articles AMC AIME geometry USA(J)MO USAMO MATHCOUNTS

What do you guys do, specifically, to avoid dumb mistakes? I've read the article in the resources section. Just wondering if there's anything I should be doing other than:

1. Sleep. This is really important for me, at least, though I know it isn't as important for some.
2. Checking work. The issue is I still don't catch some of my mistakes.
3. Writing neatly, etc.

Consciously working at a slower pace than normal will cause you to make less errors, in my experience.
And of course, reading the damn problem and having a strategy (i.e. when to check, etc.) help.

http://www.artofproblemsolving.com/Resources/AoPS_R_A_Mistakes.php

With respect to the organizing work part of this article, this year I had at most one problem's work on a sheet of paper (though a problem could have multiple pages), with that problem's number at the top left. This was the first year I did this; now let's look at my AIME timeline of problems missed by stupid mistakes.

2005: 4
2006: 3
2007: 3
2008: 0

There are also other checking methods specific to problems.
Counting: Best way by far is to find two ways to count the stuff.
Geometry: Compute various quantities using the given information, and then recompute it using the answer you got. (i.e. if the problem asks for a side length, use that side length to recompute)
Algebra: Plug everything back in. Test various steps/equalities in your solution with set values of numbers.
Number Theory: Varies. Often the algebra methods work.

Not really. I only got around 5.5 hours of sleep the night before the AIME, hence I was very tired. We took the NEML (Math League) contest 6 right before, and I ended up getting one wrong because I was so tired (I basically did the whole thing in 5 minutes and tried to sleep for the other 25).

However, when we started the AIME, my mad adrenaline rush kept me going through all three hours, and I did better than I would have expected.

[quote="boo"]Do you guys feel that drinking coffee/red bull/energy drink helps with taking a competition in the morning?[/quote]

For me definitely since I'm addicted to the stuff  :D

For the OP's question, I personally work on a problem, reach a point which I'm stuck on slightly, and then come back later because I know I'll make a mistake. Check things like recursion sequences to see if your recursion was actually right all the way through. Plug numbers back in to see if they work. Don't go super fast also.

I always spend the last 5 minutes checking my answer bubbles.

I tried using one paper per problem for this AIME.  Since I have no previous AIMEs to compare with, I don't have much to judge by except for practice AIMEs.  Although some mistakes still slipped through, I think this strategy helped significantly.

[quote="boo"]Do you guys feel that drinking coffee/red bull/energy drink helps with taking a competition in the morning?[/quote]

Definitely not.  The best thing for the brain is sleep, as I find sometimes I work out mathematical problems and just generally feel more content in the mind with a good night's sleep, and drinking energy drinks make me less focused and jittery.  Also, It helps to label your problems and keep them somewhat organized, so that when you look over your work, you can go through it pretty fast, looking for mistakes.  Leaving yourself 10-15 minutes when you're done to simply go over your work is a good strategy, as you're more likely to get a point back from catching a mistake than solving another problem in that time. (unless you're the kind of person that gets 12+)

For me last year, I made 4 mistakes that cost me USAMO (I only got 4 others right).  This year I made 2 stupid mistakes, but I answered more correctly.  I decided that I would answer as many as possible in the first two hours (I hoped to answer around 8) and then check the last hour.  I did this and found 2 mistakes in my previous answers.  I would suggest completely redoing the problems so that you don't make the same mistakes again.

When I check my answers, I try to do the problem a different way (if it does not take long) or if it's the kind of problem where you can just plug something in, I plug it in to make sure it works.... It also helps to wait a while before going back and checking your work; otherwise (for me at least) it's easy to just make the same stupid mistake again.

This year, on the fourth problem ($ x^2\plus{}84x\plus{}2008\equal{}y^2$), I missed a negative sign, so I had $ (x\plus{}42\minus{}y)(x\plus{}42\plus{}y)\equal{}244$ instead of $ \minus{}244$, so I got $ 20$ and $ 60$. Luckily, they still sum to $ 080$. But I checked my answer with one minute remaining, and confirmed that it was $ 18$ and $ 62$.

I hurry myself too much on the AIME, which causes me to make mistakes. This is because I'm not used to taking 3 hours long on a 15-question exam.
so basically, don't hurry yourself, and practice more

I guess I shouldn't be saying anything, since I made 4 stupid mistakes on the AIME, but I've definitely gotten better at not making careless errors.  The one thing that I see different is...

CONFIDENCE

Sure, sleeping and eating are a good way to get your brain ready, but I think you have to be mentally prepared in other ways.  To me, I think that if you go into a test thinking that you will do well, then you will.  Of course, this method hasn't always worked for me.  Take March 18, 2008.  Not saying that 9 is bad, but since I only discussed the answers to 1-10 with my teacher, and within that range, I had only 5 right, I was definitely worried (2 were guesses...)

But yeah, in general, just have confidence.  That might mean doing tons of math, but I think it's worth it.

Leave [b]A LOT[/b] of time to double check all your work, even the problems you think there is absolutely no way you could be wrong. It is sometimes better to do 9-10 problems but get them right for sure, then do 14-15 but end up getting only half of them right. Another thing is to make sure you answer what the problem is asking you. During this AIME, I almost got number 1 wrong because I misread the question, and found out only 20 min before the end.
Practice helps a lot too.

[quote="Anaxerzia"][quote="Smartguy"]actually, sleeping enough is the best way to prepare for anything[/quote]

Eat a good breakfast too and bring food to the test. Helps a lot for me :)[/quote]
I brought food to the test, but I never had the chance to eat it because the teacher brought donuts to everyone like every 30 minutes or something.

[quote="Darklegacy52"][quote="Zmastr"]I did the exact same thing...got 412 on question 1 and then looked over my questions and realized that i didn't have 58% girls after adding the 20 boys with like 20 minutes left.  If I didn't catch that mistake I would have a 117/7, which is significantly worse than a 117/8 USAMO-wise.[/quote]

Good job, you actually caught your mistake. Me, on the other hand, checked it over and still didn't see that i forgot to add the extra 20 at the end. =( I would have gotten an 8 also. BAH![/quote]

7 and 8 on the AIME are critical scores

For me its extremely hard to catch mistakes. In fact I don't think I ever have in my whole life. On this AIME I made 2 dumb mistakes, then I was just being dumb when doing 2 other problems (like, majorly flawed approach).
For the dumb mistakes issue, do lots of practice and be extremely careful. Take plenty of time on easy problems, even if they seem obvious. Plug everything back in to see if it works, even if its a computation nightmare.

One really important thing is reading questions correctly, and interpreting them correctly. For example, this year's number 12 was very easy, but I panicked at the word "maximum", since the numbers seemed random, and I didn't try it. My friend read number 7 incorrectly and put 292. These happen way too often, and lots of really beastly people get brought down by them and their scores massacred.

Caffeine actually does help, but for me if I have too much caffeine I panic at easy problems a lot...which isn't good (at mathcounts state last year I drank 3 16 oz. monsters, for a total of I believe 300 mg of caffeine? Yea my brain got messed up).

EDIT: and for those of you who make just an insane number of computation errors, what you need is a good dose of mathcounts sprint rounds :D as much as I hate mathcounts.

[quote="The Zuton Force"]Since this is AoPS, I would like to share the arithmetic sequences that arise when considering my AIME performances. Read down the columns!  :D
[b]
9th grade:[/b] 4 correct, 5 stupid mistakes, 6 guesses
[b]10th grade:[/b] 5 correct, 6 stupid mistakes, 4 guesses
[b]11th grade:[/b] 6 correct, 7 stupid mistakes, 2 guesses
[b]12th grade:[/b] 7 correct, 8 stupid mistakes, 0 guesses

[i]To clarify the above, I've never used a flawed approach in an AIME problem; all the ones I got wrong, I either randomly guessed or made a stupid mistake. I don't know why I make stupid mistakes, I write my work out neatly and everything, and I never catch them when I check =\[/i][/quote]

Sounds very similar to me. I always make stupid mistakes but get the right method. I think the reason that you don't spot them when going over them is you don't actually check them, you used the same residual logic that you used the first time running through your head. So it look right no matter how stupid

For that reason, looking over your work is not the best way to find mistakes. If you got the answer to a problem wrong, then you could probably use it to derive a blatant contradiction. i.e. in algebra problems, substituting values for variables gets you $ 1 \equal{} 0$ or something. Perhaps the only exception to this is counting problems which conveniently lend themselves well to being solvable by multiple methods.

One of my previous posts in this thread has more examples of how to apply this method of checking.

I dont have a problem with careless mistakes...

Its just that I always give up on a problem that I cant do...

Then after the test, I think about it for 10 mins and am able to find a very easy method

[quote="randomdragoon"]My advice for the AIME: Do trivial calculations slowly.[/quote]

That's probably the best advice. Just take your time to get each step right the first time, checking the final answer by alternate solutions whenever possible. This has always worked well for me, especially since I don't have the patience or interest to re-do calculations that I've already done. Part of it is a test of mathematical ability as well. Having a lot of experience solving harder and more involved problems yields several advantages: you can come up with simpler solutions, you won't get confused in your reasoning once you have a solution method, and you'll notice a lot of extra equations or other relationships that can help you check or streamline your work. I think people especially underestimate these factors when they make logical mistakes--a very important part of getting better at math is being solid on logic and not just coming up with good general ideas.

Here's what I've learned (after making 4 stupid mistakes on AIME last year):
1: Kill the environment, and by that I mean use lots of paper that you keep organized by problem.
2: Write big. Very big.
3: Write down every calculation you make, especially those you make mentally. I mean every single calculation in the problem.
4: Read every question twice before beginning to work it.
5: After finishing a problem, work it backwards and see if you can retrace your progress back to the original problem statement.
6: At the beginning of the test, take a sheet of paper and write down these steps to remind yourself what to do.

On one of the questions this year, the answer was in m(root)n form. and you had to give your answer as m+n. So, I got 12rt2, but somehow, I thought that 12+2=16..and got it wrong cause of that.[/img][/code]


function
geometry
geometric transformation
reflection
calculus
calculus computations

What is the inverse graph of this function

[img]http://img210.imageshack.us/img210/5272/alrubeh4672setsect71pro.jpg[/img]

choose one of these colored graphs

[img]http://img297.imageshack.us/img297/5272/alrubeh4672setsect71pro.jpg[/img]

Reflect the graph with respect to the line $ y\equal{}x$, so the answer is ...

That's right. :lol:

Question:

Express the definition of the inverse function and the condition for which the inverse function exists?


The minister of fiance has decided that the country should have coins with values 33 and 60. What is the smallest amount that can be payed using these coins? (Both sellers and buyers have unlimited source of change)

A) 1
B) 2
C) 3
D) 6
E) 7

If $g(x) = 1-x^2$ and $f(g(x)) = \frac {1-x^2}{x^2}$ for $x\neq0$, then $f(\frac {1}{2})$ equals:

A) 3/4
B) 1
C) :sqrt: 2
D) 3
E) :sqrt: 2/2

The least integer value of k so that the equation $x(k-x)$ = 4 has no real solution is:

A) -3
B) 3
C) 0
D) -5
E) -4

[hide="Number 7"]60-33=27
33-27=6
Using the 6, we can create any multiple of six.
33-30=3
When adding and subtracting multiples of three, you always get a multiple of three. So 1 and 2 can't be paid.[/hide]

[hide="Number 8"]$g\left(\frac{\sqrt{2}}{2}\right)=\frac{1}{2}$

$f\left(g\left(\frac{\sqrt{2}}{2}\right)\right)=1$[/hide]

[hide="Number 9"]Expanding:
$kx-x^2-4=0$
$x^2-kx+4$
Discriminant:
$k^2-16<0$
$|k|<4$
So -3 is the smallest integer.[/hide]

All right. Very good.

But for number two, you can just find this:

[hide]Since g(x) = 1-x^2, you can figure out what f(x) equals. This should tell you that f(x) = x/(-x+1).  Then putting in there for 1/2, you'll get 1.[/hide]

People, try these problems. It should help you on AMC


algebra
polynomial
combinatorics proposed
combinatorics

I'm working on a project and have a expansion problem and got stuck. Could anyone help me out?

$ %Error. "displaytype" is a bad command.
\prod_{i<j}^{n}(1-x_i x_j )$
where $ x_i^k=x_i$

I expected it would be smtlike this:

$ %Error. "displaytype" is a bad command.
\prod_{i < j}^n (1 - x_i x_j) = 1 - %Error. "displaytype" is a bad command.
\sum_{i < j}^n x_i x_j + 2 %Error. "displaytype" is a bad command.
\sum_{i < j < k}^n x_i x_j x_k - 3 %Error. "displaytype" is a bad command.
\sum_{i < j < k < l}^n x_i x_j x_k x_l + 4 %Error. "displaytype" is a bad command.
\sum_{i < j < k < l < m}^n ... + (-1)^{n-1}(n-1)%Error. "displaytype" is a bad command.
\prod_{i < j}^n  x_i x_j$

Okay, so you meant $ x_i^2 \equal{} x_i$ rather than $ \exists k: x_i^k \equal{} x_i$ (as I understood it first). Then your guess is right, and you can easily prove it using the following lemma:

[color=blue]Let $ R$ be a commutative ring with unity, and let $ P$ and $ Q$ be two polynomials in $ R\left[X_1,X_2,...,X_n\right]$. If $ P\left(x_1,x_2,...,x_n\right) \equal{} Q\left(x_1,x_2,...,x_n\right)$ for every $ \left(x_1,x_2,...,x_n\right)\in\left\{0,1\right\}^n$, then $ P\left(x_1,x_2,...,x_n\right) \equal{} Q\left(x_1,x_2,...,x_n\right)$ for every $ \left(x_1,x_2,...,x_n\right)\in I^n$, where $ I$ is the set of all idempotent elements of $ R$.[/color]

  darij

Could you explain how to go from the lemma to my problem?

P/S: "If $ P\left(x_{1},x_{2},...,x_{n}\right) \equal{} Q\left(x_{1},x_{2},...,x_{n}\right)$ for every $ \left(x_{1},x_{2},...,x_{n}\right)\in\left\{0,1\right\}^{n}$ " [b]is actually what I need to prove[/b].

So you need to solve your problem only in the case $ \left(x_{1},x_{2},...,x_{n}\right)\in\left\{0,1\right\}^{n}$. So let $ I\equal{}\left\{i\in\left\{1,2,...,n\right\}\mid x_i\equal{}1\right\}$. Then, $ \prod_{i<j}^n \left(1\minus{}x_ix_j\right)$ is $ 1$ if $ \left|I\right|<2$ and $ 0$ otherwise. On the other hand, the right hand side of your equation is $ 1\minus{}\binom{\left|I\right|}{2}\plus{}2\binom{\left|I\right|}{3}\minus{}3\binom{\left|I\right|}{4}\pm ...\plus{}\left(\minus{}1\right)^{n\minus{}1}\left(n\minus{}1\right)\binom{\left|I\right|}{n}$. Now you only have to prove a combinatorial identity.

  darij

Turn out that
$ 1\minus{}\binom{\left|I\right|}{2}\plus{}2\binom{\left|I\right|}{3}\minus{}3\binom{\left|I\right|}{4}\pm ...\plus{}\left(\minus{}1\right)^{n\minus{}1}\left(n\minus{}1\right)\binom{\left|I\right|}{n} \equal{} 0$

take me quite long to proof, but I can final my proof. Thanks darij!


Κοιτουσα την παρακατω ασκηση και δεν εχω καταληξει σε τιποτα αξιολογο...
Εστω $ f: (0,\infty) \rightarrow \Re$. Αν ισχυει $ f(xy)=f(x)+f(y)$ για καθε $ x,y \in(0,\infty)$ και $ f(x)$ συνεχης στο 1 τοτε $ f(x)=lnx$.

Μεχρι στιγμης εχω βγαλει τα εξης:[hide]
η f(x) παντου συνεχης (θετοντας στο $ \lim_{x\to\(x_0}\(f(x)$, οπου x to $ x_0h$ ... πραξεις ... $ \lim_{x\to\(x_0}\(f(x)=f(x_0)$)

τις περισσοτερες ιδιοτητες των λογαριθμων με αντικαταστασεις στη σχεση (τυπου για x=1/y, y=1 κλπ...)
και ουσιαστικα ενα βασικο προβλημα που προεκυψε ηταν το εξης: γνωριζοντας μια συναρτ.σχεση τετοιου τυπου και καταληγοντας με αντιστοιχη διαδικασια με την παραπανω να αποδειξεις ολες τις βασικες ιδιοτητες μιας συναρτησης(ιδιως σε θεμελιωδεις οπως η $ lnx$), μπορεις να καταληξεις πως η σχεση περιγραφει την συγκεκριμενη συναρτηση εκμεταλλευομενος τη συνεχεια? η σκεφτηκα μεγαλη βλακεια :blush: ? μπορειτε να δωσετε καποιο αντιπαραδειγμα αν δεν ισχυει κατι τετοιο? δε μου'ρθε κανενα που να αναφερεται και σε συνεχη συναρτηση...
Ευχαριστω ![/hide]

Είναι μια χαρακτηριστική συναρτιασκή σχέση ...Θέτουμε $ x \equal{} e^u$ και $ y \equal{} e^v$ και $ f(e^u) \equal{} g(u)$

Τότε $ g(u \plus{} v) \equal{} g(u) \plus{} g(v)$  ή $ g(x \plus{} y) \equal{} g(x) \plus{} g(y)$  .

Τώρα σε αυτή εύκολα βγάζουμε ότι $ g(x) \equal{} cx$ για κάθε ρητό αριθμό x .

Μετά αποδεικνύουμε ότι η g είναι συνεχής σε όλο το πεδίου ορισμού της (άσκηση για γ λυεκίου  :P )
Τέλος λόγω πυκνότητας του συνόλου Q υπάρχει ακολουθία ρητών $ x_n$ με όριο έναν άρρητο αριθμό x .
Tote $ g(x) \equal{} \lim_{x_n\rightarrow x} g(x_n) \equal{} \lim_{x_n\rightarrow x} cx_n \equal{} cx$

Αρα τελικά $ g(x) \equal{} cx$ για κάθε x του πεδίου ορισμού οποτε μετά βρίσκουμε και την f . :)

[quote="iakovos1729"]πολυ ωραια! :)  δεν πολυκαταλαβα το σημειο που αναφερεσαι στην πυκνοτητα του Q και την υπαρξη οριου. Μπορεις να μου πεις καποιο link ή τιτλο βιβλιου σχετικα με την πυκνοτητα/συμπαγεια συνολου ?
εψαξα σχετικα αλλα δεν βρηκα και πολλα...  thnx[/quote]

[color=red][b] Σε επίπεδο ολυμπιάδων μπορείς να παίρνεις έτοιμο ότι , [/b][/color]

 Αν η $ f$ είναι συνεχής και $ f(x\plus{}y)\equal{}f(x)\plus{}f(y) , \forall x,y \in \mathbb R$ ,  τότε  $ f(x) \equal{} ax$ για κάποιο $ a \in \mathbb  R$.

[quote="silouan"]κ . Μπάμπη συγγνώμη για την αμφισβήτηση αλλά δεν το νομίζω . Από ότι μας είχαν πει φέτος στα μαθήματα προετοιμασίας πρέπει να γίνεται απόδειξη . Η συναρτισιακή του Cauchy δεν συμπεριλαμβάνεται στα well-known  :([/quote]

Μα στην ολυμπιάδα εκείνο που εξετάζεται είναι αυτό που έκανες εσύ στην απόδειξη: Να θεωρήσεις δηλαδή μετασχηματισμό και να πέσεις στην Cauchy.Το να ξαναλύσει ένας διαγωνιζόμενος την Cauchy είναι χάσιμο χρόνου , διότι την ξέρουν και ... οι πέτρες , μια και όλοι τη διδάσκονται.Εξάλλου μπορεί αλλού να συμπεριλαμβάνεται και σε σχολικά βιβλία ως συμπέρασμα.Επίσης στις ΙΜΟ αποφεύγεται όριο  συναρτήσεων(όχι όμως πάντα!) ή ανώτερη ανάλυση.Αν λοιπόν σας είπαν ότι την Cauchy να την αποδεικνύετε κάθε φορά που τη χρησιμοποιείτε , τότε [color=red]να το κάνετε [/color]. Νομίζω όμως ότι αν είναι έτσι , τότε ποτέ δε θα βάλουν άσκηση σε ΙΜΟ αυτού του είδους,με το σκεπτικό ο διαγωνιζόμενος πρέπει και να κάνει τους μετασχηματισμούς και να αποδείξει τη βασική εξίσωση του Cauchy.Είναι κυρίως χρονοβόρο θέμα.
Τελειώνω όμως με το εξής :

[color=red]Αφού στην ΕΜΕ πιστεύουν ότι χρειάζεται και η απόδειξη , τότε να τη μάθετε για τις περιπτώσεις

    α) που η f είναι συνεχής β) που f είναι γνησίως μονότονη[/color]

 Σε ευχαριστώ όμως για την επισήμανση και την παρέμβαση, διότι είναι κρίμα να χαθούν μονάδες από παράλλειψη γνωστών πραγμάτων.
 [i](Πάντως κατανοώ την αγωνία σου διότι και αλλού έχω διαβάσει ότι - κυρίως οι αμερικάνοι - είναι του πολύ μπλα-μπλα στις παρουσιάσεις των λύσεων , αιτιολογώντας ακόμα και σχεδόν προφανή πράγματα.Στις άλλες όμως χώρες αυτό που μετράει είναι η λύση και όχι η πλατεια αιτιολόγηση γνωστών θεωρημάτων. Φαντάσου για παράδειγμα το  θέμα στη συνδυαστική με που λύνονταν με το Nullstellensatz.Δε νομίζω ότι οι  teamleaders των χωρών που οι μαθητές τους το έλυσαν με αυτόν τον τρόπο ότι επέτρεψαν να κοπούν μονάδες , έστω και αν αυτό αποτελεί μια προχωρημένη γνώση στα Combinatorics.Ξέρω όμως τι θα μου απαντήσεις :wink:  και σταματώ εδώ !)[/i]

[quote="silouan"]Θα σας πω την προσωπική μου άποψη . Όταν μια άσκηση μετά από δύο σειρές καταλήγει σε Cauchy προφανώς θα πρέπει να την αποδείξεις . Αν όχι τότε μπορείς να πεις ότι αυτό θεωρείται γνωστό . Αλλά μιλάμε για κάτι που αποδεικνύεται σε 5 λεπτά.
Είναι ρουτίνα καθαρά (εκτός από την περίπτωση που είναι φραγμένη)
Όσο αφορά το Nullstelensatz αν το έπαιρνες χωρίς απόδειξη έπαιρνες ένα μεγαλοπρεπέστατο 2 στην καλύτερη περίπτωση .
Και αυτό λογικό μου ακούγεται αφού αποτελούσε άμεση εφαρμογή ....
Τέλος πάντων ας φτάσουμε στις εξετάσεις σε κάτι γνωστό και μετά βλέπουμε  :D[/quote]

Πολλές φορές η επιτροπή θέλοντας να διασφαλίσει το δικό της κύρος  για κάτι που ίσως της διέφυγε επιλέγοντας ένα θέμα και από την άλλη για να ανταμείψει τους μαθητές που βρήκαν λύση με το ελάχιστο επίπεδο γνώσεων  είναι αρκετή σκληρή σε ορισμένες αποδείξεις που βασίζονται σε πιο προχωρημένες γνώσεις. Με αυτό συμφωνώ και γω και αυτός είναι ο λόγος που προτρέπω τους μαθητές να ασχολούνται με την εξάσκηση του νου και όχι με τη συσσώρευση γνώσεων και αποστήθηση σπάνιων θεωρημάτων.
 Στο αγγλικό φόρουμ μερικές φορές γίνεται υπερβολική αναφορά σε ειδικά θεωρήματα που παρασύρουν πολλούς να νομίζουν ότι δεν έχουν θεωρητικό υπόβαθρο.Αυτό είναι επικίνδυνο. Το ''παραπάνω'' χρειάζεται μόνο για την περίπτωση που θα βρεις μια λύση ή ένα αποτέλεσμα και με βάση αυτό να καταφέρεις να προσαρμόσεις τη λύση σε πιο στοιχειώδη επίπεδα.
 Τέλος πάντων , το συμπέρασμα είναι ένα:

[b] Να προετοιμάζεται κανείς με δύσκολα θέματα που θέλουν πολύ απλές γνώσεις ![/b]


linear algebra
matrix
algebra
polynomial
floor function
linear algebra unsolved

Let $ M \in \mathbb{R}^{n \times n}$ a matrix such that $ M^{3}\minus{}4M^{2}\plus{}2M\plus{}3I\equal{}0$, how many possibles values has de determinant of $ M$? and how many if  $ M \in \mathbb{Q}^{n \times n}$?.

Factor $ x^3\minus{}4x^2\plus{}2x\plus{}3 \equal{} (x\minus{}3)(x^2\minus{}x\minus{}1)$, then use that the minimal polynomial divides this, etc ...

ok, this is to try to figure out what are the possible eigenvalues right? and considering their products you get the possible values of the det(M), but, can you write a detailed solution so i can see it quite well, please? i am not familiar with linear algebra so i will be very happy if somebody post a full detailed solution

Ok, so the possible eigenvalues are $ \phi$, $ \minus{}\frac {1} {\phi}$ and $ 3$ (where $ \phi \equal{} \frac {1 \plus{} \sqrt{5}} {2}$ is the (irrational) golden number, the positive root of $ x^2\minus{}x\minus{}1\equal{} 0$). Therefore
If $ M$ has elements in $ \mathbb{R}$, the Jordan blocks can only be the $ 1 \times 1$ matrices made of these eigenvalues, so in general $ \det M \equal{} \phi^{\alpha}(\minus{}\frac {1} {\phi})^{\beta}3^{\gamma}$, with $ \alpha, \beta, \gamma$ non-negative integers with $ \alpha\plus{}\beta\plus{}\gamma\equal{}n$, so there are in all $ \frac {(n\plus{}1)(n\plus{}2)} {2}$ distinct values possible.
If $ M$ has elements in $ \mathbb{Q}$, the irrational eigenvalues have to come in (conjugate) pairs, in $ 2 \times 2$ Jordan blocks of determinant $ \minus{}1$, while $ 3$ appears as a $ 1 \times 1$ Jordan block, so in general $ \det M \equal{} (\minus{}1)^{\alpha}3^{\gamma}$, with $ \alpha, \gamma$ non-negative integers with $ 2\alpha \plus{} \gamma \equal{} n$, so there are in all $ \left \lfloor \frac {n\plus{}2} {2} \right \rfloor$ distinct values possible.
(Notice that such $ 2 \times 2$ Jordan blocks do exist: rows $ (0,1)$ and $ (1,1)$.)


geometry
3D geometry
similar triangles
Pythagorean Theorem

I'm not positive about this, but it seems to work...

[hide="solution"]
By using similar triangles, we can conclude that the height of the original cone would have been 36 cm. We want to find the slant height of the cone. We use pythagorean theorem to find that the slant height, which is the radius of the desired circle, is $ \sqrt {36^2 \plus{} 15^2} \equal{} 39 cm$ (It's a multiple of {5,12,13}). The diameter of this circle on the plane is then $ 78 cm$, or $ (D)$
[/hide]

EDIT:sorry I can't describe what I"m saying with a picture, but try to draw it out if you don't know what i"m saying.

EDITED again for clarity.

[quote="xxxxx"]The frustrum of a right circular cone has an altitude of 12 cm. The diameter of one base is 20cm, while the diameter of the other base is 30cm. When rolled on its side on a place, the larger base describes a circle on this plane. What is the diameter of the circle in cm?

a. 36
b. 39
c. 72
d. 78
e. 49[/quote]

[hide="Solution"]

The diameter is $ D \equal{} 2 s$, where $ s$ is the slant height of the full cone.

The radii of the two bases form like sides of similar right triangles. Let $ \ell$ be the remaining length to form the altitude of the full cone. Therefore,

\[ \frac{10}{\ell} \equal{} \frac{15}{12 \plus{} \ell} \]

\[ \implies 120 \plus{} 10\ell \equal{} 15\ell \]

\[ \implies 5\ell \equal{} 120 \]

\[ \implies \ell \equal{} 24\]

\[ s \equal{}\sqrt{\left(12 \plus{}\ell\right)^2 \plus{} \left(15\right)^2} \equal{}\sqrt{36^2 \plus{}15^2} \equal{}\sqrt{1521} \equal{} 39\]

\[ \boxed{\therefore D \equal{} 2 s \equal{} 2\cdot 39 \equal{} 78}\]

[/hide]


Consider the following system of equations:

$a+8b+27c+64d=1$
$8a+27b+64c+125d=27$
$27a+64b+125c+216d=125$
$64a+125b+216c+343d=343$

These four equations determine $a, b, c, d$. Show that

$a(x+1)^{3}+b(x+2)^{3}+c(x+3)^{3}+d(x+4)^{3}$ is identical to $(2x+1)^{3}$. Use this to show that $a+b+c+d=8$ and $64a+27b+8c+d=729$.

I changed some of the variables for easy typing's sake.

hello, solving this system we get $ a\equal{}\frac{35}{2},b\equal{}\minus{}\frac{35}{2},c\equal{}\frac{21}{2},d\equal{}\minus{}\frac{5}{2}$
and it follows $ \frac{35}{2}\minus{}\frac{35}{2}\plus{}\frac{21}{2}\minus{}\frac{5}{2}\equal{}8$,  $ 64\cdot\frac{35}{2}\plus{}27\cdot\left(\minus{}\frac{35}{2}\right)\plus{}8\cdot\frac{21}{2}\minus{}\frac{5}{2}\equal{}729$ and $ \frac{735}{2}(x\plus{}1)^3\minus{}\frac{35}{2}(x\plus{}2)^3\plus{}\frac{210}{2}(x\plus{}3)^3\minus{}\frac{5}{2}(x\plus{}4)^3\equal{}(2x\plus{}1)^3$.
Sonnhard.

[hide="Without using Maple"]
Consider the polynomial $ p(x) \equal{} a(x \plus{} 1)^3 \plus{} b(x \plus{} 2)^3 \plus{} c(x \plus{} 3)^3 \plus{} d(x \plus{} 4)^3 \minus{} (2x \plus{} 1)^3$. Notice that the degree of $ p(x)$ is at most 3. However, by the system of equations in the problem, $ p(x)$ has at least 4 roots, which are at $ x \equal{} 0,1,2,3$ respectively. We conclude then that $ p(x) \equal{} 0$. The first result follows.
Then from $ a(x\plus{}1)^3\plus{}b(x\plus{}2)^3\plus{}c(x\plus{}3)^3\plus{}d(x\plus{}4)^3\equal{}(2x\plus{}1)^3$, the coefficient of $ x^3$ on the LHS is $ a\plus{}b\plus{}c\plus{}d$ while the coefficient of $ x^3$ on the RHS is $ 8$. The second result follows.
Also, $ p( \minus{} 5) \equal{} \minus{} (64a \plus{} 27b \plus{} 8c \plus{} d) \equal{} \minus{} 9^3 \equal{} \minus{} 729$, hence the third result follows.[/hide]


I don't know I just get stuck on this system of equations please help me on it
$ x^2\minus{}4xy\plus{}y^2\equal{}13$,  $ x\plus{}y\equal{}1$

Thanking you for your help  :)

[hide="solution"]Taking the second equation, $ x \equal{} 1 \minus{} y$, so substitute that into the first.

Therefore,
\[ (1 \minus{} y)^{2} \minus{} 4(1 \minus{} y)(y) \plus{} y^{2} \equal{} 13\]

\[ y^{2} \minus{} 2y \plus{} 1 \plus{} 4y^{2} \minus{} 4y \plus{} y^2 \equal{} 13\]

\[ 6y^{2} \minus{} 6y \minus{} 12 \equal{} 0\]

\[ y^{2} \minus{} y \minus{} 2 \equal{} 0\]

\[ (y \minus{} 2)(y \plus{} 1) \equal{} 0\]
So $ y\equal{}2, \minus{}1$. Putting that into the first equation, the two solutions are $ ( \minus{} 1,2)$ and $ (2, \minus{} 1)$[/hide]

$ x^2 \minus{} 4xy \plus{} y^2 \equal{} 13$

$ (x\plus{}y)^2 \minus{} 6xy \equal{} 13$

$ 1 \minus{} 6xy \equal{} 13$

$ xy \equal{} \minus{}2$

$ x \equal{} \minus{}\frac{2}{y}$
Now plug back into $ x\plus{}y \equal{} 1$.

$ \minus{}\frac{2}{y} \plus{} y \equal{} 1$

$ \minus{}2 \plus{} y^2 \equal{} y$

$ y^2 \minus{} y \minus{} 2 \equal{} 0$

$ (y\minus{}2)(y\plus{}1) \equal{} 0$

$ y \equal{} 2$

$ y \equal{} \minus{}1$

Plugging into any equation, we get $ x \equal{} \minus{}1$ and $ x \equal{} 2$. Therefore the solutions are $ \boxed{(\minus{}1,2)}$ and $ \boxed{(2,\minus{}1)}$.


quadratics
algebra
polynomial
sum of roots

[quote="Elemennop"]A very useful technique is if you are ever expected to factor a fourth degree polynomial, and you can't pull any linear terms out of it, expand the expression $(x^2+ax+b)(x^2+cx+d)$, and plug in integer values for $bd$ (the constant term) that'll work for that term, and see if you can get solutions to fit all the coefficients...It's a fairly useful tactic if you are absolutely lost on something like this.[/quote]

Nice!

[quote="Elemennop"]A very useful technique is if you are ever expected to factor a fourth degree polynomial, and you can't pull any linear terms out of it, expand the expression $(x^2+ax+b)(x^2+cx+d)$, and plug in integer values for $bd$ (the constant term) that'll work for that term, and see if you can get solutions to fit all the coefficients...It's a fairly useful tactic if you are absolutely lost on something like this.[/quote]
Wow. Awesome.

People need to post tactics like this more often.

did you come up with that by yourself, or is there a nice book somewhere?

Well  usually  like in cubic polynomials, w give the equation as $(x-a)(x-b)(x-c)=....$ and then equating the coefficents we can find many things like sum of roots,etc.Similarly  its done for a polynomial of degree $4$.

okay, so an application would be (on this problem):
first, negate equation:
$x^4-5x^2-6x-5=0 \rightarrow -x^4+5x^2+6x+5=0$
$bd$ = 5
WLOG, $b=1, d=5$
$da + bc = 6$ (x terms)
$\therefore 5a+c=6$ and $c+a=0$ (from x^3 term)
so $c=1.5$, $a=-1.5$
this gives:
$-(x^2-1.5x+1)(x^2+1.5x+5)=0$
so $(x^2-1.5x+1)(x^2+1.5x+5)=0$
what went wrong/?

[quote="Elemennop"]A very useful technique is if you are ever expected to factor a fourth degree polynomial, and you can't pull any linear terms out of it, expand the expression $(x^2+ax+b)(x^2+cx+d)$, and plug in integer values for $bd$ (the constant term) that'll work for that term, and see if you can get solutions to fit all the coefficients...It's a fairly useful tactic if you are absolutely lost on something like this.[/quote]

Actually you can use $(x^2+ax+b)(x^2-ax+d)$  (IOW You can assume c=-a) as long as the coefficient of $x^3$ is zero. (It's a standard method, of course)
Hope that helps.

okay, I think I see it. the way I wrote it ($x^4-5x^2-6x-5=0 \rightarrow -x^4+5x^2+6x+5=0$)
we have:
$(x^2+ax+b)(-x^2-ax+d)$
with $d-b-a^2=5$, $a(d-b)=6$, and $bd=5$
so we can solve these:
$bd=d-b-a^2, bd-d+b=-a^2 \rightarrow 5-6/a=-a^2$
$\therefore a^3+5a-6=0$ now on this equation we can apply the tricks again (it has no a^2 term):
$(a^2+la+m)(na-ln)$ anyway it kind of works in theory...

[quote="Gyan"]
Actually you can use $(x^2+ax+b)(x^2-ax+d)$  (IOW You can assume c=-a) as long as the coefficient of $x^3$ is zero. (It's a standard method, of course)
Hope that helps.[/quote]

Yes that's true.Did you get that from "H.K"   mr.Gyan?

[quote="Gyan"]Don't know what is "HK". The method, I believe , is called Descartes' Method.[/quote]

Yes,that's what is given here too in "hK"(hall and..., a book)


The difference between the squares of two positive integers is 244 and their positive difference is as small as possible. What is the value of their positive difference?

[hide]2[/hide]

Well..... the positive difference cant be one cause the difference between sqaures of consecutive integers is always odd, soooooo
(x + 2)^2 - x^2 = 244...
x^2 + 4x + 4 - x^2 = 244
4x = 240
x = 60...
This works out, b/c 62^2  = 3844 and 60^2 = 3600, and 3844 - 3600 = 244... you don't really need to check it, but there you go...

[quote="ZhangPeijin"]Please read the fourm guidelines.
If you have multiple questions POST THEM ALL IN ONE ENTRY, not 7 entries with 1 question each.
Thank you.[/quote]

1) This is a PM matter.  Actually, this is as well, but w/e.  You do not need to make this a public affair.
2) I have not seen them in the forum guidelines, and I just read them.

just so this isn't spam, here's another way to look at it

[hide]
We can express a perfect square $ n^2$ as $ 1\plus{}3\plus{}5\plus{}...\plus{}(2n\minus{}1)$.  If you play around a bit, you should find that if you express the difference of two perfect squares as the sum of the fewest possible number of odd numbers, then the number of odd numbers used is the least possible difference between the square roots of the two perfect squares.  well, i think so...


Prove that if a,b are positive real numbers then

[(a^1+b^1)^2][(a^2+b^2)^2]......[(a^n+b^n)^2] \geq [a^(n+1)+b^(n+1)]^n

This is very simple and I have a very simple solution.

Two solutions:
   First Solution:
  Induction. It reduces to proving that (a^n+b^n)^(n+1)>=(a^n+1+b^n+1)^n, whch is easy.
  Second solution:
  Use the fact that (a+b)^2(a^n+b^n)^2, (a^2+b^2)^2(a^n-1+b^n-1)^2,...>=(a^n+1+b^n+1)^2,


vector
linear algebra
linear algebra unsolved

For a given positive integer n>1, show that there exist subspaces X1, X2, ......., Xn of R^m for some integer m>n and a linear transformation
T: R^m → R^m such that,
i) dim X_k =k, k=1,2,.......,n.
ii) for i≠j, Xi ∩ Xj ={0}, where 0 is the zero vector of R^m,
iii) T(X_k)= X_(k-1), k=1,2,........,n, where X0={0}.
Also find Rank of T.

You could for example take $ m \equal{} n^2$, with a basis $ \{e_{i,j}: 1\leqslant i,j\leqslant n\}$ for R^m. Let $ X_k$ be the subspace spanned by $ \{e_{i,j}: i\plus{}j\equal{}k\plus{}1\}$, and define $ T$ by $ Te_{i,j}\equal{}\begin{cases}e_{i,j\minus{}1}&(j>1),\\0&(j\equal{}1).\end{cases}$

can you explain little elaborately sir please?


AMC
AMC 12
AMC 12 A
AIME
Asymptote
LaTeX
\/closed

The AMC and AIME pdf's all have problems... whoever made them was sloppy!

[quote="pleurestique"]still no response, huh? careless people are a dime a dozen around here.[/quote]
Dude, they are being provided for free and are autogenerated...

The generator seems to have issues with asymptote diagrams and some LaTeX stuff. Be patient. I've already talked to an admin about it.

Here's how you can help:

1. Learn Asymptote.

2. Do us all a favor, regenerate the images and send them to an admin for placement in the new PDF files.

[quote="pleurestique"]oh i'm patient... the problems can be viewed pretty clearly in html form... but what's the point of a pdf file if it's sloppy? like it or not, that's what the AMC and AIME files have been for the last couple of years. and nothing is totally auto-generated... down the line, some incompetent person is responsible ;)[/quote]

Bugs happen. Just be patient and wait for the admins to fix it. If you think it's sloppy, don't use it. As far as I know the use of Asymptote specifically in diagrams is relatively new (only having been added to the site a few months ago, and I seem to be the only one using it), so we're just now getting into the first competition season with Asymptote being used for diagrams.

Seriously, patience is key. Maintaining a piece of software as big as this one can't be easy. The bug is in the software, not someone who just randomly did a careless job of making a file.

The Contests section works like this:
1. Someone posts a problem on the forums. If that problem needs a diagram, they have four options: link to an external site, attach it to the post, create it in LaTeX's picture environment, or use Asymptote. Linking to an external site creates problems when the files are inevitably taken down, attachments don't show up in the contests section, and the picture environment is very limited, so Asymptote is generally the best bet.
2. An RManager takes the problem and puts it into the Contests section - each problem in there is linked to one post on the forums, as is each solution.
3. A PDF file is autogenerated for that contest on those dates.

That's basically it. Actually the asymptote images will be added to the resource section coding at some point during the next couple of months. This will work retroactively: ie all the problems that have currently asymptote code in them will be displayed nicely on the PDF as well.

I recommend you start using asymptote rather than any of the other three methods described above.


I read in a book that Field inside a conductor is zero only under "ELECTROSTATIC CONDITIONS". Anybody care to explain what that means.. I lost mark in some part test cause of this. :(

that means if and only if there is no current inside the conductor the electric field is 0
thsi follows from the mere definition of electrostatics ...however even if there are currents but if they are steady that is $ (\frac {\delta \rho(\vec r,t)}{\delta t})_{\vec r} \equal{} 0$ then u have $ \nabla.\vec E \equal{} 0$
but not $ \vec E \equal{} 0$ necessarily

The divergence of the electric field is zero in the presence of a steady current (that would include electrostatic condition as well). This follows from the continuity equation.
Sathej

yes that's what i wrote ....and yes this equation obviously follows for electrostatic cond.s where $ \vec E$ is always 0 or as sathej pointe out the 'continuity' equation


My parents won't let me go (>_<) because they think it's too far away and I'm only 13.

>_<
>_<
>_<
>_<

Dr.Kevin any help convincing them?

What is your location?

Our youngest campers were in 5th grade, 10 or 11 years old.  Thirteen-year-olds are common.  Our staff are responsible for airport pick-up, all-time supervision, and airport send-off.  As mentioned in another thread, as a 13-year-old you can usually travel by yourself, but you can also travel as an unaccompanied minor and let the airline take care of you all the way.  We also keep close contact with parents about how the students are doing during the camp.  Our staff enforce the daily schedules (get up, breakfast, classes, recreation time, study time, bed time).

If parents worry about the academics: I pay personal attention to each camper and work with them to get the best out of the camp.  All students report a big leap forward in their math and problem solving skills.

That being said, I want to emphasize that your parents know what's best for you.  So if they feel that you are not ready, I would respect their decision.  We will be happy to have you at the camp, if not this year, then next year is also great.


calculus
integration
trigonometry
function
calculus computations

The integral of (sinx)^n from 0 to pi/2 with respect to x is equal to The integral of (cosx)^n from 0 to pi/2 with respect to x

Legendary symmetric trick:

$ \varphi \equal{} \int_0^{\pi /2} {\sin ^n x\,dx} .$

Define $ u \equal{} \frac {\pi }{2} \minus{} x,$

$ \varphi \equal{} \int_0^{\pi /2} {\sin ^n x\,dx} \equal{} \int_0^{\pi /2} {\cos ^n u\,du} \equal{} \int_0^{\pi /2} {\cos ^n x\,dx} .$

Ohhh... when posting I hadn't noticed mlok's post.

$ \int_0^{\frac{\pi}2}\sin x\,dx\equal{}\int_0^{\frac{\pi}2}\cos x\,dx\equal{}1.$ (That's easy.)

Since $ \int_0^{\frac{\pi}2}\cos^2x\,dx\equal{}\int_0^{\frac{\pi}2}\sin^2x\,dx$ but

$ \int_0^{\frac{\pi}2}\cos^2x\,dx\plus{}\int_0^{\frac{\pi}2}\sin^2x\,dx\equal{}
\int_0^{\frac{\pi}2}1\,dx\equal{}\frac{\pi}2,$ then

$ \int_0^{\frac{\pi}2}\cos^2x\,dx\equal{}\int_0^{\frac{\pi}2}\sin^2x\,dx\equal{}\frac{\pi}4.$

In my calculus class, I call this the "yin-yang lemma" (draw the picture!) and quote the result as "half the box." In many, many problems involving squares of sines or cosines integrated over a whole number of quadrants, this is a great time-saver.

Now: show that

For $ n$ even, $ \int_0^{\frac{\pi}2}\cos^nx\,dx\equal{}\int_0^{\frac{\pi}2}\sin^nx\,dx\equal{}
\frac{(n\minus{}1)(n\minus{}3)\cdots3\cdot1\cdot\pi}{n(n\minus{}2)\cdots4\cdot2\cdot2}.$

For $ n$ odd,$ \int_0^{\frac{\pi}2}\cos^nx\,dx\equal{}\int_0^{\frac{\pi}2}\sin^nx\,dx\equal{}
\frac{(n\minus{}1)(n\minus{}3)\cdots4\cdot2}{n(n\minus{}2)\cdots3\cdot 1}.$

These latter formulas are the result of recursion using "reduction formulas" for the trigonometric power integrals. The appropriate reduction formulas can be proved by a combination of integration by parts and algebra.

(And yes, there are other arguments using double integrals and Gaussians that allow us to use fractional powers and express things in terms of the gamma function, but I was trying to keep this post at a Calculus II/Calculus BC level.)

In general, $ \int_0^{\pi/2}f(\sin x)\,dx \equal{} \int_0^{\pi/2}f(\cos x)\,dx$.


Hola!

Debo decir que no se español muy bien. Perdon. Pero intentaré escribir un poquito de español... Ojalá yo aprenda a hablar y escribir español más correctamente.

Creo que seria de mucha utilidad para todos colgar links para sites con materiales para olimpiadas. Por ejemplo, todos los problemas de las últimas Olimpiadas Brasileñas de Matematicas estan (en Portugues) en el site oficial de la OBM: [url=http://www.obm.org.br/]http://www.obm.org.br/[/url]

En el site de la OBM hay tambien muchos artículos y todas las ediciones de la Revista Eureka!, que tienen más artículos y competencias de Matematicas.

Otro site muy bueno que conosco es de la Olimpiada Argentina (OMA): [url=http://www.oma.org.ar/]http://www.oma.org.ar/[/url]

Ah, y debo colgar los problemas de la OBM en portugues para ustedes?

Buenas noches Prof. Shine...

Seria muy bueno .. que usted sea nuestro contacto en Brasil
con lo que respecta a todas las actividades de la sociedad Brasileira de Matematica
con lo que respecta a las olimpiadas, los proceso de seleccion, los test de seleccion, las semanas olimpicas,
las publicaciones de la revista EUREka, las listas de entrenamiento en teorema, etc.

en http://www.obm.org.br, podemos encontrar la informacion ..
pero me parece que no muy actualizado esta brecha podria cubrirlo ??? asi tendriamos
informacion actual  ....

Carlos Bravo
Lima-PERU

Bueno, para continuara a lo que sugiere el Profesor Shine!!

Aqui cuelgo algunas web sites que seria bueno que revisen los interesados...

Web Oficial de las Olimpiadas Iberoamericanas de Matematica:
http://www.campus-oei.org/oim/

Web de Portugal:
http://www.mat.uc.pt/~opm/

Web de España:
http://platea.pntic.mec.es/~csanchez/olimmain.htm

Web de Peru:
http://www.pucp.edu.pe/~matemat/olimpiadas/

Y una web que nos podria servir de catalogo de referencias es
la pagina mas popular de las IMOs:
http://imo.math.ca/

Carlos Bravo :)
Lima-PERU

La pagina de Colombia es:

http://olimpia.uanarino.edu.co/default.asp

JAS


Let $ a,$ $ b$ and $ c$ are non-negative numbers and
$ u=\frac{a+b+c}{3},$ $ v=\sqrt{\frac{a^{2}+b^{2}+c^{2}}{3}},$ $ w=\sqrt[3]{abc}.$
Prove that $ 9v+2w\geq11u.$
See also here:
http://www.artofproblemsolving.com/Forum/viewtopic.php?t=156224


What is the rating for where you live? How the darkest you have seen?
Reference [url=http://www.mathlinks.ro/viewtopic.php?t=157030]here[/url].
The scale reading for where I live is about $ 6$ or $ 7$, thanks to New York City. :mad:
As for the darkest skies I have seen, it is about 5 or even 4.5 on some nights.  New York City still pollutes the sky there even though it is about an hour away.

$ \text{EDITTED EDIT}^{2}$  :|  :huh: Poll Messed Up.

I think Lake Tahoe is about 4 (maybe less?), and where I live is about 5.

Cool.  :)

[url=http://www.cleardarksky.com/lp/LkThNvSPNVlp.html?Mn=astronomical]ClearDarkSky[/url] says Lake Tahoe is about a $ 4.5$. though right around it quickly goes to $ 1$.
[img]http://www.cleardarksky.com/c/LkThNvSPNVcs0.gif[/img].

After checking the map, I'm pretty sure I live in ~4.75. :)

Same here AstroPhys, I have a 6 or 7 with Dallas.

The lowest is for me was Camp Constantin,Possum Kingdom Lake, Tx.

Sorry about the poll. :oops:
Tried to edit it and now it is messed up. I put a post in the Support forums hopefully someone will tell me how to fix it.

hunter34 , I guess there is someone else that has a horribly polluted sky. :wink: Even if you have a big scope in a polluted area it does not really help to see object with lower magnitude.

Are there really places with $ 1$ on the scale? Even if there are, Mauna Kea and Cerro Paranal must be the only ones.  :D

According to the light pollution map, there are places with a Bortle Scale reading of $ 1$.


pigeonhole principle
calculus
calculus computations

For fixed $ x,y\in R$ such that $ \frac{x}{y}$ is irrational, prove or disprove that for any real number $ r,\epsilon$, we can find $ a\in N,b\in Z$, such that $ F(a,b)\equal{}|ax\plus{}by\minus{}r|<\epsilon$

We may assume $ y\equal{}1$; otherwise divide by $ y$. The statement becomes: as $ a$ runs through positive integers, the fractional parts of $ ax$ are dense in $ [0,1)$. Which is true.

[quote="mlok"]We may assume $ y \equal{} 1$; otherwise divide by $ y$. The statement becomes: as $ a$ runs through positive integers, the fractional parts of $ ax$ are dense in $ [0,1)$. Which is true.[/quote]

Yes!! I too thought something like that, but how do you prove they are dense? I think that was the intent of the question  :maybe:

By the pigeonhole principle. Divide $ [0,1)$ into $ N$ equal subintervals. One of them will contain at least two fractional parts, say, $ \{a_1x\}$ and $ \{a_2x\}$. The sequence $ \{k|a_1 \minus{} a_2|x\}$ has an element within each subinterval of length $ 1/N$.

I think this is what made pigeonholing a celebrity. [url=http://planetmath.org/encyclopedia/DirichletsApproximationTheorem.html]Dirichlet[/url]


limit
logarithms
trigonometry
calculus
calculus computations

find $ \lim_{x\minus{}>1} \frac{ln(x\minus{}1)\plus{}tan{\pi x/2}}{ctan{\pi x}}$

Because of the logarithm, this only makes sense as a right-hand limit.

I'll also assume that denominator is supposed to be the cotangent. So the problem should be

$ \lim_{x\to1^\plus{}}\frac{\ln(x\minus{}1)\plus{}\tan\left(\frac{\pi x}2\right)}{\cot(\pi x)}$

In a neighborhood of $ 1,$ $ \tan\left(\frac{\pi x}2\right)$ looks like $ \frac{\minus{}2}{\pi(x\minus{}1)}$ and $ \cot(\pi x)$ looks like $ \frac{1}{\pi(x\minus{}1)}.$ The logarithm is insignificant against that rate of growth. Hence the limit is $ \minus{}2.$

Use the l'Hospital's Rule:

${ lim_{x\rightarrow1}\frac{(log(x-1)+tan(\frac{\pi x}{2}))^{'}}{(cot(\pi x))^{'}}=lim_{x\rightarrow1}\left(\frac{sin(\pi x)^2}{\pi (1-x)}-\frac{sin(\pi x)^2}{1+cos(\pi x)}\right)=}$
$ =\frac{0}{\infty}-lim_{x\rightarrow1}\left(\frac{sin(\pi x)^2}{1+cos(\pi x)}\right)$

use the l'Hospital's Rule again:

$ -lim_{x\rightarrow1}\left(\frac{sin(\pi x)^2}{1+cos(\pi x)}\right)=-lim_{x\rightarrow1}\left(\frac{2\pi cos(\pi x) sin(\pi x)}{- \pi sin(\pi x)}\right)=2lim_{x\rightarrow1}{cos(\pi x)}=-2$

[quote="Hong Quy"][color=darkred]Find[/color] $ L\equiv\lim_{x\searrow 1} \frac {\ln\ (x \minus{} 1) \plus{} \tan\ \frac {\pi x}{2}}{\cot\ (\pi x)}$[/quote]

[color=darkblue][b][u]Proof.[/u][/b] I"ll use the substitution $ t \equal{} x \minus{} 1$, i.e. $ x \equal{} t \plus{} 1$ . Thus,

$ L \equal{} \lim_{t\searrow 0}\frac {\ln t \plus{} \tan\left[\frac {\pi}{2}\cdot (t \plus{} 1)\right]}{\cot \left[\pi (t \plus{} 1)\right]} \equal{}$

$ \lim_{t\searrow}\frac {\ln t \minus{} \cot\frac {\pi t}{2}}{\cot (\pi t)} \equal{}$ $ \lim_{t\searrow 0}\left[\ln t\cdot\tan (\pi t) \minus{} \frac {\tan (\pi t)}{\tan\frac {\pi t}{2}}\right] \equal{}$

$ \lim_{t\searrow 0}\left[t\ln t\cdot\frac {\tan (\pi t)}{t} \minus{} \frac {\tan (\pi t)}{\pi t}\cdot\frac {\frac {\pi t}{2}}{\tan\frac {\pi t}{2}}\cdot\frac {\pi t}{\frac {\pi t}{2}}\right] \equal{}$

$ 0\cdot\pi \minus{} 1\cdot 1\cdot 2\implies$ $ L \equal{} \minus{} 2\ .$

I used only the remarkable limits $ \lim_{x\searrow 0}t\ln t \equal{} 0$ and $ \lim_{x\to 0}\frac {\tan x}{x} \equal{} 1\ .$

[b]Remark.[/b] Using the substitution $ t\equal{}\frac 1x$, i.e. $ x\equal{}\frac 1t$ obtain $ \lim_{x\searrow 0}x\ln x\equal{}\minus{}\lim_{t\to\infty}\frac {\ln t}{t}\equal{}0\ .$[/color]


abstract algebra
group theory
superior algebra
superior algebra unsolved

[quote="Portland"]I am looking for an abelian infinite group which has two subgroups of order $ n$ and $ m$ ($ n$ and $ m$ being coprime), but no subgroup of order $ nm$.[/quote]

The subgroup generated by those two subgroups has order nm.  Whether or not those subgroups are contained in an *infinite* abelian group makes no difference, only the abelian part is relevant.

if $ H$ and $ K$ are subgroups of an abelian group, then $ HK$ is always a subgroup too: for $ x,y\in H, z,w\in K$, (xz)(yw)=(xy)(zw)\in $ HK$. if $ |H|\equal{}m$ and $ |K|\equal{}n$ are coprime, then $ H\cap K$ must be trivial (by Lagrange's Theorem). so if $ x,y\in H, z,w\in K$, if $ xz\equal{}yw$, then $ y^{\minus{}1}x\equal{}wz^{\minus{}1}$, but $ y^{\minus{}1}x\in H, wz^{\minus{}1}\in K$, so we must have $ y^{\minus{}1}x\equal{}wz^{\minus{}1}\equal{}e$, i.e. $ x\equal{}y,w\equal{}z$. this shows that the $ mn$ products $ xz$, where $ x\in H, z\in K$, are distinct, and thus $ |HK|\equal{}mn$.

Thanks, that explains why I couldn't find such a group :oops:


N is the largest integer such that the decimal expansions of N and 7N have 100 digits each.

What is the digit in the 50th place( counting from the left) of N?

  An error has been corrected (read 7N not &N ). My apologies for typiong error

[quote="Michael Niland"]N and [b]&N [/b][/quote]
what is &N ? Is-it $8*N$ ?

If yes this is $9$ (50th digit of $10^{100}/8 - 1$)


Hmm, isn't it just $ \sqrt[10]{256} \implies \sqrt[10]{2^8} \implies \sqrt[5]{16}$? Or am I missing something?

I think you are right. I don't know if it can be more simplified but I posted a strange solution that I'm sure it's wrong. It goes like this
$ \sqrt[10]{2^8} \implies 0.8^2$  :|

[quote="hasan4444"]I think you are right. I don't know if it can be more simplified but I posted a strange solution that I'm sure it's wrong. It goes like this
$ \sqrt [10]{2^8} \implies 0.8^2$  :|[/quote]

$ \sqrt [10]{1} \equal{} 1$

$ \sqrt [10]{1\plus{}} > 1$

You could use the Binomial Theorem if you really wanted to...

$ (1\plus{}1)^{\frac{4}{5}}\equal{}\sum_{k\equal{}0}^{\infty}\binom{4/5}{k}\equal{}1\plus{}\frac{4/5}{1}\plus{}\frac{4/5(\minus{}1/5)}{2}\plus{}\frac{4/5(\minus{}1/5)(\minus{}6/5)}{6}\plus{}\cdots$

$ \approx 1.75$

Which is pretty close to the real value, which is slightly above $ 1.74$

It is most definitely not equal to $ (0.8)^2\equal{}0.64$. In fact, there is no way that it's rational. grn_turtle has a good approximation. (Hassan, I believe you meant to say $ 2^{0.8}$  :wink:

Here's what I would have done.

$ \sqrt[10]{256}\equal{}\sqrt[10]{2^8}\equal{}2^{\frac{8}{10}}\equal{}2^{\frac{4}{5}}\equal{}\sqrt[5]{16}$, which is what isabella has.

I know it's wrong but I state it just to let you know about the answer I saw for this problem.

Thanks a lot  :)

But this isn't simplified, because there's a radical in the denominator.

Uh, I believe the answer here is irrational. You're not going to get any nice answer here. The best you can do, I think, is manipulate the square roots, or use an approximation like I did.

[quote="boohsunil"]$ \sqrt [10](256)\implies\sqrt [10]{2^8}\implies\sqrt [2]2\implies\ 1.414\cdots$[/quote]Nope, that's incorrect twice. First, that's not how you simplify radicals. Second, that's not how you use implication signs; you should use equality signs.

$ \sqrt [10](256)\equal{}\sqrt [10]{2^8}\equal{}\sqrt [5]{2^4}$ and that's as much as you can simplify, as noted by previous AoPSers.


1> Prove that x^4 + 4 is prime for only one x in N.

2> If in a triangle ABC, AD, the median from A is perpendicular to AB, find the numerical value of tan A + 2 tan B [/hide]

[hide="#1"]
$ x^4\plus{}4\equal{}x^4\plus{}4x^2\plus{}4\minus{}4x^2\equal{}(x^2\plus{}2)^2\minus{}(2x)^2\equal{}(x^2\plus{}2x\plus{}2)(x^2\minus{}2x\plus{}2)$
If this number is prime, then as $ x^2\plus{}2x\plus{}2>x^2\minus{}2x\plus{}2$, it must be $ x^2\minus{}2x\plus{}2\equal{}1$. That gives us: $ x^2\minus{}2x\plus{}1\equal{}0$ which implies $ x\equal{}1$. Checking, we get $ 1^4\plus{}4\equal{}5$ which is prime. [/hide]


linear algebra
matrix
algebra
polynomial

Let $ A,B,C\in{M_2(R)}$ so that $ AB\equal{}BA, AC\equal{}CA, BC\equal{}CB$ and $ det(AB\plus{}BC\plus{}CA)<\equal{}0$. Prove that $ det(A^2\plus{}B^2\plus{}C^2)>\equal{}0$.

Consider the polynomial  $ P(X) \equal{} \det(A^2 \plus{} B^2 \plus{} C^2 \plus{} X(AB \plus{} BC \plus{} CA)) \equal{} aX^2 \plus{} bX \plus{} c$ , where $ a \equal{} \det(AB \plus{} BC \plus{} CA)$ and $ c \equal{} \det(A^2 \plus{} B^2 \plus{} C^2)$ .Try to prove that $ P(\minus{}1) \geq 0$ and $ P(2)\geq 0$.These two conditions are equivalent to :

$ a\minus{}b\plus{}c\geq 0$
$ 4a\plus{}2b\plus{}c\geq 0$   .Multiplying the first one with 2 and adding to the second one we get :
$ c\geq \minus{}2a \geq 0$ .The desired result .

I can not explain why $ P(\minus{}1)\geq 0$. Please help me to explain....

[quote="igochess"]I can not explain why $ P( \minus{} 1)\geq 0$. Please help me to explain....[/quote]
Using the identity $ (A \plus{} \epsilon B \plus{} \overline{\epsilon}C)(A \plus{} \overline{\epsilon} B \plus{} \epsilon C) \equal{} A^2 \plus{} B^2 \plus{} C^2 \minus{} AB \minus{} BC \minus{} CA$ we get that
$ \det(A^2 \plus{} B^2 \plus{} C^2 \minus{} AB \minus{} BC \minus{} CA) \equal{} \det (A \plus{} \epsilon B \plus{} \overline{\epsilon}C)\cdot \overline{\det (A \plus{} \epsilon B \plus{} \overline{\epsilon} C)} \equal{} |\det(A \plus{} \epsilon B \plus{} \overline{\epsilon} C)|^2\geq 0$

where $ \epsilon$ is the root of the equation $ \epsilon ^2 \plus{} \epsilon \plus{} 1 \equal{} 0$ .

Another solution is the following :

$ A^2\plus{}B^2\plus{}C^2\plus{}AB\plus{}BC\plus{}CA\equal{}A^2\minus{}A(B\plus{}C)\plus{}\left(\frac{B\plus{}C}{2}\right)^2\minus{}\left(\frac{B\plus{}C}{2}\right)^2 \plus{}B^2\plus{}C^2\minus{}BC\equal{}\left(A\minus{}\frac{B\plus{}C}{2} \right)^2 \plus{} \left( \frac{\sqrt{3}}{2} (B\minus{}C)\right)^2$
 and now just use this http://www.mathlinks.ro/viewtopic.php?t=174477


What do you folks think is the best way to define and use lemmas in proofs (for olympiads)? Should they be placed before or after the main bulk of the proof? How should they be labelled (Lemma 1, Lemma 2, etc..?) ?

I usually used them in the middle of the proof. That is, I'd go along for a while and then write

Lemma 1. <Insert your favorite lemma here.>

Proof of Lemma 1. <Insert your favorite proof here.>

Then I'd write some more stuff and then

Lemma 2. <Insert your second-favorite lemma here.>

Proof of Lemma 2. <Insert your favorite proof here.>

and continue on like that.

If you're using more than one lemma, you should definitely number them to avoid confusion.

Oh, so you'd prove each lemma right after you use it? For example, like this?

By Lemma 1, it is clear that [insert statement] is true.

Lemma 1. [stuff]

Proof of Lemma 1.
[proof]

Now, we have... [continuing solution]

Yeah, that seems to be the convention. On the other hand, it seems that claims are stated and then proved at the end, but I have never used claims.

"claims"? Is this a technical term? Aren't lemmas claims themselves?

Yes, one can use "claim" as a technical term.  I actually prefer it to "lemma."

Claim: 2^x :equiv: 0 mod 2 for all integral x greater than 0.
Proof:  Well, duh.

Just as one would use a lemma.

I think they tend to be used differently. Lemmas have proofs right below them, and claims have proofs at the end. At least, that's what I have seen rather frequently.

I wrote a proof that was about a page long of (mainly) english.  There were three different cases, but it was difficult for the reader to keep track of things.  So I rewrote it, putting the guts of the proof into three lemmas.  Then the proof of the theorem essentially said: "We've already proved each of the three cases as a lemma."  This may be a bit nonstandard, but it dramatically improved the readability of the proof, which (IMHO) is a problem in many articles.

--Dan


Hello people,
I still haven't known what to call this:
Recently while tarvelling in a 'matatu' (Kenyan public transport) I noticed a peculiar behaviour after observing the reg plates of the vehicles. I would add the numbers horizontally after breaking them up into singles and vertically and their answers when summed up again would lead to the same answer. In whatever bases, it was more fascinating when this was base 10 and above.

(I am not a mathematician but I like thinking things abstactly, just a business administrator with a strong liking for thinking.)

I noticed:  Pick any two numbers base 10, assume ab:cd

This goes horizontal addition when the numbers are sepaerated follows as

ab:cd = a+b+c+d = ef = e+f = Z

When the numbers are added horizontal in the base form unseparated the answer would be the same as follows:

    ab
 + cd
 ------
    gh  = g + h =  Z
====

For example : Pick 65:78

Horizontally when you disable the bases to single; 6 + 5 + 7 + 8 = 26 = 2 + 6 = 8
                                                                                ==
Vertically in base 10;

   65
+ 78
-------
  143 = 1 + 4 + 3 = 8
====                   ==

Assume another example; abc:efg

a + b + c + e + f + g = rstu = r + s + t + u = vw = v + w = Z

again;

   abc + efg = ijk = i + j + k = Z

Try it !

Please discuss the above and lets find out why the numbers behave this way. How can we use this?

thanks to all

Kenneth Kyalo

partity mod 9 is always preserved.

but our sum must be between 0 and 9 inclusive due to how you add the digits

but we cannot have 0, unless we allow 00...00:00...00, which is easy to handle

but 1,2,3,...9 is all residues mod 9, so we are done


group theory
abstract algebra
geometry
geometric transformation
reflection
linear algebra
matrix

If H is subgroup of GL_n(p) with p odd prime and (|H|,p)=1, is H abelian?

Of course not. $ GL_n(F)$ contains an isomorphic copy of $ S_n$ for any $ F$, embedded as the permutation matrices with a one in each row and column. If $ n\ge 3$, this is not abelian.

More specifically, every $ GL_n(F)$ contains an isomorphic copy of the dihedral group on the square if $ n\ge 2$ and the characteristic of $ F$ isn't $ 2$. The reflection $ \begin{bmatrix}1&0\\0&\minus{}1\end{bmatrix}$ and the rotation $ \begin{bmatrix}0&1\\\minus{}1&0\end{bmatrix}$ don't commute.


inequalities
trigonometry
geometry proposed
geometry

Hi!

Let $ \triangle EFG$ be a triangle and $ EH$ the altitude to $ FG$. Prove that if $ {{EF}\over{EG}}\equal{}{{1}\over{\sqrt{3}}}$ then $ EH\leq {{\sqrt{3}}\over{2}}FG$.

[color=darkblue]Using the standard notations, the your problem becomes :[/color]

[quote="yuyang"][color=darkred]In $ \triangle\ ABC$ exists the implication $ b \equal{} c\sqrt 3\implies 2h_a\le a\sqrt 3$ .
Have equality if and only if $ A \equal{} 30^{\circ}$ .[/color][/quote]
[color=brown][b][u]Lemma.[/u][/b] $ \left|b^2 \minus{} c^2\right|\le 2Ra$ with equality iff $ |B \minus{} C| \equal{} 90^{\circ}$ .[/color]

[color=darkblue][b]Proof.[/b] Indeed, from the [b]Sinus' theorem[/b] $ \frac {a}{\sin A} \equal{} \frac {b}{\sin B} \equal{} \frac {c}{\sin C} \equal{} 2R$

obtain : $ \left|b^2 \minus{} c^2\right|\le 2Ra$ $ \Longleftrightarrow$ $ \left|\sin ^2B \minus{} \sin^2 C\right|\le \sin A$ $ \Longleftrightarrow$ $ \sin |B \minus{} C|\le 1$ .[/color]

[quote="Virgil Nicula"][color=darkred][b][u]An easy extension.[/u][/b] $ b \equal{} m\cdot c\implies \left|m^2 \minus{} 1\right|\cdot h_a\le m\cdot a\ \ (*)$ .
Have equality if and only if $ \tan A \equal{} \frac {\left|m^2 \minus{} 1\right|}{2m}$ .[/color][/quote]
[color=darkblue][b]Proof.[/b] For $ m \equal{} \frac bc$ we have $ \left|m^2 \minus{} 1\right|\cdot h_a\le m\cdot a$ $ \Longleftrightarrow$ $ \left|b^2 \minus{} c^2\right|\cdot h_a\le abc$

$ \Longleftrightarrow$ $ \left|b^2 \minus{} c^2\right|\cdot h_a\le a\cdot 2Rh_a$ $ \Longleftrightarrow$  $ \left|b^2 \minus{} c^2\right|\le 2Ra$ , what is the upper lemma.

You can show easily the case of equality in the relation $ (*)$ .

[b]Particular case.[/b] $ m \equal{} \sqrt 3\implies$ the [b]Yuyang[/b]'s proposed problem.[/color]


jajaja estos son los errores que cuestan medallas de oro... :D

$7\times 3= 7\times 4$ si y sólo si $3=4$ si y sólo si $1=0$ que es un resultado bien conocido.
 :lol:


Prove that the nth fibonacci number is a factor of the a(nth) fibonacci number where a and n are integers.

hint:[hide]express your numbers after the nth fibonacci number in that modular base[/hide]

Hi Scrambled,

My attempt based on your hint : (long but explanatory attempt)

(I tried to put the solution in spoiler, but it didn't happen. Sorry about that :(
Doesn't LaTeX code get hidden due to spoiler?)

[tex]F_{n+1} = F_n + F_{n-1}[/tex] $\Rightarrow$ $F_{n+1} \equiv F_{n-1} \pmod{F_n}$

$F_{n+2} = F_{n+1} + F_n$ $\Rightarrow$ $F_{n+2} \equiv F_{n+1} \equiv F_{n-1} \pmod{F_n}$

$F_{n+3} = F_{n+2} + F_{n+1}$ $\Rightarrow$ $F_{n+3} \equiv 2 * F_{n-1} \pmod{F_n}$

$F_{n+4} = F_{n+3} + F_{n+2}$ $\Rightarrow$ $F_{n+4} \equiv 3 * F_{n-1} \pmod{F_n}$

$F_{n+5} = F_{n+4} + F_{n+3}$ $\Rightarrow$ $F_{n+5} \equiv 5 * F_{n-1} \pmod{F_n}$

and it goes on like this.
(Notice a sort of Fibonacci pattern in this decomposition ; i.e., for $F_{n+k}$, we have contributions coming from $F_{n+k-1}$ and $F_{n+k-2}$ )

Thus, lets guess that : $F_{n+k} = F_k * F_{n-1} \pmod{F_n}$ --------- (1)
where $k \geq 0$
($F_0 = 0$ by definition of Fibonacci numbers)

Induction : $F_{n+0} \equiv 0 \equiv F_0 * F_{n-1} \pmod{F_n}$ (Base case)
Let the statement (1) be true for all $0 \leq m < k$
Then,
\[F_{n+k} = F_{n+k-1} + F_{n+k-2}
\Rightarrow F_{n+k} \equiv F_{k-1} * F_{n-1} + F_{k-2} * F_{n-1} \pmod{F_n}
\Rightarrow F_{n+k} \equiv ( F_{k-1} + F_{k-2} ) * F_{n-1} \pmod{F_n}
\Rightarrow F_{n+k} \equiv F_k * F_{n-1} \pmod{F_n}\]

So, our statement (1) is correct.
Now using (1), we get :
\[ F_{an} = F_{n + (a-1)n} \equiv F_{(a-1)n} * F_{n-1}
\equiv F_{(a-2)n} * F_{n-1} * F_{n-1} \equiv \cdots \equiv F_n * (F_{n-1})^{a-1}  \equiv 0 \pmod{F_n} \]

Thus, $F_n$ is a factor of $F_{an}$.

******************************************************************
Sambit

and why did you have to make a statement about m when it appears nowhere in your proof? (perhaps you meant n)

Hi Scrambled,

When I mentioned "Let the statement (1) be true for all $0 \leq m < k$", I meant that the statement is assumed to be true for all non-negative integers less than $k$. And hence by assumption, it is true for $k-1$ and $k-2$, which I used in the next step for the induction proof.

Sambit


calculus
derivative
real analysis
real analysis unsolved

Problem: Show that if $ f: (a,b) \rightarrow R$ is differentiable and $ f'$ is monotonic on $ (a,b)$, then $ f'$ is continuous on $ (a,b)$

I think you can do this by using the fact that $ f'$ satisfies the intermediate value theorem (eliminating "simple discontinuities") and the fact that monotonicity eliminates other types of discontinuities. But is there a more direct approach that may involve say epsilons or the Mean Value Theorem? Thanks.

To see how to employ the Mean-Value-Theorem, try here

 http://www.mathlinks.ro/Forum/viewtopic.php?t=314194

[quote="Snipez"]Problem: Show that if $ f: (a,b) \rightarrow R$ [/quote]


Let n be a postive integer and let $ a_n$ denote the number of positeve integers which can be formed who digits are chosen from 1,3,4 and the sum of whose digits are equal to n. Prove that $ a_{2n}$ is a perfect square for every n.

[hide="hint"]
Prove $ a_n \plus{} a_{n \plus{} 2} \equal{} F_{n \plus{} 4}$ where $ F_k$ is the $ k$th Fibonacci number with $ F_1 \equal{} 0$, $ F_2 \equal{} 1$.  Or, equivalently, prove $ a_{n\plus{}2}\equal{}a_{n\minus{}2}\plus{}a_{n\minus{}1}\plus{}a_{n\plus{}1}$
[/hide]

This problem has been posted and solved before.  Let $ A_n$ denote the set of such integers (so $ |A_n| \equal{} a_n$) and note that that deleting the last digit of $ a \in A_n$ gives a member of $ A_{n\minus{}1}, A_{n\minus{}3}, A_{n\minus{}4}$, hence we have the recurrence

$ a_n \equal{} a_{n\minus{}1} \plus{} a_{n\minus{}3} \plus{} a_{n\minus{}4}$

which has characteristic equation $ x^4 \minus{} x^3 \minus{} x \minus{} 1 \equal{} (x^2 \minus{} x \minus{} 1)(x^2 \plus{} 1) \equal{} 0$.  We therefore have $ a_n \equal{} A \phi^n \plus{} B \varphi^n \plus{} C i^n \plus{} D (\minus{}i)^n$ for some constants $ A, B, C, D$, but the computations are tedious from here.


[quote="Lena"]$\left ( 5x^{2}-x^{4}\right ) \left ( 1-x \right )^{-3}$

find the coefficient of $x^{16}$

[/quote]

Are you sure?  :lol:

[quote="Lena"]$\left ( 5x^{2}-x^{4}\right ) \left ( 1-x \right )^{-3}$
find the coefficient of $x^{16}$ i came up with 58 am i correct?[/quote]
Perhaps 509. Let $n\in{\mathbb N}$  and $|x|<1 .$ Then
\[\frac{1}{(1-x)^{n+1}}=\sum_{k=0}^{\infty}{n+k\choose k}x^{k}\]
implies
\[\frac{5x^{2}-x^{4}}{(1-x)^{n+1}}=\sum_{k=2}^{\infty}A_{k}(n)x^{k}\]
where
\[A_{k}(n)=\left\{\begin{array}{lcl}5{n+k-2 \choose k-2}\; &, & \; k\in \{2,3\}\\ &&\\ 5{n+k-2 \choose k-2}-{n+k-4 \choose k-4}\; &,& k\ge 4 . \end{array}\right. \]
Further find $A_{16}(2) .$

yes i am sure

its $x^{4}$ not 4

how did you get to 509 ?


quadratics
probability
graph theory
number theory
real analysis

Best things are made by people;very taluented researchers make universitie's fame.So according to me, make a choice ,and don't look behind you.beleive in it and put all your energy in it.You'll become for sure an attractive person for others like poeple who are making this forum for example. We all know that some of the problem solvers here are the' poles ' around whom we turn.For sure veRy famous universities , resarchers and engeneering companies are hunting them ,because they are  the persons in whom we beleive that they will make the difference ,the fame and the money.A career is not all the time maneged by a diploma, but in fact by how long you can "retain your breath",how long you can wait ,and how deep is your beleif in your tools.Diploma is a wearing,a coat;you need it for a social position in the mind of the poeple,it's only the first reference for them.After this,they will take time to look at who you really are:is really Zorro behind his mask.Mit for example giveS zorro's mask to zorros only.

[quote="senouy"]Mit for example giveS zorro's mask to zorros only.[/quote]Well... this university gives zorros mask to anyone that has ever weared a black coat. (like most normal universities)

But I'm sometimes a little afraid that for example our math education is too weak to build up a really good research carreer. I see that I'm way better than most of the other students from my class, but I see that most of the other college students on mathlinks are way better than me. So I'm in doubt on my skills. (next to that, I've heared for doing lifetime math research in belgium, you need to become a professor, and that for becoming a professor you need to have a "political color" - but that is quite unclear to me since of course no professor will want to talk about that if it's true...)

Peter, perhaps you've heard of Prof Ingrid Daubechies. She received both her Bachelors and PhD degree in Physics from Vrije Universiteit (Free University) in Belgium. She is one of the most famous mathematicians in the world right now, with her specialty being in the theory of Wavelet Analysis.

So it is possible to major in a science or engineering field, and to still get a solid mathematics education. I know many people with formal physics or engineer degrees who know more math than people with formal math degrees. In college I read a paper by an electrical engineer who proved some interesting theorems about the distribution of quadratic residues, and how they relate to cryptography.

You shouldn't worry about the specific title of your degree as long as you can demonstrate that you are grounded in the core courses of mathematics as they relate to your field of interest - probably graph theory, discrete mathematics, number theory, real analysis, probability theory, etc.

One thing you might do, if you are really concerned, is to look at the Curriculum Vitae of people who are successful in your field of interest and see what Universities they went to for graduate school and what their degrees are in. Often times you can find Curriculum Citaes on professors' webpages.

Never heard of her no. :) but your post is quite good news to me.

[quote="gauss202"]You shouldn't worry about the specific title of your degree as long as you can demonstrate that you are grounded in the core courses of mathematics as they relate to your field of interest - probably graph theory, discrete mathematics, number theory, real analysis, probability theory, etc.[/quote]Add algebra to that and you know my fields of interest :D so I conclude from that I picked the right course? Great! :)

Oh, is anyone able to complete the ... in gauss202's list of graph theory, discrete mathematics, number theory, real analysis, probability theory,...?

I would like to know which subject I should not skip lightly... (since many cross-programs contain holes I might be tempted to abuse :P)

[quote="Peter VDD"]What's the best diplome for doing research in branches that are on the border of math-informatics? (like coding theory, graph theory, ...)

I'm currently doing engineer-compuscience and mathematics, though if there's any better thing I'd be glad to change. It may be important to know that our univ is very strong in engineering but pretty weak in math (on international level)... I'm currently taking 2 directions at once, but that will get very hard in the later years, so I'm still in doubt what to do...[/quote]Hehe... I am thinking how to convince my advisor to do some research in coding more than in quantization!  :D .

Any way, you have to choose, what is what you like the more. I personally discovered that I liked most the side of math than computer science. Nevertheless, the formal education that you can have also have its own weight. If you like math more than CS, well, I don't think your university has a very strong engineering department and a very very weak math dept. hence, I would recomend to stay in math, otherwise, well, depends on a lot of circumstances.

By the way, both in Coding (group in Electrical Engineering School) and in Graph Theory (group in Math) there are good people working here at Gatech, if you would like to apply for the grad school  :D .


hi can anyone help me with this
i need the name of software or a software that can spit out combination and permutation of numbers
so that i can connect it through USB to my IPAQ hp pocket pc to open the password. it asks for 4 digits numbers, i forgot the password and i have very very important fles inside it i dont want to reset it. i know the permutation is $ 10^4$. but thats very long and time comsuming if i keep trying each one myself. is there such software to try each of $ 10^4$ ?

Such a program, if one exists, would allow anyone to break any Pocket PC's password, so obviously nobody here's going to tell you if they knew of one.

So you don't remember a [i]single digit[/i] in your own password? If you remembered even one digit in your password, and not even which place it was in, you'd have to try fewer than 3500 combinations, and could finish in four and a half hours.


Additionally, mention if you've never gone below 1200.

Can we have a "1" option?

And also, PI-Dimension hit 666.

Oh wow, the last option got cut off. Arg. It was 1-899.

[quote="james4l"]1189. However that was the first FTW scoring system which was admittedly quite different from current rating system.[/quote]

Was that when you could win one game and then go up like 200?

That would be too cheap. I believe you could get 35-40 rating for beating a similar-ranking person. (Now at about 15-20)

ummm.... 1184, new system, from 8 mo. ago, through my main account.

my bad account, which is technically my parent's account went down to 1154 soon after FTW began.

Well i joined around the christmas time... and lowest rating was 1016  :blush:
YES i know its pretty bad.... but right now my rating is 1712... which is
currently my highest rating.

rating died at school cuz teacher made me disconnect  :bomb:

LOLz ppl. minewentdown to 938  :blush: mostly b/c I was trying to get a 1 like mew but got lazy  :wink:

IDK how I even got to 1560.

ARRGH posted on my cousins acc.  :!:  When did he log on to my laptop and save his account?!

Little cousins.... :mad:

Hmm, I dropped to 928 :blush:  And, here's the cool part: I was actually trying to get a good rating!

Yes.... Meh.... I remember that 1st day on a hot summer day last year ......

I happened to click this link that said "For The Win"......

It looked very intriguing!.....

And I remember that back then the comp that I was on was vvvvveeeerrrrryyyy lllllaaaaggggyyyyy aaaannnnddddd ssssssllllooooowwwwww.................

Which turned out to be a GOOD thing because when I answered a question in 20 seconds it for some reason I answered it in 1.5 seconds!  :D

Okay bottom line is I achieved 1080 - my lowest rating - on my first day on AoPS FTW!!!!

Wow How Far I've Gone.........................................

On my current username (professormom), it's 1200.  On my abandoned username (this one), it was 1096.

If its abandoned, then why are you using that account.

I use professordad for typing on the forums, Alcumus, and AoPS Classes.  I use professormom for FTW only...because I registered on MathLinks as professormom and decided to give my sister a head start, but since my rating went sky high, I changed my mind.

Lowest Rating: 1041
Highest Rating: 1200
Games Played: 50
Games Won: 8

[size=167][color=red]FAIL[/color][/size]

Lowest:1160'ish (too lazy to look now. Its like 11 pm)
Highest: 1324
Current:1160'ish (darn it im in a losing spree now)
Played/loss: Something like 13/34 or close to there

I play FTW so infrequently because I *stink* at it

Played: 132
Rating: 1040
Lowest: 1030
Highest: 1200  :(
Won: 7

My lowest rating was 1198.
My highest was 1351 :(

Eh well, I don't really play FTW that much these days...

Highest 1917, lowest 1200.

Your lowest rating will remain 1200 if you're decent when you join and don't lose the first few games.

lowest: 1136 (bombing streak)

highest:1357 (got there by accident due to a glitch my brother and i didnt realize until we noticed that my rating was going up by 2 everytime my brother beat me.....)

now: 1337 (GO LEET)

games played: 757 or somewhere around there
games won: 232

Most of the games i played were last summer

lowest: 1088
highest: 1217
Now, I'm at about 1203.  But I'm still on the monthly leaderboard anyway.  :D

Highest:  1471.
Lowest:  1200.
Current:  1400.

I only play new games, never countdowns (until the school year starts...I want to make better of the chapter countdown  :P )  But I'm still not on leaderboard!   :wallbash_red:

Highest:1736
Lowest:1200
Current:1637

I fail.


Hi,
I was noticing that in some forums there is a field that allows you to post the problem's source.  (usually appears below the subject text field)

I am wondering why this is not available in the Getting Started forum.

Thanks

It's not available there to make it easier for younger students to input questions. There will be soon an option that you can enable/disable to have the source field present or not.

I believe the source will appear if you request it to admins. At least for me, I have source fields down to Mathcounts Forum.

[quote="Valentin Vornicu"][quote="Silverfalcon"]I believe the source will appear if you request it to admins.[/quote]I am an admin, and I've just said above that there will be an option to turn on/off that per each user soon.

Thus I will not manually enable it anymore ;)[/quote]

Heh..  :D


need help on solving this system of 2 linear equations in 3 variables. im not sure whether specific roots can be extracted. please help fast... thanks.

5x - 4y + 3z = 1
3x - 5y + 7z = 11

Subtract three times the first one from five times the second one and rearrange to get $y=2z-4$.
Substitute into the first or the second to get $x=z-3$.

So $(x,y,z)=(z-3,2z-4,z)$.


For each real number $a\in [0,2]$ define the function $f_{a}(x,y)=x^{2}+axy+y^{2}$ and $\tilde f_{a}(x,y)=\min\{f_{a}(x-m,y-n)|m,n\in\mathbb Z\}$. Prove that for each $(x,y)\in\mathbb R^{2}$ \[\tilde f_{a}(x,y)\leq\frac1{2+a}\]

No idea ?
[hide="Hint:"]
Consider the function $d((x_{1},y_{1}),(x_{2},y_{2}))=\sqrt{f_{(}x_{1}-x_{2},y_{1}-y_{2})}$. For each point $(m,n)\in\mathbb Z^{2}$ consider the points $(x,y)$ which $\forall(m',n')\in\mathbb Z^{2}, d((m',n'),(x1,y))\geq d((m,n),(x,y))$
[/hide]


[quote="shobber"]I think that there are lots of theorems about the areas of triangle, can you post some of it?[/quote]

Let's start. Suppose $\triangle ABC$ has sides $a,\ b$ and $c$ in the usual notation. Therefore

$(ABC)=\frac12ab\sin C=\sqrt{s(s-a)(s-b)(s-c)}=\frac{abc}{4R}=rs$

Some other formula?

For two triangles($ABC$ and $A'BC$) which have a common side $BC$, let the intersection of $BC$ and $AA'$ be $P$,we will have:
\[\frac{[ABC]}{[A'BC]}= \frac{AP}{A'P}\]

Follow my last post, we can imply that if line $AA'$ and $BC$ don't intersect, then $[ABC]=[A'BC]$.

These one is for the triangles who have a common side, now we shall consider the trangles with common(equal) angles.

[b]Theorem[/b] In triangle $ABC$ and $XYZ$, if $\angle{ABC}=\angle{XYZ}$, or if $\angle{ABC}=180^{o}-\angle{XYZ}$, then:
\[\frac{[ABC]}{[XYZ]}=\frac{AB \cdot BC}{XY \cdot YZ}\]

Proof. Put the two triangles togather, let $X$ be on $AB$, and $Z$ be on $BC$.
Then follow the theorem in my last post, we have:
$\displaystyle \frac{[ABC]}{[XYZ]}=\frac{[ABC]}{[XBC]} \cdot \frac{[XBC]}{[XYZ]}= \frac{AB \cdot BC}{XY \cdot YZ}$

also for a triangle ABC the area is .5ABtimes the sinC (opposite side C)

[quote="pgpatel"]... area is .5ABtimes the sinC (opposite side C)[/quote]
What does $.5$ mean??

It means $0.5 = \dfrac 12$. Usually people omit the 0 when writing $.123$ :)


AMC
AIME
USA(J)MO
USAMO
percent
MATHCOUNTS
AMC 8

In short, I would like to analyze the scores from 1977 to see if it was easier, harder, or similarly difficult to earn a score over 120 then as opposed to recently.  Further, if I could find this information for all years since 1950, that would be even better.  And, then if my wish list really gets met, why not an analysis of AIME scores back then also as opposed to now? (I believe the test writers have purposefully raised the average AIME score in the past few years, but I am not certain that is correct)

Thanks!

I took the AHSME back in 1978 and 1979, and the test then was harder than current AMC 12's.  That's mainly because there was no AIME back then. (The first AIME was 1983.)  To qualify for USAMO, you had to be among the top 100 students on the AHSME.

For example,  in 1979, there were 377,764 students who took the AHSME.  Only 830 students (about 2 percent) made Honor Roll, which was for scores of at least 100.  I got a score of 129, which was good enough for top 100 (about 0.03 percent of students).  That score was also 7th in the VA-MD-DC region, which had 17,591 takers.

In contrast, in 2009, it looks there were about 117,030 students who took the AMC 12.  About 7 percent got a score of 100 or better (compared to the 2 percent from 1979).  About 465 of them (0.4 percent) got a score of 129 or better (compared to the at most 0.03 percent from 1979).

It's sad to see that the number of participants has gone down over time.  Even when you count the AMC 10 and 12, it looks like there were only about 221,000 students last year, compared to 378,000 students from 1979.

[quote="Ravi B"]It's sad to see that the number of participants has gone down over time.  Even when you count the AMC 10 and 12, it looks like there were only about 221,000 students last year, compared to 378,000 students from 1979.[/quote]

And, in fact, it was actually less than 221,000 students in 2009, due to multiplicity (i.e., same student taking an AMC test on A and B date.)  Back in 1979 (and in fact prior to 2002), students could only take the contest one day per year, rather than having the option of taking it twice.

Ravi, does your statistics source for 1979 give any information that would allow us to see how much of the decline can be attributed to fewer schools offering the contest vs. how much can be attributed to fewer students per school taking the contest?

On the other hand, I do think many of the high school students taking the contest today are much better prepared than they were back in your day.  There were no AoPS books, classes, or forums then.

Also, there was no MATHCOUNTS or AMC8 (or AJHSME) back then, so very few students encountered math contests prior to 9th grade.  MOEMS only started up in the late 70s, so no high school students in 1979 would have been exposed to it.

If your 1979 statistics give grade-level data, it would be interesting to see how many of the students taking the AMC contests were below 9th grade back then.

My data came from two letters that are 30 years old: one from Sam Greitzer (chair of the Olympiad committee) inviting me to USAMO and one from Ed Bender (chair of VA/MD/DC region) for doing well in the region and awarding me a 10-dollar prize (!).  My parents kept the letters in a file of my other academic stuff, and I recently went through it.  (Does it mean I have no life if I'm looking at stuff from so long ago?)

Anyway, Prof. Bender's letter says that 6425 schools in the United States and fourteen foreign countries participated in the contest.  Last year, there were around 3600 schools in the US that participated in the AMC 10 or 12.  (We sent a letter to each of those schools for the Math Prize for Girls.)  So the number of participating schools seems to have gone down.

I agree that kids today are better prepared, thanks to middle-school competitions, AoPS, and the internet.  My first math contest was not until 10th grade, and my first AHSME was in 11th grade.  Schools like Stuyvesant and Bronx Science used to dominate contests, because they had an oral tradition of handing down techniques.  Now information is more democratized.

Sorry, I don't have any info about grade-level distribution in the old days.


Hey, I was just wondering about something. A lot of people say that community college credits don't transfer beyond state boundaries, but I was wondering what if I got my Associates of Arts Degree, a 2 year degree, (in Mathematics) before I graduate? Then, having a degree, would my credits hold any more weight (I mean, the community college I go to is a really good one - ranked 1st in the nation for technology and with credentials like that you do get some really good teachers when compared to other C.C.s).

For Ivies/1st Tier Schools:
I'm just wondering this mainly for prereqs and getting the humanities/history reqs out of the way (and maybe some of the calculus classes, but as I've been seeing, taking the school's actual classes seems to help one get a feeling of how future classes will be, and although I may "know" the stuff, I most likely won't know it at as much depth as they want.)

For Out of State (Florida) - 2nd/3rd Tier Schools
Will they let me start off as a junior (or maybe even higher depending what scores I get on my IB/AP exams?)

For Instate:
I'm almost 100% they'll give me at least junior standing for my A.A. so that's all good.

Because I'm already halfway through the credits that I'll need for my A.A., and if all goes as planned I'll have it by this time next year. So any thoughts and opinions would be very welcome (I'll end up getting my A.A. whether or not it will be beneficial to me, mainly because I've already put in so much work for it, and it does help my GPA)

EDIT: Oh yeah, and its FREE.

That depends upon the policies of the next college you go to. If you apply as a TRANSFER student to top-tier colleges, your probability of admission may decrease greatly. (Princeton's current policy is not to accept undergraduate transfer applicants, for example.)

Whether or not your credits are accepted for any purpose also depends on the policies of the college you apply to next. A fairly typical top-tier answer to an applicant who applies as a first-year undergraduate candidate with previous credits for "high school" earned at a community college is that NONE of those credits will be accepted as credits toward graduation, but some may be accepted as evidence of achieving prerequisite qualifications for higher-level courses at the top-tier college. Most colleges have policies for awarding advanced placement for new undergraduates who have certain AP test scores, and most have placement tests to ensure placing students at an appropriate level in math, foreign languages, and other subjects in which students may have previous study experience.

Yes, as you note, your state (and many other states) may allow full transfer of all community college credits to a state university. Whether or not that is a good idea in your case depends on how well the state university fits your needs for an undergraduate degree program.

Note that list price and actual price adjusted for value are two different concepts. It may be worthwhile to apply to a college with a high list price as a first-year admissions candidate, apply for financial aid, and see how much attending there for four years will REALLY cost out-of-pocket with the financial aid offer. And consider that if the degree really is more valuable from a more prestigious school (a debatable issue, but a possible truth), then it even makes economic sense to pay more in the beginning to get a better college degree.

Hope this helps!

If you get the AA degree, make sure that the courses that you take count towards the degree that you might seek from a university in the Florida system.  While the CC folks like to say that you'll transfer as a "junior", that doesn't mean that you will graduate in 2 additional years with a 4 year degree.  You'll still need to meet all course requirements for the 4 year degree, many of which may be embedded in the first 2 years of the curriculum.  You'll probably be ok if you pursue a B.A/B.S in Math or related discipline.  If you want Engineering, Compter Science or similar discipline, you may need much more than 2 years, especially if you haven't taken the appropriate preparatory courses.

Hopefully, your CC course catalog indicates which courses will transfer within the Florida system.  These will be taught by qualified faculty (at least 18 graduate hours in the discipline.)  If you can't find a list of transfer courses, you'll need to contact a transfer admissions officer (or possibly a department chair) at one or two 4-year Florida institutions of interest.  Usually, only courses listed at the freshman- or sophomore-level at the university will transfer.  Compare the tranferable courses offered at the CC to the list of course requirements for the 4-year degree(s) of interest.  Make sure that you understand how many additional credit hours will be required to get the 4 year degree in specific majors of interest.  Again, "junior" status does not imply that you can graduate in 2 additional years.

In many states, AA degree holders will be automatically exempt from taking the base-level general education "core" curriculum courses at a 4 year in-state institution.  If you attend a private or non-Florida institution, courses will probably transfer on a "course-by-course" basis.  The more information you can provide the university person tasked with evaluating transfer credit, the more credits you'll receive.   Keep your course outlines/syllabi etc. in case they are needed for this process.

As to the question about whether to

1)  attend a "top-tier" institution for your undergrad degree for 4 years without transfer credits or

2) attend a state institution in Florida for perhaps as little as 2 years

- it's hard for me to say.

You should consider the possibility of using the time and money saved at the Florida institution in a top-tier graduate program.  Of course, admission into such a program may be impacted by the perceived quality of your undergraduate background.

[quote="WindSlicer"][allegedly] community college credits don't transfer beyond state boundaries, but I was wondering what if I got my Associates of Arts Degree, a 2 year degree, (in Mathematics) before I graduate? Then, having a degree, would my credits hold any more weight (I mean, the community college I go to is a really good one - ranked 1st in the nation for technology and with credentials like that you do get some really good teachers when compared to other C.C.s).   [/quote]

The question is unclear.  If you mean a mathematically advanced high school student taking enough courses at local college to qualify for a 2-year degree, but ultimately heading for a fancier top-tier university, it doesn't matter except as a free credential in case of dropping out of college later on.   Nobody will be impressed by the CC degree if you do something bigger and better later on, so it would be just a formality (or a vanity). If you mean how to transfer to a more selective school from a community college, it will depend strongly on what pair of universities you are talking about.

You mention Florida;  the New School of USF is (now) reputable, and for mathematics in particular it was put on the map after William Thurston went there as undergraduate(and was rejected at some grad schools that had never heard of his undergrad institution).

[quote="fleeting_guest"][quote="WindSlicer"][allegedly] community college credits don't transfer beyond state boundaries, but I was wondering what if I got my Associates of Arts Degree, a 2 year degree, (in Mathematics) before I graduate? Then, having a degree, would my credits hold any more weight (I mean, the community college I go to is a really good one - ranked 1st in the nation for technology and with credentials like that you do get some really good teachers when compared to other C.C.s).   [/quote]

The question is unclear.  If you mean a mathematically advanced high school student taking enough courses at local college to qualify for a 2-year degree, but ultimately heading for a fancier top-tier university, it doesn't matter except as a free credential in case of dropping out of college later on.   Nobody will be impressed by the CC degree if you do something bigger and better later on, so it would be just a formality (or a vanity). If you mean how to transfer to a more selective school from a community college, it will depend strongly on what pair of universities you are talking about.[/quote]

Well, that's the thing - I don't know what would be better for me (if I get accepted into a prestigious top-tier school, that is). Whether I should just put that AA on my resume and never really use it again (except for the knowledge that I get there - what I'm referring to is the actual credit), or whether I should maybe go to a less prestigious university as a transfer student, get my Bachelors two years early and head off to the same prestigious school of which, presumably, I chose not to go to.  The problem I have with this is whether that same prestigious school will accept me as a graduate student from a less-prestigious university (hopefully with top marks).

[quote="fleeting_guest"]
You mention Florida;  the New School of USF is (now) reputable, and for mathematics in particular it was put on the map after William Thurston went there as undergraduate(and was rejected at some grad schools that had never heard of his undergrad institution).[/quote]

Interesting, I've never heard USF mentionedto me before, so thanks for the idea. But would USF be a better school to go to in-state than UF or UofMiami? Because if I'm going to stay in-state I'm going to do my best to get into the best school that will accept many (if not all) of my Community College credits (as well as I.B. and A.P. but the latter shouldn't matter at any college).

Also, just wanted to thank the other two posts for their input (I did read them in the vicinity of when they were posted) because they did help clear up a lot of the questions that I had, but I still don't know what I should do about going to top-tier as freshman or in-state as junior or higher.

Given 4 years or so and sufficient effort, it is possible to qualify for top-level graduate schools almost without regard to where you spent the time.   It is of course easier if some famous professor at MIT writes recommendation letters on your behalf, but you would have had in the meantime to deal with the heavier workload and to compete with smart and ambitious classmates.   The tradeoffs between different paths through the university system are not so directly connected to grad school admission as one might imagine.

After posting, I learned that the Florida school changed its name in the past few years to "New College of Florida", formerly "New College of the University of Southern Florida" in Sarasota FL.   This was originally a private college that went out of business several decades ago and was turned into a magnet school campus, to borrow the phrase, of the public USF system.  They figure regularly in the annual ratings of US colleges (I believe US News lists them under "bargains"), and I think they have scholarships for valedictorians or otherwise high-credentialed high school students.   See website at:
http://ncf.edu

I thought that name rang a bell for some reason, and as I was getting something to drink in the kitchen I found a letter from New College (The Honors College) of Florida in Sarasota, Florida. Pretty coincidental.

Anyways, I'll talk to my guidance counselor about it and put that on my list of colleges to visit during next summer. And any other insight is welcome, I thank you for your input.


geometry
3D geometry
pyramid
integration

This problem was in a worksheet:

Find the center of mass of a pyramid of height h and square base of length L. (Height of the center of mass)

My answer, computed geometrically is $ (1 \minus{} \sqrt [3]{\frac {1}{2}})h$. This gives equal mass on both sides of the horizontal dividing plane.

The answer my teacher provided was $ \frac {h}{4}$, and I can't figure our what is wrong with her solution:

Setting the origin at the vertex of the pyramid for convenience:
$ Mass \equal{} \displaystyle\int^h_o (\frac {Lz}{h})^2\,dz \equal{} \frac {L^2h}{3}$. Here making sense - that's the formula for volume of a pyramid we all know and love.

$ Z_{cm} \equal{} \frac {1}{m} \displaystyle\int^h_o z\,dm \equal{} \displaystyle\int^h_o z(\frac {Lz}{h})^2\,dz \equal{} \frac {1}{m} * \frac {L^2h^2}{4} \equal{} 3h/4$
Center of Mass is sum of the product of the position of each particle and it's mass divided by the total mass. In this case a "particle" is a horizontal slice of the pyramid, its mass $ dm$.

This is the distance from the top, so distance from the bottom is $ \frac {h}{4}$.

What is wrong with this reasoning?

Thanks!

There is nothing wrong with the reasoning, that is correct. :D

Your reasoning is wrong.

equal masses on either side of the plane does not guarantee that the centroid is on that plane. For example, take an equilateral triangle.
the centroid is at 2/3 the median from any vertex.
if the equilateral triangle's area is A, the part above the centroid is 4/9*A.
what you want is equal sums of moments on each side of the plane.


search
inequalities unsolved
inequalities

let $x;y;z \geq 0 ;x+y>o ;y+z>o; z+x>o $.prove that:
$\frac {x}{\sqrt{x+y}}+\frac {y}{\sqrt{x+y}}+\frac {z}{\sqrt{x+y}} \leq \frac{5}{4} \sqrt{x+y+z}$

This has been posted a VERY long time ago. Search for a post by Fedor Bakharev.

Here is the link
[url]http://www.mathlinks.ro/Forum/viewtopic.php?t=1459[/url]


[quote="N.T.TUAN"]Find the largest possible number of rooks that can be placed on a $3n \times 3n$ chessboard so that each rook is attacked by at most one rook.[/quote]

1) for $n=1$, the requested number is $M_{1}=4$ :
1.1) $4$ is possible : put the rooks in the squares $(1,1),(1,2),(2,3),(3,3)$
1.2) $4$ is the maximum : easy to see (as soon as one line has 2 rooks, this forbid 5 other squares, and it remains just two free squares).

2) For $n>1$, the requested number is $M_{n}\geq 4n$
Just use the configuration described in 1.1 above on each $n$ $3 \times 3$ squares on the diagonal of the $3n \times 3n$ chessboard.

3) For $n>1$, the requested number is $M_{n}\leq M_{n-1}+4$
First notice that, in a $3n \times 3n$ maximum configuration, we have at least one line with 2 rooks (since $M_{n}\geq 4n>3n$ and, for the same reason, at least one column with 2 rooks.
Then notice that, in a maximum configuration, we always can swap two lines ou two columns and we have another maximum configuration. Swap then the lines in order to have in line $1$ one "two-rooks-line". Switch then the columns in order to have these two rooks in positions $(1,1)$ and $(1,2)$. Notice then that none of the two columns one or two contains 2 rooks. So it exist another column with 2 rooks. This colum may be swapped in order to be in column 3. Obviously, no rook is in position $(1,3)$
Then it is possible to swap lines in order to have the two rooks of column 3 in position $(2,3)$ and $(3,3)$.
No rook may exist in lines 1,2,3 except the four in columns 1,2,3.
No rook may exist in columns 1,2,3 except the four in lines 1,2,3.
So the remaining $3(n-1) \times 3(n-1)$ square contains $M_{n}-4$ rooks and we have $M_{n}-4\leq M_{n-1}$. Q.E.D.

1), 2) and 3) imply immediatly $M_{n}=4n$

--
Patrick