, since $ 0\leq u<m<m\plus{}1$, we have $ s_{m\plus{}1}\equal{}u$
And $ \sum_{k \equal{} 1}^{m\plus{}1}s_{k}\equal{}mu\plus{}u\equal{}(m\plus{}1)u$

And it is obvious (by induction for example) that $ s_{p}\equal{}u$ $ \forall p\geq m$
(and $ \sum_{k \equal{} 1}^{p}s_{k}\equal{}\sum_{k \equal{} 1}^{m}s_{k}\plus{}\sum_{k \equal{} m\plus{}1}^{p}s_{k}\equal{}mu\plus{}(p\minus{}m)u\equal{}pu\equal{}0\pmod{p}$[/hide]


Triangle ABC is isosceles.  AB = AC.  If CD is perpendicular to AB, show that angle DCB = 1/2angle A.

:what?: Why have you put it to Advanced Fields Unsolved Problems?
I will move it to Getting Started.

I think this is it.

[hide]
$\angle ABC = \beta$
$\angle DCB = \alpha$
$\beta + \alpha + 90 = 180 \Rightarrow \beta + \alpha = 90 \Rightarrow \alpha = 90-\beta$
$\angle BAC = 180-2\beta$(Iscoceles triangle), and
$\angle DCB = 90-\beta$

$180-2\beta=2(90-\beta) \Rightarrow \angle BAC = 2\angle DCB$[/hide]

I'm pretty sure this question was already asked in the intermediate section.  I solved it there.

[hide]we can call angle B and angle ACB x since they are equal. therefore, angle A is 180-2x
since the perpindicular forms two right triangles, and we know angle CDB=90, we know that Angle DCB=90-x
and 90-x is the same as (180-2x)/2[/hide]

[hide]
\[\angle{DCB}+\angle{B}=90=\angle{A}+\angle{ACD}\]
\[\angle{A}+\angle{C}-\angle{BCD}=\angle{BCD}+\angle{B}\]
Since $\angle{B}=\angle{C}$
So we get
\[2\angle{DCB}=\angle{A}\]
That is:
\[\angle{DCB}=\frac12\angle{A}\][/hide]

Just for future reference, I think Intermediate Forum is just at the right level for most elementary proof problems in Euclidean geometry.


pigeonhole principle
inequalities unsolved
inequalities

[quote="various"]$ E={ 1, \frac{1}{2},\frac{1}{3},\frac{1}{4}...,\frac{1}{n} }$

$ {a_1,a_2,...,a_n}$  and ${b_1, b_2,...,b_n}$ are two permutations of E such that:
$a_1+b_1 >= a_2+ b_2 >=...>=a_n+b_n $
 prove that:$ a_m+b_m <= \frac{4}{m}$  for each m in$  [1,n] $

thanx..[/quote]

any ideas ?
..this makes me think about 'v. Proizvolov' equality..

take n=3
E=1,1/2,1/3
1+1>=1/2+1/2>=1/3+1/3
this is one of many possibilities,
but 1/2+1/3>=1+1/2>=1/3+1/2 is not possible....
how to determine that  generally

[quote="Leva1980"]Suppose that we have $ a_{m} + b_{m} > \frac{4}{m} $. Then for every $ 1 \leqslant j \leqslant
m $ we have $ a_{j} + b_{j} > \frac{4}{m} $. If we write $ a_{j} = \frac{1}{x_{j}} $
and $ b_{j} = \frac{1}{y_{j}} $ and take $ z_{j} = min(x_{j},y_{j}) $, then we have
$ \frac{2}{z_{j}} \geqslant \frac{1}{x_{j}} + \frac{1}{y_{j}} = a_{j} + b_{j} > \frac{4}{m} $,
for every $ 1 \leqslant j \leqslant m $, so $ 1 \leqslant z_{j} < \frac{m}{2} $,
$ j=1,2, \ldots , m $. Therefore by Dirichlet principle, there exist some different
$ 1 \leqslant i,j,k \leqslant m $, such that $ z_{i} = z_{j} = z_{k} $, and this is impossible.

I hope it is correct.[/quote]
it seems nice..
thanx

Can you explain this?

" For every $1 \leq j \leq m$, we have $1 \leq z_j < \frac{m}{2}$.  Therefore, by the Dirichlet principle, there exist some $1 \leq i,j,k \leq m$ such that $z_i = z_j = z_k$."

Because there are less than $ m/2 $ possibilities for $ m $
numbers $ z_{1},z_{2}, \ldots , z_{m} $, therefore we have some
value which appears at least $ 3 $ times.


induction
Euler
algebra
polynomial
algorithm
function
Pascal\u0027s Triangle

Consider the triangular array of numbers with 0,1,2,3,... along the sides and interior numbers obtained by adding the two adjacent numbers in the previous row. Rows 1 through 6 are shown.
[code]
          0
        1   1
      2   2   2
    3   4   4   3
  4   7   8   7   4
5  11  15  15  11   5
[/code]

Let f(n) denote the sum of the numbers in row n. What is the remainder when f(100) is divided by 100?

[hide]the recursion for f(n) is f(n)=2(f(n-1)+1). we now use induction to show that f(n)=2^n-2. Thus, we want 2^100-2 (mod 100). well, euler doesn't help to much, we're still left with 2^20-2, but we can work this down to 24^2-2:equiv:76-2:equiv:74, methinks. correct?[/hide]

No; your formula for f(n) is (slightly) wrong. Keep in mind that 0 is Row 1 (I think that's where you blundered)

It would actually make more sense to call row 1 row 0, wouldn't it?

Is this problem your own creation?  I really like variations on Pascal's triangle -- have you tried to figure out an explicit form for the terms?

No, this is from the 46th AHSME (1995), problem 27.

Ah, okay.  Well, in that case, let me tack on a variation of my own:

Let me redefine the rows so that the first row is 0.  In that case, find an explicit form for R[n, k], which we will define as the kth term from the left in the nth row, the leftmost term being the 0th.  Thus, in the table, R[0, 0] = 0, R[4, 2] = 8, R[5, 1] = R[5, 4] = 11.

This problem is doable, but hard.  A first hint:

[hide]Try comparing the terms of this table and some terms of Pascal's Triangle.[/hide]

A second (big) hint:  (Don't use this until you've played around for at least a half-hour with the first hint)

[hide]The table he gave us looks a lot like part of Pascal's triangle if you add diagonals that are all 1's on the outside of the table confuted gave us.  Hint, hint.[/hide]

[hide]well the number in row n+1 if (row n  +2)*2-2 because if we add 1's to the outsides we get pascals triangle, or close. so the 100th row will be (row 2)*2^98 -2. (each time we multiply by 2 take 2 away and for next row add two again). so row 100 will have a sum of 2^99-2. now someone needs to finish this off with mods, i am stuck.[/hide]

Yes, he had earlier mis-numbered his rows, starting with row 0 at the top instead of row 1.  Anyone care to try my problem?

i will attempt it, so dont post the solution just yet.  :D

it looks interesting.  i dont have much time to do it tonight, though, but i dont have school tues. so i'll work on it tomorrow.

[quote="cats..."]huh?  I thought mystic had the right answer?  The recursion is correct, and 2^100-2 is congruent to 74(mod100), so it should be 74 i think...

Edit: unless if he had a different answer before he edited, which he prob. did.[/quote]
Correct.

[hide]I modified the original table: chopped off the top row, and add diagonals of 1's on the outside. now, each number (except for the ones on the outside) can be represented as a sum of 4 binomial coefficients. we see this in 2 ways: we can make the informal note that it looks awfully like 4 pascal's triangles smushed together, or we use a block-walking argument, where we can start from 4 places. [/hide]

Code:            1    1    1    1
         1    2    2    2    1
      1    3    4    4    3    1
   1    4    7    8    7    4    1
1    5    11   15   15   11     5     1

I think it actually looks nicer if you write it as the difference of binomial coefficients, not the sum.  Fewer of them involved.  Now try a table in which the top-most number is 1, and the outer diagonals are the triangular numbers.  Then go for the square numbers.

(I don't get you, Mystic -- everyone else is solution-averse.  You're perfectly happy to post solutions, but you steadfastly refuse to attach answers to them :))

[code]

           1
        3    3
      6   6    6
   10  12  12  10
  15  22  24  22  15
21 37  46  46  37  21
[/code]
While you're at it, why not find an expression for the sum of each row in this chart?

With what little programming ability I have, I've coded a program that will output the chart in which the outer rows are any polynomial function.  So, I challenge you to:

1) Do that for some nice polynomials.  I did it by hand for the triangular numbers and the square numbers, as well as the problems originally posted here, before I went on to the computer.

2) Prove that if you have the table where the square numbers make up the outer diagonals, and the table where the numbers 1, 2, 3, 4, etc. make up the outer diagonals, and the actual Pascal's triangle (where 1, 1, 1, etc. is the outer diagonal), then you can generate a table where the outer diagonal is given by a*n:^2: + b*n + c for any constants a, b, and c.

3) Describe an algorithm that you could use to find the full table where the outer diagonal is any polynomial.

4) What function should the outer diagonals be so that every row is a constant?  (For example, this works in the first 2 rows of Pascal's triangle and the first 3 rows in the table with triangular numbers on the outer diagonal.)

[quote="JBL"]I think it actually looks nicer if you write it as the difference of binomial coefficients, not the sum.  Fewer of them involved. [/quote]

sorry...I hadn't read your hints.

how exactly would you write it as a difference?

It should come out looking either like C(n + 4, k + 2) - C(n, k) or
C(n + 4, k + 2) - 2C(n, k), depending on which problem you're doing.  (This will give you the value at the same place as C(n, k) gives you in Pascal's Triangle.)


someone cans help-me ragards the theorem of theory of groups?
Thanks
sastra

DO you mean Group theory? Here is a website I found, hope it helps. :)
http://en.wikipedia.org/wiki/Group_theory

or look on aops or other math related websites ..

You can also purchase the book 'Theory of groups' from Ilmi Kitab Khana if you are in Lahore.
It is a nice book to relax with.


Put the sci-fi show "Eureka" onto the poll. It's great.

I vote LoTR.

Two comments, though:

1) Personally, I think calling [i]Gulliver's Travels[/i] 'modern' is a little bit of a stretch.  The same goes for [i]A Tale of Two Cities[/i].

2) Where is [url=http://en.wikipedia.org/wiki/Discworld]Discworld[/url]?

tom sawyer is good. and so is the shadow children series

id have to say HP.
bartemaeus trilogy rocks too

I like the His Dark Materials books but I also agree with junggi with the bartemaeus trilogy.

[quote="nutz_for2.718281828"]COME ON!!!

Eureka has really good ratings nd really good episodes. unfortunately, there are only 12 or so episodes per season. The first season was just finished.[/quote]
Umm I think we are talking about books.

[quote="anirudh"]I don't exactly like HP. Sorry about it :maybe:[/quote]

The fifth and sixth book were disappointing in my opinion

Bartimaeus Trilogy is.

I'd have to go for The Catcher in the Rye for best books though. You have a very limited scope here, almost all fantasy/sci fi and no really great literary books.

HP is so overrated now. They were nice in the beginning, but then you get a little tired of it and in the later books, there is less contributing to the main plot.

[quote="neelnal"]So, Which book/trilogy/sequence of books do all of you consider the best/favorite?[/quote]

you forgot to add Star Wars...

anyway, Lord of the Rings and Harry Potter PWNs everything else....unless something better is released

HP PWNS. ok justa little. i also like the series of unfortunate events

1984 by George Orwell is from 1948.  I'd say it's pretty modern, especially when LOTR is considered as well.

It's really a good book.

Too bad it's so unknown that most of the amazingly obsessed V for Vendetta fans note (at best) that there might be "some similarities". :mad:

[quote="nutz_for2.718281828"]Why didn't he say sequence of books?     :mad:    :oops:[/quote]

Well, you can't call Harry Potter a trilogy.

I should have called it series, I know.

LOTR of course. :lol:

Have anyone read Whell of Time series by Robert Jordan or Song of Ice and Fire by G.R.R. Martin?

Discworld really should be on the list...

The Wheel of Time is a little... bleh. Mixed.

His Dark Materials didn't have any impression on me.

I voted LOTR just because I grew up on it.  :D

Ignite--great literary books? When we're mostly a bunch of kids with too much time on our hands, hanging out on the brainless portion of a math forum? You might be asking a bit much...

I haven't read all that many literary classics, but I have been wondering what accounts for literary greatness... Maybe you (or anyone) could enlighten me? Is it the quality of writing? Entertainment? The message? Characterization and convincingness? A perfect balance of all of these, or going out on a limb to stand out from the rest?

why is lotr and harry potter listed first and second, respectively? :roll:

[quote="anirudh"]the first 4 were good, the 5th and 6th were a letdown[/quote]

the first and third are very good, 2nd and 4th okayish, 5th tolerable, but it was the 6th that took sucking to an artistic level.

I'd say Inheritance Trilogy is way better. It has somewhat gone close to the LOTR.

Personally I'd also call the act of putting "A Tale of Two Cities" on the same platform with HP an act of sheer callousness, if not a desecrating and blasphemous one.

[quote="moogra"]i also like the series of unfortunate events[/quote]
i have seen that book in our school library. strange sort of a book.

[quote="funcia"]and where is AF?[/quote]

Ah ... I am yet to read Opal Deception  :(

I just got done of the red one after the opal deception but i forgot it's name already

What about the five-book [i]Hitchhiker's Guide to the Galaxy[/i] trilogy?

:D

[quote="dts"][quote="Koda"]Personally I'd also call the act of putting "A Tale of Two Cities" on the same platform with HP an act of sheer callousness, if not a desecrating and blasphemous one.[/quote]

Oh?  And why so?

It's not as though AToTC was [i]always[/i] considered a classic; when Dickens was alive, his work was considered a relatively base form of entertainment (this from my ENGL 101 teacher).[/quote]

Harry Potter will be forgotten in a hundred years. Now compare that to A Tale of Two Cities. Dickens was, after all, Dickens.

[quote="mathenrd314"]What about the five-book [i]Hitchhiker's Guide to the Galaxy[/i] trilogy?
[/quote]

That would have been another blasphemy.
Anyway it is actually a five-book trilogy in four parts.  :rotfl:


Let a,b,c,d real positive numbers with (d+1)3 \sqrt 3 \geq (a+b+c).

Prove that

 (b+cd) 2 /a  + (c+ad) 2 /b +(a+bd) 2 /c \geq abc

Well, unfortunately, this one is trivial. Indeed, we obviously have
 sum (b+cd)^2/a>=(a+b+c+ad+bd+cd)^2/(a+b+c)=(a+b+c)(d+1)^2>=1/27*(a+b+c)^3>=abc


calculus
derivative
function
inequalities proposed
inequalities

First some warm up problems!

1. Prove that for any $ f: R\to R$ such that
$ \frac {f''(x)}{x^2} \minus{} \frac {f'(x)}{x^3}$
is increasing for $ x > 0$,
$ f(5) \plus{} f(6) \plus{} f(11) \ge f(1) \plus{} f(9) \plus{} f(10)$.

2. Prove that for any function $ f: R\to R$ such that $ f( \minus{} x) \equal{} f(x)$ and $ f^{(6)}(x)\ge 0$ for all $ x$,
$ f(5) \plus{} f(6) \plus{} f(11) \ge f(1) \plus{} f(9) \plus{} f(10)$.

Now for the real thing:

3. For any nine real numbers $ a_1, ..., a_9$ such that $ \sum_{i \equal{} 1}^9 a_i \equal{} \sum_{i \equal{} 1}^9 a_i^3 \equal{} 0$, $ \sum_{i \equal{} 1}^9 a_i^2 \equal{} 8$, and for any function $ f: R\to R$ such that $ f^{(4)}(x) \ge 0$ for all $ x$,
$ f(2) \plus{} 7f(0) \plus{} f( \minus{} 2) \ge \sum_{i \equal{} 1}^9 f(a_i) \ge 4f(1) \plus{} f(0) \plus{} 4f( \minus{} 1)$.

Of course, just proving special cases of this is cool too (for instance, try proving it when $ f(x) \equal{} x^4$, or when $ a_{10 \minus{} i} \equal{} \minus{}a_i$).

No ideas? Are they too hard or just uninteresting?

Maybe I should give some hints:

[hide="Problem 1"]The sum isn't the same on both sides... maybe a change of variables?[/hide]
[hide="Problem 1"]What's another way to write $ \frac{d}{xdx}$?[/hide]

[hide="Problem 2"]This isn't true if the function isn't even, for instance, the sum isn't the same on both sides. But if we replace $ f(x)$ by $ \frac{f(x) \plus{} f(\minus{}x)}{2}$...[/hide]

[hide="Problem 3"]In the special case $ f(x) \equal{} x^4$, Lagrange Murderpliers work out in a really cool way. Well, I thought it was cool...

I don't actually know if assuming $ a_{10\minus{}i} \equal{} \minus{}a_i$ helps. :/[/hide]


What is the topic of question like this?
It is from a contest in turkey. but this contest is to decrease the number of contestants for Turkish MO which is a classical exam forming by 6 problems like IMO.
The questions are multiple choice question type. So where is the location of this type problems in this forum.

Q1.
We will change the location of numbers beginning from.
12345
(e.g 12345 -> 32145 -> 42135)
After even steps ( 2n), which of the numbers cant be get?
a)13425
b)21435
c)35142
d)43125
e)53124

Q2.
The sides of an quadrilateral is 1,4,7,8. So what is the maximal area of this quadrilateral.

Either in the [url=http://www.mathlinks.ro/Forum/index.php?f=217]Olympiad Section[/url] or in the High-School >> Pre Olympiad section.


How many ways are there to distribute 90 identical books to 10 people if everyone gets at least 1 book?

[hide]
This is the same as distributing 80 identical books to 10 people with no restrictions, which is $\binom{89}{9}$ by encoding/ball-in-urn formula/blah blah.[/hide]

what is this encoding/ball-in-urn fomrula?

If you have $x$ different urns and $n$ identical balls, you can distribute the balls into the urns in __ ways. Try it with lower cases and make a conjecture. (I don't give an answer because I don't know the formula  :roll: ) What is it, agolsme?

does neone have another approach to the qsnt?

heyy....i rkn the answer is $89C9$....cos we have $90$ books, which makes 89 spaces, in which we can place $9$ dividers in order to separate the $99$ books into $10$ different objects....so it should $89C9$

[hide]Let the books be represented by b:

bbbbb ...... bbbb

Now we put 9 dividers anywhere between the b's so that there are 10 segments of b's. That represents how many books each person gets.

Since there are 89 spaces to choose from, 9 dividers, the answer is ${89\choose9}$[/hide]

[hide]In one of the other threads (I'm not even sure which now), someone said the formula was $\frac{(n+r-1)!}{r!(n-1)!}$, where $n$ is the number of containers and $r$ is the number of objects.  Since everyone has at least 1 book, 10 of them have already been used, so it is like 80 books to 10 people, just like agolsme said.  Plugging this in, we get $\frac{(80+10-1)!}{80!(10-1)!}$, which gives us $\frac{89!}{80!9!}$, or ${89\choose9}$[/hide]


1. If p and q are positive integers and p + q < 10, how many different values can the product pq take?

(A) 36
(B) 20
(C) 12
(D) 15
(E) 16

2. If the line y = 2x + 3 is reflected in the line y = x + 1, then what does its equation become?

(A) 2x - y = 0
(B) x - 2y + 1 = 0
(C) x - 2y = 0
(D) x - 2y - 2 = 0
(E) x - y + 1 = 0

3. In triangle PQR, S is a point on PQ. PR = 35, PS = 11, and RQ = RS = 31. Find the length of SQ.

(A) 10
(B) 11
(C) 12
(D) 13
(E) 14

EDIT: Solved 3 using Stewart's Theorem, but never thought we were supposed to know that.

4. The vertices of a square are (x, 0), (0, y), (a, b) and (c, d). If a + b = 19 and c + d = 14, then x + y is

(A) 11
(B) 9
(C) 17
(D) 15
(E) 13

THANKS so much if you can help me with these!

#1
[hide="solution"]
We can make a systematic list to solve this.
If $ p\equal{}1$, then $ 1\le q \le8$, giving 8 values for the product pq.
If $ p\equal{}2$, then $ 1\le q \le7$, giving 3 more values of pq: 10, 12, and 14. (anything less is already covered)
If $ p\equal{}3$, then $ 1\le q \le6$, giving 3 more values of pq: 9, 15, and 18. (12 is already covered)
If $ p\equal{}4$, then $ 1\le q \le5$, giving 2 more values of pq: 16 and 20.
Now, if p was more than four, we would just get the same results except with p and q switched, so the answer is $ 8\plus{}3\plus{}3\plus{}2\equal{}\boxed{16}\rightarrow E$
[/hide]

Thanks for the solution andersonw, can anyone help with 2 and 4?

[hide]For number two, all you need to do is draw a diagram.  You find that the two lines intersect at (-2,1) and that one of the points of the image is (2,1).  If we find the equation of the line that passes through these two points, we get that $ y\equal{}\frac {1}{2}x$.  Thus, the answer is $ \boxed {C: x\minus{}2y\equal{}0}$.[/hide]

Hope that helps!

[quote="Math Geek"][hide]For number two, all you need to do is draw a diagram.  You find that the two lines intersect at (-2,1) and that one of the points of the image is (2,1).  If we find the equation of the line that passes through these two points, we get that $ y \equal{} \frac {1}{2}x$.  Thus, the answer is $ \boxed {C: x \minus{} 2y \equal{} 0}$.[/hide]

Hope that helps![/quote]

The intersection point is (-2,-1)


does anyone have the answers to the '03 nationals sprint and target?

PM me  :ninja:

EDIT: this is my 314th post.  my $\pi$ post.  which, to the nearest hour, i posted on the 14th hour [eastern time, my time] of the third day after pi day. how insane is that???


Let $a\geq b\geq c\geq d > 0$. Prove that
$\displaystyle \frac{a}{a+b}+\frac{b}{b+c}+\frac{c}{c+d}+\frac{d}{d+a} \geq 2$.

Cool problem ! Believe it or not, it is equivalent (if my computations arent wrong!) with $ (a-c)(b-d)(ac-bd)\geq 0$. Indeed, substract $\frac{1}{2} $ from each fraction in the left-hand side and bring it to $ \frac{a-b}{a+b}+\frac{b-c}{b+c}+\frac{c-d}{c+d}\geq\frac{a-b+b-c+c-d}{a+d} $. Now, substract from the right-hand side three fractions ( look at their numerators to see which corresponds to each) and reduce it to
  $\frac{ (a-b)(d-b)}{a+b}+\frac{(b-c)(a-c+d-b)}{b+c}+\frac{(a-c)(c-d)}{c+d}\geq 0$. This is of course the same as $ (a-c)[\frac{b-c}{b+c}+\frac{c-d}{c+d}]\geq (b-d)[\frac{ a-b}{a+b}+\frac{b-c}{b+c}] $. Now, clear denominators inside each square paranthesis and reduce it to the form stated in the beginning of the solution. Very nice problem! .

Harazi, a very nice solution! I just wanted to point out an equivalent formulation of the inequality:

If we denote $\displaystyle x=\frac{b}{a}$, $\displaystyle y=\frac{c}{b}$ and $\displaystyle z=\frac{d}{c}$, then $a\geq b\geq c\geq d>0$ implies $x\leq 1$, $y\leq 1$ and $z\leq 1$. We have $\displaystyle \frac{a}{a+b}=\frac{1}{1+b/a}=\frac{1}{1+x}$, similarly $\displaystyle \frac{b}{b+c}=\frac{1}{1+y}$ and $\displaystyle \frac{c}{c+d}=\frac{1}{1+z}$, and finally $\displaystyle \frac{d}{d+a}=\frac{d/a}{d/a+1}=\frac{xyz}{xyz+1}$, so the initial inequality is equivalent to

$\displaystyle \frac{1}{1+x}+\frac{1}{1+y}+\frac{1}{1+z}+\frac{xyz}{xyz+1}\geq 2$.

Using $\displaystyle 2-\frac{xyz}{xyz+1}=1+\frac{1}{1+xyz}$, this can be rewritten as

$\displaystyle \frac{1}{1+x}+\frac{1}{1+y}+\frac{1}{1+z}\geq 1+\frac{1}{1+xyz}$.

I think this is a very nice inequality (of course, holding for $x\leq 1$, $y\leq 1$ and $z\leq 1$ as I have said above). It is equivalent to fuzzylogic's original inequality, since one can just reverse the substitution, but it also has a direct proof:

$\displaystyle \frac{1}{1+x}+\frac{1}{1+y}+\frac{1}{1+z}-\left( 1+\frac{1}{1+xyz}\right) =\frac{\left( 1-yz\right) \left( 1-zx\right) \left( 1-xy\right) }{\left( 1+x\right) \left( 1+y\right) \left( 1+z\right) \left( 1+xyz\right) }\geq 0$.

  Darij

Both of your solutions are wonderful! Here is another one

Let $\displaystyle A=\frac{a}{a+b}+\frac{b}{b+c}+\frac{c}{c+d}+\frac{d}{d+a}$ and
$\displaystyle B=\frac{b}{a+b}+\frac{c}{b+c}+\frac{d}{c+d}+\frac{a}{d+a}$, then it's clear that $A+B=4$.

So it suffices to prove that $A\geq B$ which is equivalent to
$\displaystyle C=\frac{a-b}{a+b}+\frac{b-c}{b+c}+\frac{c-d}{c+d}\geq \frac{a-d}{a+d}$ ----- (1)
(up to this point, this is similar to harazi's solution).

By Cauchy-Schwarz, we have

$\displaystyle (a^2-d^2)C=((a^2-b^2)+(b^2-c^2)+(c^2-d^2))\left(\frac{a-b}{a+b}+\frac{b-c}{b+c}+\frac{c-d}{c+d}\right)\geq (a-d)^2$ which implies (1).

Q.E.D.

Note: this can obviously be extended to n variables.


Let $x,$ $y$ and $z$ are non-negative numbers such that $xy+xz+yz=9.$ Prove that
$(1+x^{2})(1+y^{2})(1+z^{2})\geq64.$

[hide="sorry i need to outline the steps for myself"]
\[LHS=1+x^{2}y^{2}z^{2}+x^{2}y^{2}+y^{2}z^{2}+z^{2}x^{2}+x^{2}+y^{2}+z^{2}=(x+y+z-xyz)^{2}+64 \]
\[\iff 1+x^{2}y^{2}z^{2}+\sum x^{2}y^{2}=2(\sum xy)+x^{2}y^{2}z^{2}-2(xyz)(x+y+z)+64\]
\[\iff \sum x^{2}y^{2}=81-2xyz(x+y+z)\]
\[\iff \sum (xy)^{2}+\sum 2(xy)(xz)=81\]
which is finally clear from the given.[/hide]
How did you come upon the relation? Simple homogenizing?

To Can: it is easier than it looks. To me@home, we need the idenity
\[(a+b+c)^{2}=a^{2}+b^{2}+c^{2}+2(ab+bc+ca).\]