mathleague.org ARML college MATHCOUNTS AMC AIME geometry

Wow, we had some crazy high scores at the contest.  Congrats to Alumen with his perfect 200 =)

Sorry, but after seeing the type of test used to select the ARML team, I no longer want to go.

I was planning to up until today :/ .

Time to go play SC...

^Yes, and I can understand why he doesn't want to go to any more tryouts.
However, this is a good tryout for ARML? because we need people who won't make silly mistakes in a testing environment

Yeah, it's better than last year.  Although last year's preliminary rounds got me in the A team, they weren't exactly the greatest indicators of mathematical ability.

Also, I disagree with the general format and organization. I wasn't able to go to the first 2 contests because of conflicts, and I don't want to go to the Berkeley ones. That makes me ineligible to qualify for ARML, supposedly, which is pretty ridiculous in my opinion. Even if I could go to them, I don't many any other MOPpers would want to spend three Saturday mornings doing ridiculous contests (read: crud) like this - and wasn't the point to ATTRACT top-level math students? Seriously, if I wanted to do boring unoriginal problems, I would pull out a random MathCounts. Oh, sorry, even MathCounts has creative problems. I guess I'll go to an Algebra II teacher from school and get a test from him.

Seriously, halfway through the Sprint I stopped and solved an ISL problem instead.

Sorry for the rant. I came in today with very low expectations already, but this contest was even worse than what I imagined. Well, at least there were a lot of people there.

Wait Albert, why the hell did you even want to go in the first place? Did I not tell you enough stories?

Does anyone have the answers to the sprint round? I left early so I didn't get my score.

[quote="Ubemaya"]
Seriously, halfway through the Sprint I stopped and solved an ISL problem instead.
[/quote]
You probably weren't alone?  Several people were probably done by the halfway mark, and I doubt they felt like checking their answers.

Maybe our expectations are too high? after being fed with high-quality AMC problems, mostly high-quality AIME problems, and well-chosen olympiad problems?

[quote]Even if I could go to them, I don't many any other MOPpers would want to spend three Saturday mornings doing ridiculous contests (read: crud) like this - and wasn't the point to ATTRACT top-level math students?[/quote]

^I love the elitism in that statement, is taking a normal math contest really that demeaning for MOPpers?  Just to let you know, MOPpers didn't do any better than the rest of the A team at ARML last year...

[quote="DataBox"]

^I love the elitism in that statement, is taking a normal math contest really that demeaning for MOPpers?  Just to let you know, MOPpers didn't do any better than the rest of the A team at ARML last year...[/quote]
Now are you trying to get cocky? His argument was, if this is to attract high level people, then it miserably fails because he doesn't see how any MOPpers would want to take it.
Also I would argue that mathleague<<<<<<<<<<<<"normal math contest".

^Haha, definitely not trying to come across as being cocky, but seriously, Tim's contests attract enough math talent as it stands (seeing as how we had like 7 MOPpers on the team last year).  Also, if it's [b]that[/b] easy, why don't we see [i] any [/i] of your so called "pro people" getting perfect scores...(And don't use the BS reasoning of "it was to boring/disparaging for them")

Oh man if I had gone I'd probably get like 25/200 or whatever. What were the types of problems?

[quote="archimedes1"]
Maybe our expectations are too high? after being fed with high-quality AMC problems, mostly high-quality AIME problems, and well-chosen olympiad problems?[/quote]

It's true that my expectations are very high, but I feel as if the problems should have been at LEAST as good as average MathCounts problems.

[quote="DataBox"]
^I love the elitism in that statement, is taking a normal math contest really that demeaning for MOPpers?  Just to let you know, MOPpers didn't do any better than the rest of the A team at ARML last year...[/quote]

I think it is. After taking 4 problem/4 hour tests every other day for three weeks, normal math tests just aren't the same anymore. The ARML and the AMCs are the only really fast tests that are still interesting. Also, how are you sure of that? And even if the scores were about the same individually, I think that the MOPpers probably contributed a lot more to the team tests, especially the power round. Anyways, just count the participation today. I saw two MOPpers there from last year, out of about a dozen, and one of those two is not coming again unless with very very good reason. The point is, there is going to be a deficiency of MOPpers on the ARML team this year unless they all sign up at the last minute and Tim shoves them onto the A team.

(A new post came up while I was typing).
Last year, apparently a bunch of MOPpers just came at the last minute and got put onto the A team. Also, that is part of the point I'm trying to make - these tests are skewing your senses of what's good and bad. They have little correlation with a strong ARML score or math knowledge. The ARML is not a speed test where [b]everyone[/b] knows how to do all the problems and the sole focus is on avoiding careless mistakes. On the ARML, the easy problems are such that you are given enough time to do them, and the hard problems single out the top participants. Also, the team contests are even more difficult. A perfect score on this contest doesn't really mean anything.

I'm not trying to pass off as arrogant or anything, I'm just trying to show why I think this contest is about as stupid as it can get for ARML selection.

Honestly, by halfway through the target, I was already getting irritated with the problems. Someone seriously misadjusted the difficulty level today (read: wtf was this easy crap?) If anything, today was "don't make stupid mistakes" day. (Yeah, I know it's been mentioned) IMO, this is not really a good indicator of who's ready or not ready for ARML, especially considering that interesting system where MOPpers just get thrown into A team. Seems to make tryouts pointless to begin with.

Seriously though, there was a 200, two 195's, like 5 or 6 190's...

@Ubey: This was by far the easiest tryout, the others actually weren't this crappy.

In case anyone cares, I won't be going to ARML either this year.

EDIT: Oh, and sprint answers were:

BCBBE
CEDDD
BACCB
ADCDD
BAACD
ADBCC

[quote="DataBox"][quote] Last year, apparently a bunch of MOPpers just came at the last minute and got put onto the A team.  [/quote]

Correct me if I am wrong, but the only MOPpers on the ARML team last year were Vishal (red, A team), John (red, A team), Kevin Y (red, A team), Kevin S (red, A team), Bayley (red, S team), Taylor (blue, A team), and Evan (blue,black, imo, A team) and of those all but Evan dutifully attended the qualifying rounds....[/quote]
Taylor was either on the verge of going to B team or already on B team, from missing practices, and he got stuck in A. Kevin Yang did not make MOP.
Albert is trying to point out the exceptional crappiness of this contest, and he's given lots of valid reasoning for it. I say it's time you stop your fanboyism.

^Wait, who are you kidding?  I'm fairly certain that Kevin Yang DID make Red MOP in 2007....

Interesting vocabulary, I'm not saying that these contests are great and I agree with him and Austin with regard to the overall quality of the last one, the only point that I'm trying to make is that some very talented students did take the time to take the qualifying rounds last year, and there is no reason that students of similar caliber should think that they are "too pro" to take them.

Kevin Yang didn't make MOP in 08, and usually when we say MOP we mean the one of the current year. Besides, Kevin is special because he likes doing ridiculous problems.
Now tell me, how many people from your little list are missing from these "contests"? Hmm..I see a lot.

Last year a lot of people went because the team selection method was new. I even went, but I left after a few of them because I had had enough of that ridiculous bs. Some people decided that ARML was worth going to even though they had to do all those ridiculous problems, so they stayed. I didn't think it was worth it, so I left. Some decent people stay anyways because they still want to go to ARML, but I am very sure most people there find the "contests" a waste of their time.

Also, didn't Albert say he thought people would not find it worth their time, rather than that they were "too pro"? Sure that could also work, but I think he had more emphasis on the "not worth the time" part. As I said before, I think nearly everyone can agree on that.

You can be the exception.

[quote="xscapezaer"]EDIT: Oh, and sprint answers were:

BCBBE
CEDDD
BACCB
ADCDD
BAACD
ADBCC[/quote]

Thanks. :)

All the ones I got wrong were due to careless mistakes. What a surprise... If anything, these contests make me feel stupid every time I make a careless mistake on an easy problem.

Why does Tim insist on writing his own low-quality problems for these contests? It was better when we used old ARMLs. (or TRMLs :o)

Anyway, I agree with Albert in that the problems were uncreative. and evan should totally be kicked off because he doesnt come to practices  :roll:

[quote]and evan should totally be kicked off because he doesnt come to practices Rolling Eyes[/quote]

lol that would be bad to say the least, haha evan did like a third of the power round by himself last year  :!:

[quote="bowei"](or TRMLs :o)[/quote]

Were those the Taiwan ones because those were craaaaazy. (Maybe because I was used to the easy ones...)

[quote="DataBox"]lol that would be bad to say the least, haha evan did like a third of the power round by himself last year  :!:[/quote]

Well putting Evan in the A team at the last minute isn't exactly fair, since the rest of us have to go through these "training sessions."

[quote="DataBox"]^True, but Evan is one of the best mathematicians the Bay Area has seen since 2001, so turning him down wouldn't be in Tim's best interests.  (I really doubt we would have pulled off the perfect power/team last year if it were not for Evan)[/quote]
The point is, if you create a system, you should follow it, even if it's totally screwed up. Bumping people who do well on the AMC series isn't following the system. Why do you think they don't just tell the IMO team hey, you don't have to worry about AIME and stuff next year, you guys automatically qualify for USAMO cuz you're so beast.

Y'know, it's not that easy to run an ARML team.  Get off Mr. Sander's case.


[quote="mdk"]Find the minimum value of $ a^{2}+4b^{2}-4ab-18a-36b$.[/quote]

Are there any constraints?

[hide="If not,"]then the answer is just $ -\infty$, because
$ a^{2}+4b^{2}-4ab-18a-36b=(a-2b)^{2}-18(a+2b)$

If you want this to be equal to $ N$, let

$ (a,b) = (-N/36,-N/72)$,

so we conclude that this can attain any value.[/hide]

[deleted]

[size=59]The message is too small. Please make the message longer before submitting.[/size]


real analysis
pigeonhole principle
induction
topology
real analysis unsolved

We say that two real numbers are equivalent iff their difference is a rational number. Let $\mathcal C$ be the collection of all the sets which contain one and only one real from $[0,1]$ in every equivalence class of this equivalence relation. Prove that the set $\{m^*(S)\ |\ S\in\mathcal C\}$ is dense in $[0,1]$, where $m^*$ is the Lebesgue outer measure on the real line.

Hi Grobber,

You surely mean that the elements of $C$ are all included in $[0,1]$ as well right ? Otherwise it's trivial ...

Of course. Sorry if that wasn't clear. Everything in this problem takes place in $[0,1]$.

Let $f(a)=m^*([0,a]\cap S)$. It seems like $f$ is 1-Lipschits, hence it takes all intermediate values.

Hi,

Fedor I agree with your statement but I do not understand what you were trying to prove ... remember that S must be included in $[0,1]$.

Lemma 1. Suppose that for any $ \rho<1 $ there exists a Vitali set $ S $ of outer measure more than $ \rho $. Then th claim holds.

Indeed, consider sets $ S*r $ for rationals $ r<1 $. They have outer measures $ m^{*}(S)\cdot r $, which are dense in $ [0,m^{*}(S)] $ and hence in $ [0,\rho] $. SInce $ \rho $ is arbitrary, we are done.

Lemma 2. for $ a<b<c $ and arbitrary set $ S\subset \mathbb{R} $ we have $ m^{*}([a,b]\cap S)+m^{*}([b,c]\cap S)=m^{*}([a,c]\subset S) $

Clear.

Lemma 3. Let $ S $ be arbitrary Vitali set and $ \delta\subset [0,1] $ be an interval such that $ m^{*}(\delta\cap S)>\rho \cdot |\delta| $. Then there exists a Vitali set $ S' $ of outer measure more than $ \rho $.

Proof. Divide an interval $ \delta $ onto many small intervals of kind $ [a/N,(a+1)/N] $ for large natural $ N $ and naturals $ a $ and two intervals of kind $ [x,a/N] $ and $ [b/N,y] $, where $ x $ and $ y $ are endpoints of $ \delta $.
Using lemma 2 and Dirichlet pigeonhole principle, we find $ a $ and $ N $ such that $ m^{*}([a/N,(a+1)/N]\cap S)>\rho/N $. Then denote $ S_1=[a/N,(a+1)/N]\cap S $ and consider a set $ N\cdot (S_1-a/N) $. It is a subset of some Vitali set and has outer measure more than $ \rho $.

Lemma 4.  For any set $ S\subset [0,1] $ of positive outer measure and any $ \rho<1 $ there exists an interval $ \delta $ such that $ m^{*}(S\cap \delta)>\rho \cdot |\delta| $.

Proof. Assume a contrary. Denote $ m^{*}(S)=a $ and consider a collection of intervals $ \delta_i $ such that $ S\subset \cup \delta_i $ and $ \sum |\delta_i|<a /\rho $. Any interval $ \delta_i $ may be replaced to a collection of interval with total measure $ <\rho |\delta_i| $. Make it for any $ \delta_i $ and get a contradiction.

Lemma 5. Any Vitali set has positive outer measure.

Well-known.

So, the statement is hopefully proved.

Thank you for the interest, guys :). I did not read it, Fedor, but I bet it's Ok. I had a more complicated approach:

Suppose you want a Vitali set with outer measure between $ a $ and $ b $, with $ 0<a<b<1 $. Also take some $ \varepsilon>0 $ s.t. $ a<b-\varepsilon<b $. Consider an interval $ I_1\subset I $ of length $ b-\varepsilon $ in $ I=[0,1] $, and another interval $ I_2\subset I $, disjoint from $ I_1 $, of length $ \varepsilon $.

Now let $ \alpha $ be the smallest continuum ordinal, and let $ (U_{\beta})_{\beta<\alpha} $ be an indexing of the sets of measure $ \le a $ which are countable unions of open intervals with rational endpoints included in $ I_1 $. We can construct a set by transfinite induction as follows:

For each $ \beta<\alpha $, the set $ I\setminus U_{\beta} $ has continuum cardinality (because it's a closed set of positive measure), and we have only chosen elements in $ <c $ ($ c $ is the continuum cardinality) equivalence classes up until now, so we can choose an element in this set from a different equivalence class. The set thus constructed contains at most one element from each equivalence class, it's included in $ I_1 $ which has length $ b-\varepsilon<b $, and it's not covered by any sequence of intervals of measure $ \le a $, so it has outer measure $ \in[a,b-\varepsilon] $.

It may be the case that some equivalence classes are not represented by points in this set. We take one point in $ I_2 $ from each such equivalence class. The union of the two sets (the one in $ I_1 $ and the one in $ I_2 $) is now the required set.


Harold had a bag of marbles. He gave $ 40\%$ of the marbles to his best friend Steve. He gave his sister a fourth of the remaining marbles. There were then 18 marbles left in the bag. How many marbles did Harold have originally?

When he empties his bag of $ 40\%$ of his marbles, he has $ 60\%\equal{}\frac{3}{5}$ left. Using this reasoning, we have the equation $ m\cdot\frac{3}{5}\cdot\frac{3}{4}\equal{}\frac{9m}{20}\equal{}18$, where $ m$ is the number of marbles he started out with. Solving for $ m$, we have $ m\equal{}\boxed{40}$.

I'm not sure if this is faster or slower but...
Working backwards, $ 18$ marbles is $ 3/4$ of $ 60$% of his original marbles. so $ 24$ is $ 60$% of his original marbles. divide that by $ 3$ to get $ 20$% of his marbles, then multiply by 5 to get $ 100$% of his old marbles which is $ 40$


I was reading over the course information for Algebra 3, and I notice in one place it says there are 4 challenge sets, but then it also says under due dates:
Due Dates: Challenge Set 1: July 2, 2009
Challenge Set 2: July 23, 2009
Challenge Set 3: August 13, 2009
Challenge Set 4: September 3, 2009
Challenge Set 5: September 24, 2009
Challenge Set 6: October 15, 2009
Challenge Set 7: November 5, 2009
Challenge Set 8: November 26, 2009

I really hope that this is a typo...

Can you let me know where it says 4 Challenge Sets?  I need to change that.  There are 8, but they're half as long as the old ones, so it's the same amount of homework.  We just spread it out more in the hopes that shorter, more frequent Challenge Sets would help keep the students on track better.

My Classes=>Algebra 3=>Course Introduction=>Bottom of page 2, under course assignments it says "This class includes 4 Challenge Sets..."


geometric transformation
ratio
geometry
power of a point
radical axis
geometry proposed

in two non intersecting circles, one of common external tangents touches them at A, B.
then, their radical axe meets line through two centers at C.
let their insimilar center be D.
prove that A,B,C,D are cyclic.

I'm afraid of not being good post. :blush:

Let the circle that $ A$ lies on it have the center named $ O$. we must only prove $ \Delta OAD \sim \Delta MCB$.where M is the midpoint of $ AB$ and it lies on the radical axe obviously.For this we should only prove
$ \frac {OA}{MC} \equal{} \frac {OD}{MB}$.the rest is solved easily by computing.

Let $O_1,O_2$ denote the centers of the given circles, $ A \in (O_1)$ and $ B \in (O_2).$ Radical axis of $ (O_1),(O_2)$ goes through the midpoint $ M$ of the common external tangent $\overline{ AB}.$ Let $ E \equiv AB \cap O_1O_2$ be the exsimilicenter of $ (O_1),(O_2).$ Since cross ratio $ (O_1,O_2,D,E)$ is harmonic, its orthogonal projection $(A,B,L,E)$ on the external tangent is also harmonic $\Longrightarrow$ $ EA \cdot EB \equal{} EL \cdot EM.$ But the quadrilateral $ LMCD$ is cyclic on account of the right angles $ \angle DLM$ and  $ \angle MCD,$ then $ EL \cdot EM \equal{} ED \cdot EC$ $ \Longrightarrow$ $ EA \cdot EB \equal{} ED \cdot EC $ $\Longrightarrow$ Quadrilateral $ ABCD$ is cyclic.

denote E is the intersection of AB and CD
ED/EO2=EB=EH=2R1/(R1+R2), where R1,R2 is the radii  of O1 and O2


It's cool if it's for 1 or 2 days. Not much longer though, 10 days is a torture! :)

[quote="JSRosen3"][quote="yif man12"]Ugh, I remember the last time I went hiking...  :maybe:
They didn't mention that in the Mathcounts Nationals brochure.[/quote]

How about in New Hampshire? That was fun, no? Although 9 miles in one day in the rain/drizzle was a bit strenuous...[/quote]

That's what I meant. I actually have gone hiking since but not that much distance or slope.  :lol:
It was pretty funny at the end, when we thought we'd never find the exit, and how we mixed up our orders at the restaurant.


Kalispera se olous,

1) Stelnw mia askisi akolouthiwn kathws afenos exei ksexastei teleiws to sigekrimeno thema afou pleon oi akolouthies den didaskontai men stin sxoliki ili (opws palia) kai afeterou einai polles fores thema se megalous diagwnismous

2) Na anaferw oti opws paratirisa to forum mas paei PARA poli kala kai einai 3o stin PAGOSMIA katataksi tou mathlinks. To roumaniko forum mas pernaei gia 200-300 dimosieuseis. Kathws vlepw ptwsi tis teleutaies meres, kati prepei na kanoume na anevoume stin 2h thesi toulaxiston kai meta na arxizoume na kontrarizomaste me ton Kanada (pou exei ksefigei teleiws exontas panw apo pente xiliades dimosieuseis).

Pame loipon:

Να βρεθεί ο ελάχιστος όρος της ακολουθίας θετικών ακεραίων $a_{n}, \ \ n=1,2,\ldots$ που ορίζεται ως εξής:

${{a_{n+1}=\left\{\begin{array}{ll}{\frac{a_{n}}{2}}&{\alpha\nu \ \ a_{n}\ \ \alpha\rho\tau\iota o\varsigma}\\{a_{n}+7}&{\alpha\nu \ \ a_{n}\ \ \pi\epsilon\rho\iota\tau\tau o\varsigma}\end{array}\right.}$

όπου $a_{1}=1993^{1994^{1995}}$

Alexandros

λοιπον ας ασχοληθουμε ξανα με το ελληνικο,εχει δικιο ο αλεξανδρος.εγω εχω  χρονο  :D

Αν $a_{n}>7$ τότε, αν ο $a_{n}$ είναι άρτιος θα ισχύει $a_{n+1}=\frac{a_{n}}{2}<a_{n}$ , ενώ αν ο $a_{n}$ είναι περιττός θα ισχύει $a_{n+2}=\frac{a_{n}+7}{2}<a_{n}$. Με άλλα λόγια η ακολουθία αποκτά όλο και μικρότερους όρους και ο μικρότερος όρος της ακολουθίας θα είναι ένας ακέραιος μεταξύ του $1$ και του $7$ (συμπεριλαμβανομένων), γιατί αν είναι μεγαλύτερος σύμφωνα με τα προηγούμενα θα υπάρχει κάποιος μικρότερος όρος από αυτόν (άτοπο). Ο μικρότερος όρος δεν μπορεί να είναι $5$ ή $6$ γιατί αυτοί καταλήγουν σε έναν όρο ίσο με $3$. Επίσης δεν μπορεί να είναι $2$ ή $4$ γιατί αυτοί καταλήγουν σε έναν όρο ίσο με $1$. Άρα ο ζητούμενος μικρότερος όρος είναι ή $1$ ή $3$ ή $7$.

Αν ο $a_{1}$ είναι πολλαπλάσιος του $7$ τότε κάθε όρος της ακολουθίας θα είναι πολλαπλάσιος του $7$ και συνεπώς ο ζητούμενος μικρότερος όρος είναι ο $7$.

Αν $a_{1}\equiv 6 \pmod 7$ τότε $a_{2}\equiv 3 \pmod 7$ , $a_{3}\equiv 5 \pmod 7$ , $a_{4}\equiv 6 \pmod 7$, κ.λ.π. , δηλαδή τα υπόλοιπα των όρων αν τους διαιρέσουμε με το $7$ εμφανίζουν μια περιοδικότητα και επειδή οι όροι γίνονται όλο και μικρότεροι, ο ζητούμενος μικρότερος όρος της ακολουθίας θα είναι ο $3$.
Όμοια και αν $a_{1}\equiv 5 \pmod 7$ ή $a_{1}\equiv 3 \pmod 7$ , o μικρότερος όρος της ακολουθίας θα είναι ο $3$.

Σε κάθε άλλη περίπτωση ο μικρότερος όρος της ακολουθίας θα είναι ο $1$.

Επειδή $1993^{1994^{1995}}\equiv (-2)^{1994^{1995}}\equiv 2^{1994^{1995}}\equiv 2^{2^{1995}}\equiv 2^{2}\equiv 4 \pmod 7$ , ο ελάχιστος όρος της δοσμένης ακολουθίας θα είναι ο $1$.

Στέλιος...


Prove the inequality
$ \frac {1}{x_1} \plus{} \frac {x_1}{x_2} \plus{} \frac {x_1 x_2}{x_3} \plus{} \frac {x_1 x_2 x_3}{x_4} \plus{} ...\frac {x_1 ... x_n}{x_{n \plus{} 1}} \ge 4(1 \minus{} x_1 ... x_{n \plus{} 1})$
holds for all positive real numbers $ x_1 , . . . , x_{n \plus{} 1}$.


floor function
number theory theorems
number theory

$ \pi(x)\equal{}\pi(\sqrt{x})\minus{}1\plus{}\sum_{d|P_x}\mu(d)\big\lfloor\dfrac{x}{d}\big\rfloor$
Where $ x>1$, $ \pi(x)$ is the number of primes $ \le x$, and $ P_x$ is the product of all primes $ \le \sqrt{x}$, $ P_2\equal{}1$.

Can you prove this?

Let $ p_1, ..., p_{\pi(\sqrt {x})}$ denote the primes $ \leq \sqrt {x}$. By PIE, we have that $ \sum_{d|P_x} \mu(d)\lfloor \frac {x}{d} \rfloor \equal{} \sum_{1\leq i_1 < \cdots < i_k \leq \pi(\sqrt {x})} ( \minus{} 1)^{k} \lfloor \frac {x}{p_{i_1} \cdots p_{i_k}} \rfloor$, counts the number of integers $ \leq x$ that are divisible by only primes $ > \sqrt {x}$. But each of these integers must clearly be prime. The result follows immediately.


group theory
abstract algebra
Ring Theory
superior algebra
superior algebra unsolved

Assume $ R$ is a ring, $ |R|<\infty$. If $ M$ is a non-trivial maximal ideal of $ R$. Has it to be true that $ M$ is a maximal subgroup too?

No, that's not true. Consider a ring where $ R/M~=\mathbb{F}_{q}$ with a nonprime q. Then there are subgroups between $ M$ and $ R$.


Hi! :)

You mustn't use the same value of $n$ with different keys $d_1$ and $d_2$ if you send the message $m$ to two people.
Indeed, let us assume that $d_1$ and $d_2$ are coprime.

i) Then, show that, if a hacker manages to get $m^{d_1}$ and $m^{d_2}$ and bets that it's the same message, he can find the message.

ii) Give an example.

iii) We have $n=187$; $d_1 = 3$; $d_2 = 7$.
iii) a) the pirate manages to get $m^3 = 140$ and $m^7 = 106$. Find $m$.
iii) b) the pirate manages to get $m^3 = 128$ and $m^7 = 8$. Find $m$.

[i]Nota Bene[/i]  : $1 \le m \le 127$.
Obviously, do not find directly the private key!! ;)

 :D

Nobody wants to try this one? :)
If you want a hint, just ask.
If you think it's too hard for Pre-Olympiad, feel free to move it into the Number Theory section (Proposed and Own problems).

;)

Ok then. ;)

[hide="Hint"]Since $d_1$ and $d_2$ are coprime, we know, thanks to [url=http://www.mathlinks.ro/Forum/viewtopic.php?t=76610]Bezout[/url], that there exists $u$ and $v$ in $\mathbb{Z}$ such that : $ud_1 + vd_2 = 1$.
[/hide]

[hide="Ok, second hint..."]
From the previous hint, we have that : $m^{ud_1+vd_2} = m$.[/hide]

:)

So, noone? Even with these two hints?  ;)


Find $ x;y;z \in Z$ such that
$ 2^x \sqrt{y\plus{}1}\plus{} 3^x \sqrt{y\plus{}2}\plus{} 4^x \sqrt{y\plus{}3}\equal{} 2^x \sqrt{y\plus{}2}\plus{} 3^x \sqrt{y\plus{}3}\plus{} 4^x \sqrt{y\plus{}1}$

By Rearrangement Ineq, we have

$ 2^x \sqrt {y \plus{} 1} \plus{} 3^x \sqrt {y \plus{} 2} \plus{} 4^x \sqrt {y \plus{} 3} \geqslant 2^x \sqrt {y \plus{} 2} \plus{} 3^x \sqrt {y \plus{} 3} \plus{} 4^x \sqrt {y \plus{} 1}$

Hence for the equality to hold we must have either $ \sqrt{y\plus{}1}\equal{}\sqrt{y\plus{}2}\equal{}\sqrt{y\plus{}3}$ or $ 2^x\equal{}3^x\equal{}4^x$. The former is impossible, and the latter implies $ x\equal{}0$. Hence all the solutions are $ (0,y)$ where $ y\geqslant \minus{}1$ (in order for all the roots to be defined).


let $ a;b;c\geq 0$ ; prove that:
$ a^5b\plus{}ab^5\plus{}b^5c\plus{}bc^5\plus{}c^5a\plus{}ca^5\plus{}6a^2b^2c^2\geq a^4b^2\plus{}a^2b^4\plus{}b^4c^2\plus{}b^2c^4\plus{}c^4a^2\plus{}c^2a^4\plus{}2abc(a^3\plus{}b^3\plus{}c^3)$

[quote="litbon"]let $ a;b;c\geq 0$ ; prove that:
$ a^5b \plus{} ab^5 \plus{} b^5c \plus{} bc^5 \plus{} c^5a \plus{} ca^5 \plus{} 6a^2b^2c^2\geq a^4b^2 \plus{} a^2b^4 \plus{} b^4c^2 \plus{} b^2c^4 \plus{} c^4a^2 \plus{} c^2a^4 \plus{} 2abc(a^3 \plus{} b^3 \plus{} c^3)$[/quote]
It's wrong. Try $ a\equal{}1$ and $ b\equal{}c\equal{}2.$ :wink:


summer program
Mathcamp
AMC
AIME
MathPath
email

MathCamp sounds really cool. I love math and doing it but I am not sure if I am good enough for the camp. I had a 67 through 3 rounds of USAMTS. I could have qualified for the AIME if I did not have the flu on the test day :( . What are people who get accepted to MathCamp like? I am asking this because I don't want to bother my teachers if the chance is unrealistic.

Thanks.

Wouldn't the first thing to do be to try the application quiz? You're probably better than you think you are, but the way to know is to find the quiz on the MathCamp Web site and work on it for a while.

People who get accepted to MathCamp are people who like and have an aptitude for math.  I would say definitely apply! I went to mathcamp last year and would definitely go back this year except that to get my parents to agree to let me go to Harvey Mudd for college next year (YAY!) I had to agree to work all summer :/.  You do not have to know tons of complex math to get in, you mainly to show your desire and ability to learn it (at least that was my impression).  Thought process seemed to be just as important as getting the right answer.

Several people have mentioned MathPath as a good, less challenging alternative for MathCamp.

FYI, Mathpath is restricted to middle-school students (defined as currently in 6th, 7th, 8th grades).

All of the programs mentioned in this thread are challenging in different ways.  You might check into all the programs listed under Resources (Programs tab), because they're all different; many have their entrance exam questions postes, and some are free to apply to, so you can taste-test most of them in some way, shape, or form.

I would say keep trying to solve the rest of the problems and make sure the solutions for the ones that you already solved are rigorous, make sense and are logically sound (although 7 problems seems like a solid amount of work) I already submitted my application, but in my opinion, I think I solved the first 9 problems with full solutions and the last one with a partial solution. Hopefully it is enough, but you never know...

[quote="xxreddevilzxx"]does writing a good qualifying quiz solution count?[/quote]

yes, when we grade the qualifying quizzes we consider the quality of the solutions.

[quote="mcalderbank"]If I am a first-timer, and I can solve 7 of the quiz problems absolutely and partially for the rest, do I have a shot?[/quote]

it's hard to say for sure, since we haven't had a chance to calibrate this year's quiz against previous years yet, and since we don't know how many people will apply.  however, i can tell you that in previous years, 7 absolutely correct solutions was generally enough to get in.

fanaticsm is right, the application deadline isn't until next month, so there's no reason not to keep working on the rest of the problems to try to do even better.

True, I will. Thanks for the input. (maybe If I was a girl.......well, lets not go there :blush: )

is it a lot easier to get in to mathcamp as a girl?

Well, if you are a girl I don't think you will have a [b]significant[/b] advantage of getting accepted over boys, but MathCamp does encourage girls to apply. In fact, according to their website, you might even be eligible for special scholarships as well, so it seems that girls might have a slight advantage of getting in.

[quote]Mathcamp seeks out and actively encourages applications from girls and from minority students. These applicants are given special consideration for [url=http://www.mathcamp.org/fin.php#scholarships]scholarships[/url]. You can read about some girls' experiences at Mathcamp [url=http://www.mathcamp.org/girls.php]here[/url].[/quote]

I have split off the topics discussing other programs and placed them in the 'Other US Contests & Programs'.  Please keep discussion in the Mathcamp forum focused on Mathcamp, and use the 'Other Programs' forum to discuss relative merits of various programs.

[quote="Jimmy"]What is the website for math camp, and how do you apply?
Thanks.[/quote]

Mathcamp is: http://www.mathcamp.org

You can fill out an application online at http://www.mathcamp.org/application
Application instructions are at http://www.mathcamp.org/apply.php

this is a little bit off topic, but when will we know what we got on the qualifying quiz?

They actually grade the qualifying quiz??? I thought it was just another means of choosing people...

now I am afraid that i will get a 0... Ive looked at the problem, and they are certainly not my kind of problems... :(

we'll have a preliminary round of admissions decisions by the end of the first week of may, though there are always a bunch of people who get in off the waitlist after that.  we don't generally give out the scores on the qualifying quiz, although if there are particular questions where you're curious about how you did, send us an email in may once you hear about admission.

does Qualifying Quiz make up a huge part on admission decision??? Becuase I cant seem to get the solutions started, because i have never faced these kinds of problems before.

Here's what ztbb wrote in another thread:

[quote="ztbb"]i can give you some exact numbers: last year mathcamp had 131 new applicants, of whom 86 were accepted and 77 came to the program.  so the acceptance rate was just under 2/3.  (there were 34 alums at mathcamp as well, but alumni can automatically return if they want to.  once you're in you're in.)

eve is correct that some people get into mathcamp having solved around half the quiz, but there are also people who solve half the quiz and aren't admitted.  the two biggest things that can make a difference aren't recs and contest results, though -- it's the personal statement, and whether the five problems are solved *well*.[/quote]

(note that that post contains statistics regarding admissions  to mc2003, not mc2004.  applications were up last year, so the acceptance rate was correspondingly lower.)

Is there a way I can find quizzes from past years?  I kinda like the problems even if I have seen several of them before.:D

I think they use the same test each year...but I am not positive.

No, but some problems do get reused. At least the last two years only have one problem in common.

Nope, you have an option of sending things through the mail or online...

[quote="juggle7"]Do they prefer everything online?  I haven't talked with those that might write recommendation letters yet, so they may or may not prefer to use postal mail.  It would be easier for me to just send in a PDF document for the quiz solutions.[/quote]

Online is more convenient for us: the graders don't all live in Boston.  But if your recommenders prefer to use postal mail, then it's perfectly alright to upload your PDF quiz and still send your recs by mail.

[quote="lorddude"]the only problem with submitting results online is any diagrams i drew
what could I do?
I don't really have a huge amount of time left[/quote]

Two possibilities: you could scan your diagrams and upload the picture; or you could fax it to us at (617) 812-6339.

[quote="bubala"]if one of my solutions is really long text (#4), then should I use white space and make it ever longer (which might make you think my solution was too looooooooooooooooooooooong), or should I make it one big paragraph (which might make you think I can't write)?[/quote]

You should definitely use white space.

k, it is 2009 now, and I am applying to MAthCamp for the first time. I am in grade 11, and I solved the first 7 out of 8 problems. I am still wrapping my head around #8 (doesn't seem impossible, just a bit hard). So, from what I gather, they look at your solution presentation skills, your personal statement and your highschool above and beyond activities. Am i right about this? And how are other people doing on the qualifying quiz so far?

7 out of 8 problems is really really goood :xx
I'm applying and I've only gotten 3 without full proof >< arghhh.

Oh yeah , and if you read the top part of the qualifying quiz, Mathcamp mentions it accepts people who just get 4 problems right, so I think you're fine. :)

Yep, all of that is right. I would say that "our highschool above and beyond activities" really only means ones associated with math. Essentially, they want to see whether you're [i]interested[/i] in math.


algebra
polynomial
abstract algebra
group theory
Galois Theory
superior algebra
superior algebra unsolved

:?  hi
sorry for bothering you with perhaps simple problems , but I'd really
like to understand them :

determine which of these rational polynomials can be inbedded (are  - up to isomorphism - part of) in some cyclotomic field (a splitting field of
x^n-1 for some natural n) over Q

x^3-27*x-4    its discriminant is 30^2*29
x^3+27*x-4   its discriminant is -36*733
x^3-39*x+65
the last one i solved myself (i think) because it is the minimal polynomial
of w+w^5+w^8+w^12 with w some root of x^13-1=0 (a nonone root)

all i know is that the splitting fields of both polynomials are of degree
six, and that the first one can be inbedded in the reals, and the second
one can't

perhaps an idea  :?  for the second one is nonabelianness of the galois group? or is this worthless?

note that my problem doesn't require n to be prime, so the
galois group of the splitting field of x^n-1 isn't necessarily cyclic

thx

fred

i made a tiny mistake

i of course don't mean, which rational polynomials can be inbedded

i meant : of which of these rational polynomials can the splitting field over Q be inbedded in some cyclotomic field extension

does anyone have any ideas

or is this problem in fact trivial? :?

[quote="fredbel6"]:? no one has ideas? ;)[/quote]just a remainder: next time post in the right forums to spare us the trouble of moving the posts ;) thanks.

sorry  :D  indead my problems are problems i encountered at college

just check the galois groups :)
if they are abelian, one can always embed them (kronecker-weber theorem) and if not, one can't since they would have to be a subgroup of an abelian group.

;)  thanks peter
that's quite an interesting theorem, i've never seen it explicitly

but what exactly do you mean subgroup, i've heard talk about taking quotients or something? :?

i suppose as soon as your galois group is abelian but not cyclic, your root of unity will not be a root x^p-1 with p prime?

fred

oh sorry, i mean quotient of course, since there is the correspondence between subgroups of the galois groups and the subfields given by taking the fixed field of the subgroup; therefore the galois groups of the subfields are the quotient of the galois group of the large field by the subgroup of automorphisms where the subfield is fixed. anyway, they got to be abelian.

however, i don't want to calculate the galois groups of these polynomials :D


Let p,q be positive integers (p,q>2004), and let f(n) be the greatest x, such that x | n and (x, p)=1. Prove that if we denote S=f(1) + f(2) +... +f(p^q) , then S can be written as a product between a sum of  p-1 consecutive positive integers and a sum of q perfect squares.

Firstly, note that the right statement involves the sum \[S=f(1)+f(2)+f(3)+\dots+f(p^{q}-1).\] Secondly, note if $q>2004$ then we only need to extract the multiplier which is a sum of $p-1$ consecutive numbers, since second multiplier can be represented as a sum of 4 perfect squares in any case :)

I checked the statement for the case $p$ is prime. It is true! :D
The general case requires some nice idea I think.

P.S. Just to level-up this nice problem after more than 3 years ;)


AOPS is full of so many really smart people.. so i am just wondering: what is your IQ?
take the test at http://www.IQTest.com to find out
mine is 148 and i thought that was good but im sure most people on these forums will have even higher IQ's
IQ Score Cognitive Designation Possessors

Below 75 Learning difficulty

75 - 85 Below Average

85 - 114 Average Middle school graduates

115 - 124 Above average High school graduates

125 - 134 Gifted University graduates

135 - 144 Highly gifted Professionals

145 - 154 Genius  Professors/Researchers

155 - 164 Genius Nobel Prize Winners

165 - 179 High genius

180 - 200 Extraordinary genius

i dont think these are right tho cuz im not a genius  :P
edit: maybe make it a poll

wow. on this site, my 130 IQ seems to be below average  :?
that's disturbing. maybe i'm not as smart as i thought i was  :(

[quote="djimenez"]First... Didn't we have this same topic less than two months ago. I didn't look for it, but I am almost sure.
Second... The IQ is actually very trainable! you can, if you want, to train yourself to get the highest scores on the regular IQ tests, most of the questions on that kind of tests only measure a specific learned set of skills. There is a theory about seven intelligence areas (multiple intelligence as it is called sometimes) and the IQ tests measure just one of those seven!
Finally... Those tests are pretty unconsistent. During the years I have taken a few of those, and always the resoult is different, going from 140 to 190! (I have never train for those tests, :P )[/quote]
yea thats prob true, but its fun to take them anyways  :lol:
sry i didnt know there was one a month ago, or if i did i forgot  :blush:

[quote="Annika"]Sorry, but I have the feeling that people who care about IQ seriously misunderstood some things.

Anyway, internet IQ tests are not reliable at all (they want to sell you your "Complete Intelligence Profile" or stuff like that, so what is their aim? To make you proud of your "IQ 160" or whatsoever so that you want to buy their certificate to show off with it).
You get better IQ tests (properly calibrated) from psychologist and at Mensa, but I'm conviced anyone can score really high in an IQ test if he or she does training for the question types.[/quote]Great point... and I agree 100%, even when sheepwarrior is right, even when they are not accurate or reliable, the kind of questions they have are funny to solve... I have taken both, internet tests and psychologist administrated tests, and in both cases, the truth is that they are a lot of fun any way!

My IQ is uhhh..... 68.

IQ doesn't really matter  :)

There have been not one, but at least three threads with the exact same topic in recent history. Please search before you post. Here's an example:

http://www.artofproblemsolving.com/Forum/viewtopic.php?highlight=IQ+what+is&t=17041

Closing...and to be deleted.


linear algebra
matrix
analytic geometry
combinatorics unsolved
combinatorics

How many $ n$-by-$ n$ matrices are there if the entries $ x_{i,j}$ of the matrix are integers satisfying $ 0\leq x_{i,j}\leq q$ and the order of the coordinates does not matter? (i.e. $ (0,0,1,3)$ and $ (1,0,3,0)$ are considered to be the same)

The fact that they are a matrix is clearly irrelevant which means that this is just the number $ \left(\!\!\binom{q}{n^2}\!\!\right) \equal{} \binom{q \plus{} n^2 \minus{} 1}{n^2}$ of multisets with $ n^2$ elements from a set with $ q$ elements.


When I watch this year's national countdown round, mathlets like Bobby, David, Max, ..... etc buzzed in really fast.
How can you manage to get the answer so fast?
Do you read the whole question carefully or you just scan the question fast and match a type you know?
Do you memorizing more numbers so you can get the answer fast?

I feel if you read the question as a new question and then work on it, you have no chance to beat these mathlets.
It seems to me they know some of the problems and they know the answers to these problems.
They didn't spend time to read the question and compute the answer.

What is the average buzz in time for the top national finishers when they get the correct answer with first buzz in
(not after someone missed first)?

[quote="pythag011"][quote="AOPS23"]Well, I think practice counts.
As they say: Practice makes Perfect
The National Mathcounts Winners Probably Practiced A LOT, so they may have found some problems similar to problems they have done before.[/quote]

Actually, David didn't really do mathcounts practice this year.

However, he uses the amazingly useful strategy of buzzing in before you know the answer.

Somehow though, he lost in first round at state cd?[/quote]
I guess he pracitced a lot after that. Also, you don't have to read everything, like names. You can just think "it".

Well my dad got all the cds from past years, so I'm practicing now, and I'm wondering how long on average it takes for the people to get an answer to the questions.

[quote="reebokballa96"][quote="pythag011"][quote="AOPS23"]Well, I think practice counts.
As they say: Practice makes Perfect
The National Mathcounts Winners Probably Practiced A LOT, so they may have found some problems similar to problems they have done before.[/quote]

Actually, David didn't really do mathcounts practice this year.

However, he uses the amazingly useful strategy of buzzing in before you know the answer.

Somehow though, he lost in first round at state cd?[/quote]

hmmmmm.........I wonder what David's AOPS username is?   :P[/quote]

...if you want to try and be a fake nike, you should at least learn how to copy my avie correctly.

[quote="reebokballa96]
hmmmmm.........I wonder what David's AOPS username is?   :P[/quote]

I thought this is well known to a lot people here.
David is very active on this forum.
He is a smart and helpful person who is not afraid to speak out.

[quote="AOPS23"]Well, I think practice counts.
As they say: Practice makes Perfect
The National Mathcounts Winners Probably Practiced A LOT, so they may have found some problems similar to problems they have done before.[/quote]

I would like to disagree with the phrase "Practice Makes Perfect."

Practice can only make better. Perfection is perhaps achieved only through passion.

[quote="rd5493"][quote="pythag011"]
Actually, David didn't really do mathcounts practice this year.

However, he uses the amazingly useful strategy of buzzing in before you know the answer.

Somehow though, he lost in first round at state cd?[/quote]

Sometimes though when you buzz in early, you can't figure the answer out in the three seconds that buy yourself.  it ends up being awkward silence followed by an extra 30 seconds for the other person to figure it out.  be patient, i mean, you would hate to lose because you buzzed in a second too early[/quote]

Yes but under normal conditions one would expect to be able to calculate 125-64 in 3 seconds...

Actually if you watch the webcast David Yang used this strategy a lot?

Well, you have to have talent and practice. Or else you will fail in nats cd. Like I did.  :mad:

Seriously, let's just say MC is not important at all compared to .... (Fill this in pythag...)

Anyways, I failed to use the strategy of buzzing in a second early, so I lost epicly.

[quote="dnkywin"]Well, you have to have talent and practice. Or else you will fail in nats cd. Like I did.  :mad:

Seriously, let's just say MC is not important at all compared to .... [color=olive]proving the Riemann Hypothesis[/color]

Anyways, I failed to use the strategy of buzzing in a second early, so I lost epicly.[/quote]

Also, for anyone preparing for nationals, I think it's best to do minimal cd practice unless you think you will PROBABLY will make cd (and if you think you always will, then don't do cd practice)

Learn to read the question first and then go back and scan the question for numbers and directions. This is my method and it works quite well in Oklahoma. But I need to get into the habit of buzzing in before I know the answer because I usually buzz in when I get the answer but I still beat a lot of people.

[quote="jonathanchou711"]Learn to read the question first and then go back and scan the question for numbers and directions. [/quote]

I guess this will not work at Nationals.
I watched this years national CD at home.
Sometimes before I finish reading the question (I try to read fast, faster than the announcer who reads on stage) someone had already buzzed in.
This make me thinking maybe the contestants don't read the question in whole, just scan the key elements.

Anyone has any tips on how to read a CD problem fast?
What are key elements in a CD problem?

hmm..
I just look at important stuff like the operations, numbers, stuff like that.
P.S. I also mem  :)

In my state (Michigan) the 4th person on the team got 5th written and moved up. Their was a person a couple years ago who got 10th or 9th written and moved up to 1st. Also, Texas and Missouri have official CD (I think) and they still did pretty good.

[quote="PowerOfPi"]It also sucks to get 5th written and get 4th in an unofficial countdown. I think they should be official because, as Urc said, they can balance stupid mistakes and luck.[/quote]

Sure, but generally countdown problems are even more based upon luck. They can probably balance stupid mistakes however.

Case in point: in states, NorCal has top-bottom format. I seeded against #15 and lost (2-1). [of course, this may be due to the fact that: 1) the tiebreaker shoelaced so I intuitively shoelaced as well, 2) the fact that I shoelaced a (declared) square.] Note, #1 seeded #16 and lost, so countdown is hugely luck.

... really?
huh.
i thought all states had 10...

Depending on the type of question given.... if you are familiar to that specific one.... then your mind can process the solution much much faster...

     - Algebra is where math revolves, without algebra math simply would not be math.

[quote="Mewto55555"][quote="Urc"][quote="PowerOfPi"]In my state, there are only 8 people in countdown. In half of California, there are 16. Big difference. Also, I agree that countdown is luck.[/quote]

maybe it depends on the cd format...ca's is quite...unorthodox, compared to how i hear most states (officially) run it.

off topic (ignore this): omg! i used a vocab word. yay me.[/quote]

Haha not compared to MO:

Match 1: 5v6
2: 4v7
3: 3v8
4: 2v9
5: 1v10
6: Winner of 1 v Winner of 2
7: winner of 6 v winner of 3
8: winner of 7 v winner of 4
9: winner of 8 v winner of 5[/quote]

i take back what i said.

[quote="genius"]
Match 1: 5v6
2: 4v7
3: 3v8
4: 2v9
5: 1v10
6: Winner of 1 v Winner of 2
7: winner of 6 v winner of 3
8: winner of 7 v winner of 4
9: winner of 8 v winner of 5[/quote]

i'm still trying to process this. is cd unofficial (i'm sure it is).

also, did bobby win written and cd in TX?

yes to nike...

so (@ max) this means the 10th seed can win in 2 rounds in your countdown?

I am actually wondering about that myself. It seems to change year to year. Sometimes they do that, other times, after the first round, they reorganize by seed # and then do ladder (which means #6 could possibly win in 2 cds!)

This is what they do in ND (only the top 8 get to countdown):

CD 1: 8 v 7
CD 2: W(1) v 6
CD 3: W(2) v 5
CD 4: W(3) v 4
CD 5: W(4) v 3
CD 6: W(5) v 2
CD 7: W(6) v 1

In ND countdown is unofficial, though. I think ND does it a fair way.

PS: When I wrote W(x), I meant "Winner of CD x".

Yeah that is the standard ladder format.

i dont get the point of any format other than ladder in an mc competition besides nats...

Does anyone know what is the lowest seed in written round ended in winning the national title in Countdown round?


S_(ABCD) shows the area of the trapezoid... But the RHS of the equality is given in the question...[/hide][/code]


Well I am really glad I found this website.  I have shifted my career from computer programming to high school mathematics trainer.  (We are actually training teachers strategies in teaching and not the content.) Though, I have been an olympiad contestant during my high school days (i.e. 10 years ago), I think I have already forgotten those math stuff and AOPS is a good site where I can review my Basics.

Mathematics in our country is not that rigid and our high school here is only up to year 10. Well that's our curriculum and I can really do nothing about it.

I have to admit that my foundations in math is not that strong so I think I will be here often.

Ok, enough of this gibberish. Its really good to be here.  AOPS ROCKS!!!

I find AoPS a little too strict for my liking, but yeah, I'm not gonna scare away "potential customers"...

I really like this site. Other than the fact that the range of problems in each section is too broad, everything is great!

Well , AoPS is seriously amazing , no doubts about that...By the way , could someone move this to some other section ? ( This doesn't exactly fit in Getting started forum.)

I'd say that it is Mathlinks that is magical.
For our guys who do not have the text books......never get anything involved with AoPS......

[quote="math92"]I love this site because it has lots of math to do. :)[/quote]

I've seen that youl like posting a lot :P

I like this site because I got a $lot$ better at math than I was... :lol:  :lol: [/i]

Yeah, same here.  I only got a 2 on the AIME this year  :blush:  :(  But I've been coming here since and I've learned so much... I'll do a lot better next year.  I'm going to try to beat our state's undisputed math champ, who is a senior next year... And that won't be out of reach if I finally sit down and read thru both AoPS books.

this site's great. i just bought AOPS volumes 1 and 2  and i find them really good as well. the problems are challenging, yet solvable. if u don't understand a problem, just post it on the forum and you'll get a great solution in no time. :)

[quote="SnowStorm"]Too strict? How so?[/quote]I think Treething is reffering to our strict policy regarding the language  ;)

HaHa.

You know, I'll move this to the Round Table.

Are Aops Classes really good for their cost??
i use a Mathlinks a lot to do math, and i was thinking to enroll Woot.. but it is kind of $ and i dont know if it is better to study by myself more hours a day and train by independent study.
I mean in my country there is no much help for me and for mymath and i've study independently a lot but i would like a good professor to so that's why i thought of Woot for next year's IMO.. but i still don't know.. i think i can do it by me self...
can somebody give me an advice?
without impulsing me to enroll woot so that AoPs can grow richer..  i mean weren't the classes once free?
isn't this like prostituing math in some sense? math spirit should be taught for free!
maybe im wrong, and maybe i don't know why it is so costly so im sorry if i dont understand the mathlinks and aops idea!

[quote="manuel"]isn't this like prostituing math in some sense? math spirit should teached for free!
maybe im wrong, and maybe i don't know why it is so costly so im sorry if i dont understand the mathlinks and aops idea![/quote]There are 3 of us (rrusczyk, MCrawford, and me) who work full-time at AoPS.  We need to eat, pay our rent, put gas in our cars, and so on.  From a company standpoint, we also need to pay our office rent, pay for computer equipment and internet access, etc.  We all left higher paying jobs to come to work at AoPS, because this is what we love to do.  But it does cost money!

We provide a lot of stuff for free: the forums and the Math Jams are probably the two that stand out the most.  But I hardly feel that we are out of line that we charge fees for some of our more premium services, like our classes and textbooks, or that we solicit sponsorships so that we can create programs like WOOT (and trust me, if we did not have outside sponsorship, there is no way that we could have possibly afforded to start WOOT).

It may be nice to say something like "math should be taught for free", but it is naive to think that the world works that way.

Think about it....teachers at school get payed quite a bit more for providing LESS of a service to their students. Despite the fact that classes meet 5 days a week in school while only once a week for 12 weeks on AoPS, I feel that here you are able to fully understand a mathematical concept, from the basic proofs to the advanced applications while in most schools you are taught simply to memorize a formula, and thus learn nothing. As Mr. Patrick said, this is a full time job for the instructors here and they need to profit in order to maintain their everyday lives. I have always liked math, even before I discovered AoPS; however, AoPS has helped me reach goals I never thought attainable and has further helped me see the beauty of mathematics. To all the AoPS staff, keep up the great work!

Isn't saying 'prostituting' math a little harsh?

I mean, if you look at it from a business standpoint, the administrators of this forum could be charging so much more - because their devotion and service to our education is so much more than most of you will get anywhere else.

Think about all our options here at AOPS.

I don't think anyone means math to be a business, nor is Aops a sales pitch. AOPS has always been here to help us develop our skills at what we know and teach us what we don't know. Without AOPS, I don't think I would have gone far in my pursuit of mathematics. I think the community here is the best tool - if you use it wisely.

For my interest in mathematics, I thank all the AOPS staff for this wonderful site; and I hope it continues to grow in the future.

Haha, prostituting math. I think it's a bit harsh as well, but it's pretty funny.

The classes are just like normal school education. Since AoPS isn't funded by the goverment, shouldn't they charge for classes?

[quote="manuel"]Are Aops Classes really good for their cost??
i use a Mathlinks a lot to do math, and i was thinking to enroll Woot.. but it is kind of $\$ and i dont know if it is better to study by myself more hours a day and train by independent study.
I mean in my country there is no much help for me and for mymath and i've study independently a lot but i would like a good professor to so that's why i thought of Woot for next year's IMO.. but i still don't know.. i think i can do it by me self...
can somebody give me an advice?
without impulsing me to enroll woot so that AoPs can grow richer..  i mean weren't the classes once free?
isn't this like prostituing math in some sense? math spirit should be taught for free!
maybe im wrong, and maybe i don't know why it is so costly so im sorry if i dont understand the mathlinks and aops idea![/quote]A few comments I'd like to add: first of all it does not cost that much. Most Romanian students pay much more for tutoring for BAC than that. And Romania is very far from being a rich country.

You can try to find sponsorship to enter the program if you can't afford it. I did just this 5 years ago when I didn't have money to take the SATs.

And finally, this program was created for students that do not have (permanent) access to good math olympiad teachers that can guide them. It is indeed possible to succeede on your own, but it's much harder than if you have professional guidance. I only wish I would have had such an option available to me when I was a student ...
 :roll:  :(

OK you convicend me! :)
i'll just have to work a bit because my parent's say that i can study by my own and i think they won't pay  the whole $500. but there is no good tranning in my country..
anyways i can afford it..  ;)
Anyways, so far the only spanish speaking students who are entering the program are marko and me. do you know others, valentin?

note:hey guys, i don't have anything against the aops community!! i love it and it has helped me a lot!! only that i had my doubts about the classes because they are abit expensive for me! but anyways i just sounded  a little 'harsh' because i wanted other people to defend and point the good points of woot so i can know them since i have to pay 500 bucks..
but, i'll probably enroll and hopefully enjoy and learn..

[quote="manuel"]Anyways, so far the only spanish speaking students who are entering the program are marko and me. do you know others, valentin?[/quote]There are almost 100 students (counting the MOPpers) enrolled in WOOT at this time. I do not know if there are any more spanish speakers among them, but probably there are :)

[quote="manuel"]but anyways i just sounded  a little 'harsh' because i wanted other people to defend and point the good points of woot so i can know them since i have to pay 500 bucks..
but, i'll probably enroll and hopefully enjoy and learn..[/quote]You can quit the class after 1 test (which will not be graded if you quit) and 1 lecture and get a full refund. But I doubt that will happen :)


All the icons in the MathLinks view disappeared. What hapened.

:alien: Space aliens are eating our icons!! :alien: WHAAAH  :yoda:

No seriously, they're still there on my screen...

they just came back after 3 days. Something strage is going on. they disappear and then they reappear like tiis consistently for 3 days continuesuly

What browser, computer, etc. do you use? :) (just for the record, it has never happened to me). Also it might have something to do with your internet connection ...


Problem 14.

A mail carrier delivers mail to the nineteen houses on the east side of Elm Street. The carrier notices that no two adjacent houses ever get mail on the same day, but that there are never more than two houses in a row that get no mail on the same day. How many different patterns of mail delivery are posssible?

So, here's how I did it, and can someone please help me with my reasoning? I don't understand what I did wrong.

We can immediately see that the minimum number of houses that must receive mail is 6, and that the maximum number of houses that can receive mail is 10.

Let receive mail be desginated A, and no mail designated B.

For 6 A's, each A must be separated by at least one B.

So far, we have the arrangement: ABABABABABA

Now, here are the possible spaces to put 8 more B's to obtain the 19 spots/houses.

_ _ AB_AB_AB_AB_AB_A_ _

There are 9 spots to put 8 B's. So 9C8.

For 7 A's.

ABABABABABABA

So, _ _AB_AB_AB_AB_AB_AB_A_ _
There are 10 spots to put 6 B's. So 10C6.

Now the same for 8 through 10, and you get 11C4, 12C2, and 13C0 respectively.

That added up is not the book answer. Why not? Where did my counting overlap?

[quote="tetrahedr0n"]If A and B are mail and no mail, you cannot have AA or BBB anywhere in the sequence.[/quote]

I don't have AA or BBB... what do you mean?

I was responding to the poster above, not to your problem. But I'll take a look at it right now.


limit calculus integration logarithms function geometry rectangle

$ \lim_{n\rightarrow\infty}
\frac{1\plus{}\frac{1}{2}\plus{}\ldots\plus{}\frac{1}{n}}{log(n)}$

What am I suppose to assume is dx?

This is a tricky application of integral (one that doesn't convert directly into a Riemann sum).
[hide]
For $ 1\leq k\leq n$, express $ \frac{1}{k}$ as $ \int_k^{k\plus{}1} \frac{dx}{k}$. This can be written as
\[ \int_k^{k\plus{}1} \frac{dx}{x}\plus{}\int_k^{k\plus{}1} \frac{x\minus{}k}{kx}\ dx\]
Summing from $ k\equal{}1$ to $ n$, we get
\[ \sum_{k\equal{}1}^n \frac{1}{k}\equal{}\int_1^{n\plus{}1} \frac{dx}{x}\plus{}\sum_{k\equal{}1}^n \int_k^{k\plus{}1} \frac{x\minus{}k}{kx}\ dx\]
The first integral is $ \ln (n\plus{}1)$. For the sum of small integrals, the $ k$th integral is nonnegative and at most
\[ \int_k^{k\plus{}1} \frac{x\minus{}k}{k^2}\ dx\equal{}\frac{1}{2k^2}\]
since $ x\geq k$ over the interval $ [k,k\plus{}1]$. But $ \sum_{k\equal{}1}^\infty \frac{1}{2k^2}$ converges, so there is a constant $ \gamma$ such that
\[ \ln (n\plus{}1)\leq \sum_{k\equal{}1}^n \frac{1}{k}\leq \ln (n\plus{}1)\plus{}\gamma\]
from which it follows that
\[ \lim_{n\rightarrow \infty} \frac{1}{\ln n}\sum_{k\equal{}1}^n \frac{1}{k}\equal{}\boxed{1}\]
[/hide]

Same idea in different form: since the function $ f(x) \equal{} 1/x$ is decreasing for $ x > 0$, the leftpoint Riemann sum $ \sum_{i \equal{} 1}^{n \minus{} 1} f(i)$ (with $ \Delta x \equal{} 1$) [b]overestimates [/b]the integral $ \int_1^{n} f(x)\,dx$ (The sum can be represented by the area of $ n$ rectangle which together cover the graph of $ f(x)$ over the interval $ [1,n \plus{} 1]$). By a similar argument, the rightpoint Riemann sum $ \sum_{i \equal{} 2}^{n} f(i)$ [b]underestimates [/b]$ \int_1^n f(x)\,dx$. Hence
\[ \sum_{i \equal{} 2}^{n} \frac {1}{i}\le \ln n\le \sum_{i \equal{} 1}^{n \minus{} 1} \frac {1}{i}
\]
which can be rearranged as
\[ \ln n \plus{} \frac {1}{n}\le \sum_{i \equal{} 1}^{n} \frac {1}{i}\le \ln n \plus{} 1
\]
Divide the latter inequality by $ \ln n$ and apply the Squeeze theorem.

[quote="mlok"]Same idea in different form: since the function $ f(x) \equal{} 1/x$ is decreasing for $ x > 0$, the leftpoint Riemann sum $ \sum_{i \equal{} 1}^{n \minus{} 1} f(i)$ (with $ \Delta x \equal{} 1$) [b]overestimates [/b]the integral $ \int_1^{n} f(x)\,dx$ (The sum can be represented by the area of $ n$ rectangle which together cover the graph of $ f(x)$ over the interval $ [1,n \plus{} 1]$). By a similar argument, the rightpoint Riemann sum $ \sum_{i \equal{} 2}^{n} f(i)$ [b]underestimates [/b]$ \int_1^n f(x)\,dx$. [/quote]

If this is not convincing enough, consider the expressions

$ \int_{1}^{n} \frac{1}{ \lceil t \rceil} \, dt < \int_{1}^{n} \frac{1}{t} \, dt < \int_{1}^{n} \frac{1}{\lfloor t \rfloor} \, dt$

Let $ x_n\equal{}\displaystyle 1\plus{}\frac{1}{2}\equal{}\ldots\plus{}\frac{1}{n}$ and $ y_n\equal{}\ln n$.
$ y_n$ is ascending and unbounded. We apply Stolz-Cesaro lemma.
$ \displaystyle \lim_{n\to\infty}\frac{x_{n\plus{}1}\minus{}x_n}{y_{n\plus{}1}\minus{}y_n}\equal{}\lim_{n\to\infty}\frac{\frac{1}{n\plus{}1}}{\ln\frac{n\plus{}1}{n}}\equal{}\lim_{n\to\infty}\frac{1}{\ln\left(1\plus{}\frac{1}{n}\right)^{n\plus{}1}}\equal{}\frac{1}{\ln e}\equal{}1$.
So $ \displaystyle\lim_{n\to\infty}\frac{x_n}{y_n}\equal{}1$.


vector
induction
combinatorial geometry
real analysis
real analysis unsolved

$E$ a normed vectorial space and $C$ a dense convex subset of $E$. Is $E=C$ ?

Here's something very similar (if not the exact same thing :)) to a counterexample I gave in another thread started by you (alekk):

We can even take $C$ to be a dense subspace of $E$.

Let $E$ be the real vector space of real sequences which determine absolutely convergent series (i.e. sequences $(a_n)_n$ s.t. $\sum_n|a_n|<\infty$; take the norm $\|(a_n)_n\|=\sum_n|a_n|$), and let $C$ be the subspace of $E$ consisting of real sequences which are eventually $0$.

I found the same counter-example  :)  (and in finite dimension ...)

[quote="alekk"]and in finite dimension ...[/quote]

In finite-dimensional spaces it does follow that $C=E$. The intersection of $C$ with some hyperplane is also convex and dense in that hyperplane, so we can use induction on dimension.

or construct an increasing set of polytopes $P_i$ with vertices in $C$ such that their union is $E$ (possible since $C$ is dense).
Conclusion follows.

Does a dense convex subset of a Banach space always contain a dense affine subspace (i.e. a shifted linear subspace)?

I remembered this (ancient :)) topic and question by accident, and decided to think about it. As it turns out, in infinite-dimensional Banach spaces, one can construct sets pathological enough to get a negative answer. Here's a brief sketch:

First, find a dense linearly independent set $e_{i},\ i\in I$ in our space $X$. I think I can prove this is always possible for first-countable topological vector spaces over $\mathbb R$. I don't know how important first-countability is (not very, I suspect :)), and I did not bother to try to extend it to all topological vector spaces. The convex set $C$ that we're looking for will be the convex hull of the $e_{i}$. It's dense, because it contains the dense set $\{e_{i}\}$. Now consider (discontinuous) linear functionals $f_{i},\ i\in I$ dual to the $e_{i}$, in the sense that $f_{i}(e_{j})=\delta_{ij},\ \forall i,j\in I$ ($\delta_{ij}$ is the Kronecker delta). On $C$, each $f_{i}$ takes values in $[0,1]$. It follows now that $C$ can't even contain a ray, i.e. a set of the form $\{x+ty\ |\ t\ge 0\}$ for some $x\in X,\ y\in X\setminus\{0\}$: otherwise $y\ne 0$ would belong to $C-C$ (the set of differences $u-v,\ u,v\in C$), and we would be able to find $i\in I$ with $f_{i}(y)\ne 0$; on the other hand, $ty\in C-C,\ \forall t\ge 0\Rightarrow tf(y)\in f(C-C),\ \forall t\ge 0$, which contradicts the fact that $f_{i}$ is bounded on $C-C$.


topology
advanced fields
advanced fields unsolved

Prove that an metric space M is compact $ \Leftrightarrow$ every non empty chain of closed subsets in M has no empty intersection

I believe that should be "non-empty intersection".

The first direction follows from the definition of compactness.  Suppose that every nested sequence of nonempty closed sets in $ M$ has nonempty intersetcion, and let $ (x_{n})$ be a sequence in $ M$.  Then there is some point $ x$ in the nonempty set $ \bigcap_{n\equal{}1}^{\infty}\overline{\{x_{k}: k\geq n\}}$.  It follows that some subsequence of $ (x_{n})$ must coverge to $ x$.


On the bottom of page 213 in AoPS Vol. 2, I don't get how they write $ \sqrt{2}$ as a continued fraction.  Could someone please show me?  Thanks!

As a sidenote, I thought irrational numbers couldn't be expressed as fractions.  Is this not correct?

irrational numbers cannot be expressed in terms of a fraction with integers a, b such that irrational number$ \equal{}\frac{a}{b}$

if you look at a continued fraction for an irrational number, it goes on indefinitely (because irrational numbers are transcendental)

so you know that $ \sqrt{2}$ greater than $ 1$

Here, you can use a shortcut (I don't know if I'm following the book because I don't have it atm)

so $ \sqrt{2}\equal{}1\plus{}\frac{1}{x}$
Where you don't know x.
However, you can find the continued fraction recursively, much in the way you would solve a continued fraction that has a definite pattern (If you haven't done this, or forgot how to, speak up!)

Solve for x, and you get $ x\equal{}\frac{1}{\sqrt{2}\minus{}1}\equal{}\sqrt{2}\plus{}1$

and you are solving for $ \sqrt{2}$ so you can solve for $ \sqrt{2}$ from earlier and get that $ x\equal{}1\plus{}\frac{1}{x}\plus{}1\equal{}2\plus{}\frac{1}{x}$

Now, you can finish this thing. $ \sqrt{2}\equal{}1\plus{}{\frac{1}{2\plus{}\frac{1}{2\plus{}\frac{1}{...}}}}$

Extra extension, solve backwards going from $ 1\plus{}{\frac{1}{2\plus{}\frac{1}{2\plus{}\frac{1}{...}}}}$
See if you get $ \sqrt{2}$
[hide="hint"]set it equal to x, then manipulate until you see something recursive![/hide]

[quote="zapi2007"]
if you look at a continued fraction for an irrational number, it goes on indefinitely (because irrational numbers are transcendental)
[/quote]
 :roll:  careful with the terms, transcendental has nothing to do with the problem here $ \sqrt{2}$ is irrational but definitely not transcendental.

whoops yeah

haha

my terminology needs a little brushing up...


induction
inequalities unsolved
inequalities

Please forgive the easy questions but I am just learning how to write proofs.

a) Suppose that 0 < X1 < 2 and Xn+1  = √ (2 + Xn) for n ∈ N.
Prove that  0 < Xn  < Xn+1 < 2 holds for all n ∈ N

( In part a the x terms  should be read:
X1 means x subscript 1,
Xn+1 means x subscript n+1 and
Xn means x subscript n)

b) Prove that 2^n + 3^n is a multiple of 5 for all odd n ∈ N

a)Quite easy
You can use induction
$ 1 < x_{n\plus{}1}< 2$ for all n
$ x_{n\plus{}1}\minus{}x_{n}\equal{}\sqrt{x_{n}\plus{}2}\minus{}x_{n}> 0$
So done.
b)
$ a\minus{}b|a^{n}\minus{}b^{n}$


$\alpha$ and $\beta$ are the roots of the equation:-

$asin\theta+bcos\theta=c$

 find:-

$cos(\alpha-\beta)$

in terms of $a,b,c$

Sorry,this should be shifted to the G.S forum

Not seeing an obvious start, I crunched some numbers and it seems that I 'easily' found a counter-example.  Am I missing something?

Somewhat randomly, I let A=2, B=5, and theta=0.4 in radian mode.
Asin(0.4)+Bcos(0.4)=5.38414...

Then I let theta = cos(0.4) and stored it in T.

Asin(T)+Bcos(T)=4.61736...; not C=5.38414...

Just to clarify, are you saying the following?

$a\sin\alpha +b\cos\alpha =c$

$a\sin\beta +b\cos\beta =c$

oh sorry,i totallly gave the wrong question(i made it up from a question .I've edited the question  to a more correct one,sorry


isnt this problem worded realy bad. our team got like 3.85 or something. the answer is simply 4.

Four circular chips are each entered on one of four adjacent
squares on checkerboard such that the centers of the chips are
the four vertices of a square. What is the area of this square if
the checkerboards squares each measure two inches on side?

I think it's worded fine.  You know the diameter of the circles because they told you the side length of the checkerboard square, and you thus know the side length of the square connecting the centers of the circles.

but then all you have to do is 2*2. our team didnt get that

i don't see what the problem with the wording is. after reading it the first time, i already had the answer u mentioned. where did u get mixed up?

I thought this was perfectly good problem. What do you propose was wrong with the wording?

I think it has to do with the fact that it is a Nat'ional Team Round Problem, which are supposed to be the hardest problems made by MATHCOUNTS.

I wouldn't know, I never went to state. :mad:
Since abacadaea just gave the year I assumed it was from nationals, but you're probably right.

yeah its states. well appearently we just got stupid according to you guys
your probably right. we woke up at 5:00 in the morning haveing a pillow fight


Excuse me, but shouldn't there be a possibility that $ (p \minus{} 1)^2 \equal{} q^3$, as stated in the original problem (N. N. Osipov's problem no. 1787 in Kvant 2001: http://kvant.info/zkm_2001.htm )

EDIT: Sorry, Osipov's statement is for all $ p,q$ not just primes as you pointed out :D

[quote="Elio (n)"]Prove that if $ p$ and $ q$ are prime  number and:
$ p|(q^{3} \minus{} 1)$

$ q|(p \minus{} 1)$

than $ p \equal{} q^{2} \plus{} q \plus{} 1$[/quote]

Hi.

[hide="solution"]
It is clear that $ p>q$.so since $ p|(q\minus{}1)(q^2\plus{}q\plus{}1)$ then $ p|(q^2\plus{}q\plus{}1)$ follows.

We can let $ q^2\plus{}q\plus{}1\equal{}kp$,where $ k$ is a positive integer.Because $ q|(p\minus{}1)$ we have $ p\equiv 1\pmod q$ ,

then $ k\equiv 1\pmod q$ follows.But if we suppose $ k\ge q\plus{}1$,then we have $ q^2\plus{}q\plus{}1\equal{}kp\ge (q\plus{}1)^2$: contradiction!

so possible $ k$ is only $ k\equal{}1$ and we get $ p\equal{}q^2\plus{}q\plus{}1$.

[/hide]

Proof :
[hide]

$ \text{Clearly } p\ge q , \text {let  }q^3-1=pk  => (q-1)(q^2+q+1)=pk ,  q^2+q+1\equiv 0\pmod  p , p\equiv 1\pmod q \text{ so that }  q^2+q+1=pd ,d\ge 1,\text {and  } d\equiv 1\pmod q .$

$ \text {Suppose } d=u_1q +1 , p=u_2q+1 ,$

$ pd=u_1u_2q^2+(u_1+u_2)q+1
=>u_1u_2=1 \text{ and }   u_1+u_2=1 , \text {but }p\ge q , u_2>=1$

$ Thus ,(u_1-u_2)^2=(u_1+u_2)^2-4 u_1u_2=-3 \text{(impossble ) }
=> d=1 \text{ and }$

$ p= q^2+q+1(only)$[/hide]


Equilateral triangle $DEF$ is inscribed in equilateral triangle $ABC$ such that $\overline{DE} \perp \overline{BC}$. The ratio of the area of $\triangle DEF$ to the area of $\triangle ABC$ is
[asy]
size(180);
pathpen = linewidth(0.7); pointpen = black; pointfontpen = fontsize(10);
pair B = (0,0), C = (1,0), A = dir(60), D = C*2/3, E = (2*A+C)/3, F = (2*B+A)/3;
D(D("A",A,N)--D("B",B,SW)--D("C",C,SE)--cycle); D(D("D",D)--D("E",E,NE)--D("F",F,NW)--cycle); D(rightanglemark(C,D,E,1.5));[/asy]
$\textbf{(A)}\ \dfrac{1}{6}\qquad
\textbf{(B)}\ \dfrac{1}{4} \qquad
\textbf{(C)}\ \dfrac{1}{3} \qquad
\textbf{(D)}\ \dfrac{2}{5} \qquad
\textbf{(E)}\ \dfrac{1}{2}$

Hey, I was just working with something like this earlier today. :lol:

[hide="Answer"]If an equilateral triangle is inscribed in another equilateral triangle such that one side of the inscribed triangle forms a right angle with a side of the circumscribed triangle, then all points where they touch form 30-60-90 triangles. Therefore, each side of the inner triangle has side length $\frac{1}{\sqrt{3}}$ that of the outer triangle, and the ratio of the areas is $\frac{1}{3}\Rightarrow \boxed{C}$.[/hide]


$ 10$ must be on top row. The biggest number on the second row can be $ 9$. The biggest number on the third row can be $ 7$ (Cannot be $ 8$, because we've used $ 1$). The biggest number on the last row can be at most $ 4$.
Now let's observe this case.

[code]
9 10 3 8
 1  7 5
   6 2
    4
[/code]

and

[code]
8 10 3 9
 2  7 6
   5 1
    4
[/code]

So the answer is $ D$.

[quote="xeroxia"]The biggest number on the third row can be $ 7$ (Cannot be $ 8$, because we've used $ 1$).[/quote]

That part doesn't exactly work.

If $ 8$ is at the third row, then $ 1$ and $ 9$ must be at second row. If $ 1$ is at the second row, then $ 9$ must be at the first row. But we've used $ 9$.

If $ 7$ is at the third row, its neighbour must be $ 1$ or $ 2$ to get $ 5$ at the bottom row. At the condition $ (7,2)$, its parents must be $ (8,1)$. If $ (8,1)$ is at the second row, parents of $ 8$ cannot be $ (10,2)$ (because we've used $ 2$) or $ (9,1)$ (because we've used $ 1$).

If the third row is $ (6,1)$, parents of $ 6$ cannot be $ (7,1)$ , $ (8,2)$ (because to get $ 8$ we must use $ (10,2)$ or $ (6,1)$) , and $ (9,3)$ (because one of the parents of $ 9$ must be $ 1$, but it is used).

Let $a$ be the number in the lower box.
Let $b$ be the smaller number on $a$'s top. Then the greater one will be $a+b$.
Let $c$ be the smaller number on $a+b$'s top. Then the greater one will be $a+b+c$.
Let $d$ be the smaller number on $a+b+c$'s top. Then the greater one will be $a+b+c+d\leq 10$.
Since $a, b, c, d$ are all different, if $a=5$ then the sum of other three will be at least $1+2+3=6$. This makes $a+b+c+d=11$. So $a$ is less than $5$.
In fact, there are four possible configurations.
[code]
9 10 3 8
 1  7 5
   6 2
    4
[/code]
[code]
8 10 3 9
 2  7 6
   5 1
    4
[/code]
[code]
6 1 10 8
 5 9  2
  4 7
   3
[/code]
[code]
6 10 1 8
 4  9 7
   5 2
    3
[/code]


A glider was pulled down an air track by a thin string attached to a pulley and some weights.  This was done several times with different weights.  Two sensors were set up on the air track; a timer started when the glider was released at the first sensor, and it stopped when the glider passed through the second sensor.  The force of the hanger, $ 9.8 \times \text{mass}$, versus measured acceleration is plotted (force is on the y-axis and acc. is on the x-axis).  The measured acceleration is found with the formula $ a \equal{} 2x/t^2$.  A line of best fit is passed through this scatterplot.  Should the y-intercept be 0?  Thanks in advance!

Yeah, theoretically, since $ x \equal{} {v}_{0}t \plus{} \frac{1}{2}a{t}^{2}$.  Since $ {v}_{0}$ appears to be zero, the y intercept should be zero.

On the other hand, since this sounds like an experiment, it's possible that it's not zero because of some experimental error.


inequalities
trigonometry
inequalities proposed

If $ A$, $ B$ and $ C$ are measures of angles of triangle $ \triangle ABC$ then prove the inequality:

$ \cos^{3}(A) \plus{} \cos^{3}(B) \plus{} \cos^{3}(C) \plus{} 5\cos(A)\cos(B)\cos(C)\leq 1$

[quote]Are we supposed to discuss this before the deadline? I don't know...[/quote]

Deadline was until January 19 so it is over.

[quote]yes, january 19 was the deadline so i think there's no problem to discuss about it... that inequality is mine, hope you had enjoyed it... if no one publishes a solutions (which i doubt) i'll post mine in a couple of days...[/quote]

Thank you for this problem that looks beautiful but unfortunately I have not solved it
So if no one post solution please post your

Salem

[hide="{just a hint}"]$ (1\minus{}\cos A)(1\minus{}\cos B)(1\minus{}\cos C)\geq \cos A\cos B\cos C$.[/hide]

I have a partial solution, but it only works for acute triangles, wich is easier to show.

It uses Jensen and AM-GM.

Maybe someone can extend it.

[hide]
$ \sum{\cos^3(A)}\le 3\left(\cos(\frac{A\plus{}B\plus{}C}{3})\right)^3 3(\frac{1}{2})^3\equal{}\frac{3}{8}$

$ 5\cos(A) \cos(B) \cos(C) \le 5 \left( \frac{\cos(A)\plus{}\cos(B)\plus{}\cos(C)}{3}\right)^3\le  5(\frac{1}{8})^3\equal{}\frac{3}{8}$

[/hide]

 :)

i think there's a mistake because it seems that $ \cos^3 x$ is not a concave function... and also note that the equality holds also if $ (A,B,C)\equal{}(90,90,0)$, so i think that breaking the inequality in parts would not help...

try again  :lol:

Congratulations for this beautiful inequality, [i]campos[/i]! I have to say your hint was a big one! :)

Let $ x \equal{} \cos A$, $ y \equal{} \cos B$ and $ z \equal{} \cos C$. By [i]campos[/i]'s hint, we have $ xyz\le (1 \minus{} x)(1 \minus{} y)(1 \minus{} z)$. Using the identity $ x^2 \plus{} y^2 \plus{} z^2 \plus{} 2xyz \equal{} 1$, the inequality to prove rewrites as
$ x^3 \plus{} y^3 \plus{} z^3 \plus{} 3xyz\le x^2 \plus{} y^2 \plus{} z^2$.
This is $ 3xyz\le x^2(1 \minus{} x) \plus{} y^2(1 \minus{} y) \plus{} z^2(1 \minus{} z)$, which follows by [i]AM-GM[/i] and [i]campos[/i]'s hint. We do not have to distinguish two cases (when $ x,y,z$ are positive and when one of them is $ < 0$).

And for completeness, here's a proof for $ (1 \minus{} \cos A)(1 \minus{} \cos B)(1 \minus{} \cos C)\ge \cos A\cos B\cos C$:

Clearly, this inequality is obvious if the triangle is right or obtuse. Assume it is acute angled.

Let $ c \equal{} \frac {\cos A\cos B}{\cos C}$, $ b \equal{} \frac {\cos A\cos C}{\cos B}$ and $ a \equal{} \frac {\cos B\cos C}{\cos A}$. The identity mentioned above, $ \cos^2 A \plus{} \cos^2 B \plus{} \cos^2 C \plus{} 2\cos A\cos B\cos C \equal{} 1$, becomes $ ab \plus{} bc \plus{} ca \plus{} 2abc \equal{} 1$. Hence we can make the substitutions $ a \equal{} \dfrac x{y \plus{} z}$, $ b \equal{} \dfrac y{x \plus{} z}$ and $ c \equal{} \dfrac z{x \plus{} y}$, for some $ x,y,z > 0$ (different from the $ x,y,z$ used when proving the original inequality).
Then $ \cos A \equal{} \sqrt {bc} \equal{} \sqrt {\frac {yz}{(x \plus{} y)(x \plus{} z)}}$.
Coming back to $ (1 \minus{} \cos A)(1 \minus{} \cos B)(1 \minus{} \cos C)\ge\cos A\cos B\cos C$, we see that it becomes $ \displaystyle\prod \left[\sqrt {(x \plus{} y)(x \plus{} z)} \minus{} \sqrt {yz}\right]\ge xyz$.
Well, now it suffices to use $ \sqrt {(x \plus{} y)(x \plus{} z)}\ge x \plus{} \sqrt {yz}$, which is true by Cauchy.

finally a solution :D  i hope you had enjoyed it!

I think your hint is equivalent to the problem.

Actually, from your hint one can get the result;
on the other hand, denote

$ x\equal{}\cos A\plus{}\cos B\plus{}\cos C$
$ y\equal{}\cos A\cos B\plus{} \cos B \cos C\plus{} \cos C \cos A$
$ z\equal{}\cos A \cos B \cos C$

we have

$ x^2\minus{}2y\plus{}2z\equal{}1\ \left(1 \le x \le \frac{3}{2} \right)$

And

With the identity, the problem is equivalent

$ x(x^2\minus{}3y)\plus{}4(1\plus{}2y\minus{}x^2) \le 1$

which is

$ y \le \frac{(x\minus{}1)(3\plus{}3x\minus{}x^2)}{8\minus{}3x}$

If it's right, we can prove your hint

$ y \le x(x\minus{}1)$

Actually it suffices to show

$ \frac{3\plus{}3x\minus{}x^2}{8\minus{}3x} \le x$

Which is equivalent to

$ (2x\minus{}3)(x\minus{}1) \le 0$

It follows from $ 1 \le x \le \frac{3}{2}$.
Well I have to say your inequality is nice! :lol:

wow, i haven't noticed this, it's a very cool fact  :D  in fact i have another similar inequality, which is very nice and shows an stronger inequality than the hint i proposed, although this part is easier than the problem at Math Reflections... hope you enjoy it too!

Let $ ABC$ be a triangle, prove that

$ \cos^3A\plus{}\cos^3B\plus{}\cos^3C\plus{}5(1\minus{}\cos A)(1\minus{}\cos B)(1\minus{}\cos C)\geq 1$.

From your hint, we have

$ y \le x^2\minus{}x$

And what we're going to prove is

$ (3x\minus{}3)y \le (x\minus{}1)^2(x\plus{}3)$

So it suffices to show

$ (3x\minus{}3)(x^2\minus{}x) \le (x\minus{}1)^2(x\plus{}3)$

which is equivalent to

$ (x\minus{}1)^2(2x\minus{}3) \le 0$

So they're all equivalent I think.  :wink:

are you sure? because this inequality has equality cases $ (60,60,60)$ and $ (90,\alpha,90\minus{}\alpha)$ for $ \alpha\in(0,90)$  :D

[quote="campos"]are you sure? because this inequality has equality cases $ (60,60,60)$ and $ (90,\alpha,90 \minus{} \alpha)$ for $ \alpha\in(0,90)$  :D[/quote]

You mean the following inequality?

$ \cos^3 A\plus{}\cos^3 B\plus{}\cos^3 C\plus{}5(1\minus{}\cos A)(1\minus{}\cos B)(1\minus{}\cos C)\ge 1$

But I don't think the equality occurs for $ (90,\alpha,90\minus{}\alpha)$ for $ \alpha \in (0,90)$  :maybe:

:blush:  i wanted to say $ (0,\alpha, 180\minus{}\alpha)$ for $ \alpha\in(0,180)$  :D

[quote="campos"]:blush:  i wanted to say $ (0,\alpha, 180 \minus{} \alpha)$ for $ \alpha\in(0,180)$  :D[/quote]

Well, there isn't any problem. You know we use $ x \le 1$ i.e.

$ \cos A\plus{}\cos B\plus{}\cos C \ge 1$

in which the equality occurs when $ (A,B,C)\equal{}(0,\alpha,180\minus{}\alpha)$  :wink:

ok, i know that the "hint" follows from both inequalities since  $ \cos^3A\plus{}\cos^3B\plus{}\cos^3C\plus{}5(1 \minus{} \cos A)(1 \minus{} \cos B)(1 \minus{} \cos C)\geq 1$

and  $ 1 \geq \cos^3A\plus{}\cos^3B\plus{}\cos^3C\plus{}5\cos A\cos B\cos C$

what i don't understand is how the last inequality $ \cos^3A\plus{}\cos^3B\plus{}\cos^3C\plus{}5(1 \minus{} \cos A)(1 \minus{} \cos B)(1 \minus{} \cos C)\geq 1$ follows from the hint, can you explain a little bit more?

[quote="shalex"]From your hint, we have

$ y \le x^2 \minus{} x$

And what we're going to prove is equivalent to

$ (3x \minus{} 3)y \le (x \minus{} 1)^2(x \plus{} 3)$

So it suffices to show

$ (3x \minus{} 3)(x^2 \minus{} x) \le (x \minus{} 1)^2(x \plus{} 3)$

which is equivalent to

$ (x \minus{} 1)^2(2x \minus{} 3) \le 0$

So they're all equivalent I think.  :wink:[/quote]

Here is what I explained for the way to get

$ \cos^3 A \plus{} \cos^3 B \plus{} \cos^3 C \plus{} 5(1 \minus{} \cos A)(1 \minus{} \cos B)(1 \minus{} \cos C) \ge 1$

from the hint  :wink:


I hope this is the right place to put this!
I am stuck on the second part of a question. The first part was to prove that the set of binary sequences was uncountable which i have done, the second part is to use this results to prove that the set P(N+) is uncountable (P(N+) being the power set of the natural positive numbers). this is the part i am stuck on! I would be grateful for any help! I think the easiest way to do this is to find a bijection between the sets? But I'm not sure how to go about this! Thank you!

Take any binary sequence $(a_n)_{n \in \mathbb{N}}$ and map it to the subset $N \subset \mathbb{N}$ that contains $n$ iff $a_n=1$.
This is a bijection and the result follows.

[quote="hcast"]This topic should be moved to real analisys, don't you think?[/quote]
OK, goes straight to "Solved Problems" :)


calculus integration trigonometry limit symmetry quadratics logarithms

For any $ \alpha$ such that $ 0<\alpha<2$, find
$ \int_{0}^{\infty}\frac{x^{\alpha\minus{}1}}{1\plus{}x^{2}}\ dx$

Find
$ \int_{0}^{\infty}\frac{dx}{1\plus{}x^{3}}$

I'm interested in the solutions, not so much the answers, since I know the answers.  I know how to solve these using residues and complex integrals, but those are boring.  I'f like to see a real-variable solution.

Let $ I\equal{}\int_0^{\infty}\frac{x^{\alpha\minus{}1}}{1\plus{}x^2}\,dx.$

Note that $ \frac1{1\plus{}x^2}\equal{}\int_0^{\infty}e^{\minus{}(1\plus{}x^2)t}\,dt.$

$ I\equal{}\int_0^{\infty}\int_0^{\infty}x^{\alpha\minus{}1}e^{\minus{}t}e^{\minus{}tx^2}\,dt\,dx.$

Switch the order of integration so that the $ dx$ integral is inside. Look at that inside integral and make the change of variables $ x\equal{}\sqrt{\frac{u}{t}}.$

$ \int_0^{\infty}x^{\alpha\minus{}1}e^{\minus{}tx^2}\,dx\equal{}\frac12t^{\minus{}\alpha/2}\int_0^{\infty} u^{\frac{\alpha}2\minus{}1}e^{\minus{}u}\,du\equal{}\frac12t^{\minus{}\alpha/2}\Gamma\left(\frac{\alpha}2\right).$

So $ I\equal{}\frac12\Gamma\left(\frac{\alpha}2\right)\int_0^{\infty}t^{\minus{}\alpha/2} e^{\minus{}t}\,dt\equal{}\frac12\Gamma\left(\frac{\alpha}2\right)\Gamma\left(\frac{2\minus{}\alpha}2\right).$

Checking this with the one easily verified value $ \alpha\equal{}1$ does give the correct integral.

Note that if we let $ I_{\alpha}\equal{}\int_0^{\infty}\frac{x^{\alpha\minus{}1}}{1\plus{}x^2}\,dx,$ then the substitution $ u\equal{}\frac1x$ gives us that $ I_{\alpha}\equal{}I_{2\minus{}\alpha}.$

Furthermore, $ \lim_{\alpha\to0^\plus{}}I_{\alpha}\equal{}\infty,$ and the same should be true (by the symmetry above) for the limit as $ \alpha\to2^\minus{}.$ Does your previous computation agree with these conditions?

For the second problem, we could be completely unimaginative and do it as in a regular Calculus II class: factor $ x^3\plus{}1\equal{}(x\plus{}1)(x^2\minus{}x\plus{}1),$ pull the fraction apart by partial fractions, complete the square in the irreducible quadratic, and compute the integral as a combination of logarithms and arctangents. As I said, completely unimaginative.

[quote="JRav"]Interesting.  Using complex contour integrals, we derived a solution using cosines and sines.  I would provide it but I have a feeling either I copied it incorrectly or there was an arithmetic mistake.[/quote]
I just repeated the contour integral calculation and got the value $ \frac {\pi}{2\sin\left(\frac {\alpha\pi}2\right)}$

This does satisfy the limit and symmetry properties that called for above, and does equal $ \frac {\pi}2$ for $ \alpha \equal{} 1,$ as it should. I see no problems with either computation. Hence, we have as a side benefit derived a Gamma function identity. Let $ \beta \equal{} \frac {\alpha}2$ and we have the following:
\[ \boxed{\text{For }0 < \beta < 1,\ \Gamma(\beta)\Gamma(1 \minus{} \beta) \equal{} \frac {\pi}{\sin(\pi \beta)}}
\]
Of course, by analytic continuation, this identity is actually true for all $ \beta\in\mathbb{C}\setminus\mathbb{Z}.$

That's very similar to what we derived in class, except there was a cosine in there.  We must have just made a small error.

For the second problem, we deliberately made it more complicated by modifying the function to $ f(z)\equal{}\frac{\log{z}}{1\plus{}z^{3}}$, and then taking a "keyhole" type contour along the branch cut.


Find all function $f: \mathbb{R}\to\mathbb{R}$ satisfying:$(x+1)f(x)+2f(-x)=sinx$  $\forall x\in\mathbb{R}$

We know that $(x+1)f(x)+2f(-x) = \sin x$, so $(1-x)f(-x)+2f(x) = \sin(-x) =-\sin x$. Adding those two equations, we get $(x+3)f(x)+(3-x)f(-x) = 0$. Now let $g(x) = (x+3)f(x)$. The previous equation yields $g(x)+g(-x) = 0$, so $g(-x) =-g(x)$.

We return to $(x+1)f(x)+2f(-x) = \sin x$. Multiply both sides with $x^{2}-9$. We obtain \[\begin{aligned}(x^{2}-9)\sin x & = (x-3)(x+1)(x+3)f(x)-2(x+3)(3-x)f(-x) = (x-3)(x+1)g(x)-2(x+3)g(-x) \\ & = (x^{2}-2x-3)g(x)+2(x+3)g(x) = (x^{2}+3)g(x) = (x^{2}+3)(x+3)f(x)\end{aligned}\]Therefore, for all $x\neq-3$, we get $f(x) = \frac{(x-3)\sin x}{x^{2}+3}$. We now just need to find $f(-3)$.

We know that $f(3) = 0$, so $\sin(-3) =-2f(-3)+2f(3) =-2f(-3)$, from which we get $f(-3) = \frac{1}{2}\sin 3$, which satisfies $f(x) = \frac{(x-3)\sin x}{x^{2}+3}$ as well.

We can check that $f(x) = \frac{(x-3)\sin x}{x^{2}+3}$ is a solution to the initial equation, and therefore, it's the only one.

Substitute $-x$ for $x$ and solve the two equations obtained for $f(x)$.  You then obtain
\[f(x)=\frac{(x-3)\sin x}{x^{2}+3}. \]
[b]Solution.[/b] $x\rightarrow-x$ implies
\[\\ (x+1)f(x)+2f(-x)=\sin x\\ (1-x)f(-x)+2f(x)=-\sin x \]
Multiply the first equation by $-\frac{1-x}{2}$,
\[\frac{(x^{2}-1)}{2}f(x)-(1-x)f(-x)=\frac{x-1}{2}\sin x. \]
Adding the last two equations, one obtain
\[\\ \frac{(x^{2}-1)}{2}f(x)+2f(x)=\frac{x-3}{2}\sin x\\ f(x)=\frac{(x-3)\sin x}{x^{2}+3}, \]
as required.$\blacksquare$


calculus
derivative
function
inequalities proposed
inequalities

Find the maximal value of \[ \left( x^3+1 \right) \left( y^3 + 1\right) ,  \] where $x,y \in \mathbb R$, $x+y=1$.

[i]Dan Schwarz[/i]

why wouldn't it just be $\frac{81}{64}$ ?  $f(x,y)=f(y,x)$ in this case, so it's symmetric.  Intuitively, I'd say $x=y=\frac{1}{2}$.  If you can use calculus ( :roll: ), just substitute $y=1-x$, set $V=(x^3+1)((1-x)^3+1)$ and just take the derivative and find the zeroes.

[quote="perfect_radio"]Find the maximal value of \[ \left( x^3+1 \right) \left( y^3 + 1\right) ,  \] where $x,y \in \mathbb R$, $x+y=1$.[/quote]
$(x^3+1)(y^3+1)\leq4\Leftrightarrow(xy+1)^2(xy-2)\leq0.$ This is true, since $xy-2<0.$
Equality holds when $x+y=1$  and $xy=-1.$ :)

[quote="Elemennop"]I got a minimum of 1/2... dunno about max.[/quote]

$\frac{1}{2}$ is still special.  Eh, it was one of them :P

:first: He's one of the many who will get the "Gold Strawberry Prize"(a.k.a. Capsunica De Aur) probably ;)

I think we can get the answer without using Calculus.
Let $f(x,y)=(x^3+1)(y^3+1)$.
After some simplification, we will have $f(x,y)=x^3y^3-3xy+2=(xy-1)^2(xy+2)$.
Clearly, if we want to obtain the maxiumum of f(x,y), we only need to consider $xy+2\geq 0$.
Apply AM-GM inequality(Note that $1-xy\geq 0$):
$f(x,y)={\frac 1 2}(1-xy)(1-xy)(2xy+4) \\ \leq {\frac 12}(\frac{(1-xy)+(1-xy)+(2xy+4)}{3})^3=4$
Equality is attained when $1-xy=2xy+4$ if and only if $xy=-1$.

[quote="gnosis"]Or a less elementary proof  ;)  :

We have that $E(x,y)=(x^3+1)(y^3+1)=(xy)^3+(x+y)(x^2-xy+y^2)+1=(xy)^3+1-3xy+1$, which together with $xy\leq \left(\frac{x+y}{2}\right)^2=\frac{1}{4}$ resumes with finding the maximum of the function $f(\alpha)=\alpha^3-3\alpha+2$ on $(-\infty,\frac{1}{4} ]$, but since $f$ is increasing on $(-\infty,-1)$ and decreasing on $(-1,\frac{1}{4}]$ as it can be seen from $f'(\alpha)=3\alpha^2-3$ we find that $f(-1)=4$ is the maximum, for which $x+y=1=-xy\Leftrightarrow\{x,y\}=\{\frac{1+\sqrt{5}}{2},\frac{1-\sqrt{5}}{2}\}$.[/quote]

Your answer is correct. :)

[quote="niyo"]IS IT HOLDER? (x^3+1)(1+y^3)(1+1)>=(x+y)^3=1 answer is 1/2 correct? :)[/quote]

This can't be correct:
if $x=-10$ and $y=11$ we have a neg. thing and there couldn't occur equality.

Simplifying we have that $(x^3+1)(y^3+1)=-(x^2-x-2)(x^2-x+1)^2$.  Setting $a=x^2-x+1$, we want to find the maximum value of $-(a-3)(a^2)=3a^2-a^3$.  Taking the first derivative and setting it equal to $0$ we have that is either $0$ or $2$ simple substitution shows that it is $2$.   Therefore the answer is $3a^2-a^3=3(2^2)-2^3=4$

[quote="mavropnevma"]How many errors can one make in a couple lines?

Simplifying we have that $(x^3+1)(y^3+1)=-(x^2-x \ \color{red}{+}$  $2)(x^2-x+1)^2$, instead of the correct $x^2 - x - 2$.
Setting $a=x^2-x+1$, we want to find the maximum value of $-(a-3)(a^2)=3a^2-a^3$.
Taking the first derivative and setting it equal to $0$ we have that is either [color=#FF0000]0[/color] $0$ or $2$ simple [color=#FF0000]substiution show [/color] substitution shows that it is $2$. Therefore the answer is $3a^2-a^3=3(2^2)-2^3=4$.[/quote]

sorry  :blush:

Take $f(x)=\left(x^3+1\right)\left((1-x)^3+1\right)=-x^6+3x^5-3x^4+x^3+3x^2-3x+2$.
We could easily see that $f(x)=(2-x)(x+1)(x^2-x+1)^2$.
Next we get $f'(x)=-3(2x-1)(x^2-x-1)(x^2-x+1)$ which has two complex roots and 3 real roots.
These are $-0,5$ where we obtain a local minimum, $\phi$ and $\phi'$ where we obtain the global maximum $4$.


linear algebra
matrix
abstract algebra
superior algebra
superior algebra unsolved

In $ M_{2}(R)$, let $ e$ be a non-stupid idempotent (i.e., neither the 0-matrix nor the identity matrix).  Determine whether the two-sided ideal generated by $ e$ is a proper ideal in $ M_{2}(R)$. The two-sided ideal generated is $ \left\langle ReR\right\rangle$, which is all possible linear combinations $ r_{1}es_{1}\plus{}r_{2}es_{2}\plus{}\cdots\plus{}r_{n}es_{n}$ where $ r_{i},s_{i}\in R$.

I don't see what relevance being an idempotent has.  Is the inclusion proper?

If R is an associative, unital ring, e in R, ReR = R, and ee=e, then e is called a full idempotent.  Such idempotents are important in defining an equivalence of module categories between R and eRe.

If S is an associative, unital ring, then the ideals of R=M_2(S) are exactly M_2(I) where I is an ideal of S.  In particular, if S is a field, then every nonzero idempotent of M_2(S) is full.  If S is not a field, then of course this need not be the case, since S itself could have nonzero non-full idempotents.


please and thank you =]

one is each of the letters a b c is replaced by a distinct digit 0 through 9 make the following product true what is the greatest possible value of a +b +c.
ab*b=cab

[hide="Solution"]Your equation is \[ ab \cdot b \equal{} cab\] We need to have the largest possible sum of $ a\plus{}b\plus{}c$ which means that we need the largest possible, distinct digits for $ a,b,c$. Let's take a look at only the units digits so we can figure out what $ b$ is. From the equation, we get the that when $ b$ is squared, the units digit of that is $ b$ \[ b^2 \equal{} ...b\]. Let's list out all the numbers: \[ \boxed{0 \cdot 0 \equal{} 0}\]\[ \boxed{1 \cdot 1 \equal{} 1}\]\[ {2 \cdot 2 \equal{} 4}\]\[ {3 \cdot 3 \equal{} 9}\]\[ {4 \cdot 4 \equal{} ...6}\]\[ \boxed{5 \cdot 5 \equal{} ...5}\]\[ \boxed{6 \cdot 6 \equal{} ...6}\]\[ {7 \cdot 7 \equal{} ...9}\]\[ {8 \cdot 8 \equal{} ...4}\]\[ {9 \cdot 9 \equal{} 1}\] The ones that fit our conditions are boxed. Thus, the possible values of $ b$ are \[ 0, 1, 5, 6\] But we aren't there yet; we have to see if this value of $ b$, $ 6$ will work with $ a$. When we multiply $ b \cdot b$ and we plug in $ 6$ for $ b$ (remember we're trying to get the highest possible sum which means highest possible digits), we have to add $ 3$ when we multiply $ 6 \cdot a$. Thus, \[ 6a\plus{}3 \equal{} ...a\] We need to plug all numbers for $ a$ and see what case this works for. We have \[ 1,9\]\[ 2,5\]\[ 3,1\]\[ 4,7\]\[ 5,3\]\[ 6,9\]\[ 7,5\]\[ 8,1\]\[ 9,7\]\[ 0,3\]Man! None of them work!
Now continue from this, plugging in $ 5$ for $ b$, checking whether there is a case that will work with $ a$, and then find all of them. Out of these cases, find the one that will give you the highest value. I started you off so now you have to finish it (this gives you good practice, if I give you the whole solution, it won't be as effective to learning as doing it yourself).
If you have any questions, or can't finish it, don't hesitate to PM me![/hide]

[hide="Finishing it off..."]
Like akhil said, the last digit must be 0, 1, 5, or 6.  Testing, like he did, yields a=7, b=5, c=3 so a+b+c=15.[/hide]

[hide="Algebraic Approach"]
As noted in the previous posts, $ b \equal{} \{0,1,5,6\}$.

Also, $ (10a \plus{} b)\cdot b \equal{} 10ab \plus{} b^2 \equal{} 100c \plus{} 10a \plus{} b$
$ \implies10ab \minus{} 10a \plus{} b^2 \minus{} b \equal{} 10a(b \minus{} 1) \plus{} b(b \minus{} 1) \equal{} (10a \plus{} b)(b \minus{} 1) \equal{} 100c$.

Since $ c$ is positive, this quickly eliminates $ b \equal{} 0$ and $ b \equal{} 1$, so now we can only have $ b \equal{} \{5,6\}$. However, to be safe, I will check the case $ b \equal{} 1$ ($ b \equal{} 0$ is definitely eliminated because $ c$ definitely has to be nonnegative; I'm just not sure if the problem means that $ \overline{cab}$ is a three-digit number).

[hide="Case: b=1"]
This means that $ 0 \equal{} 100c\implies c \equal{} 0$. This leaves no restrictions on $ a$, so the highest we can go is $ \overline{cab} \equal{} 091\implies0 \plus{} 9 \plus{} 1 \equal{} 10$.
[/hide]

[hide="Case: b=5"]
This means that $ (10a \plus{} 5)(5 \minus{} 1) \equal{} 20(2a \plus{} 1) \equal{} 100c\implies2a \plus{} 1 \equal{} 5c\implies1\equiv c\pmod2$, so $ c$ is odd. Also, since $ 0\le a\le9$ by the definition of a digit, $ 1\le5c\le19\implies1\le c\le3$. From these two conditions, both $ c \equal{} 1\implies a \equal{} 2\implies\overline{cab} \equal{} 125\implies1 \plus{} 2 \plus{} 5 \equal{} 8$ and $ c \equal{} 3\implies a \equal{} 7\implies\overline{cab} \equal{} 375\implies3 \plus{} 7 \plus{} 5 \equal{} 15$ work.
[/hide]

[hide="Case: b=6"]
This means that $ (10a \plus{} 6)(6 \minus{} 1) \equal{} 10(5a \plus{} 3) \equal{} 100c\implies5a \plus{} 3 \equal{} 10c\implies3\equiv0\pmod5$, contradiction; there are no solutions in this case.
[/hide]

Now, we can conclude that $ \boxed{15}$ is the largest.
[/hide]

[b]Note:[/b] akhil0422's argument is not rigorous in that it assumes the higher $ b$, the higher the sum $ a \plus{} b \plus{} c$. Here, it is true, but in other situations, a higher $ b$ may result in lower $ a$ and $ c$ which could result in a lower sum.


trigonometry
inequalities
geometry
circumcircle
geometry solved

Prove that in any triangle ABC the following inequality holds!

$\sum\frac{bc}{b+c}<3R$,

where R is the circumradius of triangle ABC.

We have that

$\sum\frac{bc}{b+c}\leq\sum\frac{b+c}{4}=\frac{a+b+c}{2}=R\left(\sin A+\sin B+\sin C\right)\leq R\cdot\frac{3\sqrt3}{2}<3R$.

We obtained a stronger inequality!

[b]Solution 1:[/b]

because $\left(\sin\frac{A}{2}\right)^2\leq\frac{a^2}{4bc}$ we get that $\sin\frac{A}{2}\leq\frac{a}{2\sqrt{bc}}$
also we have that $\left(\cos\frac{A}{2}\right)^2<\frac{\left(b+c\right)^2}{4bc}$ and thus $\cos\frac{A}{2}<\frac{b+c}{2\sqrt{bc}}$.

we multiply the last two ineqs and get that $\sin A<\frac{a\left(b+c\right)}{2bc}$ which gives:
$\frac{bc}{b+c}<\frac{a}{2\sin A}$ so $\frac{bc}{b+c}<R$
the same with the rest and voila the ineq! :)

[b]Solution 2:[/b]

it seems that today I am lucky.. I found another solution :)

using the sinus theorem and the fact that $\frac{xy}{x+y}\leq\frac12\sqrt{xy}$ we get that
$\sum\frac{bc}{b+c}=2R\left(\sum\frac{\sin B\sin C}{\sin B+\sin C}\right)\leq R\left(\sum\sqrt{\sin A\sin B}\right)<3R$ because $0<\sin A\leq 1$, but the inequality is not strict, since we cannot have sin A = sin B = sin C = 1.

cheers! :D