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n points are chosen randomly on the surface of a sphere. What is the probability that all n points lie on some hemisphere of this sphere?

[i]The word "randomly" must be interpreted in the obvious way: given a region whith area A of the surface of a sphere with area S, each point is chosen in this region with probability A/S.[/i]

No ideas at all?  :(

I'll give you a big hint: instead of considering all the points on the sphere, just consider $n$ of them, plus their opposites wrt the center of the sphere. Suppose wlog they're all distinct (because this case has probability $1$), and now try to figure out how many of the $2^n$ possible choices of "north" and "south" for each couple of antipodal points give raise to a set of $n$ points in a same hemisphere.

Note also that the case n=4 is a famous 1992 Putnam problem:

4 points are chosen randomly on a sphere. What is the probability that they are vertices of a tetrahedron containing the center of the sphere?

There are many solutions for this problem, but not all of them can be generalized...

I once struggled for days to show that the solution to that problem was $\frac 18$, and finally managed to do it, but I don't think that solution can be generalized.

Let $N$ points be chosen at random on the surface of the unit sphere in a $n$-space.
Let $p(n,N)$ be the probability that all the points lie on some hemisphere.
It is known (see reference below) that :
$p(n,N) = 2^{-N+1} \sum_{k=0}^{n-1} C_{N-1}^k.$

Reference :
J.G.Wendel, 'A problem in geometric probability', Math. Scan., 1 (1962), p.109-111.

Pierre.

All righty...   :D

Seriously now, why not trying to concentrate on the 3D case, before generalizing it? There's a simple and elementary proof for this case, come on!!   ;)

Hi! Maybe it is time to give some stronger hints about this problem?
I have googled the paper that Pierre kindly provided, but the first and only useful occurence is the one on ML  :?
Btw, Pierre, do you remember all the references you write by heart, or you just have a very efficient way to find them :) ?

[quote="DusT"]Btw, Pierre, do you remember all the references you write by heart, or you just have a very efficient way to find them :) ?[/quote]
Maybe he's giving random references, thinking that nobody will check them.
Sorry, just kidding!   ;)

Anyway, consider n couples of points on the sphere (n points plus their n opposites). We suppose wlog they're 2n distinct points, as I stated earlier (last year  :lol: ).
Then, among the $2^n$ choices of "north-south", just $n^2-n+2$ identify a set of n ponts on an hemisphere. To see this, count in how many regions a sphere is divided by n distinct equators.
So, since the choices of the n points (and their n opposites) are "all alike", the resulting probability is $\frac{n^2-n+2}{2^n}$.

I hope this will clarify things.

:first:

That technique is profoundly clever.

Sorry MindFlyer, but I'm sort of just not seeing why counting the regions on a sphere helps you come up with that n^2 - n + 2 number.  Care to elaborate?

Alex

Sure!
First of all, to prove the formula about the regions of the sphere, just use simple induction. Every new equator intersects any previous equator twice, hence adding as many regions.
Then, given n couples of antipodal points, consider the n corresponding equators. Roughly speaking, an equator corresponds to a couple of antipodal ponts iff it has the maximal distance from these points. Equivalently, the two points lie on the "poles" of the hemispheres wrt the given equator. So, we say that one point in each couple is "north", the other is "south".
We must choose north or south for each couple so that we get n points in a same hemisphere. It's easy to prove that this happens iff the n corresponding hemispheres have a non-empty intersection. Now, since there are $n^2-n+2$ possible regions on the sphere, each of which can be an intersection of n hemispheres, and that different north-south choices determine different intersections, we have proved what we needed.

[quote="MindFlyer"]So, since the choices of the n points (and their n opposites) are "all alike", the resulting probability is $\frac{n^2-n+2}{2^n}$.[/quote]
Ok, someone may be disappointed by this conclusion.
All the infinitely many choices of n points on the sphere can be partitioned into subsets of cardinality $2^n$, and within each subset, just $n^2-n+2$ chioces are successful. However, this is of course insufficient to deduce that the overall probability is $\frac{n^2-n+2}{2^n}$, since the subsets are infinitely many.
To prove this properly, some sharper argument is required, but the main and interesting idea behind the proof remains the same.

[quote="Phelpedo"]Apologies for reviving this old topic, but I don't understand some of the things you are trying to say. First off, why would $n$ "equators" divide the sphere into $n^{2}-n+2$ regions? It seems to me like the number should be just $2n$, like $n$ diameters of a circle divide the circle into $2n$ regions.

Second, I don't really see how you are using these regions to arrive at your answer.[/quote]

Haha I didn't even realise this topic existed. But anyway for an answer to your first question, you forgot that this is in 3 dimensions, not 2. Just consider the case $n=3$ by cutting the sphere into 8 parts. And for your second question, the number of ways of choosing one point of each of $n$ pairs of opposite points so that all $n$ chosen points lie in one hemisphere is the number of ways you can divide the sphere into two hemispheres with a different set of points on one hemisphere. This is the number of regions that the equators of all the pairs divide the surface of the sphere, because you can take any point in a region and select the hemisphere bounded by its equator that includes it.

Here is the exact reference that Pierre has been meant to mention:
 J. G. Wendel, A Problem in Geometric Probability, Math. Scand., 11 (1962) pp 109-111.
Make  a Google search and you will see what you want ! :lol:


inequalities
calculus
inequalities unsolved

[quote="cvix"]Prove that for all $x\in (0,1)$ it is true that
$2\cdot \sqrt{1-x^2}+x\leq \sqrt{3}+0.5$[/quote]

it is wrong . Take $x=\frac{1}{\sqrt{5}}$ and RHS=$\sqrt{5}>\sqrt{3}+0.5$

in fact , we can prove that the max is $\sqrt{5}$ by calculus .But here is a method using cauchy

$(\sqrt{4}\cdot \sqrt{1-x^2}+1\cdot \sqrt{x^2})^2\leq (4+1)(1-x^2+x^2)=5$

$\Longleftrightarrow 2\cdot \sqrt{1-x^2}+x\leq \sqrt{5}$


probability
quadratics
MATHCOUNTS
geometry
rectangle
absolute value

Chapters (and schools) are up:[url]http://www.mathcounts.org/webarticles/anmviewer.asp?a=510&z=47[/url]

How'd you guys do?

I got a 45 but don't know why I didn't get a perfect... :(   It was way too easy this year.

got a 41 :blush: 25/16 Target was too easy.

a bit on the easier side? I dont think a perfect this year says much.

According to the poll, someone got a perfect.  Who was that?

[quote="Math Geek"][quote="xpmath"]Yeah, I somehow subtracted wrong on #30.[/quote]

Oh, I got some earlier problem wrong according to the score sheet that they gave my teacher, but I don't know which one.  They probably couldn't read my handwriting or something, cuz I was 99.999999% sure that I knew the answers to each one.  My imperfect score is depressing though... :(   Oh well, fortunately, my chapter is crappy enough for me to get first place anyways, so I guess it doesn't matter too much...[/quote]
I came first too. Honestly, I don't care about chapter and school as long as you make state.

[quote="#H34N1"]Forgot to put my name on one of the targets :o[/quote]
Wow, serious bad luck, dude.  It's okay.  Like unimpossible said, perfects don't mean much, simply that one doesn't tend to make careless mistakes.

i got a 45, too, missed number 29. i got like 8 points more on the chapter than on the school.

45  :(
missed either 28 or 29. I can't remember :P

45, I killed the midpoint of the line segment problem (#17).

[quote="alanchou"]46 :)

for some stupid reason I blanked on the sprint round at the beginning[/quote]

lol this is ur 2^10th post.

28/16 much better than my 0/0 last year
(didn't do it)

[quote="pascal_1623"]Jeez, am I like the only one who got a 44? :(

But I agree with xpmath, I didn't care about getting first as long as I made state.[/quote]

Don't worry, I'm stupid too.

I missed the last target because of an reading error =(

THIS YEAR:

30/14 (missed pirate one)-3rd

LAST YEAR

25/16 -1st place!!!

Is it just me, or am I just stupid.

Look at this
last year grade rankings
1.7
2.8
3.8
4.8
5.8
6.7
7.6
8.6

HOW DID I GET TO 3rd?!?!?!?1

Because you purposely lost to me in the tiebreaker round, letting me get a free TI-84 plus silver edition. :D  :D  :D

25/16 = 41.  :)  I think I did better than last year, although I missed some easy questions on the sprint.

I STILL haven't gotten my score. It's probably around 42 though. It's so sad that I got first in chapter with a score like that, when there's so many people in the chapter...

academics, how did you do at state? I was there.

/me agrees with that, especially when it's in sprint.

lol I'm not actually in Connecticut...
I'm going to boarding school there next year. Look at my sig.
For me, States is tomorrow... :jump:  I hope I do as well as you did! :clap:

[quote="gauss1181"][quote="ZhangPeijin"]Don't you guys hate it when some Mathcounts question gives you a question that has an answer with radicals, but says to express your answer to the nearest tenth?
It's making it so that people w/o calculators will have a disadvantage.[/quote]
Yes. That's totally true. I guess people should just bring calculators to mathcounts even if they feel they don't have to. Cuz you would never know what is dangerous to expect. :wink:[/quote]

Well, the rules do say that calculators are "assumed," not just allowed.

i cant believe my luck...
i got a 23/16 on the mathcounts.org chapter test online

sprint was semi-difficult
target was easy-peesy

im a 9th grader and i couldnt figure out some of the geometric problem on the sprint...

i got a 32
missed some easy problems on the target

I got a 46 but I forget what problem 29 was. Can someone tell me?

I think NJ had tons of perfects. I would guess the three people ahead of me in written at states all got 46s, so I have confidence for NATS.

Predictions for NATS

Top 4 teams (in no particular order)

TX
NJ
MA
IN


Εξετάστε αν για κάθε θετικό ακέραιο $ n$, μπορούμε να βρούμε 8 μη αρνητικούς ακεραίους  $ a$, $ b$, $ c$, $ d$, $ e$, $ f$, $ g$, $ h$ ώστε $ n \equal{} \frac {2^a \minus{} 2^b}{2^c \minus{} 2^d}\cdot\frac {2^e \minus{} 2^f}{2^g \minus{} 2^h}$

Η πρώταση είναι ισοδύναμη με το να δείξουμε ότι κάθε φυσικός γράφεται ως:

$ 2^m \plus{} 2^n \minus{} 2^o \minus{} 2^p \equal{} 2^{m \minus{} o} \plus{} 2^{n \minus{} p}$ που διαιρείται με το 2, εκτός αν $ n \equal{} p$ ή $ m \equal{} o$

απότε οι αριθμοί της μορφής $ 2^{m \minus{} o}\plus{}1$ θα είναι περιττοί, αλλά δε θα εκφράζουν το σύνολο των περιττών

(δεν είμαι απόλυτα σίγουρος  :blush: )

Τελικά ρε Σιλουανέ τί απάντηση έχεις γι αυτή???? Είναι σωστός ο Maniopas? :lol:

Θεωρόυμε $ a>b$, $ c>d$, $ e>f$, $ g>h$. Τότε έχουμε ότι

$ n=\frac{2^{a}-2^{b}}{2^{c}-2^{d}}\cdot \frac{2^{e}-2^{f}}{2^{g}-2^{h}}=\frac{2^{b}(2^{a-b}-1)\cdot 2^{f}(2^{e-f}-1)}{2^{d}(2^{c-d}-1)\cdot 2^{h}(2^{g-h}-1)}$. Άρα $ (2^{a-b}-1)\cdot (2^{e-f}-1)=2^{d+h-b-f}\cdot n\cdot (2^{c-d}-1)\cdot (2^{g-h}-1)$.  $ (1)$

Το γινόμενο $ (2^{c-d}-1)\cdot (2^{g-h}-1)$ και $ (2^{a-b}-1)\cdot (2^{e-f}-1)$ είναι περιττοί αριθμοί και ο μόνο;ς άρτιος είναι ο $ 2^{d+h-b-f}$. Άρα θεωρούμε τον $ n$ περιττό, οπότε αναγκαστικά $ 2^{d+h-b-f}=0\Rightarrow d+h-b-f=0$.

Θεωρούμε επίσης, χωρίς βλάβη της γενικότητας, ότι $ d=h=b=f$. Οπότε τώρα η σχέση $ (1)$ παίρνει τη μορφή $ (2^{a}-1)\cdot (2^{e}-1)=n\cdot (2^{c}-1)\cdot (2^{g}-1)\Leftrightarrow n=\frac{(2^{a}-1)\cdot (2^{e}-1)}{(2^{c}-1)\cdot (2^{g}-1)}=\frac{(2^{a}-1)}{(2^{c}-1)\cdot (2^{g}-1)}\cdot(2^{e}-1)$

 Όμως εφόσον $ n$ ακέραιος πρέπει $ (2^{c}-1)\cdot (2^{g}-1)| (2^{a}-1)\Rightarrow c|a$ ή $ g|a$. Επειδή όμως έχουμε γινόμενο στον αριθμητή θεωρούμε επίσης τον $ n$ πρώτο ώστε να μην είναι της μορφής $ 2^e-1$. Για $ e=4,5$ δίνει σύνθετο και πρώτο αντίστοιχα. Οπότε ο αριθμός $ n$ που ψάχνουμε είναι ο $ 19$.

 Άρα αρκεί τώρα να αποδείξουμε ότι η ισότητα $ 19=\frac{2^{a}-1}{(2^{c}-1)\cdot (2^{g}-1)}$ δεν ισχύει.
Πρέπει $ 2^{a}-1\geq 19\Rightarrow a\geq 18$.

Για $ c=g=\frac{a}{2}$ έχουμε ότι $ 19=\frac{2^{a}-1}{(2^{\frac{a}{2}}-1)^{2}}\in Z$

Για $ c,g\leq \frac{a}{4}$ προκύπτει ότι:$ c=\frac{a}{2}$ και $ g=\frac{a}{3}$ παίρνουμε $ \frac{2^{\frac{a}{6}}+1}{2^{\frac{a}{6}}-1}\notin Z$

Θεωρώ ότι η λύση σου δεν είναι σωστή .... :(
Η παράσταση του αριθμού αρκεί να γίνεται για μια και μόνο οκτάδα , οπότε αυτό που λες εκεί χωρίς βλάβη της γενικότητας , όχι μόνο επιρρεάζει τη γενικότητα αλλά την εκμηδενίζει ....

silouan, για δες εδω [url]http://forum.math.uoa.gr/viewtopic.php?t=3496[/url]  :roll:

Ίσως δεν έμαθα ακόμα να καταλαβαίνω πότε κάποιος κάνει λάθος ....
Τα ίδια βέβαια έλεγαν και προ διετίας κάποιοι όπου χρειάστηκε 18 ποστ για να εξηγήσω το λάθος τους .
Θα προσπαθήσω να το εξηγήσω για τελευταία φορά και από εκεί και πέρα μπορείτε να θεωρείται ότι θέλετε.
Εάν μεν ίσχυε ότι ton κάθε αριθμό $ n$ μπορούσαμε να τον παραστίσουμε σε αυτή τη μορφή (και θέλαμε αυτό να αποδείξουμε)
τότε λογικό θα ήταν να πάμε κατασκευαστικά δηλαδή να ΕΠΙΛΕΞΟΥΜΕ εμείς τους αριθμούς $ a,b,ktl$ και τότε κάλιστα θα μπορούσες να τα πάρεις ίσα αν με αυτό τον τρόπο αποδείκνυες ότι κάθε ακέραιος $ n$ εκφράζεται σε αυτή τη μορφή .

Όταν όμως  όπως εδώ στην άσκηση αυτή θέλουμε τελικά να αποδείξουμε ότι κάποιοι αριθμοί δεν γράφονται σε αυτή τη μορφή τότε αυτό πρέπει να αποδειχτεί για όλες τις δυνατές τιμές των 8-αδων .
Θα δώσω ένα παράδειγμα για να γίνω πιο κατανοητός
έστω ότι μας δίνουν να εξετάσουμε αν ισχύει η ανισότητα $ x^2\plus{}y^2\plus{}z^2\geq xy\plus{}yz\plus{}zx$ για κάθε χ,y,z
Αν βρούμε έστω και μια τριάδα (x,y,z) ώστε να μην ισχύει τότε τελειώναμε (εδώ δεν υπάρχει αλλά λέμε ...)
Όμως για να αποδείξουμε ότι ισχύει δεν μπορούμε να πάρουμε και τα τρία ίσα και να πούμε τότε ισχύει άρα ισχύει και γενικά...
Για να το κλείσω ...
Όταν στο τέλος ας πούμε θέλεις να αποδείξεις ότι ο 19 δεν παριστάνεται στη μορφή
$ 19\equal{}\frac{2^a\minus{}1}{(2^c\minus{}1)(2^g\minus{}1)}$ τότε  πρέπει να  αποδείξεις ότι δεν υπάρχει τριάδα αριθμών (a,c,g) ώστε ο 19 να γράφεται σε αυτή τη μορφή .Εσύ απλά μου παίρνεις την περίπτωση που $ c\equal{}g\equal{}\frac{a}{2}$ οπότε δεν το αποδεικνύεις για κάθε τριάδα....

Είπα να ασχοληθώ  με το πρόβλημα. Αλλά επειδή δεν έχω ασχοληθεί με θεωρία αριθμών (μισό μάθημα στο πανεπιστήμιο πριν 15 χρόνια και αυτά τα έχω ξεχάσει) δείξτε λίγο επιείκεια  :oops:
Παραθέτω μια προσπάθεια απόδειξης βασισμένη μόνο στην κοινή λογική.

[b]Αν ο n είναι πρώτος τότε αναγκαστικά το ένα κλάσμα θα είναι ίσο με n και το άλλο με 1 (αλλιώς θα γράφαμε σαν τον πρώτο σαν γινόμενο 2 αριθμών διαφορετικών από τον εαυτό του και τη μονάδα, άτοπο).[/b] Υπάρχει βέβαια περίπτωση να χρειαστεί και μια ανταλλαγή παρονομαστών αλλά η ουσία του επιχειρήματος δεν αλλάζει.

Τα 2 κλάσματα είναι τις ίδιας μορφής και άρα παράγουν τις ίδιες ακριβώς ποσότητες.

Άρα θα πρέπει να δείξουμε ότι ένα κλάσμα της μορφής (2^α – 2^β)/(2^γ – 2^δ) δεν παράγει όλους τους πρώτους. Θα δείξω ότι δεν είναι δυνατόν να παράγει τον 13.

Προτιμώ να δουλέψω στο δυαδικό σύστημα αρίθμησης λόγω των δυνάμεων του 2 που εμφανίζονται. Ότι λέω παρακάτω αφορά αναπαραστάσεις στο δυαδικό σύστημα αρίθμησης.

Ένας αριθμός της μορφής 2^α παράγει αριθμούς που στο δυαδικό σύστημα αρίθμησης αποτελούνται από έναν άσσο και μετά μόνο μηδενικά, δηλαδή είναι της μορφής 10000…0000.

Άρα ο αριθμητής σαν διαφορά 2 τέτοιων αριθμών  παράγει αριθμούς που έχουν μπροστά όλους τους άσσους και πίσω όλα τα μηδενικά, δηλ 11…1100…00.
Αυτό αποδεικνύεται εύκολα (αποφεύγω τις λεπτομέρειες για συντομία και περιγράφω στα γρήγορα το σκεπτικό) αν σκεφτούμε ότι αν βάλουμε τους αριθμούς κάθετα και κάνουμε την κλασσική αφαίρεση τότε πίσω θα έχουμε αφαιρέσεις 0 από 0 και άρα θα μένει 0. Κάποια στιγμή καθώς προχωράμε προς τα εμπρός θα συναντήσουμε τον μοναδικό άσσο του μικρότερου θα προκαλέσει έναν άσσο στη διαφορά και ένα κρατούμενο που λόγω των μηδενικών του μεγαλύτερου αριθμούς από τους 2 θα διαδίδεται συνεχώς προς τα ψηφία ανώτερης τάξης προκαλώντας άσσους στη διαφορά.

Τόσο ο αριθμητής όσο και ο παρονομαστής λοιπόν παράγουν ποσότητες που στο δυαδικό σύστημα έχουν τους άσσους μπροστά και τα μηδενικά πίσω (σε αυτή την κατηγορία ανήκει και το 0 καθώς και οι αριθμοί με μόνο άσσους όπως εύκολα αποδεικνύεται).

Θα δείξω ότι διαιρώντας 2 αριθμούς τέτοιας μορφής δεν μπορεί να παραχθεί το 1101 (=13).

Αρκεί να δείξω ότι αν πολλαπλασιάσω το 1101 (πηλίκο) με οποιονδήποτε αριθμό της μορφής 111…000 (διαιρέτης) δεν γίνεται να παραχθεί αριθμός της μορφής 111…000 (διαιρετέος).

Θα χρησιμοποιήσουμε το γνωστό κλασσικό κάθετο πολλαπλασιασμό του δημοτικού  (που πολλαπλασιάζεις κάθε ψηφίο του ενός με τον άλλο και προσθέτεις τα ολισθημένα κατά ένα ψηφίο αποτελέσματα) .

Τα μηδενικά στο τέλος των 2 αριθμών που πολλαπλασιάζονται θα δώσουν τα μηδενικά στο τέλος του γινομένου πράγμα που δεν μας χαλάει την επιθυμητή μορφή. Μπορούμε λοιπόν να αγνοήσουμε τα μηδενικά και να δείξουμε ότι αν πολλαπλασιαστεί οποιοσδήποτε αριθμός της «όλο άσσοι» με το 1101 θα παράγει αναγκαστικά κάποιο μηδενικό μεταξύ των άσσων του γινομένου (και άρα το γινόμενο δε θα είναι της επιθυμητής μορφής). Παρακάτω βάζω τελίτσες για τη στοίχηση

1101
*  1
------
1101   (υπάρχει μηδενικό ανάμεσα στους άσσους)

1101
 * 11
---------
...1101
.1101
----------
100111 (υπάρχει μηδενικό ανάμεσα στους άσσους)

   1101
* 111...111
  -------------
......1101
....1101
. 1101
1101
κλπ
----------------
 ........[b]0[/b]11

Από δω και πέρα όσο και να μεγαλώνει ο δεύτερος παράγοντας πάντα θα μένει το 0 στην τρίτη θέση από το τέλος αφού απλά θα έχουμε στην πρόσθεση ολισθήσεις του 1101 που δε θα επηρεάζουν το αποτέλεσμα.
Άρα δεν είναι δυνατόν πολλαπλασιάζοντας το 1101 (=13) με κάποιο αριθμό να κατασκευάσεις αριθμό με όλους τους άσσους μαζεμένους μπροστά και όλα τα μηδενικά πίσω. Και ισοδύναμα δεν είναι δυνατό διαιρώντας 2 αριθμούς της μορφής «οι άσσοι στην αρχή και τα μηδενικά στο τέλος» να φτιάξεις το 1101.

Και επειδή οι αξίες των αριθμών είναι ανεξάρτητες από το σύστημα αρίθμησης που τους αναπαριστούμε, δεν είναι δυνατόν το κλάσμα να (2^α-2^β)/(2^γ – 2^δ) να παράγει το 13.

Ελπίζω να μη μου ξέφυγε κάτι.  Πείτε μου αν υπάρχει λάθος.

Ο αριθμός 13 παράγεται από αυτή τη μορφή

Πραγματι $ 13\equal{}\frac{2^{12}\minus{}1}{2^{6}\minus{}1}\cdot\frac{2^2\minus{}1}{2^4\minus{}1}$   :)

Πράγματι έτσι είναι. Το λάθος μου είναι στη φράση που έκανα bold παραπάνω. Αυτό που έδειξα είναι ότι το ένα κλάσμα δεν μπορεί να φτιάξει το 13 αλλά αυτό δεν είναι αρκετό.
Μπορεί κάποιος  πρώτος να γραφτεί σαν γινόμενο 2 αριθμών που κανένας να μην είναι 1 αλλά να είναι κλάσματα.


The people in a club are planning their summer vacations. Trips $t_1,...,t_n$ are avaibale, but trip $t_i$ has capacity $c_i$. Each person likes some of the trips and will travel on at most one. Interms of which people like which trips, derive a necessary and sufficient condition for being able to fill all trips (to capacity) with people who like them.

For every set S, let $p_S$ be the set of people that desire some element in $\{ t_x | x \in S\}$

Observe the following condition $(*)$:

For every subset $S$ of $\{1, 2, \dots, n\}$, we have

$|p_S| \geq \sum_{x \in S} c_x$

---------

I think this is the condition, but im having trouble proving its sufficent..

its halls theorem.
i can prove suffiency if you give up, but keep trying it is a good excerise


quadratics
inequalities
algebra
inequalities unsolved

Let  $ a,b \in R$ such that
$ 9a^2\plus{}8ab\plus{}7b^2 \leq6$

Prove that :$ 7a\plus{}5b\plus{}12ab \leq9$

[hide]Let $ x = 9a^2 + 8ab + 7b^2$, $ y = 7a + 5b + 12ab$, and $ z = 13a + 11b$.  We must show that if $ x \le 6$, then $ y \le 9$.

From the equation $ z = 13a + 11b$, $ b = (z - 13a)/11$, so
\begin{align*}
x &= 9a^2 + 8ab + 7b^2 \\
&= 9a^2 + 8a \cdot \frac{z - 13a}{11} + 7 \left( \frac{z - 13a}{11} \right)^2 \\
&= \frac{1128a^2 - 94za + 7z^2}{121},
\end{align*}
so $ 128a^2 - 94za + 7z^2 - 121x = 0$.  As a quadratic equation in $ a$, this has a soltuion in $ a$ (given $ z$ and $ x$) if and only its discrminant is nonnegative, i.e. $ (94z)^2 - 4(7z^2 - 121x)(1128) = 22748(24x - z^2) \ge 0$.  Hence, $ z^2 \le 24x \le 144$, so $ |z| \le 12$.

Also,
\begin{align*}
y &= 7a + 5b + 12ab \\
&= 7a + 5 \cdot \frac{z - 13a}{11} + 12a \cdot \frac{z - 13a}{11} \\
&= -\frac{156a^2 - 12za - 12a - 5z}{11},
\end{align*}
so $ 156a^2 - (12z + 12)a + 11y - 5z = 0$.  Again, as a quadratic equation in $ a$, this has a solution in $ a$ (given $ y$ and $ z$) if and only if its discriminant is nonnegative, i.e. $ (12z + 12)^2 - 4(156)(11y - 5z) \ge 0 = 48(3z^2 + 71z - 143y + 3) \ge 0$, so
\[ y \le \frac{3z^2 + 71z + 3}{143}.\]
The maximum value of $ (3z^2 + 71z + 3)/143$ for $ -12 \le z \le 12$ occurs when $ z = 12$, in which case
\[ y \le \frac{3 \cdot 12^2 + 71 \cdot 12 + 3}{143} = 9.\]

We check that when $ a = b = 1/2$, $ 9a^2 + 8ab + 7b^2 = 6$ and $ 7a + 5b + 12ab = 9$.[/hide]

[quote="nsato"][hide]Let $ x = 9a^2 + 8ab + 7b^2$, $ y = 7a + 5b + 12ab$, and $ z = 13a + 11b$.  We must show that if $ x \le 6$, then $ y \le 9$.

From the equation $ z = 13a + 11b$, $ b = (z - 13a)/11$, so
\begin{align*} x & = 9a^2 + 8ab + 7b^2 \\
& = 9a^2 + 8a \cdot \frac {z - 13a}{11} + 7 \left( \frac {z - 13a}{11} \right)^2 \\
& = \frac {1128a^2 - 94za + 7z^2}{121}, \end{align*}
so $ 128a^2 - 94za + 7z^2 - 121x = 0$.  As a quadratic equation in $ a$, this has a soltuion in $ a$ (given $ z$ and $ x$) if and only its discrminant is nonnegative, i.e. $ (94z)^2 - 4(7z^2 - 121x)(1128) = 22748(24x - z^2) \ge 0$.  Hence, $ z^2 \le 24x \le 144$, so $ |z| \le 12$.

Also,
\begin{align*} y & = 7a + 5b + 12ab \\
& = 7a + 5 \cdot \frac {z - 13a}{11} + 12a \cdot \frac {z - 13a}{11} \\
& = - \frac {156a^2 - 12za - 12a - 5z}{11}, \end{align*}
so $ 156a^2 - (12z + 12)a + 11y - 5z = 0$.  Again, as a quadratic equation in $ a$, this has a solution in $ a$ (given $ y$ and $ z$) if and only if its discriminant is nonnegative, i.e. $ (12z + 12)^2 - 4(156)(11y - 5z) \ge 0 = 48(3z^2 + 71z - 143y + 3) \ge 0$, so
\[ y \le \frac {3z^2 + 71z + 3}{143}.
\]
The maximum value of $ (3z^2 + 71z + 3)/143$ for $ - 12 \le z \le 12$ occurs when $ z = 12$, in which case
\[ y \le \frac {3 \cdot 12^2 + 71 \cdot 12 + 3}{143} = 9.
\]
We check that when $ a = b = 1/2$, $ 9a^2 + 8ab + 7b^2 = 6$ and $ 7a + 5b + 12ab = 9$.[/hide][/quote]
The solution is very interesting

[quote="vuthanhtu_hd"]Let  $ a,b \in R$ such that
$ 9a^2 \plus{} 8ab \plus{} 7b^2 \leq6$

Prove that :$ 7a \plus{} 5b \plus{} 12ab \leq9$[/quote]
By AM-GM Inequality, we see that
$ 7a\plus{}5b\plus{}12ab \le 7\left( a^2\plus{}\frac{1}{4}\right) \plus{}5\left( b^2\plus{}\frac{1}{4}\right) \plus{}12ab \equal{}(9a^2\plus{}8ab\plus{}7b^2) \minus{}2(a\minus{}b)^2\plus{}3 \le (9a^2\plus{}8ab\plus{}7b^2) \plus{}3 \le 6\plus{}3\equal{}9.$
Equality holds if and only if $ a\equal{}b\equal{}\frac{1}{2}.$
:)

[quote="can_hang2007"][quote="vuthanhtu_hd"]Let  $ a,b \in R$ such that
$ 9a^2 \plus{} 8ab \plus{} 7b^2 \leq6$

Prove that :$ 7a \plus{} 5b \plus{} 12ab \leq9$[/quote]
By AM-GM Inequality, we see that
$ 7a \plus{} 5b \plus{} 12ab \le 7\left( a^2 \plus{} \frac {1}{4}\right) \plus{} 5\left( b^2 \plus{} \frac {1}{4}\right) \plus{} 12ab \equal{} (9a^2 \plus{} 8ab \plus{} 7b^2) \minus{} 2(a \minus{} b)^2 \plus{} 3 \le (9a^2 \plus{} 8ab \plus{} 7b^2) \plus{} 3 \le 6 \plus{} 3 \equal{} 9.$
Equality holds if and only if $ a \equal{} b \equal{} \frac {1}{2}.$
:)[/quote]

Very nice,thanks you!

[quote="nsato"][hide]Let $ x = 9a^2 + 8ab + 7b^2$, $ y = 7a + 5b + 12ab$, and $ z = 13a + 11b$.  We must show that if $ x \le 6$, then $ y \le 9$.

From the equation $ z = 13a + 11b$, $ b = (z - 13a)/11$, so
\begin{align*} x & = 9a^2 + 8ab + 7b^2 \\
& = 9a^2 + 8a \cdot \frac {z - 13a}{11} + 7 \left( \frac {z - 13a}{11} \right)^2 \\
& = \frac {1128a^2 - 94za + 7z^2}{121}, \end{align*}
so $ 128a^2 - 94za + 7z^2 - 121x = 0$.  As a quadratic equation in $ a$, this has a soltuion in $ a$ (given $ z$ and $ x$) if and only its discrminant is nonnegative, i.e. $ (94z)^2 - 4(7z^2 - 121x)(1128) = 22748(24x - z^2) \ge 0$.  Hence, $ z^2 \le 24x \le 144$, so $ |z| \le 12$.

Also,
\begin{align*} y & = 7a + 5b + 12ab \\
& = 7a + 5 \cdot \frac {z - 13a}{11} + 12a \cdot \frac {z - 13a}{11} \\
& = - \frac {156a^2 - 12za - 12a - 5z}{11}, \end{align*}
so $ 156a^2 - (12z + 12)a + 11y - 5z = 0$.  Again, as a quadratic equation in $ a$, this has a solution in $ a$ (given $ y$ and $ z$) if and only if its discriminant is nonnegative, i.e. $ (12z + 12)^2 - 4(156)(11y - 5z) \ge 0 = 48(3z^2 + 71z - 143y + 3) \ge 0$, so
\[ y \le \frac {3z^2 + 71z + 3}{143}.
\]
The maximum value of $ (3z^2 + 71z + 3)/143$ for $ - 12 \le z \le 12$ occurs when $ z = 12$, in which case
\[ y \le \frac {3 \cdot 12^2 + 71 \cdot 12 + 3}{143} = 9.
\]
We check that when $ a = b = 1/2$, $ 9a^2 + 8ab + 7b^2 = 6$ and $ 7a + 5b + 12ab = 9$.[/hide][/quote]
The idea is nice.But why we let $ z=13a+11b$.Is it a trick?

[quote="dulldrake"]
The idea is nice.But why we let $ z \equal{} 13a \plus{} 11b$.Is it a trick?[/quote]

If you graph the equations $ 9a^2 \plus{} 8ab \plus{} 7b^2 \equal{} 6$ and $ 7a \plus{} 5b \plus{} 12ab \equal{} 9$, you find that they are tangent at the point $ (1/2, 1/2)$, and the tangent line is $ 13a \plus{} 11b \equal{} 12$.


I guess this doesn't go here bu I couldn't exactly find where this goes...

How do I convert rectangular to polar coordinates using my calculator. I used to do it the regular way until I read in my precalculus textbook:

"Most graphing and scientific calculators come equipped with a polar to rectangular conversion key and vice versa"

What is this key and how do I use this function

One thing is obvious - I can't see it.

(I am using a Ti-84 SE)

I know almost everything about the TI-83+ series calculators and as far as I know, there is not a built in function to convert coordinates.  However, you can graph in different styles, etc.  I would check the manual; if there is such a function, it would be listed in the manual.

I could make a program for you to do the conversions, if you want...


function
inequalities unsolved
inequalities

You can find it in "Functions of one complex variable II" by Conway, published in the Springer series "Graduate Texts in Mathematics". It is obviously not an olympiad section material.


Hey!
Is anyone here participating in the IChO?
I have a question concerning the Problem Set Evaluation.
When I was submitting the answers for the first three problem sets, I noticed that I could enter the answers for ore than once!
Are they taking the scores I entered the first time or the highest scores?

Probably only a glitch in the system. Only the first one should count.

But if you really aspire to make the IChO team, you should be aware that the multiple choice really doesn't count THAT much in the end.


geometry
analytic geometry
graphing lines
slope
rotation
trigonometry

Consider the points $A(0,12)$, $B(10,9)$, $C(8,0)$, and $D(-4,7)$. There is a unique square $S$ such that each of the four points is on a different side of $S$. Let $K$ be the area of $S$. Find the remainder when $10K$ is divided by $1000$.

[hide="Solution"]Consider a point $E$ such that $AE$ is perpendicular to $BD$, $AE$ intersects $BD$, and $AE=BD$. E will be on the same side of the square as point $C$.

Let the coordinates of $E$ be $(x_E,y_E)$. Since $AE$ is perpendicular to $BD$, and $AE=BD$, we have $9-7=x_E-0$ and $10-(-4)=12-y_E$
The coordinates of $E$ are thus $(2,-2)$.

Now, since $E$ and $C$ are on the same side, we find the slope of the sides going through $A$ and $C$ to be $\frac{-2-0}{2-8}=\frac{1}{3}$. Because the other two sides are perpendicular, the slope of the sides going through $B$ and $D$ are now $-3$.

Let $A_1,B_1,C_1,D_1$ be the vertices of the square so that $A_1B_1$ contains point $A$, $B_1C_1$ contains point $B$, and etc. Since we know the slopes and a point on the line for each side of the square, we use the point slope formula to find the linear equations. Next, we use the equations to find $2$ vertices of the square, then apply the distance formula.

We find the coordinates of $C_1$ to be $(12.5,1.5)$ and the coordinates of $D_1$ to be $(-0.7,-2.9)$. Applying the distance formula, the side length of our square is $\sqrt{\left( \frac{44}{10} \right)^2 + \left( \frac{132}{10} \right)^2}= \frac{44}{\sqrt{10}}$

Hence, the area of the square is $K=\frac{44^2}{10}$. The remainder when $10K$ is divided by $1000$ is $936$.
  [/hide]

Me and my professor worked on this problem over the past few days, and came up with a different solution.

[hide="solution"]
The sides of the square will lie on the sides $\ell_{1}$, $\ell_{2}$, $\ell_{3}$, and $\ell_{4}$, with $\ell_{1}\parallel \ell_{3}$ and $\ell_{2}\parallel \ell_{4}$. $\ell_{1}$ and $\ell_{3}$ are of the form $\ell_{i}=mx+b_{i}$ and $\ell_{2}$ and $\ell_{4}$ are of the form $\ell_{i}=-m^{-1}x+b_{i}$. In order to find the area of the square, we need only find the distance between the lines, $b_{3}-b_{1}$ and $b_{4}-b_{2}$. However, we must normalize:
$\frac{b_{3}-b_{1}}{\sqrt{m^{2}+1}}$ and $\frac{b_{4}-b_{2}}{\sqrt{m^{-2}+1}}=\frac{m*(b_{4}-b_{2})}{\sqrt{m^{2}+1}}$
Since this is a square, the distance between the parallel lines will be equal, that is:
$\frac{b_{3}-b_{1}}{\sqrt{m^{2}+1}}=\frac{m*(b_{4}-b_{2})}{\sqrt{m^{2}+1}}$.
$\frac{1}{\sqrt{m^{2}+1}}$ cancels out and we have: $b_{3}-b_{1}=mb_{4}-mb_{2}$. Now we need to find expressions for the $y$-intercepts in terms of $m$. We'll have $\ell_{1}$ pass through $A$ and $\ell_{2}$ through $B$. Then we have $y_{1}= mx+12$, $y_{2}=-\frac{x}{m}+\frac{10}{m}+9$, $y_{3}= mx-8m$, $y_{4}=-\frac{x}{m}-\frac{4}{m}+7$, so $b_{1}=12$, $b_{2}=\frac{10}{m}+9$, $b_{3}=-8m$, and $b_{4}=-\frac{4}{m}+7$, and our equation becomes:
$-8m-12 = m(\frac{-4+7m-(10+9m)}{m}\Rightarrow 12+8m=14+2m\Rightarrow m=1/3$. Then $b_{3}=8*(\frac{1}{3})$ and our side is $\frac{-\frac{8}{3}-12}{\sqrt{\frac{1+9}{9}}}= \frac{-44}{\sqrt{10}}$ (the negative sign doesn't really matter as we'll be squaring it anyway). The area of the square is $\frac{44^{2}}{10}$, and 10 of these is $44^{2}=1936$, and the remainder when divided by 1000 is $936$.
[/hide]

I always use complex coordinates for these types of problems  :D (Mr. Green).

Set up the points on a complex coordinate system.
Our goal is to rotate the points and the square by an angle of $ \theta$ so the sides are parallel to the axes.
Then, $ Im(A' \minus{} C') \equal{} Re(B' \minus{} D')$=side length of the square.
So we have
$ Im(12i cis \theta \minus{} 8 cis \theta) \equal{} Re((10 \plus{} 9i) cis \theta \minus{} ( \minus{} 4 \plus{} 7i) cis \theta)$
$ 12 cos \theta \minus{} 8 sin \theta \equal{} 10 cos \theta \minus{} 9sin \theta \plus{} 4 cos \theta \plus{} 7 sin \theta$
$ 2 cos \theta \equal{} \minus{} 6 sin \theta$, so $ \theta \equal{} \arctan \minus{} \frac {1}{3}$
Subbing back in we have
$ Im(12i cis \theta \minus{} 8 cis \theta)\equal{}
12 cos  (\arctan \minus{} \frac {1}{3}) \minus{} 8 sin  (\arctan \minus{} \frac {1}{3})\equal{} \newline
12\cdot \frac{3}{\sqrt{10}}\minus{}8\cdot \frac{\minus{}1}{\sqrt{10}}\equal{}\frac{44}{\sqrt{10}}$
So the side length of the square is $ \frac {44}{\sqrt {10}}$.

How did you simplify the last part... I'm making a careless mistake somewhere:

$ 12 \cos \theta \minus{} 8 \sin \theta \equal{} (\minus{}4\cos \theta \minus{} 7 \sin \theta) \minus{} (10 \cos \theta \minus{} 9 \sin \theta)$
$ 26 \cos \theta \equal{} 10 \sin \theta$
$ 13 \cos \theta \equal{} 5 \sin \theta$
$ \tan \theta \equal{} \frac{13}{5}$

Yes, I realized right before you posted that I copied it down wrong, I subtracted in the wrong order. I have fixed it.

Now how do you plug that $ \arctan \minus{} \frac {1}{3}$ back into the original equation... things don't work out that nicely as far as I can tell?

EDIT:  Never mind I figured it out. :roll:  :blush:

Nice problem and nice solution.  http://www.contestmath.com/mathdocs/Problem_20_Complex_Numbers.pdf

[quote="dgreenb801"]Thank you for making my solution into a nice PDF document!  :)
By the way, you don't need to use that $ sin^2x \plus{} cos^2x \equal{} 1$.
Instead, knowing that arctan $ \theta \equal{} \minus{} \frac {1}{3}$, you can draw a picture of a triangle in the fourth quarant with vertices (0,0), (3,0), and (3,-1) and use socahtoa from there (that's how we did it in school, I know it's actually the same thing).[/quote]

That's what I did too! haha but I didn't feel like explaining that.  That's part of a series of [url=http://www.contestmath.com/solved.html]solved problems[/url] from AoPS...

I'm going to shamelessly bump this thread to present a solution to the general problem of constructing a square on whose sides 4 given points lie (each point lie on a different side).

 [hide]
 [geogebra]1123c32b3331b1b14bda28be6bde23ab4f3b940e[/geogebra]
Given points $ A,B,C,D$, we attempt to construct square $ EFGH$ such that $ HE$ contains $ A$, $ EF$ contains $ B$, and so on.
Then we have $ \angle AEB\equal{}\angle BFC\equal{}\angle CGD\equal{}\angle DHA\equal{}90$. Thus $ E$ lies on the circle with $ AB$ as diameter, and similarly for all other points.

Also, since $ EG$ bisects angles $ HEF$ and $ FGH$, $ EG$ must go through the midpoints of the arcs $ AB$ and $ CD$ (the arcs that are inside the square). Thus connect them and let the line be $ l$. Then the intersection of $ l$ and the circle with diameter $ AB$ is $ E$, and that with the circle with diameter $ CD$ is $ G$. Similarly we can find $ H$ and $ F$.
[/hide]

Eleven year bump, whoops.  Here's a solution with minimal computation.
[asy]
size(200);
defaultpen(linewidth(0.6)+fontsize(10));
pair A = (0,12), B = (10,9), C = (8,0), D = (-4,7);
pair E = (-5.1,10.3), F = (8.1,14.7), G = (12.5,1.5), H = E+G-F;
pair X = foot(C,E,F), Y = foot(B,E,H);
draw(C--X^^rightanglemark(A,X,C,16)^^B--Y^^rightanglemark(B,Y,H,16),gray(0.6));
draw(E--F--G--H--cycle);
dot("$A(0,12)$",(0,12),NW);
dot("$B(10,9)$",(10,9),NE);
dot("$C(8,0)$",(8,0),SE);
dot("$D(-4,7)$",(-4,7),SW);
dot("$X$",X,NW,gray(0.6));
dot("$Y$",Y,SW,gray(0.6));
[/asy]
Let $X$ be the projection of $C$ onto the side of the square containing $A$, and let $Y$ be the projection of $B$ onto the side of the square containing $D$.  Observe that, since the given figure is a square, $CX = BY$.

Consider the transformation $\Phi$ of the plane defined by rotating $90^\circ$ counterclockwise about the origin and translating by the vector $\langle -1,10\rangle$.  Then $\Phi(B) = C$ by construction.  Since $\Phi$ is an orthogonal transformation of the plane (i.e. does not change distances), line segment $\overline{BY}$ is taken to line segment $\overline{CX}$, so $\Phi(Y) = X$ as well.  It follows that the side of the square containing $A$ must also contain the point $\Phi(D) = (6,14)$, meaning that the slope of that side, as a line, is $\tfrac 13$.

From this, the coordinates of $X$ are of the form $(8-t,3t)$ for some $t > 0$.  The line connecting $A$ and $X$ has slope $\tfrac 13$, so
\[
\frac{3t - 12}{8 - t} = \frac 13,
\]
and so $t = \tfrac{22}5$.  It follows that the side length of the square is $K = \tfrac{22}{5}\sqrt{10} = \tfrac{44}{\sqrt{10}}$, and the requested answer is $\boxed{936}$.


Putnam geometry LaTeX videos Alcumus algebra function

There is a ISBN for every published book. There are databases of information about books. Maybe not all of the books, but clearly a lot of them. Like Amazon.

So I thought, what if there is a unique id for every problem published? and a database to hold information about all those problems. Then it's uniquely identify any problem.

By information about the problem, it can hold relations between problems, for example, if 2 problems are identical, or one is the generalization of the other.
It can hold information of the source of the problem, like from Putnam 1997.
For problems without copyright, the database can even have the source of the problem.

Many possible applications could come out of this. It is possible to filter problem out of the entire set of problems by specify which qualities the problem should have. Like only return numbery theory or geometry IMO problems.

There can even be API's, where user can take advantage of it and create different kind of useful applications. One of the possible ones is aggregate the latex source of the problem, and automatically create worksheets.

The database can even be wiki like, where users can do modifications.
A community can even base around it. Where people can discuss the problem, write their own solutions.

I have thought about this system for a long time(like started last year). Technology is fit enough to create a database like this 10 years ago. but no such thing is really come to about. AoPS wiki is close, but a lot of objectives are really hard to achieve though Mediawiki.

The closet I have come is made a very inefficient and incomplete prototype with Drupal. The video featured how user can create their own problem sets out of problems, and generate latex source for the entire problem set. how user can filter so only some problems shows up.

[youtube]hy0a_Lf9lCA[/youtube]

The prototype is not well designed and have many problems to it. But it offers an insight of what could be possible.

I have talked to a few people about this, and they all seem enthusiastic about something like this. but why such things are never made? It will be awesome if AoPS take this project.

Something kind of along those lines is [url=http://www.projecteureka.org/problem]ProjectEureka[/url]. Alcumus can also be thought of as a sort of organized problem database. But I think your idea has some merit. The biggest problem would be copywrite issues for problems from copywritten sources.

ProjectEureka is pretty much a trimmed down version of what can be achieved. It's using a subset of problems with quantitative solutions. It use a tagging system for classification. The test system pretty much what I implemented as problem sets.

Alcumus is quite close, but it is designed for teaching thus have some restrictions. Like it is not a wiki where everyone can edit and add more problems.

True, copyright can be a problem. It will stop a lot of problem entering the database. Those problems can be just placeholders. I believe there are enough problems in the public domain to create a decent database.

To me, I think it is hard to design the system well enough to get it started.
How can the problems be categorized? Currently I think there are a 4 mutually exclusive data for any general problem: topics(like number theory), source(imo 2009), types(multiple choice, quantitative answer, programming problem...), misc(the rest, like incomplete, problem has typo, contain latex source).
Within each general problem, different type have type specific data, which is also useful in filtering out results. For programming problems, one might specify problems with memory limit larger than 100kB. for multiple choice problems, one might specify problems with only 4 choices.
How do we categorize though source? tags are useful, but they are not strctured. If someone tagged all IMO problems with IMOXXXX(where xxxx is the year number), if someone just search for problems tagged with IMO, nothing will show up. That's not too much of a challenge. Add a ability to assign parent child relationship between tags will fix it. so someone can create the tag IMO as parent of all IMO problems. tag 2009 as parent of all problems from 2009's contest. It creating a direct acyclic graph for categorization.
Type related data, is usually faceted metadata, thus a complete different way to categorize them.

That's enough decisions for one to spend a long time on and there are a lot of detail omitted, like what if two tags means the same thing, if someone tagged a problem with both a parent tag and it's child tag, how should it be handled.
What should be done when there is duplicate problems? How should trivial variation of a problem be treated?

These are not hard to resolve, but such problems keep popping up one by one. sometimes one might realize some decision have hindered some required features.


algebra
polynomial
superior algebra
superior algebra unsolved

Please see whether the following is correct and suggest better arguments. I like to prove that $ f(x) = (x - a_1)\cdots(x - a_n) + 1$ where $ a_i$ are distinct odd integers is irreducible over $ \mathbb Z$ whenever n>2.

Given $ f(x) = (x - a_1)\cdots(x - a_n) + 1$ where $ a_i$ are distinct odd integers. If it were reducible over $ \mathbb Z$, then $ f(x) = g(x)h(x)$, where $ g,h\in\mathbb Z[x]$. Now, for all $ a_i$, we'd have that $ g(a_i)h(a_i) = 1$, i.e., $ g(a_i) = h(a_i) = \pm 1$. This means, any $ a_i$ is a root of both $ g(x) + 1,h(x) + 1$ or of both $ g(x) - 1,h(x) - 1$. So,
\[ g(x) - 1 = (x - a_1)\cdots(x - a_r)\bar g(x);
\]

\[ h(x) + 1 = (x - a_{r + 1})\cdots(x - a_n)\bar h(x).
\]
If any of the $ \bar g,\bar h$ had positive degree, then $ g(x)h(x)$ would have degree greater than $ f(x)$, impossible. Again, the coefficient of $ x^n$ is one in $ f(x)$, so, $ \bar g\bar h = 1$. Similarly,
\[ h(x) - 1 = a(x - a_1)\cdots(x - a_r);
\]

\[ g(x) + 1 = b(x - a_{r + 1})\cdots(x - a_n).
\]
where $ ab = 1$. So,
\[ g(x)h(x) = a\bar g\{(x - a_1)\cdots(x - a_r)\}^2 + (a + \bar g)(x - a_1)\cdots(x - a_r) + 1.
\]
So it means that $ n = 2r$ and $ a\bar g = 1$. So, $ a + \bar g = \pm 2.$ Equating $ g(x)h(x)$ with $ f(x)$, we get that
\[ (x - a_1)\cdots(x - a_r)\{(x - a_{r + 1})\cdots(x - a_n) - (x - a_1)\cdots(x - a_r)\pm 2\} = 0.
\]
Which gives,
\[ (x - a_{r + 1})\cdots(x - a_n) - (x - a_1)\cdots(x - a_r)\pm 2 = 0\cdots\cdots (1)
\]
. Now, if $ r > 1$, then it is impossible. Because, putting $ x = a_1$, the part $ (x - a_{r + 1})\cdots(x - a_n)$ involves product of more than one even numbers, which would always be greater than 2 in principal value; and the equation (1) can never hold.

Someone please pay a little attention to it! I re-emphasise the problem:

Given distinct odd integers $ a_1,\cdots, a_n$, the polynomial \[ f(x)\equal{}(x\minus{}a_1)\cdots(x\minus{}a_n)\plus{}1\] is irreducible over integers (equivalently, over rationals).

[quote="saurav"]$ g(a_i) \equal{} h(a_i) \equal{} \pm 1$.[/quote]
This gives $ n$ roots of $ g(x) \minus{} h(x)$ with degree $ < n$.
So $ g(x) \equal{} h(x)$ and $ f(x)\minus{}1 \equal{} g(x)^2 \minus{} 1 \equal{} (g(x) \minus{} 1)(g(x) \plus{} 1)$.
Then $ n \equal{} 2m$ even and after some ordering (assume WLOG $ g(x)$ is monic)
$ g(x) \minus{} 1 \equal{} (x \minus{} a_1)\cdots (x \minus{} a_m)$ and $ g(x) \plus{} 1 \equal{} (x \minus{} a_{m \plus{} 1})\cdots (x \minus{} a_{2m})$.
Then $ a_i \minus{} a_j\mid g(a_i) \minus{} g(a_j) \equal{} 2$ for all $ i\le m,j > m$,
and therefore the maximal distance between any two $ a_i,a_j$ is at most $ 4$.
This gives $ \{a_1,\ldots, a_n\}\subseteq \{\alpha,\alpha \plus{} 1,\alpha \plus{} 2,\alpha \plus{} 3,\alpha \plus{} 4\}$ for some $ \alpha\in\mathbb{Z}$ and $ n \equal{} 2$ or $ 4$.
Little case work gives the only two exceptions
$ f(x) \equal{} (x \minus{} \alpha)(x \minus{} \alpha \minus{} 2) \plus{} 1 \equal{} (x \minus{} \alpha \minus{} 1)^2$ and
$ f(x) \equal{} (x \minus{} \alpha)(x \minus{} \alpha \minus{} 1)(x \minus{} \alpha \minus{} 2)(x \minus{} \alpha \minus{} 3) \plus{} 1 \equal{} ((x \minus{} \alpha)^2 \minus{} 3(x \minus{} \alpha) \plus{} 1)^2$
with some $ \alpha\in\mathbb{Z}$. (forget about odd)
Though, your proof is also correct.

Thank you very much olorin, your result is much more general! With an exception for n=2 or 4, it shows that such polynomials are irreducible, no matter whether or not $ a_i$ are odd.

By the way, why did you say "only two" exceptions? Is there no other exception, I mean one of the form not included in those two?

[quote="saurav"]By the way, why did you say "only two" exceptions? Is there no other exception, I mean one of the form not included in those two?[/quote]
Yes exactly.
$ \{a_1,\ldots, a_n\}\subseteq \{\alpha,\alpha \plus{} 1,\alpha \plus{} 2,\alpha \plus{} 3,\alpha \plus{} 4\}$
gives only finitely many exceptions for each $ \alpha$. (there are only $ 2^5$ subsets)
And a little more case work gives only the two exceptions I posted.

This can in fact be generalized to
$ f(x)\equal{}c(x\minus{}a_1)\cdots (x\minus{}a_n)\plus{}1$ for $ c\in\mathbb{Z},c\not\equal{}0$
with only one more exception
$ f(x)\equal{}4(x\minus{}\alpha)(x\minus{}\alpha\minus{}1)\plus{}1\equal{}(2(x\minus{}\alpha)\minus{}1)^2$
for some $ \alpha\in\mathbb{Z}$.


geometry
perimeter
AMC
USA(J)MO
USAMO
geometry proposed

ABC is a triangle such that $ Â=2B$ (angles)

and $ C$ is an obtuse angle and $ BC$, $ AB$, $ AC$ are natural numbers.

find the least value possible of the perimeter of ABC.

Where this problem was proposed, please ?

It's from USAMO, 1991:
[url]http://www.artofproblemsolving.com/Forum/viewtopic.php?t=58215[/url]


group theory
abstract algebra
number theory
relatively prime
superior algebra
superior algebra solved

1.
If $N\lhd G$, $[G:N]$ finite, $H<G$, $|H|$ finite, and $[G:N]$ and $|H|$ are relatively prime, then $H<N$.
2.
If $N\lhd G$, $|N|$ finite, $H<G$, $[G:H]$ finite, and $[G:H]$ and $|N|$ are relatively prime, then $N<H$.

$(1)$:

$N\cap H\lhd H$ and $H/N\cap H\cong NH/N$, which means that $|H/N\cap H|=|NH/N|\ |\ |G/N|$. Also, $|H/N\cap H|\ |\ |H|$, which means that $|H/N\cap H|=1\Rightarrow H\le N$, as desired.

$(2)$:

We have $[G:H]=[G:NH][NH:H]$, so $[NH:H]\ |\ [G:H]$. It is also readily seen that $[NH:H]\ |\ |N|$, so $[NH:H]=1$, and the conclusion follows.


trigonometry inequalities calculus derivative

Solve the equation: $ \sin^8x\plus{}\cos^8x\equal{}64(\sin^{14}x\plus{}\cos^{14}x)$

hello, your equation is equivalent to
$ 1681\plus{}1974\cos(4x)\plus{}363\cos(8x)\plus{}14\cos(12x)\equal{}0$
this equation has no real solutions.
Sonnhard.

[quote="Dr Sonnhard Graubner"][hide]hello, your equation is equivalent to
$ 1681 \plus{} 1974\cos(4x) \plus{} 363\cos(8x) \plus{} 14\cos(12x) \equal{} 0$
this equation has no real solutions.
Sonnhard.[/hide][/quote]

It would be nice if you could show the manipulation for this and why it doesn't have real solutions, as it doesn't appear to be obvious.

Alternate solution:

[hide]Because the period of each expression on each side of the equation is $ \pi$, we only need to consider values of x less than this and greater than 0, inclusive.

Apply AM-GM to $ \sin ^{14}{x}$ and $ \cos ^{14}{x}$ to yield:

$ \sin ^{14}{x} \plus{} \cos ^{14}{x} \geq 2\sin ^{7}{x}\cos ^{7}{x}$

The product on the RHS is maximized when $ x \equal{} \frac {\pi}{4}$, and equality in our AM-GM inequality holds here. At this value, $ \sin ^{14}{x} \plus{} \cos ^{14}{x} \equal{} \frac {1}{64}$ Thus, the minimum value of the original RHS expression of the equation is 1.

Take the derivative of the LHS of the equation to find all maxima and minima in the desired range. The derivative of $ \sin^{8}x \plus{} \cos^{8}x$ is, after factoring:

$ 8\cos{x}\sin{x}(\sin^6{x} \minus{} \cos^{6}{x})$

Thus, it is maximized or minimized at values of $ x \equal{} 0, \frac {\pi}{4} , \frac {\pi}{2} , \frac {3\pi}{4}, \pi$
Testing all of these, we find that it reaches a maximum value of 1 at $ x \equal{} 0, \frac {\pi}{2}, \pi$. Since this is the minimum value of the RHS of the equation as established before, all solutions must occur at these values. Testing these values, we find that none are solutions. Thus, the original equation has no solutions.[/hide]

[quote="phuong"]Solve the equation: $ \sin^8x \plus{} \cos^8x \equal{} 64(\sin^{14}x \plus{} \cos^{14}x)$[/quote]

$ \frac{sin^8 x\plus{}cos^8 x}{64} \equal{}(sin^{14} x\plus{}cos^{14} x)(sin^2 x\plus{}cos ^2 x) \ge (sin^8 x\plus{}cos^8 x)^2$ by Cauchy Schwarz

$ \frac{1}{64} \ge sin^8 x\plus{}cos^8 x$

$ \sqrt[4]{\frac{1}{128}} \ge \sqrt[4]{\frac{sin^8 x\plus{}cos^8 x}{2}} \ge \frac{sin^2 x\plus{}cos^2 x}{2}\equal{}\frac{1}{2}$ by power mean.

But $ \sqrt[4]{\frac{1}{128}} < \frac{1}{2}$. So there are no real solutions.


trigonometry
logarithms
quadratics
algebra
complex analysis

Im trying to compute arccos z in terms of log z. really this was covered in class and I have it written but I just don't understand what I wrote.

$w = \arccos z$
$\displaystyle z = \cos w = \frac{e^{iw}-e^{-iw}}{2}$
$0 = e^{2iw} - 2ze^{iw} + 1$
$e^{iw} = z + \sqrt{z^2-1}$
$w = -i \log (z + \sqrt{z^2-1})$

where the heck did I come up with $0 = e^{2iw} - 2ze^{iw} + 1$ ?

thanks for any help

You got it from $z=\cos w=\frac{e^{iw}+e^{-iw}}2.$

Note that you've written the defnition of the cosine wrong in your post.

yes sorry, I meant the + sign.

I know that I got it from that line, but why do I set it to zero? how do I get from one line to the other.

You're trying to solve the equation $z=\frac{e^{iw}+e^{-iw}}2$ for $w.$

You multiplied by $2e^{iw}$ and then moved everything to one side (with zero on the other side) for the sake of familiarity: so you could recognize it as a quadratic equation for $e^{iw}$ and use the quadratic formula.


[b]24[/b]

hmm... more boys than girls, but girls are PWNING... wierd :huh:

[b]-1[/b]

Haha girls win. Where are all the boys?   :huh:


geometry
perimeter
function
algebra proposed
algebra

Here is an interesting little problem which I seen solved,quite cleverly, on another site.

"Find a function that inputs the perimeter of a right triangle with area $25 ft^{2}$ and gives length of the hypoteneuse".

Maybe someone can come up with another clever method. It's not a killer problem, just fun.

[hide="Solution"]
We have $ab = 50, a^2 + b^2 = c^2$ and we wish to solve for $c$ given $a+b+c = p$.  Squaring, we have $a^2 + b^2 + c^2 + 2ab + 2bc + 2ac = 2c^2 + 100 + 2c(a+b) = 2c^2 + 100 + 2c(p - c) = 100 + 2cp = p^2$.

From this we easily derive $c = \frac{p^2 - 100}{2p}$.
[/hide]

Cool!. That's exactly how it was done on the other site. Kudos.


Solve this system in $ \mathbb R$:

xy(x-y)=-2
yz(y-z)=12
zx(z-x)=6

[quote="trbst"]Solve this sistem in R:

xy(x-y)=-2
yz(y-z)=12
zx(z-x)=6[/quote]

Let $ y\equal{}x\plus{}u$ and $ z\equal{}y\plus{}v\equal{}x\plus{}u\plus{}v$. The equations become :

$ ux^2\plus{}u^2x\equal{}2$
$ vx^2\plus{}(2uv\plus{}v^2)x\equal{}\minus{}12$
$ (u\plus{}v)x^2\plus{}(u\plus{}v)^2x\equal{}6$

And, subtracting the first and the second from the third : $ 0\equal{}16$ and so, as Dr Sonnhard previously said : no solution (real or complex).

trbst asked me thru pm to check my  post.

And, ...;  :blush:  there is a mistake :

[quote="pco"] Let $ y \equal{} x \plus{} u$ and $ z \equal{} y \plus{} v \equal{} x \plus{} u \plus{} v$. The equations become :

$ ux^2 \plus{} u^2x \equal{} 2$
$ vx^2 \plus{} (2uv \plus{} v^2)x \equal{} \minus{} 12$[/quote]
It should be $ vx^2 \plus{} (2uv \plus{} v^2)x \plus{}uv(u\plus{}v)\equal{} \minus{} 12$

And, subtracting the first and the second from the third no longer gives contradiction.
Sorry.

The system seems to have a solution.
Try (1,-1,3)

hello, your solution is correct, since
$ xy(x\minus{}y)\equal{}\minus{}1(1\plus{}1)\equal{}\minus{}2$
$ yz(y\minus{}z)\equal{}\minus{}3(\minus{}1\minus{}3)\equal{}12$
$ zx(z\minus{}x)\equal{}3(3\minus{}1)\equal{}6$
Sonnhard.


When I push the build button, I get errors. One of them is "premeable.tex not found".
In my preamble I have written "\input{premeable.tex}".

I am working in technixcenter. Please help.

Make sure that premeable.tex is in the same directory as the tex file you are building.


How to add it where?

Did you check the MathType documentation?  (MathType is not LaTeX, so you may not find help here -- this site uses LaTeX almost exclusively.)

I believe you can translate from MathType language to LaTeX using the translater :)


the braking distance for a truck is given by the formula d=0.2v 2 +0.015v where d is in metres and v is in metres per second.. find its stopping distance at a speed of 60 kilometres/hour

[quote="3X.lich"]This is more like physics than math.

About the formula, did you mean $d=0.2v^2+0.015v$?

Next, convert $60\,\mathrm{km}/\mathrm{hr}$ into $\mathrm{m}/\mathrm{sec}$ with conversions:
$1\,\mathrm{km}=1000\,\mathrm{m}$
$1\,\mathrm{hr}=3600\,\mathrm{sec}$[/quote]

yea I did, but I got the answer already. thanks
[color=blue]it's 56.03 m[/color]

[quote="michaeln"][hide]I came up with 55.80555555555555555555... meters.

The first step is to decipher how many meters per second this truck is going. To do this, find out the meters per hour (60 kilometers is 60,000, so 60,000 per hour) and divide that by the number of seconds in an hour (3600 seconds in an hour.) after doing that, you should have ~16.66666666667 meters per second. then you simply plug that number into the equation and you get d=55.805555555555...

the equation IS suposed to be d= instead of d+, right? ;)[/hide][/quote]

yeah :D
sorry about that, I typed + instead of =
I guess there is some difference in the answers because I rounded stuff up along the way ;)


I have to leave the country due to an emergency. I will be back on the morning of the AMC B so I will miss taking the exam with the rest of my classmates. If I were to take it in the afternoon, after the rest of my classmates, but still on the same day would my score count as official?

It should.  Usually for the AMC all that matters is the day...

The AMC has special rules for students at the same school taking the tests at different times.  Contact them for details.


The question is:

What is the remainder when $ 19^{2003}$ is divided by $ 23$?

I have tried using Euler-Fermat (Totient) on it, but it does not work out.

[hide="this is what I have so far"]23, of course, has 22 coprime integers less than 23, so I get:

$ 19^{22}\equiv1(\mod 23)$
${ 19^{2003} = 19^{({22}^{90} + 21})}$
     $ = (19^{22})^{90}\times19^{21}$
$ \equiv1^{90}\times19^{21}$
$ 19^{21}\equiv x(\mod 23)$[/hide]

How do I simplify from here?

The problem asks for the remainder upon division by $ 6$.  Why are you trying to find the remainder upon division by $ 23$?

[hide="If 6"] $ 19 \equiv 1 \bmod 6$. [/hide]
[hide="If 23"] $ 19 \equiv \minus{}4 \bmod 23$.  Then

$ (\minus{}4)^{21} \equiv (\minus{}64)^7 \equiv 5^7 \equiv 5 \cdot 125^2 \equiv 5 \cdot (\minus{}13)^2 \equiv 5 \cdot 169 \equiv 5 \cdot 8 \equiv 40 \equiv \boxed{17} \bmod 23$ [/hide]
[hide="Discussion"] For reasonably small moduli, there is a more efficient algorithm to compute large modular powers known as [url=http://en.wikipedia.org/wiki/Exponentiation_by_squaring]exponentiation by squaring[/url].[/hide]

[quote="brianb"]The question is:

What is the remainder when $ 19^{2003}$ is divided by $ 6$?

I have tried using Euler-Fermat (Totient) on it, but it does not work out.

[hide="this is what I have so far"]23, of course, has 22 coprime integers less than 23, so I get:

$ 19^{22}\equiv1(\mod 23)$
${ 19^{2001} = 19^{({22}^{90} + 21})}$
     $ = (19^{22})^{90}\times19^{21}$
$ \equiv1^{90}\times19^{21}$
$ 19^{21}\equiv x(\mod 23)$[/hide]

How do I simplify from here?[/quote]
Didn't I give a solution on your post with 19 problems?  Or do you not understand it?  Please tell me if you don't.   Here's the link anyways: http://www.artofproblemsolving.com/Forum/viewtopic.php?t=177417

Actually, repeated squaring is a good method for exponentiation in general, even not modulo a number. Notably, it's used in RSA encryption where both the exponent and modulus are huge.

Huh....what's wrong with
[hide="this"]
$ \phi(23)\equal{}22$
$ 19^{2003 \pmod{22}} \equal{} 19^1\equal{}\boxed{19}$

[/hide]

[quote="mysmartmouth"]Huh....what's wrong with
[hide="this"]
$ \phi(23) \equal{} 22$
$ 19^{2003 \pmod{22}} \equal{} 19^1 \equal{} \boxed{19}$

[/hide][/quote]
Nothing, if it's in mod 6. But if it's in mod 6, it's also congruent to 1...

Oops.  I used brianb's calculation, which was faulty.  He replaced $ 2003$ with $ 2001$ in one of his lines.

[quote="brianb"]My bad. I made the 2 corrections.

The QuattoMaster 6000: No, I do not understand what you did after the point I reached:

$ 19^{21}\equiv k\bmod 23\implies 1\equiv 19^{22}\equiv 19k\bmod 23\implies 19k\equiv 1\bmod 23$

Where did you get the one from?[/quote]

Exactly what you did in your first line.  $ 19^{22} \equiv 1 \bmod 23$ by Euler's Totient Theorem / Fermat's Little Theorem.


An equilateral triangle whose side is $ 2$ is divided into a triangle and a trapezoid by a line drawn parallel to one of its sides.  If the area of the trapezoid equals one-half of the area of the original triangle, the length of the median of the trapezoid is:

$ \textbf{(A)}\ \frac{\sqrt{6}}{2} \qquad
\textbf{(B)}\ \sqrt{2} \qquad
\textbf{(C)}\ 2\plus{}\sqrt{2} \qquad
\textbf{(D)}\ \frac{2\plus{}\sqrt{2}}{2} \qquad
\textbf{(E)}\ \frac{2\sqrt{3}\minus{}\sqrt{6}}{2}$

[hide]
Label the original triangle as ABC.
Let the points where the segment intersects AB and AC be D and E, respectively.
[img]http://img4.imageshack.us/img4/6802/trianglemj6.png[/img]
Let $ x$=$ DE$
Let $ h$=the height of the smaller triangle formed.

Since $ \bigtriangleup ADE$ is equilateral, its altitude is $ \frac {x\sqrt {3}}{2}$.
The altitude of trapezoid $ BCDE$ is then $ \sqrt {3} \minus{} \frac {x\sqrt {3}}{2}$, or $ \frac {(2 \minus{} x)\sqrt {3}}{2}$.
The area of trapezoid $ BCDE$ is $ \frac {(DE \plus{} BC)}{2} \cdot$(altitude of trapezoid $ BCDE$).
Thus, the area of trapezoid $ BCDE$ is $ \frac {(x \plus{} 2)}{2} \cdot \frac {(2 \minus{} x)\sqrt {3}}{2} \Longrightarrow \frac {(4 \minus{} x^2)\sqrt {3}}{4}$
This is equal to half the area of $ \bigtriangleup ABC.$
The area of $ \bigtriangleup ABC$ is $ \sqrt {3}$.
$ \frac {\sqrt {3}}{2} \equal{} \frac {(4 \minus{} x^2)\sqrt {3}}{4}$
$ 2 \equal{} 4 \minus{} x^2$
$ x^2 \equal{} 2$
$ x \equal{} \sqrt {2}$

The median of a trapezoid is the average of the two bases. Therefore, the median of trapezoid $ BCDE$ is
$ \frac {2 \plus{} \sqrt {2}}{2}, \boxed{D}$
[/hide]

27r's solution can be shortened considerably
[hide="Short solution"]Since $ \triangle ADE\sim\triangle ABC$, $ \frac{[\triangle ADE]}{[\triangle ABC]}\equal{}\frac12\equal{}\frac{DE^2}{BC^2}$. We know $ BC\equal{}2$, so we find that $ DE\equal{}\sqrt2$ and the median can be found easily.[/hide]

Hey guys,

Here is an alternative solution. Let the small equilateral triangle's side length be x and the big triangle's side length y. Then, after you use that the area of an equilateral triangle is x squared times square root 3 divided four, you get that x squared = y squared divided by two. Therefore, x = y sqrt 2 divided by two. After you plug in the area of a trapezoid, and then use the corresponding numbers. Then, you get that the median of the trapezoid is the average of the bases, which is: (2 + sqrt 2)/2. Sorry, but I don't know how to do Latex.

rts2007


Use a 36-72-72 degree triangle to prove that sin 18 = $ (\sqrt{5}\minus{}1)/4$ and cos 36 = $ (\sqrt {5} \plus{}1)/4$

[hide="Hint"]
Draw the $ 36 \minus{} 72 \minus{} 72$ triangle, and then try to draw other $ 36 \minus{} 72 \minus{} 72$ triangles in your triangle. Make use of the isosceles triangles in your drawing.
[/hide]


Not that big of a deal, but I was in the AIME prob series B board, and I scrolled to the bottom, and it said no one was logged onto that forum.
Then when I went to the next page it was fixed, so I hit the back button.

[quote="#H34N1"]are you hidden?[/quote]
Maybe you should read my post again ;)
No, I'm not hidden as when I clicked the next page it was fine.
Also, it would still say 1 hidden user...


[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=59802]www.artofproblemsolving.com/Forum/viewtopic.php?t=59802[/url]

I don't get this exactly...

Are we supposed to post multiple threads with one problem per thread or only post one problem in total, or post them on the resources page, or what?

Some moderator, like Miths or Silverfalcon will post individual problems in the appropriate forums - one problem with source noted and by itself. Then, in the resource section, there will be links to the respective problems.

Mostly correct. You can post problems all you want. When you do it might be good (not mandatory, since we do post searches before posting complete competitions) to drop me a PM. You can also contact Silverfalcon, who's also doing some work on AMC problems at the moment. :-) We do need them to be one problem per thread, and the source field should clearly state what competition they're from as well as the year and other important info.

:)

And topic locked. Unless Valentin drops some more detail on this thread.


function
real analysis
real analysis solved

Is f(x) = xlnx uniformly continuos for x  >=  1?

I have looked at
Abs(f(x+h)-f(x)) (h>0)
= Abs((x+h)ln(x+h)-xlnx)
= Abs(xln(x+h/x)+hln(x+h))
= xln(x+h/x) + hln(x+h) (since x  >=  1, h > 0)
...

Dear Antilopen (what is a "Lopen"?),

we have by  the middle-value property $f(x+h)-f(x)\geq h\cdot f'(x)$ for $h>0$ and $x>1$ (note that $f'(x)$ is everywhere increasing and $f(x)$ is increasing for $x>1/e$). But since $f'(x)=1+\log(x)\to \infty$ as $x\to\infty$ the function $f(x)$ is not uniformly continuous for $x>1$.

Isn't it,
Leon

$x_n=e^n+ \frac{1}{\sqrt n}^$
$y_n=e^n$

Obviously $x_n-y_n \rightarrow 0$

but

$f(x_n)-f(y_n)=e^n\ln(1+\frac{1}{e^n\sqrt{n}})+\sqrt n +\frac{1}{\sqrt n}\ln(1+\frac{1}{e^n\sqrt{n}}) \rightarrow +\infty$

For x >= 1 and h > 0
Abs((x+h)ln(x+h) - xlnx) > (x+h)lnx - xlnx = hlnx which appr. :inf:  when x appr. :inf:.


abstract algebra
function
group theory
superior algebra
superior algebra unsolved

Prove that every finite abelian group occurs as $ S$-ideal class group of a number field and as the ideal class group of a function field.

$ \mathrm{Pic}(\mathcal{O}_S)$. (take the open subscheme $ \mathrm{Spec}(\mathcal{O}_K) \setminus S$)


calculus
integration
complex analysis
calculus computations

Find integral
\[ \int_{0}^{\infty}{\left(\frac{1}{1\plus{}x^{\alpha}}\minus{}\frac{1}{1\plus{}x^{\beta}}\right)\frac{1}{x}\;\text{d}}x\]
where $ \alpha,\beta\in\mathbb{R},\;\alpha\beta>0$.

In the last line, does that say that the product of $ \alpha$ and $ \beta$ is positive, or did you forget a comma and they are both positive?

Do you know any complex analysis?  This problem becomes little more than an exercise in algebra with the residue theorem.

I mean product in the last line. It can be solved without complex numbers ;).

Put $ x\to\frac1x$ and the answer is zero.


Let $AA_1,BB_1,CC_1$ be the medians of the non-isosceles and non-rightangled triangle $ABC$ and let $O$ be its circumcenter. Prove that the circumcircles of the triangles $AOA_1,BOB_1,COC_1$ have another common point, except $O$.

If $R$ is the circumradius of $ABC$, an inversion of pole $O$ and power $R^2$ will turn the three mentioned circles into the symmedians of $ABC$ which are concurrent. As a consequence, the preimages of the symmedians, which are the three circles, have a second common point. It's the inverse of the Lemoine point of $ABC$ in its circumcircle.

This is soooo old... http://www.mathlinks.ro/Forum/viewtopic.php?t=19720 .

  darij


function
topology
symmetry
vector
advanced fields
advanced fields unsolved

Consider the real valued function f(x,y,z)=[2-(x^2+y^2)^(1/2)]^(2)  + z^2 on R^3-{(0,0,z)}.

a) Show that 1 is a regular value of f.
b) Identify the manifold M=f^(-1) (1)
c) Given that N = {(x,y,z) in R^3 | x^2 + y^2=1}. Show that M is not transverse to N.
d) Is M intersect N a manifold??

I can only show part (a). I couldn't continue afterwards...can someone help?

(b) So you have the equation $ (2 \minus{} (x^2 \plus{} y^2)^{1/2})^2 \plus{} z^2 \equal{} 1$
Notice that $ z$-axis is the axis of symmetry of $ M$, because $ x$ and $ y$ appear only in $ \sqrt {x^2 \plus{} y^2}$, which is the distance from the $ z$-axis. Consider the trace of $ M$ in the halfplane $ y \equal{} 0$, $ x > 0$: it is $ (2 \minus{} x)^2 \plus{} z^2 \equal{} 1$, which is the circle of radius $ 1$ centered at $ (x \equal{} 2,z \equal{} 0)$. The manifold is obtained by rotating this circle around the $ z$-axis. It's a torus.

(c) is geometrically obvious: the cylinder $ N$ is internally tangent to the torus $ M$. You can also show this analytically: the normal vectors (i.e. gradients) are parallel at every point of $ M\cap N$.

(d) Yes, and not a very complicated one. Moral: transversality is sufficient, but not necessary for the intersection to be a manifold.

Thx, I have a much clear picture now.
for (c), do I need to show that for every point of M intersects N? Or just one is enough???

and for (d), I believe it's a circle in the plane right???

Strictly speaking, one point would be enough, but I'd show it at every point of the intersection. And yes, it's a circle.


$ x = \frac {a}{b}$
then the inequalities equivalent :
$ 4(|x|^3 + 1)\geq(|x + 1|^3 + |x - 1|^3$
[u]Case 1[/u] $ x\in( - \infty, - 1)$
$ 4(1 - x^3)\geq - (x + 1)^3 - (x - 1)^3$
$ \Leftrightarrow (x + 1)^2(x - 2)\leq 0$ (true)
[u]Case 2[/u] $ x\in [ - 1,0)$
$ \Leftrightarrow 4(1 - x^3)\geq (x + 1)^3 - (x - 1)^3$
$ \Leftrightarrow (x + 1)^2(2x - 1)\leq 0$ (true)
 [u]Case 3[/u] $ x\in [0,1)$
$ 4(x^3 + 1)\geq (x + 1)^3 - (x - 1)^3$
$ \Leftrightarrow (x - 1)^2(2x + 1)\geq 0$ (true)
[u]Case 4[/u] $ x\in [1,\infty)$
$ x^3 - 1 > 0$ (true)
So we has inequlities.
Is there other method?


Show that there exist infinitely many composite positive integer $ n$ which divide $ 2^{n\minus{}1}\minus{}1.$

we will use the following lemma:
[b]lemma:let $ m\mid n$ and let $ a\in\mathbb{Z} - \{1\}$ then $ a^m - 1\mid a^n - 1$[/b]
now using this lemma we obtain that:
\begin{eqnarray*}n\mid 2^{n - 1} - 1\implies 2^n - 1\mid 2^{2^{n - 1} - 1} - 1\implies 2^n - 1\mid \left(2^{2^{n - 1} - 1} - 1\right)\left(2^{2^{n - 1} - 1} + 1\right)\implies 2^n - 1\mid 2^{2^n - 2} - 1\end{eqnarray*}
now let $ 2^{n} - 1 = t$ we get that:
$ t\mid 2^{t - 1} - 1$
so if $ n\mid 2^{n - 1} - 1$ then $ t\mid 2^{t - 1} - 1$ so its sufficient to find a natural number $ n$ such that $ n\mid 2^n - 1$ just put $ n = 1$...

For each $ k$ show that there exist an integer $ n$ having exactly $ k$ prime factors such that $ n$ divides $ 2^{n\minus{}1}\minus{}1$ .

It is not true. $ n\equal{}p^kq^m$ (with 2 prime factors p,q) is not solution $ n|2^n\minus{}1$.

You are not right Rust. Of course, there is no positive integer $ n$ other than 1 dividing $ 2^n\minus{}1$. But here we deal with the composite integers dividing $ 2^{n\minus{}1}\minus{}1$. There exists such $ n$ of the form $ p\cdot q$, for example as found Boliya $ n$ can be $ 11\cdot 31$, because of $ 11|2^{31\minus{}1}\minus{}1\equal{}32^6\minus{}1$ and $ 31|2^{11\minus{}1}\minus{}1\equal{}32^2\minus{}1$.

[color=green]Same problem :

Show that there exist infinitely many composite positive integer $ n$ which divide $ 2^n \plus{} 1$
[hide="hint"]$ \longrightarrow x_0 \equal{} 1,x_{k \plus{} 1} \equal{} 2^{x_k} \plus{} 1$ or $ n \equal{} 3^k$ so that the condition  :) [/hide][/color]

Odd $ n=p_1^{k_1}...p_m^{k_m},p_1<p_2<...<p_m$ solution $ n|2^n+1$ if and only if, for any $ p_j$ exist $ l_i\le k_i,i<j$, suth that \[ p_j|2^{p_1^{l_1}..p_{j-1}^{l_{j-1}}+1.}\]
Therefore minimal prime $ p_1=3$, and $ n=3^k$ is solution.

I don't understand why these 2 problem is the same?


Your inequality seems to be easy  as I think Cauchy Schwartz might do it but I haven't yet completed my solution.
WIll try my best to. Bit confident. How exactly did you get this inequality nguoivn?

[quote="Agr_94_Math"]Your inequality seems to be easy  as I think Cauchy Schwartz might do it but I haven't yet completed my solution.
[/quote]
If you use Cauchy- Schwarts, you'll obtain:
$ (9 \plus{} ab \plus{} bc \plus{} ca)^2 \geq\ 24(3 \plus{} a^2 \plus{} b^2 \plus{} c^2)$
And this last ineq is not trues  :wink:
Of course, my ineq often easy but not too easy as you think, my friend  :)
[quote="Agr_94_Math"]
How exactly did you get this inequality nguoivn?[/quote]
I created it by my secret way (by classic ineq) but now, I don't want to say  :oops:

[quote="nguoivn"]Given $ a, b, c \geq\ 0$ satisfy $ a \plus{} b \plus{} c \equal{} 3$.

Prove that: $ \frac {(1 \plus{} a)^2(1 \plus{} b)^2}{1 \plus{} c^2} \plus{} \frac {(1 \plus{} b)^2(1 \plus{} c)^2}{1 \plus{} a^2} \plus{} \frac {(1 \plus{} c)^2(1 \plus{} a)^2}{1 \plus{} b^2} \geq\ 24$
 :)[/quote]
Sorry,I didn't see how you created it. :oops: Can you show me your nice proof? :)
Here is my ugly proof:

Sorry,I made a typo in the forth line.$ RHS$ should be $ \sum (1\plus{}a)^2(1\plus{}b)^2(1\plus{}c^2)$ :blush:


[b]AIME 1999, #11.[/b]  Find sin 5 + sin 10 + sin 15 + ... + sin 175. You may express the answer as tan(a/b).

[b]AIME 1984, #14.[/b]  What is the largest even integer that cannot be written as the sum of two odd composite positive integers?

Kalva's solutions are pretty unclear...can anyone offer some help with these? Thanks!  :D

AIME 1999, #11. Find sin 5 + sin 10 + sin 15 + ... + sin 175. You may express the answer as tan(a/b).

If a + b = 180 :^o:, then sin a + sin b = 2 sin (a + b)/2 cos (a - b)/2.

That should help a bit.

the first one:

[hide]

notice that sin(180-x)=sinx

so the statement can be rewritten as

2(sin5+sin10+sin15+....+sin85)+sin90

we'll work with the first portion of that expression first

distribute the 2, so we get

2sin5+2sin10+2sin15+....+2sin85

now we multiply that by sin2.5/sin2.5

so now our statement is

(2sin5sin2.5+2sin10sin2.5+2sin15sin2.5+....+2sin85sin2.5)/sin2.5

notice how 2sinAsinB=cos(A-B)-cos(A+B)

so now you get (cos2.5-cos7.5+cos7.5-cos12.5+cos12.5 +..... +cos82.5-cos87.5)

that telescopes nicely to give you (cos2.5-cos87.5)/sin2.5

now notice that cos(a)=sin(90-a)

now the expression becomes (cos2.5-sin2.5)/sin2.5

this is just the left portion of the expression, we still need to add sin90

so (cos2.5-sin2.5)/sin2.5+sin90

this simplifies down to cot2.5, or cot(5/2), which is equivalent to tan(90-5/2)=tan(175/2)

[/hide]

/edit:  ok that answer's wrong.  lemme see what i did

/edit2: it's fixed.

little error at  the end, nosepuppy.

Second one (minor hints)--- [hide]try to simplify it. For example, what is the largest integer that can't be expressed as 3(2a+1)+5(2b+1) for positive integers a and b. (or if you'd rather work with non-negative integers, 3(2a+3)+5(2b+3)). Something that might help is the chicken mcnugget theorem, which says that the largest number that can't be written in the form px+qy for non-negative p and q (given x and y which are relatively prime) is xy-x-y.[/hide]


geometry
inequalities
inequalities proposed

Prove that for x in (0, pi/2) we have 2cos(x)/(1+cos(x))<sin(x)/x

this ineq is equivalent to
2x < sin x + tan x

since x/2 < tan x/2
it is sufficient to prove that 4 tan x/2 < sin x + tan x
tan x/2 = (sin x)/(1+cos x)
we need only to prove that
4sin x/(1+cos x) < sin x + sin x/cos x
<==>
4cos x < (cos x + 1)^2
which holds obviously.

in fact,
1/(1/(sin x) + 1/(tan x)) = sin x/(1+cos x) = tan x/2 > x/2
and by AM-GM-HM ineq,
(sin x + tan x)/2 > 2/(1/(sin x) + 1/(tan x)) > x
i.e.,
2x < sin x + tan x

and there's a geometric proof for
x < (sin x + tan x)/2

Suppose there's a unit circle O and AB is an arc on it with angle AOB = x.
Let the tagent line passing through A meet the extension of OB at D, and the tangent lines passing A and B intersect at point C. Thus, the area of sector OAB, triangle OAB and triangle OAD are
x/2, sin x/2 and tan x/2
respectively. then we should prove
S_{sector OAB} < (S_{triangle OAB} + S_{triangle OAD})/2

The quadrilateral OACB includes the sector OAB. it is sufficient to prove that
S_{OACB} < (S_{triangle OAB} + S_{triangle OAD})/2
<==>
S_{OACB} - S_{triangle OAB} < S_{triangle OAD} - S_{OACB}
<==>
S_{triangle ACB} < S_{triangle CDB}.

The two triangles, ACB and CDB, have a common vertex B, their edges AC and CD are on the same line, hence they have the same height. To proof the last inequality, we need only to prove that AC<CD. This is correct, because in the right triangle CDB, CD(hypotenuse)>CB.

Remark.  :D  :D  :D
Proving x < (sinx + tanx)/2 is a problem in Hungrian MO 1909. :D

We can prove a little bit stronger inequality
2sinx + tgx > 3x for  0 < x < Pi/2.
Proof: Let consider f(x) = 2sinx + tgx - 3x
Then f'(x) = 2cosx + 1/(cosx)^2 - 3 >=0 (by AM-GM 2cosx + 1/(cosx)^2 = cosx + cosx + 1/(cosx)^2 >=3)
So f(x) >= f(0) = 0.

Namdung

well, this one is absolutely ok.
but i think the problems posted here need elementary solutions.

[quote="Namdung"]We can prove a little bit stronger inequality
2sinx + tgx > 3x for  0 < x < Pi/2.
[/quote]
i saw this one as an exercise on a book named 'Problem-solving Through Problems' by Loren C. Larsen (1983 Springer-Verlag).This one is called 'Huygens's Ineq'.
and there is another exercise:
Prove that (2+cos x)x > 3sin x


I want to have the solutions to the booklet "Olympiad Inequalities" by Prof Mildorf. Can anyone provide? A special request to Prof. Mildorf himself if he can guide.

I'm not sure he wrote solutions to go along with that, and he's not a prof :).

He (by the way, who is he , then ?) might not have written, but somebody else might have.

well that's his personal info, but he just finished his undergrad last spring. is doing research, but not in university.

What about solutions to "Inequalities through Problems" by Hojoo Lee ?

Here is some of it http://www.mathcenter.net/forum/showthread.php?t=1465 for H.Lee's Problem.

Mr.Mildorf's problems can be found in the old topic. :roll:


Isn't my first post clear enough??  :blush:

[quote="Humpty_Dumpty"]Hello!
I've translated this problem from italian... so i hope it will be clear enough..   :blush:
[i]A car's wheel with tread few centimeters thick can cover distances of the order of 50000km before the tread completely wears out. Find the order of magnitude of the thickness of the tread eaten away in each turn and put this number in relation with a plausible description of the microscopic behaviour regulating this phenomenon (that of the wearing out of the tread).[/i][/quote]

What is the radius of the car's wheel from axel to pavement?

What is your definition of "a few centimeters?" as in 'with tread [a] few centimeters thick.'

$2\pi r N =5 \times 10^7 \text{ meters}$

$\lambda t = N \Rightarrow \lambda = N/t$

$\lambda = \frac{5 \times 10^7}{2 \pi r t} (\text{ turns/meter of tread})$


let $a,b,c$ be positive real   such that :
$(a+b-c)(\frac{1}{a}+\frac{1}{b}-\frac{1}{c})=4$
 find minimal of expression P
                    $P=(a^4+b^4+c^4)(\frac{1}{a^4}+\frac{1}{b^4}+\frac{1}{c^4})$
 I want to have a solution complete

"such that (a+b-c)(\frac{1}{a}+\frac{1}{b}-\frac{1}{c})"
What is this ?

"such that $(a+b-c)(\frac{1}{a}+\frac{1}{b}-\frac{1}{c})$"
What is this ?


function
inequalities
algebra proposed
algebra

Find all functions $f:R->R$ such that $f(2x-5)\leq2x-3\leqq(2x)-5$

[quote="cezar lupu"]Find all functions $f:R->R$ such that $f(2x-5) \leq 2x-3 \leq f(2x)-5$[/quote]

Tricky....

[hide]set $a=2x\ =>$

$f(a-5) \leq a-3 \leq f(a)-5$

$Left\ Inequality$

$f(a-5) \leq a-3 $

set $b=a-5\ <=>a=b+5$

$f(b) \leq (b+5)-3 <=>f(b) \leq b+2\ \ \ (1)$

$Right\ Inequality$

$a-3 \leq f(a)-5 <=>$

$a-3+5 \leq f(a) <=>$

$a+2 \leq f(a) \ \ \ (2)$

So we have together $f(x)\leq x+2 \ $ and $f(x)\geq x+2 \ $ hence $f(x)=x+2$[/hide]

\[ f(x+m)\le x+n\le f(x)+m, \ for\ any \ x\in R\ \ \rightarrow\ \ f(x)=x-m+n\] and in this case we have equality evertime.


$\begin{cases} 4x^2+yz=A \\ 4y^2+zx=B \\ 4z^2+xy=C \end{cases}$
If any of $x,y,z$ is $0$ then the problem becomes easy. Assume they are all positive real numbers.
$\Rightarrow\begin{cases} \frac{4x^2}{yz}=\frac{A}{yz}-1 \\ \frac{4y^2}{zx}=\frac{B}{zx}-1 \\ \frac{4z^2}{xy}=\frac{C}{xy}-1 \end{cases}$
Multiplying these equations gives
$64=(\frac{A}{yz}-1)(\frac{B}{zx}-1)(\frac{C}{xy}-1)$
Now let $p=\frac{A}{yz}-1, q=\frac{B}{zx}-1, \frac{64}{pq}=\frac{C}{xy}-1$ where $p,q$ are positive real parameters.
$\Rightarrow \begin{cases} yz=\frac{A}{p+1} \\ zx=\frac{B}{q+1} \\ xy=\frac{Cpq}{64+pq} \end{cases}$
$\Rightarrow \begin{cases} x=\sqrt{\frac{BCpq(p+1)}{A(q+1)(64+pq)}} \\ y=\sqrt{\frac{ACpq(q+1)}{B(p+1)(64+pq)}} \\ z=\sqrt{\frac{AB(64+pq)}{Cpq(p+1)(q+1)}} \end{cases}$

P.S. This is not an open problem at all. Next time please put your unsolved problem in the unsolved section. :)


i remember that someone posted a website with old AMC/AIME problems a while back. It's on the Kansas site or something. does anyone know the site?

yes, i know aops wiki has it but the kansas site had it on pdf, the pdf version doesnt work on the wiki section. thanks.

[quote="azjps"][url=http://www.math.ksu.edu/main/handbook/ProblemSets]Here[/url], though the "Contests" resource section also has pdfs (the pdfs should also be linked to from the wiki pages).[/quote]

The pdfs in the contest section don't display diagrams or dollar signs correctly so they only have limited use. You have to print out the diagrams separately.

If there is an AMC/AIME that is not in the link that azjps gave above, it's better to print the test directly from the AoPSWiki in printable view, not as a pdf. This way the dollar signs and images will all be there. You could then copy it into a Word document/etc. if you wanted to.

the thing is that i have the old microsoft word so the images wont copy unless i save image as for all the images and then load them up. it will take way too long to do that. does anybody remember the site?

Also [url=http://agmath.com/]my teacher's website[/url] has quite a bit of resources.


Well, I would like to ask something.If it has been discussed before please give me the link.
Any help appreciated!! :)

[color=blue]Problem[/color]:
Find necessary and sufficient conditions for matrices $A$ and $B$ so that $AB=BA$!

Thank you!

This question doesn't make sense to me. It seems unlikely that a "nice" answer will appear.

Any thoughts, in general?
Will be helpful!!

 :)  :)

Sufficient: $A,B\in M_n(\mathbb{C})$ and both $A$ and $B$ have $n$ distinct eigenvalues.  Furthermore, $A$ and $B$ have the same eigenvectors.  Then $AB=BA$.


$1-a=b+c$, etc, so the inequality becomes:

\[(b+c)(c+a)(a+b)\ge 8abc\]
$(b+c)\ge 2\sqrt{bc}$; $(c+a)\ge 2\sqrt{ca}$; $(a+b)\ge 2\sqrt{ab}$ by the AM-GM inequality.

Multiply those to get the desired result. Equality occurs when $a=b=c$ due to the equality case of the AM-GM inequality.

[quote="me@home"]What are $a,b,c$? Reals or positives or what?[/quote]
It works for any real numbers that satisfy $a+b+c=1$. If the problem is read it doesnt put any constraints on them.

[quote="Kouichi Nakagawa"]If $a, \ b, \ c \in \mathbb{R}^{+}$ then,
$\left ( \because \begin{array}{l}\text{If}\ (a,b,c)=(-1,-1,3) \ \text{then}\ (1-a)(1-b)(1-c)=-8, \ 8abc=24, \\ \text{so}\ (1-a)(1-b)(1-c) \leq 8abc \end{array}\right )$

$(1-a)(1-b)(1-c) \geq 8abc$
$\iff 1-(a+b+c)+ab+bc+ca-abc \geq 8abc$
$\iff ab+bc+ca \geq 9abc$
$\iff (a+b+c)(ab+bc+ca) \geq 9abc$
[/quote]

You can use A.M.-G.M. directly.

[quote="me@home"]Here's the thing: you can't use am-gm unless you have a sum-of-squares proof, since the problem is only for reals.[/quote]

Haven't said Kouichi Nakagawa that If $a,\ b,\ c\in{\mathbb{R^{+}}}$? I replied to that case.  :wink:

Is that what his first line was saying? Sorry, I didn't understand what exactly he was saying (... too much latex and not enough tex[b]t[/b] :P )
Sorry for the inconvenience...

[hide]Rearranging $a+b+c=1$, we have $1-a=b+c,1-b=a+c, 1-c=a+b.$
Thus we have $(1-a)(1-b)(1-c)=(a+b)(a+c)(b+c)\ge{8abc}$.
By AM-GM, $a+b\ge{2\sqrt{ab}},a+c\ge{2\sqrt{ac}},b+c\ge{2\sqrt{bc}}$.
Multiplying the three expressions together gives us the desired inequality.[/hide]

You are right. Anyway, this problem should be add that $a>0,\ b>0,\ c>0$.
Please edit this.

but maybe its true for some $a>0,b<0,c<0$ otherwise one should give a counterexample...

[quote="Altheman"]$1-a=b+c$, etc, so the inequality becomes:
\[(b+c)(c+a)(a+b)\ge 8abc \]
$(b+c)\ge 2\sqrt{bc}$; $(c+a)\ge 2\sqrt{ca}$; $(a+b)\ge 2\sqrt{ab}$ by the AM-GM inequality.

Multiply those to get the desired result. Equality occurs when $a=b=c$ due to the equality case of the AM-GM inequality.[/quote]

Not true if $a,b,c$ are not positive reals... but then the inequality proposed itself is false... check out $a = b =-1, c = 3$  :wink:

yeah your probably right but what about when only one of them was negative,I mean with the condition $b,c>0$ and $a<0$... :maybe:
is it still wrong!?

Yes it still is wrong if one of them is negative. A counter-example (which may be a bit contrived): $a = \frac{9+\sqrt{79}}{2}, b = \frac{9-\sqrt{79}}{2}, c =-8.$

Let us try to prove that $(1-a)(1-b)(1+c) \ge-8abc$ where $a+b = c+1$ and $a,b,c \in \mathbb{R}^{+}.$ This is the same thing as proving that $(1-a)(1-b)(1-c) \ge 8abc$ where $c < 0$ and $a,b,c \in \mathbb{R}^{+}.$ So, if we expand and simplify, we get $9abc+a(b-c)-a-b(c+1)+c+1 \ge 0.$ If we replace all $c$ by $a+b-1,$ we get, after simplifying, $(a+b)(9ab+1) \ge (a+b)^{2}+8ab.$ Now, let $ab = k,$ where $k$ is a fixed, positive constant. In that case, if we replace $ab$ by $k$ and $a+b$ by $c+1,$ we get $k \ge \frac{c(c+1)}{9c+1}\Leftrightarrow ab \ge \frac{c(c+1)}{9c+1}.$ Now, this is not always true, as we can let $ab = 1/2$ and $c = 8,$ for instance. If we do this, we then get $a+b = 9,$ and solving linearly gives the above. Basically, in order for $a, b> 0, c < 0,$ we must have that $ab \ge \frac{c(c+1)}{1+9c}.$ Once we get an appropriate value for $c,$ which is positive in the inequality, take it's inverse, and BAM, there you have a counterargument.

thanx...
ok then according to what [b]karth[/b] said the inequality is wrong for negative numbers,so kunny your right...


calculus
derivative
trigonometry
calculus computations

$x^3y^2+x\cos y = 3 \Rightarrow 3x^2y^2+2x^3y\dfrac{dy}{dx} + \cos y - x\dfrac{dy}{dx}\sin y = 0$

$\Rightarrow \dfrac{dy}{dx} = \dfrac{3x^2y^2+\cos y}{x\sin y - 2x^3y}$


Let  $ a_{0,0} \equal{} 0$, $ a_{m,0} \equal{} \dfrac{1}{2^{m}}$ for m =1,2,3,..., $ a_{0, n} \equal{} \minus{}\dfrac{1}{2^{n}}$ for n=1,2,3..., and $ a_{m, n} \equal{} \dfrac{1}{2^{m\plus{}n}}\left(\dfrac{m\minus{}n}{m\plus{}n}\right)\left(\begin{array}{c}
m\plus{}n\\ n \end{array}\right)$ for m,n = 1,2,3,.... Show that $ \sum _{m \equal{} 0}^{\plus{}\infty}\sum _{n \equal{} 0}^{\plus{}\infty} a_{m,n} \equal{} \minus{}1$ and
$ \sum _{n \equal{} 0}^{\plus{}\infty}\sum _{m \equal{} 0}^{\plus{}\infty} a_{m,n} \equal{} 1$.


algebra
polynomial
function
inequalities
inequalities proposed

Let $ f(a,b,c)$ be a symmetric polynomial function of degree 6 such that

$ f(a,b,c) \ge 0$

for any nonnegative real numbers $ a,b,c$, with equality for $ (a,b,c)\equal{}(1,1,1)$, and again for $ (a,b,c)\equal{}(1,2,3)$, or any permutation thereof.
Find an example of such an inequality.

[quote="Vasc"]Let $ f(a,b,c)$ be a symmetric polynomial function of degree 6 such that

$ f(a,b,c) \ge 0$

for any nonnegative real numbers $ a,b,c$, with equality for $ (a,b,c) \equal{} (1,1,1)$, and again for $ (a,b,c) \equal{} (1,2,3)$, or any permutation thereof.
Find an example of such an inequality.[/quote]
$ f(a,b,c) \equal{} 359(a\plus{}b\plus{}c)^6 \plus{} 289413 \cdot (abc)^2 \minus{} 20412 \cdot (ab\plus{}bc\plus{}ca)^3$. has $ f(1,1,1)\equal{}f(1,2,3)\equal{}0$. I have absoluttely no idea if $ f(a,b,c) \ge 0$ holds... :)

[quote="Mathias_DK"][quote="Vasc"]Let $ f(a,b,c)$ be a symmetric polynomial function of degree 6 such that

$ f(a,b,c) \ge 0$

for any nonnegative real numbers $ a,b,c$, with equality for $ (a,b,c) \equal{} (1,1,1)$, and again for $ (a,b,c) \equal{} (1,2,3)$, or any permutation thereof.
Find an example of such an inequality.[/quote]
$ f(a,b,c) \equal{} 359(a \plus{} b \plus{} c)^6 \plus{} 289413 \cdot (abc)^2 \minus{} 20412 \cdot (ab \plus{} bc \plus{} ca)^3$. has $ f(1,1,1) \equal{} f(1,2,3) \equal{} 0$. I have absoluttely no idea if $ f(a,b,c) \ge 0$ holds... :)[/quote]
It does not hold for a=1 and b=c=2.

[quote="arqady"]$ (a \minus{} 2b \plus{} c)^2(b \minus{} 2c \plus{} a)^2(c \minus{} 2a \plus{} b)^2\geq0.$[/quote]
Nice, Arqady.
Can you also find an inequality for the case where equality holds for (a,b,c)=(1,1,1) and (a,b,c)=(0,1,3)?

This inequality holds for the equality case $ (1,1,1)$, $ (0,1,k)$  for any positive $ k$.

\[ (a\plus{}b\plus{}c)\cdot abc \cdot (a^2\plus{}b^2\plus{}c^2\minus{}ab\minus{}bc\minus{}ca) \geq 0
\]

 :)

I hope i didn't make any stupid mistake.

[quote="shaaam"]This inequality holds for the equality case $ (1,1,1)$, $ (0,1,k)$  for any positive $ k$.
\[ (a \plus{} b \plus{} c)\cdot abc \cdot (a^2 \plus{} b^2 \plus{} c^2 \minus{} ab \minus{} bc \minus{} ca) \geq 0
\]:) [/quote]
Nice, Shaaam.
Can you also find an inequality for the case where equality holds for (a,b,c)=(1,1,1) and (a,b,c)=(1,2,4)?

\[ (2a^2\plus{}2b^2\plus{}2c^2\minus{}3ab\minus{}3bc\minus{}3ca)^2 \cdot (a^2\plus{}b^2\plus{}c^2\minus{}ab\minus{}bc\minus{}ca) \geq 0
\]

 :)

It is correct, Shaaam.
This is an inequality with only p and q. Can you find a similar inequality with p, q and r?

Actually I am interested of other related problem.
Can somebody invalidate (with a particular example) the following

[i]Proposition. [/i] Let  $ f(a,b,c)$ be a symmetric polynomial function of degree 6 such that $ f(1,1,1)\equal{}0$, $ f(0,b,c) \ge 0$ and $ f(1,b,b) \ge 0$ for all nonnegative (or real) $ b$ and $ c$. Then $ f(a,b,c) \ge 0$ for all nonnegative (or real) $ a,b,c$.

[quote="No Reason"]How about this?
$ (r \minus{} \frac {p^3}{36})^2 \plus{} p^2(q \minus{} \frac {11p^2}{36})^2 \minus{} \frac {r^3}{\omega}$ when $ \omega \rightarrow infinity$
Take $ a \equal{} 1,b \equal{} 2,c \equal{} 3$ to see that it's wrong.[/quote]
What abaut the condition $ f(1,1,1) \equal{} 0$ ?

In the above Proposition, $ f(a,b,c)$ is homogeneous, too.

[quote="Vasc"] Can somebody invalidate (with a particular example) the following

[i]Proposition. [/i] Let  $ f(a,b,c)$ be a symmetric and homogeneous polynomial function of degree 6 such that $ f(1,1,1) \equal{} 0$, $ f(0,b,c) \ge 0$ and $ f(a,1,1) \ge 0$ for all nonnegative $ b$ and $ c$. Then $ f(a,b,c) \ge 0$ for all nonnegative $ a,b,c$.[/quote]

Indeed, for

$ f(a,b,c)\equal{}(abc\plus{}\frac 2{27}p^3\minus{}\frac 1{3}pq)^2\minus{}(p^2\minus{}\frac {36}{11}q)^2(\frac {41}{10}q\minus{}p^2)\plus{}\frac 9{110}q^3$,

where $ p\equal{}a\plus{}b\plus{}c$ and $ q\equal{}ab\plus{}bc\plus{}ca$, we have $ f(1,1,1) \equal{} 0$, $ f(0,b,c) \ge 0$ and $ f(a,1,1) \ge 0$ for all nonnegative $ b$ and $ c$, but $ f(1,4,8)<0$.

[quote="me@home"]how did you find this? It seems quite nontrivial, how did you figure out the form?[/quote]
Indeed, I found this result after much effort, using EV Theorem and Computer. :wink:  :idea:


function
inequalities
calculus
derivative
real analysis
real analysis theorems

A function is said to be increasing on $ [a,b]$ if for every $ x_{1} ,x_{2}\in(a,b)$ where $ x_{2} > x_{1}$,then $ f(x_{2}) > f(x_{1})$.

I read in a book that function $ y = f(x)$ is increasing on $ [a,b]$ if $ f'(x)\geq 0$ on the interval $ (a,b)($but f'(x) should not become zero on any sub-interval of $ (a,b)))$

This is given in  standard texts

So what exactly does strictly increasing mean i.e when is a function said to be strictly increasing on the interval $ [a,b]$.

The definition of strictly increasing was given as follows:

The  function $ y = f(x)$ is strictly increasing on $ [a,b]$ if $ f'(x)> 0$ on the interval $ (a,b)$.

So what exactly is the definition of increasing function and what exactly is the definition of strictly increasing function

When a function is strictly increasing, all the relevant inequalities should be strict, i.e. $ f(x_2) > f(x_1)$.  Without qualification, "increasing" means weakly increasing, i.e. $ f(x_2) \ge f(x_1)$.  (Note that the derivative conditions are sufficient, not necessary; the derivatives may not exist.)

The exact meaning depends on the textbook or on the person who is speaking. There are several schools that disagree about how the words should be used. I belong to the one that uses "non-decreasing" and "increasing" instead of t0rajir0u's "increasing" and "strictly increasing" but I cannot say I am always consistent. In any case, these words normally relate to the comparision of values. The definition
[quote]
$ f(x)$ is strictly increasing on $ [a,b]$ if $ f'(x)> 0$ on the interval $ (a,b)$.
[/quote]
you cited is certainly non-standard.

For the practical impact of that: is $ x^3$ strictly increasing on the whole real line? I say it is. The $ f'>0$ definition says it isn't.


function
complex analysis
Dirichlet Series
zeta function
Convolution

Prove that $ \zeta(z)\zeta(z\minus{}1)\equal{}\sum_{n\equal{}1}^{\infty}\frac{\sigma(n)}{n^z}$, for $ Rez>1$, where $ \sigma(n)$ is the sum of the divisors of $ n$.

This isn't really complex analysis, although I'm not sure where it does belong.

Multiplying two functions given by Dirichlet series convolves their series, with the convolution defined over the multiplication of natural numbers; $ f*g(n) \equal{} \sum_{d|n}f(d)g(n/d)$.
Of course, $ \zeta(z) \equal{} \sum_n \frac1{n^z}$ and $ \zeta(z \minus{} 1) \equal{} \sum_n \frac1{n^{z \minus{} 1}} \equal{} \sum_n \frac {n}{n^z}$. Convolve those, and $ \zeta(z)\zeta(z \minus{} 1) \equal{} \sum_n \frac1{n^z}\sum_{d|n} d\cdot 1 \equal{} \sum_n \frac {\sigma(n)}{n^z}$.

OK, that's convergence as a formal series. For convergence as a numerical series, that product formula requires both parts to converge absolutely, which is true here when the real part of $ z$ is greater than $ 2$.
When the real part is between $ 1$ and $ 2$, it diverges.


counting
derangement
combinatorics proposed
combinatorics

If $ n$ people hand in their coat before they go into the theatre. Then when they come out they each receive a coat. How many ways are there with each man receiving a different coat?

D(n)=no. of derangement of n article= n!(e^-1 up to n+1th term only)

n is the number of persons.Each one has n-1 of chance to get a diferent clothe,I am not sure,but I think that there are (n-1)^n possible ways of each one get another clothe.

A1,A2,A3,A4,.........Am  are sets  and

A=A1UA2UA3U...UAm

n(A)  =  n(A1)+ n(A2) +n(A3)+.......... n(Am)
            - Σn(Ai∩Aj) ,  1≤i<j≤m
            +Σn(Ai∩Aj∩Ak)   ,   1≤i<j<k≤m
              and  so on.......

This  is  called  PIE ( principle  of inclusion  and exclusion)

now,           we  can  see

             if  A1,A2,A3............Am  are  subsets  of  AA  containing N  elements,  then

                     n(A1`∩A2`∩...........∩Am`)

           =  N  -  Σ n(Ai)  +Σ  n(Ai∩Aj)    -       Σ  n(Ai∩Aj∩Ak)+......
                       i                      1≤i<j≤m      1≤i<j<k≤m

I'm not sure(again),but I think that the answer is $ (n\minus{}1)*(n\minus{}2)^{n\minus{}2}$.

Number of ways = $ n!\left[1\minus{}\frac{1}{1!}\plus{}\frac{1}{2!}\minus{}\frac{1}{3!}\plus{}\cdots\plus{}(\minus{}1)^{n}\frac{1}{n!}\right]$


Can you rearrange the terms of the following series to get the summation S equals to 2?

$ S=1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}...\infty$

No, because it's not true. In fact it is true that $ S\equal{}\sum_{n\equal{}1}^{\infty}(\minus{}1)^{n}\frac{1}{n}\equal{}ln2$.
It is known as alternating harmonic sequence, and the equality can be proved using Taylor series on ln :wink: I don't think that there exists a fully elementar proof.

If you sum the terms [i]in that particular order[/i], you get $ \ln 2$.  If you rearrange the terms (as the question suggests) you can get any limit you desire.  There is also a standard process: if you want to get a positive limit $ a$, take positive terms of your sequence until you pass $ a$ going up; then take negative terms until you pass $ a$ going down; then take positive terms going up, etc., etc.  So, for example, to get a sum of $ 2$ you would take $ 1 \plus{} 1/3 \plus{} 1/5 \plus{} \ldots \plus{} 1/15 \minus{} 1/2 \plus{} 1/17 \plus{} \ldots \plus{} 1/41 \minus{} 1/4 \plus{} \ldots$.  (If you want to get a negative limit, you start with negative terms.  If you want to get 0, it doesn't matter, and if you want either $ \infty$ or $ \minus{}\infty$ you have to change your approach slightly (consider that a challenge question).)  This works for any conditionally convergent series.

Aah, sorry, I misunderstood the question, I thought that [b]ashwinrk_jain[/b] asks for some rearrangement as in the sum $ \sum_{n \equal{} 1}^{\infty}\frac {1}{n(n \plus{} 1)}$ :wink:
EDIT: You're right about the method, but how would you formally define such a sequence?


In triangle $ ABC$, $ AB \equal{} 5$, $ BC \equal{} 4$, and $ CA \equal{} 3$.

[asy]defaultpen(1);

pair C=(0,0), A = (0,3), B = (4,0);
draw(A--B--C--cycle);

label("\(A\)",A,N);
label("\(B\)",B,E);
label("\(C\)",C,SW);[/asy]

Point $ P$ is randomly selected inside triangle $ ABC$.  What is the probability that $ P$ is closer to $ C$ than it is to $ A$ and $ B$?

We divide this into multiple parts, find the areas of these, and find the area of the intersection.

I. Inside the triangle.

II. To be closer to C than A, a point must be below the line parallel to BC through the midpoint of AC.

III. To be closer to C than B, a point must be left of the line parallel to AC through the midpoint of BC.

The intersection of these areas is a rectangle of area $ 1.5 \times 2 \equal{} 3$. The full area of the triangle is $ \frac{3 \times 4}{2} \equal{} 6$.

The final probability is $ \frac{3}{6} \equal{} \boxed{1/2}$.

[hide] This is also easy to coordinate bash, but letting $P$ be $(x, y)$ and computing bounds on each coordinate separately. [/hide]

[hide=answer]Notice, that in triangle ABC, that if we pick any point "P" on DG, then triangle APD will be congruent to triangle CPD by SAS. Thus CP will be  equal to AP. But, for any point "P" that we select inside triangle CDG, then point "P" will be further from A - and certainly further from B - than "P" is from C. This is so because angle CAP will be less than angle ACP. And in any triangle, the greater angle is opposite the greater side. Thus, AP will be greater than CP.

And for the same reasons, any point "P" on GE, will make triangles BPE and CPE congruent. But, any point "P" inside triangle CEG will be further from B - and certainly further from A - than "P" is from C.

Thus....all points in rectangle CDGE will be futher from A or B than they are from C. And the area of this rectangle is just (2)(1.5)  = 3 sq units.

And the area of triangle ABC  = (1/2)(4)(3)  = 6 sq units.

So the probability that some point "P" is clser to C tham to either A or B  =  3 / 6   = 1 / 2.[/hide]

I wonder why the problem doesn't just say 3-4-5 triangle


John has four American coins, none of which has a value greater than 25 cents. What is the number of cents in the positive difference between the greatest and least amounts money John could have?

[hide="solution"]
The least amount of money John could have is if all the coins were pennies. Then he would have $ \$0.04.$

The greatest amount of money John could have is if all the coins were quarters (remember, it said the values were no greater than $ \$0.25,$ so it can be $ \$0.25$). Then he would have $ \$1.00.$

Finally, the difference is $ \$1.00\minus{}\$0.04\equal{}\$0.96\equal{}\boxed{96\text{ cents}}.$[/hide]


Which one of the following combinations of given parts does not determine the indicated triangle?

$ \textbf{(A)}\ \text{base angle and vertex angle; isosceles triangle}$
$ \textbf{(B)}\ \text{vertex angle and the base; isosceles triangle}$
$ \textbf{(C)}\ \text{the radius of the circumscribed circle; equilateral triangle}$
$ \textbf{(D)}\ \text{one arm and the radius of the inscribed circle; right triangle}$
$ \textbf{(E)}\ \text{two angles and a side opposite one of them; scalene triangle}$

[hide="Solution?"]Given a base of $ 1$ and a vertex angle of $ 30^\circ$, you could either have an isosceles triangle or a 30-60-90, or anything in between.
So the answer must be $ \textbf{(B)}$.[/hide]

I dislike problems like these...  :(

Would it not be $ \boxed{(A)}$ because you cannot determine any triangle just by its angles? (I assume the question refers to a unique triangle.)

sunehra, given two angles 30 and 75, you can figure out the third angle to be 75, and it must be an isosceles triangle.

But for $ \textbf{(C)}$, you wouldn't be able to determine that it was an equilateral triangle.  I think the question means something like "If you are given that a certain triangle is the indicated triangle, and you are also given the parts named, which of the below triangles would you be unable to solve?" in which case the answer would be $ \textbf{(A)}$.

Oh sorry, I misread the question. (I was answering the question of which of the choices can determine a unique triangle.)

The actual answer is $ \boxed{\text{(A)}}$, but I agree with what tenniskidperson3 said about not being able to determine that it was an equilateral triangle. It would be so much more convenient if the old AHSME writers clearly wrote their problems...