geometry Mafia algebra polynomial articles

Those (e. g. Santosh and Karthik) who have been at this program, could I get your comments please? Any information would be helpful, including format of classes, daily schedule, free time activities, paths from building to building with the most shade...

Governor's Honors ('07 here, wootz) was basically the greatest summer ever. You spend four hours every morning (8:00-12:00) Monday-Saturday in your major, which I'm just assuming is math... you also select a minor area once there, which you'll go to for two-and-a-half hours after lunch Monday-Friday. The rest of the time is basically yours until 10:30, at which time you'll have to be on your hall and your RA will conduct a hall check. They hold random seminars every day at all hours all throughout the week, anything from playing Guitar Hero in a lecture hall to making and eating baklava, playing large group games like Mafia, whatever, and you'll be stuck at VSU with 700 of the most awesome people in Georgia, so you'll always have something to do. The different majors will also hold events and concerts the entire time you're there, which you should not miss out on for any reason. There's no excuse for not seeing the best musicians in the state for free. Some of the classes are mostly lecture format, and some are based more around individual and group problem solving and the like. You'll also take a programming course, different ones depending on your abilities. The professors are awesome and brilliant teachers. You're bound to learn something about something nearly every day.

It'll be hot. Shade is scarce, and gnats are plentiful. The food is apparently better than it was in the summers before I was there, but it's still hit or miss. There's always a salad bar, a sandwich bar, and amazing frozen yogurt in addition to the main meal, though, and there's a small store where you can buy snacks and stuff, not to mention the pizza place that's a little pricey, but great food. Your whole hall will share only a handful of showers, so there may be some waiting in line going on. You'll also have to keep your room relatively neat, as your RA will check it often. The staff don't have too many rules, and they're not hard to follow, but they enforce them strictly, and breaking them repeatedly may get you sent home early. Happened to two or three people while I was there, so don't be like them. All of the staff are great people, and most of them have been in your shoes before, participated in GHP themselves, so they know everything that you'll be going through, good or bad. They'll always be there to help out if you ever need anything.

I could ramble all day about GHP, so I'll cut it out here. Make sure you enjoy it to the fullest, because you'll spend the next year or so wishing you could go back. And by the way, congratulations.

Hmm, DannyKim, is it just me, or are all of your posts pointless and spammy and completely nonsensensical? :wink:

who is "Criticalline1859" and "DaNnYkIm"?

Hi, Boru! How's your GHP going? (Or is it started yet?)

I think Criticalline1859 is Andrei. I don't know the other one, but I'm sure he is not from Georgia.

Hence his location saying "Tallahassee, FL" :P !

PS:  Hi, Boru, Tim, and maybe-Andrei!

Same from me!

I didn't get into GHP because I'm not beast/balanced like that.

Yeah, I'm Andrei (my username is 50% reference to math stuff-Riemann Hypothesis, by the way, 30% not thinking of anything else, and 20% a failed attempt at poking fun at other usernames). The second week of GHP is coming to an end, and Boru will not tell me how his GHP is going so I can put it into my post and be efficient (we're in the same computer lab working on a project).

[quote="llolloll"]Hey Andrei, Tianqi, and Mathgeek.

How are you?

GHP is pretty fun...it is slightly slower than I would like, but we're still learning lots :)

For example, I learned that i^2 = -1

(Who is JohnJimJoeBob?)[/quote]

Um, I'm Sitan. :wink:

Wait, you didn't know that i^2=-1???????????????????????? :rotfl:

Sarcasm is hard to detect, you know...

Boru is taking a complex numbers class (mainly about the connections to analytical geomety, I think), hence the joke.

Sitan, I guess I did know that i^2= -1....

Who is Iniquitus?

By the way, since this is a GHP forum, maybe I should post some GHP problems here?

I'm pretty sure Iniquitus is Oliver the |_|83R 1337 beaaaaaaaaaaaaaaaaaast.

[quote="Arvind_sn"]It's just you.  :wink:[/quote]

Ah...  That explains the fact that at least the first 10 pages of your posts from your profile are from GFF. :P

Hmm, this thread is getting quite spammy and non-GHP related. :ninja:

Hi guys I'm Eric Chen

I'm at GHP with Andrei and Boru

Hey Eric,

It's nice to meet you. Since this is a GHP forum, maybe you would like to share some GHP (math) problems?

Hey Andrei,
That article was pretty funny.

Please choose another name, chenxiyuan. "Eric" has already been claimed. Thanks.

[quote="Criticalline1859"][quote="Arvind wrote"]He is andrei's long lost twin.[/quote]
?
Anyway, here is something I found funny.
[url]http://en.wikipedia.org/wiki/Proof_by_intimidation[/url][/quote]

Arvind wrote wrote?

Anyways, that means that half of Karthik's proofs are proof by intimidation. :|  :rotfl:

Also, @Eric M., yeah I thought he was talking to you for a sec, and I'm like, wait, Eric hasn't posted much in this thread, and then I'm like, oh, yeah, Eric Chen. :P

Now I know I shouldn't post at 10:00.
But seriously, Karthik's ability to see through those long series of steps is pretty impressive.

Fortunately for mere mortals, the methods in his proofs by intimidation are mostly unnecessary - he makes problems much harder than they need to be :).

Mathematicians have long judged proofs not only by their correctness, but also by their elegance. An "ugly" proof will frequently be dismissed purely because it is ugly.

Hi Mr. Fulton!!!!!!!!!

Anyways, ugly proofs would include computer proofs, right?  Like the Four Color Theorem's proof?  Well, sometimes, the problem is too complicated to call for an elegant proof, so we have no choice.

Unless if, of course, we're talking about Karthik's proofs.   :| :rotfl:  jkjkjkjkjkjkjkjk

Hi Sitan,

I don't believe that all computer proofs are necessarily ugly. The four color map proof was certainly a unique proof at the time. I don't know if anyone has given any better proof even 30 years later.

Tom

Hi,

Does anyone know how to do that problem where we have a 4 digit number and each digit is less than or equal to the digit to its left...well, does anyone know how many such numbers there are? Can this be done without casework?

Yes, I'm sure that someone knows how to do that problem, and that someone knows how many numbers there are.  :wink:
Yes, it can be done without casework.

[hide="hint"]Try picking which digits you are going to use before arranging them in order.[/hide]

[hide]
First note that each group of chosen digits can be arranged in exactly one way. Let $ x_i$ denotes the number of digit $ i$ used. We have
\[ x_0\plus{}...\plus{}x_9\equal{}4
\]
By balls-and-urns, this equation has $ \binom{13}{4}\equal{}715$ solutions, so there are $ 715$ such 4-digit numbers. But we also counted $ 0000$, so the answer is $ 714$.
[/hide]


I seem to have lost my ability to post an avatar.  My other account can, but this account has lost the whole avatar control panel.  The whole control panel disappeared.  Any chance an admin may have removed it?

Thanks.

go to your profile and check if you enabled avatars?


modular arithmetic
superior algebra
superior algebra unsolved

Hi, please help!

B(subscript42) = {0,1,7,15,21,22,28,36}. I need to find all atoms of B(subscript42).

I understand that all of these are (mod42). Don't know what do to next!

Many thanks in advance!

An atom of a Boolean algebra is a nonzero element a that
cannot be written in the form

a = b $ \vee$ c with

a $ \neq$ b and

a $ \neq$ c.

Hello again!

Are there any other ideas as to how solve this?

Many thanks!

Consider divisibility by each of the prime factors of 42. Each such prime will appear in a product if and only if it appears in one of the factors.

Thank you!

So, the prime factors of 42 are: 2, 3, 7

Forgot to mention that B(subscript42) is a set of idempotents.

So, B(subscript42) = {0,1,7,15,21,22,28,36}

0 divides 2,3,7
1 does not divide any of the prime factors of 42
7 divides 7
15 divides 3
21 divides 3 and 7
22 divides 2
28 divides 2 and 7
36 divides 2 and 3

So, which ones are the atoms of B(subscript42)?

I could not quite understand this: "Each such prime will appear in a product if and only if it appears in one of the factors". Does it mean that the only atom is 0?

Thanks!

Quite the opposite: 0 is not an atom since $ 7\cdot 36\equal{}0$. As 7 is divisible by 7 and 36 is divisible by 2 and 3, their product is divisible by 42 so must be 0.

In a similar way, $ 36\equal{}22\cdot 15$ is not an atom, but 7 is.

Hi, sorry but I am struggling to understand it.

I have just done all 64 multiplications.

How do we define an atom?

7*36=0 is not an atom because it divides 2,3,7 and its product is divisible by 42

How about 7*15=21 i.e. 7/7, 15/3, 21/ (7 and 3), but not 2. So understand it's an atom, as the product is not divisible by 42?

Or 7*28=28, so 7/7, 28/ (2 and 7), 28 / (2 and 7), but not 3 - is it again an atom?

Or 7*22=28, so 7/7, 22/2, 28/ (7 and 2), but not 3 - it is an atom.

All these products are divisible by either 7,2,3 or all of them.

Boolean algebra is a new territory for me.  So, apologies, that it's taking me a long time to understand it.

Thanks for your help!

Umm... [i]you[/i] defined an atom for us, by giving the definition. An atom is a number $ a$ that can't be written as a product of two elements other than $ a$. So:

- 0 is not an atom because it equals $ 7\cdot 36$ and neither 7 nor 36 is equal to 0.

- Saying that $ 7\cdot 15\equal{}21$ and that $ 7\cdot 22\equal{}28$ means 21 and 28 aren't atoms either, since neither 7 nor 15 equals 21 and neither 7 nor 22 equals 28.

- Saying that $ 7\cdot 28\equal{}28$ doesn't say anything about whether something is an atom. You wrote 28 as a product of two numbers, but one of them is 28. (But we know 28 isn't an atom from the above paragraph.)

In this case, I think I should also explain why 7 is an atom. This is more complicated because you can't just use a single equation to verify this.

You could go through the entire multiplication table. That's a lot of work, but if you really did multiply every pair of numbers, you can check that the only times you get 7 as a result of multiplying two numbers is when at least one of the numbers is 7.

However, a more elegant way is this: Suppose $ 7\equal{}bc$ where $ b,c$ are elements of $ B_{42}$. Because $ bc\equiv 7\pmod{42}$, $ bc$ is divisible by 7 so either $ b$ or $ c$ is.

WLOG say $ b$ is divisible by 7, so we want to show $ b\equal{}7$. If $ b$ were not equal to 7, then $ b$ would be divisible by either 2 or 3 (or both). But then so would their product (i.e. if $ b$ were even then $ bc$ modulo 42 would be even), so we would not get a product of 7, a contradiction. Therefore, $ b\equal{}7$. So the only factorizations of 7 are the "trivial" ones where at least one element equals 7, making 7 an atom.


i think you meant to say $ f(3)\equal{}5$, not $ f(4)$.

the characteristic polynomial of the recurrence $ t^4\minus{}4t^3\plus{}6t^2\minus{}4t\plus{}1\equal{}(t\minus{}1)^4$, whose only root, 1, has multiplicity 4. the solution thus takes the form $ f(n)\equal{}c_0\plus{}c_1n\plus{}c_2n^2\plus{}c_3n^3$. to get the $ c_i$, use the initial values. the solution is $ f(n)\equal{}1\plus{}\frac{4}{3}n\minus{}\frac{1}{2}n^2\plus{}\frac{1}{6}n^3$.


Παιδιά θα ήθελα όλοι να απαντήσετε σε αυτην την δημοσκόπηση με reply για το όνομα κι την ομάδα σας εκτός από την ψήφο σας καλό ειναι να γνωριχζουμε τ πάντα ο ένας για τον άλλο κι τι καλύτερο από την ομάδα σας
 :rotfl:  :rotfl:  :D οκ λοιπόν πρώτος εώ Είμαι οπαδός του δικέφαλου αετού περιμένω απαντήσεις σας

Το είχα δει να έρχεται ...  :D

Αεκάρα είμαι , δεν πολυασχολούμαι όμως   :)
Του βορρά ή του νότου  Αμβρόσιε ???   :P

Ολυμπιακός μέχρι το κόκκαλο...... :D

[u][b]PANATHINAIKOS FOOTBALL CLUB[/b][/u]

[hide]Na lynoume omws kai kamia askisi pou kai pou.......[/hide]

Παναθηναικοςςς :omighty: ... :D

πάντως οι περισσότεροι στα προφορικά πέρισι ήταν [color=green]ΠΑΟ[/color]

Oπα sorry +εγω ψηφισα αλλα δεν εστειλα μνμ.....Ε,μην λεμε τα ιδια ...Οτι και οι τελευταιοι ή καλυτερα οποιοι ειναι περισσοτεροι...(βλ,results...)

 :D  :D  :D

Ειμαι ολυμπιακος οπως και ο Σιλουαν αλλα δεν ασχολουμαι πολυ με ποδοσφαιρο!Θα προτιμουσα να ξοδεψω τον χρονο μου με μια παρτιδα σκακι η εναν αγωνα μπιλιαρδου....... :)

Μεγάλωσα στο Κερατσίνι και έτσι είμαι Ολυμπιακός, αλλά πρώην, γιατί δεν ασχολούμαι με το ποδόσφαιρο. Δεν ξέρω αν μετράει αυτό, στη δημοσκόπηση.

Κώστας Βήττας.

Ολα μετράνε αγαπήτοί μου φίλοι
Ζήτω η τιμημένη περιοχή κι οι αυτές του Πειραιά που έχουν την τιμημενη φτωχολογιά το μεδούλι του λαού μας

ζητω η μπαλλα ακι ζητω η ΑΕΚ.ΠΑΜΕ ΓΙΑ ΤΟ ΔΙΠΛΟ ΑΜΒΡΟΣΙΕ

ΟΛΥΜΠΙΑΚΟΣ- ΑΕΚ   1-0          :lol:

[quote="silouan"]ΟΛΥΜΠΙΑΚΟΣ- ΑΕΚ   1-0          :lol:[/quote]

Ναι είδαμε μεγάλη μπάλα πάλι  :D

ΝΑΙ ΦΙΛΟΙ ΜΟΥ ΑΛΛΑ ΤΕΤΟΙΑ  ΝΙΚΗ ΜΕ ΑΥΤΟΓΚΟΛ ΚΙ ΣΑΣ ΠΑΙΖΑΜΕ ΜΟΝΟΤΕΡΜΑ ΚΙ ΜΑΣ ΑΔΙΚΗΣΕ Ο ΒΑΣΑΡΑΣ ΝΑ ΤΟ ΧΑΙΡΕΣΤΕ ΤΟ ΠΡΩΤΑΘΛΗΜΑ ΣΑΣ ΚΙ ΤΗΝ ΣΟΥΠΕΡ ΚΙΛΚΑ Ε ΣΥΓΝΩΜΗ ΛΙΓΚΑ

Ποίος ΜΑΓΚΑΣ Σπάει το κατεστημένο κι διάλεξε τον τιμημένο Εργοτέλη ????
Φαντάζομαι ο Αλέξανδρος ο Κρητικός  Μπράβο σε αυτόν που το εκανε κι θα ήθελα να το πεί δημόσια

Εργοτελερα!!! μ'αρεσει. κανενας Παοκτζης? υπαρχει? Αμβροσιε καλο το παιχνιδι χθες αλλα η ισοπαλια δε βολευει.

Pragmati egw ipostirizw ton Ergoteli (kai amesws meta fisika ton OFH- parolo pou mas ta xei kanei mantara apo tote pou efige o Gerald(akis)). Den asxoloumai sxedon katholou me mpala. Eimai omws tis apopsis na ipostirizoume tis omades tou topou mas... Diladi apo poion prepei na perimenei o Ergotelis na ton ipostiriksei? Apo ton Serraio 'h ton Salonikio? (Doksa tw thew iparxei kai o Panserraikos, o Paok, o Aris, h Kalamaria...)

Alexandros

εχει δικιο ο Αλεξανδρος .και εγω υποστηριζω δυο ομαδες (η μια ειναι ο ΠΑΟ). η δευτερη που ειμαι φανατικος δεν υπαρχει σε αυτες που προτεινει ο Αμβροσιος. :roll:

Διότι δεν ηταν δυνατό από το φόρουμ να βάλω τόσο πολλές επιλογές

το ξερω ρε Αμβοσιε.ετσι και αλλιως ειμαστε Β εθνικη. :blush:


inequalities
geometry
circumcircle
angle bisector
geometry proposed

$r,R$ are inradios and circumradius
Prove :
$\frac{R}{r}\ge \frac{a}{b}+\frac{b}{a}$

See

http://www.mathlinks.ro/Forum/viewtopic.php?t=15808
http://www.mathlinks.ro/Forum/viewtopic.php?t=15807
http://www.mathlinks.ro/Forum/viewtopic.php?t=49433

  Darij

[hide=" Here is the shortest solution what I know for this nice and yet simple inequality."]Denote $l_a,\ h_a$ the lengths of the angle-bisector and altitude from the vertex $A$.
$l_a=\frac{2}{b+c}\sqrt{bcp(p-a)}\Longrightarrow l^2_a=\frac{4bc}{(b+c)^2}\cdot p(p-a)\le p(p-a)\ .$ Therefore,
$h_a\le l_a,\ l_a\le p(p-a)\Longrightarrow h^2_b+h^2_c\le l^2_b+l^2_c\le p(p-b)+p(p-c)\Longrightarrow$
$h^2_b+h^2_c\le ap\Longrightarrow 4S^2\left(\frac{1}{b^2}+\frac{1}{c^2}\right)\le ap\Longrightarrow4S^2\left(\frac cb+\frac bc\right)\le abcp\Longrightarrow$
$4S^2\left(\frac cb+\frac bc\right)\le 4pRS\Longrightarrow \frac cb+\frac bc\le \frac{Rp}{S}\Longrightarrow \frac cb+\frac bc\le \frac Rr\ .$[/hide]


Doua cercuri C1 si C2 sunt tangente intern intr-un punct N astfel incat C2 se gaseste in interiorul lui C1. Punctele C,S,T apartin lui C1 si au proprietatea ca CS si CT sunt tangentela C2 in M si K.Demonstrati ca UCVW este un paralelogram , unde W este al doilea punct de intersectie dintre cercurile circumscrise triunghiurilor UMC si VCK.

:blush:  :oops:
Am uitat sa spun ca U-mij arcului CS si V-mij arcului CT

Ma scuzati!! :oops:

O sa formulez problema inversa (obtinuta prin inversiunea problemei initiale in raport cu polul $C$ si puterea $CM^2$). Pentru orice punct $X$, imaginea lui $X$ prin inversiune va fi numita $X'$ (si, in general, imaginea figurii $\mathcal F$ va fi $\mathcal F'$). Si o sa mai sar peste o gramada de pasi :).

Se dau punctul $C$ si cercul $c_2'$ astfel incat $CK',CM'$ sunt tangente la $c_2'$. Se iau $U',V'$ pe o dreapta $c_1'$ tangenta la $c_2'$ astfel incat $S'C=S'U',T'C=T'V'$, unde $S',T'$ sunt intersectiile dreptelor $CM',CK'$ cu $c_1'$ ($U',T'$ sunt de parti opuse ale lui $CS'$, si $V',S'$ sunt de parti opuse ale lui $CT'$). Se mai ia $W'=V'K'\cap U'M'$.

Sa se arate ca $W'U'C$ si $W'CV'$ sunt asemenea (in ordinea asta a varfurilor).

E usor de vazut ca putem obtine concluzia daca aratam ca $\angle U'W'V'=\pi-\angle S'CT'\ (*)$ si ca $C$ se afla pe bisectoarea unghiului $U'W'V'$.

Fie $P$ punctul unde $c_1'$ atinge $c_2'$, si fie $P_u=U'W'\cap CP,P_v=V'W'\cap CP$. Se vede ca $P$ este mijlocul lui $U'V'$. $P_uCU',P_vCV'$ sunt isoscele, deci distanta de la $C$ la $U'W'$ este egala cu distanta de la $U'$ la $CP$, si, la fel, distanta de la $C$ la $V'W'$ este egala cu cea de la $V'$ la $CP$. Din moment ce $P$ este mijlocul lui $U'V'$, iese de aici ca distantele de la $C$ la $U'W',V'W'$ sunt egale, deci $C$ e pe bisectoarea lui $\angle U'W'V'$.

Mai trebuie acum sa obtinem $(*)$, ceea ce este usor, pentru ca unghiurile $\angle WP_uP_v,\angle WP_vP_u$ ale lui $WP_uP_v$ sunt egale cu $\pi-2\angle PCU'$ si $\pi-2\angle PCV'$ respectiv.

Imi pare rau ca nu am demonstrat tot, dar ar fi devenit plictisitor, si oricum, o figura bine facuta ar trebui sa clarifice tot :). Sunt sigur ca iese repede si problema initiala, dar din moment ce primul lucru pe care l-am facut a fost sa o inversez, nu m-am mai chinuit sa o rezolv direct, fara inversiune.

Am gasit si eu o rezolvare sintetica…
 :roll: In esenta se demonstreaza ca (U, M, N);  (V, K, W); (M, W, K) sunt coliniare si de aici se rezolva usor.
O2K si VO1 sunt perpendiculare pe CT => sunt paralele. (N, O1, O2) coliniare si NO2/NO1 = O2K/O1V => N, K, W coliniare. Analog U, M, V coliniare.
Din patrulaterele UCWM si CVKW inscriptibile => m(MWC) = 180 – m( MUC) = 180 – m( NUC) si m(CWK) = 180 – m( CVK) = 180 – m(CVN). Dar CVNU inscriptibil => m( CVN) + m( NUC) = 180 => m(MWC) +m(CWK)= 180 => (M, W, K) coliniare.
Acum m(CWU) = m(CMU) = m(SMN). Deoarece SM tangenta la cercul circumscris lui MNK => m(SMN)= m(MKN). m(MKN)=m(WCV) => m(CWU)=m(WCV) => UW║CV. Analog UC║WV => UCVW paralelogram. q.e.d
 :D


Solve and write the demonstration outline, and the program outline.

P=>-Q, -R=>Q, P  l- R[/b][/hide][tab]

Wow, do I ever not know what's going on there.

This belongs in Other Problem Solving Topics. It really is a shame that not very many people know this stuff; I was fortunate enough to have logic classes in 6th and 7th grade, so I actually do know this stuff reasonably well, although it was a long time ago.

I recognized the last line as a statement in propositional logic, but I have never heard of a demonstration outline or program outline.

I now recognize the symbolic logic -- I have no idea what the first line is asking about, though.  Also, I don't know what |- means.

The turnstile P=>-Q, -R=>Q, P |- R means you are given P=>-Q, -R=>Q, and P, and you have to prove R.  Note that P |- R is subtly different from P => R.

I'll follow ComplexZeta's suggestion and move this to Other Problem Solving Topics (although I would otherwise have moved it to Intermediate). It should be accessible to high schoolers with an unusual background in logic, but it is definitely not a Getting Started problem. I have also retitled the thread for clarity.

[quote="ballerkidd4"]Solve and write the demonstration outline, and the program outline.

P=>-Q, -R=>Q, P  l- R[/quote]

Demonstration

1. P by hypothesis
2. P => ~Q by hypothesis
3. ~Q by Modus Ponens on 1 & 2
4. ~R => Q by hypothesis
5. ~Q => R by Law of Contrapositive on 4
6. R by Modus Ponens on 3 and 5.

Q.E.D.

Modus Ponens? Doesn't that mean 'way of placing' in latin? (I take latin :)) What does it mean in math?

I don't know Latin, but in logic, it means that if you have A and A => B, then you can infer B.

By the way, in gauss's argument, I think we can collapse steps 5 and 6 into one step, by using modus tollens.  ("If you have ~B and A => B, then you can infer ~A.")  Unless we consider going from ~~R to R a separate step.


topology
real analysis
real analysis unsolved

[quote="Kalle"]Let $X$ be a compact Hausdorff space with a sequence of closed, connected subspaces $C_1 \supseteq C_2 \supseteq C_3 ...$. Is it true that $C = \cap^\infty_{n=1} C_n$ is connected?

Is this the right forum? I'm sorry if it's not.[/quote]

C is connected.
Proof:
If it is not, we can find open sets $U_1$ and $U_2$ such that ${U_1}\cup{U_2}\supseteq{C}$ and ${U_1}\cap{U_2}=\emptyset$.
Now the union of countably infinite open sets $U_1$, $U_2$, and $(C_n)^c (n=1,2,...)$ equals X.
Since X is compact, we can choose from them finite sets that still cover X.
Obviously, because C must be covered, $U_1$ and $U_2$ must be included in those finite sets.
And notice that $(C_1)^c\subseteq(C_2)^c\subseteq...$, so $(C_n)^c\cup{U_1}\cup{U_2}=X$ holds for some n.
Thus ${U_1}\cup{U_2}\supseteq{C_n}$, which contradicts with the fact that $C_n$ is connected.

Nice proof. But do you really need the Hausdorff property here?

may I know how can the existence of $U_1, U_2$ be guaranteed?

[quote="AlvaroRecoba"]
If it is not, we can find open sets $U_1$ and $U_2$ such that ${U_1}\cup{U_2}\supseteq{C}$ and ${U_1}\cap{U_2}=\emptyset$.
[/quote]
Do you skip a whole bunch of steps here? I could work your reasoning out like this:

Let $C = A \cup B$ be a disjunction of $C$, where $A$ and $B$ are open subsets of $C$. $A$ and $B$ must also be closed in $C$, and since $C$ is closed in $X$, $A$ and $B$ must also be closed as subsets of $X$. Since $X$ is compact and Hausdorff it is normal and we can find open, disjoint sets $U_1$ and $U_2$ in $X$ such that $A \subseteq U_1$, $B \subseteq U_2$

Is this what you mean or is there a much easier argument since you omitted it?

[quote="AlvaroRecoba"]
Thus ${U_1}\cup{U_2}\supseteq{C_n}$, which contradicts with the fact that $C_n$ is connected.[/quote]
I also do not follow this contradiction... Please clarify

Kalle is right.

If $C$ is not connected, we can find closed sets ${V_1}\neq{\emptyset}$ and ${V_2}\neq{\emptyset}$ in $C$ such that ${V_1}\cup{V_2}=C$ and ${V_1}\cap{V_2}=\emptyset$. And since $C$ is closed, $V_1$ and $V_2$ are also closed sets in $X$.
Because compact Hausdorff space has the $T_4$ property, there are open sets ${U_1}\supseteq{V_1}$, ${U_2}\supseteq{V_2}$ such that ${U_1}\cap{U_2}=\emptyset$.

At the end of my proof, we conclude that ${U_1}\cup{U_2}\supseteq{C_n}$, then ${U_1}\cap{C_n}\neq{\emptyset}$ and ${U_2}\cap{C_n}\neq{\emptyset}$ are two open sets in ${C_n}$, their union equals ${C_n}$, and their intersection is $\emptyset$.
This is impossible since ${C_n}$ is connected.


function
geometry
geometric transformation
reflection

a shining metal ball with a small black dot on its surface is heated to a very high temperature and then taken to a dark room

a)  the spot appears brighter than ball
b)  they both appear equally bright
c)  the spot appears darker than ball
d)  both are invisible in the dark room

[quote="kubus"]I guess the whole ball emits about the same spectrum of frequencies, but the black dot emits more energy. Therefore it looks brighter, if observed in a right frequency range. If that's what you mean. (What is a "dark room"?)[/quote]Now why is that? Could you explain the whole radiation thing a little bit? (I know, we learned that in school - I only wish I could remember any of it :) .)

Every surface has its own emissivity - that is a constant between 0 and 1 (0 is a perfect mirror, 1 is a black body)(well, in fact, it is not a single constant, but a function of frequency) that describes what fraction of the Planck-law radiation the body really emits. This emissivity is exactly the same for absorbing radiation. You can prove this by contradiction, because if the emissivity factor could be different for emission and absorbtion, you could violate 2nd law of thermodynamics (I guess;). It depends on the very structure of the material, its interaction with EMG waves, resonance in frequencies, ...

Now clearly the black dot has higher emissivity than the rest of the ball and it radiates more energy.

Nice sol kubus,didnt see ya for a long time. :lol:
SO u needed to know this emissivity thin for this question,i thought that the metal ball shines a bit in the dark room but the black dot looks more prominent in the whole room :blush:  :oops:
IS that right or lame? :?

:rotfl:  :rotfl:  :rotfl:  :rotfl:  :rotfl:  :rotfl:(hey might be true too ,  :lol:  :P  :lol:  :P )

dont start overacting again  please!

However, the "emissivity thing" (I don't know how it is called properly either) is given in the problem statement. What was kubus trying to say is that if something reflects for example 75 % light (= is shiny) then it only radiates 25 % of what would a normal object at the same temperature radiate. If something's completely black (= reflects 0 % of light) it radiates 100 %, if something reflects 100 % it won't emit any radiation at all, etc. - it always has to add up to [b]exactly[/b] 100 %. So a shiny ball will only radiate 25 % (or whatever fraction) of what the black dot radiates and therefore will be darker.

[quote="shadysaysurspammed"]i thought that the metal ball shines a bit in the dark room but the black dot looks more prominent in the whole room :blush:  :oops:
IS that right or lame? :?[/quote]I'm sorry, that doesn't mean that it also looks [i]brighter[/i] :) .


geometry
perimeter
algebra unsolved
algebra

i'm having a very hard time figuring out the answers to these problems so if anyone can figure it out, please help

Write 3 equations in 3 variable for each problem

The sum of the digits of a three digit number is 11.  The hundreds digt exceeds the sum of the tens digit and the units digit by 1.  When the digits are reversed, the new number is 396 less than the original number.  Find the number.  Write the three equations

One side of a triangle is 3 inches longer than another side of the riangle.  The sum of the length of the two sides less one inch equals the length of the the 3rd side of the triangle.  If the perimeter is 29 inches, find the length of each side.  Write the three equations

[quote="yvette4848"]
One side of a triangle is 3 inches longer than another side of the riangle.  The sum of the length of the two sides less one inch equals the length of the the 3rd side of the triangle.  If the perimeter is 29 inches, find the length of each side.  Write the three equations[/quote]
I wonder if this question should be put in the intermediate forum instead.

[quote="yvette4848"]
The sum of the length of the two sides less one inch equals the length of the the 3rd side of the triangle.[/quote]
Do you mean 'the sum of the lengths of the two sides is one inch greater than the third side'?
$
\begin{cases}
x=y+3 \\
x+y-1=z \\
x+y+z=29
\end{cases}
$
Substitute the first and second equations into the third one,
$(y+3)+y+(x+y-1)=29$
$y+3+y+(y+3)+y-1=29$
$4y+5=29$
$y=6$
Correspondingly, $x=9, z=14$


[quote="Ligouras"]Let $ a, b, c$ non-negative real numbers such that $ abc \equal{} 1$. Prove or disprove that

$ a^2 \plus{} b^2 \plus{} c^2 \plus{}\frac{a}{b} \plus{}\frac{b}{c} \plus{}\frac{c}{a} \plus{} 9 \ge 2(a \plus{} b \plus{} c \plus{} \frac{1}{a} \plus{} \frac{1}{b} \plus{} \frac{1}{c})$

_______
 :?:  :?:[/quote]
_____
 This is the statement correct?
Where is the equality?
____
Marin Sandu

[quote="Ligouras"]Let $ a, b, c$ non-negative real numbers such that $ abc \equal{} 1$. Prove or disprove that

$ a^2 \plus{} b^2 \plus{} c^2 \plus{}\frac{a}{b} \plus{}\frac{b}{c} \plus{}\frac{c}{a} \plus{} 9 \ge 2(a \plus{} b \plus{} c \plus{} \frac{1}{a} \plus{} \frac{1}{b} \plus{} \frac{1}{c})$
[/quote]
Let $a = \frac{x}{y}$, $b = \frac{y}{z}$, $c = \frac{z}{x}$, then WLOG $a = \min\{a,b,c\}$, $b = a + u$, $c = a + v$, $u,v \geq 0$ and the inequality becomes \[u^3 v^3 +
 u^2 v^4 + (u^3 v^2 + 5 u^2 v^3 + 2 u v^4) x + (u^4 - 3 u^3 v +
    12 u^2 v^2 + 5 u v^3 + v^4) x^2 + (2 u^3 + 5 u^2 v + 13 u v^2 +
    2 v^3) x^3 + (6 u^2 + 9 u v + 6 v^2) x^4 + (6 u + 6 v) x^5 + 3 x^6 \geq 0\] which is obvious, e.g. by AM-GM $3x^2u^3v \leq xu^3v^2 + x^2u^4 + x^3u^2v$ or by checking positivity of $f(u) = 1 + 5 u + 12 u^2 - 3 u^3 + u^4$.

[quote="Ligouras"]Let $ a, b, c$ non-negative real numbers such that $ abc \equal{} 1$. Prove or disprove that

$ a^2 \plus{} b^2 \plus{} c^2 \plus{}\frac{a}{b} \plus{}\frac{b}{c} \plus{}\frac{c}{a} \plus{} 9 \ge 2(a \plus{} b \plus{} c \plus{} \frac{1}{a} \plus{} \frac{1}{b} \plus{} \frac{1}{c})$
[/quote]
Because $a^2+b^2+c^2+3\geq a+b+c+ab+ac+bc$ and by Schur  $\frac{a}{b} \plus{}\frac{b}{c} \plus{}\frac{c}{a}+3\geq a+b+c+ab+ac+bc$


Does anyone have access to older (complete) AHSME tests on the internet, specifically those given before 1980?

Check Tipton's bookcase; she has several copies of Contest Problem Books 1-4, which contain the AHSME's from 1950 to approximately 1990.

lol

We go to the same school, and our math teacher is named Tipton.

Sorry that nobody else really has the opportunity to browse her bookcase.

Thanks Seva, I am aware of the greatest source of knowledge outside of the internet, but I was wondering if anyone knew of online resources containing them.


A convex quadriteral $ ABCD$ is given with $ \angle ABC \equal{}$ $ \angle ADC$ and $ M$ and $ N$ are points on $ BC$ and $ CD$ such that $ AM \perp BC$ and $ AN \perp CD$. $ K$ is a point where $ MD$ and $ NB$ meet. Prove that $ AK \perp MN$

[quote="mr.danh"][quote="vedran6"]A convex quadriteral $ ABCD$ is given with $ \angle ABC \equal{}$ $ \angle ADC$ and $ M$ and $ N$ are points on $ BC$ and $ CD$ such that $ AM \perp BC$ and $ BN \perp CD$. $ K$ is a point where $ MD$ and $ NB$ meet. Prove that $ AK \perp MN$[/quote]
I'm sure that you made a mistake, the condition $ AN\perp CD$, [b]not[/b] $ BN\perp CD$[/quote]
 you were right, I have editted it :)

$ \triangle ADN\sim\triangle AMB \Rightarrow\frac{AN}{DN}\equal{}\frac{AM}{MB}\equal{}k$
Let AH denote the altitude of the triangle AMN. Let E in the opposite ray of AH such that $ AE\equal{}k.MN$
We have $ \frac{AE}{MN}\equal{}\frac{AN}{DN}, \hat{EAN}\equal{}\hat{MND}$, hence $ \triangle EAN\sim\triangle MND$
$ \Rightarrow EN\perp MD$.
Similar, $ EM\perp BN$
Hence, three altitudes $ MD,NB,EH$ of $ \triangle EMN$ are concurrent. So, $ AK\perp MN$.


not a "bug", rahter a weird cosmet(h?)ical thing.

I know, I don't mind it :) It's because of the space in the middle of my username. If you have a simple css solution for it, please post it.

Of course I have... as I posted in the other post, it's because phpbb standardly sets spaces around your name.

In index_body.tpl (or whatever it's called in the new version), where it would generate

[color=red] [/color]
Valentin Vornicu[color=red] [/color]

Just remove the 2 red [color=red] [/color]'s *and* the space right after the first one. That should make it perfectly aligned. :)


Dear moderators,
I am a newbie in using Latex, and I would like to create the document like this... But I dont know how, could you upload the tex file of this document for me to know how to create it? Thank you very much.

I'll re-type up three for you here:

Prove that there are infinitely many positive integers $ n$ such that $ n^{2}\plus{}1$ has a prime divisor greater than $ 2n\plus{}\sqrt{2n}$.

To type this problem up, there is minimal syntax.  The text is just text, and the math is pretty basic.  If you scroll over what I typed, it should reveal the syntax.

[quote="JRav"]I'll re-type up three for you here:

Prove that there are infinitely many positive integers $ n$ such that $ n^{2} \plus{} 1$ has a prime divisor greater than $ 2n \plus{} \sqrt {2n}$.

To type this problem up, there is minimal syntax.  The text is just text, and the math is pretty basic.  If you scroll over what I typed, it should reveal the syntax.[/quote]
No, maybe you dont understand me. I meant how to create the file like this.

There is a nice tutorial on basic $ \text{\LaTeX}$ on AoPSWiki, which, surprisingly enough, is named [[LaTeX]] and has a link to it (quite unexpectedly, with the same name) clearly visible on both skins :P. Just read it. If you have any particular questions after that, feel free to ask here :).


For $0^\circ \leq \theta <180^\circ,$ consider the function $f(\theta)=3\cos 2\theta+4\sin\theta.$
Let $\alpha$ be the angle such that $0^\circ <\alpha <90^\circ$ and $f(\alpha)=3.$

(1) Find the maximum and minimum value of $f(\theta).$
(2) Find the value of $\sin (\alpha +30^\circ).$

[quote="dengmi"]The second one:
[hide]$\cos(2\alpha)=1-2\sin^2\alpha$ so we have $3-6\sin^2\alpha+4\sin\alpha$.  We want to find such an $\alpha$ where $f(\alpha)$ is 3.  $3=3-6\sin^2\alpha+4\sin\alpha$.  We cancel the constants and we factor the remains.  $0=\sin\alpha(-6\sin\alpha+4)$.  Obviously, $\alpha=0$ is a solution to the equation so $\sin(\alpha+30)=1/2$ is one answer.  The other solution has $\sin\alpha=2/3$.  Therefore, when we add 30 degrees, we have $\sin\alpha*\cos\30+\sin30*\cos\alpha$.  Since $\sin\alpha=2/3, \cos\alpha=\sqrt(5)/3$.  We have $2/3*\sqrt(3/2)+1/2*\sqrt(5)/3=\frac{2\sqrt(3)+\sqrt(5)}{6}$.  That's the other solution.[/hide][/quote]

The answer is the latter part of your answer. :)


geometry
AMC
AIME
inequalities
AMC 12
USA(J)MO
USAMO

I was wondering how i could determine my level with respect to this forum. Since i never took any contest(s) in high school i am having trouble figuring out where i stand (i will of course try problems in both sections but for curiousities sake). I was hoping there is some sort of test (if not, wouldnt it be a great idea?) to determine where i should be reading most actively.
  Anyway i have done 2 problems in USSR Olympiad book with double stars, 3 with single stars, and a few dozen of the regular ones. My geometry is really bad (well in the sense that i cant yet draw lines very well), i have worked out a few of the easier AIME problems (non-geometry) and can do most ASHME problems(particulary inequalities and counting). Other than that i am completely self taught (didnt take any classes or have any type of instruction, besides of course my books) for all classes after algebra I (which i actually failed). Hope this might give you guys a better view so you can tell me what is best.

Even though I am not the best to tell because I am not nearly as good as you are but you probably belong to the advanced section if you can solve olympiad type problems. Also, I think there are AIME problems in the amc form and intermediate.

By the way don't some ussr problems look even harder then usamo/imo problems, for some reason they look like that to me.  :P

I would say that you are likely "Pre-Olympiad", though don't be discouraged into looked at problems in the "Advanced" section.

If you can do most AMC-12/AHSME problems and a fair number of AIME problems, then you are certainly well beyond "Getting Started" and probably at the high end of "Intermediate" if not beyond.

There is some degree of randomness to which problems end up in which forums, as it is up to the initial poster to decide in which forum to initially post their problem.  If the moderators and/or administrators feel that a problem is significantly too hard for its forum, we will move it to a more appropriate forum.

pre-olympiad? really?

i think im on about the same level as you - i got a 6 on the AIME, then fixed some stupid errors and figured out some more when i got the test back...

...and i find that the intermediate forum suits me just fine for challeges, and the getting started forum has plenty of easier problems to practice with. there r too many theories and concepts in pre-olympiad that ive never heard of, let alone understand... :(

Here's the official description of the Intermediate Problem Solving forum:

[quote="rrusczyk"]The Intermediate Topics board is for problems and questions that are beyond the basics, but not quite olympiad level. These are problems that might be on the AMC, AIME, Mandelbrot, or ARML. If you're not sure what level your problem or question is, you can generally use the other posts in the board as a guideline.

Generally, the Intermediate Topics board will include topics of high school math such as trigonometry, polynomials, conditional probability, logarithms, etc., as well as some broader beginning problem solving techniques such as basic problems in induction or the pigeonhole principle.
[/quote]

If you can do "most" of the AMC12 problems and "some" of the AIME problems, then you're not too far from a USAMO-qualifying score, hence I recommend "Pre-Olympiad".  But as I said above, placing problems into a forum by difficulty is not an exact science; many times a problem in the "Intermediate" forum will be harder than a problem in the "Pre-Olympiad" forum.

Generally, you should try to work on problems that are as hard as you can manage.  Working on easier problems can be good for practice, but you'll only really learn if you try to solve hard problems.

I agree with DPatrick.  You are definitely past Beginner.  It sounds like you will have a shot at USAMO.  I suggest that you browse the forums however.  There might be something of worth in a forum that you don't usually use.

One problem with inspecting the forums for examples is that people are still getting used to the new five-level division of the problem-solving forums, and MOST forums have too many problems that are too hard for their intended level.

hello:

  Yeah i hear that alot but the USSR olympiads, but the USSR problem book isnt all hardcore. Some of the problems are even ASHME level, but all of them are interesting (except for one of them that required too much calculation).

L_Li:

  Nah your probally much better then me, 6 questions on a single AIME is not yet reality for me (although i am looking at some of the older ones so i dont know how those compare).

 I will most likely try to do all the ASHME's (untill i get bored) for the next 3 weeks, its been a long time since i solved problems so i need to get back into the groove. I will post my progress as often as possible perhaps along with problems/solutions i found interesting.[/quote]


algebra
polynomial
quadratics
function
algorithm
Ring Theory
geometry

Here's the question: (rephrased)

The remainder when P(x) is divided by x-99 = 19.
---------------ditto----------------- x-19 = 99.
What's the remainder when P(x) is divided by (x-19)(x-99) in terms of x?

For those of you who want to check, the answer is [hide] 118-x [/hide]

EDIT!!!: HUGE typo mistake, sorry everybody, the original problem was indeed actually with (x-99) and (x-19) instead of what I wrote originally. Sorry!

[quote="erdogankerem123"]Here's the question: (rephrased)

The remainder when P(x) is divided by x+99 = 19.
---------------ditto----------------- x+19 = 99.
What's the remainder when P(x) is divided by (x+19)(x+99) in terms of x?

For those of you who want to check, the answer is [hide] 118-x [/hide][/quote]

the remainder will be in form ax+b

$ P(x)=(x^{2}+118x+1881)q(x)+ax+b$

so $ P(-99)=19=(0)(q(x))-99x+b$ and
    $ P(-19)=99=(0)(q(x))-19x+b$

$ 19=-99a+b$
$ 99=-19a+b$        subtract the two and you get

$ -80=-80a$
$ a=1$ therefore $ b=118$
so the remainder is $ \boxed{x+118}$?????

In everything I say in this post, I'm assuming that every polynomial is monic - that is, the coefficient of its highest order term is $ 1.$

The remainder when $ P(x)$ is divided by $ x-a$ is $ P(a).$ That's an extremely important part of the theory of polynomials.

Now consider dividing $ P(x)$ by $ D(x),$ where $ D$ is a monic polynomial of degree $ k.$ Then $ P(x)=Q(x)D(x)+R(x),$ where $ R(x)$ is the remainder. Certainly, $ R$ has degree $ \le k-1.$ Now, suppose $ D(a)=0.$ Then $ P(a)=Q(a)\cdot 0+R(a),$ so $ R(a)=P(a).$

Now for the problem at hand. The first line tells you that $ P(-99)=19.$ The second line tells you that $ P(-19)=99.$ Finally, we divide by the quadratic polynomial $ (x+19)(x+99).$ From what I said above, the remainder $ R(x)$ is a first degree (or smaller) polynomial, which is to say a linear function, and $ R(-19)=99$ and $ R(-99)=19.$

At this point, I think you've misquoted the problem. I don't feel like chasing down the 1999 AHSME right now, but weren't we dividing by $ x-99,\ x-19,$ and $ (x-19)(x-99)?$

[quote="K81o7"][quote="sonny"][quote="K81o7"]ie, there are f(x) and g(x) such that

g(x)(x+99)+19=P(x)=f(x)(x+19)+99

By the Chinese Remainder Theorem for polynomials (well, this is one thing which extends to polynomials), there is a unique linear h(x) such that P(x)=(x+19)(x+99)(stuff)+h(x).  If I'm not mistaken, though, it should be x+118...[/quote]

I got that at first and then I realized my mistake, um did you just estimate and come up with x+118?[/quote]

?
Is not x+118   19 mod x+99   and   99 mod x+19?[/quote]

Uh yeah, much like Mr.Merryfield I got confused between the posted and the real.

Sorry everybody for misquoting the problem. I get it anyways, though, simply by just using same method, different numbers. Thanks everyone!

[quote="Altheman"][quote="K81o7"]ie, there are f(x) and g(x) such that

g(x)(x+99)+19=P(x)=f(x)(x+19)+99

By the Chinese Remainder Theorem for polynomials (well, this is one thing which extends to polynomials), there is a unique linear h(x) such that P(x)=(x+19)(x+99)(stuff)+h(x).  If I'm not mistaken, though, it should be x+118...[/quote]

err...do you have anything else to add about CRT with polynomials? Is there some more general result that corresponds to a field extension of the integers?[/quote]

Hmm...polynomials behave a little differently from other field extensions.  For example, if p(x) is an irreducible polynomial, then Z[x]/p(x) is not necessarily a field (remember that because the constant terms and x terms and irreconcilable, a polynomial residue class has an infinite number of elements, and cannot be a field unless we take it modulo some prime).  On the other hand, the Gaussian integers behave almost identically to the integers.  Certain field extensions of the form $ \mathbb{Z}[\sqrt{-d}]$ also behave nicely for squarefree integers d, but as you might imagine, the behavior of prime factorization and division algorithm and intimately tied to the norm function, which is tied to the solvability of $ x^{2}+dy^{2}=p$ over primes p, which is tied to $ \left(\frac{-d}{p}\right)$.  For example, if you consider the subring of the integers from which you form when you take the norms of all different numbers in the field extension, that itself must have unique prime factorization.  Even when there is no unique prime factorization in a field extension, we can use what are called "ideals" of the extension.  This is a group theory term, and although I can define it, I'm not as experienced in that area, so you may want to ask someone else like, say, JSteinhardt.

But try proving things about primes and residue classes with $ \mathbb{Z}[i]$.  Because of the nice behavior of the equation $ x^{2}+y^{2}=p$, this ring behaves nicely.

[quote="Altheman"]Is there some more general result that corresponds to a field extension of the integers?[/quote]

More general than that, actually:

http://en.wikipedia.org/wiki/Chinese_remainder_theorem#Statement_for_general_rings


geometry
rectangle
perimeter
geometry unsolved

Given an $ a \times b$ rectangle with $ a > b > 0,$ determine the minimum side of a square that covers the rectangle. (A square covers the rectangle if each point in the rectangle lies inside the square.)

It has to be the least perimeter of the square possibly.

solution $\pm$

first you show all $4$ points of the rectangle are on the boundary of the square.

then you have the angle between rectangle and square $0$ or $45$ degrees.

so $min (a, \frac{\sqrt2} 2 (a+b) )$

[hide= Potential Solution?]
Let $R$ denote the rectangle, and let $S$ denote the square with minimum length that covers $R$ . Let $s$ denote the length of a side of $S$. We claim that $R$ is inscribed in $S$. We also claim that $R $can only be inscribed in the two ways shown below. Let $S_1$ and $S_2$ denote the squares shown on the left-hand and right-hand sides, respectively. For $S_2$ , the sides of $R$ are parallel to the diagonals of $S2$. We can see that $s = a$ if $S = S_1$.

[img] https://lh3.googleusercontent.com/8IBseHsKPR3t0qdwjmS_YCmzdjC7tevE6s-KzyEQKCojYni3RwjDSLTX1sSWPXAqziqrSA=s170 [/img]

It is easy to see that $s=\frac{\sqrt{2}(a+b)}{2}$ if $S=S_2$. Taking the minimum value of $(a,\frac{\sqrt{2}(a+b)}{2})$, we conclude that

$s=\[   \left\{
\begin{array}{ll}
      a & a<(\sqrt{2}+1)b \\
      \frac{\sqrt{2}(a+b)}{2} & a\geq (\sqrt{2}+1)b \\
     \end{array}
\right. \]$

Now we prove our claim that these are only two ways. Let $R = ABCD$ and $S = XYZW$. WLOG, we place $XY$ horizontally. By the minimality of $S$, we can assume that at least one vertex, say $A$ , of $R$ lies on one side of $S$ , say $WX$ (see the left diagram). If neither $B$ nor $D$ lies on the sides of $S$, we can then slide $R$ down vertically so that one of them, say $B$, lies on side $XY$ (see the middle figure). If neither $C$ or $D$ lies on the sides of $S$ , then we can apply an enlargement, centered at $X$ with scale less than $1$ , to $S$ such that the image of $S$ still covers $R$ .
This violates the minimality of $S$. Therefore, at least one of $C$ and $D$ lies on the sides of $S$, that is, three consecutive vertices of $R$ lie on the sides of $S$ (see the right-hand side figure). WLOG assume that they are $A, B, C$. If any of these three vertices coincide with any of the vertices of $S$, then we clearly have $S = S_1$. Hence we may assume that $A , B, C$ are on sides $WX$, $XY$ and $YZ$, respectively. By symmetry, we may also assume that $AB =a>b=BC$.

[img] https://lh3.googleusercontent.com/lX6YAPjUs7xeaT73NSstZRW_31vZ4-Rui5zM7ZRqOvhw-W_DrjH3AHEOdhmGxa9R7-d9Mw=s170 [/img]

[hide= Note] All rectangular-like shapes are indeed rectangles [/hide]

If $D$ does not lie on line segment $ZW$, then we can slide $R$ up so both $B$ and $D$ lie in the interior of $S$ (see the left-hand slide) Let $O$ be the center of $R$. We can then rotate $R$ around $O$ with a small angle so that all four vertices lie inside $S$ (see the middle figure shown below). It is easy to see that we can use a smaller square to cover $R$ (by applying an enlargement centered at $O$ with a scale less that $1$) , violating the minimality of $S$. This means our assumption is wrong, and D must lie on side $ZW$ (see the righthand side figure shown below) , which is the case when $S = S_2$.

[img] https://lh3.googleusercontent.com/j6nT3PMMp8rueVTHgUEE_B-m-ZHc99H1uG2L4e6-E_Qfi_AKxCUBaUYFn7HOnP8_2NUp=s170 [/img]

[hide= Note] All rectangular-like shapes are indeed rectangles [/hide]

We finish our proof that if $S = S_2$, then the sides of $R$ are parallel to the diagonals of $S_2$. By symmetry, it suffices to show that $AX = XB$. It is not difficult to see that $\angle XAB = \angle YBC = \angle ZCD= \angle WDA$. So triangles $ABX$, $BCY$, $CDZ$ and $DAW$ are similar. Set $AX = ax$ and $XB = ay$. Then $BY = bx$ and $CY = by$. Also, $DW = bx$ and $WA = by$, which gives $by + ax = WA + AX = WX = XY = XB + BY =ay+bx$, implying that $(a-b)x= (a-b)y$, or $x=y$, as desired $\blacksquare$ [/hide]


Note that 1 + 2 + 3 + 45 + 6 + 78 + 9 = 144. In how many [i]other[/i] ways is it possible to make a total of 144 using only 1, 2, 3, 4, 5, 6, 7, 8,  and 9 in that order and addition signs?

Source: Australian Math Competition, J30 (I30, S26), 2002

Here's what I got.  I'm new to these boards, by the way.  I'm glad to be a part of this.

My solution is written out longhand, the important things are bolded to make it easier to read.

Answer: 3 (12 + 34 + 5 + 6 + 78 + 9; 12 + 3 + 45 + 67 + 8 + 9; 1 + 2 + 34 + 5 + 6 + 7 + 89)

Solved by:

STEP 1:    The sum of the numbers from 1 through 9 is 45, so in order for the sum of the new numbers (with certain numbers promoted to the tens place) to end with a four, [b]the total of the numbers promoted must end in a 1[/b].  [b]No number may be promoted to the 100's place[/b] as this will always result in a number greater than 144.  9 cannot be promoted as there is no number after it.

STEP 2:    The total of all promoted numbers times ten is less than the total number of the sum of all nine numbers.  Thus, the total of all promoted numbers cannot be greater than 14.4.  The only two numbers the promoted numbers can now add up to are 1 and 11, as they are less than 14.4 and end with a 1.  The only way the total of promoted numbers to equal 1 is when we have 12 + 3 + 4 + 5 + 6 + 7 + 8 + 9, which equals 54, a number less than 144.  [b]Only series in which the promoted numbers equal 11 can equal 144.[/b]

STEP 3:    When we promote a total of 11, we add 110 to the total and subtract 11.  This gives us 110 + 45 - 11 = 144.  [b]Thus, any series of numbers in which the sum of promoted numbers equals 11 has a total sum of 144[/b].

STEP 4:    If we promote numbers 5 and greater only, we will not be able to reach a total of promotions that equals 11.  This is because 5 + 6 is the only way to get 11 from the remaining numbers, but 6 can't be promoted: to do so would promote 5 to the 100's place, which can't be done.  [b]All promotions must contain a number from 1 - 4.[/b]

STEP 5:    [b]If we promote 1, we have to choose numbers from 3 to 8 to promote to reach a total of 10 (to reach 11.)  This can only be done with the non-adjacent pairs 3 & 7 and 4 & 6, giving a total of 2 combinations[/b].  We cannot promote 2 - doing so would require a pair of 4 & 5, which are adjacent and thus cannot both be promoted.  [b]If we promote 3, we need to also promote 8[/b].  A total of 8 can not be reached in any other way.  [b]This gives a final total of 3 combinations[/b], as the combination using a promoted 4 was given in the problem.

Correct.. and welcome to this place, it's great!  :)

Good job cpp - interesting solution.  (in the future, please hide your solution)

My solution

[hide]
$1 + 2 + 3 \cdots + 9 = 45$

If we combine to get a three-digit number, the only possible way to keep the number under 144 is to combine 1-2-3, which gives us $123 + 4 + 5 \cdots + 9 = 162$, which does not work.  Thus, we can only combine numbers into other 2-digit numbers.

When we combine, say, a 3 and a 4, we subtract 3 from our sum and add 30, essentially adding $9 * 3$.  $144 - 45 = 99$, so the sum of the numbers we combine that are in the tens place must be 11.

If we combine 1-2, then we can also do 3-4 and 7-8, or we can do 4-5 and 6-7.

If we combine 2-3, we have no options.  We cannot use 3-4 and 6-7, since 3 is already combined.  We cannot use 4-5 and 5-6, since they are too close.  We cannot use 9 since no number follows it.

If we combine 3-4, then we can also combine 8-9.  (We have already counted 1-2, 3-4, and 7-8).

If we combine 4-5, we can use 7-8, but this is given in the example.

If we combine anything greater than 4, either we have already counted it or the numbers are too close.

Thus, we have three other options.
12 + 34 + 5 + 6 + 78 + 9 = 144
12 + 3 + 45 + 67 + 8 + 9 = 144
1 + 2 + 34 + 5 + 6 + 7 + 89 = 144
1 + 2 + 3 + 45 + 6 + 78 + 9 = 144 (example)
[/hide]

Thank you, your solution was very good too. : )


I'm proving something about unitary operators, and at one point I want to show that the eigenvalues have absolute value 1.

Anyway, I've reached the point where I've devised that $ {\gamma}^2 \equal{} \lambda$ and $ |\lambda|\equal{}1$, and I want to show that $ |\gamma|\equal{}1$, where $ \gamma , \lambda \in \mathbb{C}$

So, $ {\gamma}^2 \equal{} \lambda$
$ \Rightarrow |{\gamma}^2| \equal{} |\lambda| \equal{} 1$
$ \Rightarrow {|\gamma|}^2 \equal{} 1$
$ \Rightarrow |\gamma| \equal{} \sqrt{1}\equal{}1$

Is this correct? It seems very trivial, I'm just concerned about pulling out the power in line 2 to line 3. Thanks.

$ |zw|\equal{}|z|\,|w|$ for all $ z,w\in\mathbb{C}.$

That's more than enough to justify the step you're worried about.


geometry
circumcircle
inradius
incenter
IMO Shortlist
reflection
Inversion

Let $ABC$ be a triangle, $\Omega$ its incircle and $\Omega_{a}, \Omega_{b}, \Omega_{c}$ three circles orthogonal to $\Omega$ passing through $(B,C),(A,C)$ and $(A,B)$ respectively. The circles $\Omega_{a}$ and $\Omega_{b}$ meet again in $C'$; in the same way we obtain the points $B'$ and $A'$. Prove that the radius of the circumcircle  of $A'B'C'$ is half the radius of $\Omega$.

Please post your solutions. This is just a solution template to write up your solutions in a nice way and formatted in LaTeX. But maybe your solution is so well written that this is not required finally. For more information and instructions regarding the ISL/ILL problems please look here: [url=http://www.mathlinks.ro/Forum/viewtopic.php?t=15623]introduction for the IMO ShortList/LongList project[/url] and regarding[url=http://www.mathlinks.ro/Forum/viewtopic.php?t=15623]solutions[/url]  :)

I remember posting this a long time ago :).

Consider the inversion wrt the incircle. It invariates the incircle and $\Omega_a,_b,_c$, so it turns $X$ into $X'$, for any $X\in\{A,B,C\}$. At the same time, if the incircle touches $BC,CA,AB$ in $D,E,F$, then the inversion turns $A$ into the midpoint of $EF$ and the like, so $A',B',C'$ are the midpoints of $EF,FD,DE$ respectively, meaning that the circumcircle of $A'B'C'$ is the nine-point circle of $D EF$, thus having half its circumradius, which is the inradius of $ABC$.

[b]OFFICIAL SOLUTION[/b]

Denote by $I$ the incenter, by $r$ the inradius, by $D,E,F$ the contacts of the incircle with $BC,CA,AB$ respectively and by $P,Q,R$ the midpoints of the segments $[EF],[FD],[DE]$. We will prove that $\Omega_{a}$ is the circle $(B,C,Q,R)$. We firstly notice that from the right-angled triangles $IBD$ and $IDQ$ we get $IQ \cdot IB = ID^{2} = r^{2}$ and in the same way $IR \cdot IC = r^{2}$, therefore the points $B,C,R,Q$ are on a circle $\Gamma_{a}$. The points $Q$ and $R$ belong to the segment $(IB)$ and $(IC)$, so $I$ is exterior to the circle $(B,Q,R,C)$ and $I's$ power with respect to the circle $(B,Q,R,C)$. In the same way $\Gamma_{b}$ is the circle $(C,R,P,A)$ and $\Gamma_{c}$ is the circle $(A,P,Q,B)$. It follows that $A',B',C'$ coincide with $P,Q,R$ and the required conclusion is not obvious.

[b]Another Solution[/b]:

Let $O_{a}$ be the center of $\Omega_{a}$ and $M$ be the midpoint of $(BC)$. Denote, using oriented segments, $\overline{M,O_{a}} = x$ (the positive sense on the perpendicular bisector of $(BC)$ being $\overrightarrow{ID}$. The radius of $\Omega_{a}$ is $x^{2} + \frac{a^{2}}{4}$ and

\[IO^{2}_{a} = DM^{2} + (\overline{ID}+ \overline{MO_{a}})^{2} =
\frac{a}{2} - \left(\frac{p-b}{2}\right)^{2} + (r+x)^{2}.\]

The condition of $\Omega$ and $\Omega_{a}$ being orthogonal is $O_{a}I^{2} = O_{b}B^{2}+r^{2}$, that is

\[x^{2} + \frac{a^{2}}{4} + r^{2} = \left(\frac{b-c}{2}\right)^{2} +
(r+x)^{2} \Leftrightarrow \frac{(p-b)(p-c)}{2r} \Leftrightarrow X
= 2R \sin \frac{A}{2} \cos \frac{B}{2} \cos \frac{C}{2}.\]

It follows that

\[OO_{a} = \overline{OM} + \overline{MO_{a}} = 2R \sin \frac{A}{2}
\cos \frac{B}{2} \cos \frac{C}{2} = R + \frac{r}{2}.\]

Therefore $O_{a},O_{b},O_{c}$ are on the circle of the center $O$ and radius $R + \frac{r}{2}$. Let $\{N\} = O_{a}O_{c} \cap BC$. The angle $\widehat{O_{a}OO_{c}}$ has measure $\pi - b$ (regardless of the position of $B$), so $\widehat{O_{a}OO_{c}} = \frac{B}{2}$ and

\[MN = x \tan \frac{B}{2} = 2R \sin \frac{A}{2} \cos \frac{B}{2}
\cos \frac{C}{2} = \frac{p-c}{2},BN = \frac{a}{2} - \frac{p-c}{2}
= \frac{p-b}{2}.\]

Hence $BN = \frac{BD}{2}$ and, because $MN < BM, N \in (BM)$. The same holds for the common point of $AB$ and $O_{a}O_{c}$, therefore the reflection of $B$ in $O_{a}O_{c}$ is on $DF$. This proves that $B'$ - the second common point of $\Omega_{a}$ and $\Omega_{b}$ - is the midpoint of $(DF)$. The conclusion follows easily.

Just for the sake of completeness, the other thread where the problem was discussed is http://www.mathlinks.ro/Forum/viewtopic.php?t=4592 .

  darij

Let $r$ be the radius of $\Omega$, then we have that the power of $I$ with respect to $\Omega_a, \Omega_b, \Omega_c$ is $r^2$. Thus, $I$ is the radical center of $\Omega_a, \Omega_b, \Omega_c$, so $AA', BB', CC'$ concur at $I$. Also, \[r\cdot AB = 2[AIB] = IA \cdot IB \cdot \sin \angle AIB\] where $[AIB]$ is the area of triangle $AIB$. But \[\angle AIB = 180^{\circ} - \angle IAB - \angle IBA = 180^{\circ} - \angle IAC - \angle IBC. \] By the similarity of $ICA$ and $IA'C'$ and of $ICB$ and $IB'C'$, this equals \[180^{\circ} - \angle IC'A' - \angle IC'B' = 180^{\circ} - \angle A'C'B',\] so \[r\cdot AB = IA \cdot IB \cdot \sin \angle A'C'B'.\] By the similarity of $IAB$ and $IB'A'$, \[A'B' = \frac{IB' \cdot AB}{IA} = \frac{r^2 AB}{IA \cdot IB}\] (because the power of $I$ with respect to $\Omega_a$ is $r^2$). And now, we use the (extended) law of sines: Let $R$ be the circumradius of $A'C'B'$. Then \[R = \frac{A'B'}{2 \sin \angle A'C'B'}=\frac{r^2 AB}{2IA \cdot IB \sin\angle A'C'B'}.\] Since $r\cdot AB = IA \cdot IB \cdot \sin \angle A'C'B',$ this is equal to $r/2$ and we are done.

This problem also has a solution like 2009 ISL G3's solution.

[b]Solution[/b]

Let $O_a$, $O_b$ and $O_c$ denote the centres of circles $\Omega_a$, $\Omega_b$ and $\Omega_c$, respectively. Let $A_1$ and $A_2$ denote the intersections of $\Omega_a$ and $\Omega$. Define $B_1$, $B_2$, $C_1$ and $C_2$ analogously. Let $D, E$ and $F$ denote the intersections of $\Omega$ with $BC$, $CA$ and $AB$, respectively. Let $P$, $Q$ and $R$ denote the midpoints of $EF$, $DF$ and $DE$, respectively.

By the definition of orthogonality, lines $O_aA_1$ and $O_aA_2$ are tangent to circle $\Omega$. Since $O_aB=O_aA_1=O_aA_2=O_aB$, it follows that $O_a$ lies on the radical axes of circles $B$ and $\Omega$ and circles $C$ and $\Omega$. Note that these radical axes are the midlines of the triangles $\triangle{BDF}$ and $\triangle{CDE}$. By the same argument, $O_b$ lies on the midlines of triangles $\triangle{CDE}$ and $\triangle{AEF}$ and $O_c$ lies on the midlines of triangles $\triangle{BDF}$ and $\triangle{AEF}$. Reflecting points $A$, $B$ and $C$ in the midlines of triangles $\triangle{AEF}$, $\triangle{BDF}$ and $\triangle{CDE}$ respectively yields that $A'=P$, $B'=Q$ and $C'=R$. Therefore, $\triangle{A'B'C'}$ is the medial triangle of $\triangle{DEF}$ and has a circumradius equal to half of that of circle $\Omega$.

Let $D,E,F$ be the points of tangency of $\Omega$ with $BC,CA,AB$. Do an inversion at $\Omega$. Notice that $\Omega_i$ maps to itself $\forall i \in \{ a,b,c \}$. Then, $ \{C',C\} = \Omega_a \cap \Omega_b \rightarrow \Omega_a \cap \Omega_b = \{ C', C\} $. However, $C$ is not its own inverse and thus $C'$ is the inverse of $C$, i.e. the midpoint of $DE$. An analogous thing occurs with $A$ and $B$. Notice

$ R(A'B'C')=R(DEF)/2 = R(\Omega)/2$

since $A'B'C'$ is the medial triangle of $DEF$. Done.

Inversion wrt $\Omega$ maps $A,B,C$ to $A',B',C',$ so $A'B'C'$ is the medial triangle of $ABC-\text{tangential triangle},$ thus done.

Take an inversion with respect to $\Omega$. We have that $\Omega_A$, $\Omega_B$, and $\Omega_C$ stay constant from the orthogonal condition. This means that $(A'B'C')$ is the inverse of $(ABC)$ wrt to $\Omega$. Let $(ABC)=\omega$. Let $OI \cap \omega = P,Q$ with $P$ and $I$ on opposite sides of $O$. Let $R$ and $S$ be the intersection of $\omega$ with the line through $I$ perpendicular to $OI$. We have that $PI=\frac{r^2}{R-\sqrt{R^2-2Rr}}$, $QI=\frac{r^2}{R+\sqrt{R^2-2Rr}}$, and $RI=SI=\frac{r^2}{\sqrt{2Rr}}$. Let $X$ be the center of $(PQRS)$. We have that $X$ lies on $PQ$, so $XI=\frac{r\sqrt{R^2-2Rr}}{2R}$. This means the radius of the new circle is $$\sqrt{XI^2+RI^2}=\sqrt{\frac{r^2(R^2-2Rr)}{4R^2}+\frac{r^3}{2R}} = \sqrt{\frac{r^2R^2-2Rr^3}{4R^2}+\frac{2r^3R}{4R^2}} = \frac{r}{2},$$ as desired so we are done.

Let $A_1B_1C_1$ be the contact triangle of $ABC$. Now, we redefine the points $A', B', C'$ so that $A'B'C'$ is the medial triangle of $A_1B_1C_1$.

Because $A$ and $A'$ are clearly inverses wrt $\Omega$, we know $A'$ belongs to $\Omega_b$ and $\Omega_c$. Using analogous reasoning allows us to recover the original definitions of $B'$ and $C'$ as well. Thus, since $(A'B'C')$ is the Nine-Point Circle of $A_1B_1C_1$, the desired result follows readily. $\blacksquare$

Consider the inversion about $\Omega$. Since $\Omega_{a},\Omega_{b},\Omega_{c}$ are orthogonal, they are fixed. Now, intersections are preserved, so $A\to A'$ and $A'\to A$. Thus, $A'B'C'$ is the medial triangle of the contact triangle. We are done.


Find, with proof, all triples of real numbers $(a, b, c)$ such that all four roots of the polynomial $f(x) = x^4 +ax^3 +bx^2 +cx+b$ are positive integers. (The four roots need not be distinct.)

I did this by hand, so I hope I didn't make any mistakes:

$\left(x-r_{1}\right)\left(x-r_{2}\right)\left(x-r_{3}\right)\left(x-r_{4}\right)=x^{4}-\left(r_{1}+r_{2}+r_{3}+r_{4}\right)x^{3}+\left(r_{1}r_{2}+\left(r_{1}+r_{2}\right)\left(r_{3}+r_{4}\right)+r_{3}r_{4}\right)x^{2}-\left(r_{1}r_{2}\left(r_{3}+r_{4}\right)+r_{3}r_{4}\left(r_{1}+r_{2}\right)\right)x+r_{1}r_{2}r_{3}r_{4}$

Then we can set this equal to $x^{4}+ax^{3}+bx^{2}+cx+b$ to say:

$a=-\left(r_{1}+r_{2}+r_{3}+r_{4}\right)$

$b=r_{1}r_{2}+\left(r_{1}+r_{2}\right)\left(r_{3}+r_{4}\right)+r_{3}r_{4}=r_{1}r_{2}r_{3}r_{4}$

$c=-\left(r_{1}r_{2}\left(r_{3}+r_{4}\right)+r_{3}r_{4}\left(r_{1}+r_{2}\right)\right)$

...and then I guess it would require some more work to find all values.  Is there a quicker way maybe?

This solution received a 5, but it is rather heavy on casework and very long.

[hide]In a quartic equation, the coefficient of the $x^2$ term is the second symmetric sum of the roots, while the constant term is the product of the roots.  Since they are equal, we have

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$pqrs=pq+pr+ps+qr+qs+rs$

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where p, q, r, and s are the four roots and are, as specified, positive integers.  If we divide both sides by pqrs we get

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$1=\displaystyle\frac{pq+pr+ps+qr+qs+rs}{pqrs}=\displaystyle\frac{1}{rs}+\displaystyle\frac{1}{qs}+\displaystyle\frac{1}{qr}+\displaystyle\frac{1}{ps}+\displaystyle\frac{1}{pr}+\displaystyle\frac{1}{pq}$

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We may assume that $rs\leq qs, qr, ps, pr, pq$. We can also assume $r\leq s \leq q \leq p$. So, if rs=2, r=1 and s=2.  Substituting these values in, we have

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$1=\displaystyle\frac{1}{2}+\displaystyle\frac{1}{2q}+\displaystyle\frac{1}{q}+\displaystyle\frac{1}{2p}+\displaystyle\frac{1}{p}+\displaystyle\frac{1}{pq}\Rightarrow \displaystyle\frac{1}{2}=\displaystyle\frac{1}{2q}+\displaystyle\frac{1}{q}+\displaystyle\frac{1}{2p}+\displaystyle\frac{1}{p}+\displaystyle\frac{1}{pq}$

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It is clearly seen that, because of the $\displaystyle\frac{1}{q}$ term, $q>2$ so that we do not have a negative number equal to a sum of positive numbers. When q=3 we have

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$\displaystyle\frac{1}{2}=\displaystyle\frac{1}{6}+\displaystyle\frac{1}{3}+\displaystyle\frac{1}{2p}+\displaystyle\frac{1}{p}+\displaystyle\frac{1}{3p}\Rightarrow 0=\displaystyle\frac{1}{2p}+\displaystyle\frac{1}{p}+\displaystyle\frac{1}{3p}$

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This is also impossible, so $q>3$. When q=4,

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$\displaystyle\frac{1}{2}=\displaystyle\frac{1}{8}+\displaystyle\frac{1}{4}+\displaystyle\frac{1}{2p}+\displaystyle\frac{1}{p}+\displaystyle\frac{1}{4p}\Rightarrow \displaystyle\frac{1}{8}=\displaystyle\frac{7}{4p}\Rightarrow p=14$

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Since 14 is a positive integer, $(r,s,q,p)=(1,2,4,14)$ is a possible set of roots. If q=5, we have

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$\displaystyle\frac{1}{2}=\displaystyle\frac{1}{10}+\displaystyle\frac{1}{5}+\displaystyle\frac{1}{2p}+\displaystyle\frac{1}{p}+\displaystyle\frac{1}{5p}\Rightarrow \displaystyle\frac{1}{5}=\displaystyle\frac{17}{10p}\Rightarrow p=\displaystyle\frac{17}{2}$

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This is not a positive integer, so $q\neq 5$. If q=6 we get

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$\displaystyle\frac{1}{2}=\displaystyle\frac{1}{12}+\displaystyle\frac{1}{6}+\displaystyle\frac{1}{2p}+\displaystyle\frac{1}{p}+\displaystyle\frac{1}{6p}\Rightarrow \displaystyle\frac{1}{4}=\displaystyle\frac{5}{3p} \Rightarrow p=\displaystyle\frac{20}{3}$

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Again, we do not have a positive integer.  We move on to q=7:

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$\displaystyle\frac{1}{2}=\displaystyle\frac{1}{14}+\displaystyle\frac{1}{7}+\displaystyle\frac{1}{2p}+\displaystyle\frac{1}{p}+\displaystyle\frac{1}{7p}\Rightarrow \displaystyle\frac{2}{7}=\displaystyle\frac{23}{14p}\Rightarrow p=\displaystyle\frac{23}{4}$

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This time, not only do we not have an integer, $p<q$, contradicting the earlier assumption that $p\geq q$.  We notice that if we continue to make q larger, p will become continually smaller to compensate. Thus we cannot go any further with rs=2.

Our next case is rs=3. Here we have r=1 and s=3, so that we have

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$1=\displaystyle\frac{1}{3}+\displaystyle\frac{1}{3q}+\displaystyle\frac{1}{q}+\displaystyle\frac{1}{3p}+\displaystyle\frac{1}{p}+\displaystyle\frac{1}{pq}\Rightarrow \displaystyle\frac{2}{3}=\displaystyle\frac{1}{3q}+\displaystyle\frac{1}{q}+\displaystyle\frac{1}{3p}+\displaystyle\frac{1}{p}+\displaystyle\frac{1}{pq}$

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To avoid a negative number equal to a positive number (and because $s\leq q$), we must start with q=3. In that case:

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$\displaystyle\frac{2}{3}=\displaystyle\frac{1}{9}+\displaystyle\frac{1}{3}+\displaystyle\frac{1}{3p}+\displaystyle\frac{1}{p}+\displaystyle\frac{1}{3p}\Rightarrow \displaystyle\frac{2}{9}=\displaystyle\frac{5}{3p}\Rightarrow p=\displaystyle\frac{15}{2}$

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This result not being a positive integer, we move to q=4:

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$\displaystyle\frac{2}{3}=\displaystyle\frac{1}{12}+\displaystyle\frac{1}{4}+\displaystyle\frac{1}{3p}+\displaystyle\frac{1}{p}+\displaystyle\frac{1}{4p}\Rightarrow \displaystyle\frac{1}{3}=\displaystyle\frac{19}{12p}\Rightarrow p=\displaystyle\frac{19}{4}$

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Still without a positive integer, we examine q=5:

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$\displaystyle\frac{2}{3}=\displaystyle\frac{1}{15}+\displaystyle\frac{1}{5}+\displaystyle\frac{1}{3p}+\displaystyle\frac{1}{p}+\displaystyle\frac{1}{5p}\Rightarrow \displaystyle\frac{2}{5}=\displaystyle\frac{23}{15p}\Rightarrow p=\displaystyle\frac{23}{6}$

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We now have $p<q$ and p is not an integer, so we stop with rs=3.

Our next case is rs=4.  There are two possibilities now:r=2 and s=2, and r=1, s=4. Using the first, we get

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$1=\displaystyle\frac{1}{4}+\displaystyle\frac{1}{2q}+\displaystyle\frac{1}{2q}+\displaystyle\frac{1}{2p}+\displaystyle\frac{1}{2p}+\displaystyle\frac{1}{pq}\Rightarrow \displaystyle\frac{3}{4}=\displaystyle\frac{1}{q}+\displaystyle\frac{1}{p}+\displaystyle\frac{1}{pq}$

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We may begin with q=2, getting

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$\displaystyle\frac{3}{4}=\displaystyle\frac{1}{2}+\displaystyle\frac{1}{p}+\displaystyle\frac{1}{2p}\Rightarrow \displaystyle\frac{1}{4}=\displaystyle\frac{3}{2p}\Rightarrow p=6$

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Here, we have a positive integer. This tells us that $(r,s,q,p)=(2,2,2,6)$ is a successful set of roots. We continue with q=3:

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$\displaystyle\frac{3}{4}=\displaystyle\frac{1}{3}+\displaystyle\frac{1}{p}+\displaystyle\frac{1}{3p}\Rightarrow \displaystyle\frac{5}{12}=\displaystyle\frac{4}{3p}\Rightarrow p=\displaystyle\frac{16}{5}$

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We do not have an integer, and move on to q=4, getting

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$\displaystyle\frac{3}{4}=\displaystyle\frac{1}{4}+\displaystyle\frac{1}{p}+\displaystyle\frac{1}{4p}\Rightarrow \displaystyle\frac{1}{2}=\displaystyle\frac{5}{4p}\Rightarrow p=\displaystyle\frac{5}{2}$

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This case is now exhausted, for $p>q$ and p is not an integer.

We next consider r=1, s=4, so that

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$1=\displaystyle\frac{1}{4}+\displaystyle\frac{1}{4q}+\displaystyle\frac{1}{q}+\displaystyle\frac{1}{4p}+\displaystyle\frac{1}{p}+\displaystyle\frac{1}{pq}\Rightarrow \displaystyle\frac{3}{4}=\displaystyle\frac{1}{4q}+\displaystyle\frac{1}{q}+\displaystyle\frac{1}{4p}+\displaystyle\frac{1}{p}+\displaystyle\frac{1}{pq}$

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Here, due to the assumption that $s\leq q$, we must start with q=4:

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$\displaystyle\frac{3}{4}=\displaystyle\frac{1}{16}+\displaystyle\frac{1}{4}+\displaystyle\frac{1}{4p}+\displaystyle\frac{1}{p}+\displaystyle\frac{1}{4p}\Rightarrow \displaystyle\frac{7}{16}=\displaystyle\frac{3}{2p}\Rightarrow \displaystyle\frac{24}{7}$

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This is not an integer and already we have $p<q$, so we end this case.

Our next case is rs=5.  Here, r=1 and s=5. We get

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$1=\displaystyle\frac{1}{5}+\displaystyle\frac{1}{5q}+\displaystyle\frac{1}{q}+\displaystyle\frac{1}{5p}+\displaystyle\frac{1}{p}+\displaystyle\frac{1}{pq}\Rightarrow \displaystyle\frac{4}{5}=\displaystyle\frac{1}{5q}+\displaystyle\frac{1}{q}+\displaystyle\frac{1}{5p}+\displaystyle\frac{1}{p}+\displaystyle\frac{1}{pq}$

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So that $s\leq q$, we start with q=5:

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$\displaystyle\frac{4}{5}=\displaystyle\frac{1}{25}+\displaystyle\frac{1}{5}+\displaystyle\frac{1}{5p}+\displaystyle\frac{1}{p}+\displaystyle\frac{1}{5p}\Rightarrow \displaystyle\frac{14}{25}=\displaystyle\frac{7}{5p}\Rightarrow p=\displaystyle\frac{5}{2}$

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This is not an integer and $p<q$.  We could consider the case where rs=6, but then $6\leq qs, qr, ps, pr, pq$, and the only way that six unit fractions with denominators $\geq 6$ can sum to 1 is if all the denominators are 6, in which case $r=s=q=p=\sqrt{6}$, which is not an integer.  It is impossible for six unit fractions with denominators $\geq 7$ to sum to 1, so we are done.  The only possible sets of roots are (1,2,4,14) and (2,2,2,6).

We must now determine the triples of real numbers (a,b,c).  a is the negative of the sum of the roots, b is the product of the roots (and the second symmetric sum), and c is the negative of the third symmetric sum of the roots.  Thus for the first set,

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$a=-(1+2+4+14)=-21$

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$b=1*2*4*14=112$

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$c=-(1*2*4+1*2*14+1*4*14+2*4*14)=-204$

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and for the second set,

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$a=-(2+2+2+6)=-12$

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$b=2*2*2*6=48$

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$c=-(2*2*2+2*2*6+2*2*6+2*2*6)=-80$

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So our two triples are $(-21, 112, -204)$ and $(-12, 48, -80)$.
[/hide]

[quote="chess64"][size=59]Look at your own signature, nah just kidding. :D[/size]

BTW, instead of using all those \displaystyle tags you can just use [code]\[ \] [/code][/quote]

...or all the $\LaTeX$ on the forums is already rendered in displaystyle mode.


inequalities
geometry
incenter
circumcircle
geometry proposed

Given $\triangle ABC$ with incenter $I$, circumcenter $O$ orthocenter $H$ and circumradius $R$ prove that
$IA^{3}+IB^{3}+IC^{3}+2R.OH^{2}\ge 3R^{3}$
 :)  :)

Your inequality is not homogeneous......

oh sorry I have  some mistake I have fixed it! thanks! :)


[hide="Brute Force Solution"]The sum of all the terms, $S$ satisfies the following:
$S=98a_{1} + \frac{97 \cdot 98}{2}=137$.

Therefore, $S=\frac{137-\frac{97 \cdot 98}{2}}{2}$, which we will call $Q$, for the sake of sanity.

The sum of all even terms is equal to $\frac{Q}{2}+49^2=93$.[/hide]


The rays $[OX$ and $[OY$  and the positive real number $k$ are given. Let $A$ be an arbitrary point on the ray $[OX$ and $B$ be an arbitrary point on the ray $[OB$ such that $|OA| + |AB| + |BO| = k$. Show that there is a constant point in the interior part of the $\angle XOY$ distinct from $O$ such that the $\angle APB$ is constant while $A$ and $B$ vary in above condition.

Let $s = \frac{OA + OB + AB}{2} = \frac k 2$ be the constant semiperimeter of the triangle $\triangle OAB$. Let $(P)$ be the excircle of the triangle $\triangle OAB$ against the vertex $O$ and $S, T$ its tangency points with the rays $OX, OY$, respectively. The excircle $(P)$ is centered on the internal bisector of the angle $\angle AOB \equiv XOY$ and the distance of the tangency points $S, T$ from the vertex $O$ is constant: $OS = OT = s = \frac k 2$. Thus the excircle $(P)$ is fixed and the line $AB$ is a tangent to this fixed circle. The excircle $(P)$ is also centered on the bisectors of the external angles $\angle SAB, \angle TBA$ of the triangle $\triangle OAB$.

$\angle SAB = \angle AOB + \angle ABO,\ \ \angle TBA = \angle AOB + \angle BAO$

$\angle APB = 180^o - \frac{\angle SAB + \angle TBA}{2} = 180^o - \left(\angle AOB + \frac{\angle ABO + \angle BAO}{2}\right) =$

$= 180^o - \left(\angle AOB + \frac{180^o - \angle AOB}{2}\right) = 90^o - \frac{\angle AOB}{2} = 90^o - \frac{\angle XOY}{2}$

and this angle is constant.


search
MATHCOUNTS
geometry
inequalities
ratio
email
similar triangles

Hi, I'm looking for a way to practice for the qualifying exam for Bronx Science incoming freshmen! I heard examples from NYCIML and NYML are good. Does anyone have a link to those or other problems or suggestions how to practice?

Thanks,

Edna

You can find 6 problems from the (NY) Math League at
[url]http://www.mathleague.com/contests.htm#High_School[/url].

Lower down on this NYC forum (more than a year ago), I think I posted a few problems from NYCIML.

[quote="Bictor717"]I don't know of any place online where the NYML and NYCIML questions are available. You can try the Search, and might be able to dig up a few random questions.
Fortunately, you can find many good practice problems on this forum. I don't think the qualifying exam questions are at the level of those in the Mathcounts section, so that and the Middle School forums may be good places to lurk.

Welcome to AoPS![/quote]

Thanks! The MathCounts is interesting, you choose your own types of problems (team or algebra, etc.) This looks good! Did you take a qualifying test when you first joined a math team?

[quote="JBL"]Bictor was the star of the Bronx Science math team a couple of years ago, if I'm not mistaken.  (He's now a sophomore at college.)[/quote]

Ah! So maybe Bictor can tell me what the qualifying exam experience is like!...?

[quote="Bictor717"]The exam is split into three half-hour long parts, each with 10 questions. Most of them are word problems, and it's not multiple-choice. The topics include geometry (area, similar triangles, Pythagorean theorem), algebra (solving equations and inequalities, absolute value, square roots) and miscellaneous things like ratios and rate$\times$time$=$distance.
I've proctored and graded the freshman qualifying exam for three years, but it's been such a long time since I actually took it. In my freshman year, the exam was split over three days in September, after class began, but they have this whole orientation day now that I didn't have, so the math department decided it would be a good idea to have the placement and math team qualifying exams after the orientation stuff is over. Yes, it will be a long day, so bring water and snacks. Also bring pencils and a hard surface to write on.[/quote]

Hi Bictor,

Thanks so much, this is great info! On the Bronx Science website it says this test is the [i]first [/i]part of the Math Team qualifying test. Do you know what else is given and when?

[quote="Bictor717"]If you don't pass the first exams, math team is removed from your schedule, but if you do, it's added. I think there is a second series of tests for people who get in, but they're conducted after the school year begins, during the math team period. They're mainly for those people who barely got in, but I'm not sure how that works; I don't think the coach kicks people off after the year begins, barring people with behavior issues (even then, she usually just deals with it :) ). Maybe they're for selecting the starting team for competitions, but there are plenty of opportunities to make a good impression, and everyone participates in those anyway.
So if I interpreted the use of [i]first[/i] correctly, you don't need to worry about any additional exams. It could also be outdated, a typo, a reference to some alternate test day, or something about which I have no idea. If you want to know for certain, call the department office, email the coach, Joanne Strauss, or wait for me to visit the school and ask for you (sometime next week).[/quote]

OK thanks so much for all your help!!!

I'm gonna be senior at Bronx Science in the fall, and I can safely say that if you know what you're doing to any extent the test should be a breeze, the problems aren't from NYML or NYCML they are more straight forward (at least they were 3 years ago) factoring - computation, or analyzing - computation.

wow BX Science is hard core... I would have never made it onto the math team if they required that at my HS...


Let S be the set {0, 1}. Given any subset of S we may add its arithmetic mean to S (provided it is not already included - S never includes duplicates).Show that we can eventually include any rational number between 0 and 1.

Claim follows inductively from the existance of binary expansion of every rational number involving only finite number of ones.


Denote $ M$ be the set of all rational numbers in the intervel $ (0,1)$ . Does there exists a set $ A$ which is a subset of $ M$ such that each element of $ M$ can be represented uniquely into one or several (pair wise distinct) elements in $ A$ .

represented uniquely as a *sum*? well that's easy: reciprocals of powers of 2

Oh sorry! You are right. " represented as sum... ". Can you please write a full solution for me?

let $ x\in M$
we take $ (a_i)$ such that:
$ a_{i\plus{}1}\equal{}min(\{k\in\mathbb{N}^*|\ 2^{k}(x\minus{}\sum_{j\equal{}1}^{i}\frac{1}{2^{a_i}})\ge 1\})$ if $ x\minus{}\sum_{j\equal{}1}^{i}\frac{1}{2^{a_i}}\neq 0$ and if $ x\minus{}\sum_{j\equal{}1}^{i}\frac{1}{2^{a_i}}\equal{} 0$ we stop the seqence.
then we have $ a_{i\plus{}1}\ge a_i\plus{}1$
if $ (a_i)$ is a finite sequance then $ x\equal{}\sum_{i\equal{}1}^{p}\frac{1}{2^{a_i}}$
if $ (a_i)$ is a infinite sequance then $ x\equal{}\sum_{i\equal{}1}^{\plus{}\infty}\frac{1}{2^{a_i}}$
so $ x$ can be represented into one or several (pair wise distinct) elements in $ a\equal{}\{\frac{1}{2^k}|\ k\in\mathbb{N}^*\}$
if there exists (b_i) such that $ b_{i\plus{}1}\ge b_i\plus{}1$ and $ x\equal{}\sum_{i\equal{}1}^{p}\frac{1}{2^{b_i}}\ or\ \sum_{i\equal{}1}^{\plus{}\infty}\frac{1}{2^{b_i}}$
then $ min(\{k\in\mathbb{N}^*|\ 2^{k}(x\minus{}\sum_{j\equal{}1}^{i}\frac{1}{2^{b_i}})\ge 1\})\equal{}b_{i\plus{}1}$ so $ (a_i)\equal{}(b_i)$
so $ x$ can be represented uniquely into one or several (pair wise distinct) elements in $ a\equal{}\{\frac{1}{2^k}|\ k\in\mathbb{N}^*\}$

I'm not sure if the author really wanted to allow infinite sums, I think he wants finite ones...

And the posts above mine do nothing else than writing the rationals in base $ 2$. But it's wrong that this is unique: $ 0.1 \equal{} 0.01111...$.


An isosceles trapezoid has bases of length a and b with a<b.  THe line joining the midpoints of the two equal legs divide the interior of the trapezoid into areas A1 and A2 with A1<A2.  Then the ratio A1/A2 is...

[hide]a and b are the bases where b is bigger.  The line joining the midpoints of the legs is parallel to the base and is the average of the two bases.  The height of the original trapezoid is also bisected by this line.  So you have h/2(a+[a+b]/2)/2 as A1 and h/2(b+[a+b]/2)/2 as A2.  A1 equals (3a+b)h/8 and A2 equals (3b+a)h/8.  The ratio of these two is (3a+b)/(3b+a).[/hide]

[quote="Iversonfan2005"][quote="riddler"]Do you want integer values?[/quote]
[hide]
That's what I did, I called $a=2$ and $b=4$, therefore the line joining midpoints is $3$, from there you can call the height something like $2$, so therefore $A1=\dfrac{5}{2}$ and $A2=\dfrac{7}{2}$, therefore the ratio is: $\boxed{\dfrac{5}{7}}$.[/hide][/quote]
I dont think you are ALLOWED to do that because if you take any other value you get something different.

[img]http://www.artofproblemsolving.com/Forum/album_pic.php?pic_id=292[/img]

[hide="Since you guys do not believe dengmi..."]Let $AB=a$, and $CD=b$. Then $EF=\dfrac{a+b}{2}$.

$[ABEF]=(1/2)(EF+AB)(AG)$
$[ABEF]=\left(\dfrac{AG}{2}\right)\left(\dfrac{3a+b}{2}\right)$
$[ABEF]=\dfrac{AG(3a+b)}{4}$

$[EFCD]=(1/2)(CD+EF)(GH)$
$[EFCD]=\left(\dfrac{GH}{2}\right)\left(\dfrac{3b+a}{2}\right)$
$[EFCD]=\dfrac{GH(3b+a)}{4}$

Because $\angle DAH = \angle EAG$ and $\angle AGE = \angle AHD = 90^\circ$, we have $\triangle AGE \sim \triangle AHD$. This implies $\dfrac{AE}{AD}=\dfrac{AG}{AG+GH} \implies AG+GH=2AG \implies AG=GH$.

Therefore, we have

$[ABEF]/[EFCD]=\dfrac{\dfrac{AG(3a+b)}{4}}{\dfrac{GH(3b+a)}{4}}\\ =\dfrac{AG(3a+b)}{GH(3b+a)}\\ =\boxed {\dfrac{3a+b}{3b+a}}$[/hide]

I never said I dont believe him, I just wanted to know if the question asked for numerical values.

Apparently Iversonfan didn't :roll:

And also, that's as simplified as you can get. ;)


bache ha man start mizanam.inja dar morede musighi harf mizanim.
man hamoon tor ke goftam POPular ro kheili doos daram.shoma ham biyayd nazar bedin :wink:
inja age beshe ham site moarrefi mikonim ham music attach mikonim ham flash :D  :)

What the heck man "oogra moogra loolo", "yasirban kirkuvanu refungara tulapandaragoo" If you are going to post something put in english please!
 :rotfl:


floor function
combinatorics open
combinatorics

In how many ways can n people, numbered from 1 to n, be seated around a circular table such that person 1 sits next to at least one of person 2 or person n, person 2 sits next to at least one of person 1 or person 3, and so on?

I've had this problem for a while, and it has thus far resisted all my efforts at a solution. I've worked it out in the cases n = 5 and n = 6 (the answer is of course (n-1)! for n<5) but no clear pattern emerges in those cases. In fact, I can't even find reasonably good bounds on the answer for general n, though I can show (it isn't hard) that it grows faster than any exponential function. (So don't be fooled into thinking it's less than a constant times 4^n...)

Have you tried looking up your values for $n=1$ to $6$ in Sloane's Encyclopedia of Integer Sequences?


calculus
integration
real analysis
trigonometry
function
calculus computations

Does $ \int_0^\infty \cos(t^3/3 \plus{} xt) dt$ converge for $ x \in \mathbb{R}$?

Yes, it does converge. (As an improper integral - certainly not as a Lebesgue integral.)

As a first step, choose some positive $ b$ and then find $ a$ such that $ \frac {t^3}3 + xt > b$ for $ t > a.$ (Note that $ a$ might depend on $ x.$) Observe also that $ t^2 + x$ is positive and bounded away from zero for $ t > a.$

$ \int_0^a\cos(\frac {t^3}3 + xt)\,dt$ is the integral of a continuous function over a compact interval; its convergence is not in doubt. That leaves the integral from $ a$ to $ \infty.$ For that, we're going to integrate by parts.

$ \int_a^{\infty}\cos(\frac {t^3}3 + xt)\,dt = \int_a^{\infty}\frac1{t^2 + x}\,\cos(\frac {t^3}3 + xt)(t^2 + x)\,dt$

$ = \frac1{t^2 + x}\,\sin(\frac {t^3}3 + xt)\right_a^{\infty} + \int_a^{\infty}\frac {2t}{(t^2 + x)^2}\sin(\frac {t^3}3 + xt)\,dt$

Since $ t^2 + x$ is positive and bounded away from zero on $ [a,\infty)$ and since the required limit at infinity is zero, the boundary terms evaluate to a finite number. The final integral converges absolutely by comparison to

$ \int_a^{\infty}\frac {2t}{(t^2 + x)^2}\,dt.$

The statement that this doesn't converge as a Lebesgue integral is nothing more than the statement that the original integral doesn't converge absolutely.

Next question: is the quantity the integral converges to a differentiable function of $ x?$


calculus
derivative
analytic geometry
conics
parabola

Dear everyone,

       Can you please help me with the following problem?

a) Find the equations of all lines that are tangent to both y = x^2 + 2 and y = -2 - x^2. Indicate the points where the tangent line intersects with the curve.

b) A man 2m tall walks at 5/3 meters / s along one edge of a straight road that is 10m wide. On the other side
    of the road is a light atop a pole that is 6m tall. How fast is the length of the man's shadow (on the
    ground) increasing when he is 15m beyond the point directly across the road from the pole?

       Thank you so much!

Best,
Angel

This belongs in the high school basics forum, or the calculus forum

1. You have 2 coordinates for $ f(x)$ and two coordinates for $ f'(x)$ so sub them in a make four simultaneous equations

2. ship A is $ 20 \minus{} 15t$ miles from the Origin and ship B is $ 15 \minus{} 20t$ miles
Now just apply Pythagoras and differentiate

EDIT: did you change your questions???

[quote="angel_09"]

a) Find the equations of all lines that are tangent to both y = x^2 + 2 and y = -2 - x^2. Indicate the points where the tangent line intersects with the curve.Angel[/quote]

The two parabola are symmetric for $ x$ axis and for the origin. Thus we can set the tangent lines as $ y \equal{} \pm mx$.
The condition for which the parabola $ y \equal{} x^2 \plus{} 2$ touches to the tangent is $ x^2 \minus{} mX \plus{} 2 \equal{} 0$ has double root.
$ D \equal{} m^2 \minus{} 8 \equal{} 0\Longleftrightarrow m \equal{} \pm 2\sqrt {2}$, yielding the tangent lines are $ y|\equal{}\pm \sqrt{2}x$ and  the tangentpoints are $ (\pm \sqrt {2}, 4)$.

Hi Ocha,

    I am sorry about that, cos I just realized that I posted the wrong Q.

    I was trying to PM you to say thanks to you.  But since I am a new member, it forbids me to do so within the following 2 hrs.

    Sorry about that.

Thanks,
Angel


[quote="pvthuan"]Let $x,y$ be real positive numbers such that $x^2+y^2=2$. Prove that
\[ (1-x)(1-y)+(1-x-y)^2\leq1.  \][/quote]

equinalent to $(1-x)(1-y)leq 0$  :?

Then square......

is it ok??

First, we break the square $(1-x-y)^2=1-2(x+y)+x^2+y^2+2xy$, and then multiplying $(1-x)(1-y)$ out to have $1-x-y+xy$. Then the desired inequality reduces to
\[ 3-3(x+y)+3xy\leq 0.  \]
Put $s=x+y$, then $s\leq2$, and $xy=\tfrac12s^2-1$. The rest is for you.


Among all the triangles which have a fixed side $l$ and a fixed area $S$, determine for which triangles the product of the altitudes is maximum.

[hide]The altitude to the side $l$ is fixed, because the length of that altitude is twice $\frac{S}{l}$.  Hence, the other two altitudes, if the angle opposite $l$ is $\theta$ and the two other sides are $a$ and $b$, are $a\sin{\theta}$ and $a\sin{theta}$, so the product of those two altitudes is $ab\sin^2{\theta}$.  However, the area is fixed, and the area is $\frac{ab\sin{\theta}}{2}$.  So, the product of the altitudes is $\frac{2S}{l}2S\sin{\theta}=\frac{4S^2\sin{\theta}}{l}$.  Thus, we want to maximize $\sin{\theta}$ by getting it as close to a right angle as possible.[/hide]


[quote="Nameless"]$x,y,z\ge0$. Prove or disprove:
$\sum\sqrt{3x^2+3xy+3xz+y^2+z^2-2yz}\ge3(x+y+z)$[/quote]
This is wrong: $x=0,y=z=1.$ ;)

What about this:
$\sum\sqrt{4x^2+4xy+4xz+y^2+z^2-2yz}\ge3(x+y+z)$


binomial coefficients
algebra
binomial theorem

Hello Ms. Sexy,

These two problems are at most 'Intermediate' and should be placed in that forum. However, getting back to the two problems:

$\left(x^3-{2x^{-2}}\right)^{4}$ = $x^{12}-8x^7+24x^2-\frac{32}{x^3}+\frac{16}{x^8}$

Therefore, the coefficient of $x^{-8}$ is $16$...

For the second problem, you can use the binomial coefficients, but the cofficient of $x^{20}$ will be $125970$...

$\left( {x^3  - x^{ - 2} } \right)^{20}  = \sum\limits_{k = 0}^{20} {\left( \begin{array}{l}
 20 \\
 k \\
 \end{array} \right) \cdot \left( {x^3 } \right)^{20 - k}  \cdot } \left( { - 1 \cdot x^{ - 2} } \right)^k  = $
$ = \sum\limits_{k = 0}^{20} {\left( \begin{array}{l}
 20 \\
 k \\
 \end{array} \right)\left( { - 1} \right)^k x^{60 - 5k} } $
then:
$60 - 5k = 20 \Leftrightarrow k = 8$
then coefficient of  $x^{20} $  will be  $\left( \begin{array}{l}
 20 \\
 8 \\
 \end{array} \right) = \frac{{20!}}{{8! \cdot 12!}}$

You should know the binomial theorem before you even attempt these problems.  If you do you can see that for the first one $x^{-8}$ will happen on the final term of the expansion.  Thus, your coeffecient should be $(-2)^4 = 16$.

For the second one you have to find the term where $x$ is raised to the 20th.  If the max degree of x is 60, you can set $20 = 60-5r$ r must be 8 so it is the eigth term.  20 nCr 8 = 125970 which should be your final coefficient.


On the same set of axes are drawn the graph of $ y\equal{}ax^2\plus{}bx\plus{}c$ and the graph of the equation obtained by replacing $ x$ by $ \minus{}x$ in the given equation.  If $ b \neq 0$ and $ c \neq 0$ these two graphs intersect:

$ \textbf{(A)}\ \text{in two points, one on the x\minus{}axis and one on the y\minus{}axis}\\
\textbf{(B)}\ \text{in one point located on neither axis} \\
\textbf{(C)}\ \text{only at the origin} \\
\textbf{(D)}\ \text{in one point on the x\minus{}axis} \\
\textbf{(E)}\ \text{in one point on the y\minus{}axis}$

[hide="Solution"]Let $ f(x)\equal{}ax^2\plus{}bx\plus{}c\equal{}ax^2\minus{}bx\plus{}c\implies bx\equal{}\minus{}bx\implies 2bx\equal{}0$.
Because $ b\neq 0$, $ x\equal{}0$.

So the answer is $ \boxed{\textbf{(E)}}$.[/hide]


inequalities
combinatorics unsolved
combinatorics

5 problems were given at a math contest and it turned out that every problem was solved by at least $ 2/3$ of the contestants. Porve that there is a set of 3 problems solved by at least $ 1/5$ of the contestants.

let the number of the contestants $ m$
and $ a_i$ is the number of the problems that $ i$th student solved.

Want to show: $ \displaystyle\sum_{i \equal{} 1}^{m}\dbinom{a_i}{3} \ge 2m$

it is easy to check for positive integers $ a$ and $ b$, ($ a\ge b \plus{} 2$), $ \dbinom{a}{3} \plus{} \dbinom{b}{3} \ge \dbinom{a \minus{} 1}{3} \plus{} \dbinom{b \plus{} 1}{3}$
using this truth and trivial inequality that $ \displaystyle\sum_{i \equal{} 1}^{m}a_i \ge \dfrac{10}{3}m$,
we can know that the minimum of $ \displaystyle\sum_{i \equal{} 1}^{m}\dbinom{a_i}{3}$ appears
when  $ \dfrac{2}{3}m$ of $ a_i$'s are 3 and $ \dfrac{1}{3}m$ of $ a_i$'s are 4.
and in this case, $ \displaystyle\sum_{i \equal{} 1}^{m}\dbinom{a_i}{3}$ is $ 2m$

So we proved 'Want to show'......

Because there are total 10 pairs of 3 problems and a pair can appear only one time from one contestant,
there is a pair which are solved by at least $ \dfrac{2m}{10} \equal{} \dfrac{m}{5}$ people.(Because of 'Want to show')

QED


I looked at my MiKtex package manager for the first time this morning and saw "Obsolete" on all but one of my packages. I realized I still have MiKtex 2.4.

I just installed 2.6, but I don't know how to have TeXnicCenter switch programs from 2.4 to 2.6 without deleting any of my .tex files.

Help please?
Thanks!

Build -> Define Output Profiles and change the [i]Path to the (La)TeX compiler[/i] to MiKTeX 2.6's directory.


Sa se constriasca corpuri cu 4, 16, 25 elemente.

 Edit:

Ca sa construiesti un corp cu $p^{n}$ elemente(cu $p$ numar prim) este suficient sa gasesti un polinom ireductibil $f$ de grad $n$ peste $Z_{p}$ si sa consideri corpul $Z_{p}[X]/(f)$. Concret in cazurile cerute de tine putem avea \[\begin{array}{l}Z_{2}[X]/(X^{2}+X+1)\,;\\ \ \\ Z_{2}[X]/(X^{4}+X+1)\,;\\ \ \\ Z_{5}[X]/(X^{2}+2) \,. \end{array}\]


abstract algebra
vector
superior algebra
superior algebra unsolved

Find all the finite groups $ G$ with the property that any automorphism different from $ 1_{G}$ has an unique fix point.

Consider the inner automorphisms, this is: those of type $ x \mapsto axa^{\minus{}1}$ for some fixed $ a$.
This will give you that the group is abelian. Now continue.

For $ x\in G\setminus Z(G)$ we have the automorphism $ G\to G,y\mapsto xyx^{\minus{}1}$
with different fixed points $ 1,x$ and $ \not\equal{}\text{id}_G$. Contradiction!
So $ G\equal{}Z(G)$ is abelian, and using the [url=http://en.wikipedia.org/wiki/Fundamental_theorem_of_finitely_generated_abelian_groups#Classification]classifikation of all finite abelian groups[/url]
we get $ G\equal{}\langle x\rangle \times H$ with some $ H\le G$ and some $ x\in G$ of order $ \text{exp}(G)$.
Then for $ H\not \equal{}1$ and $ 1\not\equal{}y\in H$ we get the well-defined automorphism
$ G\to G, x^nz\mapsto x^ny^nz$ for all $ n\in\mathbb{Z},z\in H$
with $ x\mapsto xy\not\equal{}x$ and fixed points $ H\not\equal{}1$. Contradiction!

So $ G\equal{}\langle x\rangle$ is cyclic and for $ |G|>2$ we have the non-trivial automorphism
$ G\to G,g\mapsto g^{\minus{}1}$ with no fixed points $ \not\equal{}1$, i.e. no involutions in $ G$, i.e. $ |G|$ is odd.
Let $ p$ be the smallest prime divisor of $ |G|$ and $ y\in G$ of order $ p$.
Then $ p\plus{}1\equal{}2\cdot {p\plus{}1\over 2}$ is a product of primes $ <p$ and therefore coprime to $ |G|$.
This gives the automorphism $ G\to G,g\mapsto g^{p\plus{}1}$ with different fixed points $ 1,y$.
Then by assumption $ g^{p\plus{}1}\equal{}g$ and $ g^p\equal{}1$ for all $ g\in G$, so $ |G|\equal{}p$.

This gives the only solutions $ G\equal{}1$ and $ G\cong \mathbb{Z}/p\mathbb{Z}$ with $ p$ prime.

What I had in mind:
Consider $ x \mapsto x^{ \minus{} 1}$, this gives that all elements $ \neq e$ have order $ 2$. This gives that the group is isomorphic to $ (\mathbb Z / 2 \mathbb Z)^n$ (as we have a $ \mathbb Z / 2 \mathbb Z$ vector space). Now it's trivial.

[quote="ZetaX"]
Consider $ x \mapsto x^{ \minus{} 1}$, this gives that all elements $ \neq e$ have order $ 2$. [/quote]
Only if $ |G|$ is even.


modular arithmetic
algebra
polynomial
function
number theory

[quote="countdownmath"]Let $ p$ be a prime number. Prove that $ (A_{1}+A_{2}+...+A_{n})^{p}\equiv A_{1}^{p}+A_{2}^{p}+...+A_{n}^{p}\pmod{p}$.[/quote]
This is just Fermat's little theorem: for any $ a \in \mathbb{Z}$ and any prime $ p \in \mathbb{N}$, it's $ a^{p}\equiv a \bmod p$. Hence, if $ A_{1}, \ldots, A_{n}\in \mathbb{Z}$: $ (A_{1}+\ldots+A_{n})^{p}\equiv A_{1}+\ldots+A_{n}\equiv A_{1}^{p}+\ldots+A_{n}^{p}\bmod p$.

this solution might be wrong,anyway:
we know that for every $ a \in\mathbb{Z}$ we have:
$ a^{p}\equiv a (\bmod p)$
where $ p$ is a prime number
so we get that:
$ LHS \equiv (A_{1}+A_{2}+...+A_{n}) (\bmod p)$
and also:
$ RHS \equiv (A_{1})+(A_{2})+...+(A_{n}) (\bmod p)$
thus we get that:
$ LHS \equiv RHS \equiv A_{1}+...+A_{n}(\bmod p)$...

You could even expand it out combinatorially, but Fermat's Little Theorem seems like the quickest way.

same as s.tringali,Im not fast enough in posting the solutions...  :oops:

[quote="t0rajir0u"]What if $ A_{1}, A_{2}, ... A_{n}$ are polynomials instead of numbers?  ;)[/quote]
[b]lemma:assume the polynomial $ f(x)$ with integer coefficients where at least one of them is not divisible by $ p$ where $ p$ is a prime number and $ deg(f)=n$ modula $ p$,then the equation
$ f(x) \equiv 0 (\bmod p)$
has at most $ n$ roots...[/b]  :wink:

I want to mention a useful result.

Let $ F$ be a [i]finite field[/i] or prime charachteristic $ p$ then $ (a+b)^{p^{n}}= a^{p^{n}}+b^{p^{n}}$.

The actual proof of this simply depends on Fermat's Little Theorem with Binomial Expansion as mentioned above.

[quote="sylow_theory"]The actual proof of this simply depends on Fermat's Little Theorem with Binomial Expansion as mentioned above.[/quote]

Using Fermat's little Theorem here is somewhat enlightening in that it relates to group structure, but you could even just use a simple binomial expansion, and then show $ p^{n}\nmid k \implies p^{n}|\binom{p^{n}}{k}$ using a counting/floor function argument, if I'm not mistaken.

For example, let $ F=\mathbb{Z}_{p}$. We want to show $ (a+b)^{p}= a^{p}+b^{p}$ for all $ a,b\in F$. This is a special case of what I just said, but the idea is the same.

Notice that since $ F$ is a field the commutative laws for addition and multiplication hold so:
$ (a+b)^{p}={p\choose 0}\cdot a^{p}+{p\choose 1}\cdot a^{p-1}b+...+{p\choose{p-1}}\cdot ab^{p-1}+{p\choose p}\cdot b^{p}$

Since $ F$ has $ p$ as its [i]charachteristic[/i] it means $ p\cdot x = 0$. Where the operation $ \cdot$ is defined at $ p\cdot a = \underbrace{a+...+a}_{p}$ and similarly note the $ \cdot$ in the binomial expansion.

Now first and last elements of this binomial are $ a^{p}$ and $ b^{p}$ respectively. We argue that all the inner terms are $ 0$. We can do this by showing that $ {p\choose k}$ is divisible by $ p$ for all $ 1\leq k\leq p-1$. And then appeal to the notion of a field charachteristic mentioned above. Notice that $ {p \choose k}= \frac{p(p-1)...(p-k+1)}{k!}$. Now if we divide that fraction by $ p$ we are let with $ \frac{(p-1)(p-2)...(p-k+1)}{k!}$ we can show that this fraction is an integer. There are different ways to approach it, one way is with floor functions like you mentioned. But this shows that all inner terms dissappear.

[quote="K81o7"]$ p^{n}\nmid k \implies p^{n}|\binom{p^{n}}{k}$ using a counting/floor function argument[/quote]

The given statement is false, ($ \binom{9}{3}= 84$, for example) but it's also stronger than you need.  In a field of characteristic $ p$ there may be more than the $ p$ elements of the integers mod $ p$, but it's still true that $ p = 0$.  So all we actually need is that $ 0 < k < p^{n}\implies p | \binom{p^{n}}{k}$.

Sylow_theory, you made that argument sound much harder than it is.  We know that $ \binom{p}{k}$ is an integer from combinatorial concerns.  We know that its numerator is divisible by $ p$ and that its denominator (if $ 0 < k < p$) is not so the integer is.  However, this doesn't extend immediately to cover the $ p^{n}$ case.

[quote="JBL"][quote="K81o7"]$ p^{n}\nmid k \implies p^{n}|\binom{p^{n}}{k}$ using a counting/floor function argument[/quote]

The given statement is false, ($ \binom{9}{3}= 84$, for example) but it's also stronger than you need.  In a field of characteristic $ p$ there may be more than the $ p$ elements of the integers mod $ p$, but it's still true that $ p = 0$.  So all we actually need is that $ 0 < k < p^{n}\implies p | \binom{p^{n}}{k}$.
[/quote]

Whoops  :D  : I messed up the floor summation.  I suppose what it should be is $ p^{i+1}\nmid k \implies p^{n-i}|\binom{n}{k}$.  But the main idea I was trying to communicate was that $ p$ divides the binomial coefficient if the largest power of $ p$ which divides the denominator is smaller than that which divides the numerator...