Contest:

Anyone may participate. All other directions are included in the pdf.

Just make a post to sign up or pm me wigh the answers. Registration is open again :)

Download the pdf file in order to finish the contest.

in order to signup, you must make at least a post here, signup on http://www.4aunlimited.co.nr , or pm me.

so far:

:arrow: [hide="people who signed up"]
Anirudh
236Factorial
DrunkenMath
Math92
Chess64
Xantos C. Guin
G-Unit
piano1231
ynorman
Plinus
[/hide]

:arrow:[hide="Who made it to the next round"]
Anirudh
Plinus
236Factorial
Drunken Math
anirudh
G-Unit
Xantos c. Guin
piano1231
Chess64
[/hide]

:arrow:[hide="Final participants for round 3"]Anirudh
Plinus
236Factorial
Drunken Math[/hide]

Round 2 is here:

All the directions are included in the pdf. If you have any doubts, ask on this thread. Also, pm me the answers. Lets's see who advances.  :D

[quote="math92"]for #1 you say "I choose 3 letters from the word MATHCOUNTS. What is the probability that it is a
vowel?"
Do you mean that what is the probablitly that all 3 are vowels or 1 is a vowel?
And for #2 you want us to find the answer.  Does that mean the answer in terms of x?[/quote]

if at least 1 is a vowel
For #2 it is simplify.

[b][color=black][size=200]ps. a new version of the round is now available.[/size][/color][/b]

For #3, the height to which side?

Is it like this?: "An isosceles triangle $\triangle ABC$ has sides $AB=AC=2$. The altitude $AD$ has length $2$. If the area of $\triangle ABC$ is $n$, then find $2n$."

If so, then it's not a triangle, it's a line, because by Pythagorean, we have $CD=DB=0$, because $AC=AB=AD$. Can you maybe provide a diagram or fix the problem?

Thanks for making this competition though. I see you liked my format! :)

[quote="Durnken math"]What do we do with the answers?

How many rounds are there?

Do we need to PM the answers to you?

How many people are in it?

Is it a competition by elimination?

Will the points you get matter for anything besides getting to the next round?

For Number 2, what do you mean "the answer"? Simplify?
:(  :(  :(  [/quote]

You private message the answer to me.

There will be as many rounds as necessary to extract out thewinner.

The points won't matter, except for getting to the next round.

For number 2, i mean simplify. Download the new version also--it fixes typos in the previous version.

[quote="Chess64"]For #3, the height to which side?

Is it like this?: "An isosceles triangle $\triangle ABC$ has sides $AB=AC=2$. The altitude $AD$ has length 2. If the area of $\triangle ABC$ is $n$, then find $2n$."

If so, then it's not a triangle, it's a line, because by Pythagorean, we have $CD=DB=0$, because $AC=AB=AD$. Can you maybe provide a diagram or fix the problem? [/quote]

Well, thanks for proving me wrong. I was just in a hurry to post some prooblems. You can download the newest version in ordder to finish the round 1.

[quote="math92"]Do we have to show work?[/quote]

Read the directions...

DON'T SPAM!  :mad:

[quote="math92"]sorry, i didn't see that the directions were in the document.[/quote]

Read the second line of the first post  ;)

[quote="G-UNIT"]In question 1, is each letter distinct.  For example, are the two "T's" considered the same or are they considered different.[/quote]

read the next version. the question has changed.

Can you make a hide tab thingie that shows the people who made it to the next round?

it's a tricky question, but if i explain anything more, i will give it away. anyway, even if you don't get it, you gcan advance even if you don't answer it.  :) . also, edit, not double psot.

What is the dead-line for round 2 solutions?

Get it in 5 days---Also, I will post  round 1's solutions, since chess64 was evicted.

Round 1 solutions are here!

Click on this link below to download

[url=http://dasarathy0.tripod.com/AoPS/round1answers.pdf]Click here to download[/url]

except the title should be 'solutions to round 1' :lol:

Anirudh:

#10: doesn't make sense
#2: you can't factor it
#5: should we round
#3: does it mean all primes up to 100 are in there?

$q^2-(q-1)^2=2q-1$

Total possible points = $10^2=100$
You must get at least $80$

you can factor number 2  ;)

I need a response from anirudh about my answers

well, that's better than leaving it as it is, right?  ;)

First of all, where did i put my last name  :(  :blush:  :?:  :?:  :?:  :?:  :mad:  :mad:  :mad:  :mad:  :blush:

[quote="chess64"]#10: doesn't make sense
#2: you can't factor it
#5: should we round
#3: does it mean all primes up to 100 are in there? [/quote]

For number 10, you can skip it. If you attempt and get it right ( I will accept 3 answers), you get bonus points.

For number 2, a hint is that (a-b) is equal to -(b-a).

For number 5, round off

For 3, all the primes under 100 are there, e.g. 2, 3, 5... and u have to find the probability of you picking an evemn number.

Well, that's your problem. Please edit that post immediately because it might give a hint to the answer. The round is due Aug. 17

[quote="G-UNIT"]When are round 2 problems due?  I dont want to do these problems until the day-before.

EDIT: Theres no rounding in question 5.................it comes out to a whole number...........[/quote]

Actually, 5 is technically not an integer, although it is very close. But it is close enough to the integer to round to an integer.

OK now. WE will now skip round 2 and in another day or 2, i will start round 4(since some people thought this was round 2). For that round, I'll  use real MATHCOUNTS Problems. All the other challenges that i posted may be used as preparation. Thanks.


algebra
polynomial
function
algebra unsolved

Determine all polynomials $p$ with real coefficients for which $p(0)=0$ and $f(f(n))+n=4f(n)$ for all $n\in N$, where $f(n)=[p(n)]$

I suspect it's the floor function (the integer part).

[quote="Beat"]Yes, I think that it`s the integer part, too[/quote]

On the other hand, if it IS the integer part, then $f(f(n))$ doesn't make sense, since it trivially simplifies to $f(n).$

hmm..  :read: strange!
Maybe you are right..
See problem 3 from 3-rd and 4-rd grades: http://www.imo.org.yu/othercomp/Yug/YugMO05.pdf

Now that makes sense. Any ideas? or even solutions?

For some $a\in \R*$ and $d\in \N$, $f(x) = a x^d + o(x^d)$.
We get
\[ f(f(n))+n = a(an^d)^d+n + o(n^{d^2}) = 4 f(n) = 4 a n^d + o(n^d), \]

so \[ a^{d+1} n^{d^2} - 4 a n^d + n + o(n^{d^2})=0 \]

Thus $d^2=d$, therefore $d=0$ or $d=1$.

Now for $d=0$, since $p(0)=0$, $p(x)=0$ for all $x$.
If $d=1$; $p(x)=ax$. It suffice to check these, which I will not do tonight : its late.


Solve the equation
$\frac{1}{x-a}+\frac{1}{x-b}+\frac{1}{x-c}=\frac{1}{a}+\frac{1}{a}+\frac{1}{a}$
which a,b,c are three sides of a triangle.

I think you mean that
$\frac{1}{x-a}+\frac{1}{x-b}+\frac{1}{x-c}=\frac{1}{a}+\frac{1}{b}+\frac{1}{c}$

Anyway, I think something wrong with this problem. :?


Loipon kirioi kai kiries,

Me IDIAITERH XARA sas anakoinwnw oti h xwra mas pleon EPISIMA simmetexei ston diethni diagwnismo KANGAROO (Ton MEGALITERO mathimatiko diagwnismo ston kosmo me simmetoxi (perisina stoixeia) peripou 3.950.000 mathitwn) o opoios fetos tha dieksaxthei sta teli tou Marti 2007. Mporeite na to diastaurwsete kai stin episimi istoselida tou diagwnismou  pou einai

[url]http://www.mathkangaroo.org[/url]

O sigekrimenos diagwnismos einai aneksartitos tis EME kai apla exei ekproswpous se oles tis xwres pou simmetexoun. Merikes leptomereies diadikastikes menoun akoma gia na sistathei o sillogos tou diagwnismou stin Ellada. Exoun katatethei ola ta aparaitita xartia stous antistoixous foreis kai anamenetai i apofasi eggrisis idrisis tou sillogou gia ton diagwnismo!!!

Simmetexoun, (opws tha deite kai stin parapanw istoselida pou sas edwsa)  42 xwres ( o diagwnismos ksekinise apo USA) kai fetos mpikan gia prwti fora h Ellada, i Kipros, o Kanadas, Ekouador, India, Kirgistan.

Stin Ellada o ekproswpos tou diagwnismou einai o kos Mixalis Lamprou apo to Panepistimio tis Kritis o opoios me ekdilo to endiaferon gia tous mathimatikous diagwnismous (eksou kai auta pou evala tin teleutaia fora stin istoselida gia talantouxous mathites Gimnasiou- Likeiou), pire tin apofasi na valei tin Ellada ston diagwnismo KANGAROO pou sas anefera kai na ksekinisei siga siga tin prospatheia diadosis tou se olous tous mathites...

O Diagwnismos exei katapliktiki vasi opws mou eksigise kai apo konta o kos Lamprou. Katarxin OLOI oi mathites pou tha simmetasxoun ston diagwnismo tha paroun merika dwrakia pou aforoun ta mathimatika (mplouzakia me to logotipo tou diagwnismou,kataskeues, paixnidia kai diafora alla pou tha tous proselkisoun to endiaferon. Polla apo auta ta eida ki egw kai einai pragmatika APISTEUTA kai ELKISTIKOTATA! ). To kalitero omws einai oti o diagwnismos autos apeuthinetai se mathites apo

2a DHMOTIKOU EWS KAI 3h Likeiou  (Endiaferon etsi???)

Epeidi tha iparksei oikonomiki simmetoxi twn mathitwn pou tha lavoun meros ston diagwnismo (tha sas eksigisw kai parakatw ta xrimata auta pou pigainoun), h opoia anerxetai girw sta 10-15 eurw den mporei distixws gia arxi toulaxiston (mexri na doume ti tha ginei sto mellon) na dieksaxthei o diagwnismos se dimosia sxoleia... Edw stin Kriti sigoura tha ginei se kapoio idiwtiko ekpaideutirio ('h sto Panepistimio) , opws episis kai stin Thes/niki tha ginei se kapoio idiwtiko ekpaideutirio (opws me pliroforise i ka Eleni Mitsiou , mathimatikos kai amesa endiaferomeni gia tetoious diagwnismous).

O Diagwnismos anamenetai kathe xrono na auksanetai sinexws se plithos mathitwn kathws ta themata einai poli poli elkistika! Tha exoun tin morfi erwtisewn pollaplis epilogis kai tha einai peripou 30 ston arithmo kai tha prepei na apantithoun girw stin 1:30 - 2:00 wres. Exei apodeixthei oti oi mathites pou vraveuontai ston sigekrimeno diagwnismo, vraveuontai kai stous ipoloipous megalous diagwnismous tipou IMO, BMO, USAMO ktl ktl... Oi erwtiseis einai klimakoumenis diskolias kai sigoura KANENAS mathitis den tha figei apogoiteumenos (kai logw tou oti tha exei parei kapoia dwrakia pou exoun sxesi me ta mathimatika, kai logw tou oti tha apantisei sigoura kapoia thematakia. Pantws den tha figei xwris na asxolithei ligaki, fainomeno pou sixna distixws sinantousame stous diagwnismous tis EME logw tou oti eprepe na einai liga kai kala (kai den to katigorw auto) ).  Twra pleon kai oi mikroteroi mathites tha exoun kinitro gia na diavasoun mathimatika! Fantasteite ena pitsiriki 2a Dimotikou pou ksekinaei, den ta paei kai toso kala kai ksekinaei tin epomeni xronia me megaliteri oreksi kai diavazei... Anarwtitheite loipon pws tha ftasei na einai stin A-B Likeiou an ksekinisei kai tou dwthei to erethisma apo toso mikro!!!

To kalo loipon einai oti kapoios mathitis mporei na ksekinisei apo poli mikros apo autous tous diagwnismous ki etsi siga siga na auksanei tis gnwseis tou kai na exei kinitra gia na diavasei mathimatika kai na auksithei etsi to mathimatiko ipovathro tis xwras mas kathws episis kai i parousia mas stous diethneis diagwnismous. (Kati pou ews twra den ipirxe afou se poli elaxista parartimata stin Ellada kai me diki tous prwtovoulia, diorganwnan eswterikous diagwnismous mathimatikwn gia to dimotiko).

Osoi diakrithoun stin prwti fasi tou diagwnismou, iparxei kai deuteri...

Ta xrimata twn simmetoxwn twn mathitwn (opws anefera parapanw girw sta 10-15 eurw) sigentrwnontai ston sillogo KANGAROO tis xwras, me ton elenxo PANTA tou diethnous simvouliou KANGAROO, kai apo auta kaliptontai OLES oi anages tou eksoplismou alla kiriws ta vraveia kai ola auta pou apokomizoun oi mathites pou diakrinontai stous diagwnismous (kathws episis kai ta dwrakia pou pairnoun OLOI stin arxi tou diagwnismou).  Episis gia tous diakrithentes mathites tha iparxoun vivlia kai tha diorganwnontai THERINA SXOLEIA MATHIMATIKWN  (!!!!!), (gia paradeigma stin Roumania osoi diakrinontan ston diagwnismo KANGAROO tous mazeuoun to kalokairi kai tous kanoun proponiseis gia mathimatikous diagwnismous, diorganwnoun wraies ekdilwseis ktl san vraveusi twn mathitwn gia ton kopo tous kai tin epidosi tous) ta eksoda twn opoiwn (kai sigoura tha einai arketoutsika) tha kaliptontai apo auta ta xrimata tis simmetoxis twn diagwnizomenwn (kai isws kai apo alles xorigies...). Oi kathigites vevaia pou tha didaskoun tha to kanoun apolitws ANIDIOTELWS. SIGOURA loipon, ta esoda auta tha epistrafoun sto megisto stous mathites mas... Kai gia na prolavw kapoious pou tha skeftoun oti mporei ta xrimata auta na pane se kapoies tsepes, exw na pw oti o diagwnismos autos einai toso diadedomenos sto ekswteriko pou iparxei apo PANTOU elenxos. Toses xwres simmetexoun.... (Epanalamvanw einai o megaliteros se simmetoxi mathitwn diagwnismos se OLOKLHRO ton kosmo). Skefteite posa pragmata tha allaksoun rizika morfi me tin eisagwgi autou tou diagwnismou o opoios einai diethnous kirous... Einai loipon ligo diskolo xwris tin parapanw oikonomiki simmetoxi twn mathitwn na iparksoun auth i prosfora gnwsis ti stigmi kata tin opoia ta perissotera xrimata pou plirwnoun oi mathites, tous epistrefetai pisw ipo tin morfi kapoiwn dwrwn 'h ean diakrithoun ipo tin morfi vivliwn, therinwn sxoleiwn ktl ktl...

Mporeite na deite palaiotera themata stin selida
[url]http://www.kangurusa.com/new/pastproblems.htm[/url].

Tautoxrona o kos Lamprou metafrazei auto ton kairo deigmata apo autous tous diagwnismous sta Ellinika etsi wste na apostaloun se diaforous nomous se oli tin Ellada kai na paroun oloi oi sinadelfoi mathimatikoi kathws kai oi mathites mia idea gia to ifos tou diagwnismou...

Episis i epilogi twn thematwn ginetai 2-3 mines PRIN apo ton diagwnismo me tin simmetoxi kai tis protaseis olwn twn xwrwn, gia na iparxei o xronos na ftiaxtoun kai na diatipwthoun oso to dinaton KALITERA ta themata tou diagwnismou (Kai etsi den iparxei to provlima pou antilamvanomaste oloi kai pou anefere se proigoumeno post o kos Stergiou gia to poso diskolo einai na ftiakseis themata pollaplwn epilogwn otan asxolountai me auto poli liga atoma. Edw tha proteinoun OLES oi xwres opote tha iparxoun arketa kai kala themata.) Etsi iparxei auto to apotelesma pou vlepete sta parapanw palaiotera themata tou diagwnismou.

DIADWSTE loipon ta kala nea se olous osous gnwrizete kathws episis kai na anazitisete pithana meri sta opoia mporei na ginei o diagwnismos stis poleis sas. Sigoura prepei na anazitisoume stin Athina, Patra kapoia eksetastika kentra tou diagwnismou kathws episis kai stin Eparxia...

Ante to prwto vima egine, kai nomizw oti itan ARKETA MEGALO VHMA gia tin Ellada i prosthiki allou enos diagwnismou mathimatikwn pou epivalotan. Kathe arxi kai diskoli... Se liga xronia tha einai thesmos autos o diagwnismos stin Ellada me terastia antapokrisi opws eksalou ginetai se olo ton kosmo!!

Perimenw OPWSDIPOTE ta sxolia sas gia to thema!!!
(Kwsta sou aresan ta nea pou sou eipa oti tha anaferw prin apo merikes imeres???)

Auta gia twra.... Tha sas enimerwsw sxetika kai me tin akrivi imerominia tou Diagwnismou apo to idio Topic. Ews twra einai gnwsto oti tha ginei telos Martiou.

Tha ftiaxtei kai sxetiki istoselida me ta nea tou diagwnismou kai tou sillogou KANGAROO stin Ellada...

KANGAROO loipon DYNAMIKA!!

Alexandros

Αλέξανδρε ,

 ΚΑΛΗ ΑΡΧΗ !!!

 Πρόκειται για πολύ καλό διαγωνισμό! Τον παρακολουθώ χρόνια. Πρίν δύο χρόνια μου είχε προτείνει η υπεύθυνη του γερμανικού διαγωνισμού, καθηγήτρια στο πανεπιστήμιο του Βεριλίνου , κυρία Noack , να παρευρεθώ στν προεοιμασία, αλλά δε μπορούσα να πάω γιατί είχα μικρό. Μου είχε πει μάλιστα να διαδώσω το διαγωνισμό και στην ελλάδα. Αυτό ήταν δύσκολο για μένα, γιατί δεν είχα τη δυνατότητα να φτιάξω τέτοιο φορέα. Να λοιπόν που έγινε  και χαίρομαι !!!. Πρέπει να τον επικυρώσουμε στο Υπουργείο και να έχουμε έγκριση διεξαγωγής του στα σχολεία και μάλιστα καθημερινή !!!
 Θα τα πούμε από κοντά. Έχω και λύσεις των θεμάτων από το 2000 , περίπου ,  στα γερμανικά που μου έστειλε η Γερμανίδα καθηγήτρια.
  Εύχομαι ο διαγωνισμός αυτός να βρει ανταπόκριση . Θα είναι μια γιορτή !!!

ΣΥΓΧΑΡΗΤΗΡΙΑ ΣΕ  ΟΛΟΥΣ ΣΑΣ !!! - ΜΠΑΜΠΗΣ

Αλέξανδρε

 πολύ καλά νέα μας έφερες .Τα καλά της παγκοσμιοποίησης  :)
 Οι διαγωνισμοί πρέπει να ξεκινάνε από μικρές ηλικίες .
 Μπράβο στον δραστήριο κ.Λάμπρου .

αγαπητοί μου φίλοι χαιρετώ
Σίγουρα αυτό το γεγονός είναι τεράστια επιτυχία για την Ελλάδα μας να συμμετέχει σε αυτόν τον διαγωνισμό προτέινω προφανώς το Κολλέγιο ΑΘηνών ως το καλύτερο εξεταστικό κέντρο των Αθηνών ανεπιφύλακτα κι θα ήθελα επειδη θα ενημερώσω τον αρμόδιο καθηγητή για τέτοια θέματα να μου δώσετε συγκντρωτικά ¨ΟΛΑ ΤΑ λινκ να ττα δει κι να προτείνει προτάσεις για τα εξεταστικά κέντρα αλλά κι για ότι άλλο θέλετε μια ερώτηση έχω μόνο
Οποιος θέλει πάει στον διαγωνισμό αυτό ή υπαρχει επιλογή κι πώς γίνεται αυτή ???
ευχαριστώ καλή αρχή

[quote="stergiu"]Πρέπει να τον επικυρώσουμε στο Υπουργείο και να έχουμε έγκριση διεξαγωγής του στα σχολεία και μάλιστα καθημερινή !!!
 Θα τα πούμε από κοντά. Έχω και λύσεις των θεμάτων από το 2000 , περίπου ,  στα γερμανικά που μου έστειλε η Γερμανίδα καθηγήτρια.
  Εύχομαι ο διαγωνισμός αυτός να βρει ανταπόκριση . Θα είναι μια γιορτή !!!

ΣΥΓΧΑΡΗΤΗΡΙΑ ΣΕ  ΟΛΟΥΣ ΣΑΣ !!! - ΜΠΑΜΠΗΣ[/quote]

Sigxaritiria se ekeinon ton anthrwpo pou asxoleitai toso energa me to thema auto twn diagwnismwn kai ton ponaei...

Egw apla metedwsa tin pliroforia auti!! Sigoura omws tha voithisw oso mporw stin diadosi tou kai stin dieksagwgi tou...

Makari na epikirwthei apo to Ypourgeio Paideias! Sigoura tha to kinigisei o kos Lamprou ap'oti mou eipe...

K. Stergiou euxaristoume gia ta kala logia. Diadwste to oso mporeite! An eidika vrethei kai ena eksetastiko kentro gia ton Nomo Evoias tha einai euxis ergon. Euxomai ki egw na vrei antapokrisi siga siga o diagwnismos!

[quote="ambros"]
Οποιος θέλει πάει στον διαγωνισμό αυτό ή υπαρχει επιλογή κι πώς γίνεται αυτή ???
ευχαριστώ καλή αρχή[/quote]

Amvrosie opoios thelei mporei na parei meros ston diagwnismo apo B Dimotikou mexri 3h Likeiou, arkei na iparxei kapoio eksetastiko kentro sto opoio na mporoun na apostalthoun ta themata...

Rwtise loipon ton kathigiti sou gi'auto kai diavase analitika to email pou esteila parapanw sto opoio grafw tin episimi istoselida tou diagwnismou kathws episis kai to link me palaiotera themata! (Sintoma pisteuw oti tha iparxei kai istoselida gia ton diagwnismo stin Ellada me oles tis plirofories). Apo ekei kai pera mporeis na vreis mono tis selides kathe kratous ksexwrista pou den nomizw na voithaei kai poli!!!

Φοβερά νεα. ειναι μια καινοτομια στη χωρα μας. σιγουρα θα  βοηθησει την μαθηματικη παιδεια μας.
=> το κυρος του διαγωνισμου ειναι μεγαλο κινητρο
=>ειναι μια δευτερη ευκαιρια για οποια παιδια  που εχουν ταλεντο για καποιο λογο δεν τα καταφερνουν στου διαγωνισμους της ΕΜΕ
ειναι σημαντικο να το ξερει αυτο ενας διαγωνιζομενος ,τον απελευθερωνει και δεν βλεπει τους διαγωνισμους σαν ια 3-ωρη μαχη που αν αποτυχει τελειωσε το μαθηματικο ταξιδι  :)
Αλεξανδρε ειναι φανταστικα τα νεα αυτα, μπραβο σε ολους το πετυχαν και φυσικα στο κ Μιχαλη Λαμπρου.

Αλέξανδρε είναι εκπληκτικό .Μπράβο σε όλους και ιδιαίτερα στον κο Λάμπρου .
Αν βρούμε εξεταστικό κέντρο εδώ στα Τρίκαλα (είναι πολλά τα παιδιά που έρχονται και στους διαγωνισνούς της ΕΜΕ ) θα μπορούν τα θέματα να έρχονται και εδώ ?

Και αν δε βρούμε εξεταστικό κέντρο εγώ θα μπορώ να πάρω μέρος (ή όποιος άλλος θέλει) σε ένα άλλο εξεταστικό κέντρο ?

Περιμένω τις απαντήσεις σου .
Ευχαριστώ.

Silouan eimai sigouros oti o diagwnismos autos tha einai arketa eueliktos kai apla mallon tha dilwneis simmetoxi kai tha mporeis na diagwnisteis kai se kapoio allo eksetastiko kentro!!

Gi'auto to anefera se esas apo twra gia na to exete ipopsin sas kai na koitaksete se poia meri mporoume na vroume eksetastika kentra! Sigoura Kriti kai Thes/niki exoume vrei idi... Prepei toulaxiston na vroume kai se alla megala kentra (Athina, Patra, isws se kapoia megala nisia, kai opou allou mporoume...) O Diagwnismos einai kainourios kai isws ton antimetwpisoun ligo distaktika... Euelpistoume ola na pane kala!!

Tha sas enimerwnw taktika gia opoiodipote neotero se sxesi me tin dieksagwgi tou diagwnismou!! Prepei na palepsoume gia na ton diatirisoume (Twra pleon mporei na enimerwthoun gia ton diagwnismo oxi mono ta gimnasia+Likeia alla kai ta Dimotika!! :) ) kai oi prwtoi pou prepei na voithisoume eimaste oloi oi mathimatikoi kai ola ta paidia ekeina pou asxolountai sovara me tous diagwnismous kai agapane ta mathimatika! Gi'auto eksallou to anaferw edw...

Alexandros

Πρέπει πιστεύω να σταλλεί χαρτί σε όλα τα σχολεία με όλα αυτά που θα περιγράφει τα δώρα και όλες τις ελκυστικές πληροφορίες.
Εξάλλου πως κάθε βδομάδα μας διαβάζουν για διαγωνισμούς εκθέσεων -ποίησης κτλ .Ας ακουστεί επιτέλους και κάτι μαθηματικό  :)
Πάντως θεωρώ ότι εδώ θα βρούμε χώρο για το διαγωνισμό .

Αλέξανδρε ,

 θέλω να συζητήσεις με τον  Κ. Λάμπρου τα έξής:
  α) Μήπως ο διαγωνισμός , σιγά σιγά, μπορεί να γίνεται στα σχολεία . Πχ στο τελευταίο τρίωρο στις 24 Μάρτη (αν δεν είναι Σ.ή Κυριακή) ή κάποια άλλη κατάλληλη μέρα. Θα είναι μεγάλη επιτυχία να μπορέσουμε να επιβάλουμε - καθιερώσουμε το διαγωνισμό ως ένα μαθηματικό άθλημα σε όλη την εκπαίδευση.Φυσικά η συμμετοχή θα είναι προαιρετική. Με τον τρόπο αυτό θα δραστηριοποιήσουμε όλους τους εκπαιδευτικούς.
   β) Το κάθε σχολείο μπορεί να κάνει τη δική του βράβευση σε μια ειδική (πιθανόν πολύπλευρη)εκδήλωση στο τέλος του χρόνου και να θεσπίσει και δικά του βραβεία.
   γ) Το διαγωνισμό πρέπει να τον αναλάβει οικονομικά κάποιος γερός χορηγός, πχ η εθνική ή κάποια άλλη μεγάλη τράπεζα. Μόλις τον διαγωνισμό τον οργανώσουμε καλά , μπορούμε να το ζητήσουμε. Να ξέρεις ότι πολλοί θα θέλουν να βοηθήσουν. Σε τελευταία ανάλυση , το κέρδος από την διάδοση των μαθηματικών θα τον έχουν πρώτες οι τράπεζες και η βιομηχανία. Άρα πρέπει να πληρώσουν. Θα τους το πούμε ξεκάθαρα : Εσείς κερδίζετε εσείς θα πληρώσετε !!!.
 Αφού λοιπόν όλη η κοινωνία αρέσκεται με την πλήρη ιδιωτικοποίηση των πάντων , πρέπει να καταλάβει ότι τη γνώση θα την πληρώσει .Και για να είμαι δίκαιος , θα πρέπει να πω ότι τη μαθηματική γνώση που είναι η σπουδαιότερη και η δυσκολότερη θα την χρυσοπληρώσει. Γιατί ο κόσμος πάει μπροστά μόνο με τα μαθηματικά και τις εφαρμογές του. Εμείς οι δάσκαλοι θα βοηθήσουμε από καθήκον προς την πατρίδα και τους μαθητές μας.Όμως το οικονομικό κόστος θα το αναλάβουν αυτοί που έχουν τη μερίδα του λέοντος.
 Φυσικά ,  η σκέψη ότι οι συμμετέχοντες πρέπει , αρχικά τουλάχιστον ,  να επιβαρύνονται με ένα μικρό συμβολικό χρηματικό ποσόν , με βρίσκει σύμφωνο.Δε γίνεται διαφορετικά.
 Λοιπόν , επειδή δεν ξέρω πότε θα συναντήσω τον Κ. Λάμπρου  για να τα πούμε από κοντά, σε παρακαλώ να τυπώσεις αυτό το κείμενο και να του το δώσεις  προσωπικά. Εχω πολλά ακόμα , αλλά ας μη γίνω κουραστικός.Προτίθεμαι αν βοηθήσω από όποια θέση μπορώ.
 ΜΠΑΜΠΗΣ

Φίλοι μου αποτελεί μεγάλο ατού του διαγωνισμού η απλή θέληση για συμμετοχή κι όχι η επιλογή έχετε το λόγο μου ότι θα προσπαθήσω όσο μπορώ με τον καθηγητή μου να καθιερώσουμε το Κολλέγιο Αθηνών ως εξεταστικό κέντρο για τον διαγωνισμό
κι όσον αφορά τους χορηγούς θεωρώ ότι αυτοί θα βρεθούν με λίγη προσπάθεια αλλά επίσης πρέπει το κόστος συμμετοχής των παιδιών να είναι συμβολικό άν υπάρχει διαφορετικά δεν θα προσελκύσουμε Κόσμο που είναι το ζητούμενο
κι επίσης θα πρέπει τα βραβεία να είναι ενιαία κι έξυπνα βραβεία
πχ δωρεάν παρακολούθηση θερινού camp μαθηματικών  :wink:  ή κατι άλλο για να έρθει κόσμος
αυτά περιμένω σχόλια σας θα επανέλθω

[quote="stergiu"]Αλέξανδρε ,

 θέλω να συζητήσεις με τον  Κ. Λάμπρου τα έξής:
  α) Μήπως ο διαγωνισμός , σιγά σιγά, μπορεί να γίνεται στα σχολεία . Πχ στο τελευταίο τρίωρο στις 24 Μάρτη (αν δεν είναι Σ.ή Κυριακή) ή κάποια άλλη κατάλληλη μέρα. [/quote]

Πιστεύω ότι αυτό πρέπει να γίνει . :D
Θα είναι τρομερό.  :lol:

ο χρόνος του διαγωνισμού κατά την γνώμη μου δεν επηρεαζει καθόλου αλλά μια καλή χρονική τοποθέτηση προφανώς κι θα δώσει ώθηση για συμμμετοχή καλή ώρα θα είναι το πρωί κάποιας μέρας με όχι βεβαρημένο σχολικό πρόγραμμα η Σαββατοκύριακο ώστε να μην επεισέρχεται αρνητικά ο παράγων σχολείο κι μαθήματα
αυτά χαιρετώ

[quote="stergiu"]Αλέξανδρε ,

 θέλω να συζητήσεις με τον  Κ. Λάμπρου τα έξής:
  α) Μήπως ο διαγωνισμός , σιγά σιγά, μπορεί να γίνεται στα σχολεία . Πχ στο τελευταίο τρίωρο στις 24 Μάρτη (αν δεν είναι Σ.ή Κυριακή) ή κάποια άλλη κατάλληλη μέρα. Θα είναι μεγάλη επιτυχία να μπορέσουμε να επιβάλουμε - καθιερώσουμε το διαγωνισμό ως ένα μαθηματικό άθλημα σε όλη την εκπαίδευση.Φυσικά η συμμετοχή θα είναι προαιρετική. Με τον τρόπο αυτό θα δραστηριοποιήσουμε όλους τους εκπαιδευτικούς.
[/quote]

Fisika, nomizw oti auto tha einai to prwtarxiko melima gia ton diagwnismo! Na mporesei na ginetai se ola ta sxoleia tis xwras 'h estw na mporoun ola ta sxoleia na enimerwthoun gia tin dieksagwgi tou diagwnismou mesw tou ipourgeio opws ginetai kai gia ton diagwnismo tis EME!! To mono provlima einai oti epeidi iparxei i oikonomiki simmetoxi tou mathiti den kserw to kata poson auto ginetai (Den kserw tin nomothesia panw se auto to thema). Den kserw an einai efikto na ginei kati tetoio apo fetos (giati den to exw rwtisei) alla ap'oti mou eipe kai o kos Lamprou tha einai apo ta prwta prwta pragmata pou prepei na taktopoiithoun.

[quote="stergiu"]
 Φυσικά ,  η σκέψη ότι οι συμμετέχοντες πρέπει , αρχικά τουλάχιστον ,  να επιβαρύνονται με ένα μικρό συμβολικό χρηματικό ποσόν , με βρίσκει σύμφωνο.Δε γίνεται διαφορετικά.
[/quote]

Distixws opws eipate ki eseis den ginetai diaforetika... Gi'auto sto prwto email anefera otan i ekseuresi xorigwn gia ton diagwnismo tha itan i kaliteri lisi! Auto akrivws pou proteinete ki eseis... Pisteuw oti gia arxi prepei na isxiropoiithei o diagwnismos wste meta na mporesei kapoios na xtipisei tin porta tis X trapezas kai na zitisei xorigia... Diladi me kapoio tropo prepei na diafimistei kai i trapeza, kai o kaliteros tropos gia na ginei auto einai ean iparxei arketos kosmos pou simmetexei.

[quote="stergiu"]
 Λοιπόν , επειδή δεν ξέρω πότε θα συναντήσω τον Κ. Λάμπρου  για να τα πούμε από κοντά, σε παρακαλώ να τυπώσεις αυτό το κείμενο και να του το δώσεις  προσωπικά. Εχω πολλά ακόμα , αλλά ας μη γίνω κουραστικός.Προτίθεμαι αν βοηθήσω από όποια θέση μπορώ.
 ΜΠΑΜΠΗΣ[/quote]

O kos Lamprou vrisketai auti ti stigmi stin Agglia ws episkeptis kathigitis se kapoio panepistimio. Ola auta mou ta anefere prin apo merikes imeres pou eixe katevei gia liges meres stin Kriti gia na taktopoiisei kapoies doulitses!

Tha epikoinwnisoume omws ilektronika kai tha tou prwwthisw oles tis erwtiseis kai tis apories sas. Ki egw den gnwrizw perissotera apo auta pou anefera... Thelei poli organwsi twra stin arxi kai tha doume pws tha voithisoume!!

Alexandros

Αλεξανδρε και Μπαμπη μπορουμε να βοηθησουμε και εμεις ?σε οτιδηποτε...

Αγαπητοί φίλοι, χαιρετίσματα.

Συγχαρητήρια για την δραστήρια  συμμετοχή σας στο Forum.

Ζητώ συγνώμη που δεν επικοινώνησα  νωρίτερα με το Forum, αλλά αυτό τον καιρό βρίσκομαι στο εξωτερικό και έχω φορτωμένο πρόγραμμα. Ο Αλέξανδρος πάντως επικοινωνεί μαζί μου, και έχω μελετήσει με προσοχή όλες σας τις προτάσεις.
Θέλω να σας ευχαριστήσω δημοσία όλους, ιδίως τον Αλέξανδρο και τον κ. Στεργίου για την ηθική τους στήριξη και τις ιδέες τους. Ευχαριστώ.

Ας πω μερικά λόγια για το Ελληνικό Καγκουρό:

Μπορεί να πάρει μέρος ο καθένας. Ελπίζουμε ότι αργά ή γρήγορα ο διαγωνισμός θα γίνεται στα σχολεία, για να μην μετακινούνται σε αγνώριμο περιβάλλον οι διαγωνιζόμενοι. Προς το παρόν, όμως, η γραφειοκρατία μας περιορίζει κάπως.
Αν κάποιο σχολείο δεν μπορεί να φιλοξενήσει τον διαγωνισμό, ή ο Καθηγητής του δεν επιθυμεί να τον στηρίξει, τότε οι μαθητές του εν λόγω σχολείου μπορούν να πάνε όπου αλλού θέλουν. Φυσικά θα τους κατατοπίσουμε που να πάνε, για να μην ψάχνουν μόνοι τους.
Σχεδιάζουμε τον διαγωνισμό για Σάββατο 31 Μαρτίου, αλλά αυτό δεν είναι ακόμη οριστικό. Όλοι θα ειδοποιηθούν εγκαίρως.

Η συμμετοχή στον διαγωνισμό είναι 10 ευρώ. Ας σημειώσω ότι τα χρήματα αυτά δεν επαρκούν για τα έξοδα του διαγωνισμού (τοπικούς οργανωτές, επιτηρητές, εκτύπωση θεμάτων, δώρα για όλους τους μαθητές, βραβεία για τους πρωτεύοντες και πολλά ακόμη). Ελπίζουμε ότι η διαφορά θα είναι μικρή. Θα καλυφθεί πάντως από δύο-τρία άτομα.
Ευελπιστούμε ότι στο μέλλον ο διαγωνισμός θα έχει αρκετή επιτυχία ώστε να βρεθούν επίσημοι χορηγοί. Ότι κέρδος υπάρχει, στο μέλλον, θα επενδύεται πίσω στους διαγωνιζόμενους ως βραβεία.

Πάντως, ο κάθε διαγωνιζόμενος θα παίρνει δώρα εμπορικής αξίας εφάμιλλης των 10 ευρώ που θα πληρώσει. Παραδείγματος χάριν σχεδιάζουμε να πάρει ο καθένας που θα συμμετέχει

1) ένα βιβλίο 96 σελίδων, τετραχρωμία και σχήματος, Α4, με α) τα θέματα όλων των τάξεων με συστηματικές λύσεις (όχι μόνο απαντήσεις), β) πολλά άρθρα με μαθηματικό περιεχόμενο για όλες τις τάξεις, γ) άρθρα πάνω στα διασκεδαστικά μαθηματικά, δ) γρίφους και μαθηματικά "μαγικά τρυκ",  και πολλά ακόμη.

2) ένα CD με μαθηματικά, παιχνίδια, φωτογραφίες, video και άλλα,

3) κατασκευή από χαρτόνι με κάποιο ενδιαφέρον μαθηματικό θέμα. Είναι δύσκολο να περιγράψω τι ακριβώς εννοώ, αλλά στέλνω ένα δείγμα στον Αλέξανδρο, για το αναρτήσει στην ιστοσελίδα του. Από εκεί θα μπορείτε να το κατεβάσετε ελεύθερα, για να πάρετε μια ιδέα. (Πρόκειται για τα λεγόμενα ξυλάκια του Genaille με τα οποία μπορεί κανείς να κάνει πολλαλασιασμούς χωρίς κόπο).

4) άλλα!

Τα βραβεία στους πρωτεύοντες είναι άλλη υπόθεση, και την περιέγραψε ο Αλέξανδρος. Άλλωστε ελπίζουμε το κάθε σχολείο χωριστά, ή η κάθε περιοχή χωριστά, να δίνουν (επιπλέον) τα δικά τους συμβολικά βραβεία.

Αυτά για την ώρα. Καλές γιορτές.

Φιλικά,
Μιχάλης Λάμπρου.

Κύριε Καθηγητά,
σας  ευχαριστούμε για αυτή την τόσο σπουδαία πρωτοβουλία που αφορά στα μαθηματικά και την εκπαίδευση γενικότερα. Ευχόμαστε όλα να πάνε καλά και η μαθηματική κοινότητα να βρει επιτέλους κάποια στιγμή τη δύναμη να αναλάβει τα ηνία και να βγάλει την παιδεία από το τέλμα.
  [color=red]Μερικές σκέψεις ακόμα :[/color]
  [b] α) [/b]Η προσπάθεια πρέπει να εστιαστεί στην αρχή στην ενημέρωση των μαθηματικών και κυρίως αυτών που είναι στα σχολεία , διότι αυτοί επηρρρεάζουν περισσότερους μαθητές.Πρέπει με κάθε τρόπο να καταλάβουν ότι πρόκειται για κάτι τελείως νέο και επαναστατικό για την Ελλάδα. Οι μαθηματικοί στα σχολεία είναι αρνητικοί στους διαγωνισμούς , διότι έχουν στο μυαλό τους τους διαγωνισμούς της ΕΜΕ , οι οποίοι είναι τελείως διαφορετικής φιλοσοφίας από τον ΚΑΝΚΟΥΡΩ.Σκεφτείτε ότι στα σχολεία , οι μαθηματικοί αποτρέπουν τους διευθυντές να ενημερώνουν τους μαθητές για τον ΘΑΛΗ πχ. Αυτό δείχνει το μέγεθος του προβλήματος.
[b]   β)[/b] Μαζί περίπου με την έγκριση του διαγωνισμού από το ΥΠΕΠΘ , πρέπει να σταλλεί  στα σχολεία και υλικό για ενημέρωση , αλλά και ένα τουλάχιστον φύλλο ερωτήσεων με απαντήσεις, ανά τάξη.Αυτό μπορούν οι μαθηματικοί να το φωτοτυπίσουν και να το δώσουν στους μαθητές ή να το κάνουν στην τάξη για εξάσκηση.
 [b]  γ)[/b] Δεν ξέρω σε πόσες φάσεις θα τελειώσει ο διαγωνισμός , αλλά νομίζω ότι οι 50 καλύτεροι μαζί με τους  50 καλύτερους του Αρχιμήδη πρέπει να αποτελέσουν τους μαθητές από τους οποίους μετά από 3 τεστ επιλογής θα επιλεγούν οι ομάδες για την εκπροσώπιση της Ελλάδας στις διεθνείς υποχρεώσεις. Με άλλα λόγια ο διαγωνισμός ΚΑΓΝΚΟΥΡΩ , αν και διαφορετικός , δεν πρέπει να καθιερωθεί ως διαγωνισμός β' κατηγορίας , μετά δηλαδή από τους αντίστοιχους της ΕΜΕ. Ο Καγκουρώ πρέπει και αυτός να οδηγεί τελικά στις ΙΜΟ. Θα υπάρξουν ίσως κάποιες αντιδράσεις στην αρχή , αλλά όλα θα ξεπεραστούν.Προφανώς , ο διαγωνισμός θέλει και στήριξη θεσμοθετημένου φορέα , ώστε το Υπουργείο να καταστήσει με διάταγμα  ισάξους τους  δύο διαγωνισμούς ( ΘΑΛΗΣ κλπ - Καγκουρώ ).Με αυτόν τον τρόπο θα συμμετέχουν οι ταλλαντούχοι μαθητές  σε όλους τους διαγωνισμούς.Βέβαια , αυτό θα γίνει αφού πρώτα ο διαγωνισμός βρει το βηματισμό του και αποκτηθεί και η σχετική εμπειρία.
  Αυτά για τώρα. Ελπίζω να τα πούμε και από κοντά.
                      ΚΑΛΑ ΧΡΙΣΤΟΥΓΕΝΝΑ !!!

Kalispera se olous,

Anevasa to arxeio pou mou esteile o kos Lambrou stin gnwsti selida sto section me tous diagwnismous. Sintoma exw sto mialo mou na ftiaksw ena ksexwristo section gia mikra arthra pou aforoun ena poli sigekrimeno kommati twn mathimatikwn, kathws episis ena section me spazokefalies, trik kai diafora "magika" pragmata twn mathimatikwn mesa sta opoia perilamvanetai kai auto pou mou esteile o kos Lambrou simera.

Filika,
Alexandros

Αλέξανδρε τι θα γίνει τελικά με αυτό το διαγωνισμό ? :maybe:

Silouane o diagwnismos ap'oti kserw, tha ginei to Savvato 31 Martiou. Perissoteres leptomereies den gnwrizw dioti o M. Lambrou eleipe stin Kipro ws proedros tis epitropis gia ton 1o diethni diagwnismo gia foitites (O opoios "off the record" tou xronou mallon tha ginei sto Panepistimio Kritis). Mou eipe oti twra pou teleiwse ekei o diagwnismos tha organwsei ton diagwnismo edw o opoios arxika den tha exei poli simmetoxi logw tis kathisterisis kai to oti arketoi den to gnwrizoun. Siga siga omws me ta xronia pisteuw tha arxisei na anaptisetai...

Alexandros

Αλέξανδρε ευχαριστώ για την πληροφορία .Πέφτει μαζί με το Μεσογειακό όμως..... :!:  . Εκτός αν γράψουμε πρώτα Μεσογειακό και μετά Kangaroo... :wink:

Molis enimerwthika oti o diagwnismos tha dienergithei stis 31 Martiou imera Savvato kai wra 9 to prwi. Mporeite na enimerwtheite apo tin episimi selida tou diagwnismou stin Ellada :

http://www.kangaroo.gr

Stin istoselida tha mathete arketes plirofories gia ton epikeimeno diagwnismo. Kalo tha itan na peisete ton mathimatiko sas na dieksagei ton diagwnismo sto sxoleio sas. Den tou kostizei tipote i diorganwsi (nomizw oti tha plirwthei kiolas, 1,5 eurw kata simmetoxi mathiti) kai i apostoli twn thematwn tha ginei me fax 'h email opote den tha xreiastei na metakinitheite.

Deite posoi simmetexoun kathe xrono apo diafores xwres tou kosmou... Aksizei na simmetasxete se mia nea prospatheia kathierwsis tou diagwnismou stin Ellada.

Episis aksizei na deite posa dwra dinontai se kathe diagwnizomeno aneksartita apo to an grapsei 'h oxi kala stin eksetasi. Exei entelws diaforetiki filosofia apo tous gnwstous mas diagwnismous... Enimerwste olous tous simmathites sas kai tous kathigites sas gia auto. Tha deite oles autes tis plirofories stin parapanw istoselida.

Ean tha pate stin katigoria "Deigma Thematwn" mporeite na deite perissotera themata sta Ellinika. Ean episis mpeite kai stin episimi istoselida

http://www.mathkangaroo.org

tha vreite ola ta palaiotera themata tou diagwnismou sta Agglika.

Telos einai o monadikos diagwnismos stin Ellada (pera apo tous diagwnismous pou dieksagoun me diki tous prwtovoulia ta diafora parartimata tis EME) pou afora mathites Dimotikou sxoleiou.

Ean telika den mporesete na peisete ton mathimatiko sas na dieksagei ton diagwnismo, mporeite na pate se OPOIODIPOTE sxoleio pou dienergei ton diagwnismo se opoiodipote meros. Gi'auto as grapsoume edw opoiodipote meros pou kseroume oti tha dienergithei o diagwnismos gia na dieukolinthoun ta paidia pou endiaferontai. Sigoura stin Xalkida apo ton ko Stergiou fantazomai, sto Hrakleio apo ton ko Lambrou, sti Thes/niki apo tin ka Mitsiou kai vlepoume... Tha enimerwsoume sxetika gia ta meri pou tha dieksaxthei me sigouria oso pio sintoma mporoume gia na mporesete na pate sto kontinotero meros.

[color=red]KALH ARXH[/color]

Filika,
Alexandros

Y.G. H kathisterisi ofeilotan kiriws se grafeiokratia. Twra pisteuw oti ola tha einai OK kai xrono me ton xrono oi simmetexontes tha auksanontai.

Ωραία όλα αυτά αλλά η ημέρα διεκσαγωγής δεν προσέχθηκε καλά .Όπως ανέφερα και παραπάνω την ίδια ώρα και μέρα θα γίνεται το ΤΕΣΤ επιλογής -Μεσογειάδα για τη χώρα μας .
Είναι κρίμα νομίζω εκείνοι που θα γράφουμε εκείνη τη μέρα το Μεσογειακό να χάσουμε τον άλλο διαγωνισμό.
Δεν μπορεί να γίνει τίποτα για αυτό ?
Εκτός αν τα θέματα του Καngaroo χρησιμοποιηθούν ως τεστ επιλογής  :D

H νοοτροπία του διαγωνισμού όσον αφορά στην επιλογή θεμάτων είναι η ίδια με αυτή στους διαγωνισμούς της ΕΜΕ ή των Ολυμπιάδων? Γενικά είναι δυσκολότερος διαγωνισμός από τον Αρχιμήδη, για παράδειγμα? Επίσης είναι σε φάσεις ή μία κι έξω?

Paidia distixws den mporei na ginei allagi tis imerominias dieksagwgis tou diagwnismou Kangaroo, kathws einai epivevlimeno na ginei kai stis 41 xwres pou simmetexoun ston diagwnismo mexri to telos martiou.

Episis ta themata tou diagwnismou den exoun tin idia filosofia me ta themata tis EME kai kat'epektasin den mporoun na siggrithoun. Omws exei apodeixthei ta 10+ xronia pou dieksagetai, oti ta idia atoma pou simmetexoun stous diethneis diagwnismous mathimatikwn einai kai ekeinoi pou pianoun tis prwtes theseis ston diagwnismo Kangaroo.

H fisi tou diagwnismou kangaroo (opws leei kai stin episimi istoselida parapanw) exei na kanei me 30 erwtimata pollaplwn epilogwn (28 gia to prwto epipedo) ta opoia exoun mia diavathmisi diskolias. Ta prwta 8-10 einai (sxetika) eukola kai apo ekei kai pera exei klimakoumeni diskolia. Opote genika einai diskolo mesa sto xroniko perithwrio tis 1,5 wras tou diagwnismou na lithoun ola ta themata teleia. Omws exei mia simantiki diafora apo ta themata tis EME. Kaneis den feugei disarestimenos psixologika alla kai ilika. Psixologika dioti apokleietai na min mporesei na apantisei se kapoies erwtiseis (estw kai liges - auto einai simantiko gia ton simmetexonta) kai deuteron tha figei me arketa dwra (twn opoiwn - kalo einai na tonistei - h oikonomiki aksia einai megaliteri apo ta 10 eurw pou iparxei ws simmetoxi mono kai mono gia ta eksoda tou diagwnismou) alla kai me vevaiwsi simmetoxis ston diagwnismo. Auto kanei ton diagwnismo Kangaroo poli prosito se olous tous mathites se sxesi me tous diagwnismous tis EME stous opoious kapoios pou den exei proaitoimastei katallila isws na apogoiteutei. Vevaia exei kai ta thetika tou kai den mporei kaneis na to amfisvitisei. Outws 'h allws prokeitai gia ena diagwnismo ston opoio pagosmiws simmetexoun para polloi mathites kai gia tin Ellada einai POLI simantiko gegonos oti iparxei kai deuteros diagwnismos mathimatikwn. Aksizei loipon na ton ipostiriksoume...

Oson afora ti mesogeiada, distixws den mporei na ginei kati. tou xronou isws na einai pio organwmena ta pragmata kai oi dio foreis dieksagwgis twn diagwnismwn (EME kai Kangaroo) na sinenoithoun metaksi tous gia na min simpesoun metaksi tous.

Filika,
Alexandros

Ευχαριστώ cretanman. Είναι όντως κρίμα που συμπίπτει με τη Μεσογειάδα, αλλά νομίζω ότι είναι μεγαλύτερη προτεραιότητα τώρα η Μεσογειάδα. :wink:

Μπες στο site που δίνει ο cretanman. Εξαρτάται από τα ίδια τα σχολεία να στείλουν καταστάσεις με τα ονόματα των μαθητών που θέλουν να συμμετέχουν. Καλή επιτυχία. :wink:

[quote="DenPoNaEi"]opios gnorizei ama 8a uparksei sxoleio sthn kozanh   opou 8a  ginei o diagonismos as me enhmerosei parakalo polu .  :)[/quote]

Agapite file,

To diaforetiko kai poli asinithisto me auto ton diagwnismo einai oti mporei na diorganwthei se opoiodipote sxoleio arkei o kathigitis 'h o dieuthintis tou sxoleiou na theloun na erthoun gia 1,5 wra to Savvato to prwi (oso diladi i diarkeia tou diagwnismou). Opote ean ton peiseis na dieksagei ton diagwnismo sto sxoleio sou, mporei na simmetasxeis esi kathws episis kai opoiosdipote apo opoiodipote sxoleio tis Kozanis ('h ekei girw). H alli epilogi pou iparxei einai na pas se kapoio allo kontino sxoleio sto opoio tha dieksaxthei o diagwnismos. Akoma kai 1 atomo na simmetasxei mporei o diorganwtis na steilei ta themata eite ilektronika eite me courier eite me fax. Einai poli eueliktos o tropos dieksagwgis. Arkei loipon to sxoleio sou na dilwsei simmetoxi sta tilefwna pou dinontai stin episimi selida tou diagwnismou stin ellada http://www.kangaroo.gr

Auta...

Alexandros

Katarxin kali epitixia se olous sti Mesogeiada! Gia osous den lamvanoun meros se auti na anaferw oti o diagwnismos KANGAROO tha dieksaxthei SIGOURA to Savvato 31/3/2007 stis 9 to prwi stis poleis:

THESSALONIKH

ATHINA

HRAKLEIO KRHTHS

Opoios endiaferetai na simmetasxei mporei na mpei stin episimi selida tou diagwnismou

http://www.kangaroo.gr

kai na mathei perissoteres plirofories gia to meros dieksagwgis tou diagwnismou. Fisika mporoun na simmetasxoun kai oloi osoi mporoun na metakinithoun se autes tis poleis. Kalitera omws na parete tilefwno sta tilefwna pou iparxoun sti parapanw istoselida wste na mathete perissotera apo tous ipeuthinous tou diagwnismou...

Kali arxi...

Alexandros

Τι έγινε με το καγκουρό. Έδωσε κανένας να μας πει? :?:

[quote] Έδωσα εγώ αλλά Κύπρο, με λίγο διαφορετικούς κανονές νομίζω (1 ώρα αντί 1,5). Τα θέματα ήταν καλά αν και ξεμείναμε από χρόνο. Έδωσε κάποιος άλλος;[/quote]

Bellerofront, μπορείς να κάνεις upload τα θέματα σας;

Θα προσπαθήσω όταν βρω χρόνο...

Αν και νομίζω ότι σε Ελλάδα και Κύπρο ήταν τα ίδια τ θέματα...

Μπορείς να μου πεις ενδυκτικά ποιες είναι οι πρώτες ερωτήσεις του  υψηλότερου επιπέδου στην Ελλάδα για να δουμε αν είναι πράγματι έτσι;

Παιδιά βγήκαν τα αποτυελέσματα του Καγκουρώ!!!!!!!!!!!!!!!!!! [hide="click text"]δείτε το site : http://www.kangaroo.gr[/hide]

Ομολογουμένως δεν ήταν και πολύ δύσκολος διαγωνισμός. :wink:

το πρόβλημα δεν ήταν ότι ήταν δύσκολος αλλά ότι είχαμε μόνο μιαμιση ώρα για 30 θέματα. Εγώ προσωπικά
δεν τα πρόλαβα παντως :blush:

Τα έλυσα κι εγώ από το σπίτι βέβαια με ηρεμία. :P  Δεν ξέρω τι τάξη είσαι αλλά στην Α Λυκείου τα έλυσα μέσα σε μία ώρα και ένα τέταρτο και είχα δύο λάθη απροσεξίας (28). Πιστεύω αν χρησιμοποιούσα αυτό το τέταρτο θα είχα πιάσει 29 μπορεί και 30 :wink: . Αυτοί οι διαγωνισμοί όπως και ο AMC (o Θαλής των Αμερικάνων φυσικά πολύ πιο δύσκολος) χρειάζονται ορισμένες τεχνικές. :wink:

Βασικά ειμαι Β Λυκείου και τα θέματα ήταν καπως πιο δύσκολα...... :ninja:
Πάντα βέβαια είναι αλλιώς να τα βλέπεις απο το σπίτι με ηρεμία..... :)
Σίγουρα πάντως τα πιο δύσκολα από αυτά δεν ξεπερνούσαν σε επιπεδο
ούτε τον Ευκλείδη. :wink:


inequalities
combinatorics unsolved
combinatorics
algebra

There is a row that consists of digits from $ 0$ to $ 9$ and Ukrainian letters (there are $ 33$ of them) with following properties: there aren’t two distinct digits or letters $ a_i$, $ a_j$ such that $ a_i > a_j$ and $ i < j$ (if $ a_i$, $ a_j$ are letters $ a_i > a_j$ means that $ a_i$ has greater then $ a_j$ position in alphabet) and there aren’t two equal consecutive symbols or two equal symbols having exactly one symbol between them. Find the greatest possible number of symbols in such row.

i find with an ugly way that the answer is 73 am i right?

anybody solved this problem?

Yes, the answer is 73.
I prove it as follows:

Consider some sequence (not necessarily of maximal length). When looking just at its digits (or at its letters), we obtain non-decreasing sequences.

Let $b_i$ be the number of occurrences of the digit $i$ in the sequence, as well as $c_{i,j}$ the number of letters written between the occurrences $j$ and $j+1$ of the letter $i$, for $j < b_i$. W.l.o.g., we can suppose that $b_i \geq 1$, and we must have $c_{i,j} \geq 2$. Moreover, let $c_{-\infty}$ be the number of letters written before the first occurrence of $0$, and $c_\infty$ the number of letters written after the last occurrence of $9$. We get $32 \geq c_{-\infty} + \sum_{i=0}^9 (-1+\sum_{j=1}^{b_i-1}c_{i,j}) + c_{\infty}$, i.e. $42 \geq c_{-\infty} + \sum_{i=0}^9 \sum_{j=1}^{b_i-1}c_{i,j} + c_{\infty}$, and the sequence has a length $L = c_{-\infty} + \sum_{i=0}^9 (b_i + \sum_{j=1}^{b_i-1}c_{i,j}) + c_\infty \leq 42 + \sum_{i=0}^9 b_i$. Moreover, $42 \geq c_{-\infty} + \sum_{i=0}^9 \sum_{j=1}^{b_i-1}c_{i,j} + c_{\infty} \geq 0 + \sum_{i=0}^9 \sum_{j=1}^{b_i-1}2 + 0 = 2 \sum_{i=0}^9 b_i - 20$, and therefore $ \sum_{i=0}^9 b_i \leq \frac{42+20}{2} = 31$, so that $L \leq 42 + 31 = 73$.

Conversely, the sequence 0АБ01БВ12ВГ23ГҐ34ҐД45ДЕ56ЕЄ67ЄЖ78ЖЗ89ЗИ9ІЇ9ЙК9ЛМ9НО9ПР9СТ9УФ9ХЦ9ЧШ9ЩЬ9ЮЯ9 is of length 73. (Of course, I used equality cases in the above inequalities to find such an example)

thanks for your solution i do the same but had a little more cases

there is something I didn't get...why did you use $ 32 $ in your inequality instead of $ 33 $?  :maybe:

When counting the repetitions, you are counting one too many, so you subtract 1 from 33


modular arithmetic
induction
number theory unsolved
number theory

If natural numbers $ x$, $ y$, $ p$, $ n$, $ k$ with $ n > 1$ odd and $ p$ an odd prime satisfy $ x^n \plus{} y^n \equal{} p^k$, prove that $ n$ is a power of $ p$.

first solution: by Zsigmondy theorem, $ x^n \plus{} y^n$ has at least 2 different prime factors (a prime factor of $ x \plus{} y$, and a primitive prime divisor of $ x^n \plus{} y^n$) - a contradiction, so the problem is trivial.

second solution: by the identity $ (a \plus{} p^kl)^{p^s t} \minus{} a^{p^s t} \equiv alt p^{k \plus{} s} \pmod {p^{k \plus{} s \plus{} 1}}$ for $ k > 0, s \ge 0$ (can be proved by induction on $ s$), we deduce the "Lifting the Exponent" lemma:  $ ord_p{x^n \plus{} y^n} \equal{} ord_p{(x \plus{} y)} \plus{} ord_p{n}$ (for odd n, when x+y is divisibly by p). x+y must be divisible by p in our case (because $ x \plus{} y | x^n \plus{} y^n \equal{} p^k$ ), so we conclude: $ k \equal{} ord_p{(x \plus{} y)} \plus{} ord_p{n}$. Let $ m \equal{} ord_p{n}$. We have, by the same lemma, $ p^k | x^{p^m} \plus{} y^{p^m}$. We must have $ n \equal{} p^m$, else:
$ x^n \plus{} y^n > x^{p^m} \plus{} y^{p^m} \ge p^k$, a contradiction.

[Can someone please post a proof of Zsigmondy's Theorem?]

[quote="bambaman"]Can someone please post a proof of Zsigmondy's Theorem?[/quote]
Sure :) http://www.ams.org/proc/1997-125-07/S0002-9939-97-03981-6/S0002-9939-97-03981-6.pdf Theorem 3.


Hi guys!
I have some patterns I'm confused with. I can't seem to find them... Please help! I would also like an explanation to go with it since I'm supposed to under stand it
1- J, F, M, A, M, J, J, A, S, __, __, __
2- S, J, I, U, X, N, T, D, E, R, E, __, __ ,__
Thanks! :D

I figured that one out  :blush:
Meh, its the other one.
Does I found something to do with As and Zs. I think that the X has to do with something perhaps? I'm thinking about it that way because its like an outlier.


inequalities
induction
number theory solved
number theory

Here is a nice problem that I found, but its just an adaptation  after a well-known property. Anyway, I like it. It says that if the perfect number n has k positive divisors, then $ n<2^{2^k} $.

I remember that i read the paper about it it is not olympiadic problem.i will find it.but i distinctly remember that if N be odd perfect number and k be thenumber of distinct prime factor then $N<2^{4^k}$

Well, problems only arise when $n$ is odd, because when it's even it must be of the form $2^{p-1}(2^p-1)$, where $2^p-1$ is a Mersenne prime, so the inequality becomes really weak: $n$ has exactly $2p$ divisors, and it's obvious that $n<2^{2p}$ (not to mention $2^{2^{2p}}$).

I say it's quite an olympiad problem. I know the problem you read about, samn. That one is not an olympiad problem, though a very related problem was given as an inequality in a Chinese TST. And to grobber: what about n odd?

[quote="harazi"] the famous egiptean theorem of Erdos: if $ 1/x_{1}+...+1/x_{k}<1$ then $ 1/x_{1}+...+1/x_{k}\leq 1/a_{1}+...+1/a_{k}$, where $ a_{1}=2$ and $ a_{n+1}=a_{n}^{2}-a_{n}+1$. [/quote]
That is http://www.mathlinks.ro/Forum/viewtopic.php?p=234089#234089  :wink:

Sorry ; but I think the upbound is too weak

    We can strengthened it to $ 2^{2^{k-2}-1}$ ( I think ) or at least $  2^{2^{k-1}-1}$ ( I have proved)


Frequently Asked Question # 4
http://www.unl.edu/amc/f-miscellaneous/faq.shtml

Steve Dunbar
MAA Director, American Mathematics Competitions

[quote="AMC site"]Check our web pages for a list of participating colleges. [/quote]

I'm having trouble finding the site with the list.  Could someone post the URL there?

[quote="jmerry"]If your school offers it, take it there. If your school doesn't, try to convince them. If you can't, look for somewhere on [url=http://www.unl.edu/amc/b-registration/b1-archive/2008-2009/CU2009/2009-CU-list.shtml#CU]this list[/url].[/quote]

It's right here alex.


What do you guys think?
In Belgium we have Azerty.  At my university of Ghent, in order to keep international profile, everything is Qwerty.  Easy for foreign students, difficult for me (I am no hero in informatics) as I prepare certain exams at home in Azerty, but then have to do them in Qwerty at Uni.

My personal experience is that for a mathematician, Azerty is actually better.  Once in a while, you need to write () for instance,.  In Azerty ( is left above the t, along with the five key, but it's for the five that you need to use caps.
In qwerty it's the other way around, caps for the (, no caps for the number.  The result is when using qwerty for me I constantly get  x = 9x+1 0^2 and stuff like that.  It makes no sense to me to get an abundant easy nocaps numerical zone, and difficult to reach ( and )

So this is something I find strange
What do you thinK? :oops:

Well, in AZERTY at least the semicolon isn't on the home row, so it's better than QWERTY, maybe.
I like to use Dvorak because the letters aren't just laid out at random.

I'm too used to qwerty to try anything else, lol

I use QWERTY, but I would probably like more efficient diesigns that allow for faster typing better (I have never used anything else).

It would take me a long time to switch from QWERTY (I get closer to 150 than 100 WPM)

It seems many people indeed haven't heard yet of Azerty

Of course, why would you use that if Qwerty is universal?

It just bugs me that at our university we adapt so much to foreign standards (if they think highly enough of us to adapt in a matter of weeks to Qwerty, why can't they the opposite with foreigners?)
while I think Azerty is actually better for mathematicians!

But let me give this topic another boost by bringing up another element.  Netherlands has Qwerty, France has Azerty, because Azerty is a French style keyboard.  Because the southern half of Belgium, bordering France, is French speaking, all of Belgium is Azerty.
Although I live in the northern Dutch speaking part bordering the Netherlands, my name Frédéric has a French spelling (although a typical Flemish way would be Frederik)
It's much more difficult to do those accents on Qwerty (I even know people asking me how I did that accent when they replied in email).  You may say big deal, write it as you like.  But!  In official documents you need to spell your name completely correct.  My sister has an accent too and she has been in serious trouble at the same university because her work was submitted in a different spelling than in other records...

Well, my point is not really that all of the universe should use azerty, I just find it interesting to debate it with others

[quote="fredbel6"]It just bugs me that at our university we adapt so much to foreign standards (if they think highly enough of us to adapt in a matter of weeks to Qwerty, why can't they the opposite with foreigners?)
while I think Azerty is actually better for mathematicians![/quote]

Do you mean that Azerty is better for French or English mathematicians? I've never heard of it--though my computer does support Dvorak and it's [b]really[/b] weird. But back to Azerty: English speakers have no use for a keyboard that has accents and a pound key, not to mention... what is this... mu???

They can't do the opposite with foreigners for the same reasons that they can't teach French to everybody. Qwerty is the universal keyboard layout and is arranged to be able to write English as quickly and with as little hand movement as possible.

Also, I think I use numbers just as much as parenthesis. I wouldn't want to hit shift every time I needed a number (I guess you would use caps lock for math--again, we aren't used to this).

[quote="jb05"]They can't do the opposite with foreigners for the same reasons that they can't teach French to everybody. Qwerty is the universal keyboard layout and is arranged to be able to write English as quickly and with as little hand movement as possible.[/quote]

Actually, Qwerty is designed to be as inefficient as possible.  It was designed in the days when typewriter keys still got stuck when they were pressed too quickly, and so they needed to limit the speed of peoples' typing.

Yes, it was designed to keep keys from sticking, but it was certainly not designed to be inefficient. It was designed to combine the above with efficiency.

[url]http://en.wikipedia.org/wiki/Qwerty[/url]
"Frequently-used pairs of letters were separated.... QWERTY also attempted to alternate keys between hands, allowing one hand to move into position while the other hand strikes home a key. This sped up both the original double-handed hunt-and-peck technique and the later touch typing technique;..."

Please help me, my troubles have gone from bad to worse

Now I find myself experiencing trouble in replying in some threads, because I should write in qwerty
so if i type blindly , i make mistakes

It appears to be random?  Like right now it is no problem but in superior algebra a few minutes ago it still was?

Well, I've QWERTZ ($\neq$ qwerty) here... and I find them OK.
But spanish (mexican¿) ones are good too, you can make some math signs without shift etc.

Uhm, but the problem I am reporting now, is more annoying than usual
I have a decent azerty keyboard but mathlinks misinterprets it? :huh:

[quote="Arne"]I am using the signs \ { and } a lot (LaTeX) and I have to press the Alt button (on my azerty keyboard) the whole time to type those, which is pretty annoying. Is this also the case on a qwerty keyboard?[/quote]
No, in qwerty, { and } use Shift, and \ doesn't need any special key.

[quote="Arne"]I am using the signs \ { and } a lot (LaTeX) and I have to press the Alt button (on my azerty keyboard) the whole time to type those, which is pretty annoying. Is this also the case on a qwerty keyboard?[/quote]No, this is one of the reasons I like qwerty, they're also used a lot in programming.

The signs [,],\,| are easier to type. I am using qwerty at home and in university, it is much handier in every way. Too bad, our university seems to switch slowly to azerty. (at least that's what I think, 2 new computer classes have azerty... I hope it's just an accidental mistake.)

[quote="ZennyK"]Who decided where to put each of the letters on a QWERTY keyboard in the first place?[/quote]Christopher Sholes. See [url=http://en.wikipedia.org/wiki/QWERTY]this[/url].


linear algebra
matrix
linear algebra theorems

Prove that the reduced row echelon form of a matrix is unique.

Different proofs are given for this theorem and some of which are available on the web. I normally use an inductive approach due to Thomas Yuster in my class, which, in my opinion, is one of the simplest proofs at hand.

The Reduced Row Echelon Form of a Matrix is Unique: A Simple Proof
Thomas Yuster
Mathematics Magazine, Vol. 57, No. 2 (Mar., 1984), pp. 93-94
A sample proof on the web: [url]http://www.cs.uleth.ca/~holzmann/notes/reduceduniq.pdf[/url]


Here is a problem that I am getting stuck at.

Two circles with centre X and Y intersect each other at the points A,B.The point A is joined with S the mid-point of XY.The perpendiculars to SA through the point A intersects the circles at the points P and Q .Prove that PA=AQ.

I tried this problem , but I do not understand what the problem is.

My idea:

[hide]I tried to prove that P and Q lie on the same circle, or that PS=SQ.[/hide]

Thank you.

Well....My English is so poor....I hope U can understand what I mean.
And my solution given to U is just part of my thought. It's not excellent  at all... :(

Now, let's begin.
According to the question, we already know that XS=SY because S is the mid-point of XY.
[color=orange]Then if u draw 2 lines through the point X and Y which are seperately perpendicular to PQ.[/color]
Therefore, we know 3 perpendiculars which is XC, SA and YD.
Thusly, we can get that [color=red]2AS=XC+YD[/color]
Afterwards, it's easy to know that AC=AD.
Next step is that we can understand that PC=AC and AY=YD according to the properties of perpendiculars.

Finally, we can conclude that AP=AQ.

What do U think of my explanation? Pretty bad, right?    :oops:

draw $ XL\perp AP$ and $ YM\perp AQ$
then, $ AL \equal{} \frac{PA}{2}$ and $ AM \equal{} \frac {AQ}{2}$
also, $ LX\parallel AS\parallel MY$
now since $ XS \equal{} SY$. it  implies on transversal $ LM$ that $ A$ is mp. of $ LM$.
so $ LA \equal{} AM$
thus, $ PA \equal{} AQ$
and by congruency $ PS \equal{} SQ$.


vector
calculus
derivative
geometry
geometric transformation
reflection
inequalities

An ant is traveling from the point $(8, 15)$ to the point $(5, 12)$, but it wants to touch both axes along the way. Let $d$ be the minimum distance it has to travel. Find $d^{2}$.

If I understand your problem.

the ant need to touch the x axis and the y axis, in points differents or the same, or best, the more fast way

${\text{A1}=\{8,15\}}$
${\text{A2}=\{5,12\}}$
${\text{B1}=\{a,0\}}$
${\text{B2}=\{0,b\}}$

Because $A2$ is more near to $B2$ of all vectors

${d=\|\text{B1}-\text{A1}\|+\|\text{B2}-\text{B1}\|+\|\text{A2}-\text{B2}\|}$
${d=\sqrt{(b-12)^{2}+25}+\sqrt{(a-8)^{2}+225}+\sqrt{a^{2}+b^{2}}}$

We need to take the partial derivatives of $d$ and equal it to zero. Let's just make a thing

${\frac{\partial \left(\sqrt{(b-12)^{2}+25}+\sqrt{(a-8)^{2}+225}+\sqrt{a^{2}+b^{2}}\right)}{\partial a}=0}$
${a=\frac{8 b}{b\pm 15}}$

${\frac{\partial \left(\sqrt{(b-12)^{2}+25}+\sqrt{(a-8)^{2}+225}+\sqrt{a^{2}+b^{2}}\right)}{\partial b}=0}$
${b=\frac{12 a}{a\pm 5}}$

The solution of this is just when

${(a=0\land b=0)\lor \left(a=\frac{7}{9}\land b=\frac{21}{13}\right)\lor \left(a\text{=}-57\land b\text{=}\frac{171}{13}\right)\lor (a=-7,b=7)\lor \left(a\text{=}\frac{19}{3}\land b\text{=}57\right)}$

Testing this values the best is

${\left(a=\frac{7}{9}\land b=\frac{21}{13}\right)}$

${d=\sqrt{898}}$
${d^{2}=898}$

i'm wrong?

if i'm wrong, what is the question, can you remake?

[hide]
reflect the point (5,12) across the origin to (-5,-12). Any path from (8,15) to (-5,-12) is the same distance as the path described in the problem, since the ant would pass through both axes on the way to (-5,-12). The shortest distance to (-5,-12) is $\sqrt{(8+5)^{2}+(15+12)^{2}}$

So the answer is $(8+5)^{2}+(15+12)^{2}=\boxed{898}$
[/hide]

[quote="rnwang2"]This problem reminds me of one that's in ACoPS.  :wink:[/quote]

What is ACoPS???

[quote="mathgeniuse^ln(x)"]What is ACoPS???[/quote]
[i]The Art and Craft of Problem Solving[/i]...  :)

[quote="Thales418"]
${d=\sqrt{(b-12)^{2}+25}+\sqrt{(a-8)^{2}+225}+\sqrt{a^{2}+b^{2}}}$

We need to take the partial derivatives of $d$ and equal it to zero. [/quote]

Actually, at this point with the reasonable assumptions $b \le 12, a \le 8$, we can use [hide="magic"] Minkowski's Inequality, which gives the same result as tjhance. [/hide]


trigonometry geometry inradius Euler circumcircle

Prove that in any triangle

1) $ \tan{\alpha\over 2}\tan{\beta\over 2}\tan{\gamma\over 2} \equal{} {r\over s}$

2) $ {\varrho_a\varrho_b\varrho_c\over r} \equal{} s^2$

where $ \alpha,\beta,\gamma$ are the internal angles, $ \varrho_a,\varrho_b,\varrho_c$ are the exradii, $ r$ is the inradius and $ s$ is the semiperimeter.

hello, we have
$ \tan(\frac{\alpha}{2})\equal{}\frac{r}{s\minus{}a}$
$ \tan(\frac{\beta}{2})\equal{}\frac{r}{s\minus{}b}$
$ \tan(\frac{\gamma}{2})\equal{}\frac{r}{s\minus{}c}$
from here we get
$ \tan(\frac{\alpha}{2})\cdot\tan(\frac{\beta}{2})\cdot\tan(\frac{\gamma}{2})\equal{}\frac{r^3s}{(s\minus{}a)(s\minus{}b)(s\minus{}c)s}\equal{}\frac{r^3s}{A^2}\equal{}\frac{r^3s}{s^2r^2}\equal{}\frac{r}{s}$ with $ s\equal{}\frac{a\plus{}b\plus{}c}{2}$.
Sonnhard.

hello, we have
$ \rho_a\equal{}\frac{A}{s\minus{}a}$
$ \rho_b\equal{}\frac{A}{s\minus{}b}$
$ \rho_c\equal{}\frac{A}{s\minus{}c}$
and from here we get $ \rho_a\rho_b\rho_c\equal{}\frac{A^3s}{s(s\minus{}a)(s\minus{}b)(s\minus{}c)}\equal{}\frac{A^3s}{A^2}\equal{}As\equal{}rs^2$.
Sonnhard.

In my opinion, this topic is better at the setion "Intermediate topics".

I add two identities and I wish more proofs of them :

$ 1\blacktriangleright\ ab\plus{}bc\plus{}ca\equal{}s^2\plus{}r^2\plus{}4Rr$ .

$ 2\blacktriangleright\ r_a\plus{}r_b\plus{}r_c\equal{}4R\plus{}r$ .

[quote="Virgil Nicula"]I add two identities and I wish more proofs of them :

$ 1\blacktriangleright\ ab + bc + ca = s^2 + r^2 + 4Rr$ .

$ 2\blacktriangleright\ r_a + r_b + r_c = 4R + r$ .[/quote]

Note: $ P$ denotes the area of the triangle.

[b]Identity 1.[/b]

\begin{eqnarray*}r^2s & = & {P^2\over s} \\
& = & (s - a)(s - b)(s - c) \\
& = & s^3 - (a + b + c)s^2 + (ab + bc + ca)s - abc \\
& = & s^3 - 2s^3 + (ab + bc + ca)s - abc \\
& = & - s^3 + (ab + bc + ca)s - abc\end{eqnarray*}

From there, $ ab + bc + ca = {s^3 + r^2s + abc\over s} = s^2 + r^2 + {abc\over s}$, but $ {abc\over s} = {4RP\over s} = 4Rr$, so we finally get $ ab + bc + ca = s^2 + r^2 + 4Rr$

[b]Identity 2.[/b]

From $ \varrho_a = {P\over s - a},\varrho_b = {P\over s - b},\varrho_c = {P\over s - c}$ we get

\begin{eqnarray*}\varrho_a + \varrho_b + \varrho_c & = & P\cdot\frac {(s - a)(s - b) + (s - b)(s - c) + (s - c)(s - a)}{(s - a)(s - b)(s - c)} \\
& = & P\cdot\frac {3s^2 - (2a + 2b + 2c)s + ab + bc + ca}{{P^2\over s}} \\
& = & P\cdot\frac {3s^2 - 4s^2 + ab + bc + ca}{Pr} \\
& \stackrel{\mathrm{Identity\,1}}{ = } & {-s^2 + s^2 + r^2 + 4Rr\over r} \\
& = & r + 4R\end{eqnarray*}

Another linear-angled identities in a triangle $ ABC$ (standard notations).

$ 1\blacktriangleright\ \cos A\plus{}(\cos B\plus{}\cos C)\equal{}$ $ 1\minus{}2\sin^2\frac A2\plus{}2\cos\frac {B\plus{}C}{2}\cos\frac {B\minus{}C}{2}\equal{}1\plus{}2\sin\frac A2\left(\cos \frac {B\minus{}C}{2}\minus{}\cos\frac {B\plus{}C}{2}\right)\equal{}$

$ 1\plus{}4\sin\frac A2\sin\frac B2\sin\frac C2\equal{}$ $ 1\plus{}4\cdot\sqrt {\frac {(s\minus{}b)(s\minus{}c)}{bc}\cdot\frac {(s\minus{}c)(s\minus{}a)}{ca}\cdot\frac {(s\minus{}a)(s\minus{}b)}{ab}}\equal{}$

$ 1\plus{}\frac {4(s\minus{}a)(s\minus{}b)(s\minus{}c)}{abc}\equal{}$ $ 1\plus{}\frac {4sr^2}{4Rrs}\equal{}$ $ 1\plus{}\frac rR$ $ \implies$ $ \boxed {\ \cos A\plus{}\cos B\plus{}\cos C\equal{}1\plus{}\frac rR\ }\ (1)$ .

$ 2\blacktriangleright\ \sum a(s\minus{}a)\equal{}2Rr\cdot \sum \sin A\cot\frac A2\equal{}$ $ 2Rr\cdot\sum \left(2\sin\frac A2\cos\frac A2\cdot \frac {\cos\frac A2}{\sin\frac A2}\right)\equal{}$ $ 2Rr\cdot\sum 2\cos^2\frac A2\equal{}$

$ 2Rr\cdot\sum (1\plus{}\cos A)\stackrel{(1)}{\ \equal{}\ }$ $ 2Rr\left(4\plus{}\frac rR\right)\equal{}$ $ 2r(4R\plus{}r)$ $ \implies$ $ \boxed {\ \sum a(s\minus{}a)\equal{}2r(4R\plus{}r)\ }\ (2)$ .

$ 3\blacktriangleright\ 2\cdot\sum (s\minus{}b)(s\minus{}c)\equal{}$ $ 2\cdot\sum [(s\minus{}b)\plus{}(s\minus{}c)](s\minus{}a)\equal{}$ $ \sum a(s\minus{}a)\stackrel {(2)}{\ \equal{}\ }$ $ 2r(4R\plus{}r)\implies\boxed {\ \sum (s\minus{}b)(s\minus{}c)\equal{}r(4R\plus{}r)\ }\ (3)$ .

$ 4\blacktriangleright\ s(s\minus{}a)\plus{}(s\minus{}b)(s\minus{}c)\equal{}bc\ \implies\ \sum bc\equal{}s^2\plus{}\sum (s\minus{}b)(s\minus{}c)\stackrel {(3)}{\ \equal{}\ }$ $ \boxed {\ ab\plus{}bc\plus{}ca\equal{}s^2\plus{}r(4R\plus{}r)\ }$ .

$ 5\blacktriangleright\ r_a\plus{}r_b\plus{}r_c\equal{}\frac 1r\cdot\sum rr_a\equal{}\frac 1r\cdot\sum\frac {S^2}{s(s\minus{}a)}\equal{}$ $ \frac 1r\cdot\sum (s\minus{}b)(s\minus{}c)\stackrel{(3)}{\ \equal{}\ }4R\plus{}r$ $ \implies$ $ \boxed {\ r_a\plus{}r_b\plus{}r_c\equal{}4R\plus{}r\ }$ .

[color=darkred][u]The [b]Euler[/b]'s theorem[/u]. Let $ ABC$ be a triangle with the circumcircle $ C(O,R)$ and the incircle $ C(I,r)$ .

Denote the distancies $ x$ , $ y$ , $ z$ of the circumcenter $ O$ to the sides $ [BC]$ , $ [CA]$ , $ [AB]$ respectively.

Then $ \boxed {\ x + y + z = \{\begin{array}{ccc} R + r & \mathrm{if} & \triangle ABC\mathrm{\ is\ acute} \\
 \\
r_a - R & \mathrm{if} & \triangle ABC\mathrm{\ is\ A - obtuse}\end{array}\ }$ .[/color]

[color=darkblue][b][u]Proof[/u].[/b] Denote the midpoints $ M$ , $ N$ , $ P$ of the sides $ [BC]$ , $ [CA]$ , $ [AB]$ respectively.

Apply the [u][b]Ptolemy[/b]'s relation[/u] in the cyclic quadrilaterals $ ANOP$ , $ BPOM$ , $ CMON$ .

$ \blacktriangleright$ [u]Case[/u] $ \triangle ABC$ [u]is acute[/u]. $ \|\begin{array}{cc} ANOP\ : & cy + bz = aR \\
 \\
BPOM\ : & az + cx = bR \\
 \\
CMON\ : & bx + ay = cR \\
 \\
*** & ax + by + cz = 2pr\end{array}\|\bigoplus\ \implies$

$ (a + b + c)(x + y + z) = (a + b + c)(R + r)\ \implies\ \boxed {\ x + y + z = R + r\ }$ .

In this case  $ \|\begin{array}{c} x = R\cdot\cos A \\
 \\
y = R\cdot\cos B \\
 \\
z = R\cdot\cos C\end{array}\|\ \Longrightarrow\ \boxed {\ \cos A + \cos B + \cos C = 1 + \frac rR\ }$ .

$ \blacktriangleright$ [u]Case[/u] $ \triangle ABC$ [u]is[/u] $ A$-[u]obtuse[/u]. $ \|\begin{array}{cc} ANOP\ : & cy + bz = aR \\
 \\
BPOM\ : & - az + cx = - bR \\
 \\
CMON\ : & bx - ay = - cR \\
 \\
*** & - ax + by + cz = 2(p - a)r_a\end{array}\|\bigoplus\ \implies$

$ ( - a + b + c)(x + y + z) = ( - a + b + c)(r_a - R)\ \implies\ \boxed {\ x + y + z = r_a - R\ }$ .

[b][u]Remark[/u].[/b] If $ A = 90^{\circ}$ , then $ R + r = r_a - R\ \Longleftrightarrow\ r_a = 2R + r\ \Longleftrightarrow\ p = a + (p - a)$ , what is truly.[/color]

[color=darkred][b][u]Proposed problem[/u].[/b] Let $ ABC$ be a triangle. Then $ \boxed {\ \begin{array}{ccc} A = 90^{\circ} & \Longleftrightarrow & r_a = 2R + r \\
 \\
90^{\circ}\in\{A,B,C\} & \Longleftrightarrow & p = 2R + r\end{array}\ }$ . [/color]

$ \blacktriangleright\ r_a = 2R + r\Longleftrightarrow$ $ p\tan\frac A2 = 2R + (p - a)\tan\frac A2$ $ \Longleftrightarrow$ $ a\cdot \tan\frac A2 = 2R$ $ \Longleftrightarrow$ $ \tan\frac A2\sin A = 1$ $ \Longleftrightarrow$

$ \frac {\sin A}{1 + \cos A}\cdot\sin A = 1$ $ \Longleftrightarrow$ $ \frac {1 - \cos^2A}{1 + \cos A} = 1$ $ \Longleftrightarrow$ $ 1 - \cos A = 1$ $ \Longleftrightarrow$ $ \cos A = 0$ $ \Longleftrightarrow$ $ A = 90^{\circ}$ .

$ \blacktriangleright\ \|\begin{array}{c} r_a + r_b + r_c = 4R + r \\
 \\
r_ar_b + r_br_c + r_cr_a = p^2 \\
 \\
r_ar_br_c = p^2r\end{array}\|$ , i.e. $ f(x)\equiv x^3 - (4R + r)\cdot x^2 + p^2\cdot x - p^2r = 0\ \Longleftrightarrow\ x\in \{r_a,r_b,r_c\}$ .

Therefore, $ 90^{\circ}\in \{A,B,C\}\Longleftrightarrow$ $ (2R + r)\in\{r_a,r_b,r_c\}$ $ \Longleftrightarrow$ $ f(2R + r) = 0\Longleftrightarrow$

$ (2R + r)^2[(2R + r) - (4R + r)] + p^2[(2R + r) - r] = 0$ $ \Longleftrightarrow$ $ 2R + r = p$ .


if $p$(greater than $5$)is a prime.then for any partition of the set ${1,2,3,...,p-1}$into $3$ subsets ,then there exist three numbers$x,y,z$ each belonging to a different subset,s.t $x+y=z(mod p)$


geometry
geometric transformation
reflection
trigonometry
integration

can you please post solution for hc verma geometrical optics 11 th problem

say the point is at $ P$ on the axis .now a ray from $ P$ strikes the air- water interface refracts then reflects in the concave mirror and then it again refracts and reaches the point $ P$. now consider the first part that is light ray ray refracting and striking the mirror say at some point $ S$ .now by principle of retracibilty of path if u would send a light ray along the path it came from $ P$ the light ray would reach $ P$.also note that a ray parallel to the axis from $ S$ goes unrefracted.hence if we were to have an object at $ S$ then it's image would be along an axis from $ S$ parallel to the principal axis.

now an actual ray of light emerging after reflection at $ S$ would say follow some path and reach $ P$.now extend the 2nd refracted ray that is the one  that leaves the  water-air interfaceafter reflection it has to intersect the axis parallel to principal axis from $ S$ at he very same point our first refracted ray intersected.but these two lines also intersect at $ P$.hence the are one and the same!

Hence the light ray traces back the path it came from.hence angle of reflection and refraction are both $ 0$ so the reflected ray was along the radius.
let the point $ P$ be at a height $ H$ then

$ (R-h) \tan r = H \tan i$
so $ H = \frac{(R-h) \tan r}{ \tan i}\approx \frac{(R-H) \sin r}{\sin i}= \frac{(R-h)}{\mu}$

a shorter proof would be since for given two points and interfaces there exists only one path te light follows by fermat's least action principle we have the light after reflection in the mirror has to trace the same path.hence the centre point of incidence on surface and point of incidence on mirror are collinear.hence
$ (R-h) \tan r = H \tan i$
so $ H = \frac{(R-h) \tan r}{ \tan i}\approx \frac{(R-H) \sin r}{\sin i}= \frac{(R-h)}{\mu}$

are you  preparing for iit jee 2008
and what is that little fermat's principle
 :)

yes :roll:  that little fermat's theorem belongs to maths
thanks for the solution

sorry my second proof isn't all correct   :rotfl:  .so better have the first one
(the statement of second is not general...it's proof is in the first)

[quote="pardesi"]yes.
and fermat's least action (and not little that's different) principle states that the path taken by light is such that it covers the distances between any two given points eperated by some interfaces in the least possible time and such a path is unique.[/quote]
is there any connection between this and mean free path??? :(   :D  :maybe:

by mean free path i think u r referring to the motion of Gas molecules.
but here i was talking about the Optical path length that is $ \int_{l} \mu dx$
a light ray always trasverses a path between two points such that the OPL is an extremum mostly minimum


calculus
derivative
function
real analysis
calculus computations

This problem seems way too easy compared to the other ones which makes me think I am doing it wrong.

Question: Let $ f$ have the following charcteristics: $ f: \mathbb R\mapsto I$ with $ I\subset \mathbb R$, $ f$ is everywhere differentiable, and $ |f'(x)|\leqslant M\,\,\forall x\in\mathbb{R}$. Prove that $ f$ is uniformly continous on $ \mathbb R$

Answer: Consider $ f$ on $ [a,b]\subset\mathbb R$, since $ f$ is differentiable on this interval it follows that $ f$ is continous on this interval. Now it is known that on compact spaces that continuity and uniform continuity are synonomous thus on any closed interval $ f$ is uniformly continous.

Now consider $ f$ on $ (a,b)\subset\mathbb R$ we can conclude that this is continous because based on the everywhere differentiability of $ f$ since we may extend continuously $ f$ from $ (a,b)$ to $ [a,b]$. (This was proven in my book)

What bothers me is that I did not use the boundedness of the derivative. Any problems?

Uniform continuity on $ [a,b]$ for all $ a$ and $ b$ is [i]not[/i] uniform continuity on $ \mathbb{R}.$ You proved the former, not the latter.

The path for this problem is to prove that having a bounded derivative implies Lipschitz, and Lipschitz implies uniform continuity.

[quote="Kent Merryfield"]Uniform continuity on $ [a,b]$ for all $ a$ and $ b$ is [i]not[/i] uniform continuity on $ \mathbb{R}.$ You proved the former, not the latter.[/quote]

Thank you...that was stupid of me.
[quote]
The path for this problem is to prove that having a bounded derivative implies Lipschitz, and Lipschitz implies uniform continuity.[/quote]
My book has not talked bout Lipschitz continuity yet, but I looked it up and it basically seems that if $ f: X\mapsto Y$ then f is Lipschitz everywhere if $ \frac {d_Y(f(x),f(y))}{d_X(x,y)}\leqslant K$. So for  $ f: \mathbb{R}\mapsto I\text{ }I\subset\mathbb R$ to be Lipschitz we would have to have that $ \frac {|f(x) \minus{} f(y)|}{|x \minus{} y|} \equal{} \left|\frac {f(x) \minus{} f(y)}{x \minus{} y}\right|\leqslant K$

I will now attempt to first prove that a bounded derivative implies Lipschitz continuity.

Consider the interval $ [a,b]$ by the hypothesis $ f$ is continuous and differentiable on this interval, therefore the Mean Value Theorem applies. So there exists a $ \xi\in(a,b)$ such that $ \frac {f(b) \minus{} f(a)}{b \minus{} a} \equal{} f'(\xi)\text{ }{\color{red}(*)}$. But once again by hypothesis $ \color{red}(*)$ implies that $ \left|\frac {f(b) \minus{} f(a)}{b \minus{} a}\right|\leqslant M$, and since $ [a,b]$ was arbitrary this completes the proof.

Now I will attempt to show that Lipschitz continuity implies uniform continuity.

We must show that $ \forall\varepsilon > 0$ there exists a corresponding $ \delta > 0$ such that $ |x \minus{} y| < \delta$ implies $ |f(x) \minus{} f(y)| < \varepsilon$. To do this we must just note that $ \left|\frac {f(x) \minus{} f(y)}{x \minus{} y}\right|\leqslant M\implies |f(x) \minus{} f(y)|\leqslant M|x \minus{} y|$. So we may take $ \delta \equal{} \frac {\varepsilon}{M \plus{} 1}$ in the definition of uniform continuity. This shows that $ f$ being Lipschitz continuous implies it is uniformly  continuous which is what was to be shown.

How does that look?

Yes, that is correct.

I suppose you could have used the Mean Value Theorem and not named the intermediate step as Lipschitz - but I think that having that name there helps.

By the way, it is possible for a function to uniformly continuous but not Lipschitz. Example: $ f(x) \equal{} x^{1/3}$ on $ \mathbb{R}.$

And it is possible for a function to be Lipschitz and not have a bounded derivative everywhere. Example: $ f(x) \equal{} |x|.$


geometry
3D geometry
sphere
combinatorics
maximization
geometric inequality
IMO Shortlist

For points $ A_1, \ldots ,A_5$ on the sphere of radius 1, what is the maximum value that $ min_{1 \leq i,j \leq 5} A_iA_j$ can take? Determine all configurations for which this maximum is attained. (Or: determine the diameter of any set $ \{A_1, \ldots ,A_5\}$ for which this maximum is attained.)


algebra
polynomial
calculus
derivative
function
analytic geometry
graphing lines

[b][size=134]Suppose $f(x)=x^3+ax^2+bx+c$ satisfies $f(-2)=-10$, and takes the extreme value $\frac{50}{27}$ when $x=\frac{2}{3}$. Find the values of a, b, and c.[/size][/b]

Buddy I didn't get the second and third solution. How did you came up from your second and third solution?

[quote="minsoens"]Derivative:
$f'(x)=3x^2+2ax+b$, $f'\left(\frac{2}{3}\right)=0$
Using synthetic division, we get
$\frac{2}{3}(2a+2)+b=0$(2)
[/quote]
I assume you don't know calculus? "Extrema" means that the function either has a relative maximum or minimum at that point, which means if you draw the tangent line to the function at that point, the slope of the function will be 0. The derivative of a function is sort of another function that allows you to calculate the slope of the tangent line to the function at a particular point, so set the derivative equal to 0 and solved.

For any polynomial $f(x)=\sum_{i=0}^{n}{a_ix^i}$, the derivative is $f'(x)=\sum_{i=1}^{n-1}{(i+1)a_{i+1}x^{i}}$

[quote="minsoens"]$f\left(\frac{2}{3}\right)=\frac{8}{27}+\frac{4}{9}a+\frac{2}{3}b+c=\frac{50}{27}$(3)
$a=-1$, $b=0$, $c=2$
[/quote]
For this one I substituted $x=\frac{2}{3}$ into $f(x)=x^3+ax^2+bx+c$

This can actually be done without calculus.  [hide="Solution"]Since $f$ takes the value $\frac{50}{27}$ at $x = \frac23$ and this is an extremum, the function $f(x) - \frac{50}{27}$ must have a double root at $x = \frac23$, so $f(x) = (x - r)(x-\frac23)^2 + \frac{50}{27}$.  Plugging in $x = -2$ to both sides, we get $-10 = (-2 - r)(-frac\83)^2 + \frac{50}{27}$, and so we can solve for $r$ to get $r = -\frac13$.  Then $f(x) = (x + \frac13)(x-\frac23)^2 + \frac{50}{27} = x^3 - x^2 + 2$, so $a = -1, b = 0, c = 2$.[/hide]

[quote="luimichael"]JBL claimed:
[size=150]Since f takes the value $\frac{50}{27}$ at $x = \frac23$ and this is an extremum, the function $f(x) - \frac{50}{27}$ must have a double root at $x = \frac23$.[/size]
How to prove this fact without using Calculus?[/quote]
Without using any concepts except continuity and basic properties of polynomials:
Suppose we have a polynomial $P(x)$ and $P(x_0) = 0$ is an extremum of $P$.  I claim $P$ has a double root at $x_0$.
Take a polynomial $P$ with root $x_0$ and suppose that $P$ does not have a double root at $x_0$.  Then $P(x) = (x - x_0)Q(x)$, $Q$ is a polynomial.  Let $x_1$ be the root of $P$ (and hence also of $Q$) closest to but different from $x_0$.  Consider values of $x$ which satisfy $x_0 - |x_1-x_0| < x < x_0 + |x_1 - x_0|$.  By the choice of $x_1$, $Q$ does not have any roots in this interval.  Because $Q$ is continuous, it always has the same sign (either positive or negative) in this interval.  However, $x - x_0$ changes sign at $x_0$, so $P$ changes sign as well.  Thus, $x_0$ is not a local extremum, since on one side of $x_0$ $P$ is positive while on the other side it is negative.  Taking the contrapositive of this statement, we have our desired result.

If a polynomial $P(x)$ has a root $r$ so that $P(r) = 0$, then there is some polynomial $Q$ such that $P(x) = (x - r)Q(x)$.  It needs to be $x - \frac 23$, not $x + \frac 23$, because when I put in $\frac 23$ it needs to spit out 0.  (I need two of them because it's a double root.)  For the other term, I could have used $x + r$ (the other root being $-r$) and the solution would have ended up working out exactly the same -- the third factor would have become $x + \frac 13$ with that method, as well.

You mean I can use $f(x)=(x+r)(x-\frac{2}{3})^2+\frac{50}{27}$ as well?I will get the same answer too?

You'll get a different value for $r$, but you'll get the same final answer.  (Your value for $r$ will be the negative of my value, but since you add $r$ while I subtract $r$, it will work out to the same function $f$.)  Work it out and see!

Buddy I tried it and studied it and it really worked. My last question is why did you used $\frac{50}{27}$ and not $-\frac{50}{27}$?

We know that $f(x) - \frac{50}{27}$ is a polynomial with a double root at $\frac23$ because it's a polynomial and $f(\frac23) - \frac{50}{27} = 0$ is an extremum (the given), and the argument I gave above.  Thus $f(x) - \frac{50}{27} = (x - r)(x - \frac23)$ and adding $\frac{50}{27}$ to both sides gives us the result.


$P,Q$- $nxn$ matrices s.t. $P^2=P,Q^2=Q$ and $1-(P+Q)$ is invertible. Show that $rankP=rankQ$

:)

The rank of $P(I_n-P-Q)$ is equal to the rank of $P$, meaning that $-PQ$ has the same rank as $P$. $(I_n-P-Q)Q$ has the same rank as $Q$, which means that $-PQ$ has the same rank as $Q$. :)

yes, it is simple(i think that you smile meant it), my solution is just a bit longer.


MATHCOUNTS
number theory
prime factorization

Let r(n) be the number of ways n can be written as the sum of two squares of integers that need not be positive, and order is important. For example, r(2)=4 because

2=
1^2+1^2=
1^2+(-1)^2=
(-1)^2+1^2=
(-1)^2+(-1)^2.

Evaluate

sum from n=1 to infinity (-1)^n*r(n)/n.

The solution to this problem can be found on the alumni forum at the Mathcounts website, http://www.mathcounts.org.  The idea behind the proof has to do with a graphical representation of what it means for a number to be the sum of squares of other numbers.  The answer is [hide]pi*ln(2)[/hide], I believe.  As I said, the mathcounts website contains a full explanation.

The explicit answer to this question can be found in an exercise in A Course in Number Theory by H.E. Rose (I highly recommend this book to you Zeta), on page 120, exercise 2.

The simplest way to describe it is to take the prime factorization of n and write it as 2^a*(p_1^h_1...p_n^h_n)*(q_1^k_1...q_m^k_m), where the p_i's are primes 1 mod 4 and the q_i's are primes 3 mod 4. Then if any of the k_i's are odd, the number of representations of n as sums of two squares is 0. Otherwise, the number of representations is 4*(h_1 + 1)(h_2 + 1)...(h_n + 1).

The sum that you give looks very similar to the average number of representations by sums of two squares (except for the (-1)^n sign). This is known as Gauss's Circle Problem. A discussion about it can be found in just about any good intermediate Number Theory book. A particularly nice and simple proof of it can be found in G.E. Andrews book Number Theory. A more advanced book that covers geometric number theory in detail is Geometric and Analytic Number Theory by Hlawka, et al. I would just recommend the Andrews book though.

Here's another description of the number of ways to represent some number as the sum of two squares taken from Davenport's book, The Higher Arithmatic:

"For two squares the rule (due to Legendre) is as follows. Count the number of divisors of n of the form 1 mod 4 and those of the form 3 mod 4. If these numbers are D_1 and D_3 respectively, then the number of representations is 4*(D_1 - D_3)."

He then goes on to give a description of the number of ways to write a number as the sum of 4 squares. He does these without proofs though.


[quote="ajai"]Six fair coins are tossed. What is the probability that 3 heads and 3 tails are tossed?[/quote]

[hide]There are $2^{6}=64$ total total outcomes.

To have 3 heads and 3 tails, we need to count how many arrangements of $HHHTTT$ are possible - there are $\frac{6!}{3!3!}=20$.

Therefore the probability equals $\frac{20}{64}=\boxed{\frac{5}{16}}$.[/hide]

if we consider the coins to be distinct, then there are $\frac{6!}{3!3!}= 20$ ways to flip 3 heads and 3 tails.  but there are $2^{6}= 64$ possible flips, hence the probability is $\frac{20}{64}= \frac{5}{16}= 0.3125$


Ive never heard of Harvey Mudd until i came to AOPS, especailly to this forum. Is it only good at Mathematics? What are some of its other strengths??

I might consider there with many other not-well-known but places with great programs.

"Harvey Mudd College seeks to educate engineers, scientists and mathematicians..."

it's one of the schools in Barron's Most competitive college book

it's in CA and it's mostly men

http://www.hmc.edu

(and the name sounds very cool)

Its part of the claremont colleges its a math science school all undergrad the people are very cool and very friendly(at least the ones I've met)

SOme people turn down Caltech for Harvey Mudd. THough "more girls" and still yet a good science/math curriculum is probably the reason.

Apparently Harvey Mudd produces more science and engineering PhDs, percentage-wise, than any other college except for Caltech ([url]http://www.earlham.edu/~ir/bac_origins_report/bac_origins.html[/url]).

wow it is almost Caltech-Harvey Mudd-MIT in every Math/Science category.

I've been impressed with what I've read about the Mudd math department.  It's a small school (~700 students) but in recent years they've done well in the Putnam, in receiving NSF Fellowships, and in placing majors in top graduate departments: http://math.hmc.edu/students.shtml .


calculus
integration
real analysis
real analysis unsolved

Show that:
$ \int\limits_0^\infty \prod\limits_{k=1}^n\frac{k^2}{k^2+x^2}
dx=\frac{n\pi}{4n-2} $
For this I have a solution but is very long...

I think this should lead to a shorter solution:
Let $f(z)=\prod_{k=1}^{n}\frac{k^2}{k^2+z^2}$.

$\int _{0}^{\infty} f(x)dx=lim_{R \rightarrow \infty} \int_{0}^{R}f(x)dx$.
$\int _{-\infty}^{\infty} f(x)dx=\int_{-R}^{R}f(x)dx$, because f(x)=f(-x) for $x\in \mathbb{R}$.

By the residue theorem we have, for R>n, $\int_d f(z)dz=2\cdot \pi\cdot i\cdot \sum_{k=1}^{n} Rezf(k\cdot i)$, where d=d1+d2, d1 is the straight path joining -R and R and d2 is the circular path joining R and -R.

It is fairly easy to show that $\int_{d2} f(z)dz \rightarrow 0,\ as \ R\rightarrow \infty$. This shows that $\int_{-\infty}^{\infty} f(x)dx=2\pi i\sum_{k=1}^n Rezf(ki)$.

Let $P(x)=\prod_{k=1}^{n}(k^2+x^2)$.
We have to prove that $\frac{n\pi}{2n-1}=2\pi i\sum_{k=1}^{n}\frac{(n!)^2\cdot 2n}{P'(ki)}$

I don't know how to do this.

I have the same ideea at the begining, with the same problem :? .

Define $ f(z)\equal{}\prod_{k\equal{}1}^n (k^2\plus{}z^2)$ then $ f'(ik)\equal{}\prod_{\ell\equal{}1}^{k\minus{}1} \Big( \ell^2\plus{}(ik)^2 \Big)\; 2ik \,  \prod_{\ell\equal{}k\plus{}1}^{n} \Big( \ell^2\plus{}(ik)^2 \Big)$ .

Because of $ \prod_{\ell\equal{}1}^{k\minus{}1} (\ell\plus{}k) \, \prod_{\ell\equal{}1}^{k\minus{}1} (\ell\minus{}k)\equal{}\frac{(2k\minus{}1)!}{k!}\; (\minus{}1)^{k\minus{}1}\, (k\minus{}1)!$ and $ \prod_{\ell\equal{}k\plus{}1}^{n} (\ell\plus{}k) \, \prod_{\ell\equal{}k\plus{}1}^{n} (\ell\minus{}k)\equal{}\frac{(n\plus{}k)!}{(2k)!} \; (n\minus{}k)!$

$ f'(ik)\equal{}(\minus{}1)^{k\minus{}1} \, \frac{i}{k} \, (n\plus{}k)! \, (n\minus{}k)! \;\;  . \qquad \left.\text{Res} \frac{1}{f}\right|_{z\equal{}ik}\equal{}\frac{1}{f'(ik)}
\equal{}\frac{1}{i}\, \frac{(\minus{}1)^{k\minus{}1}\, k}{(n\plus{}k)!\, (n\minus{}k)!}$

$ \boxed{I_n\equal{}\int_{\minus{}\infty}^\infty \frac{1}{f} \, dz\equal{}2\pi i \, \sum \text{Res} \frac{1}{f}\equal{}2\pi \, \sum_{k\equal{}1}^n \frac{(\minus{}1)^{k\minus{}1} \, k}{(n\plus{}k)!\, (n\minus{}k)!}}$

$ \int_z^\infty \frac{t^{m\minus{}1}}{(m\minus{}1)!} \, e^{\minus{}t} \, dt\equal{}\left[\minus{}\frac{t^{m\minus{}1}}{(m\minus{}1)!}\, e^{\minus{}t}\right]_z^\infty\plus{}\int_z^\infty \frac{t^{m\minus{}2}}{(m\minus{}2)!} \, e^{\minus{}t} \, dt\equal{}$$ ...\equal{}\sum_{k\equal{}0}^{m\minus{}1} \frac{z^k}{k!}\, e^{\minus{}z}$ after repeated integration by parts.

So $ \int_0^z \frac{t^{m\minus{}1}}{(m\minus{}1)!} \, e^{\minus{}t} \, dt\equal{}\sum_{k\equal{}m}^\infty \frac{z^k}{k!}\, e^{\minus{}z}$ which can also be written as $ z^m \, \sum_{k\equal{}0}^\infty \frac{1}{(m\plus{}k)!}\, z^k \; \sum_{k\equal{}0}^\infty \frac{(\minus{}1)^k}{k!}\, z^k$

and this is $ \sum_{n\equal{}0}^\infty \sum_{k\equal{}0}^n \frac{1}{(m\plus{}k)!} \, \frac{(\minus{}1)^{n\minus{}k}}{(n\minus{}k)!} \, z^{n\plus{}m} \qquad \text{Cauchy Product}$

Now differentiate both sides with respect to $ z \; : \quad \frac{z^{m\minus{}1}}{(m\minus{}1)!} \, e^{\minus{}z}\equal{}\sum_{n\equal{}0}^\infty \sum_{k\equal{}0}^n \frac{n\plus{}m}{(m\plus{}k)!}\, \frac{(\minus{}1)^{n\minus{}k}}{(n\minus{}k)!}\, z^{n\plus{}m\minus{}1}$ ,

write LHS as series $ \frac{1}{(m\minus{}1)!} \, \sum_{n\equal{}0}^\infty \frac{(\minus{}1)^n}{n!}\, z^{n\plus{}m\minus{}1}$ and compare the coefficients $ \boxed{\frac{1}{(m\minus{}1)!} \, \frac{1}{n!}\equal{}\sum_{k\equal{}0}^n \frac{n\plus{}m}{(m\plus{}k)!} \, \frac{(\minus{}1)^k}{(n\minus{}k)!}}$

For $ m\equal{}n$ we obtain $ \frac{1}{(n\minus{}1)!} \, \frac{1}{n!}\equal{}\sum_{k\equal{}0}^n \frac{2n\, (\minus{}1)^k}{(n\plus{}k)! \, (n\minus{}k)!}$

and for $ m\equal{}n\minus{}1$ we obtain $ \frac{1}{(n\minus{}2)!} \, \frac{1}{n!}\equal{}\sum_{k\equal{}0}^n \frac{(2n\minus{}1)\, (\minus{}1)^k}{(n\minus{}1\plus{}k)! \, (n\minus{}k)!} \,
\Rightarrow \,  \frac{1}{2n\minus{}1}\, \frac{1}{(n\minus{}2)!} \, \frac{1}{n!}\equal{}\sum_{k\equal{}0}^n \frac{(n\plus{}k)\, (\minus{}1)^k}{(n\plus{}k)! \, (n\minus{}k)!}$

So $ \sum_{k\equal{}0}^n \frac{\minus{}k\; (\minus{}1)^k}{(n\plus{}k)!\, (n\minus{}k)!}\equal{}\frac12 \,\frac{1}{(n\minus{}1)!} \, \frac{1}{n!}\minus{}\frac{1}{2n\minus{}1}\, \frac{1}{(n\minus{}2)!}
\, \frac{1}{n!}\equal{}\frac{1}{n!^2}\left(\frac{n}{2}\minus{}\frac{(n\minus{}1) \, n}{2n\minus{}1}\right)\equal{}$ $ \frac{1}{n!^2} \, \frac{n}{4n\minus{}2} \, \Rightarrow \, n!^2 \, I_n\equal{}\frac{2\pi n}{4n\minus{}2}$


function
floor function
number theory unsolved
number theory

Given $ p$ is a prime , $ k|(p\minus{}1)$ and $ d\equal{} \frac{p\minus{}1}{k}$,
find the cardinality of the set $ S(k)$ defined as :
$ S(k) \equal{} \{u \in \{0,1,...,k\minus{}1\} | (1\plus{} du,dk) \equal{}1\}$
(explicitly in terms of known functions like euler's phi function elc....)

I'm not sure what kind of expression for $ S(k)$ you are looking for.

I can suggest the following expression for the case of $ k > 1$ and $ (d,k) \equal{} 1$:
\[ S(k) \equal{} \sum_{t|k} \mu(t) \left\lfloor\frac {k \minus{} 1 \minus{} ( \minus{} d^{ \minus{} 1}\bmod t)}{t}\right\rfloor
\]
where $ \mu(\cdot)$ is Moebius function and the result of $ (\minus{} d^{ \minus{} 1}\bmod t)$ is taken from the set $ \{0,1,\dots,t \minus{} 1\}.$

And I'm not sure if the primality of $ kd \plus{} 1 \equal{} p$ is useful here.

Let $ (1 \plus{} du,p \minus{} 1) \equal{} l > 1$, then $ (l.d) \equal{} 1$, therefore $ l|k_1$, were $ k_1 \equal{} \frac {k}{(d,k)}$ and $ k_1|k$.
For any $ 1 < l|k_1$ we had $ k/l$ means of u, suth that $ l|1 \plus{} du$ or equavalent $ (1 \plus{} du,p \minus{} 1) \equal{} l$.
Therefore number of solutions is $ \frac {k}{k_1}\phi(k_1)\equal{}(d,k)\phi(\frac{k}{(d,k)})\equal{}k\prod_{p|k, p\not |d}(1\minus{}\frac 1p )$.

A more general problem is posed at: http://www.mathlinks.ro/Forum/viewtopic.php?t=90779
It implies that
\[ S(k)\equal{}\frac{\phi(p\minus{}1)}{\phi(d)}\]


A piece of wood with a mass of 6.88 kg is placed in fresh water (density = 1.00 g/cm^3).  What is the density of the wood if it has an apparent weight of -6.13 N?

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

If the wood floated, then all of its weight would be supported by the bouyant force of the displaced fluid.
Since we are told it has an apparent weight, then it must be submerged. (if it floated, the apparent weight would be zero)

You are given the actual weight and the apparent weight of the wood, so you can work out what the bouyant force is.

The bouyant force is equal to the weight of the fluid displaced. Since you know what the bouyant force now is, then you can work out the volume of the fluid displaced.
Since the wood is submerged, then the volume of the displaced fluid is exactly the same as the volume of the wood.

You now know the volume and weight of the wood, so you can work out its density.

I found that the buoyant force is equal to 73.6 N (67.49 + 6.13).

Then, since the buoyant force is equal to the density of the water times the volume of the displaced water times gravity, I found that the volume of the displaced water is 7.50487.
So this means that the volume of the wood is 7.50487, right?

Then I simply divided the mass of the wood by the volume of the wood to get .917 g/cm^3.  However, this doesn't make sense to me because shouldn't the density of the wood be greater than the density of the water? Is this work right?  Please help.

No, the density of the wood should be less than 1 g/cm^3 because it has negative apparent weight. If it was more dense than water, it would have some positive apparent weight but negative apparent weight means that the wood is pulled upwards, the bouyant force is greater than the weight of the wood.
Of course, this means that the wood should be floating and not be submerged - but it's still possible to grab the wood and take it down to the bottom of the lake or whatever it floats in and then observe it's negative apparent weight.


Let $ a,b,c\geq0$ s.t. $ abc\equal{}1$.Prove that :
$ (\dfrac{a\plus{}b\plus{}c}{3})^5\geq\dfrac{a^2\plus{}b^2\plus{}c^2}{3}$

it's equavilent  to :  $ \frac{abc (a^2\plus{}b^2\plus{}c^2)}{3}  \leq (\frac{a\plus{}b\plus{}c}{3})^5$ (*)
 let $ a\plus{}b\plus{}c\equal{}3$   and  $ abc\equal{}r , ab\plus{}bc\plus{}ca\equal{}q$  note that  : $ q^2 \geq 9r$
(*) becomes :  $ 9r \leq 3 \plus{}2qr$ or $ 9t^2 \leq 3 \plus{}6t^3$ or $ 3(t\minus{}1)^2(2t\plus{}1) \geq 0$


greatest common divisor
number theory
divisibility tests

When the first 44 positive integers are written in order-123456...41424344, it forms a number N, What is the remainder whne N is divided by 45?

Observe: $45=9*5$

Let $N = 1234...4041424344$ as stated.

We'll first sum all the digits to test for divisibility by $9$.  $1$ through $9$ appear four times in the units digit and $1$ through $4$ appear once more.  $1, 2,$ and $3$ appear in the tens digit each ten times while $4$ is only seen $5$ times.

$4 \cdot 9 \cdot (9+1)/2+4 \cdot (4+1)/2+10 \cdot (1+2+3)+5 \cdot 4=270$
$2+7+0=9$

Thus $N$ is divisible by nine.

$N-9 \equiv 0 \text{ (mod }5)$
$N \equiv 9 \text{ (mod }5)$
$N/9 \equiv 9/9 \text{ (mod }5)$
$N/9 \equiv 1 \text{ (mod }5)$
$\cdot 9 \text{ to all parts}$
$\Rightarrow\ N \equiv 9 \text{ (mod }45)$

Thus the remainder is $\boxed{9}$.

Yeah I probably screwed it up when I divided by 9.  And I probably should of used 45 as my mod opposed to 5.

I just felt like trying it, I just read a book on number theory and saw the problem.

yeah, 9 seems right

When skimnc did $M\equiv1 \text{ (mod }5)$  the answer should have been multiplied by $9$ to get it into $\text{ (mod }45)$

Hence, $9*1=\boxed{9}$

I edited my solution up there.  I'm not sure if you're allowed multiply the left, remainder, and modulo by a constant but it seems to work for several examples.

I also just noticed you can look at it and see what is the closest number to $N$ with either a $5$ or a $0$ on the end that is also a factor of $9$.

Let $S =$ GCF of $5$ and $9$ $< N$

We know $N$ is divisible by $9$ so

$S = N-9 \cdot k$ where k > 0 and $S$ is divisible by $5$ (ends in a $5$ or $0$)

The closest value that satisfies this is $N-9$ thus the remainder is $\boxed{9}$


function
trigonometry
real analysis
real analysis unsolved

Hi all,

How can we prove that the unique 2D function which is separable and circularly symmetric is a 2D Gaussian, i.e.
$G(x, y) = a e^{\frac{x^2+y^2}{b}}$.
Thanks in advance.

Separable (which I take to mean that we can write it as $G(x,y)=u(x)v(y)$) and radial aren't strong enough conditions to exclude nonmeasurable monsters of functions. You also need some kind of measurability or continuity condition.

To move this forward, I'm willing to assume a fairly strong condition: assume also that $G$ is continuously differentiable. Now what?

We can now find the relation between u and v

$\sin\theta \frac{u'(r\cos\theta)}{u(r\cos\theta)} = \cos\theta \frac{v'(r\sin\theta)}{v(r\sin\theta)}$

where $(r, \theta)$ is the radial coord.

it seems to me that we r in need of another differential equation to completely specify the forms of u and v.

Note that the values on the $x$ axis must be the same as the values on the $y$ axis. We can WLOG assume that $u=v.$


Mr. Troll and Mr. Little are running on a 400 meter track. They start at diametrically opposite sides of the track, and run in opposite directions. Mr. Troll runs at 6 meters per second and Mr. Little runs at 4 meters per second. When they meet, Mr. Troll immediately pushes down Mr. Little. Mr Little takes 10 seconds to recover, while Mr. Troll continues to run. Then Mr. Little continues to run in the opposite direction as Mr. Troll. In 4 minutes, how many times did Mr Little get pushed down [u]AND[/u] recover?

I think it is [hide]5 times.[/hide]

[hide]Mr. Little gets pushed down at these moments:
20s
64s
108s
152s
196s
240s

However, the last time (at 240s) Mr. Little does not have time to recover.[/hide]


I'm looking for a way to use asymptote to graph an arbitrary three-dimensional surface when given an equation describing it in x, y, and z.  What is the best way to do this?

[url=http://asymptote.sourceforge.net/doc/graph3.html]Documentation for the graph3 module[/url]


The name of the game is magic and Ive seem them all ( ok, I havent really), David Blaine, Criss Angel, and I've always wondered how do these guys peform these astounding optical illusions?  The one that intrigues me the most is the levitation act ( where the performer actually rises from the ground, sometimes a few inches, sometimes a few feet)  and I'm just dieing to know how these guys do it.  Of course there are videos that you can buy, but I'm asking if anyone can explain the levitation act to me (other tricks are just as welcome) without me having to write a big juicy check.  Thanks alot.

I'm rather sure the levitation is an optical illusion

He stands in such a way, that you cannot see one of his feet. Raises himself up by goin on the tip toes of one foot, but that is not visible due to the fact that he covers it. It sounds complicated and fake, but it can be done, and my cousin can replicate the effect of the levitation. However, the audience must be in certain position....

But I am quite sure I can see both feet raise from the ground, as I mentioned, some people can levitate several feet in the air!

In that case... clearly something is holding them up? If you're watching it on TV then it's not all that difficult. If you're watching it live, then you are looking at mirrors in key places that are blocking your view of what's really there.

Forget about the Levitation, thats obviously fake, the card tricks are something else ;) I was watching this thing about David Blaine on T.V and the card tricks were just awesome.

his card tricks are amazing! Especially when he siwtches cards in his hand. If you watch it in super slow mo, you can actually see them switching. Amazing. That isn't magic, it's awesome skill.

Nope.. everything has a way to do it. I used to think some card tricks were magical - stuff that most people still dont know how to do, one of my friends started doing some magic tricks.. its all skill and illusion...

Also, don't forget about the power of camera tricks when watchin on TV

This levitation thing is no joke.  Obviously theres some trick to it, but it looks amazing realistic!  Heres a link to someone who has actually found out how to do it, and is selling the secret.

http://www.ellusionist.com/order/king.htm

EDIT: [quote="Stupidkid"]
Back on the levitation thing, I watched an episode of Mind Freak featuring Criss Angel, and he does levitate himself a couple feet up. He also levitated a total stranger whom he put in hypnosis so the person did not remember a thing. I'm pretty sure the levitaion is not an optical illusion and hypnosis probably has something to do with it.
[/quote]

Ha ha, I've also seen that episode.  But I disagree with you that levitation is nothing more than a optical illusion, click the link above!

okay now honestly the link is pretty but see how the guy only lifts up a little bit and criss angel manages to levitate up to at least 10 feet up in the air.. i mean i don't kno but if u can lift up a random person and out of nowhere push downwards and they slowly go down.. i don't think there's any way to put that as an illusion.. there's nothing over or under her as he has proved it with his stick.. how can you have an illusion of someone lifting up and staying up in the air if u don't expect it to happen in the first place.. i mean if he tells you i'm gonna fly now then yes he could make it an illusion but if he randomly starts flying up in the air when you're expecting him to do sumthing else it's not an illusion the guy is good