Μέχρι ο Vornicu να μας φτιάξει υποφόρουμ , ας οργανώσουμε μόνοι μας λιγάκι τη δουλειά. Σε αυτό το χώρο θα στείλουμε 10 ασκήσεις  γεωμετρίας  , με τον υπότιτλο  [color=red]ΓΕΩΜΕΤΡΙΑ - 1[/color]  κλπ και κόκκινο χρώμα , με τις λύσεις τους.Με τον ίδιο υπότιτλο[color=red]ΓΕΩΜΕΤΡΙΑ - 1, ΛΥΣΗ[/color] θα στέλνουμε και τη λύση.Μόλις συμπληρώσουμε τον αριθμό 10 , θα ανοίξω νέο topic  για τη συνέχεια. Έτσι  , στο τέλος, με  το  search  θα μπορούμε να βρούμε όλες τις γεωμετρίες μαζί.
Θα κάνω το ίδιο με τη Θεωρία Αριθμών κλπ. Το ξέρω ότι θέλει λίγο υπομονή , αλλά δε γίνεται διαφορετικά.
 Για παράδειγμα , όποιος θέλει τώρα να βρει την πρώτη άσκηση γεωμετρίας που στάλθηκε στο φόρουμ, πρέπει να ανατρέξει στο search και πάει ... γυρεύοντας!

 Μπάμπης

[color=red]ΓΕΩΜΕΤΡΙΑ 1 [/color]
Συνδέουμε το τυχαίο σημείο $A$ της περιφέρειας $C_{1}$ και κέντρου $O$ με τα σημεία $M,\ N$  που ανήκουν σε διάμετρο $PD$ του κύκλου και βρίσκονται εκατέρωθεν του $O$ ώστε $OM=ON$ .
Οι $AM$ και $AN$ τέμνουν τον κύκλο στα σημεία $B,C$ αντίστοιχα . Να δείξετε ότι $\frac{AM}{BM}+\frac{NA}{NC}=\frac{2(OA)^{2}+2(OM)^{2}}{(OA)^{2}-(OM)^{2}}$

Η παραπάνω δόθηκε σε μάθημα προετοιμασίας για την ΙΜΟ .

μια επισήμανση:
ο καθένας θα είναι καλό να γράφει στην αρχή του μηνύματος σε ποια άσκηση αναφέρεται ώστε να μη γίνει χαμός

Σιλουανέ είσαι σίγουρος ότι η εκφώνηση είναι σωστή??Η αλήθεια αίναι ότι και στο μάθημα μας την είχαν δώσει με συτή την εκφώνηση-αλλά μου φαίνεται ότι δεν είναι πλήρης.Δηλαδή θέλω να πω ότι μάλλον λείπει μία συνθήκη!!Και αυτό γιατί προκύπτει ότι για σταθερά Α και Μ και για μεταβλητό Ν, το $\frac{NA}{NC}=ct$, πράγμα προφανώς άτοπο!

[b]Λοιπόν η συνθήκη που προτείνω εγώ είναι η εξής $OM=ON$.Προσπαθήστε τη έτσι...[/b]

Έχεις δίκιο .Το δεύτερο μέλος είναι ανεξάρτητο του Ν .Το διόρθωσα . :)
Περιμένω τις λύσεις σας.....

[u][color=red][b]ΓΕΩΜΕΤΡΙΑ 1-ΛΥΣΗ[/b][/color][/u]
[hide]$\frac{MA}{MB}+\frac{NA}{NC}=\frac{MA^{2}}{MA\cdotp MB}+\frac{NA^{2}}{NA \cdotp NC}=...$[/hide]

Δηλαδή ??????????? Υπόδειξη είναι αυτό ή καθαρή λύση ????????????

Καθαρή λύση (αριθμητής = πρώτο θεώρημα των διαμέσων, παρανομαστής = δύναμη σημείου στο εσωτερικού κύκλου).

(AM)·(BM) = (MP)·(MD) = (OA)² - (OM)² = (NP)·(ND) = (NA)·(NC)

Κώστας Βήττας.

Τελικά Νίκο το πρόβλημα δεν ήταν και τόσο δύσκολο.
Ήταν ουσιαστικά η ιδέα που είχα πει τότε στο μαθημα  :wink:

Μα όταν έχεις έναν κύκλο,μερικά σημεία και ευθείες και ένα μέσο ευθυγράμμου τμήματος,τι άλλο εκτός από δύναμη σημείου ως προς κύκλο και θεώρημα διαμέσων μπορεί να χρησιμοποιήσει κανείς?
Μ' αυτό το πρόβλημα όμως μπορεί κανείς να κατανοήσει γιατί η λύση κάθε προβλήματος βρίσκεται κρυμμένη στη εκφώνηση του.

Συμφωνώ απολύτως. Όταν δεν καταφέρνω να λύσω κάποιο πρόβλημα, είναι γιατί δεν βρίσκω τρόπο να χρησιμοποιήσω τις γεωμετρικές ιδιότητες, που απορρέουν από αυτήν την ίδια την εκφώνηση.

Κώστας Βήττας.

[color=red]ΓΕΩΜΕΤΡΙΑ 2- ΕΚΦΩΝΗΣΗ[/color]
Έστω κύκλος κέντρου $O$ και δύο σημεία του $P,Q$ .Οι εφαπτόμενες στα $P,Q$ τέμνονται στο $Z$ .Έστω μία χορδή $PA$ .Από το $Z$ φέρουμε παράλληλη στη $PA$ η οποία τέμνει τon κύκλο στα $B,C$ έτσι ώστε το τετράπλευρο που σχηματίζεται να είναι το $ACBP$ .Δείξτε ότι η $AQ$ τέμνει την $BC$ στο μέσον της .

[b][color=red][u]Γεωμετρία 2-Λυση[/u][/color][/b]
Το Ζ είναι η τομή της εφαπτομένης του κύκλου στο Q και της BC.Έυκολα με όμοια τρίγωνα αποδεικνύουμε ότι $\frac{ZB}{ZC}=\frac{CQ^{2}}{BQ^{2}}$,άρα αν Κ το αρμονικό συζυγές του Ζ,δηλαδή Κ η τομή της BC με την PQ τότε η QK είναι η συμμετροδιάμεσος του τριγώνου BQC.Και αφού $\angle AQC=\angle PQB$ (λόγω της παραλληλίας), συμπεραίνουμε ότι η AQ διχοτομεί την BC.

[i]Only for Silouan:[/i][hide]Ρε Σιλουανέ,λύσε μου μια απορία:Ποιά είναι μεγαλύτερη-η εκφώνηση ή η λύση??? :P  :rotfl: [/hide]

[hide]Δε νομίζω . Η λύση είναι 53 λέξεις και η εκφώνηση 45 . [/hide]

[i]Στο περιοδικό Gazeta Matematica  διαβάζω  ένα θέμα γεωμετρίας που  τέθηκε στην Εθνική Ολυμπιάδα της Ρουμανίας το 2006.Δίνω το θέμα αυτό και στο ελληνικό κοινό για δύο λόγους:
  α) Για να βλέπουμε όλοι μας  που κινούνται πάνω – κάτω τα θέματα και άλλων χωρών
  β) Για να έχουν τη δυνατότητα να δοκιμάσουν τις δυνάμεις τους  και Έλληνες μαθητές – επισκέπτες του forum-  μικρότερων τάξεων.
   Σκέφτομαι αυτή τη στιγμή μήπως είναι καλό να ανοίξουμε ένα topic με ασκήσεις για μαθητές μέχρι Α΄ - Λυκείου , ώστε εκεί μέσα να θέτουν και να λύνουν οι junior  σχετικά απλούστερα προβλήματα. Το topic να το ονομάσουμε:

 [color=blue]JUNIOR , ΑΣΚΗΣΕΙΣ 1 - 10 [/color]

 κλπ και να το εμπλουτίσουμε με ωραία διδακτικά προβλήματα επιπέδου μέχρι  αυτό της Βαλκανιάδας Νέων. Στη λύση των ασκήσεων θα βοηθάνε  οι senior και όλοι φυσικά  που μπορούν  να βοηθήσουν , μόνο που καλό είναι να κρύβουν την απάντηση για να μη φαίνεται η λύση , πριν καλά καλά οι μαθητές δουν την εκφώνηση.Γιατί είναι γεγονός ότι μέχρι τώρα  , όλα τα προβλήματα που μπαίνουν στο ελληνικό forum είναι θέματα αξιώσεων και αυτά κατά κανόνα , κάτι που είναι λογικό, δεν ενθαρρύνουν τους μικρότερους μαθητές να ασχοληθούν με τη λύση ασκήσεων. Ας μη ξεχνάμε ότι όσους περισσότερους  ''μικρούς’’ μαθητές βάλουμε στο παιγνίδι των ολυμπιάδων και των διαγωνισμών , τόσο μεγαλύτερο θα είναι το όφελος για  όλους![/i]

 [color=red]ΓΕΩΜΕΤΡΙΑ - 3[/color]

  ''Δίνεται τετράπλευρο  $ABCD$  εγγεγραμμένο σε κύκλο με ακτίνα $R$  , το οποίο έχει την ιδιότητα:
 Υπάρχει σημείο $P$ στην πλευρά $CD$   τέτοιο , ώστε $CB=BP=PA= AB$.Να αποδείξετε ότι $PD = R$ ''

Μπάμπης

[quote="stergiu"][color=red]ΓΕΩΜΕΤΡΙΑ 8[/color]

  [i]Ψάχνοντας στο δίκτυο , συνάντησα σε ένα διαγωνισμό του πανεπιστημίου της Florida [u] Μπάμπης[/u][/quote]

Πως λεγεται ο διαγωνισμος? Υπαρχει και ιστοσελιδα?
 :)

[quote="stergiu"]Για δες εδώ! Νομίζω θα ανοίξει.

http://www.math.fau.edu/MathematicsCompetition/

 Καλή Μελέτη ![/quote]

Nai οκ. πολυ καλο!! Θενκς!!

 :lol:

αν έχει ξαναλυθεί η άσκηση πρακαλώ κάποιος να την καθαρογράψει
Εστω ΑΒΓ τρίγωνο με ΑΔ ΒΕ ΓΖ ύψη τότε αν Ρ τομή ΒΓ ΕΖ  ΝΔΟ ΡΗ Κάθετη στην ΑΜ όπου Η ορθόκεντρο κι Μ μέσο ΒΓ

Ευχαρισ΄τώ θερμά τον Φίλο μου τον Νίκο Ράπανο αφού τον ρωτήαω πώς πάει?

Mia xara amvrosie!!Diavazo autes tis meres....an kai den einai prototypo!!!An exeis kanena problima me tin askisi pes mou!

Θα βάλλω τώρα μια άλυτη άσκηση (με είχε πεδέψει κάποιο διάστημα ) που μας είχε δοθεί σε μάθημα προετοιμασίας για τη Βαλκανιάδα .Νίκο αν έχεις βρει λύση βάλτην σε παρακαλώ ....
Αλλα και όποιος άλλος τη λύσει θέλω να βάλλει τη λύση του .

[color=red]ΓΕΩΜΕΤΡΙΑ 9-ΕΚΦΩΝΗΣΗ  [/color]
Έστω ΑΒC τρίωνο με γωνίες Α,Β,C  kai έστω $w$  η γωνία Brocard .Δείξτε ότι $8w^{3}\leq \hat{A}\cdot\hat{B}\cdot\hat{C}$ όπου οι γωνίες μετριούνται σε rad .

Nai tin thymamai!!Kontogiannis etsi?
Nomizo to eixame syzitise Silouane...Arketa dyskoli askisi!Nomizo pos tin exo ti lysi alla tora den prolavaino...Tha ti valo kapoia apo tis epomenes meres an den ti lysete mexri tote!

Ps. Pantos Silouane mi fovasai alyti den tha meinei.

An kai den asxolume me tin geometria oso o rapanos,o siluan, o vittasko....klp.klp ostoso merika 8emata geometrias me trelenun..... :wow: ......

An den kano la8os afto ixe pesi +se ena diagonismo kapu notioanatolika...(taiwan,tailandi,filipines,japan.......)distixos den ixa di tin lisi... :(
to mono poy den jero an boi8ai ine oti isxiiun san limata(den 8imame pu akrivos ta exo dei....malon sto f-g-m..)kai isxiun ta eksis

1)an P,P1 ine ta simia broccard,w i gonia broccard,O to perikentro tu ABC kai K to simio Lemoine,tote to trigono POP1 ine isoskeles me OP=OP1,kai episis i gonia POP1ine isi me 2wkai ta P,P1 anikun se kiklo me diametro OK.
2)gia tin gonia broccard (w) i sxii $w^{3}\leq (A-w)(B-w)(C-w)$

nomizo oti ayta prepi na boi8an stin lisi.....
a +kati alo...iparxi tipos pu na dini to a8risma ton ginomenon ton gonivn ana dio??? i to ginomeno +ton trion gonion????(sinartisi otidipote.....)

 :D

Στο φυλλάδιο που μας είχε δώσει ο Κοντογιάννης έλεγε να χρησιμοποιήσουμε την ανισότητα που έλεγες .Θυμάμαι επίσης ότι μας είχε πει ότι απλά είναι να βρούμε την κατάλληλη συνάρτηση και μετά να χρησιμοποιήσουμε την μονοτονία της και την κυρτότητά της .Αυτή που έβαλες ας πούμε είναι να κάνεις μονοτονία και Jensen για μια συνάρτηση (εσείς θα μου πείτε ποια  :wink: ) αλλά για αυτή δεν την έχω βρει ακόμα... Πάντως είμαι κοντά τώρα που την ξανακοίταξα.... :wink:

@Niko : Ήμουνα σίγουρος ότι δεν θα την είχες αφήσει άλυτη  :D

@silouan: oxi re an eite me ton enan eite me ton allo tropo tha ti vroume ti lysi!

Ήθελα να γράψω [u]δεν[/u] θα την αφήσεις άλυτη  :wink:

[b][size=100][color=red]ΓΕΩΜΕΤΡΙΑ 10 - ΛΥΣΗ[/color][/size][/b]

Νομίζω πως το πρόβλημα είναι αυτό http://www.mathlinks.ro/Forum/viewtopic.php?t=110887 και θα βάλω στο δικό μας φόρουμ, την απλή συνθετική απόδειξη για το γενικό θεώρημα, που έχω υποσχεθεί στο τελευταίο reply.

Κώστας Βήττας.

Nίκο μπορείς να βάλλεις τη λύση σου για αυτό με τη γωνία Brocard ?Α και κάτι άλλο .Στη γεωμετρία 6 με το ισοσκελές τραπέζιο θέλετε να βάλλω λύση ?


function
combinatorics unsolved
combinatorics

Let set $S=[1, 2, ... 2004]$. Denote $d_{1}$ be the number of subsets of $S$ for which the sum of the elements are $7 (mod32)$, and denote $d_{2}$ be the number of subsets of $S$ for which the sum of the elements are $14 (mod16)$. Compute the value of $d_{1}/d_{2}$.

Lemma: The number of subsets for which the sum of elements are t(mod 32) are equal. And the same is true (mod 16)

Proof: we use only that $2004 > 16$ and $2004>32$ so all the numbers $2^{0},2{1}...32$ are in S. So we can find bijection between sets giving distinct remainders (mod 32) the bijection concerns only power of 2.
Thus the lemma is proven.
From the lema $d1=2^{1999}$ and$d2=2^{2000}$ so $\frac{d1}{d2}= 0.5$

Baronov (or anyone else): I'm having trouble "seeing" this bijection -- could you provide a little further explanation?

you can use generating function to prove it
Consider $f(x)=(1+x)(1+x^{2})...(1+x^{2004)}=\sum b_{i}x^{i}$ we need to calculate $\sum_{32|(i-7)}b_{i}$
$g(x)=(1+x)(1+x^{2})(1+x^{4})...(1+x^{16})=1+x+x^{2}+..x^{31}$
$f(x)=g(x)h(x)$
$h(x)=\sum a_{k}x^{k}$
when multiply each term $a_{k}x^{k}$ with g(x) we get $a_{k}(\sum_{i=0}^{31}x^{k+i})$ there are exactly one i so that $k+i\equiv 7(mod 32)$
so the answer is $\sum a_{k}=h(1)=2^{1999}$


You could use the pgf package, it provides some really nice drawing features. Some of the features include:
[list]
-different styled arrows
-drawing in both polar and rectangular coordinates
-some advanced color controls including transitions of color
-including graphics and changing how they look right in the LaTeX file
-3-D appearence
-Nodes
[/list]

[quote="Valentin Vornicu"][quote="filletwho"]What's the best application for drawing images like triangles? Thx[/quote]Paint?!...  :?[/quote]

 That's the one I use as well :D

[quote="juggle7"]But, the *.bmp compressions have all the gunk around everything.  Mine looked absolutely horrible on round 1.  This time, they came out better.[/quote]Nobody said anything about .bmp files ... in plus they have no compressions, as they are "bit-maps" ...  :?

[quote="juggle7"]I'm not very computer saavy.  All I know is I draw an image in paint and save as a jpg, and it comes out crummy.  Maybe I can do something to prevent it?  I thought paint was a bitmap, so I assumed it compressed from bmp to whatever file you want.[/quote]Save it as a .png instead of a .jpg

http://www.mit.edu/~ibaran/kseg.html

There you go.

~Special thanks to EFuzzy for telling me about it.


[url=http://img214.imageshack.us/my.php?image=61024651za3.png][img]http://img214.imageshack.us/img214/9481/61024651za3.th.png[/img][/url]

How do I name the resultant vector?

is it $ \vec{PQ}$ or $ \vec{QP}$ and why please?

A vector generally isn't named like that.

http://home.nc.rr.com/enloephysics/vector.htm

[hide]
$ \vec{z_{1}}\minus{}\vec{z_{2}}\equal{}\vec{QP}\not \equal{}\vec{PQ}$.
[/hide]

$ Z_{1}\minus{}Z_{2}\equal{}Z_{1}\plus{}(\minus{}Z_{2})$
$ \minus{}Z_{2}$ is just $ Z_{2}$ with the head and tail switched.

[quote="414862074"]depend on (Z1-Z2) > or = or < 0[/quote]

Doesn't sound right. How can a vector be greater than or less than 0?


Let $ x,y \in R, x^2\plus{}y^2\plus{}x\plus{}y\equal{}\frac{1}{2}$. Find max and min:
$ P\equal{}\frac{x\plus{}2y}{x\plus{}y\plus{}3}$

[hide="hint?"]The condition is equivalent to $ (x \plus{} \frac {1}{2})^2 \plus{} (y \plus{} \frac {1}{2})^2 \equal{} 1$
So we must have $ \minus{} \frac {3}{2} \leq x, y \leq \frac {1}{2}$
Also: $ P \equal{} 1 \plus{} \frac {y \minus{} 3}{x \plus{} y \minus{} 3}$, set $ k \equal{} \frac {y \minus{} 3}{x \plus{} y \minus{} 3}$
[/hide]

[hide="Solution, proof"]
Note that $ k > 0$, since $ y \minus{} 3$ and $ x \plus{} y \minus{} 3$ are negative for all possible values of $ x$ and $ y$
We see $ k$ takes value 1 (x=0),
and it is easy to see $ k$ also take values greater than 1.

So for the maximun lets suppose $ k > 1$:
$ y \minus{} 3 \equal{} k(x \plus{} y \minus{} 3)$, $ (y \minus{} 3)(1 \minus{} k) \equal{} kx$ and $ y \minus{} 3 \equal{} \frac {kx}{1 \minus{} k}$,
so we must have $ \minus{} \frac {9}{2} \leq \frac {kx}{1 \minus{} k} \leq \minus{} \frac {5}{2}$ and $ x > 0$.
Working the right side: $ 2kx \ge 5k \minus{} 5$, $ 5 \ge k(5 \minus{} 2x)$ and $ \frac {5}{5 \minus{} 2x} \ge k$
it follows that the maximun of $ k$ is $ \frac {5}{4}$ when $ x \equal{} y \equal{} \frac {1}{2}$.

Analogous, for the minimun we may suppose that $ k < 1$, impliying $ x < 0$.
We get a similar relation but with inverse direction:
$ \frac {5}{5 \minus{} 2x} \leq k$
we see the minimun $ k$ can take is $ \frac {5}{8}$ when $ x \equal{} \minus{} \frac {3}{2}$ and $ y \equal{} \frac {1}{2}$.
[/hide]

[hide="answer"]
Agrouping this results:

                                  $ \frac {13}{8} \leq P \leq \frac {9}{4}$
[/hide]

 :)

_______________________________________________________________________________
Sorry I changed the + with a -, I'll post the solution in a while.

[quote="ElChapin"]The condition is equivalent to $ (x \plus{} \frac {1}{2})^2 \plus{} (y \plus{} \frac {1}{2})^2 \equal{} 1$
So we must have $ \minus{} \frac {3}{2} \leq x, y \leq \frac {1}{2}$
Also: $ P \equal{} 1 \plus{} \frac {y \minus{} 3}{x \plus{} y \minus{} 3}$, set $ k \equal{} \frac {y \minus{} 3}{x \plus{} y \minus{} 3}$

 :)[/quote]
It's wrong!
I think $ P \equal{} 1 \plus{} \frac {y \minus{} 3}{x \plus{} y \plus{} 3}$, set $ k \equal{} \frac {y \minus{} 3}{x \plus{} y \plus{} 3}$


Beth goes to school sporting her new smiley-face stamps, now with bright yellow ink!  One of her stamps is huge – a circle of radius 1 meter!  The other one is far smaller – it is an equilateral triangle of side length s.  She sneaks up to a wall when she has some free time and quickly slaps three huge smiles on the wall, each tangent to 2 others (no overlap here!).  She then decides to finish the job by placing a small triangular stamp in the middle of the three circles without any overlap, anywhere.  What is the largest value for s such that Beth can complete her artistic endeavor?

ofcourse, each of the smaller iscoceles triangles have their equal sidlength of $\frac{\sqrt{3}}{3}$. And from there, you can do it...


This is a standard question but I'm not able to figure out the last part of the argument.
[i][b]Problem:-[/b][/i]

[color=red][u]F(x) is a polynomial of degree N s.t.

F(x) = (1/x) for X =1,2,3,4 ....... N

Find F(N+1) [/u][/color]

I've got a few questions -

1) In the above case I wrote that let p(x) = xF(x) -1 . Clearly p(x) has N roots(1,2,....) .and we can factorise LHS. But ehat about the constant polynomial which is left to be determined on the RHS.?????

2)(next Qs are regarding general case)Given this much info can we determine[b] F(x+K) [/b]for K>0 (integers) . If not , how to [b]prove[/b] it?

3) Can we solve the question when [b](1/x) is replaced by an arbitrary g(x)[/b]

4) also how to solve the problem for g(x) = 2^x

[b]Thanks!@!@![/b]

1) Use that $p(0) = -1.$

2) Since you have determined $F(x)$ you can find $F(x+k)$, but the expression may be not nice.

3) I do not think it can be done in general as easily as in the original problem or in 4).

4) Let $k$ be a nonnegative integer, and let $C_x^k = \frac {x(x-1) \cdots (x-n+1)} {n!}$ (and $C_x^0 = 1$).
Let $P(x) = \sum_{k=0}^n C_x^k.$ Then $P$ is a real polynomial with degree $n$ such that $P(k) = 2^k$ for $k=0,1, \cdots , n$ (since for all integers $t \geq 0$ we have $C_t^0 + C_t^1+ \cdots + C_t^t = 2^t$ and $C_t^i = 0$ for $i>t$).
Moreover $P(n+1) = C_{n+1}^0 + \cdots C_{n+1}^n = 2^{n+1} - 1.$

Pierre.


geometry
circumcircle
trigonometry
geometry unsolved

[quote="skalgar"]A circumference $ \alpha$ intersects with circumference $ \beta$ in points $ A$ and $ B$. There is a tangent line to both circumferences $ \alpha$ and $ \beta$ which intersects them in points $ C$ and $ D$ respectively. Points $ C$, $ D$, $ B$ ($ B$ is closer to the tangent line) lie on the circumference $ \gamma$. Prove, that the radius of circumference $ \gamma$ is the geometric mean of the radiuses of the circumferences $ \alpha$ and $ \beta$.[/quote]
Let the circumcenter and the circumradius of the circles $ \alpha,\beta,\gamma$ be $ O_1,O_2,O$  and $ r_1,r_2,r$ respectively.
We shall prove that, $ r^2 \equal{} r_1 r_2$
Let $ \angle BCD \equal{} \theta_1, \angle BDC \equal{} \theta_2$

Using sin law in $ \triangle O_1BC$ we have,  $ \frac {r_1}{\cos \theta_1} \equal{} \frac {BC}{\sin 2\theta_1} \iff BC \equal{} 2r_1 \sin \theta_1$

Again, using sin law in $ \triangle O_2BD$ we have,  $ \frac {r_2}{\cos \theta_2} \equal{} \frac {BD}{\sin 2\theta_2} \iff BD \equal{} 2 r_2 \sin \theta_2$

By extended law of sine we get, in $ \triangle BDC$
\[ 2r \equal{} \frac {BC}{\sin \theta_2} \equal{} \frac {BD}{\sin \theta_1} \iff 4r^2 \equal{} \frac {BC}{\sin \theta_2}\cdot \frac {BD}{\sin \theta_1} \equal{} 4r_1r_2 \iff r^2 \equal{} r_1r_2
\]
and we are done. :)

EDIT: Fixed a typo. My 300th post :D

Thank you. But I can't understand why there is cos in $ \frac{r_{1}}{\cos\theta_{1}}\equal{}\frac{BC}{\sin 2\theta_{1}}\iff BC\equal{}2r_{1}\sin\theta_{2}$

[quote="skalgar"]Thank you. But I can't understand why there is cos in $ \frac {r_{1}}{\cos\theta_{1}} \equal{} \frac {BC}{\sin 2\theta_{1}}\iff BC \equal{} 2r_{1}\sin\theta_{2}$[/quote]
It was a typo. It should be $ \frac {r_{1}}{\cos\theta_{1}} \equal{} \frac {BC}{\sin 2\theta_{1}}\iff BC \equal{} 2r_{1}\sin\theta_{1}$

I have fixed it :)

[quote="skalgar"]Nevertheless, why do you divide r1 by $ cos\theta_{1}$[/quote]
You should think a bit before asking this. (Because here everyone writes their solution briefly)

BTW as $ CD$ is tangent we have $ \angle CO_1B\equal{}2 \times \theta_1$.Also, $ O_1B\equal{}O_1C\equal{}r_1$ i.e. $ \triangle O_1BC$ isosceles. $ \angle O_1CB\equal{}\angle O_1BC$ and $ \angle O_1CB\plus{}\angle O_1BC\equal{}180^{\circ}\minus{}2\theta \iff \angle O_1BC\equal{}90^{\circ}\minus{}\theta_1$ and we know that $ \sin (90^{\circ}\minus{}\theta_1)\equal{}\cos \theta_1$.

Never mind. I understood. Again great thanks for your help.

Let E, F and G be the centers of the three circles; since CD is tangent to the circles E and F, we get: <BCD = <BEC/2 = <BEG ( 1 ) and, similarly, <BFG = <BDC ( 2 ).

Let's now call M and N the midpoints of BC and BD (actually, the intersections BC with GE and BD with FG); obviously, GMBN is cyclic, hence <MNB = <BGM ( 3 ) and
<BGN = <BMN ( 4 ), but MN||CD, hence the triangles BEG and BGF are similar to triangle BDC, therefore similar, and from their similarity we get $ BE \cdot BF \equal{} BG ^{2}$.

[b]Note: The circles (BCD) and (ACD) are equal, hence the specification "B closer to CD than A" is not relevant![/b]

Best regards,
sunken rock[/b]

The topic can be moved to "Solved" section of the forum.


Check the results out at http://www.bprim.org
  Congratulations to all those who made it (me too!)
  Hoping to see y'all at IMOTC,
    Ashwath Rabindranath

Congrats to everyone...I am happy i made it man...I thought getting in 12th would be impossible!!!
Cheers
Bhargav

can someone from pune ([b]bhaskaracharya prathishthana[/b]) post the marks obtained by the students selected in the INMO-07 ( VIHANG PLEASE)

I made it too ! :lol:
Are the results arranged in order of rank?

CONGRATS TO ALL!!!

 :coolspeak:  :coolspeak:

IN BP SITE, IN THE FINAL RESULTS, MARKS ARE GIVEN AS WELL AS ORDER OF RANK..

DONT KNOW IF THE PRESENT RESULTS PDF TALLY IS RANKWISE.

@ASWATH
'ARE' CAN BE USED TO REPRESENT BOTH PRESENT TENSE AS WELL AS PAST TENSE.

I MEANT IT IN THE PAST TENSE AND ANYWAYZ IT ISN'T MUCH HELP.

Where are they? Can  you post a link or something. I can only find the PDF without the marks

AREN'T THE MARKS OUT...??

WHAT I MEANT WAS THEY 'USUALLY' COME OUT IN THE BP SITE.
MAYBE THIS YEAR THEY WERE TOO EAGER TO TELL U THAT U ARE SELECTED AND FORGOT TO WRITE THE MARKS..

WAIT FOR A DAY OR TWO...

@ashwath.rabindranath: Can you send to me Lecture Series for college teachers and students in http://www.bprim.org/nbhmlecseries.php  ? Here is my e-mail nttuan_qn@yahoo.com.vn Thanks!  :)

heyy

congrats to all of u!!!!!  :clap2:  :clap:

Hello,
    Congratulations to all those who made it.
Sathej

Congrats to all of those who got selected.

Now a days I don't get time to go to BP ... I'll see if I can get the marks.

Heyyy
Congrats to all u really really brainy people!! :clap2:  :thumbup:
Can any of u please tell me how you go about participating in all of this? Do you HAVE to go through your school because if the answer is yes then my school doesn't even know that INMO exists!!  NONE of the teachers or co-ordinators take interest in ANY olympiad and we always end up out of it even though so many of us want to take part. :(
So any suggestions?

contact the co ordinator of karnataka and meet him in person (if possible)or write a letter to him asking to send forms to your school somewhat in september for RMO .(it happens like this in delhi , i  dont know much about karnataka ).

Karnataka's regional co-ordinator is A.K.Nandakumaran of IISC.
You can try to contact him. His email address is given on iisc's website

Oh Ok thank you so much...will surely take a shot at it next year :D ...around what time should I do it, though?
Thanks again

krmo is usually held in november
so i guess during august or september some time


Prove that every natural number can be expresed in the same number of forms if:
there is only odd numbers
there is only even numbers(without 0). $NOTE$ the position is important, too.
Example: $4=1+1+1+1, 1+3, 3+1, 2+2$
               $4=1+1+2, 1+2+1, 2+1+1, 4$

1 1 1...1 1 (n 1s)
The number of ways to express $n$ in the sum of odd number of numbers is $\dbinom{n}{1}+\dbinom{n}{3}+\dbinom{n}{5}+...$ because we choose odd number of spaces between n 1s to insert "$+$".

The number of ways to express $n$ in the sum of even number of numbers is $\dbinom{n}{0}+\dbinom{n}{2}+\dbinom{n}{4}+...$

Both are equal to $2^{n-1}$


Where can I get past olympiad contests? If someone has them can you email them to me?

I searched online and i just found a random one.. it is attached. Are you allowed to put them on the web and stuff or is it like mathcounts?

there's a book called "Math Olympiad Contest Problems for Elementary and Middle Schools" by George Lenchner and also "Math Olympiad Contest Problems: Volume 2." I have the former, and the questions are exactly like MOEMS questions.

Yeah, like Thunder said, you could get some books from their site.
[url]http://www.moems.org/Books.htm[/url]

or u could ask for someone for them on MATHCOUNTS Resources here!

most people on AoPS don't give away competitions for free though. What if he's like me and doesnt have much of an inventory to trade with?


calculus integration trigonometry limit function probability derivative

how many methods do you know, [b][u]besides[/u][/b] squaring and transforming to
polar co-ordinates, for solving the following integral ?  :|

$ \int_{0}^\infty\;e^{-x^{2}}\;\textbf dx\;=\;\boxed{\frac{\sqrt\pi}{2}}$

[Method 1] Let $ I = \int_{0}^{\infty}e^{-x^{2}}\; dx$. Then

$ \int_{0}^{\infty}e^{-sx^{2}}\; dx = \frac{I}{\sqrt{s}}$.

Now product $ e^{-s}$ and integrate both side of this equation w.r.t $ s$ on $ [0, \infty)$.

The RHS becomes $ 2I^{2}$ with substitution $ s = x^{2}$ and the LHS can be computed as follows:

\begin{eqnarray*}\int_{0}^{\infty}\left( \int_{0}^{\infty}e^{-sx^{2}}\; dx \right) e^{-s}\; ds & = & \int_{0}^{\infty}\int_{0}^{\infty}e^{-s(x^{2}+1)}\; ds \; dx\\ & = & \int_{0}^{\infty}\frac{1}{1+x^{2}}\; dx\\ & = & \frac{\pi}{2}\end{eqnarray*}

So $ 2I^{2}= \frac{\pi}{2}$ and $ I = \frac{\sqrt{\pi}}{2}$.

[Method 2] This approach starts with the assumption that the sequence $ a(n) = \int_{0}^{\sqrt{n}}\left( 1-\frac{x^{2}}{n}\right)^{n}\; dx$ converges to the original integral.
(The reason I call it 'assumption' is that I did not prove this yet.)

Substitute $ x = \sqrt{n}\sin t$, then

$ a(n) \; = \; \sqrt{n}\int_{0}^{\frac{\pi}{2}}\cos^{2n+1}t \; dt \; = \; \sqrt{n}\frac{(2n)!!}{(2n+1)!!}$

where $ k!!$ is the double factorial. Wallis formula shows that $ (2n+1) \left( \frac{(2n)!!}{(2n+1)!!}\right)^{2}\to \frac{\pi}{2}$. Applying it to $ a(n)$, we have $ \int_{0}^{\infty}e^{-x^{2}}\; dx = \lim a(n) = \frac{\sqrt{\pi}}{2}$.

Complex analysis together with Beta and Gamma functions is complete overkill but here it goes ...

1)Let $ 0 < a < 1$ and consider the integral,
$ \int_{-\infty}^{\infty}\frac{e^{ax}}{e^{x}+1}dx$ --> This converges.

2)Consider the complex function, $ f(z) = \frac{e^{az}}{e^{z}+1}$

3)For $ R>0$ consider the contour at $ \pm R,\pm R+2\pi i$.

4)By residue theorem this integral is $ -2\pi i e^{\pi a i}$

5)There are 4 integrals to consider: the two horizontal ones + the two vertical ones.

6)If $ R$ is large then $ \left| \frac{e^{a(R+iy)}}{1+e^{R+iy}}\right| \leq \frac{e^{aR}}{e^{R}-1}\leq \frac{e^{aR}}{e^{R}-\frac{1}{2}e^{R}}= 2e^{(a-1)R}$

7)By #6: $ \left| \int_{0}^{2\pi}\frac{e^{a(R+iy)}}{e^{R+iy}+1}i dy \right| \leq 4\pi e^{(a-1)\pi}\to 0$ as $ R\to \infty$ since $ a-1<0$

8)Similarly the other vertical integral: $ \left| \int_{0}^{2\pi}\frac{e^{(-R+iy)}}{e^{-R+iy}+1}i dy \right|\to 0$ as $ R\to \infty$.

9)The bottom horizontal vertical integral is: $ \int_{-R}^{R}\frac{e^{ax}}{e^{x}+1}dx$ which is what we want.

10)The upper horizontal integral is: $ \int_{-R}^{R}\frac{e^{a(x+2\pi i)}}{e^{x+2\pi i}+1}dx = e^{2a\pi i}\int_{-R}^{R}\frac{e^{ax}}{e^{x}+1}dx$

11)Make $ R\to \infty$: $ (e^{2a\pi i}-1) \cdot (PV)\int_{-\infty}^{\infty}\frac{e^{ax}}{e^{x}+1}dx = 2\pi i e^{a\pi i}$.

12)So, (remove the Cauchy principle value by #1): $ \int_{-\infty}^{\infty}\frac{e^{ax}}{e^{x}+1}dx = \frac{\pi}{\sin \pi a}$

13)Make the subsitution $ t=e^{x}$ to get $ \int_{0}^{\infty}\frac{t^{a-1}}{t+1}dt = \frac{\pi}{\sin \pi a}$

14)Let $ \xi = \frac{t}{t+1}$ to get $ \int_{0}^{1}\xi^{a-1}(1-\xi )^{-a}d\xi = \Gamma (a, 1-a)$

15)So we obtain $ \Gamma (a) \Gamma (1-a) = \frac{\pi}{\sin \pi a}$ let $ a=\frac{1}{2}$ to get $ \Gamma \left( \frac{1}{2}\right) = \sqrt{\pi}$.

16)But #15 is the Gaussian Probability Integral when properly substituted.
Q.E.D.

[Method 3] This approach is similar to that of sylow_theory's in the point that it also evaluates the value of $ \Gamma \left( \frac{1}{2}\right)^{2}$.

Let $ 0 < z < 1$. Using Laplace transform, $ \frac{\Gamma(z)}{t^{z}}= \int_{0}^{\infty}u^{z-1}e^{-tu}\; du$ holds and

\begin{eqnarray*}\Gamma(z)\Gamma(1-z) & = & \Gamma(z)\int_{0}^{\infty}t^{-z}e^{-t}\; dt\\ & = & \int_{0}^{\infty}\int_{0}^{\infty}e^{-t}u^{z-1}e^{-tu}\; du \; dt\\ & = & \int_{0}^{\infty}\int_{0}^{\infty}u^{z-1}e^{-(u+1)t}\; dt \; du\\ & = & \int_{0}^{\infty}\frac{u^{z-1}}{1+u}\; du. \end{eqnarray*}

Plugging $ z = \frac{1}{2}$, we have

\begin{eqnarray*}\Gamma \left( \frac{1}{2}\right)^{2}& = & \int_{0}^{\infty}\frac{1}{\sqrt{u}(1+u)}\; du\\ & = & \int_{0}^{\infty}\frac{2}{1+v^{2}}\; dv \quad \quad (\text{where }u = v^{2})\\ & = & \pi. \end{eqnarray*}

Finally, $ \int_{0}^{\infty}e^{-x^{2}}\; dx = \frac{1}{2}\Gamma \left( \frac{1}{2}\right) = \frac{\sqrt{\pi}}{2}$

[quote="sos440"][Method 1] Let $ I = \int_{0}^{\infty}e^{-x^{2}}\; dx$. Then

$ \int_{0}^{\infty}e^{-sx^{2}}\; dx = \frac{I}{\sqrt{s}}$.

Now product $ e^{-s}$ and integrate both side of this equation w.r.t $ s$ on $ [0, \infty)$.

The RHS becomes $ 2I^{2}$ with substitution $ s = x^{2}$ and the LHS can be computed as follows:

\begin{eqnarray*}\int_{0}^{\infty}\left( \int_{0}^{\infty}e^{-sx^{2}}\; dx \right) e^{-s}\; ds & = & \int_{0}^{\infty}\int_{0}^{\infty}e^{-s(x^{2}+1)}\; ds \; dx\\ & = & \int_{0}^{\infty}\frac{1}{1+x^{2}}\; dx\\ & = & \frac{\pi}{2}\end{eqnarray*}

So $ 2I^{2}= \frac{\pi}{2}$ and $ I = \frac{\sqrt{\pi}}{2}$.[/quote]

cool ! :thumbup: $ \text{IuIs}$ rox! :yup:

Consider the functions:
$ f(x)\equal{}\int_0^xe^{\minus{}y^2}dy$,
$ g(x)\equal{}\int_0^1\frac{e^{\minus{}x^2(1\plus{}t^2)}}{1\plus{}t^2}dt$.
Using the theorem of differentiation under the integral sign, we get
$ g'(x)\equal{}\int_0^1\frac{e^{\minus{}x^2(1\plus{}t^2)}}{1\plus{}t^2}(\minus{}2x(1\plus{}t^2))dt\equal{}\minus{}2e^{\minus{}x^2}\int_0^1xe^{\minus{}x^2t^2}dt$.
After the substitution $ t\equal{}\frac{y}{x}$ we have
$ g'(x)\equal{}\minus{}2e^{\minus{}x^2}\int_0^xe^{\minus{}y^2}dy\equal{}\minus{}2f'(x)f(x)$.
So
$ g(x)\equal{}\minus{}[f(x)]^2\plus{}c$.
But $ g(0)\equal{}\frac{\pi}{4}$, hence $ c\equal{}\frac{\pi}{4}$.
Now
$ \lim_{x\minus{}>\plus{}\infty}(\minus{}[f(x)]^2\plus{}\frac{\pi}{4})\equal{}\lim_{x\minus{}>\plus{}\infty}g(x)\equal{}0$.
This implies
$ \lim_{x\minus{}>\plus{}\infty}[f(x)]^2\equal{}\frac{\pi}{4}$,
and therefore
$ \int_0^{\plus{}\infty}e^{\minus{}x^2}dx\equal{}\sqrt{\frac{\pi}{4}}$.


$\mbox{G}$ is centroid of triangle $ABC$ prove that:$\frac 45 \leq \frac {S_1}{S_2} \leq \frac 54$

See the EDIT in http://www.mathlinks.ro/Forum/viewtopic.php?t=43908 post #5, particularly the inequality $\frac{\left[AMNC\right]}{\left[MBN\right]}\leq\frac54$. Note that [AMNC] stands for the greater one of your two areas $S_1$ and $S_2$, and [MBN] for the smaller one, so this automatically yields $\frac45\leq\frac{S_1}{S_2}\leq\frac54$.

  darij


Solve the following equation \[\sqrt[2n]{\frac{1}{2^{n}}-x}+\sqrt[2n]{\frac{1}{2^{n}}+x}=1.\] [i]Edited by Myth[/i]

Take f(x)=sqrt2n(1/2^n-x)+sqrt2n(1/2^n+x)-1, then f is an even function,so lets study it on (0,1/2^n).A simple study shows that f is injective on this interval (being strictly monotonous), f changes sign on this interval , so it has unique solution x0, and the other solution is -x0.
But which is the solution?(The problem says solve the equation)

Maybe typo? The most evident thing is that there $\sqrt[n]{\;}$ should be.


Choose terms of the harmonic series so that the sum of the chosen terms be finite. Prove that the sequence of these terms is of density zero in the sequence
$ 1,\frac12,\frac13,\dots,\frac1n,\dots$

Let $S \subseteq \Bbb Z$ be the set of chosen terms, and let $d(n)$ be $\frac 1 n\#(S \cap [1, n])$.

Thus for each $n \geq 1$, we have $1_{n \in S} = nd(n) - (n - 1)d(n - 1)$ (where $d(0)$ can be assigned any value).

Now consider the partial sum $s(N) = \sum_{n \leq N}\frac{1_{n\in S}} n$.
Using the above expression for $1_{n \in S}$, we get $s(N) = \left(\sum_{n \leq N}d(n)\right) - \left(\sum_{n \leq N}\frac{n - 1}n d(n - 1)\right) = d(N) + \sum_{n < N} \frac {d(n)}{n + 1}$.

By assumption, the sequence $(s(N))_N$ converges.
This implies that the sequence of partial sums $\left(\sum_{n < N} \frac{d(n)}{n + 1}\right)_N$ converges, as it is increasing and bounded.
From that, we further deduce that the sequence $(d(N))_N$ converges, as it is the difference of two convergent sequences.

It is then clear that the limit $\lim_{N \rightarrow \infty} d(N)$ must be $0$: otherwise there would be some $\epsilon > 0$ such that $d(N) > \epsilon$ for all sufficiently large $N$, and thus the sequence $\left(\sum_{n < N} \frac{d(n)}{n + 1}\right)_N$ would be unbounded, as the harmonic series diverges.

[quote=Isolemma]
Let $S \subseteq \Bbb Z$ be the set of chosen terms, and let $d(n)$ be $\frac 1 n\#(S \cap [1, n])$.

Thus for each $n \geq 1$, we have $1_{n \in S} = nd(n) - (n - 1)d(n - 1)$(where $d(0)$ can be assigned any value).

Now consider the partial sum $s(N) = \sum_{n \leq N}\frac{1_{n\in S}} n$.
Using the above expression for $1_{n \in S}$, we get $s(N) = \left(\sum_{n \leq N}d(n)\right) - \left(\sum_{n \leq N}\frac{n - 1}n d(n - 1)\right) = d(N) + \sum_{n < N} \frac {d(n)}{n + 1}$.

By assumption, the sequence $(s(N))_N$ converges.
This implies that the sequence of partial sums $\left(\sum_{n < N} \frac{d(n)}{n + 1}\right)_N$ converges, as it is increasing and bounded.
From that, we further deduce that the sequence $(d(N))_N$ converges, as it is the difference of two convergent sequences.

It is then clear that the limit $\lim_{N \rightarrow \infty} d(N)$ must be $0$: otherwise there would be some $\epsilon > 0$ such that $d(N) > \epsilon$ for all sufficiently large $N$, and thus the sequence $\left(\sum_{n < N} \frac{d(n)}{n + 1}\right)_N$ would be unbounded, as the harmonic series diverges.[/quote]

[quote=Isolemma]When will I become "old" user and be able to use LaTeX?[/quote]

I think the time was one/two weeks, but I might be wrong.

[quote=GianDR]I think the time was one/two weeks, but I might be wrong.[/quote]

Thanks, I thought it was something like after posting a certain number of posts.
If it's a fixed amount of time, then perhaps I should wait that period before writing more posts.

Still it's a bit strange to disable LaTeX for "new users".

[quote=Isolemma]Still it's a bit strange to disable LaTeX for "new users".[/quote]

That's because LaTeX is rendered as images, and at some point many troll accounts caused some trouble posting lots of images (or something like that) so new accounts can't post images (unfortunately that includes LaTeX).


Let $ a,b,c>0$ with $ a\plus{}b\plus{}c\equal{}1$ Show that

$ \frac{a^{2}\plus{}b}{b\plus{}c}\plus{}\frac{b^{2}\plus{}c}{c\plus{}a}\plus{}\frac{c^{2}\plus{}a}{a\plus{}b}\geq2$

$ \sum \frac{a^2\plus{}b}{b\plus{}c} \equal{} \sum \frac{a^2 \plus{}1 \minus{} c \minus{} a}{b\plus{}c} \equal{} \sum \frac{a^2\minus{}a}{1\minus{}a} \plus{} \sum \frac{1\minus{}c}{1\minus{}a} \equal{} \minus{} \sum a \plus{} \frac{1\minus{}c}{1\minus{}a}\plus{}\frac{1\minus{}a}{1\minus{}b}\plus{} \frac{1\minus{}b}{1\minus{}c} \geq 3\minus{}1\equal{}2 \equal{}> QED$ ( AM-GM)


Se considera triunghiul ascutitunghic $\ ABC$ inscris intr-un cerc de centru $\ O$.Luam punctele $\ M,N$ astfel incat $\ BA$ sa fie bisectoarea unghiului $\ MBC$ si $\ CA$ bisectoarea unghiului $\ NCB$.Dreptele $\ BM$ si $\ CN$ se intersecteaza in $\ F$.Demonstrati ca $\ OA+OF=AF$.

@ Kosmin Pohoatza: asta mai merge mai?cum ti se pare?sper sa postezi o solutie. :D

Citesc problema asta pe la 2 noaptea asa :) dupa o debusolare :) (a batut FRANTZA :((((( )

Ceea ce trebuie demonstrat este echivalent cu a demonstra ca O, A, F sunt coliniare...

Rezultatul este evident...Intrucat observam triunghiu FBC...cu bisectoarele exterioare AB si AC ce se intersecteaza in A...deci FA este bisectoare pt FBC...si chestia aia cu centru cercului circumscris al ABC e pe AF e probabil o proprietatea ce se demonstreaza usuratic :)...o sa postez maine daka vrei si demonstratzia la asta...dar probabil ca ajea se face ;)

Mie nu pre mi se pare corect ce zici.. :? Eu am demonstat-o intr-un mod mai lung, dar sigur e corect.Explica mai detaliat, ca poate d-aia nu-nteleg.

Uite rezolvare[misto.zic eu] Notam m(ABO)=y,m(BCO)=z,m(ACO)=x.Atunci este evident ca x+y+z=90.Acum fie (BD bisectoarea unghiului FBC si cum (BA este bisectoarea unghiului MBC  rezulta ca
$m(ABD)=90^\circ \iff z+y+m(CBD)=90^\circ \Rightarrow m(BCD)=x \Rightarrow m(CBF)=2x$
.Acum aratam ca Patrulaterul BOCF este inscriptibil.Dupa acelashi ratzionament ca cel de sus avem m(BCF=2y)$\Rightarrow m(OBF)+m(OCF)=z+2x+z+2y=2(x+y+z)=180^\circ \Rightarrow$ BOCF inscriptibil rezulta ca m(CBF)=m(COF)=2x.Dar $m(COA)=180^\circ -2x$(din triunghiul ACO) rezulta ca $m(COA)+m(COF)=180^\circ$ rezulta ca A,O,F coliniare.
End of story

$\text{Solutie}$
  Fie $\ U$ si $\ V$ punctele de intersectie ale dreptelor $\ BM,CN$ cu tangenta in $\ A$ la cerc.Cu usurinta obtinem ca
$\angle UAB\equiv\angle\ ACB\equiv\angle\ ACV$,
$\angle VAC\equiv\angle\ ABC\equiv\angle ABU$, deci avem $\angle AUB=180-\angle UAB-\angle UBA=180-\angle ACV-\angle CAV=\angle AVC$, deci triunghiul $\ UFV$ este isoscel.Pe de alta parte, triunghiurile $\ ABC$ si $\ UBA$ sunt asemenea, de unde gasim $\frac{UA}{AC}=\frac{AB}{BC}$.Dar si $\ ABC$ si $\ VAC$ sunt asemenea, deci $\frac{AV}{AB}=\frac{AC}{BC}$.
   Se observa acum usor ca $\ UA=AV$ si deci $\ A$ este mijlocul segmentului $\ UV$.Drepta $\ UV$ este tangenta la cerc, asadar $\ OA\bot\ UV$.De asemenea, triunghiul $\ UVF$ este isoscel, adica avem si $\ FA\bot\ UV$.
   Coroborand cele doua relatii de mai sus, obtinem $\ A,O,F$ coliniare, deci $\ OA+OF=AF$. ;)
   Sper ca e corect.Eu asa m-am gandit.E drept, rezolvarea mi-a luat cam o ora... :blush:

Frate...E proprietate de clasa a 6-a ca...Intr-un triunghi ABC bisectoarele exterioare ale lui B si C sunt concurente in D , iar AD este bisectoare pt BAC :|...Si am zis ca faza cu O apartzine AD e o alta proprietate ce se demonstreaza usor :|

e doar desenu invers in problema ta:| e cam de cls a 6-a

[quote="pohoatza"]Frate...E proprietate de clasa a 6-a ca...Intr-un triunghi ABC bisectoarele exterioare ale lui B si C sunt concurente in D , iar AD este bisectoare pt BAC :|...[b]Si am zis ca faza cu O apartzine AD e o alta proprietate ce se demonstreaza usor :[/b]|[/quote]
poti da o solutie completa?nu de alta dar nu sunt sigur daca e bine cum am demonstrat-o eu. :oops:

O traducere a solutiei! :D
Notam cu P pct de intersectie dintre AO si cercul circumscris.
Se observa usor ca (BP si (CP sunt bis <FBC si <FCB
Deci si (FP este bis <BFC
Acum problema se rezuma in a arata ca F,P,A coliniare adica sa demonsrtam faptul ca (FA bis <BFC
Luam triunghiul FBC unde (BA si (CA sunt bis ext pt <FBC si <FCB si conform teoremei amintite de Kosmin si (FA bis <BFC :)


Well flopped as usual (and im getting quite used to it now :D )
Expecting 180's so nothing much  :(  :o

I didnt write it today.... Out of town parents, a very depressed student and board exam practicals the next day, is a very bad combination altogether... But how was the test? Straight objective or subjective too?

me....flopped very very badly in maths....
expecting around 140-150's........

@Shruthi the pattern was like that of FT 3

today i was totally exhausted
returned from bbsr today morn
was compltely awake the whole night
still i managed to appear
the paper was damn easy
i managed to get only 266

[quote="anitsahu"]today i was totally exhausted
returned from bbsr today morn
was compltely awake the whole night
still i managed to appear
the paper was - easy
i managed to get only 266[/quote]

Dei from day one on this forum you have been putting vetti scene da and nothing else  :huh: . Half my friends say that if they meet you they will surely murder you first  :D

170.. :( Smelled badly.

Paper II physics 57, my best performance.

wasted a lot of time on maths...should have done physics...wanted to improve my maths marks... :D

Dei if you had attempted physics first you would have got 200's. I hope I get 99 percentile+in paper 2.

I am expecting 151. Except I paper chemistry(64) everything down in the dumps. I didnt prepare anything and went with instincts and that led to my downfall.

Well, Chem subjectives were pathetic. Dunno why, they weren't good questions, kinda boring. And the answers are worse. I doubt anybody will get all 3 right.

[quote="madness"]arent you almost surely lying?  :huh: . And now you know how prepared i come all times; exactly the way you came today  :)[/quote]

I am positively not joking / lieing. I really flopped like hell. Please!!! No further comments until FIITJEE changes its most most pathetic solns.
 :furious:  :furious:  :censored:

As Far as i know due to the existance of a certan theorem called the [url=http://en.wikipedia.org/wiki/Uniqueness_theorem]uniqueness theorem[/url]....the intermediate mediate shouldn't amtter it's always $ F_{0}$ until otherwise the object is immersed in the medium

You read too much of theory da... My logic was that the charges in the dielectric realign itself and so there will always be a attractive force from dielectric so its either F0+x or F0-x depending on the sign of Fo.

that's because i love what i read....i think i sent an e-book address of that book to ranga ...also to u the site i sent it has the book called griffith read it it'a a classic

why da...just don't use the .linux site use lib.org.by site

Do u have math related ebooks da pardesi. If yes plz send them to me da.

Chem would be ok too  :D  :D  :D

[url=http://www.lookforbook.com]here[/url] is the site

[quote="valeriummaximum"]i am not able to get the book[/quote]

Maybe you don't know how to download???

Ya only blessed ones like our 'SHREYAS' know it. ellaya da?

Expecting in late 180's .flopped in second test
Like Shreyas(in Shivku pronounciation),everybody r laughing at me because i am different

Gautham Rang is expecting 250.  :blush:  :blush:

appadiya....

@shreyas-- i am not abook worm like you ....you always keep studying....


trigonometry
calculus
derivative
algebra
polynomial
calculus computations

Calculate value of max error using the Legrange's form of the remainder:

$\sin^{-1}(.4) \approx .4+\frac{(.4)^{3}}{2 \cdot 3}$

EDIT: If you could lay out the steps for this one it would be appreciated.. we learned this in class the other day and I was doing this one for practice and couldn't get it.

The Taylor series at zero for $\sin^{-1}x$ is given as $\sum_{n=0}^{\infty}\frac{(2n)!}{4^{n}(n!)^{2}(2n+1)}\cdot x^{2n+1}$

So you are asking how accurate is the second degree Taylor polynomial.

By the Lagrange form of the Remainder, between $.4$ and $0$ there shall exist a number $y$ strictly between them so that:

$\frac{f'''(y)}{3!}= R_{3}(.4)$

So,

$\frac{(2\cdot 3)!}{4^{3}(3!)^{2}(2\cdot 3+1)}\cdot y^{2\cdot 3+1}= R_{3}(.4)$

But $|y|<1$ so this is less than,

$\frac{(2\cdot 3)!}{4^{3}(3!)^{2}(2\cdot 3+1)}\cdot 1 = \frac{720}{64\cdot 36 \cdot 7}\approx .044$

So at least one digit is accurate.

When you say $|y|<1$, can you not assume that since $0<y<.4$, you can plug $.4$ into the remainder equation and this is the max the remainder can be?  Or does this not fit what the Lagrange form of the remainder entails?  Also (as I state below), I don't think you can assume the $\text{n}^{\text{th}}$ derivative is whatever you used in the polynomial because the polynomial is centered around $0$ and the $4^{\text{th}}$ derivative of this yields $0$ based on the polynomial.

Actually I think you got one part mixed up with another.. but I wouldn't of been able to see that unless you showed me the steps but here is how I think it should be done:
[hide]$\text{Let }f(x) = \sin^{-1}(x)$
$f(x) = \sin^{-1}(x)$
$f'(x) = \frac{1}{\sqrt{1-x^{2}}}$
$f''(x) = \frac{-x}{(x^{2}-1) \sqrt{1-x^{2}}}$
$f^{(3)}(x) = \frac{2x^{2}+1}{(x^{2}-1)^{2}\sqrt{1-x^{2}}}$
$f^{(4)}(x) = \frac{-3x(2x^{2}+3)}{(x^{2}-1)^{3}\sqrt{1-x^{2}}}$

We are trying to find how accurate the $3^{\text{rd}}$ degree polynomial is.

Thus between $0$ and $.4$ there is a number, $z$ such that:

$R_{n}(x)=\frac{f^{(n+1)}(z)}{(n+1)!}\cdot (x-0)^{n+1}$
$R_{3}(.4)=\frac{f^{(4)}(z)}{(4)!}\cdot (.4)^{4}$

Find ${f^{(4)}(z)}$ and assume the maximum error ($z=.4$):

${f^{(4)}(z)}|_{z=.4}=\frac{103750 \sqrt{21}}{64827}\approx 7.33402 \rightarrow a$

Thus

$R_{3}(.4) < \frac{(.4)^{4}}{(4)!}\cdot a \approx .0078229499$

Which is the answer the book gives as well $\boxed{R_{3}(.4) < .00782}$.

Also note that $\sin^{-1}(.4)-\left[.4+\frac{(.4)^{3}}{2 \cdot 3}\right] \approx .000850179$.[/hide]


So, this is just a problem that I made up:

In a right triangle, the arithmetic mean of the two legs is $ 3\sqrt{7}$ and the geometric mean of the two legs is $ 2\sqrt{7}$. What is the length of the hypotenuse?

It'll probably be solved pretty quickly, but ehh, I liked it.

Let $ x,\ y$ the legs, we have $ x\plus{}y\equal{}6\sqrt{7},\ \sqrt{xy}\equal{}2\sqrt{7}$, thus the length of hypotenuse is $ \sqrt{x^2\plus{}y^2}\equal{}\sqrt{(x\plus{}y)^2\minus{}2xy}\equal{}14.$

Yep. Okay another problem.

What are the last three digits of ${ 199^{{199}^{199}}}$?

And what are the last three digits of $ 199^{199^{{199}^{199}}}$?

Yeah, that's the right answer.

The way I did was a lot longer and far more cumbersome.
Could you explain what you did?

Is it $ (x\minus{}1)^a \equal{} (\minus{}1)^{a\minus{}1}(ax \minus{} 1)$ if it's mod?

It's just binomial expansion.

Higher powers of $ 200\equal{}0(mod 1000)$

$ (x\minus{}1)^{a}\equal{} (\minus{}1)^{a\minus{}1}(ax)\plus{}(\minus{}1)^a$ is  not always true . It's true only when $ x^2\equal{}0$ for the mod.

Oh wow, duh, my bad, good call on that one.


Find all positive integer satisfied:
        x|(yz+1)
        y|(zx+1)
        z|(xy+1)
I think there are only 3 solutions: (1,1,1) , (1,1,2), (1,2,3). But I can't pove it. I'm not good at number theory at all.

Let $yz+1=lx,\ zx+1=my,\ xy+1=nz$ for some $l,\ m,\ n\in\mathbb{N}$.
Multiply both sides, we have $lmn\cdot xyz=(xy+1)(yz+1)(zx+1)\Longleftrightarrow lmn=x+y+z+1+\frac{xy+yz+zx+1}{xyz}$.
Thus $lmn\in\mathbb{N}\Longleftrightarrow \frac{xy+yz+zx+1}{xyz}\in\mathbb{N}$.Set $1\leq x\leq y\leq z$, W.L.O.G. , we have $\frac{xy+yz+zx}{xyz}=\frac{1}{x}+\frac{1}{y}+\frac{1}{z}+\frac{1}{xyz}\leq \frac{1}{x}+\frac{1}{x}+\frac{1}{x}+\frac{1}{x\cdot x\cdot x}=\frac{3}{x}+\frac{1}{x^{3}}\leq 4,$
yielding $lmn\leq x+y+z+1+4$. I'm stuck on here. :(

We got xyz|(xy+yz+zx+1), and since the case x=1 is easy, suppose that $2\le x\le y\le z$.
If $xy+yz+zx+1\ge 2xyz$, then from $(x-2)(y-2)(z-2)\ge 0$ we get $2xyz>xy+yz+zx+1$
Thus xy+yz+zx+1=xyz
From now, I think it is easy to go on, I'm sleepy now. Thank you, Kunny, for nice idea!

There are many similar problems and most of them are solved by using some inequalities.

See also [url]http://www.mathlinks.ro/viewtopic.php?p=154338#154338[/url]   and
[url]http://www.mathlinks.ro/viewtopic.php?p=366273#p366273[/url]


[quote="cyberjoe"]Are you sure it's correct ? Here is something very similiar
[url=http://www.mathlinks.ro/Forum/viewtopic.php?t=43223]www.mathlinks.ro/Forum/viewtopic.php?t=43223[/url][/quote] Erm... As far as I can see, both problems are the same, just different symbols used...  :)

Yes, of course !
I was tired yesterday and didn't noticed $R(n) = \{1\leq k \leq 2n | \{\frac{n}{k}\} \geq \frac{1}{2}\}$ ;)


I remember that Brook's Theorem was posted, and i saw a problem in a book that was like an edge coloring version of the theorem but haven't succed on proving it.
here it goes:

Prove that we can color properly the edges of a graph with $d+1$ colors, where $d$ is the degree of the vertex with maximum degree.

also i saw that the bound can be improved to $\frac{3d}{2}$.
and when the graph is bipartite it is very easy to prove that it can be colored with $d$ colors.

The theorem you are looking for (the one with $d+1$) is due to Vizing, and proved also by Gupta one year later.
The improvement is due to Shannon (1949).

Proofs are quite lengthy, but I bet you may find them in any graph courses...

Pierre.


a,b and c are positive real numbers and n is an integer with n>=2 such that (a^n )+(b^n )= c^n.

For what values of k can you construct an obtuse-angled triangle with sides a^k , b^k and c^k ?

[quote="JBL"][hide="Partial Solution"]From $a^n + b^n = c^n$, we have $c > a, b$.  Thus, it follows that the function $f(x) = \left(\frac{a}{c}\right)^x + \left(\frac{b}{c}\right)^x$ is decreacing.  If $a^k, b^k, c^k$ are sides of an obtuse triangle, k positive, then $c^k$ is the largest side and so $c^k < a^k + b^k$ and $c^{2k} > a^{2k}+b^{2k}$ which implies $n \in (k, 2k)$ from the behavior of $f(x)$.  So $k \in (\frac{n}{2},n)$.  I haven't thought about negative k, but I think we'll need some condition on a, b, perhaps?[/hide][/quote]

I got that basic result from a different method, but I didn't post it... didn't know how to prove that all such triangles satisfy the triangle inequality.


calculus
integration
limit
trigonometry
real analysis
real analysis unsolved

For $t>0$ let's define :\[ g(t)=\int_{0}^{\frac{\pi}{2}}\frac{dx}{\sqrt{sin^2(x)+tcos^2(x)}} \]
Prove that:\[ g(t)=-\frac{ln(t)}{2}+2ln(2)+\epsilon (t) \] where \[ \lim_{t\to\ 0^+}\epsilon (t)=0 \]

because all the bad things appear when $x=0$, and that it is far easier to compute $h(t)=\int_0^{\frac{pi}{2}} \frac{cos(x)}{\sqrt{sin^2(x)+tcos^2(x)}} dx$, you should try to compute exactly $h(t)$, wich is possible, and then to look at $g(t)-h(t)$ that is not that big because $cos(x)=1-x^2+O(x^4)$ near $0$. But I'm not sure it will give you the second term, but for sure it will give you the $ln(t)$ term.

[quote="alekk"] But I'm not sure it will give you the second term, but for sure it will give you the $ln(t)$ term.[/quote]
Actually, it will. It is not hard to see (monotone convergence theorem) that $g(t)-h(t)=\int_0^{\frac \pi 2}\frac{1-\cos x}{\sqrt{\sin^2 x+t\cos^2 x}}\,dx\to \int_0^{\frac \pi 2}\frac{1-\cos x}{\sin x}\,dx$, which is also a computable integral. So, we can get the result up to $o(1)$ fairly easily.


I've looked up a lot of sites on how to change the margins so that the margins on all sides is 1 in., but none seem to resolve the problem of the bottom margin.

I'm using \usepackage{fullpage} now in a report class, but the bottom margin is huge (about 1.5 in.).  Is there a way to change the bottom margin so it's 1 in. as well?  Thanks in advance!

Try the [url=http://texcatalogue.sarovar.org/entries/geometry.html]geometry[/url] package. \usepackage[margin=1in]{geometry} will do precisely what you want.


Let $a,b,c\geq 0$ .Prove or disprove $\sum b^3c^3\geq abc(\sum a^3)$.

It is not true. IT can go both ways, try (a,b,c)=(100,1,1) and (1/100,1,1)

I think it is good for $abc \geq a+b+c$.Try with this condition.

Ok let's try if they are side lengths of a triangle.

No, (1.99,1,1).
What about trying to use paper/calculator to firstly think yout something?
Especially if you are trying to guess the conditions...
I belive its not so hard to fint these counter-examples for you.

think about it...

It's from the indentify:

$(c^2-ab)(b^2-ac)(a^2-bc) \le 0$

And the inequality of you true iff square of the number between the maximum and minimum $\ge$ multi of the maximum and minimum.


What is the probability that when we roll 5 fair 6-sided dice, at most 4 of them will show a 1?

when i see the words "at most", i think of complementary counting

the question asks what is the probability that at most 4 of the 5 dice will show a one
the only way that will not satisfy the condition is if all 5 of the dice show a one

so we just find the probability of them all being a one, and then subtract our result from one

what is the probability of all five dice being a one??
well, there is only 1 one on a six sided dice
therefore, the probability of getting one one is $ \frac{1}{6}$

so the probability of all five dice being ones is $ \frac{1}{6}\times\frac{1}{6}\times\frac{1}{6}\times\frac{1}{6}\times\frac{1}{6}\equal{}\frac{1}{7776}$

but we want to know the probability of at most 4 of them showing a 1, so we subtract our answer from one

$ 1\minus{}\frac{1}{7776}\equal{}\boxed{\frac{7775}{7776}}$

plz correct me if i made a mistake

I just got this problem, and I thought the answer is $ \frac {1}{1296}$ because $ \frac 16\cdot\frac16\cdot\frac16\cdot\frac16\cdot\frac66 \equal{} \frac {6}{7776} \equal{} \frac {1}{1296}$

[hide=eee]use complementary counting

there's 1/(6^5) chance that all 5 dice are 1's

so there's ((6^5)-1)/(6^5) chance that at most 4 are 1's

and 6^5=7776

so there's

7775/7776 chance[/hide]

why/how is this level 20? this got me to level 7 easily

D: I've done this problem b4 and I got it wrong :C

[hide = 'solution']Use complementary counting and count the ways that all 5 dice will show a 1 and then subtract the result from 1.
P(5 ones)  $= \frac{1}{6^5} = \frac{1}{7776}$.
P(at most 4 dice showing 1)$ = 1 - \frac{1}{7776} = \boxed{\frac{7775}{7776}}$[/hide]
pretty easy for lv 20. should be lvl 10


i think it's an old problem but i can't solve it. Thanks for your help!

Yes, the problem to find all $a$ is an open problem. I think the read-your-posts-before-sending-and-look-where-you-post conjecture on new users could help.

Sorry, this's my fisrt posts, so there're some mistakes. My posts is in previous topic, please check it to me. Thanks for ur advice!! :oops:


number theory
least common multiple
logarithms
calculus
integration
limit

Is it true or known if the sequence $ x_n=(\text{lcm}(1,2,...,n))^{\frac{1}{n}}$ converges?

I think, it converges to $e$.
Indeed, we can write
\[\mbox{lcm}(1,2,...,n)=\prod_{k=2}^{n}k^{(\pi(k)-\pi(k-1))\cdot\left[\ln n/\ln k\right]}.\]

So, using $[a]<a$ we can easily obtain upper bound for $\frac{1}{n}\ln lcm()$:
\[\frac{1}{n}\ln lcm()<\frac{\ln n}{n}\sum_{k=2}^n(\pi(k)-\pi(k-1))=\frac{\ln n}{n}\pi(n)\stackrel{n\to\infty}{\longrightarrow} 1.\]

Now for lower bound.  :)
Using $[a]>a-1$ we immediatelly obtain
\[\frac{1}{n}\ln lcm()\geq \frac{\ln n}{n}\pi(n) - \frac{1}{n}\sum_{i=1}^{\pi(n)}\ln(p_i).\]
Since $\pi(n)<(1+a)n/\ln n$ for large $n$ we conclude that $p_k\leq (1-\alpha)k\ln k$ ($\alpha>0$) for sufficiently large $k$.
Using this estimate we can easily derive
\[\frac{1}{n}\sum_{i=1}^{\pi(n)}\ln(p_i)\stackrel{n\to\infty}{\longrightarrow} 0.\]

That's all.

Unfortunatelly, we lost topic, there vess showed how one can easily (without any strong number-theory results!) prove that $\mbox{lcm}(1,2,...,n)>2^n$.
For whose, how know russian. Original article can be found in Kvant, 1995.

Myth, I am sorry, but I do not get something. I really do not get that the last limit is 0. Could you please explain a little bit more?

It is a simple integral estimate.

Why did you hide your online status?

I don" t knw why I am hidded. Do not worry, I am not afraid. Still, I insist and ask for an explanation since the integral estimate gives me a non-zero limit. Probably I am completely idiot. :(

We are to calculate $\frac{1}{n}\int_3^{n/\ln n}\ln t+\ln\ln t\,dt$, aren't we? It is something like $\frac{1}{n}t\ln t|_3^{n/\ln n}=O(\frac{1}{\ln n})$. Right?

Sorry, Myth, but I do not agree with the last estimation. I obtain 1 when I calculate what you said9 which is in fact what I did before)

I think I completed my solution, but I must go to school right now...

It is easier than I though...

We can write $lcm()=A_n\cdot B_n$, where $A_n=p_1^{a_1}p_2^{a_2}...p_k^{a_k}$, $p_i\leq \sqrt{n}$, $i\leq k$, and $B_n=p_{k+1}...p_{l}$, where $n\geq p_j>\sqrt{n}$, $j>k$.
It is clear $p_i^{[a_i/2]}\leq \sqrt{n}$ for $i\leq k$, so $\prod_{i\leq k} p_i^{[a_i/2]}\leq {\sqrt{n}}^{\sqrt{n}}$. Therefore $A_n\leq {\sqrt{n}}^{3\sqrt{n}}$, hence $A_n^{1/n}\to 1$.

Let's estimate $B$. I wrote earlier that $(1-a)s\ln s\leq p_s\leq (1+a)s\ln s$ for all large $s$, $a$ is an arbitrary small positive number.
Thus
\[(1-a)^{\pi(n)-\pi(\sqrt{n})}\prod_{\pi(\sqrt{n})< s\leq \pi(n)}s\ln s\leq B\leq (1+a)^{\pi(n)-\pi(\sqrt{n})}\prod_{\pi(\sqrt{n})< s\leq \pi(n)}s\ln s.\]
Consider logarithm and apply integral estimation.

Hope it is correct now.  :blush:

[quote="Myth"]Unfortunatelly, we lost topic, there vess showed how one can easily (without any strong number-theory results!) prove that $\mbox{lcm}(1,2,...,n)>2^n$.[/quote]

I have reposted this topic now: http://www.mathlinks.ro/Forum/viewtopic.php?p=136686 . Thanks to Prowler for providing the file!

  Darij

Hopefully I don't make a complete fool of myself like last time but I have done some work on this topic.

I was attempting to prove that: $lcm(1,2,\dots,n) < 3^n$ and I got the result that $\lim_{n \rightarrow \infty} {lcm(1,2,\dots,n)}^{\frac{1}{n}} = e$

I'll just attached the result that I sent to another website. They didn't accept it saying that the behavior of the lcm is very erradic. Maybe someone here can tell me if it is right or not.

Actually, it is the result, which I proved above.


function
real analysis
real analysis unsolved

I an interval, f:I->R derivable
J={(f(b)-f(a))/(b-a) ; a,b\in I a<b}
Prove that
a) J is an interval
b) J is included in f '(I), and f '(I)\J has at  most 2 elements
c) Using the previous result prove that f' has intermediate value property
RMO/Final round/1996/11th grade from Cristinel Mortici (web page)
-------------------------------------------------------------------
I don't have a complete solution

a)b)
We must prove that if a,b \in I with a<b then the interval with extremity
f'(a), f'(b) is included in f '(I)

Let D={(x,y) \in [a,b]^2, x>y} is convex in R^2, so connex in R^2
F D-->R, F(x,y)= f(x)-f(y))/(x-y) is continunous, F(D) is connex in R so it is an interval included in
f '(I)
For all x\in ]a,b[  we have (x,a) \in D and (f(x)-f(a))/(x-a) \in F(D)
take the limit f'(a) \in closure(F(D)), same f'(b) \in closure(F(D)) so
]f'(a),f'(b)[ is included in F(D) and included in f'(a), f'(b).

There was a similar problem:
Suppose f:[a,b]->R differentiable, then f has intermediate value property, that is, for any d which is between f'(a) and f'(b), there exists c \in (a,b) satisfying f'(c)=d.

Assume f'(b)<d<f'(a) first.
Study the function h(x)=f(x)-(f(a)+d(x-a))
we notice that h'(x)=f'(x)-d.
since f'(b)<d<f'(a), we have h'(b)<0<h'(a). Hence h(a) and h(b) are not the max value of h on [a,b]. Because h is continuous on [a,b], so there exists c \in [a,b] satisfying where h reaches its max value. Therefore, h'(c)=0, i.e., f'(c)=d.
A similar discussion works for f'(a)<d<f'(b).

The author of the problem "f'(J)" is Ion Rasa, not me! Cristinel Mortici


function
inequalities
inequalities unsolved

It is well known that for a function f(a1,a2,...,an), we can assume a1>=a2>=...>=an only when f is symmetric.
This question then comes up,
If we want to prove a^2+b^2+c^2+d^2>=ab+bc+cd+da
as a^2+b^2+c^2+d^2 is symmetric, we can assume a>=b>=c>=d, and then the remaining is trivial by rearrangement.
However, ab+bc+cd+da is not symmetric, does it mean that if we try to prove the ineq from right to left, we cannot assume a>=b>=c>=d and thus we cannot use rearrangement?

Please help to clarify my misconception.
Thank you.

[quote="Peter Scholze"]we can never assume $a\geq b\geq c\geq d$ since the inequality isn't symmetric as a whole.[/quote]
then how to prove corollary 1 and 2 using rearrangement in http://matholymp.com/TUTORIALS/Rear.pdf ?

[quote="blahblahblah"]Yes, it does not work backwards.  This is because $ab+bc+cd+da$ is not a maximal or a minimal permutation in any case.[/quote]
ab+bc+cd+da is a random sum and it should be <= direct sum all the time?

Well, this is rearrangement:

$ \sum a_ib_i \geq \sum a_ib_{\sigma(i)} \geq \sum a_ib_{n-i}$, where the sequences are of length $n$

So we can only make comparisons involving the biggest and smallest terms; e.g,. we can only start with these terms.

[quote="billzhao"]I no longer assume any order when I use rearrangement. Rather, the word [b]similarly sorted[/b] does the job nicely.

One could just say that $(a,b,c,d)$ and $(a,b,c,d)$ are similarly sorted, then the problem follows from rearrangement.[/quote]
that's nice, thanks!


let a, b, c is real number and $a^{2}+b^{2}+c^{2}=3$.
Find max value of P=a+b+c-abc
[hide="my solution"]
$a^{2}b^{2}c^{2}$ $\leqslant$ $(\frac{a^{2}+b^{2}+c^{2}}{3})^{2}=1$
$\Longrightarrow$  $-abc$ $\leqslant$ $1$ ,
$(a+b+c)^{2}$ $\leqslant$ $3(a^{2}+b^{2}+c^{2}) =9$
$\Longrightarrow$  $a+b+c$  $\leqslant$ $3$
$P_{max}$=4
True or false? Tell me your ideas.[/hide]

I actually don't think that any (a,b,c) that satisfy a+b+c=3 also satisfy abc=-1.
But I could be wrong

[hide="run away RIGHT NOW. This is not the kind of solution that typically appears in Intermediate..."]
Observe that $a=b=\frac{\sqrt{6}}{2}$ and $c=0$ produces $P=\boxed{\sqrt{6}}$. We now show that this maximum is in fact an upper bound

A nice way of writing $P$ is $a+b+c(1-ab)$. Unfortunately, the $1-ab$ is not "special" because $ab$ does not have a lower bound or upper bound of 1. But we can fix this: scale $a,b,c$ to $x,y,z$ so that $x=a\frac{\sqrt{6}}{3}$ etc., then $x^{2}+y^{2}+z^{2}=2$ and
\[P=a+b+c-abc=\frac{\sqrt{6}}{2}(x+y+z-(3/2)xyz)\leq \frac{\sqrt{6}}{2}(x+y+z-xyz) \]
so we will be done after proving the sharper bound
\[x+y+z(1-xy)\leq 2 \]
Now we can uber-pwn with Cauchy!
\begin{eqnarray*}((x+y)+z(1-xy))^{2}&\leq &((x+y)^{2}+z^{2})(1+(1-xy)^{2})\\ &=&2(1+xy)(1+(1-xy)^{2})\\ \end{eqnarray*}
Letting $t=xy$, it only remains to prove that
\[(1+t)(1+(1-t)^{2})\leq 2 \]
But
\[2-(1+t)(1+(1-t)^{2})=t^{2}(1-t) \]
$t^{2}$ is nonnegative, as well as $1-t$ because
\[t=xy\leq \frac{x^{2}+y^{2}}{2}\leq \frac{x^{2}+y^{2}+z^{2}}{2}=1 \]
Whew.
[/hide]

Wow... scorpius. Is there no other way to find the upper bound?

Well, there's always dumbassing. The inequality is equivalent to
\[((a+b+c)(a^{2}+b^{2}+c^{2})-3abc)^{2}\leq 2(a^{2}+b^{2}+c^{2})^{3}\]
After expanding out, throwing $a^{4}(a-b)(a-c)+b^{4}(b-a)(b-c)+c^{4}(c-a)(c-b)\geq 0$ or $a^{2}(a-b)^{2}(a-c)^{2}+b^{2}(b-a)^{2}(b-c)^{2}+c^{2}(c-a)^{2}(c-b)^{2}\geq 0$ (or maybe even some combination of the two) at it would probably reduce it to a bunch of AM-GM's and $a^{2}b^{2}c^{2}\geq 0$. I haven't checked if this actually works, though.

[quote="vmsvl"][quote="white_horse_king88"][quote="vmsvl"]let a, b, c is real number and $a^{2}+b^{2}+c^{2}=3$.
Find max value of P=a+b+c-abc

$a^{2}b^{2}c^{2}$ $\leqslant$ $(\frac{a^{2}+b^{2}+c^{2}}{3})=1$
$\Longrightarrow$  $-abc$ $\leqslant$ $1$.
True or false? Tell me your ideas.[/quote]

That is... $abc \leq 1$. Is it not? Thus, we can't conclude your max value.[/quote]
Why is not? $-1 \leq abc \geq 1$ . So , we have $-abc \leq$ 1[/quote]

You actually get $-1 \leq abc \leq 1$ , so that doesn't work. See scorpius's gigantic solution for why the bound is $2$.

[hide][quote="white_horse_king88"][/quote][quote="vmsvl"][quote="white_horse_king88"][quote="vmsvl"]let a, b, c is real number and $a^{2}+b^{2}+c^{2}=3$.
Find max value of P=a+b+c-abc

$a^{2}b^{2}c^{2}$ $\leqslant$ $(\frac{a^{2}+b^{2}+c^{2}}{3})=1$
$\Longrightarrow$  $-abc$ $\leqslant$ $1$.
True or false? Tell me your ideas.[/quote]

That is... $abc \leq 1$. Is it not? Thus, we can't conclude your max value.[/quote]
Why is not? $-1 \leq abc \geq 1$ . So , we have $-abc \leq$ 1[/quote][quote="white_horse_king88"]

You actually get $-1 \leq abc \leq 1$ , so that doesn't work. See scorpius's gigantic solution for why the bound is $2$.[/quote][/hide]
Ow, sorry
$-1 \leq abc \leq 1$.
Thank u.

if we are like $p(t)=t^{3}-xt^{2}+yt-z$, then $x^{2}-2y=3$, we need real roots, and we want the maximum of $x-z=m$

$p(t)=t^{3}-xt^{2}+\frac{x^{2}-3}{2}t+m-x$

now as we increase $m$, the graph moves up, since there are 3 real roots, we have a restriction, the maximum occurs when the graph is tangent to the x-axis, i.e there is a double root, so...

$2a^{2}+b^{2}=3$, and we want max $2a+b-a^{2}b=\sqrt{6-2b^{2}}+b-\frac{1}{2}b(3-b^{2})$ which we can be like omg ewwww...calculus...the max occurs when

$b\approx-.3626$, and $a=c\approx 1.197$, the expression is about $2.55\ge \sqrt{6}\approx 2.449$

[quote="10000th User"]Al, are you sure? If I plug in the values in my calculator and noting the sign on the values and carefully checking everything in detail, the expression comes out to around 2.29 or around there, which is far less than scorpius' max.[/quote]

uhh...i might be being stupid, but

(1.197)+(-.3626)+(1.197)-(-.3626)(1.197)(1.197)=2.55...

1.197^2+1.197^2+.3626^2=2.997...[rounding error...]

Uhh my bad  :blush:
Is there a more elementary solution to it though?

This [b]really[/b] wouldn't be intermediate, but this problem sets itself up to lagrange multipliers (...) :|

[quote="white_horse_king88"][/quote][quote="vmsvl"][quote="white_horse_king88"][quote="vmsvl"]let a, b, c is real number and $a^{2}+b^{2}+c^{2}=3$.
Find max value of P=a+b+c-abc

$a^{2}b^{2}c^{2}$ $\leqslant$ $(\frac{a^{2}+b^{2}+c^{2}}{3})=1$
$\Longrightarrow$  $-abc$ $\leqslant$ $1$.
True or false? Tell me your ideas.[/quote]

That is... $abc \leq 1$. Is it not? Thus, we can't conclude your max value.[/quote]
Why is not? $-1 \leq abc \geq 1$ . So , we have $-abc \leq$ 1[/quote][quote="white_horse_king88"]

You actually get $-1 \leq abc \leq 1$ , so that doesn't work. See scorpius's gigantic solution for why the bound is $2$.[/quote]
$-1 \leq abc$ most definitely implies $-abc \leq 1$

The solution is correct in the sense that it shows that the expression is $\le 4$, but you haven't shown that this value is actually attained, so you haven't shown that 4 is a maximum.

[quote]$(a+b+c)^{2}$ $\leqslant$ $3(a^{2}+b^{2}+c^{2}) =9$
$\Longrightarrow$  $a+b+c$  $\leqslant$ $3$
True or false? Tell me your ideas.[/hide][/quote]
[b]|a+b+c|[/b] $\leqslant$ 3 not [b]a+b+c[/b] $\leqslant$ 3 ([b]a, b, c[/b] is real number , lol)

[quote="Altheman"]oh just to note, the solutions to that derivative are $b=1$, $-\frac{\sqrt{12-3\sqrt{13}}}{3}$, $\frac{\sqrt{12+3\sqrt{13}}}{3}$, the first giving the maximum, $a=c=\sqrt{6\sqrt{13}+30}{6}$
for a total maximum of some ugly number...[/quote]
A small typo in that formula, a and c should be

$a=c=\frac{\sqrt{6\sqrt{13}+30}}6$

ohhh... they can be [b]negative[/b]... that messes up the $abc\geq 0$ I used initially.

Actually, the Lagrange Multiplier solution (for the case of all real numbers) doesn't appear to be [b]that[/b] scary on this problem.

[quote="vmsvl"]let a, b, c is real number and $a^{2}+b^{2}+c^{2}=3$.
Find max value of P=a+b+c-abc
[/quote]

Lagrange multipliers:

$1-bc = 2ax$
$1-ac = 2bx$
$1-ab = 2cx$
$a^{2}+b^{2}+c^{2}= 3$

$bc(1-bc) = ac(1-ac) = ab(1-ab)$

Blah.


geometry
Gauss
power of a point
radical axis
geometry unsolved

[quote="Inequalities Master"]Consider the triangle $ ABC$ and consider two variable points $ M$ and $ N$, $ M\in [AB]$ and $ N\in [AC]$. Consider the circles  $ S_1$ of diameter $ [BM]$ and the circle $ S_2$ of diameter $ [CN]$.Let $ S_1$ and $ S_2$ intersect at $ P$ and $ Q$.Prove that the line $ PQ$ passes through a fixed point that does not depend on choosing the points $ M$ and $ N$.[/quote]
I think it is not true.

We can only to say, that for every point $ N,$ as fixed one on $ AC$ and the point $ M,$ as variable one on $ AB,$ the radical axis of the circles $ (S_{1}),\ (S_{2}),$ intersects the line through $ B$ and perpendicular to $ AB,$ at the same point ( easy to prove ).

Similarly, for every point $ M,$ as fixed one on $ AB$ and the point $ N,$ as variable one on $ AC,$ the radical axis of the circles $ (S_{1}),\ (S_{2}),$ intersects the line through $ C$ and perpendicular to $ AC,$ at the same point.

Kostas Vittas.

I'm sorry, vittasko, I made a mistake. The problem is fixed now.

[hide="Solution"][quote="Inequalities Master"]Consider the triangle $ ABC$ and consider two variable points $ M$ and $ N$, $ M\in [AB]$ and $ N\in [AC]$. Consider the circles  $ S_1$ of diameter $ [BN]$ and the circle $ S_2$ of diameter $ [CM]$.Let $ S_1$ and $ S_2$ intersect at $ P$ and $ Q$.Prove that the line $ PQ$ passes through a fixed point that does not depend on choosing the points $ M$ and $ N$.[/quote]
Denote BD,CE be the altitudes of triangles ABC and they meet at the orthocenter H of ABC.
$ (S_1)$ always passes through D, but $ (S_2)$ always passes through E.
Since $ HB.HD\equal{}HC.HE$, the radical axis of $ (S_1)$ and $ (S_2)$ pass through H which is the fixed point we find[/hide].

[quote="jayme"]Dear Mathlinkers,
does any one known the origine of this problem?
Sincerely
Jean-Louis[/quote]
You referd to the origination of this problem, why are you interesting at it? It is a classic example of [b]the power[/b] (point wrt to the circle). I think this problem has no specialty.

[quote="jayme"]Dear mr. danh,
simply, I know that Bodenmiller worked on this figure. Do you know his complete name and the date of birth and death?
Sincerely
jean-Louis[/quote]
As I know, no document show his date of birth and death ( Bodenmiller live in 19th century) (maybe you want to refer to the mathematician who re-discovered the theorem about the midpoints of diagonals of a quadrilateral now also ascribed to Gauss?)


I'm kind of mixed up on this question.

Given that $x$ is directly proportional to $y$ and to $z$ and is inversely proportional to w, and that $x=4$ when $(w,y,z) = (6,8,5),$ what is x when $(w,y,z)=(4,10,9)$?

I don't understand how to set up these equations.

I thought this was easy, but I guess it's not in this question. Is there a good way to help understand to differentiate between inverse and direct?

since x is directly proportional to y and to z, then i guess x/(yz) is a constant. if x is also inversely proportional to w, then (xw)/(yz) is a constant. knowing this, you can plug in the original values of x, y, z, and w to get a constant. the rest is for you to figure out.

Here is a step-by-step way:

If x is inversely proportional to w, that means when x is multiplied by a constant k, w must be divided by k (or vice versa).

If x is directly proportional to y, that means when x is multiplied by a constant k, y is also multiplied by k (or vice versa).

So, we are given that when x=4, y=8 (ignore the other variables for now). Thus when x=5, y=10 (multiply both by 5/4).

x=5 when (w,y,z)=(6,10,5).

Now, x=9 when z=9 (multiply both by 9/5) so x=9 when (w,y,z)=(6,10,9).

Finally, x=27/2 when w=4 (multiply w by 2/3 so you have to multiply x by 3/2).

Once you understand this, you will see (xw)/(yz) is a constant and not have to go through all that work in the future.

[hide="Hint"]
You can combine all of these proportionalities:
$x = k \cdot \frac{yz}{w}$
Where k is the shared constant of proportionality.
[/hide]
[hide="Solution"]
Thus, plug in the numbers.
$4 = k \cdot \frac{8 \cdot 5}{6}$
Solve for K.
$k = \frac{24}{40}= \frac{3}{5}$

Now set up the target equation.
$x = \frac{3}{5}\cdot \frac{10 \cdot 9}{4}$
$x = \boxed{\frac{27}{2}}$
[/hide]
EDITED

from the instructions $x \propto \frac{yz}{w}$ so $x = \frac{yz}{w}* k$ where k is a constant

$4 = \frac{8 \cdot 5}{6}\cdot k$
$\frac{4}{\frac{8 \cdot 5}{6}}= \frac{4 \cdot 6}{8 \cdot 5}= k$
$\frac{24}{40}= k = \frac{3}{5}$
so we have our constant and therefor
$x = \frac{3yz}{5w}$

when we have $(4, 10, 9)$:
$x = \frac{3 \cdot 10 \cdot 9}{5 \cdot 4}= \boxed{\frac{27}{2}}$


Two sides of a triangle measure 4 cm and 9 cm. What is the greatest integer number of centimeters that could be the third side of this triangle?

By the triangle inequality, x<13.  Thus, x=12.

Note that the triangle inequality says that the sum of any two sides of a triangle can't be less than or equal to the third side, otherwise it's a degenerate triangle.

EDIT: Oh, blah, fixed. Stupid moment of the day, lol.

Sorry if you think this is spam, but what's a degenerate triangle?  Because it's IMPOSSIBLE to form it when $ a\plus{}b\ge c$...

[quote="isabella2296"]Note that the triangle inequality says that the sum of any two sides of a triangle can't be less than or equal to the third side, otherwise it's a non degenerate triangle.[/quote]

No it becomes a degenerate triangle. We want a nongenerate triangle. And AIME15 a degenerate triangle is one that follows that inequality you just wrote.

A degenerate triangle is the "triangle" formed by three collinear points.  It has no area.


limit
induction
algebra proposed
algebra

Find $ \lim_{n\rightarrow \infty}\sqrt{4+2\sqrt{4+...+\sqrt{4+2\sqrt{4+2\sqrt{4+4}}}}}$, where sign $ \sqrt$ occurs $ n$ times.

[quote="tdl"]Find $ \lim_{n\rightarrow \infty}\sqrt {4 + 2\sqrt {4 + ... + \sqrt {4 + 2\sqrt {4 + 2\sqrt {4 + 4}}}}}$, where sign $ \sqrt$ occurs $ n$ times.[/quote]
Suppose $ x=\lim_{n\rightarrow \infty}\sqrt {4 + 2\sqrt {4 + ... + \sqrt {4 + 2\sqrt {4 + 2\sqrt {4 + 4}}}}}$.
Then $ \frac{x^2-4}{2}=x$
solving gives $ x=\sqrt{5}+1$

[quote="Akashnil"][quote="tdl"]Find $ \lim_{n\rightarrow \infty}\sqrt {4 + 2\sqrt {4 + ... + \sqrt {4 + 2\sqrt {4 + 2\sqrt {4 + 4}}}}}$, where sign $ \sqrt$ occurs $ n$ times.[/quote]
Suppose $ x = \lim_{n\rightarrow \infty}\sqrt {4 + 2\sqrt {4 + ... + \sqrt {4 + 2\sqrt {4 + 2\sqrt {4 + 4}}}}}$.
Then $ \frac {x^2 - 4}{2} = x$
solving gives $ x = \sqrt {5} + 1$[/quote]

I think you need to be a little bit more rigorous.

$ 1)$ first define exactly the sequence :
$ x_{n+1}=\sqrt{4+2x_n}$, with $ x_1=4$

$ 2)$ show the sequence is a convergent one :
Induction : $ 1+\sqrt{5}<x_n$ implies $ 1+\sqrt{5}<x_{n+1}<x_n$
So $ L=\lim_{n\to +\infty}x_n$ exists, is finite and we have $ 1+\sqrt{5}\leq L<4$

$ 3)$ Establish the limit : $ x_{n+1}=f(x_n)$ with $ f(x)=\sqrt{4+2x}$
Since $ \lim_{n\to +\infty}x_n=L$ and $ f(x)$ is continuous, then $ L=f(L)$ and hence the result.


Obviously I can't post exactly what I want to write, but I have a geometric idea that I can "show" is true (by doing something that makes it intuitively obvious), but can't really prove it. So I guess my question is really, is there difference between showing something and proving something? And if so, how many points would showing it get as opposed to proving it?

Don't worry about it. If you can only "show" why it is true but can't really prove it, think about how to prove it. If you end up not being able to do so, put what you have. Knowing how many points you're going to get doesn't change how many points they're going to give you at the end.

What do you exactly mean by "show"? Often, on the USAMO, etc., "show that" and "prove that" are used interchangeably. It basically comes down to proving everything that's non-trivial.

i would think that if you show it and it is pretty obvious you would get 5.

Depends what you mean by "show".  In my experience, USAMTS graders aren't sticklers for perfect rigor, but solutions that aren't proven will get a 1.  For example, on one of last year's problems, I calculated approximate answers which appeared to be the same (which proved what I wanted).  The solution (rightfully) recieved a 1.  But you don't need to prove everything from axioms either.  Also, since you have 2 months left, work to make it better.  Do the best you can: it doesn't hurt to submit something imperfect but you can also work on it more if you are not sure (or even if you are).

what not to do: 1) draw a picture with a compass and a straight edge, and measure the answer from the picture. (perhaps a good way to get started, but not a proof!)
2) skip explanation/assume your readers know what you are thinking - such as, if you say "using whatever identities" at least state the identities you are using!
3) Be careful about geometric assumptions!  I once visualized a cross-section incorrectly, and claimed a segment was a perpendicular when it wasn't.  Another time, I assumed that some 3d figures were similar when they weren't (5/1/18)
4) Citing obscure facts without proof (I lost a point for citing that $ \sum_{n \equal{} 1}^{\infty} \frac {\phi(n)}{n^4} \equal{} \frac{\sum_{n \equal{} 1}^\infty \frac {1}{n^3}}{\sum_{n \equal{} 1}^\infty \frac {1}{n^4}}$)

What you can get away with:
assuming convergence
programs to count! I've done this too much - actual counting is much easier.
not-so-great wording - but make sure you write what you mean!!!
skipping some arithmetic/simple algebra.

One possible thing that you can do is look at solutions that others have written for the previous years to get an idea of the quality of the proofs that do well.

and how would one get access to such solutions?

They're on the USAMTS site:

http://www.usamts.org/Problems/U_ProblemsPast.php

[quote="facis"]4) Citing obscure facts without proof (I lost a point for citing that $ \sum_{n \equal{} 1}^{\infty} \frac {\phi(n)}{n^4} \equal{} \frac {\sum_{n \equal{} 1}^\infty \frac {1}{n^3}}{\sum_{n \equal{} 1}^\infty \frac {1}{n^4}}$)[/quote]What! I lost 2 points for that! :P

Seriously, though, the best suggestions are to look at solutions from past years.


How do you typeset a multiplication? Is there a package similar to longdiv except for multiplication?

No, I did not mean that.  I meant things like

**************35
*************xi29
__________________
*************315
*************700
__________________
************1015

where the "i" at the start is for spacing and the asterisks (*) are for spacing also.  (I meant things like that without the asterisks and "i"s)

Try [url=http://texcatalogue.sarovar.org/entries/xlop.html]xlop[/url]. See page 19 of the [url=http://theory.uwinnipeg.ca/scripts/CTAN/macros/generic/xlop/fr-user.pdf]documentation[/url] (in French) for examples.

can anybody please translate the aforementioned documentary?


Determine the set of all pairs (a,b) of positive integers for which the set of positive integers can be decomposed into 2 sets A and B so that $a\cdot A=b\cdot B$.

Can anyone explain me this? I don't get neither the question nor the solution :? just a practical example would really do but it seems total non-sense to me.

You did it again :? Why do you make two posts when one post is sufficient?

You need two partition the set of positive integers in two sets such that if you multiply all numbers in the first set by $a$, then you get exactly the same set (of "new numbers") as when you multiply all numbers in the second set by $b$. (Obviously you need to find all $a$ and $b$ for which this is possible...)

That's what I *THINK* it means. Am I wrong?

Yes, but that is the non-sense... I mean can you show me a practical example of such partition? It seems trivially impossible to me, but that's not the answer.

It's pretty natural to me.

First we can assume that $a$ and $b$ are coprime.

Now since 1 belongs to either $A$ or $B$, one of the numbers $a$ and $b$ divides the other one. So WLOG $a = 1$.

Now $(1, n)$ is a solution for all positive integers $n$ - for each integer $k$ define $f(k)$ as the highest power of $n$ dividing $k$. Then put $k$ in $A$ if $f(k)$ is odd and put $k$ in $B$ if $f(k)$ is even. This way we get a partition which satisfies all requirements.

Finally all solutions are given by $(d, nd)$ and $(nd, d)$ where $d$ and $n$ are positive integers.


The question is,,,

How to get eigenvector?>>>

Please answer specifilcally.,,,,,^&^

If you want a specific answer, ask a specific question. The generic and badly stated question here is impossible to work with.

I mean I want to know how to get eigenvector,,,,

If the problem gives 2*2 matrix....

So, I should get eigenvector and eigen value,,,

Have you read your textbook? If yes, what exactly wasn't clear? We can go over it with you but let's figure out where you are stuck first.

You start with eigenvalues. Suppose that your matrix is $ A\equal{}\begin{bmatrix}1&2\\4&3\end{bmatrix}$.

Question 1: what is an eigenvalue?
Question 2: what is the general equation for eigenvalues?
Question 3: what is that equation for this particular matrix?
Question 4: what are the eigenvalues of this matrix?

(if you can answer #4 immediately, just post your answer and we'll know that this part is not a problem for you. If not, start with #1 and answer as many questions as you can. Then we'll help you with the next one :) ).

[quote="fedja"]Have you read your textbook? If yes, what exactly wasn't clear? We can go over it with you but let's figure out where you are stuck first.

You start with eigenvalues. Suppose that your matrix is $ A \equal{} \begin{bmatrix}1 & 2 \\
4 & 3\end{bmatrix}$.

Question 1: what is an eigenvalue?
Question 2: what is the general equation for eigenvalues?
Question 3: what is that equation for this particular matrix?
Question 4: what are the eigenvalues of this matrix?

(if you can answer #4 immediately, just post your answer and we'll know that this part is not a problem for you. If not, start with #1 and answer as many questions as you can. Then we'll help you with the next one :) ).[/quote]

can you answer those four questions?

Yes, everyone who has commented on this thread (except perhaps for you) can, but none of us is particularly excited to do your homework for you.   :roll:

Do you think i'm just kiding?

I just want some help to solve my problems...,,

And that's exactly what you are getting. The point is to make sure that by the end of this thread you are able to do everything by yourself.

So, here is your first task: go to [url=http://mathworld.wolfram.com/Eigenvalue.html]this site[/url] and read the article. If something is not clear, ask [b]specific questions[/b]. If everything is clear, try to answer the questions I posted (all of them) [b]without[/b] looking at the article.

Thank you for you all members....

Now I understand about eigenvector,,,

Thank you~~ :roll:

Do you?  How would you compute the eigenvectors of fedja's example $ A\equal{}\begin{bmatrix}
1&2\\
4&3
\end{bmatrix}$?
If you try to compute this example, I will tell you if you are right or wrong and show you how to do it.


The Red Sox and Braves each need to win two of the remaining three games to take the World Series.  The Red Sox have probabilities $\frac{2}{5}$, $\frac{2}{5}$, and $\frac{1}{2}$ to win the fifth, sixth, and seventh games, respectively.  What is the probability that the Red Sox win the Series?

[hide]The answer I got was $\frac 25$. There are only 3 ways they can win the series:
They win twice in a row, and it won't matter if they lose or win the next game.
They lose the first game, and win the next two games.
They lose the 2nd game, and win the other two games.
The probably that any of these events occur is $\frac 25 * \frac 25 + \frac 35 * \frac 25 * \frac 12 + \frac 25 * \frac 35 * \frac 12 = \frac 25$. [/hide]

When you summed them, you had $\displaystyle \frac25 \cdot \frac35 \cdot \frac12$

Why would you need the $\displaystyle \frac25$? It doesn't matter what they do in those games, as long as they win the other two, they still get the series.

So I think it is $\displaystyle \frac{14}{25}$.

I might be wrong, I'm bad with probability.

If you think about it logically, I think it'd be less than $\frac 12$ because the probability they win any one game is less than $\frac 12$.

You guys are all wrong...my answer is correct

[quote="yif man12"]That's not a very likely situation. Everyone knows the Braves always lose in the first round.  :D[/quote]
Crappy first round.......we just get unlucky

I might be wrong but I think you all are counting the event where they win all three games more than once.

[hide]I agree with tarquin.

The probability is 2/5. They can win the series by having WWW, WWL, WLW, or LWW.

The probability of WWW is 2/5 x 2/5 x 1/2 = 4/50.

The probability of WWL is 2/5 x 2/5 x 1/2  = 4/50.

The probability of WLW is 2/5 x 3/5 x 1/2  = 6/50.

The probability of LWW is 3/5 x 2/5 x 1/2  = 6/50.

To find the probability that one of these will be the case, you add up the likelyhood of each event= 4/50 + 4/50 + 6/50 + 6/50= 20/50 = 2/5. [/hide]

Elements of math kid.  There is one problem with your line of thought.  The two scenarios WWW and WWL are virtually the same, because after they win two games, there is no need for a third game.

Ah, never mind.  I see that your answer works.

Yes, elementsofmathkid, after two wins, no third game is played.  So instead of calculating WWW and WWL, you calculate WW and add it to WLW and LWW.  However, your answer still works.  I don't know if that is coincidence or if they only work because the probability of the first two games are the same or something...

How do you hide it?

I got an anzwer but I can't hide it.

Can someone help?

Forums like this have their own code, similar to html.  To hide something, use the hide tags.  The look like this, however you must remove the *s to use it.  The *s are there so it doesn't hide it.

[*hide]
Here is where your answer goes.
[*/hide]

Make sure you put the /hide at the end, or else nothing will be hidden.

[hide]
 I thinlk it is 2/5. Is this working?
[/hide][/hide]

[hide="Click to reveal hidden content...if you dare!"]There are three ways to win, winning the next 2 games, losing-winning-winning, and winning-losing-winning. For winning the next 2 games, 2/5 * 2/5 = 4/25. For losing-winning-winning, 3/5 * 2/5 * 1/2 = 3/25. Winning-losing-winning also equals 3/25. 4/25 + 3/25 + 3/25 = 10/25 = [b]2/5[/b][/hide]


Prove that for all $a>0$ and $b>0$ holds:
$\left(a^{b}b^{a}\right)^{\frac{2}{a+b}}\leq\frac{a^{2}+b^{2}}{2}\leq\left(a^{a}b^{b}\right)^{\frac{2}{a+b}}.$

Hm...at least the left inequality is very easy:
\[\frac{a^{2}+b^{2}}{2}\geq \left(\frac{2ab}{a+b}\right)^{2}= \left(\frac{ab+ba}{a+b}\right)^{2}\geq \left(\sqrt[a+b]{a^{b}b^{a}}\right)^{2}=\left(a^{b}b^{a}\right)^{\frac{2}{a+b}}\]
by weighted AMGM.

However, the second inequality seems to be quite hard, I'll try to solve it tomorrow.

Let $f_{a}(b)=(a^{a}b^{b})^{2}-\left(\frac{a^{2}+b^{2}}{2}\right)^{a+b}$ where a is constant. for the minimum value of $f_{a}(b)$ to be reached we need both $f'_{a}(b)=f'_{b}(a)=0$ and finally $a=b$ yealding the inequality.


Fie $f:[0,1]\to[0,1]$ o functie continua in 0 si 1 ,care are limite laterale in orice punct si verifica :$f(x-0) \leq f(x) \leq f(x+0)$, pentru orice $x\in (0,1).$Aratati ca:
a)pentru $A=${$x\in[0,1]|f(x) \geq x$}, avem $supA \in A.$
b)exista $x_{0}\in[0,1]$ astfel incat $f(x_{0})=x_{0}.$

a) First of all, $A$ is a nonempty set, since $0\in A$. There is a sequence $\{x_n\}$ s.t. $x_n\in A$, $x_n\to \sup A=a$. So $x_n\leq f(x_n)$, but $\lim f(x_n)=f(a-0)\leq f(a)$. It follows that $a\leq f(a)$, i.e. $a\in A$.
b) Suppose $f(a)>a$. Note that $a<1$. Since $f(a)\leq f(a+0)$ there is $b$ s.t. $b>a$, $|b-a|<\frac{f(a)-a}{2}$ and $f(b)>\frac{f(a)+a}{2}$, then $f(b)>b$, i.e. $b\in A$. It implies $a\neq \sup A$. Contradiction.


[quote]Find the minimum possible value of $ x^2 \plus{} y^2$ given that x and y are
real numbers satisfying
$ xy(x^2 \minus{} y^2 ) \equal{} x^2 \plus{} y^2$ and $ x \neq 0$.
[/quote]

What's the main idea in this situation?   :roll:

it was discussed before.
main idea is to obtain  $ \,\left( {{{x^2  \minus{} y^2 } \over {x^2  \plus{} y^2 }}} \right)^2  \plus{} \left( {{{2xy} \over {\,x^2  \plus{} y^2 }}} \right)^2  \equal{} 1$, then using trig substitution  $ \,\,\,{{x^2  \minus{} y^2 } \over {x^2  \plus{} y^2 }} \equal{} \cos \theta \,\,,\,{{2xy} \over {\,x^2  \plus{} y^2 }} \equal{} \sin \theta$  obtain  $ \,{2 \over {\,x^2  \plus{} y^2 }} \equal{} \cos \theta  \cdot \sin \theta \,\,\,\, \Leftrightarrow \,\,\,x^2  \plus{} y^2  \equal{} {4 \over {\sin 2\theta }} \ge 4$.

By the way this is not my solution.
thanks to solver(I don`t remember name).

Hi theSA,
Thank you very much!  :D

Could you give me the link to the previous discussion please? (I tried to notice if it had been posted yet, but didn't find anything  :roll: )

Thank you once more!

I can`t remember also where is it.
but this solution was best.

Take a look at [url=http://www.mathlinks.ro/Forum/viewtopic.php?p=420940#420940]this[/url]. :)


function
linear algebra
linear algebra unsolved

Let $X\in\mathcal M_{2}(R)$ a matrix. Find $X$ such that for any $n\geq3$ we have $X^{n}+X^{n-2}=\left(\begin{array}{cc} 1&-1\\ -1&1\end{array}\right)$. :)

[quote="MoHaMeD"]$X$ is diagonalisable and its eigenvalues are $0$ and $1$ the remaind is a computation ;).[/quote]

That sounds nice, but I'm not aware of it.  :roll: [hide="My aproach"] I've got something like $X=\left( \begin{array}{cc}a&b \\ b&a\end{array}\right)$ and with $t=a+b$ a function $f(t)=t^{n}+t^{n-2}$. Now we can use derivates or I don't know what to solve this problem.. but I didin't learned them yet... :| [/hide]

i'll give u some hints then
suppose that $\lambda$ is an eigenvalue of $X$ you'll find two cases $\lambda=0$ or $\lambda=1$
u'll consider these cases:
1-$X$ is diagonalisable with 1 (or 0) as unique eigenvalue:clearly wrong
2-$X$ isn't diagonalisable but it has only one eigenvalue which is 1 so $((X-1)^2=0_n$ (or 0 so $X^2=0_n$):wrong case too (computation+trace)
3-$X$ is diagonalisable with 2 eigenvalues 0 and 1.

about 90% of these ecuations are solved quite easily using eigenvalues...it's a general method

Anyway .. could someone finish the solution with the method I've started?  :roll:

$X$ has the $0$ and $1$ as eigenvalues so from Hamilton-Cayley we get $X^2=X$ which implies $X^k=X,\forall k\in N$ so $2X=\left(\begin{array}{cc} 1&-1\\ -1&1\end{array}\right)$ from where $X=\left(\begin{array}{cc} \frac{1}{2}&-\frac{1}{2}\\ -\frac{1}{2}&\frac{1}{2}\end{array}\right)$


Here are the universities I consider and their specialities:

- UQAM (specialities: combinatorics, statistics)
- U of M (speciality: analysis)
- McGill (speciality: number theory)

I may have missed a few specialities of McGill and U of M, but I may have to factor in the quality of the female students in either of these three schools before my final decision is made. Hence the question: What university has the best women?

It's good to see someone who has their priorities straight.  :lol:

By quality I mean the mathematical talent these women have. Up to a pre-uni level (college level in the Quebecer context) I always suspected women to be better at math than men but in university, and even more so in grad school, men take the upper hand.

It's too late now... my brother's choice is made; he will attend University of Montreal.

After further search I realized that U of M's speciality is geometry (at a graduate level, topology)


[quote="Xaenir"]Find the number of solutions of
\[ \left| {\sin |x|} \right| \equal{} x \plus{} |x|
\]
when
\[ \minus{} 2\pi \leqslant x \leqslant 2\pi
\]
[/quote]

For $ \minus{}2 \pi \le x \le 0$, the equation

$ \Leftrightarrow sin(\minus{}x)\equal{}0 \Leftrightarrow sin x\equal{}0 \Leftrightarrow x\equal{}0,\minus{}\pi, \minus{}2\pi$

For $ 0<x \le \pi$, the equation $ \Leftrightarrow f(x)\equal{}2x\minus{}sin x\equal{}0$, equality holds when $ x\equal{}0$ and $ f'(x)>0$ for $ 0<x \le \pi$, so no solution in this case.

For $ \pi <x \le 2\pi$, the equation $ \Leftrightarrow g(x)\equal{}2x\plus{}sin x\equal{}0$, $ g(0)\equal{}0$ and $ g'(x)>0$ for $ \pi <x \le 2\pi$
indicates that there are no solutions again in this case.

So number of solutions $ \equal{}3$

This problem can also be solved graphically.
$ 0, \minus{}\pi, \minus{}2\pi$ are the only  solutions.


Are there any sites with extensive information on how to play pieces by certain composers or from certain styles? For example, does one play Ravel mechanically?

I'm not aware of any websites that suggest playing styles and techniques for different composers, but I wouldn't think there'd be a very comprehensive list.  If you're not getting what you want from a private instructor, then you might check your music library for resources.

To answer your example, much of ravel's pieces involve a high level of virtuosity (all three sections of Gaspard de la Nuit, for example), background texture, so a kind of nebulous sound and fluid motion is typically adopted by many professionals for playing his works.  There are exceptions, however, such as the fugue from Le Tombeau de Couperin which require a bit more of a mechanical approach to play convincingly.  Are you playing something by Ravel?

[quote="236factorial"]...does one play Ravel mechanically?[/quote]
Never. Ravel was an Impressionist, remember?  :wink:

[quote]Never. Ravel was an Impressionist, remember? Wink[/quote]

Is this sarcasm?
Because I would cite Minstrels by Debussy, Parts of the Ravel Piano Concerto, Rigaudon, Fugue, and Toccata from Le Tombeau de Couperin to the contrary...

Yeah, maybe I'll use the  :roll:  next time; the wink doesn't always cut it.  :wink:

Okay my example was awful. I’ve played Ravel’s Jeux d’eau before, so I get the idea of the impressionism.

I do have a recording in which the performer plays the song like a MIDI file.

IMO almost all music is to be played with the most expression possible
only mechanically if it is quite a technical piece
of course, it is hard to tell what emotions music expresses .


Find all digits $ (x;y)$, such that for all $ n$, number $ \underbrace{xx\cdots x}_{n}6\underbrace{yy \cdots y}_{n}4$ is a perfect square.

There are 2 solutions:
$ x\equal{}4,y\equal{}2, A\equal{}(\frac{210^{n\plus{}1}\plus{}4}{3})^2$ and\
$ x\equal{}9,y\equal{}0, A\equal{}(10^{n\plus{}1}\minus{}2)^2.$

Thanks.. But can I see a full solution....


superior algebra
superior algebra solved

[quote="coquitao"]1. Give an explicit example of an infinite field of prime characteristic.

2. Can you find a group $ \mathcal{G}$ with an infinite number of elements of finite order?

3. Prove or disprove this: There exist 2 groups, $ \mathcal{H}$ and $ \mathcal{G}$, endowed with an epimorphism $ f: \mathcal{H} \to \mathcal{G}$ that satisfies the following: $ f^{\ast}: \tau(\mathcal{H}) \to \tau(\mathcal{G})$ isn't surjective. Here, $ \tau(\mathcal{H})$ denotes the set of all alements of $ \mathcal{H}$ of finite order, and $ f^{\ast}$ is the restriction of $ f$ to $ \tau(\mathcal{H})$.[/quote]

1. The field $ F(X)$ of rational fractions in a variable $ X$ over a finite field.
2. The direct product of the $ \mathbb{Z}/n\mathbb{Z}$, for $ n \geq 1$.
3. The canonical homomorphism $ \mathbb{Z} \to \mathbb{Z}/n\mathbb{Z}$, for $ n \geq 2$ is an example.

Friendly,
P.

[color=red][b]Well done, panurge, well done... Be on the lookout for more challenges in the near future.[/b][/color]

[quote="panurge"]
2. The direct product of the $ \mathbb{Z}/n\mathbb{Z}$, for $ n \geq 1$.

Friendly,
P.[/quote]

Sorry, I wrote "direct product" instead of "direct sum" and it is too late to edit my post.
P.

Forgetting the multiplicative structure, every example for 1. is also an example for 2. owk


f(x)=x^2-6x+4
Why sometimes we need to restrict the domain of the original function?

its simple, the reason why sometimes it is necesary to restric the domain of the original funcion $f(x)$ beacuse we assume the inverse function $f^{-1}(x)$ to be a function! and in order for that to happen, the original function MUST never have the same output for any two imputs.

Except, we cannot have the $\pm$ on there, because then it ceases to be a function.

So if we restrict the domain of the original function to $[0,\infty)$, then we can have a well defined, bijective (sp?) inverse.


Show that $\left ( {\frac{a+b}{2}} \right )^2 > ab, \quad |a| \ne b$

$|a| \ne b \Longleftrightarrow |a|>b \ \vee \ |a| < b \Longleftrightarrow a>b \ \vee \ \ a<-b \ \ \vee\ -b<a<b$

$i) a >b \Longleftrightarrow a-b>0 \Longleftrightarrow (a-b)^2>0 \Longleftrightarrow a^2+b^2-2ab>0 \Longleftrightarrow a^2+b^2+2ab>4ab \Longleftrightarrow (a+b)^2>4ab \Longleftrightarrow \boxed{\left ( {\frac{a+b}{2}} \right )^2 > ab}$

Is that correct? And what about the following:

$ii) \quad a<-b$

$iii) \quad -b<a<b$

[quote="shyong"]well , if only one of $a,b$ is negative , then LHS >0 while RHS<0 and it is always true . So we only need to consider the case where both $a,b$ are positive or both negative . Both positive for $a,b$ is just a direct application of AM-GM . For both negative case, letting $a=-a',b=-b'$ where $a'.b'$ are positive then it is equaivlent to prove

$\left(\frac{a'+b'}{2}\right)^2>a'b'$ which is also true .[/quote]

Can you give a formal demonstrartion?


Column $ \text{A}$ contains all numbers that are $ 1$ in mod $ 7$, column $ \text{B}$ contains all numbers that are $ 2$ or $ 0$ in mod $ 7$, column $ \text{C}$ contains all numbers that are $ 3$ or $ 6$ in mod $ 7$, and column $ \text{D}$ contains the rest.

$ 600\equiv 5\pmod{7}$, so it's in column $ \boxed{\text{D}}$

i used a simular method and it came out pretty fast once i thought about it.

The fourth number is in column A. Also the sequence is BCDABCD. The remainder of 599/7 is 4, so if you count four spaces, you end up on A. Why is the answer D. I don't understand how that is possible.

@yaminitn: Hi! I noticed that you are new, but I think you should know that you should not revive old topics (unless you have a very good reason.)
I'll answer your question. For this problem you can't really use remainders...how did you get $ \frac {599}{7}$ anyways? In order to solve the problem, you need a basic understanding of modular arithmetic. See 5849206328x's post above.

I solved it in a little different way, it might be the same principal though. I noticed that it repeated every seven, so I divided 600 by seven, and got 85.7142...(a lot of numbers) I rounded down to the closest multiple of seven, 84, and divided by seven to get twelve. Twelve is in Column D so that was my answer. Did I solve this in an acceptable way? :huh:

From my calculations, I think the answer is $ \boxed {\text {B}} $.
I will post a solution when this is confirmed.


It seems as though Alcumus went through some changes over the night. The solution page stopped loading on every problem for me, so is this a minor problem?

I just tried alcumus, and it seems to be working fine to me.
Although there does seem to be changes on the solution page of alcumus, it is working.

xxrxx -- are you still having this problem?

I am still having this problem. Also, I think it may be something wrong with my computer, not something with the server.


[quote="Tomekk"]Prove that 25 divides: $ 3^{6n} + 7^{n} \cdot (7^{n} - 2^{n + 1} )$[/quote]

[hide="Solution"]
Expanding this we get,

$ 3^{6n} + 7^{2n} - 2\cdot 14^n$

Now we evaluate in mod 25,

\begin{eqnarray*} 3^{6n} + 7^{2n} - 2\cdot 14^n & \equiv & (3^3)^{2n} + 7^{2n} - 2\cdot 14^n\pmod{25} \\
3^{6n} + 7^{2n} - 2\cdot 14^n & \equiv & (27)^{2n} + 7^{2n} - 2\cdot 14^n\pmod{25} \\
3^{6n} + 7^{2n} - 2\cdot 14^n & \equiv & 2^{2n} + 7^{2n} - 2\cdot 14^n\pmod{25} \\
3^{6n} + 7^{2n} - 2\cdot 14^n & \equiv & (7^n - 2^n)^2\pmod{25}
\end{eqnarray*}

Now if we can show that $ 7^n - 2^n\equiv 0\pmod{5}$ we are done, and since $ 7\equiv 2\pmod{5}\implies 7^n - 2^n\equiv 2^n - 2^n\equiv 0\pmod{5}$ therefore, we are done. :ninja:
[/hide]


Im not sure how to approach this problem:

[quote]As my younger brother shouted out four consecutive integers, I divided each by my age (an integer) then added all four remainders to get 40. As he shouted out four other consecutive integers, I divided each by his age (a different integer) added all four remainders, and again got 40. What is the sum of our ages?[/quote]

 :|

Let $a$ be the age of one of the brothers and $n$ the first of the 4 consecutive numbers the younger brother shouts out. The remainder of the division $(n+i): a$ is $r_{i}$ where $i \in \{0,1,2,3\}.$ If $r_{0}< a-3$, then $r_{i}= r_{0}+i,$ hence

$r_{0}+r_{1}+r_{2}+r_{3}\;=\; r_{0}+(r_{0}+1)+(r_{0}+2)+(r_{0}+3) \;=\; 4r_{0}+6,$

which is impossible since this sum should be 40 (giving $r_{0}= 8,5$). This contradiction implies that $r_{0}\geq a-3$, which gives us the following three possible series of remainders:

\begin{tabular}{|c|c|c|c|}\hline r_{0}& r_{0}& r_{0}& r_{0}\\ r_{1}& 0 & r_{0}+1 & r_{0}+1 \\ r_{2}& 1 & 0 & r_{0}+2 \\ r_{3}& 2 & 1 & 0 \\ \hline Sum & r_{0}+3 & 2r_{0}+2 & 3r_{0}+3 \\ \hline \end{tabular}

The sum is 40, which implies that $r_{0}= 37$ (and $a|r_{1}=38$) or $r_{0}= 19$ (and $a|r_{2}=21$). Since $a > r_{0}$, the age of the two brothers are 21 and 38 years. Hence the sum of their ages is 21 + 38 = 59 years.

wow thanks for the quick and thorough reply, it all makes sense now  :lol: