albate age hamoon lagrang ro pish begirim be dast miyad:
$a(1-a^2)^2=b(1-b^2)^2=c(1-c^2)^2$
ke do javab dare yeki:$a=b=c= \frac13$
yeki :$a=b=0 ,c=1$

[quote="mohamadhosein"]:roll:
agar ba raveshe lagrang pish berim khahim did ke noghteye extermomi ke be dast miyad dar vaghe maximome. pas minimom bar rooye marzhaye tabe gharar darad ama
tabe dar faza be shekle yek mosalas ast ba rouse $(1,0,0) ,(0,1,0),(0,0,1)$ va chon marzhaye aan haman azlae an hastand va dar nahayat sabet mishavad $b \rightarrow 0$ va$a \rightarrow 0$va $c \rightarrow 1$.[/quote]
dar raveshe la grang darim: noghate extermom mitavanand noghate marzie bazeha bashand.
yani har kodam az $a ,b ,c$ maghadire marzi khod ra begirand.
pas bayad tamamie halathai ra ke $a ,b ,c$ maghadire $0,1$ migirand va jame anha $1$ shavad ra barresi konim.
ke ba barresie anha kamtarin meghdar $2$ be dast miaiad.

hala kasi mitoone nedoone lagrang va ba raveshe addi hallesh kone?

kheili sadas....
$(a^4+b^4+c^4)+(a+b+c)^4\geq2(a^2+b^2+c^2)(a+b+c)^2$
hala age ino sade konin be rabeteye badihie zir miresin:
$2(ab+bc+ca)^2+4abc\geq0$
va halate tasavi ham ine ke 2 ta shoon be sefr meyl konan va on yeki be yek ;)
albatte fek nakonin inke migam sadas az maghroorie va na be khoda man kochhiketoonam ;)

salam.
man in masalaro ba amale ghahgharaii hal kardam.filesho attach mikonam.mamnoon misham age ie negahi behesh bendazin bebinin doroste ia na.

manam ye chand ta masale mirizam inja ke shabihe hamine ke dadin:
1)$(a+b)(b+c)(c+a)=1$,$a,b,c>0$
sabet kon $ab+bc+ca\geq\frac3{4}$
2)$a+b+c=1$,$a,b,c>0$
sabet kon$a(1+b-c)^{\frac13}+b(1+c-a)^{\frac13}+c(1+a-b)^{\frac13}\leq1$
3)$ab+bc+ca=3$,$a,b,c>0$
sabet kon$a^3+b^3+c^3+6abc\geq9$
age bazam khastin bazam daram...kheili az in tip(shart dar) daram.

[quote="maryam rashin"][quote="maryam rashin"]
1)$(a+b)(b+c)(c+a)=1$,$a,b,c>0$
sabet kon $ab+bc+ca\geq\frac3{4}$ ;)[/quote][/quote]
in masale eshtebahe fekr konam manzooret in bood:
$ab+bc+ca\leq\frac3{4}$
hala javabe in masale:
az sharte masale be dast miayad $\sum{a^2}{b}+2abc=1$ ba estefade az namosavie hesabi hendesi darim$abc \leq \frac18$
midanim${(a+b)(b+c)(c+a)}^\frac13=1$ dar natije $\sum{a} \geq \frac32$
pas dar natije $1+abc \leq \frac98 \leq {\frac34}{\sum{a}}$ pas $(a+b)(b+c)(c+a)+abc \leq {\frac34}{\sum{a}}$ pas
$(\sum{ab})(\sum a) \leq {\frac34}{\sum a}$ pas hokm sabet shod.

[quote="maryam rashin"][quote="maryam rashin"]
2)$a+b+c=1$,$a,b,c>0$
sabet kon$a(1+b-c)^{\frac13}+b(1+c-a)^{\frac13}+c(1+a-b)^{\frac13}\leq1$[/quote][/quote]
midanim $(1+b-c)^{\frac13} \leq \frac{3+b-c}{3}=1+\frac{b-c}{3}$ chon$1+b-c \geq 0$
ba estefade az in namosavi va ba tavvajoh be in ke $a+b+c=1$ masale sabet mishavad.

[quote="maryam rashin"][quote="maryam rashin"]
3)$ab+bc+ca=3$,$a,b,c>0$
sabet kon$a^3+b^3+c^3+6abc\geq9$[/quote][/quote]
midanim $\sum_{a} a(a-b)(a-c) \geq 0$ ba sade kardan be dast miayad$\sum_{a} a^3 +3abc \geq \sum_{cyc} {a^2}{b}$
pas kafist sabet konim$\sum {a^2}{b}+3abc \geq 9$ ba tajzie kafist sabet konim $(\sum{ab})(\sum{a}) \geq 9$
be estefade az shart hokm be soorate zir kholase mishavad
$\sum{a} \geq 3$ amma midanim $\sum{a^2} \geq \sum{ab}=3$ pas $(a+b+c)^2 \geq 9$
hom sabet shod.


Let $ f(x,y)\equal{}x^2\plus{}xy\plus{}y^2$ where $ x,y$ are integers.
Prove that $ f(x,y)f(z,w)\minus{}>f(x,y)$ for all integers $ x,y,z,w$.

for integers $ x > y, z > w$ ($ \Rightarrow xz \plus{} yw > xw \plus{} yz$)

it is $ f(x,y) \equal{} x^2 \plus{} y^2 \plus{} xy \equal{} \frac {x^3 \minus{} y^3}{x \minus{} y}$

so , $ f(x,y)f(z,w) \equal{} f(k ,l)$

(I miscalculated k, l but I think they are what cmad has found bellow)

for integers $ x \equal{} y , z \ge w$:

$ f(x, y)f(z, w) \equal{}$

$ 3x^2(z^2 \plus{} w^2 \plus{} zw) \equal{} (2xw \plus{} xz)^2 \plus{} (2xw \plus{} xz)(xz \minus{} xw) \plus{} (xz \minus{} xw)^2$

$ \equal{} f(2xw \plus{} xz, xz \minus{} xw)$

q.e.d.  :D

While solving the above problem, I discovered that the same also holds true for all $ x ,y, z, w \in \mathbb{Z}$, as well as for

$ f(x, y) \equal{} x^2 \minus{} xy \plus{} y^2$ and $ x ,y, z, w \in \mathbb{Z}$

For the problem to be stated in a more advanced "language", we could say:

For the (Grassman) space of $ f: \mathbb{N}^2 \to \mathbb{R}, f(x,y) \equal{} x^2 \plus{} xy \plus{} y^2$
prove that multiplication is an internal operation.

... $ \zeta \equal{} \exp \frac{2\pi i}{3}$

$ \alpha \equal{} x \minus{} \zeta y \Rightarrow \vert \alpha \vert^2 \equal{} \alpha {\bar \alpha} \equal{} x^2 \plus{} xy \plus{} y^2$
$ \beta \equal{} z \minus{} \zeta w \Rightarrow \vert \beta \vert^2 \equal{} \beta {\bar \beta} \equal{} z^2 \plus{} zw \plus{} w^2$

$ f(x,y) f(z,w) \equal{} \vert \alpha \vert^2 \vert \beta \vert^2 \equal{} \vert \alpha \beta \vert^2 \equal{} \vert xz \minus{} yw \minus{} \zeta(xw \plus{} yw \plus{} yz) \vert^2$
$ \equal{} f(xz\minus{}yw, xw\plus{}yz\plus{}yw)$

Ωραία cmad . :)

Απλά νομίζω ότι είναι και $ f(a,b)f(c,d)\equal{}f(ac\minus{}bd,ad\plus{}bc\minus{}ad)$

Πολυ καλα σε ολους.... :lol:
ΑΣΚΗΣΗ:
Για ποιες ακεραιες τιμες η συναρτηση $ f(x,y)$ τετραγωνιζεται?

Ζηταω να λυσεται την εξισωση $ x^2\plus{}xy\plus{}y^2\equal{}z^2$ στους ακεραιους.
Η εξισωση εχει απειρες λυσεις...βρειτε ολες τις ακεραιες λυσεις.

Εύκολα μπορούμε να δούμε ότι για $ y \equal{} 0$ έχει λύση για κάθε $ z$ την $ x \equal{} z$

για $ y \equal{} a$ έχουμε:
$ x^2 \plus{} ax \plus{} a^2 \minus{} z^2 \equal{} 0$,
που έχει λύσεις όταν $ D \ge 0 \Leftrightarrow 3a^2 \le 4z^2$
τις $ x \equal{} \frac { \minus{} a \pm \sqrt {D}}{2}$

που είναι ακέραιες όταν $ \sqrt {D}$ είναι περιττό πολλαπλάσιο του $ a$ δηλαδή όταν

$ (2k \plus{} 1)^2a^2 \equal{} 3a^2 \plus{} 4z^2 \Leftrightarrow 4z^2 \equal{} (4k^2 \plus{} 2k \minus{} 2)a^2$ δηλαδή όταν το

$ 4k^2 \plus{} 2k \minus{} 2 \equal{} (k \minus{} 1)(4k \plus{} 2)$ είναι τέλειο τετράγωνο

εκεί έχω κολήσει (είχα βρει μια λύση αλλά είχε αριθμητικό λάθος)

Δεν νομίζω ότι αυτό οδηγεί κάπου ...

Ένα hint apo mena
[hide]Fermat's infinite descent   :wink: [/hide]

Μaniopas:
1)Γιατι ο $ \sqrt{D}$ ειναι περιττο πολλαπλασιο του $ a$?
2)Eχεις κανει λαθος στις πραξεις και μετα απο αυτο...
Silouan:
1)Mπορεις να γραψεις την λυση σου?
2)Αυτην την μεθοδο την χρησιμοποιουμε για νδο μια εξισωση δεν εχει λυσεις η οτι εχει μονο τετριμενες λυσεις...(η εξισωση εχει απειρες λυσεις)

[quote="Αρχιμήδης 6"]
2)Αυτην την μεθοδο την χρησιμοποιουμε για νδο μια εξισωση δεν εχει λυσεις η οτι εχει μονο τετριμενες λυσεις...(η εξισωση εχει απειρες λυσεις)[/quote]

Δεν είμαι σίγουρος για αυτό .  :wink:   Βασικά η μέθοδος μπορεί να σου δώσει άπειρες λύσεις και με αναδρομικό τύπο μάλιστα
(στην συγκεκριμένη περιπτωση κρατώντας σταθερό το z έχουμε ότι αν $ (x_0,y_0)$ είναι ένα ζεύγος λύσης τότε το

$ (\minus{}x_0\minus{}y_0,y_0)$ είναι λύση από όπου βρίσκουμε τον αναγωγικό τυπο  :)

Για το complete έχουme ότι οι λύσεις είναι

$ x \equal{} k(2mn \plus{} n^2)$ , $ y \equal{} k(m^2 \minus{} n^2)$ , $ z \equal{} k(m^2 \plus{} mn \plus{} n^2)$


email
percent
probability
MATHCOUNTS
AMC
AMC 8
AMC 12

Hello.

Post your AMC 8 2009 scores here if you've already recieved them. To start, I got a 20.

Official scores came today (I think) for my school. I got 25.

How id you get your official scores already? My teacher said we would get them in a couple of weeks.

Hmm...last year(AMC 8 2008)we had to wait until like after New Year's Day to recieve our scores. I wonder why some schools recieve early, some recieve late...

The AMC office began emailing official score reports for the 2009 AMC 8 to schools on Dec 3, 2009.  The first paper mail packages with scores, pins and certificates will go out with today's mail.

The email reports go out roughly in the order that we process the reports sent back from the schools and also roughly  in order of numbers of registrations within a US state.  So big population states that sent in reports early generally get processed and reported first.  Small population states and schools that sent in reports late get processed last.  Processing and reporting from start to finish takes about 3 to 4 weeks.  However there are exceptions.  In any case, scoring reports on 150,000 students from over 2200 schools in 50 US states, all US territories and over 30 countries outside the US takes time.  Please be patient.

Reporting will be done by the published deadlines.  Please consult the Teachers' Manual for the deadlines.

If you have not received an official report from your teacher yet, the likely explanation is one of the following.
(1) It has not yet been processed, scored, reported or emailed.
(2) Your teacher has not yet checked the email for the reports.
(3) The email is not sent to to the teacher, but rather to the person who registered, e.g. school secretary, or math supervisor
(4) No email address was provided to us
(5) Your teacher prefers not to detail scores yet for whatever reason

If you have not received a score yet, DO NOT request your score from the AMC office, because
(1) among 150,000 scores, sorted by school, we don't stop everything to find your one score.
(2) asking us for your score slows down the process for everyone.
(3) It is not our policy to provide scores except to the teacher or contest administrator.

Steve Dunbar
MAA Director, American Mathematics Competitions

oh..ty..
I see..we probably didn't get our scores yet..atleast that's what my teacher said

Our school received results already. Two people got perfects, and I wasn't one of them.  :(

I got a 23......  :(  :mad:

Hey guys, let's not make other people feel bad.  Don't say you failed when you got a score that places you well into the top 1/10 of a percent of test-takers.

By the way, I got a 3.

Sorry if I am rushing, but do you know in how many months the results will be posted on the AMC website?

21, but it's unofficial, maybe I misbubbled!  :lol:

[quote] Don't say you failed when you got a score that places you well into the top 1/10 of a percent of test-takers. [/quote]
How do you know if you got into the top 10%?

It's hard to qualify through AMC 10 without knowing the stuff pretty well and answering almost all the questions.

I got a 24/25 which is really annoying because the incorrect question was a careless error. ... it was the one with the 45x70 array of numbers or something.

I got a 19/25.

I'm so sad... I was 1 problem off from Distinguished Honor Roll....

UGH...When i get my amc back..:(
I'd better have gotten atleast 20 or 19..or more :P

I probably got like 19 or 20 blehhhhh.....  :)  :P

people!!!! s t o p bumping old threads!! :mad:

You seriously bumped a 12 year old thread

[quote=Mathcounts_Student_2021]was this worth to bump[/quote]

Not at all

[quote=a.shyam.25]apparently people don't like bumping old threads[/quote]
:dry:

[quote=a.shyam.25]<redacted because apparently people don't like bumping old threads.  I made a mistake, and this post can be deleted.>[/quote]

I would think that you knew this rule because you already have 1682 posts, I guess people just want to postfarm now

[quote=a.shyam.25]apparently people don't like bumping old threads.  [/quote]

Hmmm apparently the floor is made out of floor

I read the title and thought this was a troll thread lol

I am pretty sure I was in a crib when I scored a 25 on this test.


As far as I know, you can NOT take both the AIME I and II and there was a quite long topic on this last year. However, my math teacher keeps saying that people who qualified through the AMC A would take the AIME I and those who qualified through AMC B would take AMIE II. As some of you guys might already guess, he would let those who qualified through both the AMC A and B take both AIMEs.

I just want to make sure that you're not allowed to take 2 AIMEs, right? And yet, if my math teachers let us do that, would we be disqualified automatically?

um im really not sure, i dont think that you can take both, i think AIME II is like a make up dat...but i dont know seing as how i have not made AIME(but have prepared alot for it)

From my understanding, you can only take either the AIME I or the AIME II, not both.  The AIME II is the test for those who were not able to take the AIME I, and is not like the relationship between the AMC A and AMC B.  You qualify for the AIME I through the AMC A or the AMC B first.

You definitely can [b]not[/b] take the AIME on both dates.

Which AMC date you use (the A or the B) to qualify for the AIME has no bearing whatsoever on which date you take the AIME.  You're supposed to take it on the first date (one week from today) unless you have a conflict.

Your math teachers should read the [url=http://www.unl.edu/amc/d-publication/d1-pubarchive/2007-8pub/TM-AIME,2008.pdf]instruction manual[/url] that the AMC sends to every school.  And it's not like this is buried in the fine print: on [i]the first page[/i], instruction #3 is:

[b]Under no circumstances can a student take both AIMEs.[/b]

You can only take one officially, the other you could take unofficially.
I think.

I was wondering....two people from my school passed the AMC A, so my teacher only recieved 2 copies of the AIME I test. However, two or three more people from my school (in addition to the two who already passed the AMC A) passed the Amc B. Yet since my teacher only has two copies of the AIME I, are those people forced to take the AIME II? Another thing I was wondering about was if one/both of the two people who passed the AMC A happened to not pass the AMC B...and if he/she cannot take the AIME I, will a AIME II be mailed for them?

Thanks

I'm pretty sure your teacher should've received the tests by now.

My teacher has received two copies of the AIME I. Should she have received the AIME II's as well? And should she receive an AIME II for each person who passed the amc a OR the amc b? or just for those who passed the amc b?

You just got censored by AoPS.
That's sorta sad since AoPS doesn't censor a lot of words.

Where do you choose which one to take? Also, I haven't received any notification of the qualification for the AIME, and I'm 99% sure I got in, unless I bubbled 5 questions incorrectly.

Notification is sent to your Contest Manager (usually the teacher in charge), not to you directly. An yes, xenero, you did qualify for the AIME. Now go back to reading your email.

Eh, your teacher gets notification about qualifying.
Anyways, the AIME II is only for people who CANNOT take the AIME I, not because of AMC10/12 A/B qualification.

All of the questions in this thread can be answered by examining Section VII, page 5, of the AIME Teachers' Manual.    Also, see page 10 for the Alternate AIME Order Form, see Section II, page 4 for how the results are reported to your school and how the AIME contest booklets are delivered.

http://www.unl.edu/amc/d-publication/d1-pubarchive/2007-8pub/TM-AIME,2008.pdf

Steve Dunbar
MAA Director, American Mathematics Competitions


trigonometry
geometry
circumcircle
Trigonometric Identities
IMO
IMO 1960

In a given right triangle $ABC$, the hypotenuse $BC$, of length $a$, is divided into $n$ equal parts ($n$ and odd integer). Let $\alpha$ be the acute angel subtending, from $A$, that segment which contains the mdipoint of the hypotenuse. Let $h$ be the length of the altitude to the hypotenuse fo the triangle. Prove that: \[ \tan{\alpha}=\dfrac{4nh}{(n^2-1)a}.  \]

Let D be the foot of the A-altitude, AD = h. WLOG, assume DB < DC. By similar right angle triangles or by the power of the point D to the circumcircle of the right angle triangle $\triangle ABC$, $DB \cdot (a - DB) = h^2$. Let $PQ = \frac a n$ be the defined segment containing the hypotenuse midpoint, such that the points B, D, P, Q, C follow on the hypotenuse BC in this order. Denote $\phi = \angle PAD,\ \psi = \angle QAD$.

$\tan \alpha = \tan(\psi - \phi) = \frac{\tan \psi - \tan \phi}{1 + \tan \psi \cdot \tan \phi}$

$\tan \psi = \frac{QD}{h} = \frac{QB - DB}{h},\ \ \tan \psi = \frac{PD}{h}= \frac{PB - DB}{h}$

$\tan \psi - \tan \phi = \frac{QD - PD}{h} = \frac{PQ}{h} = \frac{a}{nh}$

$1 + \tan \psi \cdot \tan \phi = 1 + \frac{(QB - DB)(PB - DB)}{h^2} =$

$= 1 + \frac{QB \cdot PB - (QB + PB) DB + DB^2}{h^2} = 1 + \frac{a\frac{n + 1}{2n} \cdot a\frac{n - 1}{2n} - a DB + DB^2}{h^2} =$

$= 1 + \frac{a^2\frac{n^2 - 1}{4n^2} - DB (a - DB)}{h^2} = \frac{a^2}{h^2} \cdot \frac{n^2 - 1}{4n^2}$

As a result,

$\tan \alpha = \frac{a}{nh} \cdot \frac{4n^2h^2}{a^2(n^2 - 1)} = \frac{4nh}{a(n^2 - 1)}$

We can let each of the n segments have length 1, then a=n. Let the middle segment be XY.
The area of $ \triangle AXY$ is h/2. So
$ h \equal{} AX \cdot AY \cdot \sin{\alpha}$
By the Law of cosines on $ \triangle AXY$,
$ AX^2 \plus{} AY^2 \minus{} 1 \equal{} 2AX \cdot AY \cdot\cos{\alpha}$
Dividing, we have
$ \tan{\alpha} \equal{} \frac {2h}{AX^2 \plus{} AY^2 \minus{} 1}$
So it remains to prove
$ \frac {2h}{AX^2 \plus{} AY^2 \minus{} 1} \equal{} \frac {4h}{n^2 \minus{} 1}$ (since a=n)
$ AX^2 \plus{} AY^2 \minus{} 1 \equal{} (n^2 \minus{} 1)/2$
We now use Stewart's twice on $ \triangle ABC$ with cevians AX and AY:
$ AB^2(n \plus{} 1)/2 \plus{} AC^2(n \minus{} 1)/2 \equal{} AX^2n \plus{} (n \plus{} 1)(n \minus{} 1)n/4$
$ AB^2(n \minus{} 1)/2 \plus{} AC^2(n \plus{} 1)/2 \equal{} AY^2n \plus{} (n \plus{} 1)(n \minus{} 1)n/4$
Adding the two, and using that $ AB^2 \plus{} AC^2 \equal{} n^2$ by Pythag, we have
$ AX^2 \plus{} AY^2 \minus{} 1 \equal{} (n^2 \minus{} 1)/2$
Which is what we set out to prove.

http://www.cut-the-knot.org/wiki-math/index.php?n=MathematicalOlympiads.IMO1960Problem3

Vo Duc Dien

Consider a point $D$ on $BC$ and name the lengths like in the figure.
Let $H$ be the pedal from $D$ on $AC$. Then $\tan \Psi = \frac{m}{l} = \frac{n}{l} \tan \theta$. Let $d=\frac{a}{n}$ be be the length of a segment divided. Then we have
\[ \tan \alpha = \frac{p-q}{1+pq}\]
where $p = \frac{a+d}{a-d}\tan\theta$, $q = \frac{a-d}{a+d} \tan \theta$.
\[ \tan \alpha = \frac{4ad \sin \theta \cos \theta}{a^2-d^2} = \frac{4dh}{a^2-d} = \frac{4nh}{(n^2-1)a}\]
as desired.


try http://www.kalva.demon.co.uk you can find there old imo problems with solutions,until imo 2003.Other imo problems and national ol. with solutions you can find in this site.(use search).And also try imo official site.

[quote="xyzgenius"]Hi everyone.
I'm new to this forum so I hope I can get some instruction. Could someone show me the link to get the solutions of IMO2000-IMO2007 ? I also eager to see as many problems and solution of national Math olympiad as i can.
Thank you so much.[/quote]

 See also here

 http://www.imo.org.yu/

 but without solutions .

Of course you can press [b]RESOURCES[/b] on the top of this site to get short lists of many IMO's. There are solutions of the best problems.

[u] Babis[/u]

IMO 2000
IMO 2004
IMO 2005
http://www.mathlinks.ro/index.php?f=177

IMO 2001
IMO 2002
IMO 2003
IMO 2006
http://olympiads.win.tue.nl/imo/index.html

Thanks so much for you support i'm checking out :)


geometry
geometric transformation
homothety
cyclic quadrilateral
power of a point
radical axis

Μια και δεν εκκρεμούν άλυτες ασκήσεις στο φόρουμ , στέλνω μια που βρήκα χθες  στο site για τις ολυμπιάδες του Honk Kong. Αυτή δεν  πρόλαβα να την κοιτάξω καθόλου . Αν δυσκολεύει το σχήμα , να μου πέιτε να φτιάξω ένα με το Cabri και να το στείλλω για ευκολία.

Ωραία άσκηση, μου θύμησε κάποια άλλη από το αγγλικό φόρουμ αλλά αυτή είναι λίγο πιο δύσκολη.

Βρήκα μια λύση αλλά είμαι πολύ κουρασμένος και ελπίζω ότι δεν είναι λάθος. Θα τη βάλω αργότερα.

ΥΠΟΔΕΙΞΗ: (αυτό που μου είχε μείνει τελευταίο, για να αποδείξω)

Εάν $ Q'_{1}\equiv (O_{1})\cap P_{1}M_{1}$ και $ Q'_{2}\equiv (O_{2})\cap P_{2}M_{2},$ αρκεί να αποδειχθεί ότι το τετράπλευρο $ Q'_{1}P_{1}P_{2}Q'_{2},$ είναι εγγράψιμμο.

Κώστας Βήττας.

[b][size=100][color=DarkBlue]PROBLEM. – (Chinese team selection test 1990 – Problem 7) - Two external common tangents are drawn to two nonintesrecting circles with centers $ O_{1}$ and $ _{2}.$ One tangent touches the circles with centers $ O_{1},$ $ O_{2},$ at points $ A_{1},$ $ A_{2}$ respectively, while the other tangent touches them at points $ B_{1},$ $ B_{2},$ respectively. Let cords $ A_{1}B_{1},$ $ A_{2}B_{2},$ intersect the line segment $ O_{1}O_{2},$ at points $ M_{1},$ $ M_{2},$ respectively. A circle passing through $ A_{1},$ $ A_{2},$ intersects the circles with centers $ O_{1},$ $ O_{2},$ at points $ P_{1},$ $ P_{2},$ respectively. Prove that $ \angle O_{1}P_{1}M_{1} \equal{} \angle O_{2}P_{2}M_{2}.$[/color][/size][/b]

[b][size=100]PROOF[/size][/b]. – We denote the point $ S\equiv A_{1}A_{2}\cap O_{1}O_{2}\cap B_{1}B_{2},$ as the external homothetic center of the circles $ (O_{1}),$ $ (O_{2}),$ as well known.

From cyclic quadrilateral $ A_{1}P_{1}P_{2}A_{2},$ we have that $ \angle A_{1}P_{1}Q_{1} \equal{} \angle A_{1}A_{2}P_{2}$ $ ,(1)$ where $ Q_{1},$ $ Q_{2},$ are the intersection points of $ (O_{1}),$ $ (O_{2})$ respectively, from the line segment $ P_{1}P_{2}.$

From $ (1)$ and because of $ \angle A_{1}A_{2}P_{2} \equal{} \angle A_{2}Q_{2}P_{2}$ $ ,(2)$ we conclude that $ \angle A_{1}P_{1}Q_{1} \equal{} \angle A_{2}Q_{2}P_{2}$ $ ,(3)$

From $ (3)$ $ \Longrightarrow$ $ A_{1}P_{1}\parallel A_{2}P_{2}$ and so, we conclude that the line segment $ P_{1}P_{2},$ passes through the point $ S$ $ ($ because of $ O_{1}A_{1}\parallel O_{2}A_{2}$ $ ).$

We denote the points $ Q'_{1},$ as the symmetric of $ Q_{1},$ with respect to the line segment $ O_{1}O_{2}$ and it is easy to show it lies on the circle $ (O_{1}).$

Without using the polar theory, we can say that the line segment $ P_{1}M_{1},$ passes through the point $ Q'_{1},$ begining from $ \angle Q_{1}M_{1}A_{1} \equal{} \angle A_{1}M_{1}P_{1}$ (not hard to prove) and so, we have $ \angle Q'_{1}M_{1}S \equal{} \angle SM_{1}Q_{1} \equal{} \angle P_{1}M_{1}O_{1}.$

Similarly the point $ Q'_{2},$ as the symmetric of $ Q_{2}$ with respect to $ O_{1}O_{2},$ lies on the circle $ (O_{2})$ and the line segment $ P_{2}M_{2},$ passes through the point $ Q'_{2}.$

 It is easy to show that the line segment $ Q'_{1}Q'_{2},$ passes through the point $ S$ and let $ L$ be, the intersection point of $ P_{1}M_{1},$ $ P_{2}M_{2}.$

Because of $ \angle Q_{1}P_{1}Q'_{1} \equal{} \frac {\angle Q_{1}O_{1}Q'_{1}}{2} \equal{} \frac {\angle P_{2}O_{2}P'_{2}}{2} \equal{} \angle P_{2}Q'_{2}P'_{2},$ where $ P'_{2}\equiv (O_{2})\cap Q'_{1}Q'_{2},$

as the symmetric of $ P_{2}$ with respect to the line segment $ O_{1}O_{2},$ we conclude that the quadrilateral $ Q'_{1}P_{1}P_{2}Q'_{2}$ is cyclic and so, we have that $ (LP_{1})\cdot (LQ'_{1}) \equal{} (LP_{2})\cdot (LQ'_{2})$ $ ,(4)$

$ ($ We have $ \angle Q_{1}O_{1}Q'_{1} \equal{} \angle P_{2}O_{2}P'_{2},$ because of $ O_{1}Q_{1}\parallel O_{2}P_{2}$ and $ O_{1}Q'_{1}\parallel O_{2}P'_{2},$ from the homothety of the circles $ (O_{1}),$ $ (O_{2})$ $ ).$

From $ (4),$ we conclude that the point $ L,$ lies on the radical axis of the circles $ (O_{1}),$ $ (O_{2}),$ which (easy to prove) is the midperpendicular of the segment $ M_{1}M_{2}$ $ ($ the radical axis of $ (O_{1}),$ $ (O_{2}),$ is perpendicular to the line segment $ O_{1}O_{2}$ and also passes through the midpoint $ L'$ of the segment $ A_{1}A_{2}$ $ ).$

So, from $ LM_{1} \equal{} LM_{2},$ we have $ \angle P_{1}M_{1}O_{1} \equal{} \angle P_{2}M_{2}O_{1}$ $ ,(5)$

From the similar right triangles $ \bigtriangleup M_{1}A_{1}O_{1},$ $ \bigtriangleup M_{2}A_{2}O_{2},$ we have $ \frac {O_{1}M_{1}}{O_{2}M_{2}} \equal{} \frac {O_{1}A_{1}}{O_{2}A_{2}}$ $ \Longrightarrow$ $ \frac {O_{1}M_{1}}{O_{2}M_{2}} \equal{} \frac {O_{1}P_{1}}{O_{2}P_{2}}$ $ ,(6)$

From $ (6)$ and because of the triangles $ \bigtriangleup P_{1}O_{1}M_{1},$ $ \bigtriangleup P_{2}O_{2}M_{2},$ have $ \angle P_{1}M_{1}O_{1} \equal{} \angle P_{2}M_{2}O_{1} \equal{} 180^{o} \minus{} \angle P_{2}M_{2}O_{2},$ we conclude that $ \angle O_{1}P_{1}M_{1} \equal{} \angle O_{2}P_{2}M_{2}$ and the proof is completed.

Kostas Vittas.


algebra
polynomial
number theory
number theory unsolved

prove that every positive integer can be written as sum of not more than 53 fourth powers of integers.

[b]Hint[/b]
using [url=http://mathworld.wolfram.com/LagrangesFour-SquareTheorem.html]Lagrange's four-square theorem[/url] and [url=http://mathworld.wolfram.com/LiouvillePolynomialIdentity.html]Liouville polynomial identity[/url].

PS This problem is case $ g(4)$ of [b]Waring's Problem[/b].

Do not double post:
[url]http://www.artofproblemsolving.com/Forum/viewtopic.php?t=226213[/url]


This problem was the only one I didn't get how to do. I brute forced it and got the answer, but can anyone give a solution that doesn't require brute force?

In a bag of three marbles, there are exactly two blue marbles.
If Kia randomly chooses two marbles without replacement,
the probability of choosing the two blue marbles is one-third.
However, before Kia chooses her two marbles, additional
marbles are added to the bag. The probability of picking two
blue marbles without replacement is still one-third. What is
the least number of marbles that could be in the bag after the
additional marbles have been added?

I'm don't really think that there's another way to do it other than randomly guessing using logic...
But, that's basically what I do for a lot of problems anyways.

This was a tricky one. Actually, the only way to do this is similar to brute force. We just make it less brutal by listing out  a few quantities.

Anyways. ... [hide="my solution"]

If we analyze the problem, we can easily deduce that there are two blue marbles to start out with. We subtract one from both the numerator and the denominator to and then multiply to get 1/3.

In other words, we need a way to write $ \frac {1}{3}$
so that both the numerator AND the denominator can be factored into two consecutive numbers. Such as the original one: $ \frac {2}{6} \equal{} (\frac {(2)(1)}{(3)(2)}.$

Here, it is helpful to make a list of the products on consecutive numbers. We need to constantly check to see if any two of the products can create our 1/3.

4x5=20
5x6=30
6x7=42
7x8=56
8x9=72
9x10 = 90

AHA! $ \frac {(5)(6)}{(9)(10)}$
Thus, the two fractions are 6/10 x 5/9 = 30/90, which is a way to right 1/3.
So, there are $ \framebox{10}$ marbles needed.
[/hide]

So random guessing?
What would you do if the desired fraction was like $ \frac{13}{61}$?
You really can't list out all the products of two consecutive numbers.

Actually, that's basically what I did.
Only I used a graphing calculator to make life easier.  :lol:

We used some algebra and stuff, it was pretty tedious.  I can't recreate the solution off the top of my head though

this a hard one number 10 and 9 were the ones we missed in the real competition.

Err... Add $ x$ marbles. Assume $ b$ of them are blue

$ \frac {2 \plus{} b}{3 \plus{} x}*\frac {1 \plus{} b}{2 \plus{} x} \equal{} \frac {1}{3}$

$ 3(2 \plus{} b)(1 \plus{} b) \equal{} (2 \plus{} x)(3 \plus{} x)$

So... you can see that the terms on the right hand side shouldn't be prime as that will lead to $ b \equal{} x$, which is a no-no and it should have a factor of 3.

So, either $ x \equal{} 3,4,6,7,$ etc...
The first one that appears with no prime is when $ x \equal{} 7$

So $ 3(2 \plus{} b)(1 \plus{} b) \equal{} 90$
$ (2 \plus{} b)(1 \plus{} b) \equal{} 30$
Check it.

Two consecutive integers, $ 5$ and $ 6$. Yay.

Lol to cognos.  It's automatically hard because you missed it XD.

Ok, I guess there's no real easy solve for x sorta thing...
Oh well, thanks anyways.

Just remember this. Since this is state and you aren't allowed to use calculators, it's not going to be 13/61. If it is, you have some problems there. The reasonalble solution is to convertly start a riot, steal the answers from someone/something, then pretend to be innocent.

This was a team round question, so you actually can.

Wait, what?!? I thought... wait... I thought it was no calculators at team round. I'm confuzzled.

Rules written on the team round

"This section of the competition consists of 10 problems which the team has 20 minutes to complete.  Team members may work together in any way to solve the problems.  Team members may talk to each other during this section of the competition.  [b]This round assumes the use of calculators[/b], and calculations may also be done on scratch paper, but no other aids are allowed.  All answers must be complete, legible and simplified to lowest terms.  The team captain must record the team's official answers on his/her own problem sheet, which is the only sheet that will be scored.  If the team completes the problems before time is called, use the remaining time to check your answers."

That was written by Mr. Frost, but I'll assume it's what's on an actual team round.

I checked the actual answer to be

[hide]10 marbles[/hide]

But im not exactly sure why.


ARML
calculus
analytic geometry
graphing lines
slope
trigonometry
quadratics

These are problems that look like they require calculus (well, at least if you are an inexperienced problem solver, which sadly most people on the team are), but can be solved with precalculus methods.

1) Find the minimum value of $x^{2}-2x+2006$.

2) Compute the coordinates where the tangent line to $y=x^{2}+5x+7$has a slope of $3$.

3) Let $A$ and $B$ be the points where lines that pass through the origin are tangent to the circle $(x-12)^{2}+(y-16)^{2}=100$.  Compute the distance from $A$ to $B$.

4) Town $A$ at $(0,0)$ and Town $B$ at $(12,19)$ are separated by a river whose shores are determined by $y=4$ and $y=7$.  A bridge crossing the river from $(x,4)$ to $(x,7)$ is the be placed so that the total distance between $A$ and $B$ is minimized.  Compute $x$.

5) Compute the minimum value of $2006x+\frac{2006}{x}$ for $x>0$.

6) What is the point on the line $3x+4y=100$ closest to the origin?

7) Given a numerical input $x_{n}$, a certain machine adds half the input to the reciprocal of the input.  The result is called the output.  If the first input $x_{1}$ is $-\frac{1}{5}$, then the first output is $-5.1$.  If this $-5.1$ output, and each succeeding output, is fed back into the machine as new input, these outputs form an increasing sequence bounded above by its limiting value $a$.  What is the value of $a$?

8) The line segment connecting $(2,3)$ to $(3,6)$ is revolved three-dimensionally around the $x$-axis to produce a disk with a beveled edge.  Compute the volume of the disk.

:(
I'm only in Alg II.

I'll save theese problems for next year.  :D

[quote="I Like Pi"]Wouldn't the calculus solutions be more straightforward and faster in most cases? Numbers 1 and 2 are easy to solve using the first derivative.[/quote]

What if you didn't know calculus?

[quote="E^(pi*i)=-1"]Er, by "precalculus" I meant "pre-calculus" - anything that doesn't involve stuff at or beyond calculus.  I don't think these problems even require trig or other "precalculus" topics.[/quote]

The problems dont REQUIRE calculus. It is simply easier to some with it.

What would 2's solution w/o calculus be? Because just the definition of "slope of tangent" is the heart of calculus at this level...

This is an absurdly late response, but . . .
[hide]
We want a line $ y\equal{}3x\plus{}k$ that intersects $ y\equal{}x^2\plus{}5x\plus{}7$ at one point.  So there is one solution to $ 3x\plus{}k\equal{}x^2\plus{}5x\plus{}7\Rightarrow x^2\plus{}2x\plus{}(7\minus{}k)$.  Thus the discriminant is zero: $ 4\minus{}4(7\minus{}k)\equal{}0\Rightarrow k\equal{}6$.  When $ y\equal{}3x\plus{}6$ intersects $ y\equal{}x^2\plus{}5x\plus{}7$, $ x^2\plus{}2x\plus{}1\equal{}0\Rightarrow x\equal{}\minus{}1$. Thus $ y\equal{}3(\minus{}1)\plus{}6\equal{}3$, and the coordinates are $ (\minus{}1,3)$.
[/hide]

This is an extremely absurd later response (just found this place), but a way I find more obvious to do that e^ipi is

[hide]just subtract 3x+k from the quadratic, getting x^2 + 2x + ?. Completing the square, (x+1)^2, shows the vertex is at x = -1 so we get (-1, 3) [/hide]

uhhh this is how i did 6 but can someone tell me if its right or not and a better way possibly?

[hide]
$ 3x\plus{}4y\equal{}100$ becomes $ y\equal{}\minus{}3/4x\plus{}100$
line $ y\equal{}4/3x$ is perpendicular to that line and goes through the origin so uhhh
$ 4/3x\equal{}\minus{}3/4x\plus{}100$ solve and you get $ x\equal{}48$ so the point is $ (48,64)$.... i hope?

[/hide]

[quote="bobma99"]uhhh this is how i did 6 but can someone tell me if its right or not and a better way possibly?

[hide]
$ 3x + 4y = 100$ becomes $ y = - 3/4x + 100$
line $ y = 4/3x$ is perpendicular to that line and goes through the origin so uhhh
$ 4/3x = - 3/4x + 100$ solve and you get $ x = 48$ so the point is $ (48,64)$.... i hope?

[/hide][/quote]

Unfortunately, no. You didn't divide correctly when expressing the line in slope-intercept form.

Your solution would otherwise be correct after making that correction. Here's a different way.

[hide]The line can be expressed as $ y = -\displaystyle\frac{3x}{4} + 25$. A point on the line can therefore be expressed as $ (x, -\displaystyle\frac{3x}{4} + 25)$. The distance from the origin to such a point is therefore $ \sqrt{x^2 + (-\displaystyle\frac{3x}{4} + 25})^2$, which simplifies to $ \sqrt{\displaystyle\frac{25x^2}{16}-\displaystyle\frac{75x}{2}+\displaystyle\frac{625}{16}}$. The vertex of this parabola is at the x-coordinate $ -\displaystyle\frac{b}{2a} = 12$. Since the parabola goes off to infinity, the vertex of the x-coordinate yields the minimum value of the quadratic equation, so it is also the x-coordinate of the point that is closest to the origin because it minimizes the expression that, after being square rooted, yields the distance from the given point to the origin. Solving for $ y$, the point is $ (12, 16)$.[/hide]

oh right -.- stupidity ftw =P
well thnx, this is why you check your answers  :D

1. 2005
2. Not sure about the precalc, but really simple calc spits out (-1,3)
I'll look at the rest

[quote="cpast"]2. Not sure about the precalc, but really simple calc spits out (-1,3)[/quote]

A pre-calc solution.

[hide]Define $ f(x) \equal{} x^2 \plus{} 5x \plus{} 7$ and $ g(x) \equal{} 3x \plus{} k$. If $ g(x)$ is tangent to $ f(x)$, then their difference must have exactly one zero. Note that the difference is a quadratic, so the minimum value of $ f(x) \minus{} g(x)$ must be zero if the quadratic is to have exactly one zero. $ f(x) \minus{} g(x) \equal{} x^2 \plus{} 2x \plus{} (7\minus{}k)$. The vertex of this parabola is at $ x \equal{} \minus{}1$, so $ k \equal{} 6$ lets $ f(x)\minus{}g(x) \equal{} (x\plus{}1)^2$, which has exactly one zero. The line is therefore $ 3x\plus{}6$, and it must be tangent at $ x\equal{}\minus{}1$, so the point is $ (\minus{}1, 3)$. [/hide]

For 5, we have $ 2006x\plus{}\frac{2006}{x} \equal{} 2006(x\plus{}\frac{1}{x}).$ So, we just need to minimize $ f(x)\equal{}x\plus{}\frac{1}{x}.$ Now, $ f(x)\equal{}x\plus{}\frac{1}{x}\equal{}\frac{1}{x}\plus{}x\equal{}f(\frac{1}{x})$.  Also, $ f(2)\equal{}\frac{5}{2}>2\equal{}f(1).$ $ f(x)$ is either quadratic or cubic (not sure), and we know that it has at most 2 minimums/maximums. $ f(1)$ is a max/min, and all others come in pairs, so there are no more. Thus, since $ f(2)>f(1)$, $ f(1)$ is a minimum, and the answer is $ 2006 \cdot 2\equal{}4012$.

I'm not sure exactly how rigorous cpast's reasoning is, but...
[hide="Alternate soluton to #5"]
By AM-GM on $ x$ and $ 1/x$, we have
\[{ \frac{x+1/x}{2} \ge \sqrt{x\cdot(1/x)}} = 1\]
Multiplying by 4012,
\[ 2006x + \frac{2006}{x} \ge 4012\]
with equality resulting at $ x=1/x$, or $ x = 1$.
[/hide]


Suppose that $f_{1},...,f_{n}$ are $n$ real positive numbers and $g_{1},...,g_{n}$ are $n$ positve integers which are constant and $x_{1},...,x_{n}$ are $n$ real positive numbers. Also assume that  $f_{1}x_{1}+...+f_{n}x_{n}$ is constant. Find the maximum amount of $(x_{1}^{g_{1}})...(x_{n}^{g_{n}})$.

A solution has been posted here:
[url]http://www.mathlinks.ro/Forum/viewtopic.php?p=579931#579931[/url]


[quote="reason"]$ (x,y,z) > 0$ prove that:

$ \frac {y^{2}xz}{(y \plus{} x)^{2}(xy \plus{} z^{2})} \plus{} \frac {z^{2}xy}{(y \plus{} z)^{2}(zy \plus{} x^{2})} \plus{} \frac {x^{2}yz}{(z \plus{} x)^{2}(xz \plus{} y^{2})}\le\frac {3}{8}$

thanx.[/quote]

Ineq
$ < \equal{} > \sum\frac{1}{(a \plus{} 1)^2(b \plus{} c)} \le \frac {3}{8}$ with:$ abc \equal{} 1$
See here http://www.mathlinks.ro/Forum/viewtopic.php?t=314790

[quote="reason"]Let $ x,y,z > 0,$ prove that

$ \frac {xyz^2}{(y \plus{} z)^{2}(x^2 \plus{} yz)} \plus{} \frac {yzx^2}{(z \plus{} x)^{2}(y^2 \plus{} zx)} \plus{} \frac {zxy^2}{(x \plus{} y)^{2}(z^2 \plus{} xy)} \leq\frac {3}{8}.$[/quote]$ \frac {3}{8} \minus{} \sum_{cyc}{\frac {xyz^2}{(y \plus{} z)^{2}(x^2 \plus{} yz)}}$

$ \equal{} \sum_{cyc}{\frac {x(x \minus{} y)^2[x^3y(9y \plus{} 7z) \plus{} x^2z(11y^2 \plus{} 9yz \plus{} 36z^2) \plus{} xyz(9y^2 \plus{} 7yz \plus{} 48z^2) \plus{} yz^2(11y^2 \plus{} 33yz \plus{} 12z^2)]}{24(y \plus{} z)^2(z \plus{} x)^2(x \plus{} y)^2(x^2 \plus{} yz)}}$

$ \geq0.$

By the way, two stronger inequalities

$ \sum_{cyc}{\frac {yz(z \plus{} x)^2}{(y \plus{} z)^{2}(x^2 \plus{} yz)}}\leq\frac {3}{2}$  and  $ \sum_{cyc}{\frac {z^2(x \plus{} y)^2}{(y \plus{} z)^{2}(x^2 \plus{} yz)}}\leq\frac {3}{2}$

still hold for positive numbers $ x,y,z.$


By putting 1.80 dollars worth of gas in his Ford Ranger, the gas tank needle moves from 1/4 full to 3/8 full. What additional amount must Ben pay to finish filling the gas the tank? (Assume that the needle is always accurate.)

A. 5.40 d   B.9.00 d   C.9.90 d     D.10.80    d E.14.40 d

[hide]3/8-1/4 = 1/8 for each 1/8 tank it costs 1.80 to fill with the remaining 5/8 it would cost 5(1.80) $ 9.
Answer B)$9[/hide]

[hide]Got the same answer as Bluecarneal, sorry i''m too lazy too post it up, and the answer was B[/hide]


The goal of this game is to be the middlemost, not the best. For every round, send me a number (plz no complex numbers). The middlemost nuimber gets a point. At the end, the person with the [b]middlemost[/b] number of points wins. The number of rounds depends on the number of signups. PM to sign up. Deadline for signups is 12:00 am EST Saturday November 8.

Signups deadline is, yes, 15th. My mistake.

PS the goal is to be mediocre at being mediocre. :D

Okay i wil join.

This will be an interesting game.

Joining is by PM.
By the way numbers don't have to be integers

Currently I have:
james4l
chenhsi
cf249
pacman2812
eggylv999

[quote="cf249"][quote="aleph0"]Joining is by PM. [/quote]

*cough*[/quote]
I will not accept any signups not from PM. Signups deadline is in 2 days 6 hours 13 minutes 27 seconds.

I will join.

I am very bad at being mediocre at being mediocre.

Read post 9!

Now I have 10 people:
james4l
chenhsi
cf249
pacman2812
eggylv999
evilhamster
AIME15
pythag011
Dojo
erabel

In case you want to send in your numbers for round 1 the deadline for that is 12:00 am EST Saturday November 22.

BTW I changed the rules so when a unique middlemost number of points appears then it wins.

Also, in cases of ties between numbers, I will randomly select one (e.g. with a coin)

People who have sent in numbers:
james4l
chenhsi
pacman2812
eggylv999
kimmystar94

People who have not:
cf249
erabel
AIME15
evilhamster
pythag011
Dojo

Deadline for first round is 12:00 am EST Saturday November 22. Signups deadline is 12:00 am EST Saturday November 15. Signups by PM only.

Is there any limit to how high or low a number can be?

No limit as long as you can actually PM it to me without my computer crashing. Also no infinite numbers.

People who have sent in numbers:
james4l
chenhsi
pacman2812
eggylv999
kimmystar94
cf249
AIME15

People who have not:
erabel
evilhamster
pythag011
Dojo

Deadline for first round is 12:00 am EST Saturday November 22. Signups deadline is 12:00 am EST Saturday November 15. Signups by PM only.

People who have sent in numbers:
james4l
chenhsi
pacman2812
eggylv999
kimmystar94
cf249
AIME15
erabel

People who have not:
evilhamster
pythag011
Dojo

Signups are over.
Deadline for first round is 12:00 am EST Saturday November 22, but it would speed up the game if people PMed numbers for first round before that. People who miss the deadline will be removed.

I'd be glad to give you that point, but I don't think that would be allowed.

I have received numbers from:
james4l
chenhsi
eggylv999
cf249
jhredsox93

I have not received numbers from:
pacman2812
kimmystar94
AIME15
pythag011
Dojo
ernie

[hide="score"]
Dojo: 2
pythag011: 2
eggylv999: 2
AIME15: 2
cf249: 2
chenhsi: 1
james4l: 1
jhredsox93: 1
ernie: 0
pacman2812: 0
kimmystar94: 0
[/hide]

Round 15 is over between March 8 and 9.

[/quote]

[quote="aleph0"]I have received numbers from:
james4l
chenhsi
eggylv999
cf249
jhredsox93
pacman2812

I have not received numbers from:
kimmystar94
AIME15
pythag011
Dojo
ernie

[hide="score"]
Dojo: 2
pythag011: 2
eggylv999: 2
AIME15: 2
cf249: 2
chenhsi: 1
james4l: 1
jhredsox93: 1
ernie: 0
pacman2812: 0
kimmystar94: 0
[/hide]

Round 15 is over between March 8 and 9.

This game is going to go on for longer than I can handle and I am sorry to say that I have abandoned it.

ok I shall continue this thread.
[b]EVERYBODY RESUBMIT YOUR SUBMISSIONS TO ME![/b]

[hide="the score"]Dojo: 2
pythag011: 2
eggylv999: 2
AIME15: 2
cf249: 2
chenhsi: 1
james4l: 1
jhredsox93: 1
ernie: 0
pacman2812: 0
kimmystar94: 0 [/hide]

I have received numbers from:
cf249
james4l
eggylv999

I have not received numbers from:
Dojo
pythag011
AIME15
chenhsi
jhredsox93
ernie
pacman2812
kimmystar94

Deadline is midnight on thursday.

[quote="mathking123"]What happens if someone submits a nonreal number( like 5+3i)?[/quote]
You die.

But I imagine 1=2 would take the norm of the number.

[b]Answer Ackermann(Graham Number, Graham Number)[/b]

Just to make it harder to order the higher entries.

I'm glad to see that someone took this over.  :thumbup:


$\boxed {\ \triangle ABC\ }\ : \ A=50^{\circ}$$;\ P\in (ABC),\ PB=PC;\ E\in PB\cap AC,\ F\in PC\cap AB;$
$m(\widehat {PEF})=50^{\circ},\ m(\widehat {PFE})=30^{\circ}\Longrightarrow m(\widehat {ABE})=\ ?.$

If is possibly, I wish a nice solution, otherwise after two weeks I offer my disagreeable solution.

Although it is not an nice solution:
$P$ is the circumcenter of $ABC (<BPC=2<A=100, PB=PC)$
Denote $R=PA=PB=PC$. Now:
$ABE: \frac{AE}{sinx}=\frac{AB}{sin(x+50)}=\frac{2R.cosx}{sin(x+50)}$
$EBC: \frac{EC}{sin40}=\frac{BC}{sin(x+50)}=\frac{2R.cos40}{sin(x+50)}$
$ABC: \frac{AC}{sin(x+40)}=\frac{BC}{sin50}=\frac{2R.cos40}{sin50}$
As: $AC=AE+EC$, so:
$\frac{2R.cos40.sin(x+40)}{sin50}= \frac{2R.sinx.cosx}{sin(x+50)}+\frac{2R.sin40.cos40}{sin(x+50)}$
Finally: $x=40$

This was only an attempt, because we have found a trigonometric identity
Now the correct procedure:
$EFP: \frac{PF}{PE}=\frac{sin50}{sin30}$ ..(a)
$BPF: \frac{PF}{PB} = \frac{PF}{R} = \frac{sinx}{sin(x+80)}$..(b)
$PCE: \frac{PE}{PC} = \frac{PE}{R}=\frac{sin(50-x)}{sin(x+50)}$..(c)
Combining (a),(b),(c):
$x=30$


function
monovariant
combinatorics solved
combinatorics

In the forest 12 lonely guys lost their contacts to the outside world. There are some friendships within these guys being mutual and lasting forever. Each guy has an own house. And every house is green or brown. Each month of the year is assigned to a certain guy as visiting month where he visits his friends. At the final day of the visiting month the respective guy will change the colour of his house if more than half of his friends have a another house colour than he has.
Prove that after a finite number of month no guys has to change the colour of his house anymore !

Take a 12 vertices graph. 2 vertices i,j are connected if i and j are friends.
Each vertex has a colour (green or brown).
Notice that after a guy changed the colour of his house the number of edges of the graph that have endpoints of the same colour strictly increases.
But the number of edges with endpoints of the same colour is bounded by the total number of edges which in turn is bounded by 66 (the number of edges of a 12 complete graph).
So this process cannot be infinite.

Based on a British olympiad question :) On John Scholes' website, the statement of the problem is five times as long as the solution:
[url]http://www.kalva.demon.co.uk/bmo/bsoln/bsol912.html[/url]

The idea of my solution is the same. Making it clear just takes more  :D .

The problem was already used in different olympiads, e.g. Moscow Mathematical Olympiad. It is one of these potential questions where you have to find a property why a certain function always de-/increases by at least 1. For an example see: 1986/3.

[quote="Arne"]It's called a MONOVARIANT :)
Gabriel Caroll wrote a nice article on monovariants. Check out his site.[/quote]
where can i find this article?

Here :) : http://web.archive.org/web/20050308223109/http://www.people.fas.harvard.edu/~gcarroll/math/monovar.zip

His (archived) site : http://web.archive.org/web/20050306043340/www.people.fas.harvard.edu/~gcarroll/index.html


probability
ratio
function
percent
articles
geometry
geometric transformation

I'm creating this thread to see if people can come up with interesting mathematical analogies for God, and for people to discuss said analogies. There need be no particular religion associated with your analogy, but there can be if you want it.

Please do not insult anyone for their ideas, or turn this thread into a debate about religions. I mean no offense to atheists; you are free to contribute your thoughts as well but please don't just whine about the fact that there is no god.

Anyway, on with my idea, which sparked the creation of this thread!

One of my beliefs about religion is that religion need not conflict with rationality. Faith takes over where reason fails. I was thinking about how to relate that to mathematics in particular, and I realized that one might compare God to the natural numbers. God is what we don't know. As we understand more about the natural numbers, God sort of "moves back."

[i]But we already understand the natural numbers. Isn't this like saying we already know God?[/i]

Do we really? Think about it. Nobody can ever put God on a piece of paper. How many natural numbers do you think have been written by the hand of a human? I'm sure very few numbers over $10^{20}$ have been written down. How about named, spoken about, thought of? We have naming systems for large numbers, but only a minimal number actually have their own names, have actually been discussed. And what's the largest number anyone ever cared about? It's [url=http://tinyurl.com/36cqj]Graham's number[/url], a number so large that it's beyond any comprehension or modicum of understanding. And you know what? It's finite. If you picked a random natural number, the probability that it's larger than Graham's number is 1.

[i]But who cares about whether a particular number was actually [/i]used[i]? If you give me a any natural number, I could name it, or make new notation to name it. And we already understand the natural numbers as a class, too. [/i]

My response to that is: do we really? Then why does the field of number theory exist? I would even go so far as to claim that the only natural number that we really, truly understand, to the point of not having much else to ask about it, is $1$.

[i]All right, so the natural numbers aren't all ancient history. One last complaint: isn't this still very limiting on God? God is merely countable, even? :P [/i]

I'm just using natural numbers, one of the oldest concepts and most widely understood (such as it is) in mathematics, as an example to demonstrate the idea I mentioned at the start. There's what we know, and there's God. The natural numbers greater than $10^{20}$ are just as numerous as the natural numbers greater than Graham's number. Extending what we know doesn't reduce God. In my opinion, mathematical discovery honors and glorifies God.

[quote="Xevarion"]I mean no offense to atheists; you are free to contribute your thoughts as well but please don't just whine about the fact that there is no god.[/quote]

Ouch.

More on topic, though, this sounds like a pretty standard god-of-the-gaps idea, doesn't it?

[quote="JBL"][quote="Xevarion"]I mean no offense to atheists; you are free to contribute your thoughts as well but please don't just whine about the fact that there is no god.[/quote]

Ouch. [/quote]
I don't mean to be harsh. I just mean a comment like "this is stupid because god doesn't exist" won't contribute to the discussion at all. As it happens, I'm not religious myself. I just want to understand religion.

[quote="JBL"]More on topic, though, this sounds like a pretty standard god-of-the-gaps idea, doesn't it?[/quote]
Maybe it does to you, but I'm an uneducated boor who's just trying to figure things out for himself. What is the "standard god-of-the-gaps idea"?

from dictionary.com:
God
A being conceived as the perfect, omnipotent, omniscient originator and ruler of the universe, the principal object of faith and worship in monotheistic religions.
The force, effect, or a manifestation or aspect of this being.

math is perfect and absolute: only in math can you have [i]exactly[/i] half a pie or a [i]perfectly[/i] round circle.  math is omnipotent because it does not change depending on what people want it to be.  we cannot make a formula equal something that it isn't.  math is omniscient because all life, down to the ratios of elements or number of protons, traces back to a mathematical concept, forcing math to be always affecting everything.  in the same way, one could claim everything originates from math.  the laws of math govern physics and the functioning of the universe.  although math is not generally worshipped, a true god would not cease to be a god if he/she/it was not worshipped.  everything today is derived from, or an effect of, some core mathematical principle.

therefore, according to the dictionary, math is god.

[b]wobster109[/b]: I think that's a very attractive argument, but I have some doubts about it. My problem is this: how do we really know that mathematics governs science and nature? It seems to me (being a bit of a formalist) that mathematics is "generated," in a sense, by a set of axioms that we created to match our observations, our intuition. In fact I think if I do not tread carefully I am on a quick path to atheism -- the old "god is a social construct" idea. I don't want to dismiss this as [i]wrong[/i] (since it is in fact my belief, to some extent) but I do want to dismiss it as [i]irrelevant[/i] [JBL, this is what I meant before. I hope you were not offended; I surely didn't mean it.] However, one could still argue that mathematics as we know it is in fact divine (the axioms already existed and we figured them out).

Sorry, I guess I didn't really say much there. I hope these ideas are interesting to people other than myself...

PS [b]wobster[/b] -- math may match the definition of God, but it doesn't seem to me that the definition says an entity is God [i]if and only if[/i] it satisfies those characteristics. That is, the conditions seem to be necessary, but not sufficient. Unless you demand the uniqueness of God, too. ;)

to xevarion: very nice argument.  i'll claim the laws of physics, derived from mathematics, that the universe follows would not change whether humans generated different axioms or not.  i don't mean to be off topic, i find math a startlingly accurate representation of what people generally find a god to be.  so it's an atheist's model for god. :P

concerning the p.s., when something is defined as, say, x, doesn't that mean the condition x is both necessary and sufficient for determining the original object?  if a hominid is defined as a humanoid creature with a particular set of characteristics, and then a fossil is discovered, if the fossil fits the definition of a hominid, then it is a hominid.  on a cartesion plane, a function is defined as a relation between x and y where for every x, there is one distinct y.  if you had a relation where for every x, there was one distinct y, that is sufficient to categorize it as a function.

This is sort of off topic, but in reply to [b]wobster[/b] about definitions: How about this? A rational number is defined to be $\frac{a}{b}$ where $a$ and $b$ are integers, $b \ne 0$. But $\frac{5}{7}$ fits the definition of "real number" as well as "rational number." $11$ fits the definition of "prime" as well as "rational number." So, mathematics may fit the definition of "God" as well as some other (perhaps more specific and relevant) thing. In another vein, a hurricane is defined in terms of its wind speed. A tornado also has high enough wind speed to match the definition, but a tornado isn't really the same at all.

I guess what I mean is that a definition should be clear about whether you mean "Objects that are $X$ must have characteristics $a, b, c$" or "Objects are $X$ if they have characteristics $a, b, c$" or both. Any of those statements (in a non-mathematical context) can equally well be referred to as a definition, I think.

Back to God/math/physics/the universe: I agree with what you say about how the universe obeys the same physical laws regardless of what we say. I guess I am just not as confident that what we consider "laws of nature"  now are actual laws. Perhaps if I understood quantum physics better, I might have a different view. I don't want to start a "does God play dice" discussion here, though (maybe in another thread :)).

@ [b]Treething[/b]: Classic, but still a good one. Does anyone else have short witty ones like that? I can't think of one at the moment.

I believe Ramanujan had some idea about God and the Universe being related to the "expression" \[0 \cdot \infty\]

[quote="Xevarion"]
Anyway, on with my idea, which sparked the creation of this thread!

 I was thinking about how to relate that to mathematics in particular, and I realized that one might [b]compare[/b] God to the natural numbers. God is what we don't know. As we understand more about the natural numbers, God sort of [b]"moves back."[/b]

[i]But we already understand the natural numbers. Isn't this like saying we already [b]know God?[/b][/i].[/quote]

First I would like to say that we will never know and we mustn't try to know(absolutely everything) GOD.And also we can never compare GOD to anything,anyone etc.Actually we don't have anything to compare him with.Mathematics is just of a product a human brain which is very limited.The brain of human has a limit in understanding everything.Also we are discovering new things ,we are developing that is a very very small percent of (1-2%) what we could know,discover etc.And we can't compare GOD to anything.One little story:One computer chip asks the other : With how many volts do you work? The other says:120.And you?  "200 volts" they both ask each other "It is interesting with how many volts does  our creator works?
I think everything is obvious.
I would like to continue but:
It is 1 oclock night.I am very tired(want to sleep).Maybe I will continue later

[quote="Xevarion"][quote="spider_boy"]First I would like to say that we will never know and we mustn't try to know (absolutely everything) GOD. And also we can never compare GOD to anything, anyone etc. Actually we don't have anything to compare him with.[/quote]

I think this is true of most modern views of God. I would just like to point out that ancient mythology did not view God (or more often "the gods") as beyond human comprehension. They were supernatural, but human- (or other real-world thing-) like.

I sort of like the idea of God as "just beyond our reach" no matter what we do. I mean, I can try to understand God, and I might improve my understanding of how God is viewed by various religions and various people, but I still won't actually understand God. I don't think there is anything wrong with thinking about God. God's all about faith, right? If I think about God and conclude it* doesn't exist, I must not have had faith before when I was unsure either. If I conclude God does exist, God should be happy -- I didn't have faith before and then I would!

*I don't like how English doesn't have a good pronoun by which to refer to God. "He" is standard but pretty naive, I think.

By the way [b]spider_boy[/b] I really like your computer chip analogy. :D[/quote]
I absolutely agree with you on this:"I sort of like the idea of God as "just beyond our reach" no matter what we do. I mean, I can try to understand God, and I might improve my understanding of how God is viewed by various religions and various people, but I still won't actually understand God. I don't think there is anything wrong with thinking about God. " I also agree with you that "I don't think there is anything wrong with thinking about God. "According to muslim Allah (GOD) has asked us to pray(namaz) for him 5 times a day.And muslims say that the simple reason he has said this is because he wants us always to remember him.and to know to pray for him every time.

God is infinite. One thing I think all religions can agree on is that God created the world, if this does not comply with anyone, my bad. Since God created the world he obviously had to use some of his power. But after subtracting a number from infinity, infinity remains infinity, and God remains infinite. Is this something like what you were looking for?

It's interesting to suggest creating the world subtracted from God.
I don't know enough about Islam to know what this reference is, but I recently read:
[quote="Some Thing about Islam"] The full text of the hadīth is: “For God, the Glorious and Exalted, the world is without value; and among the creatures known to us which God has created, there is no existent more despicable to Him than the world, and since the time when He created the world, God has never looked kindly upon it.” Bihār al-Anwār, vol. 70, p. 110, The Book of Faith and Infidelity, chap. 122, hadīth 109. [/quote]
The universe as we know it is a mere compact subset of the infinite-dimensional Hilbert space that is God. :D

Before I go on here, I thought I'd point out that I am an atheist, so if any of my comments sound, well, atheistic, I did not mean to offend or oppose anyone that is religious.

[quote="wobster109"]therefore, according to the dictionary, math is god.[/quote]
[quote="Xevarion"]This is sort of off topic, but in reply to [b]wobster[/b] about definitions: How about this? A rational number is defined to be $\frac{a}{b}$ where $a$ and $b$ are integers, $b \ne 0$. But $\frac{5}{7}$ fits the definition of "real number" as well as "rational number." $11$ fits the definition of "prime" as well as "rational number." So, mathematics may fit the definition of "God" as well as some other (perhaps more specific and relevant) thing. In another vein, a hurricane is defined in terms of its wind speed. A tornado also has high enough wind speed to match the definition, but a tornado isn't really the same at all.
[/quote]

I see both of your points, but both arguments are slightly flawed.  Since mathematics fits the definition of a god, it would mean mathematics is [i]a[/i] god, not necessarily [i]the[/i] god, similar to the way that 5/7 is a real number, but not all real numbers are rational.  But, Xevarion, the tornado analogy does not apply - A hurricane must be defined in terms of much more than just wind speed, so a tornado does NOT fit the definition of a hurricane, similar to the way a rational number does not fit the definition of an irrational number, even though both are real.  So tornados and hurricanes both fit the definition of "Storm With High Wind Speed," but neither fits the definition of the other.

This means that mathematics could be considered one of many possible gods, if we are going by that definition.

However, if we are referring to God as the unknown, or the thing/being used to explain the unknown, then the obvious parts of mathematics that correspond to God are all the unknown parts - the unsolved conjectures, the open questions, the branches not yet invented.  Religion thrives on such mystery.  If everything were known, people would not feel a need to make up a god to fill in the cracks.  And this is what makes mathematics appealing as well; there is so much that is not known, that there are endless opportunities in mathematics to explore and reach out to "God".

On the other hand, if we are defining God as the creator of all things (in other words, a blanket explanation of all the things that we don't yet understand how they actually came into being) then the "god" of mathematics is our own minds.  Humans created mathematics, step by step, axiom by axiom.  It is purely abstract, and while it was developed to explain phenomena in the real world, it is not really real itself, save in our thoughts and imaginations.  Mathematics cannot exist without Human thought to create it.

[quote="hmmm"]On the other hand, if we are defining God as the creator of all things (in other words, a blanket explanation of all the things that we don't yet understand how they actually came into being) then the "god" of mathematics is our own minds.  Humans created mathematics, step by step, axiom by axiom.  It is purely abstract, and while it was developed to explain phenomena in the real world, it is not really real itself, save in our thoughts and imaginations.  Mathematics cannot exist without Human thought to create it.[/quote]
Did we really create it? Personally, I think so. I think any system of axioms generates an abstract structure, and we can explore that structure with our minds like we explore a work of art with our eyes, or music with our ears.

However, it seems like a remarkable coincidence that the simple axioms that correspond to our observations and the intuition we gain from the real world create such a rich mathematical structure. This is where we could still let God into mathematics. One could claim that the particular structure that we associate with the real world actually was "created" and existed before we did, and we have merely deduced some of its properties by observation.

The reason I don't subscribe to this is that I don't see why radically different axioms couldn't create an equally rich but totally different structure. It's just hard for us to think of it because it doesn't match our intuition.

Basically, what you said is the fundamental idea of formalism. It's a perfectly respectable school of mathematical philosophy. But it's not the only one -- don't be too quick to dismiss the others.

(Also, I don't think formalism and belief in God are mutually exclusive. There's probably just a strong correlation.)

See Pascal (probability and god) and Godel's Theorem ( infinite logical systems and god).

[quote="José"]Well, here's and idea I was teached. If we suppose that we had once nothingness, we are saying that nothing can exist now, because nothing comes from nothingness, but we are here, so there must be "something" (you can call it as you want). Hopkings call "this" the Big Bang, maybe some call "this" God, others "Alá", others "Buda", you can put "this" whatever name you want, but there MUST be "something", that ever existed, because if not, we wouldn't be here....[/quote]
Hi guys.It is good that we are discussing and everone  writes his opinion.But dear José I think there are gaps in your words.
I suppose let us don't try to think about nothingness etc. because it is just waste of time.First of all can anyone imagine nothingness?(I don't think so)As we are unable to answer this question it is obvious that we can't understand further.so let us stop here.
[quote="bubka"]Nobody said that God created the world. The only thing accepted somewhat in today's science is the existence of a prime mover. A small initial push could have set the world moving.

I can only think of one analogy to this. You could call God the basis of a proof by induction. Once you take into account his initial nudge, everything falls into place.[/quote]

I don't think so.How logical is it that "A small initial push could have set the world moving."?? :?:
Later I will provide all of you with information about the proof of existence of GOD,and why we can't see him.
I hope i am not offending anyone.

[quote="Xevarion"]Maybe it does to you, but I'm an uneducated boor who's just trying to figure things out for himself. What is the "standard god-of-the-gaps idea"?[/quote]

Overdue explanation:  your initial analogy, boiled down and stripped of the most interesting parts, sounds to me like "God is all of those things we don't know about yet."  This is the notion of a god of the gaps -- God is in everything that we don't have, say, a scientific explanation for.  It's one way of reconciling religious and naturalist beliefs about the functioning of the world that might otherwise be in conflict.

The weakness of it (from the point of view of someone who wants a strong, unapproachable god) is that there isn't any theoretical bound on our knowledge.  If god is in the things we don't yet know, that means that (a) god is constantly diminishing and (b) we might find ourselves "inducting" god out of existence.  For people of some religious persuasions (especially those who tend towards a deistic model) this is resolved simply by making god the Prime Mover -- he set everything in motion, set up whatever physical laws were necessary to make the rest follow, and then let it go.  So, in other words, the fact that evolution explains the diversity of life doesn't deny god's role in forming life, it just says that god was involved 13-15 billion years ago and hasn't bothered tinkering around since.  (This view is largely incompatible with a literal reading of any major religious text and is also prone to reject claims of things like miracles.)  The strength of this view is that it will always be compatible with observed reality -- you can't disprove a god of the gaps.  The major alternative in the United States, an activist god who literally created the world relatively recently, literally created light (and plants!) before he had created stars, literally kicked Adam and Eve out of a literal Garded of Eden, etc., has the notable deficiency that its alleged actions are in direct conflict with observed reality, and so requires very awkward contortions ("fossils were just planted to test our faith!") to reconcile its existence with what we actually know about the world.

So, that took me rather far afield.  The wikipedia article on [url=http://en.wikipedia.org/wiki/God_of_the_gaps]god of the gaps[/url] is pretty good and comes without all my unrelated blathering.

[quote="spider_boy"]Later I will provide all of you with information about the proof of existence of GOD,and why we can't see him.
I hope i am not offending anyone.[/quote]
I am not offended. I am interested. :)
[hide="An off-topic question for spider_boy"] Can you recommend for me where I can find a good English translation of the Koran?[/hide]

@[b]JBL[/b] Thanks. That's a good explanation, and your blathering, though perhaps unrelated, is still interesting. I firmly believe that religion need not contradict rationality; however, religion itself need not be rational.
(I'm saying religion but I could just as well say "ideology" or something. I just mean that there are too many questions related to how we should act and make decisions that have no firm rational answers, so one must either have some irrational beliefs or wander without direction. And religion is a very popular way to organize one's irrational beliefs.)

By the way, I should have just Googled that myself or found it on Wikipedia, but I was annoyed because I felt like you just dismissed my post as a common idea and didn't even give a reference. So I apologize for answering rudely and being lazy. No hard feelings, I hope. :)

edit: Regarding the god-of-the-gaps argument. I googled it, and I read the first few pages about it, and I found some interesting stuff. One page said

[quote="Second result"] This is not based in logic. It is simply a statement of pessimism about the future progress of science.

Down through the centuries, science has eliminated a great many of its gaps. People who had used the Gap argument were embarrassed, since their God shrank in power with each new scientific advance. [/quote]
Maybe some people view it this way, but I think this is a rather unwarranted assumption. For example, in my opinion, quantum mechanics is far more marvelous and mysterious than classical mechanics. So even though science has uncloaked some more of the mysteries of the universe, it is as though we were looking at the face of a beautiful grandfather clock, trying to figure out how it worked, and someone opened it up, and we could look inside. Now we can see what's driving the hands, but we might also be even more impressed at the skill and genius of the maker of the clock!

The second page I looked at seems to hold exactly the same view, that "god of the gaps" is foolish because it means we have an ever-shrinking (and currently pretty tiny) god. I think people who make this kind of claim are baffled by technology and don't really understand science (or infinity). You could say, "Oh, once physicists fully understand how quarks work, we're pretty much done on the small scale. Not much room for God there." But what justification is there for that? Why shouldn't, in 2000 years, people laugh at "top, bottom, up, down, strange, charm" just as we laugh at "earth, air, water, fire" now?

I guess that was the point of my initial analogy. We might feel like we understand a lot because we understand a lot more than people 50, 500, 5000 years ago. But we can't even begin to imagine how much we still don't know.

[quote="Richard Feynman"]I don't have to know an answer. I don't feel frightened by not knowing things; by being lost in the mysterious universe without having any purpose. Which is the way it really is, as far as I can tell, possibly. It doesn't frighten me.[/quote]

[quote="spider_boy"]But dear José I think there are gaps in your words.
I suppose let us don't try to think about nothingness etc. because it is just waste of time.First of all can anyone imagine nothingness?(I don't think so)As we are unable to answer this question it is obvious that we can't understand further.so let us stop here.
anyone.[/quote]

Hey wait a minute, I don't know if nothingness is the exact world (I want to say "la nada", when there's nothing).
And I can imagine nothingess, so I don't see why you make me stop my point fo view

[quote="Xevarion"]I felt like you just dismissed my post as a common idea and didn't even give a reference. So I apologize for answering rudely and being lazy. No hard feelings, I hope. :)[/quote]

Sorry, I wasn't intending to be dismissive at all but was rather curious if you thought your analogy reflected a god of the gaps type of idea -- certainly, no hard feelings.


algebra
polynomial
quadratics
Vieta
quadratic formula

[hide="is this right?"]
I got 10 numbers.
$a\in \{0; 24;-1; 25; 27;-3; 32;-8; 49;-25\}$

The idea is $a^{2}-24a = n^{2}$. Making perferct square, $(a-12)^{2}=n^{2}+144$. Change $a-12=m$. You get $(m-n)(m+n)=144$. And then find all possible natural $n$.
[/hide]

Yeah, this is correct.  :)

I had a feeling something was wrong with my solution. :D

[quote="kimby_102"][quote="4everwise"]Here's the problem: For how many real numbers $a$ does the quadratic $x^{2}+ax+6a=0$ have only integer roots for $x?$

Is my solution correct? I'm not sure if my method finds all the values of $a.$

[hide]Applying the quadratic formula, $x=\frac{-a\pm \sqrt{a^{2}-24a}}{2},$ and $a^{2}-24a=n^{2}\implies (a+n)(a-n)=24a$ for some integer $n.$

Hence, $(a+n)$ or $(a-n)$ must be a multiple of $a,$ and $a$ must be a multiple of $n.$ We now assign a new name to $a.$ Let $a=kn$ for some nonnegative integer $k.$ Substituting and simplifying, \[(kn+n)(kn-n)=n^{2}(k+1)(k-1)=24kn.\] This means that $\frac{24k}{(k+1)(k-1)}=n$ is an integer, and $(k+1)(k-1)$ are factors of $24.$ The values of $k$ that satisfy this are $k=0,2,3,5.$ Additionally, for each value of $k,$ there is exactly one corresponding value of $a,$ so our answer is $4.$ [/hide]

I have a looong way to go... :blush:[/quote]
[hide="another method"]
we write the given equation in terms of $a$:
$a=-\frac{x^{2}}{x+6}$.
even though we are not given that $a$ is an integer, we observe that the quadratic has integer roots iff $a$ is an integer.
So we use long division (or systhetic) on the above equation and we get a remainder:
$a=\frac{36}{x+6}$.
36 has 9 factors and $a$ can also take $0$, so the answer is $10$. :)
[/hide][/quote]

Nice method!

What do you use synthetic division on? I don't get where you got the $a=\frac{36}{x+6}$

well, notice that the degree of the numerator is greater before?? he basically divided it out, and since $a$ is integer, the reduced expression must also be an integer, not some fractional values.

I divided and got $a=-x+6-\frac{36}{x+6}$ :maybe:

yes, so $\frac{36}{x+6}$ should be an integer so that $a$ can be an integer...

[quote="4everwise"]Applying the quadratic formula, $x=\frac{-a\pm \sqrt{a^{2}-24a}}{2},$ and $a^{2}-24a=n^{2}\implies (a+n)(a-n)=24a$ for some integer $n.$

Hence, $(a+n)$ or $(a-n)$ must be a multiple of $a,$ and $a$ must be a multiple of $n.$ [/quote]
You are wrong. The result that $(a+n)$ or $(a-n)$ must be a multiple of $a,$ works only if $a$ is a prime.

[quote="D.P.L"]To Kimby: What did you divide with to get a=36/x+6
 :blush:[/quote]

I believe he just threw out the integer part of it.

[quote="kimby_102"][quote="D.P.L"]To Kimby: What did you divide with to get a=36/x+6
 :blush:[/quote]
sorry, i meant to say that the remainder is $\frac{36}{x+6}$, not $a$.[/quote]

Thanks for clarifying that because that was what I was wondering too.

I mean what did you -x^2/x+6 to get a remainder of 36/x+6?

Hey I was just wondering, couldn't $a$ be negative, so isn't the answer 19?

[quote="vvr1590"]How did you find that 'a' must be an integer?[/quote]
by Vieta's theorem, $-a$ is the sum of the roots.
Since the roots are integers, the sum of the integers must be an integer as well.


[quote="Fabiomr1"]Consider a convex polygon of $n$ sides and assume that it does not have two of its diagonals that are parallel nor three diagonals that concur in the same point that is not a vertex. How many are the points of intersection of the diagonals that are exterior to the polygon?[/quote]

Total number of diagonals is $d=\frac{n(n-3)}{2}$
Total number of intersections of diagonals (from distinct vertices) is $N_{Tot}=\frac{d(d-(2n-7))}{2}=\frac{n(n-3)(n^{2}-7n+14)}{8}$
Total internal intersections (from distinct vertices) is $\frac{n(n-1)(n-2)(n-3)}{24}$

Requested number is Total number of intersections of diagonals less Total internal intersections  :

$N=\frac{n(n-3)(n^{2}-7n+14)}{8}-\frac{n(n-1)(n-2)(n-3)}{24}$
$N=\frac{n(n-3)(n-4)(n-5)}{12}$

--
Patrick

[quote="Fabiomr1"]very good, mate, but could you explain better how did u find the internal diagonals from distinct vertices?
thanks[/quote]

Any set of $4$ vertices generate exactly one interior intersection. So the requested number is $\left(\begin{array}{c}n\\4\end{array}\right)$


calculus
integration
logarithms
calculus computations

Prove that:
$ \int_{2}^{\plus{}\infty} \frac{\,dx}{x(log(x))^{p}} < \plus{}\infty$
iff $ p>1$.

What can we say about $ \sum_{n\equal{}2}^{\plus{} \infty} \frac{1}{n(\ln n)^p}$?


[quote="shyce"]I got \[5x^2-40x+75=0\], although I'm not sure how the factor it to get exact answers from there.[/quote]Well first you divide by $5$ to get $x^2-8x+15$.  Now you want two numers $x$ and $y$ such that $xy=15$ and $x+y=-8$.  We can see that $-3$ and $-5$ meet this.  So we have $x^2-8x+15=(x-5)(x-3)=0$, so your answers are $5$ and $3$, assume the quadratic you have is correct.  :D

[quote="shyce"]Find the points of intersection:

\[x^2+y^2=25\] and \[2x+y=10\]

In exact answers preferably, analytically.[/quote]

It is clear that $(3, 4)$ lies on the line $2x+y=10$ (of course, on the circle too, but it is immaterial here)

So, $(3-t, 4+2t)$ represents a general point on the line.

substituting this point in the circle equation and solving for 't' will give  the required intersection points.
This method helps solving several difficult situations involving the intersection of a line with a curve.

[quote="Rushil"]Interesting method. Can you plz prove this works every time. I don't quite see why we should assume the points the way you have!!![/quote]

If $(x_1, y_1)$ is any point on the line $ax+by+c=0$, then $ax_1+by_1+c=0$.

And it is easy to check that the point $(x_1-bt, y_1+at)$ always satisfies the equation of line, showing that it represents a general point on the line.   This is a simple modification of Symmetric form of a line equation (without using the inclination of the line)

And as far as assuming the point in the given problem, it struck to me just like that, one need not have any specific reason to pick up a point on the line with this method.  It works all the time.  Still some choices will help minimise number of steps.


IF a,b,c,are the sides of a triangle prove that
a/c+a-b + b/a+b-c +c/b+c-a is greater than or equal to 3.

hello, do you meant
$ \frac{a}{c\plus{}a\minus{}b}\plus{}\frac{b}{a\plus{}b\minus{}c}\plus{}\frac{c}{b\plus{}c\minus{}a} \geq3$?
Sonnhard.

hello, your inequality is equivalent with $ \frac{2(a^3\plus{}b^3\plus{}c^3\plus{}3abc\minus{}2ac^2\minus{}2ba^2\minus{}2cb^2)}{(a\plus{}b\minus{}c)(a\minus{}b\plus{}c)(a\plus{}b\minus{}c)} \geq0$
By substituting $ a\equal{}y\plus{}z,b\equal{}x\plus{}z,c\equal{}x\plus{}y$ in the denominator we get $ yx^2\plus{}y^2z\plus{}xz^2 \geq 3xyz$ this is true because of the AM-GM inequality.
Sonnhard.

Oh,$ \sum\frac{a}{c\plus{}a\minus{}b}\geq 3\sqrt[3]{\frac{abc}{\prod(c\plus{}a\minus{}b)}}\geq 3$ :wink:

This inequality can be solved with simple subsitions for x , y and z to be the denominators of the fractions.
Then u get the AM GM form where u can argue about them being positive using the fact that a b c are sides of a triangle.


Here's a question that could have multiple answers.  The point of this question is to get you thinking.

Robert, Alec and David all got the same score on a math competition.  They want to break the tie but unfortunately are forced to settle for flipping a coin.  How should they break the tie in a fair manner?(they can flip the coint multiple times.  assume that the coin is fair)

In spoiler:

[hide]They can all flip a coin three times. Person with the most heads wins and if theres a tie between two people the two of them do it over.[/hide]

Another approach: [hide]They flip the coin twice.  HH = Robert, HT = Alec, TH = David, TT = do over.[/hide]

Edited: Put in spoiler by request.

[quote="Ravi B"]Another approach: They flip the coin twice.  HH = Robert, HT = Alec, TH = David, TT = do over.[/quote]

What happens if they had really bad luck that they always get TT? :maybe:

Yes, it is theoretically possible for that to happen, although the probability is 0.  There is no way to avoid that infinite possibility, because 1/3 can't be written as a fraction whose denominator is a power of 2.

Could you guys try and put your answers in spoiler mode, and try not to answer the questions if you're really advanced?  Since this problem is really short, I could easily see A+MATH's solution right under it.

Like I said before this problem isn't meant to have just one solution.  It is meant to make ppl think a little and be creative.  So even if you see one solution there might still be other elegant solutions out there.

Let me throw out a related problem.  Suppose you want to define an event that has probability exactly $1/\pi$.  Show how you can use a fair coin to construct such an event.

To answer the question. . . .  Alec flips with Robert, then with David, then David flips with Robert. whoever loses twice is eliminated, then the two remaining  flip once.

To answer Ravi, if no one loses twice then they do over.


Let $ a_n$ denote the integer closest to $ \sqrt{n}$. Evaluate the sum $ S\equal{} 1/a_1\plus{} 1/a_2\plus{}...\plus{}1/ a_{1980}$

See [url]http://alt1.mathlinks.ro/viewtopic.php?p=18181#18181[/url] for 2004 instead of 1980.

This would be like a number 3 or 4 AIME i think.

No, there was like a #14 on the AIME that was the sum from n=1 to something of 1/(the integer closest to the fourth root), which is even easier.

Because there is less computation, since fourth powers get larger faster than squares so you don't need to check for as many numbers. See [url]http://www.mathlinks.ro/viewtopic.php?p=394531#394531[/url]. You can do a simple brute force as I did. But today, it might be less, like a number 8.


geometry
geometric transformation
rotation

While browsing through books on Organic Chemistry, I stumbled upon something. It said that the proposed electrophilic addition mechanism of halogenation of alkenes is insufficient and cannot explain the stereochemistry of cyclic alkene halogenation.

Can someone explain this to me? Please don't just post a link because I am not very good at extracting information from long webpages.

Thanks.

hmm. seems u read a mechanism tha said electrophillic addition of $ Br_{2}$ to the double bond takes place via breaking down of $ Br_{2}$ into $ Br^{+}$ and $ Br^{-}$ nad then as usual the electrophillic addition takes place.
thsi is wrong because it can't account for the anti-stereochemistry of the reaction.
what happens is the alkene group does something of a $ S_{N}2$ substitution of $ Br_{2}$ removes one $ Br$ as $ Br^{+}$ and the intermediate is a bromonium ion intermediate now if this attack took place from the upper side of the plane conatinig alken then clerly rotation due to the cyclic bond is restricted so due to steric hindrance $ Br^{-}$ attacks from below giving vicinal di bromide.
but thsi can't be accounted by the first 'mechanism' as that wouls give an carbcation intermediate fre to rotate thus anti addition becomes meaningless

Yes that's right. The mechanism of alkene halogenation begins with the formation of a complex between the pi-bond of the alkene and the halogen molecule (this is favourable, since $ Cl_{2}$, $ Br_{2}$, and $ I_{2}$ are soft acids and C-C double bonds are soft bases), which is then converted into a cyclic halonium ion with the concomitant formation of an halide anion (this overall first step is [i]reversible[/i], that's why a low concentration of halogen favours allylic halogenation, but that's another story). The next step is an SN2 reaction, with the halide ion acting as nucleophile and attaking the cyclic halonium from the opposite side of halogen, since this is electrostatically more favourable. As the halide bonds from the opposite side of the already present halogen in the halonium ion, the overall addition is anti.

Another related and interesting issue is electrophilic addition to alkynes, which is erroneously explained in some textbooks.


[quote="mathwizarddude"]Prove that $ 2222^{5555} \plus{} 5555^{2222}$ divides 7.[/quote]
[hide]Hint:
Fermat Little's theorem[/hide]

You don't need anything that complicated...

Its not that complicated and its really fast so its advisable. Maybe.


Let ABC be a right angled triangle with <B=90 and [AB]=3,[BC]=4,[AC]=5
Take any D point interior region of ABC triangle.If [AD]=x+1, [ BD]=x and [CD]=x+2 find x=?

Hello, mmrproblem!

Quote:Let ABC be a right angled triangle with <B=90 and [AB]=3,[BC]=4,[AC]=5
Take any D point interior region of ABC triangle.
If [AD] = x + 1, [ BD] = x and [CD] = x + 2, find x.

[hide]

Draw DE perpendicular to BC.

Let BE = p, DE = q, and EC = 4-p.

Draw DF perpendicular to AB.

Then DF = p, FB = q, and AF = 3-q.

In right triangle DEB: [1] p2 + q2 = x2.

In right triangle DEC: [2] (4 - p)2 + q2 = (x + 2)2

In right triangle DFA: [3] p2 + (3 - q)2 = (x + 1)2

Expand [2]: 16 - 8p + p2 + q2 = x2 + 4x + 4

From [1], it becomes: 16 - 8p + x2 = x2 + 4x + 4

. . . which simplifies to: 4x + 8p = 12 ---> p = (3 - x)/2

Expand [3]: p2 + 9 - 6q + q2 = x2 + 2x + 1

From [1]. it becomes: x2 + 9 - 6q = x2 + 2x + 1

. . . which simplifies to: 2x + 6q = 8 ---> q = (4 - x)/3

Substituting these into [1]: (3 - x)2/4 + (4 - x)2/9 = x2

. . . which simplifies to: 23x2 + 86x - 145 = 0

. . . which factors: (23x - 29)(x + 5) = 0

. . . and has roots: x = 29/23 and x = -5

Since x is positive: x = 29/23.

[/hide]


Find all functions $f: [0,1]\rightarrow R$ such that
$f(xy) = xf(x)+yf(y)$


ABCD is a square with sides of unit length. Points E and F are taken respectively on sides AB and AD so that AE=AF and the quadrilateral CDFE has maximum area. What is this maximum area?

[quote="236factorial"]This one's strange....  wouldn't it just be one (the points are on point A)?[/quote]

Exactly what I was thinking. Unless AE=AF=0.000000000000000000000001 cm. But there isn't aany measurement!  :ninja:

wouldnt CDFE be the area of the square? because the minimum for EF is .00000000000000000000(Etc.)1 which....would round to 0?

EDIT:U=length of a side of the square

I got
[hide]

I guessed that E and F are the midpoints of AB, and AD.

$u^2-\frac{(\frac{1}{2}u)^2}{2} -\frac{\frac{1}{2}u\cdot u}{2}=$

$u^2-\frac{(\frac{1}{2}u)^2}{2}-\frac{1}{4}u^2=$

$\frac{3}{4}u^2-\frac{(\frac{1}{2}u)^2}{2} ?$
[/hide]

EDIT: New answer

[hide]
u= length of a side
x=distance from E or F to A

$u^2-\frac{x^2}{2}-\frac{(u-x)\cdot x}{2}=$

$\frac{2u^2-x^2-x^2+ux}{2}$

$\frac{u^2-x^2+ux}{2}$
now maximize, which I have no idea how to do so I'm going to assume $x=\frac{1}{2}u$

$\frac{u^2-(\frac{1}{2}u)^2+u(\frac{1}{2}u)}{2}$=
$\frac{\frac{3}{2}u^2-(\frac{1}{2}u)^2}{2}$=
$\frac{3u}{4}^2- \frac{(\frac{1}{2}u)^2}{2}$

which I haven't noticed till now is exactly what I previously got.  :(

[/hide]

If it's really how you did it, pianoforte, it is too hard for this forum.....

You should end up with a numerical answer...

solved my equation-

[hide]
$\frac{5}{8}$
[/hide]

and I found an easier way, I think :?

[hide]
F and E are the midpoints of AD and AB.

Triangle CEB is half of that half of the square=1/4

Triangle AEB is half of that quarter of the square=1/8

1-1/4-1/8=5/8[/hide]

How do you know that using the midpoints gives the maximum area?

[quote="4everwise"]Good job pianoforte. That's correct.

I'm not sure if you can assume that midpoints give the maximum area. Or can you... ?[/quote]

Yeah! :D

...and no I can't. I just guessed, it seemed right when I drew it out:?


symmetry
geometry
geometric transformation
reflection
circumcircle
perpendicular bisector
power of a point

denotefive circles c1,c2,c3,c4,c5.if we take c5 as the inversin circle, then c1 and c4 are inversly similar,c2 and c3 are also inversly similar.prove:there exists a circle tangents to c1,c2,c3,c4.
(how to use Casey theorem to get the result?)

i wonder if i've described clearly.
it's easy to calculate the tangents between c1 and c4,c2 and c3.
but i don't know how to deal with that of c1 and c3,respectively.
in fact,it's a conclusion in Modern Euclidean Geometry.as a [b]direct [/b]deduction of Casey theorem.[/b]

I don't think this has much in common with Casey's theorem. Suppose $ \mathcal C_1, \mathcal C_2$ both intersect $ \mathcal C_5.$ Since inversion preserves angles between curves, the circles $ \mathcal C_1, \mathcal C_4$ form equal angles with the inversion circle $ \mathcal C_5.$ The same holds for the circles $ \mathcal C_2, \mathcal C_3$ WRT $ \mathcal C_5.$ Invert the figure $ \mathcal C_1 \cup \mathcal C_2 \cup \mathcal C_3 \cup \mathcal C_4 \cup \mathcal C_5$ in arbitrary circle $ \mathcal O$ centered on $ \mathcal C_5.$ This inversion takes the previous inversion circle $ \mathcal C_5$ into a line $ c_5$ and the remaining circles $ \mathcal C_1, \mathcal C_2, \mathcal C_3, \mathcal C_4$ into some other circles $ \mathcal C_1', \mathcal C_2', \mathcal C_3', \mathcal C_4'.$ Again, inversion preserves angles between curves, therefore the circles $ \mathcal C_1', \mathcal C_4'$ form equal angles with the line $ c_5$ and the same holds for the circles $ \mathcal C_2', \mathcal C_3'$ WRT $ c_5.$ This means that  $ \mathcal C_1', \mathcal C_4'$ are symmetrical WRT the line $ c_5$ and $ \mathcal C_2', \mathcal C_3'$ are also symmetrical WRT $ c_5.$ By symmetry, four circles $ \mathcal K_1', \mathcal K_2', \mathcal K_3', \mathcal K_4'$ exist, all centered on the line $ c_5$ (perpendicular to this line), each tangent to all $ \mathcal C_1', \mathcal C_2', \mathcal C_3', \mathcal C_4'.$ If $ \mathcal K_1, \mathcal K_2, \mathcal K_3, \mathcal K_4$ are inversion images of $ \mathcal K_1', \mathcal K_2', \mathcal K_3', \mathcal K_4'$ in $ \mathcal O,$ then each of  $ \mathcal K_1, \mathcal K_2, \mathcal K_3, \mathcal K_4$ is tangent to all $ \mathcal C_1, \mathcal C_2, \mathcal C_3, \mathcal C_4.$

Nice solution Vladimir, I've got this approach:

Let $ P_1,P_4$ denote the variable tangency points of a circle $ \mathcal{T}_i$ with $ \mathcal{C}_1$ and $ \mathcal{C}_4.$ Let us consider the positive inversion WRT $ \mathcal{C}_5,$ although the result is still true for both positive or negative inversion WRT $ \mathcal{C}_5.$ Pairs $ P_1,P_4,$ lying on a same side of the line $ O_1O_4,$ connecting the centers of  $ \mathcal{C}_1,\mathcal{C}_4$ are inverse under such positive inversion, due to the conformity, or equivalently $ P_1P_4$ passes  through the center $ O_5$ of $ \mathcal{C}_5.$ Hence, we have a set of four circles $ \mathcal{T}_i ,$ $ i \equal{} 1,2,3,4,$ where a pair is internally and externally tangent to $ \mathcal{C}_2$ for each pair $ P_1,P_4$ lying on different sides of $ O_1O_4.$ This inversion maps $ \mathcal{C}_1 \mapsto \mathcal{C}_4,$  $ \mathcal{C}_2 \mapsto \mathcal{C}_3$ and takes $ \mathcal{T}_i$ into itself. If $ \mathcal{T}_i$ is tangent to $ \mathcal{C}_2,$ then $ \mathcal{T}_i$ is tangent to the inverse image $ \mathcal{C}_3$ as well.

I used the following lemma before, without posting a proof:

[b][color=red]Lemma:[/color][/b] Let $ (C)$ be arbitary circle, let $ (O)$ be arbitrary circle with center $ O \in (C)$ and $ c$ their radical axis. Let $ \mathbf R_{(O)}, \mathbf R_{(C)}, \mathbf R_{c}$ be reflections in $ (O), (C), c,$ respectively. Obviously, $ c \equal{} \mathbf R_{(O)}((C)).$ Let $ A$ be arbitrary point and $ B$ its reflection (inversion image) in $ (C)$. Then

$ B \equal{} \mathbf R_{(C)}(A) \equal{} \mathbf R_{(O)} \circ \mathbf R_{c} \circ \mathbf R_{(O)}(A) \equal{} \mathbf R_{(O)}^{ \minus{} 1} \circ \mathbf R_{c} \circ \mathbf R_{(O)}(A).$

[i][color=red]Proof:[/color][/i] Let $ A'$ be inversion image of the point $ A$ in $ (O),$ let $ B'$ be reflection of the point $ A'$ in the line $ c,$ and let $ B$ be inversion image of the point $ B'$ in the circle $ (O).$ By the basic properties of inversion in $ (O),$ the quadrilateral $ AA'B'B$ is cyclic, let $ (P)$ be its circumcircle. Since $ (P)$ passes through $ A$ and its inversion image $ A'$ in $ (O)$ $ \Longrightarrow$ $ (P) \perp (O).$ Since $ (P)$ is centered on the perpendicular bisector $ c$ of $ A'B'$ $ \Longrightarrow$ $ (P) \perp c.$ Since $ (P), (C)$ are inversion images of $ (P), c$ in $ (O)$ $ \Longrightarrow$ $ (P) \perp (C).$ Let polars of $ A, B$ WRT $ (O)$ (perpendiculars to $ OA, OB$ at $ A', B'$) intersect at $ X.$ The quadrilateral $ OA'B'X$ is cyclic on account of right angles at $ A', B',$ let $ (Q)$ be its circumcircle on diameter $ OX.$ Its center $ Q,$ the midpoint of $ OX,$ is also on the perpendicular bisector $ c$ of $ A'B'.$ Let $ (Q)$ cut perpendicular $ OC$ to $ c$ again at $ C'.$ Then $ XC' \perp OC$ and $ c$ is also perpendicular bisector of the segment $ OC'$ $ \Longrightarrow$ the line $ XC'$ is polar of the point $ C$ WRT $ (O).$ Since the polars of $ A, B, C$ concur at $ X,$ these 3 points are collinear. But $ A, B \in (P)$ and inversion in $ (C)$ takes $ (P) \perp (C)$ to itself $ \Longrightarrow$ the point $ B$ is the inversion image of the point $ A$ in the circle $ (C).$ $ \square$

Applying the lemma to $ \mathcal C_5$ as the circle $ (C)$ and all points $ A \in \mathcal C_1 \cup \mathcal C_2,$ (hence $ B \in \mathcal C_4 \cup \mathcal C_3$), the inversion images $ \mathcal C_1', \mathcal C_4'$ of $ \mathcal C_1, \mathcal C_4$ in $ \mathcal (O)$ are symmetrical with respect to the radical axis $ c_5$ of $ \mathcal (O), \mathcal C_5.$ So are the inversion images $ \mathcal C_2', \mathcal C_3'$ of $ \mathcal C_2, \mathcal C_3$ in $ (O).$ Regardless of $ \mathcal C_1, \mathcal C_2$ intersecting $ \mathcal C_5$ or not.


algebra
polynomial
quadratics
superior algebra
superior algebra unsolved

Construct fields of each of the following orders: $ \textbf{(a)}$ 9 $ \textbf{(b)}$ 49 $ \textbf{(c)}$ 8 $ \textbf{(d)}$ 81 (you may exhibit these as $ \frac{F[x]}{(f(x))}$ for some $ F$ and $ f$).

Relevant Theorems to use:
$ \textbf{(1.)}$ Let $ f(x)$ be a polynomial in $ F[x]$. $ \frac{F[x]}{(f(x))}$ is a field iff $ f(x)$ is irreducible.

$ \textbf{(2.)}$ $ F$ is a finite field of order $ q$ and let $ f(x)$ be a polynomial in $ F[x]$ of degree $ n \geq 1$. Then $ \frac{F[x]}{(f(x))}$ has $ q^n$ elements

Hint: 9 in 3 squared, 49 is 7 squared, 8 is 2 cubed, 81 is 3 to the 4th power. Finite fields of prime order are easy to construct (What is $ \mathbb{Z} / (p)$ when p is prime?)

All you have to do is find irreducible polynomials in the appropriate finite fields.  Those polynomials must be:  a quadratic polynomial in $ \mathbb{F}_3$, a quadratic polynomial in $ \mathbb{F}_7$, a cubic polynomial in $ \mathbb{F}_2$, and a quartic polynomial in $ \mathbb{F}_3$.


The [url=http://ipsc.ksp.sk]IPSC[/url] is happening May 21, so now is when you should probably realize that this is something you want to do and therefore sign up for. Teams of at most 3 spend 5 hours for the contest. It's a mix of coding and algorithms with math and problem solving -- you get several (more like 8) problems with corresponding "easy" and "hard" data sets that you can solve using any means you have (just one computer though.) Oh yeah... it's also fun and a great way to skip class!

It actually is not. It's closer to algorithms and general problem solving than just math -- their site has problems from past years if you want to see. But since it's fun and quite a few people here probably know enough about coding, I thought it was worthwhile to bring it up.

I actually registered... and I know a couple of ppl who has registered this... unfortunately that day is a school day and I could not skip out.

[quote="paradox"]So nobody else is doing this? Or am I just being ignored?  :)
(If you couldn't tell from my location, I'm not actually affiliated with this contest.)[/quote]

You don't have a location entered in your profile.

[quote="sandra90"]Hi! i,m from Romania but i'm in the 8th grade. Is it possible for me to participate? :)[/quote]

It doesn't seem like there are any grade limitations, except that you have to be in secondary-school are below to participate in the division.

[quote="paradox"]Oh yeah... it's also fun and a great way to skip class![/quote]

Yes, if you are able to skip class for it  :(


geometry
3D geometry
calculus
integration
geometry proposed

One day I found ( http://www.origami.ru/mod/mode/1084dm.htm ) the instructions on how to construct a glass from a square piece of paper. The unsolveable problem for me is how to calculate the volume of the glass, knowing the side of the square.

The shape of this glass is like a cone, which, however doesn't have one apex, but the "line of apexes". Trying to solve this problem I came to difficult-for-me-to-integrate triple integrals :blush: .

Intersesting. There has to be some way to do this without multiple integrals. I will try to find one.

Good luck! When you or somebody else will at last solve this problem I'll post a new one. I'm not good in geometry, that's why the second unsolvable (for me) problem is geometrical too.


Sorry guys. I've been a bit out of town this week with all the Xmas preparations and other problems. The contest will resume this Sunday (actually this is better as otherwise most of you would have been busy with holidays these two weeks :) ).

[quote="toumaf"]So everything is moved one week later, isn't it?[/quote]Yes :)

I have two questions :

- In the first problem, we have to find possible values of $R$ in term of $p$ isn't it?
- The second problem seems wrong (I took about five seconds to solve it (maybe am I wrong) ).
- Do you think my second statement was a question?

Yes, at the first question the range of $R$ depends of course of $p$.
The second question was indeed trivial, due to my mistake :)

What about problem 3 is it true?
It seems that its wrong

It seems correct to me.
Maybe there was a modification of the problem : I'm not sure it wasn't asked at first to prove that the sum of the radii equals their product instead of twice it.

[quote="amir2"]Do we have time to send solutions of 6th Round?  ;)[/quote]Yes, you can send them until tomorrow night.